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https://www.mindspore.cn/api/en/master/_modules/mindspore/nn/layer/pooling.html	[ "# Source code for mindspore.nn.layer.pooling\n\n# Copyright 2020 Huawei Technologies Co., Ltd\n#\n# you may not use this file except in compliance with the License.\n# You may obtain a copy of the License at\n#\n#\n# Unless required by applicable law or agreed to in writing, software\n# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.\n# See the License for the specific language governing permissions and\n# ============================================================================\n\"\"\"pooling\"\"\"\nfrom mindspore.ops import operations as P\nfrom mindspore.ops import functional as F\nfrom mindspore._checkparam import Validator as validator\nfrom mindspore.ops.primitive import constexpr\nfrom ... import context\nfrom ..cell import Cell\nfrom ..._checkparam import Rel\n\n__all__ = ['AvgPool2d', 'MaxPool2d', 'AvgPool1d']\n\nclass _PoolNd(Cell):\n\"\"\"N-D AvgPool\"\"\"\n\nsuper(_PoolNd, self).__init__()\n\ndef _check_int_or_tuple(arg_name, arg_value):\nvalidator.check_value_type(arg_name, arg_value, [int, tuple], self.cls_name)\nerror_msg = f'For \\'{self.cls_name}\\' the {arg_name} should be an positive int number or ' \\\nf'a tuple of two positive int numbers, but got {arg_value}'\nif isinstance(arg_value, int):\nif arg_value <= 0:\nraise ValueError(error_msg)\nelif len(arg_value) == 2:\nfor item in arg_value:\nif isinstance(item, int) and item > 0:\ncontinue\nraise ValueError(error_msg)\nelse:\nraise ValueError(error_msg)\nreturn arg_value\n\nself.kernel_size = _check_int_or_tuple('kernel_size', kernel_size)\nself.stride = _check_int_or_tuple('stride', stride)\n\ndef construct(self, inputs):\npass\n\ndef extend_repr(self):\n@constexpr\ndef _shape_check(in_shape):\nif len(in_shape) != 3:\nraise ValueError(\"The input must has 3 dim\")\n\n[docs]class MaxPool2d(_PoolNd):\nr\"\"\"\nMax pooling operation for temporal data.\n\nApplies a 2D max pooling over an input Tensor which can be regarded as a composition of 2D planes.\n\nTypically the input is of shape :math:(N_{in}, C_{in}, H_{in}, W_{in}), MaxPool2d outputs\nregional maximum in the :math:(H_{in}, W_{in})-dimension. Given kernel size\n:math:ks = (h_{ker}, w_{ker}) and stride :math:s = (s_0, s_1), the operation is as follows.\n\n.. math::\n\\text{output}(N_i, C_j, h, w) = \\max_{m=0, \\ldots, h_{ker}-1} \\max_{n=0, \\ldots, w_{ker}-1}\n\\text{input}(N_i, C_j, s_0 \\times h + m, s_1 \\times w + n)\n\nNote:\npad_mode for training only supports \"same\" and \"valid\".\n\nArgs:\nkernel_size (Union[int, tuple[int]]): The size of kernel used to take the max value,\nis an int number that represents height and width are both kernel_size,\nor a tuple of two int numbers that represent height and width respectively.\nDefault: 1.\nstride (Union[int, tuple[int]]): The distance of kernel moving, an int number that represents\nthe height and width of movement are both strides, or a tuple of two int numbers that\nrepresent height and width of movement respectively. Default: 1.\npad_mode (str): The optional values for pad mode, is \"same\" or \"valid\", not case sensitive.\nDefault: \"valid\".\n\n- same: Adopts the way of completion. Output height and width will be the same as\nthe input. Total number of padding will be calculated for horizontal and vertical\ndirection and evenly distributed to top and bottom, left and right if possible.\nOtherwise, the last extra padding will be done from the bottom and the right side.\n\n- valid: Adopts the way of discarding. The possibly largest height and width of output\n\nInputs:\n- input* (Tensor) - Tensor of shape :math:(N, C_{in}, H_{in}, W_{in}).\n\nOutputs:\nTensor of shape :math:(N, C_{out}, H_{out}, W_{out}).\n\nExamples:\n>>> pool = nn.MaxPool2d(kernel_size=3, stride=1)\n>>> x = Tensor(np.random.randint(0, 10, [1, 2, 4, 4]), mindspore.float32)\n[[[[1. 5. 5. 1.]\n[0. 3. 4. 8.]\n[4. 2. 7. 6.]\n[4. 9. 0. 1.]]\n[[3. 6. 2. 6.]\n[4. 4. 7. 8.]\n[0. 0. 4. 0.]\n[1. 8. 7. 0.]]]]\n>>> output = pool(x)\n>>> output.shape\n(1, 2, 2, 2)\n>>> output\n[[[[7. 8.]\n[9. 9.]]\n[[7. 8.]\n[8. 8.]]]]\n\"\"\"\n\nself.max_pool = P.MaxPool(ksize=self.kernel_size,\nstrides=self.stride,\nself.max_pool_with_arg_max = P.MaxPoolWithArgmax(ksize=self.kernel_size,\nstrides=self.stride,\nself.is_tbe = context.get_context(\"device_target\") == \"Ascend\"\n\ndef construct(self, x):\nif self.is_tbe and self.training:\nout = self.max_pool_with_arg_max(x)\nelse:\nout = self.max_pool(x)\nreturn out\n\n[docs]class AvgPool2d(_PoolNd):\nr\"\"\"\nAverage pooling for temporal data.\n\nApplies a 2D average pooling over an input Tensor which can be regarded as a composition of 2D input planes.\n\nTypically the input is of shape :math:(N_{in}, C_{in}, H_{in}, W_{in}), AvgPool2d outputs\nregional average in the :math:(H_{in}, W_{in})-dimension. Given kernel size\n:math:ks = (h_{ker}, w_{ker}) and stride :math:s = (s_0, s_1), the operation is as follows.\n\n.. math::\n\\text{output}(N_i, C_j, h, w) = \\frac{1}{h_{ker} * w_{ker}} \\sum_{m=0}^{h_{ker}-1} \\sum_{n=0}^{w_{ker}-1}\n\\text{input}(N_i, C_j, s_0 \\times h + m, s_1 \\times w + n)\n\nNote:\npad_mode for training only supports \"same\" and \"valid\".\n\nArgs:\nkernel_size (Union[int, tuple[int]]): The size of kernel used to take the average value,\nis an int number that represents height and width are both kernel_size,\nor a tuple of two int numbers that represent height and width respectively.\nDefault: 1.\nstride (Union[int, tuple[int]]): The distance of kernel moving, an int number that represents\nthe height and width of movement are both strides, or a tuple of two int numbers that\nrepresent height and width of movement respectively. Default: 1.\npad_mode (str): The optional values for pad mode, is \"same\" or \"valid\", not case sensitive.\nDefault: \"valid\".\n\n- same: Adopts the way of completion. Output height and width will be the same as\nthe input. Total number of padding will be calculated for horizontal and vertical\ndirection and evenly distributed to top and bottom, left and right if possible.\nOtherwise, the last extra padding will be done from the bottom and the right side.\n\n- valid: Adopts the way of discarding. The possibly largest height and width of output\n\nInputs:\n- input (Tensor) - Tensor of shape :math:(N, C_{in}, H_{in}, W_{in}).\n\nOutputs:\nTensor of shape :math:(N, C_{out}, H_{out}, W_{out}).\n\nExamples:\n>>> pool = nn.AvgPool2d(kernel_size=3, stride=1)\n>>> x = Tensor(np.random.randint(0, 10, [1, 2, 4, 4]), mindspore.float32)\n[[[[5. 5. 9. 9.]\n[8. 4. 3. 0.]\n[2. 7. 1. 2.]\n[1. 8. 3. 3.]]\n[[6. 8. 2. 4.]\n[3. 0. 2. 1.]\n[0. 8. 9. 7.]\n[2. 1. 4. 9.]]]]\n>>> output = pool(x)\n>>> output.shape\n(1, 2, 2, 2)\n>>> output\n[[[[4.888889 4.4444447]\n[4.111111 3.4444444]]\n[[4.2222223 4.5555553]\n[3.2222223 4.5555553]]]]\n\"\"\"\n\ndef __init__(self,\nkernel_size=1,\nstride=1,\nself.avg_pool = P.AvgPool(ksize=self.kernel_size,\nstrides=self.stride,\n\ndef construct(self, x):\nreturn self.avg_pool(x)\n\n[docs]class AvgPool1d(_PoolNd):\nr\"\"\"\nAverage pooling for temporal data.\n\nApplies a 1D average pooling over an input Tensor which can be regarded as a composition of 1D input planes.\n\nTypically the input is of shape :math:(N_{in}, C_{in}, L_{in}), AvgPool1d outputs\nregional average in the :math:(L_{in})-dimension. Given kernel size\n:math:ks = l_{ker} and stride :math:s = s_0, the operation is as follows.\n\n.. math::\n\\text{output}(N_i, C_j, l) = \\frac{1}{l_{ker}} \\sum_{n=0}^{l_{ker}-1}\n\\text{input}(N_i, C_j, s_0 \\times l + n)\n\nNote:\npad_mode for training only supports \"same\" and \"valid\".\n\nArgs:\nkernel_size (int): The size of kernel window used to take the average value, Default: 1.\nstride (int): The distance of kernel moving, an int number that represents\nthe width of movement is strides, Default: 1.\npad_mode (str): The optional values for pad mode, is \"same\" or \"valid\", not case sensitive.\nDefault: \"valid\".\n\n- same: Adopts the way of completion. Output height and width will be the same as\nthe input. Total number of padding will be calculated for horizontal and vertical\ndirection and evenly distributed to top and bottom, left and right if possible.\nOtherwise, the last extra padding will be done from the bottom and the right side.\n\n- valid: Adopts the way of discarding. The possibly largest height and width of output\n\nInputs:\n- input (Tensor) - Tensor of shape :math:(N, C_{in}, L_{in}).\n\nOutputs:\nTensor of shape :math:(N, C_{out}, L_{out}).\n\nExamples:\n>>> pool = nn.AvgPool1d(kernel_size=6, strides=1)\n>>> x = Tensor(np.random.randint(0, 10, [1, 3, 6]), mindspore.float32)\n>>> output = pool(x)\n>>> output.shape\n(1, 3, 1)\n\"\"\"\n\ndef __init__(self,\nkernel_size=1,\nstride=1,\nvalidator.check_value_type('kernel_size', kernel_size, [int], self.cls_name)\nvalidator.check_value_type('stride', stride, [int], self.cls_name)\nvalidator.check_integer(\"kernel_size\", kernel_size, 1, Rel.GE, self.cls_name)\nvalidator.check_integer(\"stride\", stride, 1, Rel.GE, self.cls_name)\nself.kernel_size = (1, kernel_size)\nself.stride = (1, stride)\nself.avg_pool = P.AvgPool(ksize=self.kernel_size,\nstrides=self.stride,\nself.shape = F.shape\nself.reduce_mean = P.ReduceMean(keep_dims=True)\nself.slice = P.Slice()\nself.expand = P.ExpandDims()\nself.squeeze = P.Squeeze(2)\n\ndef construct(self, x):\n_shape_check(self.shape(x))\nbatch, channel, width = self.shape(x)\nif width == self.kernel_size:\nx = self.reduce_mean(x, 2)\nelif width - self.kernel_size < self.stride:\nx = self.slice(x, (0, 0, 0), (batch, channel, self.kernel_size))\nx = self.reduce_mean(x, 2)\nelse:\nx = self.expand(x, 2)\nx = self.avg_pool(x)\nx = self.squeeze(x)\nreturn x" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5523627,"math_prob":0.9825534,"size":10442,"snap":"2020-34-2020-40","text_gpt3_token_len":3089,"char_repetition_ratio":0.13019736,"word_repetition_ratio":0.51592356,"special_character_ratio":0.33049223,"punctuation_ratio":0.2569883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99596983,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-03T09:12:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4e4ccbb3-c0d1-4c3f-af52-a46f2b92efce>\",\"Content-Length\":\"45964\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b366930-fe08-403a-ac74-5a95640e7045>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5380e05-c817-4e42-bcb6-324c82b6867f>\",\"WARC-IP-Address\":\"117.78.24.38\",\"WARC-Target-URI\":\"https://www.mindspore.cn/api/en/master/_modules/mindspore/nn/layer/pooling.html\",\"WARC-Payload-Digest\":\"sha1:N72PWESCMXQFEZUWX744NK6B7SJDHYJE\",\"WARC-Block-Digest\":\"sha1:W7CJKBJ5DNNSSGSZXESRETEFDFRBZSMQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735792.85_warc_CC-MAIN-20200803083123-20200803113123-00371.warc.gz\"}"}
https://global-sci.org/intro/article_detail/eajam/10722.html	[ "", null, "", null, "Volume 7, Issue 4\nLinear Regression to Minimize the Total Error of the Numerical Differentiation\n\nEast Asian J. Appl. Math., 7 (2017), pp. 810-826.\n\nPublished online: 2018-02\n\nCited by\n\nExport citation\n• Abstract\n\nIt is well known that numerical derivative contains two types of errors. One is truncation error and the other is rounding error. By evaluating variables with rounding error, together with step size and the unknown coefficient of the truncation error, the total error can be determined. We also know that the step size affects the truncation error very much, especially when the step size is large. On the other hand, rounding error will dominate numerical error when the step size is too small. Thus, to choose a suitable step size is an important task in computing the numerical differentiation. If we want to reach an accuracy result of the numerical difference, we had better estimate the best step size. We can use Taylor Expression to analyze the order of truncation error, which is usually expressed by the big O notation, that is, $E(h)=Ch^k$. Since the leading coefficient $C$ contains the factor $f^{(k)}(ξ)$ for high order $k$ and unknown $ξ$, the truncation error is often estimated by a roughly upper bound. If we try to estimate the high order difference $f^{(k)}(ξ)$, this term usually contains larger error. Hence, the uncertainty of $ξ$ and the rounding errors hinder a possible accurate numerical derivative.\nWe will introduce the statistical process into the traditional numerical difference. The new method estimates truncation error and rounding error at the same time for a given step size. When we estimate these two types of error successfully, we can reach much better modified results. We also propose a genetic approach to reach a confident numerical derivative.\n\n• Keywords\n\nTruncation error, leading coefficient, asymptotic constant, rounding error.\n\n65M10, 78A48\n\n• BibTex\n• RIS\n• TXT\n@Article{EAJAM-7-810, author = {}, title = {Linear Regression to Minimize the Total Error of the Numerical Differentiation}, journal = {East Asian Journal on Applied Mathematics}, year = {2018}, volume = {7}, number = {4}, pages = {810--826}, abstract = {\n\nIt is well known that numerical derivative contains two types of errors. One is truncation error and the other is rounding error. By evaluating variables with rounding error, together with step size and the unknown coefficient of the truncation error, the total error can be determined. We also know that the step size affects the truncation error very much, especially when the step size is large. On the other hand, rounding error will dominate numerical error when the step size is too small. Thus, to choose a suitable step size is an important task in computing the numerical differentiation. If we want to reach an accuracy result of the numerical difference, we had better estimate the best step size. We can use Taylor Expression to analyze the order of truncation error, which is usually expressed by the big O notation, that is, $E(h)=Ch^k$. Since the leading coefficient $C$ contains the factor $f^{(k)}(ξ)$ for high order $k$ and unknown $ξ$, the truncation error is often estimated by a roughly upper bound. If we try to estimate the high order difference $f^{(k)}(ξ)$, this term usually contains larger error. Hence, the uncertainty of $ξ$ and the rounding errors hinder a possible accurate numerical derivative.\nWe will introduce the statistical process into the traditional numerical difference. The new method estimates truncation error and rounding error at the same time for a given step size. When we estimate these two types of error successfully, we can reach much better modified results. We also propose a genetic approach to reach a confident numerical derivative.\n\n}, issn = {2079-7370}, doi = {https://doi.org/10.4208/eajam.161016.300517a}, url = {http://global-sci.org/intro/article_detail/eajam/10722.html} }\nTY - JOUR T1 - Linear Regression to Minimize the Total Error of the Numerical Differentiation JO - East Asian Journal on Applied Mathematics VL - 4 SP - 810 EP - 826 PY - 2018 DA - 2018/02 SN - 7 DO - http://doi.org/10.4208/eajam.161016.300517a UR - https://global-sci.org/intro/article_detail/eajam/10722.html KW - Truncation error, leading coefficient, asymptotic constant, rounding error. AB -\n\nIt is well known that numerical derivative contains two types of errors. One is truncation error and the other is rounding error. By evaluating variables with rounding error, together with step size and the unknown coefficient of the truncation error, the total error can be determined. We also know that the step size affects the truncation error very much, especially when the step size is large. On the other hand, rounding error will dominate numerical error when the step size is too small. Thus, to choose a suitable step size is an important task in computing the numerical differentiation. If we want to reach an accuracy result of the numerical difference, we had better estimate the best step size. We can use Taylor Expression to analyze the order of truncation error, which is usually expressed by the big O notation, that is, $E(h)=Ch^k$. Since the leading coefficient $C$ contains the factor $f^{(k)}(ξ)$ for high order $k$ and unknown $ξ$, the truncation error is often estimated by a roughly upper bound. If we try to estimate the high order difference $f^{(k)}(ξ)$, this term usually contains larger error. Hence, the uncertainty of $ξ$ and the rounding errors hinder a possible accurate numerical derivative.\nWe will introduce the statistical process into the traditional numerical difference. The new method estimates truncation error and rounding error at the same time for a given step size. When we estimate these two types of error successfully, we can reach much better modified results. We also propose a genetic approach to reach a confident numerical derivative.\n\nJengnan Tzeng. (2020). Linear Regression to Minimize the Total Error of the Numerical Differentiation. East Asian Journal on Applied Mathematics. 7 (4). 810-826. doi:10.4208/eajam.161016.300517a\nCopy to clipboard\nThe citation has been copied to your clipboard" ]	[ null, "https://global-sci.org/public/images/icon-arrow-left.svg", null, "https://global-sci.org/public/images/icon-arrow-down.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87038517,"math_prob":0.9539483,"size":3285,"snap":"2022-05-2022-21","text_gpt3_token_len":723,"char_repetition_ratio":0.14934471,"word_repetition_ratio":0.9548872,"special_character_ratio":0.20913242,"punctuation_ratio":0.10440457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.997386,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T22:36:51Z\",\"WARC-Record-ID\":\"<urn:uuid:7bbf9e3b-2689-4f73-857d-ff8eb1f94dc4>\",\"Content-Length\":\"55111\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:681b5e61-9f6c-48b7-ba9c-e247d5718bee>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c620e1f-f21c-4b02-a898-6f6fd02ce235>\",\"WARC-IP-Address\":\"8.218.69.127\",\"WARC-Target-URI\":\"https://global-sci.org/intro/article_detail/eajam/10722.html\",\"WARC-Payload-Digest\":\"sha1:LB3KY7QRYENFU4JULA36CZD2Z6IGNTUD\",\"WARC-Block-Digest\":\"sha1:CILHGQPVYRX4RXTOR4XQC4QMOKEUVU3S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522556.18_warc_CC-MAIN-20220518215138-20220519005138-00142.warc.gz\"}"}
https://support.ecotechmarine.com/hc/en-us/articles/215339608-How-do-I-calculate-head-pressure-	[ "Follow\n\n# How do I calculate head pressure?\n\nHead pressure is calculated and represented in terms of feet (ft.). In order to calculate the total head pressure of your application, you'll need to know the difference in elevation between your return pump and return outlet. You will also need to know the total number and type of fittings used as well as the total distance of pipe used.\n\nThe effect of gravity on head pressure is very simple. Every vertical foot of distance the return pump moves water equates to 1 foot of head pressure. The effects of friction on head pressure are more difficult to calculate. Roughly every 10 feet of pipe through which water is traveling adds 1 foot of head pressure. Additionally, every 90-degree bend adds 1 foot of head pressure.\n\nPipe diameter is also an extremely important factor when calculating head pressure. As a general rule of thumb, reducing the size of pipe below the return pump output will drastically increase head pressure. For minimum head pressure, using the largest diameter pipe possible is best. Avoiding things like flow meters, reactors and chillers will also help to reduce head pressure.", null, "" ]	[ null, "https://support.ecotechmarine.com/hc/article_attachments/360028765393/vectra_type.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9362454,"math_prob":0.9698417,"size":1113,"snap":"2020-10-2020-16","text_gpt3_token_len":216,"char_repetition_ratio":0.1704238,"word_repetition_ratio":0.027322404,"special_character_ratio":0.1931716,"punctuation_ratio":0.08737864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95308614,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-07T22:57:06Z\",\"WARC-Record-ID\":\"<urn:uuid:84742a6b-e2e5-4253-a995-89725890d625>\",\"Content-Length\":\"16477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b3e1a5e-e349-435d-8288-6b57db81906e>\",\"WARC-Concurrent-To\":\"<urn:uuid:03af65ae-3a91-46f6-8fcf-a117bb05c246>\",\"WARC-IP-Address\":\"104.16.52.111\",\"WARC-Target-URI\":\"https://support.ecotechmarine.com/hc/en-us/articles/215339608-How-do-I-calculate-head-pressure-\",\"WARC-Payload-Digest\":\"sha1:XSCMRHVLTZ2NL6POWNQQS577I7MSB6VN\",\"WARC-Block-Digest\":\"sha1:U5ADDVI32NH4IIYS5LYVKLC7FJB5CC4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371806302.78_warc_CC-MAIN-20200407214925-20200408005425-00326.warc.gz\"}"}
http://talkstats.com/threads/confidence-interval-of-mean-of-queue-length-m-m-1.57512/	[ "# Confidence interval of mean of queue length (M/M/1)\n\n#### Knaapje\n\n##### New Member\nGoodday,\n\nFor a simulation course I had to simulate a single server queue with Poisson processes for arrival and departure (M/M/1). Now, this really was not the issue, the issue is where I personally want to add a kind of validity check, which naturally can only be done with statistics.\n\nI want to construct a 99.5% confidence interval for the average queue length. The following information is known: total simulated 'time', list of amount of time per queue length during the duration (example: w_0 = 3 means that during the entire simulation the total amount of time the queue contained no customers is 3) and all arrival and departure times and the analytical value of the mean.\n\nNow, I know the formula for the weighted sample mean and the weighted sample variance (where the mean is known), but I have no idea how to construct the confidence interval. Any help would be greatly appreciated.\n\n#### rogojel\n\n##### TS Contributor\nhi,\na simple and pragmatic way would be to use bootstapping, as you already habe a simulation. Of course if you are interested in a formula then it is not helping.\n\nregards\nrogojel" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9479175,"math_prob":0.8372058,"size":1816,"snap":"2022-05-2022-21","text_gpt3_token_len":379,"char_repetition_ratio":0.10485651,"word_repetition_ratio":0.98381877,"special_character_ratio":0.21200441,"punctuation_ratio":0.09428571,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9799102,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T08:35:16Z\",\"WARC-Record-ID\":\"<urn:uuid:d529b9a3-1061-4c1f-b262-0586ff4eb439>\",\"Content-Length\":\"41760\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:993feaa2-1b9f-4664-8116-a3a9e3633583>\",\"WARC-Concurrent-To\":\"<urn:uuid:3bd42e68-dc67-446f-a30c-ee7518e81e41>\",\"WARC-IP-Address\":\"199.167.200.62\",\"WARC-Target-URI\":\"http://talkstats.com/threads/confidence-interval-of-mean-of-queue-length-m-m-1.57512/\",\"WARC-Payload-Digest\":\"sha1:SFPWXBUMPDILNCL4XAZJRDQ72QS5Z5AO\",\"WARC-Block-Digest\":\"sha1:GM6V2H5AOKQXAZMJ74SN2ZLBUNTXHZCA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662526009.35_warc_CC-MAIN-20220519074217-20220519104217-00548.warc.gz\"}"}
https://www.mathworks.com/matlabcentral/profile/authors/19558205?detail=cody	[ "Community Profile", null, "liu ce\n\nLast seen: 3 months ago Active since 2021\n\nStatistics\n\n•", null, "•", null, "•", null, "•", null, "Content Feed\n\nView by\n\nSolved\n\nSum all integers from 1 to 2^n\nGiven the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.\n\n4 months ago\n\nSolved\n\nMagic is simple (for beginners)\nDetermine for a magic square of order n, the magic sum m. For example m=15 for a magic square of order 3.\n\n4 months ago\n\nSolved\n\nMake a random, non-repeating vector.\nThis is a basic MATLAB operation. It is for instructional purposes. --- If you want to get a random permutation of integer...\n\n4 months ago\n\nSolved\n\nRoll the Dice!\nDescription Return two random integers between 1 and 6, inclusive, to simulate rolling 2 dice. Example [x1,x2] =...\n\n4 months ago\n\nSolved\n\nNumber of 1s in a binary string\nFind the number of 1s in the given binary string. Example. If the input string is '1100101', the output is 4. If the input stri...\n\n4 months ago\n\nSolved\n\nReturn the first and last characters of a character array\nReturn the first and last character of a string, concatenated together. If there is only one character in the string, the functi...\n\n4 months ago\n\nSolved\n\nGetting the indices from a vector\nThis is a basic MATLAB operation. It is for instructional purposes. --- You may already know how to <http://www.mathworks....\n\n4 months ago\n\nSolved\n\nCheck if number exists in vector\nReturn 1 if number _a_ exists in vector _b_ otherwise return 0. a = 3; b = [1,2,4]; Returns 0. a = 3; b = [1,...\n\n4 months ago\n\nSolved\n\nSwap the input arguments\nWrite a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...\n\n4 months ago\n\nSolved\n\nColumn Removal\nRemove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...\n\n4 months ago\n\nSolved\n\nReverse the vector\nReverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1]\n\n4 months ago\n\nSolved\n\nLength of the hypotenuse\nGiven short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<https://i.imgu...\n\n4 months ago\n\nSolved\n\nGenerate a vector like 1,2,2,3,3,3,4,4,4,4\nGenerate a vector like 1,2,2,3,3,3,4,4,4,4 So if n = 3, then return [1 2 2 3 3 3] And if n = 5, then return [1 2 2...\n\n4 months ago\n\nSolved\n\nReturn area of square\nSide of square=input=a Area=output=b\n\n4 months ago\n\nSolved\n\nMaximum value in a matrix\nFind the maximum value in the given matrix. For example, if A = [1 2 3; 4 7 8; 0 9 1]; then the answer is 9.\n\n4 months ago\n\nSolved\n\nVector creation\nCreate a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.\n\n4 months ago\n\nSolved\n\nDoubling elements in a vector\nGiven the vector A, return B in which all numbers in A are doubling. So for: A = [ 1 5 8 ] then B = [ 1 1 5 ...\n\n4 months ago\n\nSolved\n\nCreate a vector\nCreate a vector from 0 to n by intervals of 2.\n\n4 months ago\n\nSolved\n\nFlip the vector from right to left\nFlip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ...\n\n4 months ago\n\nSolved\n\nWhether the input is vector?\nGiven the input x, return 1 if x is vector or else 0.\n\n4 months ago\n\nSolved\n\nFind max\nFind the maximum value of a given vector or matrix.\n\n4 months ago\n\nSolved\n\nGet the length of a given vector\nGiven a vector x, the output y should equal the length of x.\n\n4 months ago\n\nSolved\n\nInner product of two vectors\nFind the inner product of two vectors.\n\n4 months ago\n\nSolved\n\nArrange Vector in descending order\nIf x=[0,3,4,2,1] then y=[4,3,2,1,0]\n\n4 months ago\n\nSolved\n\nDetermine if input is odd\nGiven the input n, return true if n is odd or false if n is even.\n\n4 months ago\n\nSolved\n\nPangrams!\nA pangram, or holoalphabetic sentence, is a sentence using every letter of the alphabet at least once. Example: Input s ...\n\n4 months ago\n\nSolved\n\nCounting Money\nAdd the numbers given in the cell array of strings. The strings represent amounts of money using this notation: \\$99,999.99. E...\n\n4 months ago\n\nSolved\n\nNearest Numbers\nGiven a row vector of numbers, find the indices of the two nearest numbers. Examples: [index1 index2] = nearestNumbers([2 5 3...\n\n4 months ago\n\nSolved\n\nSort a list of complex numbers based on far they are from the origin.\nGiven a list of complex numbers z, return a list zSorted such that the numbers that are farthest from the origin (0+0i) appear f...\n\n4 months ago\n\nSolved\n\nRemove all the words that end with \"ain\"\nGiven the string s1, return the string s2 with the target characters removed. For example, given s1 = 'the main event' your ...\n\n4 months ago" ]	[ null, "https://www.mathworks.com/responsive_image/150/150/0/0/0/cache/matlabcentral/profiles/19558205_1600159127372_DEF.jpg", null, "https://www.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/introduction_to_matlab.png", null, "https://www.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/community_authored_group.png", null, "https://www.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/cody_challenge_master.png", null, "https://www.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/solver.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6861306,"math_prob":0.9940376,"size":4979,"snap":"2022-05-2022-21","text_gpt3_token_len":1522,"char_repetition_ratio":0.200201,"word_repetition_ratio":0.053376906,"special_character_ratio":0.29524,"punctuation_ratio":0.16694915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964691,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T03:09:37Z\",\"WARC-Record-ID\":\"<urn:uuid:268f88f7-12a1-429e-bafe-5fdbf959741c>\",\"Content-Length\":\"101138\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ecd25b8-1374-4137-85ce-bf7d322eef44>\",\"WARC-Concurrent-To\":\"<urn:uuid:e573e699-ea3f-469e-9165-8187634893ce>\",\"WARC-IP-Address\":\"23.39.174.83\",\"WARC-Target-URI\":\"https://www.mathworks.com/matlabcentral/profile/authors/19558205?detail=cody\",\"WARC-Payload-Digest\":\"sha1:WDC7WOXLBLZPA5N573G63AIJ3W3XLBVF\",\"WARC-Block-Digest\":\"sha1:VERADDXBKFF4MSCEI46Y2WBFER2W7D5D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301670.75_warc_CC-MAIN-20220120005715-20220120035715-00498.warc.gz\"}"}
https://www.geeksforgeeks.org/maximum-squares-possible-parallel-to-both-axes-from-n-distinct-points/	[ "Maximum Squares possible parallel to both axes from N distinct points\n\n• Last Updated : 26 Nov, 2021\n\nGiven Xi and Yi co-ordinate of N distinct points in the 2-D Plane, the task is to count the number of squares that can be formed from these points that are parallel to both X and Y-axis.\nExamples:\n\nInput : X[] = { 0, 2, 0, 2 }, Y[] = { 0, 2, 2, 0 }\nOutput :\nExplanation: Only one Square can be formed using these points –\n(0, 2)(2, 2)\n(0, 0)(2, 0)\nInput : X[] = { 3, 2, 0, 6 }, Y[] = { 0, 2, 0, 6 }\nOutput :\nExplanation: No Square can be formed using these points –\n(3,0),(2,0), (0,0), (6,6)\n\nNaive Approach: Iterate over all possible combinations of four points and check if a square can be formed that is parallel to both X and Y-axis. The time complexity of this approach would be O(N4)\nEfficient Approach: We can observe that for any four points to make a required square, the following conditions must hold true –\n\n• The points lying in the same horizontal line must have the same Y co-ordinates.\n\n• The points lying in the same vertical line must have the same X co-ordinates.\n\n• The distance between these points should be same.\n\nThus the four points of any square can be written as P1(X1, Y1), P2(X2, Y1), P3(X2, Y2) and P4(X1, Y2) in clockwise direction.\nConsider any two points on the same horizontal or vertical line from the given ones. Calculate the distance between them. Based on that, form the other two points. Now check if both of the given points are present or not. If so, it ensures that a square is present. Hence, increase the count and proceed further.\nBelow is the implementation of the above approach:\n\nC++\n\n // C++ implementation of the approach#include using namespace std; // Function that returns the count of// squares parallel to both X and Y-axis// from a given set of pointsint countSquares(int* X, int* Y, int N){ // Initialize result int count = 0; // Initialize a set to store points set > points; // Initialize a map to store the // points in the same vertical line map > vertical; // Store the points in a set for (int i = 0; i < N; i++) { points.insert({ X[i], Y[i] }); } // Store the points in the same vertical line // i.e. with same X co-ordinates for (int i = 0; i < N; i++) { vertical[X[i]].push_back(Y[i]); } // Check for every two points // in the same vertical line for (auto line : vertical) { int X1 = line.first; vector yList = line.second; for (int i = 0; i < yList.size(); i++) { int Y1 = yList[i]; for (int j = i + 1; j < yList.size(); j++) { int Y2 = yList[j]; int side = abs(Y1 - Y2); int X2 = X1 + side; // Check if other two point are present or not if (points.find({ X2, Y1 }) != points.end() and points.find({ X2, Y2 }) != points.end()) count++; } } } return count;} // Driver Codeint main(){ int X[] = { 0, 2, 0, 2 }, Y[] = { 0, 2, 2, 0 }; int N = sizeof(X) / sizeof(X); cout << countSquares(X, Y, N); return 0;}\n\nPython3\n\n # Python3 implementation of the approach # Function that returns the count of# squares parallel to both X and Y-axis# from a given set of pointsdef countSquares(X, Y, N) : # Initialize result count = 0; # Initialize a set to store points points = []; # Initialize a map to store the # points in the same vertical line vertical = dict.fromkeys(X, None); # Store the points in a set for i in range(N) : points.append((X[i], Y[i])); # Store the points in the same vertical line # i.e. with same X co-ordinates for i in range(N) : if vertical[X[i]] is None : vertical[X[i]] = [Y[i]]; else : vertical[X[i]].append(Y[i]); # Check for every two points # in the same vertical line for line in vertical : X1 = line; yList = vertical[line]; for i in range(len(yList)) : Y1 = yList[i]; for j in range(i + 1, len(yList)) : Y2 = yList[j]; side = abs(Y1 - Y2); X2 = X1 + side; # Check if other two point are present or not if ( X2, Y1 ) in points and ( X2, Y2 ) in points : count += 1; return count; # Driver Codeif __name__ == \"__main__\" : X = [ 0, 2, 0, 2 ]; Y = [ 0, 2, 2, 0 ]; N = len(X); print(countSquares(X, Y, N)); # This code is contributed by AnkitRai01\nOutput:\n1\n\nTime Complexity: O(N2).\n\nAuxiliary Space: O(N)\n\nMy Personal Notes arrow_drop_up" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7455909,"math_prob":0.99792963,"size":4157,"snap":"2022-05-2022-21","text_gpt3_token_len":1340,"char_repetition_ratio":0.14206597,"word_repetition_ratio":0.1704142,"special_character_ratio":0.35386094,"punctuation_ratio":0.18221258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99900895,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T23:59:47Z\",\"WARC-Record-ID\":\"<urn:uuid:b1003942-6862-44a3-bb91-d8f0c1ce4cd9>\",\"Content-Length\":\"138186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7e03ffa-c2ff-482b-84cd-e37c990aa2c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:1aeea6f1-77ad-4e2d-80fa-a30c0a507721>\",\"WARC-IP-Address\":\"23.12.145.61\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/maximum-squares-possible-parallel-to-both-axes-from-n-distinct-points/\",\"WARC-Payload-Digest\":\"sha1:L73F6CA3OYUXLFPHZLJ6CBJKUZZSKSC7\",\"WARC-Block-Digest\":\"sha1:A2OUU3A3GCX47EKA3M24LQE7BFFWJSH3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304876.16_warc_CC-MAIN-20220125220353-20220126010353-00136.warc.gz\"}"}
http://www.numdam.org/item/?id=RSMUP_2013__130__169_0	[ "Hölder continuity for sub-elliptic systems under the sub-quadratic controllable growth in Carnot groups\nRendiconti del Seminario Matematico della Università di Padova, Volume 130 (2013), p. 169-202\n@article{RSMUP_2013__130__169_0,\nauthor = {Wang, Jialin and Liao, Dongni and Yu, Zefeng},\ntitle = {H\\\"older continuity for sub-elliptic systems under the sub-quadratic controllable growth in Carnot groups},\njournal = {Rendiconti del Seminario Matematico della Universit\\a di Padova},\npublisher = {Seminario Matematico of the University of Padua},\nvolume = {130},\nyear = {2013},\npages = {169-202},\nmrnumber = {3148637},\nlanguage = {en},\nurl = {http://www.numdam.org/item/RSMUP_2013__130__169_0}\n}\n\nWang, Jialin; Liao, Dongni; Yu, Zefeng. Hölder continuity for sub-elliptic systems under the sub-quadratic controllable growth in Carnot groups. Rendiconti del Seminario Matematico della Università di Padova, Volume 130 (2013) , pp. 169-202. http://www.numdam.org/item/RSMUP_2013__130__169_0/`\n\n E. De Giorgi, Un esempio di estremali discontinue per un problema variazionale di tipo ellitico. Boll. Unione Mat. Italiana 4 (1968), pp. 135-137. \| MR 227827\n\n M. Giaquinta, Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic Systems. Princeton Univ. Press, Princeton, 1983. \| MR 717034\n\n M. Giaquinta, Introduction to regularity theory for nonlinear elliptic systems. Birkhäuser, Berlin, 1993. \| MR 1239172\n\n Y. Chen - L. Wu, Second order elliptic equations and elliptic systems. Science Press, Beijing, 2003.\n\n M. Giaquinta - G. Modica, Almost-everywhere regularity results for solutions of non linear elliptic systems. Manuscripta Math. 28 (1979), pp. 109-158, \| MR 535699\n\n E. Giusti - M. Miranda, Sulla regolarità delle soluzioni deboli di una classe di sistemi ellittici quasilineari. Arch. Rat. Mech. Anal. 31 (1968), pp. 173-184. \| MR 235264\n\n F. Duzaar - J. F. Grotowski, Partial regularity for nonlinear elliptic systems: The method of A-harmonic approximation, Manuscripta Math. 103 (2000), pp. 267-298. \| MR 1802484\n\n L. Simon, Lectures on Geometric Measure Theory. Australian National University Press, Canberra, 1983. \| MR 756417\n\n F. Duzaar - J. F. Grotowski - M. Kronz, Regularity of almost minimizers of quasi-convex variational integrals with subquadratic growth. Annali Mat. Pura Appl. (4) 184 (2005), pp. 421-448. \| MR 2177809\n\n F. Duzaar - G. Mingione, The $p$ -harmonic approximation and the regularity of $p$ -harmonic maps. Calc. Var. Partial Differential Equations 20 (2004), pp. 235-256. \| MR 2062943\n\n F. Duzaar - G. Mingione, Regularity for degenerate elliptic problems via $p$ -harmonic approximation. Ann. Inst. H. Poincaré Anal. Non Linèaire 21 (2004), pp. 735-766. \| MR 2086757\n\n M. Carozza - N. Fusco - G. Mingione, Partial regularity of minimizers of quasiconvex integrals with subquadratic growth, Annali Mat. Pura Appl. (4) 175 (1998), pp. 141-164. \| MR 1748219\n\n S. Chen - Z. Tan, The method of A-harmonic approximation and optimal interior partial regularity for nonlinear elliptic systems under the controllable growth condition. J. Math. Anal. Appl. 335 (2007), pp. 20-42. \| MR 2340302\n\n S. Chen - Z. Tan, Optimal interior partial regularity for nonlinear elliptic systems. Discrete Contin. Dyn. Syst. 27 (2010), pp. 981-993. \| MR 2629569\n\n L. Capogna - N. Garofalo, Regularity of minimizers of the calculus of variations in Carnot groups via hypoellipticity of systems of Hörmander type. J. European Math. Society 5 (2003), pp. 1-40. \| MR 1961133\n\n E. Shores, Hypoellipticity for linear degenerate elliptic systems in Carnot groups and applications, arXiv:math/0502569, pp. 27.\n\n A. Föglein, Partial regularity results for subelliptic systems in the Heisenberg group, Calc. Var. Partial Differential Equations 32 (2008), pp. 25-51. \| MR 2377405\n\n J. Wang - P. Niu, Optimal partial regularity for weak solutions of nonlinear sub-elliptic systems in Carnot groups. Nonlinear Analysis 72 (2010), pp. 4162-4187. \| MR 2606775\n\n L. Capogna - D. Danielli - N. Garofalo, An embedding theorem and the Harnak inequality for nonlinear subelliptic equations. Comm. Partial Differential Equations 18 (1993), pp. 1765-1794. \| MR 1239930\n\n G. Lu, Embedding theorems into Lipschitz and BMO spaces and applications to quasilinear subelliptic differential equations. Publ. Mat. 40 (1996), pp. 301-329. \| MR 1425620\n\n C. Xu, Regularity for quasi-linear second order subelliptic equations. Comm. Pure Appl. Math. 45 (1992), pp. 77-96.\n\n L. Capogna, Regularity of quasi-linear equations in the Heisenberg group. Comm. Pure Appl. Math. 50 (1997), pp. 867-889. \| MR 1459590\n\n L. Capogna, Regularity for quasilinear equation and 1-quasiconformal maps in Carnot groups. Math. Ann. 313 (1999), pp. 263-295. \| MR 1679786\n\n S. Marchi, ${C}^{1,\\alpha }$ local regularity for the solutions of the $p$ -Laplacian on the Heisenberg group for $2p+\\sqrt{5}$ . Z. Anal. Anwendungen 20 (2001), pp. 617-636. \| MR 1863937\n\n S. Marchi, ${C}^{1,\\alpha }$ local regularity for the solutions of the $p$ -Laplacian on the Heisenberg group for $1+\\frac{1}{\\sqrt{5}}p\\le 2$ . Comment. Math. Univ. Carolinae 44 (2003), pp. 33-56. \| MR 2045844\n\n A. Domokos, Differentiability of solutions for the non-degenerate $p$ -Laplacian in the Heisenberg group. J. Differential Equations. 204 (2004), pp. 439-470. \| MR 2085543\n\n A. Domokos, On the regularity of $p$ -harmonic functions in the Heisenberg group. Ph. D. Thesis, University of Pittsburgh, 2004.\n\n J. Manfredi - G. Mingione, Regularity results for quasilinear elliptic equations in the Heisenberg group. Math. Ann. 339 (2007), pp. 485-544. \| MR 2336058\n\n G. Mingione - A. Zatorska-Goldstein - X. Zhong, Gradient regularity for elliptic equations in the Heisenberg group. Advances in Mathematics 222 (2009), pp. 62-129. \| MR 2531368\n\n N. Garofalo, Gradient bounds for the horizontal $p$ -Laplacian on a Carnot group and some applications. Manuscripta Math. 130 (2009), pp. 375-385. \| MR 2545524\n\n G. Folland, Subelliptic estimates and function spaces on nilpotent Lie groups. Ark. Mat. 13 (1975), pp. 161-207. \| MR 494315\n\n E. Acerbi - N. Fusco, Regularity for minimizers of nonquadratic functionals: the case $1p$ , J. Math. Anal. Appl. 140 (1989), pp. 115-135. \| MR 997847\n\n G. Lu, Embedding theorems on Campanato-Morrey space for vector fields on Hörmander type. Approx. Theory Appl. 14 (1998), pp. 69-80. \| MR 1651473" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62917626,"math_prob":0.9261684,"size":5596,"snap":"2020-24-2020-29","text_gpt3_token_len":1777,"char_repetition_ratio":0.16577253,"word_repetition_ratio":0.057387058,"special_character_ratio":0.3654396,"punctuation_ratio":0.24093023,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9744617,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T02:38:27Z\",\"WARC-Record-ID\":\"<urn:uuid:258d0756-c214-479e-a15a-5cc06207bff2>\",\"Content-Length\":\"37370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d1e58b1-6e85-4ca2-9422-42391915ad28>\",\"WARC-Concurrent-To\":\"<urn:uuid:950a52ba-1c4b-443d-a603-d7843fcde625>\",\"WARC-IP-Address\":\"129.88.220.36\",\"WARC-Target-URI\":\"http://www.numdam.org/item/?id=RSMUP_2013__130__169_0\",\"WARC-Payload-Digest\":\"sha1:67OAIMIGJ6X564WCNILI3G2JXTECQTXQ\",\"WARC-Block-Digest\":\"sha1:SJJJXZKNSXIR46IRJUYBMYDGQC74XF3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347436828.65_warc_CC-MAIN-20200604001115-20200604031115-00447.warc.gz\"}"}
https://roytuts.com/printing-pyramid-with-stars-in-unix-shell-programming/	[ "# Printing pyramid with stars in Unix Shell Programming\n\nThe below unix shell programming will show you an example on printing pyramid with stars in unix shell programming. This tutorial will also show how to use if else condition, how to use for loop in shell programming. You will also see how to take input from keyboard.\n\nI will be printing pyramid with stars in unix shell programming in two ways – one is stars exist inside and another one is stars do not exist inside the pyramid.\n\nSo let’s see the first example where stars exist also inside the pyramid.\n\nThe below shell program first takes the depth or level or height from the user for the pyramid. The program itself calculates the number of stars at each level according to the height of the pyramid. I also put the space between two stars to look good.\n\nThe below shell programming is written into the file pyramid-stars-inner.sh.\n\n``````#!/bin/bash\n\nread -r -p \"Enter depth of pyramid: \" n\necho \"You enetered level: \\$n\"\n\nspace=n\n\nfor((i=1;i<=n;i++))\ndo\nspace=\\$((space-1)) #calculate how many spaces should be printed before \n\nfor((j=1;j<=space;j++))\ndo\necho -n \" \" #print spaces on the same line before printing \ndone\nfor((k=1;k<=i;k++))\ndo\necho -n \"\" #print on the same line\necho -n \" \" #print space after * on the same line\ndone\necho -e #print new line after each row\ndone``````\n\nMake sure you change the permission of the above shell script by using the following command:\n\n``\\$ chmod +x pyramid-stars-inner.sh``\n\nTo run or execute the above shell script use the following command:\n\n``\\$ ./pyramid-stars-inner.sh``\n\nAfter running the above shell program you will get the below output. So here you see that the pyramid is filled with stars in whole area. So this is a solid pyramid where inner surface is also filled with stars.\n\nNow let’s see the example where outer surface of the pyramid is filled with stars. Here also you will get the input for level or height of the pyramid and accordingly we put stars on outer surface at each level at two furthest ends.\n\nThe following shell program is written into the file pyramid-stars-outer.sh.\n\n``````read -r -p \"Enter depth of pyramid: \" n\necho \"You enetered level: \\$n\"\n\ns=0\nspace=n\n\nfor((i=1;i<=n;i++))\ndo\nspace=\\$((space-1)) #calculate how many spaces should be printed before \n\nfor((j=1;j<=space;j++))\ndo\necho -n \" \" #print spaces on the same line before printing \ndone\nfor((k=1;k<=i;k++))\ndo\nif((i==1 \|\| i==2 \|\| i==n))\nthen\necho -n \"\" #print on the same line\necho -n \" \" #print space after * on the same line\nelif((k==1))\nthen\necho -n \"\" #print on the same line\nelif((k==i))\nthen\nfor((l=1;l<=k+s;l++))\ndo\necho -n \" \" #print spaces on the same line before printing \ndone\ns=\\$((s+1)) #as pyramid expands at bottom, so we need to recalculate inner spaces\necho -n \"\" #print * on the same line\nfi\ndone\necho -e #print new line after each row\ndone``````\n\nMake sure you again change the permission using the command:\n\n``\\$ chmod +x pyramid-stars-outer.sh``\n\nTo run the shell script use the following command:\n\n``\\$ ./pyramid-stars-outer.sh``\n\nHere is the below output you get after running the above shell script." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77144676,"math_prob":0.8175048,"size":3011,"snap":"2021-43-2021-49","text_gpt3_token_len":773,"char_repetition_ratio":0.14931826,"word_repetition_ratio":0.31548756,"special_character_ratio":0.2660246,"punctuation_ratio":0.07954545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99179167,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T06:47:53Z\",\"WARC-Record-ID\":\"<urn:uuid:2ab14082-1848-4a34-85b9-285325578a40>\",\"Content-Length\":\"49990\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d9cf3f8-795a-4783-9327-cd6a8fc6b901>\",\"WARC-Concurrent-To\":\"<urn:uuid:85c19c09-dbaa-42fa-a54d-04d7aabbc12f>\",\"WARC-IP-Address\":\"104.21.68.14\",\"WARC-Target-URI\":\"https://roytuts.com/printing-pyramid-with-stars-in-unix-shell-programming/\",\"WARC-Payload-Digest\":\"sha1:KIATHEAOGHYGRKECHB3XS3VTM4FHPKWB\",\"WARC-Block-Digest\":\"sha1:CTSFHLWITW3JHVLJGCWOBGVHYXRA2N5F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585460.87_warc_CC-MAIN-20211022052742-20211022082742-00094.warc.gz\"}"}
http://www.numbersaplenty.com/3540266303	[ "Search a number\nBaseRepresentation\nbin1101001100000100…\n…0010110100111111\n3100010201122011101012\n43103001002310333\n524222302010203\n61343144105435\n7153505524050\noct32301026477\n910121564335\n103540266303\n11155742a522\n1282976527b\n134455c7442\n14258280727\n1515ac12bd8\nhexd3042d3f\n\n3540266303 has 8 divisors (see below), whose sum is σ = 4284019872. Its totient is φ = 2856013056.\n\nThe previous prime is 3540266293. The next prime is 3540266311. The reversal of 3540266303 is 3036620453.\n\nAdding to 3540266303 its reverse (3036620453), we get a palindrome (6576886756).\n\n3540266303 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is not a de Polignac number, because 3540266303 - 24 = 3540266287 is a prime.\n\nIt is a Duffinian number.\n\nIt is a self number, because there is not a number n which added to its sum of digits gives 3540266303.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (3540266363) by changing a digit.\n\nIt is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 14874950 + ... + 14875187.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (535502484).\n\nAlmost surely, 23540266303 is an apocalyptic number.\n\n3540266303 is a deficient number, since it is larger than the sum of its proper divisors (743753569).\n\n3540266303 is a wasteful number, since it uses less digits than its factorization.\n\n3540266303 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 29750161.\n\nThe product of its (nonzero) digits is 38880, while the sum is 32.\n\nThe square root of 3540266303 is about 59500.1369998423. The cubic root of 3540266303 is about 1524.0947852619.\n\nThe spelling of 3540266303 in words is \"three billion, five hundred forty million, two hundred sixty-six thousand, three hundred three\"." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8284146,"math_prob":0.99042946,"size":2129,"snap":"2020-24-2020-29","text_gpt3_token_len":645,"char_repetition_ratio":0.17411764,"word_repetition_ratio":0.0058139535,"special_character_ratio":0.45937058,"punctuation_ratio":0.13231552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9942571,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T00:35:15Z\",\"WARC-Record-ID\":\"<urn:uuid:caabe264-7ef4-4c6e-893d-063577d5f1d1>\",\"Content-Length\":\"9224\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c063dca-01d6-4a77-bbcc-374276f6dbda>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c11dc54-9fc1-4387-a70b-247a657d7f69>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"http://www.numbersaplenty.com/3540266303\",\"WARC-Payload-Digest\":\"sha1:TJUHD34QDG3LDS43JXP3Z3UIGADDZKNO\",\"WARC-Block-Digest\":\"sha1:WGQECIENQIN2N2YIIEBH7THTN5HNH4CP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347436828.65_warc_CC-MAIN-20200604001115-20200604031115-00333.warc.gz\"}"}
http://m.maonotech.com/info/what-is-compression-ratio-27067811.html	[ "What Is Compression Ratio?\n\n- Jul 02, 2018-\n\nThe compression ratio is the ratio of the input signal gain to the output signal gain. It is generally expressed by a certain ratio. For example, a compression ratio of 1.5:1 cannot be said to be 3:2, which means that the input signal level exceeding the threshold is increased by 1.5. Decibel, the output is increased by 1 decibel. When the compression ratio is equal to infinity, the output signal will no longer increase when the input signal is increased. In this case, the compressor is used as a limiter." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9155602,"math_prob":0.9854157,"size":689,"snap":"2019-43-2019-47","text_gpt3_token_len":154,"char_repetition_ratio":0.15182482,"word_repetition_ratio":0.0,"special_character_ratio":0.20754717,"punctuation_ratio":0.15714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9692994,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T11:29:25Z\",\"WARC-Record-ID\":\"<urn:uuid:3c3bd02e-6f13-40ed-82f1-fe3f93890f51>\",\"Content-Length\":\"9430\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1a94d256-adb9-4c9a-9fe3-e85c4e5a3da1>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9b4b765-df08-448c-b388-8fc84aec4164>\",\"WARC-IP-Address\":\"104.42.26.214\",\"WARC-Target-URI\":\"http://m.maonotech.com/info/what-is-compression-ratio-27067811.html\",\"WARC-Payload-Digest\":\"sha1:POGIDM375XL5V3NMMF53H6R5I4XLSR6Q\",\"WARC-Block-Digest\":\"sha1:MR2OFPALSCDSAACH26GAPALJXOYPRCFE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986653216.3_warc_CC-MAIN-20191014101303-20191014124303-00421.warc.gz\"}"}
https://www.solvemathproblem.com/2021/08/31/1351-2/	[ "# domain and the range\n\nMidterm Exam Review MATH 178 NAME_____________________________ 1. Find the following limits: x2 − x − 2 x →−1 x 2 + 3 x + 2 a. lim 3x + 2 b. lim x→2 3×3 − 3x 2 − 6 x s →−1 x2 + x c. lim 3x 2 − 5 x → 2 x 2 − 4 e. lim d. lim x→4 1 x−4 x3 − 5 x → 2 x 4 − 4 x + 1 f. lim  x2 2. Given the following, sketch the graph and find the limits: f ( x ) =  3 − x a. Find lim− f ( x) x →2 b. Find lim+ f ( x) x →2 c. Find lim f ( x) x →2 if x2 if x2  x2 + 2 3. Given the following: f ( x) =  6 x + 9 if if x  −1 x  −1 a. Find f ( −3) b. Find f (5) c. Find f (−1) 4. Given the following graph, identify the domain and the range. 5. Find the domain for the following functions: f ( x) = 5 − 7 x b. f ( x) = x−4 2 x − 15 x + 7 2 6. Find the slope of the tangent line on the curve f ( x) = 5x 2 − 3x + 8 at the point (−2,34) 7. Find the equation of the tangent line to the curve y = 9 − 2 x 2 at the point (2, 1). Use: y = mx + b or y − y1 = m( x − x1 ) 8. Find the following derivatives: a. f ( x) = 4 x3 − 5x 2 + 8x + 2 b. f ( x) = ( x 2 + 3)( x 2 − 5) c. f ( x) = 5 x7 d. f ( x) = x+2 x −3 e. f ( x) = x2 − 5x − 1 f. g ( x) = ( x 3 − 5 ) 7 g. f ( x) = 3x ln x  x3  h. y = ln    2x − 7  i. f ( x) = 10e4 x − 3×2 j. g ( x) = k. f ( x) = ln(5x 2 − 3x + 6) l.  x +5 g ( x) =    x  2 4×2 ln x 4 9. Given f ( x) = 4 x3 − 2 x 2 − 5x + 3 find f (−2) 10. Find f (x) given f ( x) = 1 ( 2 x − 1) 3 11. A ball is thrown vertically upward with a velocity of 80 ft/sec its height after t seconds is given by the function: s(t ) = −16t 2 + 80t . Solve the following using calculus techniques discussed in class. a. What is the velocity of the ball after 2 seconds? b. What is the maximum height reached by the ball? 12. A company that produces and sells cellular phones determines that the cost of producing x telephones is C ( x) = 0.45x 2 − 15x + 250 dollars. Find the rate at which the cost is changing when 100 cellular phones are produced. 13. Given the following graph, find the limits: a. lim f ( x) = __________ x →1− d. lim− f ( x) = __________ x →2 b. lim+ f ( x) = __________ c. lim f ( x) = __________ e. lim f ( x) = __________ f. lim+ f ( x) = __________ x →1 x →−2 x →1 x →2 14. A bank advertises that it compounds interest continuously and that it will double your money in 15 years. Find the function that satisfies the equation and what is the annual interest rate? Hint: use P = Po e rt 15. How old is a Chinese artifact that has lost 60% of its Carbon-14? Carbon-14 has a decay rate of 0.01205% per year. Hint: use N = N o e − rt 16. The average cost C of a prime-rib dinner was \\$4.65 in 1962. Assuming that the exponential growth model applies, and that the exponential growth rate is 5.1%, what will be the cost of such a dinner in 2018? Hint: use P = Po e rt Solutions: 1. 2. 3. 4. 5. a) b) c) d) e) f) 8 -3 –9 DNE (does not exist) 3/2 0 a) 4 b) 1 c) DNE a) 11 b) 39 c) 3 D: (-3, 1] R: [-4, 0] a. (-inf, 5/7] b. All reals except ½ and 7 6. m = -23 7. y= – 8x + 17 8. a) f ( x) = 12 x 2 − 10 x + 8 b) −5 c) f ( x) = −35 x8 d) f ( x) = e) f ( x) = 2x − 5 2 x² − 5x − 1 f) g ( x) = 21x 2 ( x3 − 5 ) 2 3 2 − x x−7 8 x ln x − 4 x j) g ( x) = (ln x)2 i) f ( x) = 80xe4 x − 6 x 2 k) f ( x) = ( x − 3) h) y = g) f ( x) = 3 + 3ln x 9. f ( x) = 4 x ³ − 4 x 10 x − 3 5 x 2 − 3x + 6 l) g ( x) = −20( x + 5)3 x5 f (−2) = 51 10. f ( x) = 48 ( 2 x − 1) 5 11. a) 16 ft/sec b) 100 ft 12. \\$75, the cost will increase \\$75 when the 101st phone is produced. 13. a) 1 14. 4.6% b) 3 c) DNE d) 3 15. 7604 years e) 4 16. \\$80.87 f) 2 6\nPurchase answer to see full attachment\n\nJust \\$7 Welcome" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6565322,"math_prob":0.999984,"size":3545,"snap":"2022-40-2023-06","text_gpt3_token_len":1533,"char_repetition_ratio":0.18441118,"word_repetition_ratio":0.046343975,"special_character_ratio":0.5210155,"punctuation_ratio":0.119198315,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9999485,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T18:13:48Z\",\"WARC-Record-ID\":\"<urn:uuid:e657f2ea-ae87-4c15-843f-5435ef9b2adc>\",\"Content-Length\":\"50226\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21107e3f-f317-4261-a200-904b000d33c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:8539e089-f63e-48e1-86a5-0b47beed4580>\",\"WARC-IP-Address\":\"5.161.111.97\",\"WARC-Target-URI\":\"https://www.solvemathproblem.com/2021/08/31/1351-2/\",\"WARC-Payload-Digest\":\"sha1:I4V2S6H423DYSZSXDQJDFEQPXAQG3G2X\",\"WARC-Block-Digest\":\"sha1:RSEFYBZHSOD6I4YS5XDMZBDI5EHBKWXR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335362.18_warc_CC-MAIN-20220929163117-20220929193117-00621.warc.gz\"}"}
https://plato.stanford.edu/entries/counterfactuals/figdesc.html	[ "## Long descriptions for some figures in Counterfactuals\n\n### Figure 2 description\n\nA diagram consisting of eight circles labelled A through H. These are laid out in a five columns by three rows array. A is at in the first column from the left and first row from the top; B is in the same column but in the bottom row; both have arrows pointing to C which is in the second column, middle row; A's arrow is labelled $$C \\dequal A \\land B$$. D is in the third column top row and E in the same column bottom row; C has arrows going to both with the arrow to D labelled $$D \\dequal \\neg C$$ and the arrow to E labelled $$E \\dequal C$$ . F is in the fourth column top row and G in the same column bottom row; an arrow goes fro D to F labelled $$F \\dequal \\neg D$$; an arrow goes from E to G labelled $$G \\dequal E$$. H is in the fifth column middle row; arrows go to it from F and G; the arrow from F to H is labelled $$H \\dequal F \\lor G$$.\n\n### Figure 4 description\n\nThe diagram is a circle outlined in red and split into four parts by a horizontal line and a vertical line. The horizontal line is labelled ‘Antecedent’ with a $$\\phi$$ above and a $$\\neg \\phi$$ below. The vertical line is labelled ‘Consequent’ with a $$\\psi$$ to the left and a $$\\neg\\psi$$ to the right. The center of the circle is labelled $$w_0$$. The upper half of the circle is tinted blue and the upper right quadrant is shaded with a pattern. This upper half is also labelled ‘Constant Domain’. The whole diagram is labeled ‘Accessible Worlds $$R(w_0)$$. Above the diagram is a legend: “Strict Analysis of $$\\phi > \\psi$$: All φ-worlds accessible from $$w_0$$ (indicated in blue in the diagram) are ψ-worlds. I.e., shaded regions must be empty”\n\n### Figure 5 description\n\nThe diagram is three concentric circles outlined in red (so forming two rings, outer and middle, and an innermost circle) and split into four parts by a horizontal line and a vertical line. The horizontal line is labelled ‘Antecedent’ with a $$\\phi$$ above and a $$\\neg \\phi$$ below. The vertical line is labelled ‘Consequent’ with a $$\\psi$$ to the left and a $$\\neg\\psi$$ to the right. The center of the circles is labelled $$w_0$$. The upper half of the circles is tinted blue with the tint with the lightest tint in the outer ring, slightly darker in the middle ring and darkest in the innermost circle; this half is labelled ‘Variable Domain’. The upper right quadrant of the innermost circle is shaded with a pattern. In the middle ring in the upper right quadrant is a dot labelled $$w_1$$. The whole diagram is labelled ‘System of Spheres of Similarity $$\\mathcal{R}(w_0)$$’. Above the diagram is a legend: “ Similarity Analysis of} $$\\phi > \\psi$$ All $$\\phi$$-worlds most similar to $$w_0$$ (indicated in blue in the diagram) are $$\\psi$$-worlds. I.e. shaded region must be empty.\n\n### Figure 6 description\n\nThe diagram is a circle outlined in red and split into four parts by a horizontal line and a vertical line. The horizontal line is labelled ‘Antecedent’ with a $$\\phi_1$$ above and a $$\\neg \\phi_1$$ below. The vertical line is labelled ‘Consequent’ with a $$\\psi$$ to the left and a $$\\neg\\psi$$ to the right. The center of the circle is labelled $$w_0$$. The whole diagram is labeled ‘Accessible Worlds $$R(w_0)$$. The upper half of the circle is tinted blue and the upper right quadrant is shaded with a pattern. So far very similar to Figure 4; the major difference is a yellow circle outlined in black above the center of the first circle and bisected by the vertical line. Most of the yellow circle is inside the first circle, but, a small portion is outside. The yellow circle is labelled $${\\llbracket}\\phi_2{\\rrbracket}^R_v$$\n\n### Figure 7 description\n\nThis figure has two similar diagrams, a and b. Each diagram has three overlapping circles (left, center, right); however, the overlap between left and right is completely within the center circle. In the non-overlapping section of the right circle is a dot labelled w.\n\nFor diagram a the left circle is tinted yellow and labelled $${\\llbracket}\\phi{\\rrbracket}^R_v$$; the center circle is gray and labelled $${\\llbracket}\\psi{\\rrbracket}^R_v$$; the right is blue and labelled $$R(w)$$. The whole is labelled $$\\llbracket\\phi\\strictif\\psi\\rrbracket^R_v$$.\n\nFor diagram b the left circle is tinted red and labelled $${\\llbracket}\\neg\\psi{\\rrbracket}^R_v$$; the center circle is gray and labelled $${\\llbracket}\\neg\\phi{\\rrbracket}^R_v$$; the right is purple and labelled $$R(w)$$. The whole is labelled $$\\llbracket\\neg\\psi\\strictif\\neg\\phi\\rrbracket^R_v$$.\n\n### Figure 8 description\n\nThe diagram is made up of four circles (yellow, orange, blue, gray). The center of the orange circle is a point labelled $$w$$; the circle as a whole is labelled $${\\llbracket}\\phi_1{\\rrbracket}^R_v$$. The yellow circle is to the left and overlaps the orange circle (the overlap includes the $$w$$ point); the yellow circle is labelled $${\\llbracket}\\phi_2{\\rrbracket}^R_v$$. The blue circle is to the right and overlaps the orange circle (the overlap again includes the $$w$$ point); the blue circle is labelled $$R(w)$$. The orange circle is completely within the union of the blue and yellow circles. The gray circle is to the lower right of the orange circle; it includes all of the yellow/blue overlap and orange/blue overlap but not all of the orange/yellow overlap; it is labelled $${\\llbracket}\\psi{\\rrbracket}^R_v$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8959129,"math_prob":0.9980077,"size":5420,"snap":"2023-14-2023-23","text_gpt3_token_len":1408,"char_repetition_ratio":0.19405465,"word_repetition_ratio":0.2698768,"special_character_ratio":0.25701106,"punctuation_ratio":0.08293153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99930894,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T15:20:42Z\",\"WARC-Record-ID\":\"<urn:uuid:3a104f69-35a2-407d-9724-6d6aaccdeb35>\",\"Content-Length\":\"17966\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e3d3637-22de-4df4-ae43-3cbba920064b>\",\"WARC-Concurrent-To\":\"<urn:uuid:47276e45-0900-45e2-9a46-21ff56f84e37>\",\"WARC-IP-Address\":\"171.67.193.20\",\"WARC-Target-URI\":\"https://plato.stanford.edu/entries/counterfactuals/figdesc.html\",\"WARC-Payload-Digest\":\"sha1:L7JGNRLKBO7PBLBA7FNOTHFRBGS5LVJA\",\"WARC-Block-Digest\":\"sha1:IN6INCJBM2HR5LR6V2TIVAHDUBBGE6BS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948867.32_warc_CC-MAIN-20230328135732-20230328165732-00759.warc.gz\"}"}
https://instil.ca/maths/math11functions	[ "• Reset\n\n• Settings\n• Shortcuts\n• Fullscreen\n\n# Math: Functions and Relations MCR3U\n\nThis course introduces some financial applications of mathematics, extends students' experiences with functions, and introduces second degree relations. Students will solve problems in personal finance involving applications of sequences and series; investigate properties and applications of trigonometric functions; develop facility in operating with polynomials, rational expressions, and exponential expressions; develop and understanding of inverses and transformations of functions; and develop facility in using function notation and in communicating mathematical reasoning. Students will also investigate loci and the properties and applications of conics. MCR3U has 25% to 30% more content than MCF3M, requiring that the pace be much faster than in MCF3M. MCR3U requires a very high degree of commitment on the part of the student. Prerequisite: Grade 10 Principles of Math MPM2D (Academic)\n\n# Exponent Laws\n\nSpecific Topic General Topic School Date\nExponent Laws Fractions, Negatives, Roots Bayview Nov 2013\nExponent Laws Decimals, Negatives Bayview Nov 2013\n\nxa × xb = xa + b\n\nxa ÷ xb = xa - b\n\n(xa)b = xa × b\n\n## Simplify fully to an integer or fraction using positive exponents, showing your work without using a calculator.\n\n$a^{-n} = \\left( \\cfrac{1}{a^{\\ +n}} \\right) \\quad\\quad or \\quad\\quad \\left( \\cfrac{a}{b} \\right)^{\\!\\!-n} = \\left( \\cfrac{b}{a} \\right)^{\\!\\!+n}$\n\n## Write each with a single positive exponent, or an integer, or a reduced fraction, where applicable. (Without using a calculator)\n\n$\\sqrt[n]{a^m} = (\\sqrt[n]{a})^m = a^{\\frac{m}{n}}$\n\n## Simplify given two of the most basic rules for roots.\n\n\\begin{align} & \\sqrt{a^2} = a \\\\ \\\\ & \\sqrt{ab} = \\sqrt{a} × \\sqrt{b} \\end{align}\n\n# Functions\n\n## Given the general form of a quadratic, determine the equation of a quadratic function with the transformations listed below....\n\n$y = a(x - h)^2 + k$\n\n# Quadratic Functions\n\n## When a soccer ball is kicked the parabolic path of its vertical height, h, in meters is given by the equation below, where the time, t, is in seconds.\n\n$h(t) = -5t^2 + 15t + 0.1$\n\n# Trigonometry\n\n## The following acute triangle, ABC contains the known values (SSA) indicated on the diagram below.", null, "## Given the trig equation,\n\n$4\\text{cos}^2\\,\\theta - 4\\text{cos}\\,\\theta + 1 = 0$\n\n# Periodic Trig Functions\n\n## For the graphs of the parent functions of $\\text{sin}\\,\\theta$ and $\\text{cos}\\,\\theta$:", null, "## Consider a graph of the parent function between the interval: $0 \\le x \\le 3\\pi$\n\n$y = \\text{cos}\\,\\theta$\n\n## Given the trig function below:\n\nƒ(x) = -4cos 2(θ + $\\pi$) - 4\n\n## Determine the features below given the function:\n\n$P(t) = 4\\sin\\left(x - \\pi\\right) + 10$\n\n## A periodic function expresses the period as a horizontal stretch or compression, with the value k according to the formula:\n\n$k = \\dfrac{2\\pi}{period}$\n\n## The average monthly maximum temperature of a certain city can be modeled by the periodic function below where T(t) is the temperature in ˚Celsius, and t is the time in months, where t = 0 represents January 1, t = 1 represents February 1, etc.\n\n$T(t) = 7\\,\\text{sin}\\dfrac{\\pi}{6}(t - 3) + 25$", null, "## A ferris wheel starts rotating from a point at ground level, represented by the equation below where H(t) is height above the ground, in meters, and t is time, in seconds.\n\n$H(t) = -20\\,\\text{cos}\\dfrac{\\pi}{30}(t) + 20$", null, "## A ferris wheel spins around a central axis, which is 15m above the ground. The diameter of the ferris wheel is 28m. The wheel spins once around in 1.0 minute. The rotation of the ferris wheel is plotted where function H(t) is the height of the axis above ground, in meters, and t is the time, in minutes.", null, "# Exponential Functions\n\n## Solve the following exponential equations, similar to the steps shown in the example given below.\n\n\\begin{align} 3^{3x + 2} & = 243 \\\\ \\\\ 3^{3x + 2} & = 3^5 \\\\ \\\\ 3x + 2 & = 5 \\\\ \\\\ 3x & = 5 - 2 \\\\ \\\\ 3x & = 3 \\\\ \\\\ x & = 1 \\\\ \\\\ \\end{align}\n\n## A strong cup of coffee contains 200mg of caffeine. Caffeine decreases by 50% in the bloodstream every 5 hours.", null, "## Given the function for compounding below, where 'i' is initial amount, 'r' is rate of interest, 't' is time, and 'n' is compounding period:\n\n$A = i\\, \\left(1 + \\dfrac{\\left(\\tfrac{r}{100}\\right)}{n}\\right)^{n\\cdot t}$\n\n# Financial Mathematics\n\n### Calculate the monthly payments on a $300,000 mortgage over 25 years at 4% compounded monthly. Solution$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Givens: PV = 300,000 t = 25 years n = 12 i = 0.04 \\begin{align} PV & = \\dfrac{R[1 - (1 + \\tfrac{i}{n})^{-n\\,×\\,t}]}{\\tfrac{i}{n}} \\\\ \\\\ 300,000 & = \\dfrac{R[1 - (1 + \\tfrac{0.04}{12})^{-12\\,×\\,25}]}{\\tfrac{0.04}{12}} \\\\ \\\\ 300,000(\\tfrac{0.04}{12}) & = R[1 - (\\tfrac{301}{300})^{-300}] \\\\ \\\\ R & = \\dfrac{1000}{0.63151} \\\\ \\\\ R & = 1,583.51 \\\\ \\\\ \\end{align} Monthly payments are1,583.51\n\n### If the driver put $3540 down on the car, calculate the original price of the car. Solution Video$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Loans are present value (PV)... PV = Original Price - Down Payment = OP - 3540 \\begin{align} PV & = \\dfrac{R[1 - (1 + \\tfrac{i}{n})^{-n\\,×\\,t}]}{\\tfrac{i}{n}} \\\\ \\\\ OP - 3540 & = \\dfrac{500[1 - (1 + \\tfrac{0.029}{12})^{-12\\,×\\,4}]}{\\tfrac{0.029}{12}} \\\\ \\\\ OP & = \\dfrac{500[1 - (1 + \\tfrac{0.029}{12})^{-12\\,×\\,4}]}{\\tfrac{0.029}{12}} + 3540 \\\\ \\\\ OP & = 22,634.51 + 3540 \\\\ \\\\ OP & = 26,174.51 \\\\ \\\\ \\end{align} The value of the loan was22,634.51 and the original selling price of the car including the down payment was $26,174.51 ### Calculate the interest paid on the loan. Solution Video$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Interest Paid = Sum of Monthly Payments - Present Value of Loan \\begin{align} & = (500/mo)(48\\ mo) - 22,634.51 \\\\ \\\\ & = 24,000 - 22,634.51 \\\\ \\\\ & = 1,365.49 \\\\ \\\\ \\end{align} The interest paid to use the car for 48 months is $1,365.49 (Aside: part of this can be written off on taxes when the car is used only for your business) ##### Mortgage Precision of Different Compound Periods ### A mortgage on a home is$250,000 at 3.5% compounded semiannually, with a 25 year amortization period. Determine the monthly payments. Solution \\$ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Mortgages typically have semi-annual compounding, while the payments are monthly. This is an exception that is allowed. First determine the interest rate on a monthly basis (for monthly payments). A 3.5% rate is 1.75% every 6 months. Convert this rate to a monthly rate after 6 months: \\begin{align} 1 + \\tfrac{i}{n} & = (1 + i)^6 \\\\ \\\\ 1 + \\tfrac{0.035}{2} & = (1 + i)^6 \\\\ \\\\ 1.0175 & = (1 + i)^6 \\\\ \\\\ \\sqrt{1.0175} & = \\sqrt{(1 + i)^6 } \\\\ \\\\ \\sqrt{1.0175} & = 1 + i \\\\ \\\\ i & = \\sqrt{1.0175} - 1 \\\\ \\\\ i & = 0.0028956 \\\\ \\\\ \\end{align} n = 12 (monthly), t = 25, Solve for R... We have determined i = 0.0028956 so we use a slightly different equation... \\begin{align} PV & = \\dfrac{R[1 - (1 + i)^{-n\\,×\\,t}]}{i} \\\\ \\\\ PV(i) & = R[1 - (1 + i)^{-n\\,×\\,t}] \\\\ \\\\ R & = \\dfrac{PV(i)}{[1 - (1 + i)^{-n\\,×\\,t}]} \\\\ \\\\ R & = \\dfrac{250,000(0.0028956)}{[1 - (1 + 0.0028956)^{-12\\,×\\,25}]} \\\\ \\\\ R & = \\dfrac{723.906}{[1 - (1.0028956)^{-300}]} \\\\ \\\\ R & = 1,248.18 \\\\ \\\\ \\end{align}", null, "Loading and rendering MathJax, please wait...\nPercent complete:\nPRESS START\n★ WORK FOR IT & LEVEL UP\n×\n\n# CLASS PAGE SETTINGS\n\n## Streamlined Question View\n\nShow Videos\nShow Solutions\nShow Level Badges\nSound\n\n## Level of Difficulty\n\n1 2 3 4 5 6 7 8 9 10 11\n×\n\n# KEYBOARD SHORTCUTS\n\n• class settings\n• keyboard shortcuts\n• drawing tools\n• table of contents\n• fullscreen\n• close\n• previous question\n• next question\n2 15 ×\nGoogle" ]	[ null, "https://instil.ca/img/diagram/trig10diagram003.svg", null, "https://instil.ca/img/diagram/sinCosPhaseShift.svg", null, "https://instil.ca/img/publicDomainOnly/tumblr_n3tsz9iUPi1st5lhmo1_360.jpg", null, "https://instil.ca/img/publicDomainOnly/ferrisWheel_alt.jpg", null, "https://instil.ca/img/publicDomainOnly/ferrisWheel.jpg", null, "https://instil.ca/img/publicDomainOnly/coffee0283.jpg", null, "https://instil.ca/img/icon_creativeCommons/boxCrateLift256.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8543826,"math_prob":0.9995952,"size":41138,"snap":"2019-13-2019-22","text_gpt3_token_len":10455,"char_repetition_ratio":0.21532066,"word_repetition_ratio":0.15668456,"special_character_ratio":0.24624434,"punctuation_ratio":0.121392295,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9991825,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,9,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T01:42:39Z\",\"WARC-Record-ID\":\"<urn:uuid:20b46d37-b15c-467b-b74f-f4380642ee5c>\",\"Content-Length\":\"903858\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea19ddca-461c-4701-817e-dbbf9038afcd>\",\"WARC-Concurrent-To\":\"<urn:uuid:0663949f-a64b-410d-b630-5806fbb7fe1b>\",\"WARC-IP-Address\":\"68.65.122.51\",\"WARC-Target-URI\":\"https://instil.ca/maths/math11functions\",\"WARC-Payload-Digest\":\"sha1:H5NQT3KQ6KL5DR5SU5NHXPC6T3P5WFQC\",\"WARC-Block-Digest\":\"sha1:QV5EL2BWB6PAZMTTS2R2KDANJC3OAMIW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232260358.69_warc_CC-MAIN-20190527005538-20190527031538-00355.warc.gz\"}"}
http://azrefs.org/equation-will-be-the-lbs-of-food-ordered.html	[ "# Equation will be the lbs of food ordered\n\nYüklə 3.97 Kb.\n tarix 16.04.2016 ölçüsü 3.97 Kb.\n Problem #2: A group of 148 people is spending five days at a summer camp. The cook ordered 12 pounds of food for each adult and 9 pounds of food for each child. A total of 1,410 pounds of food was ordered. What is the total number of adults and total number of children in the group? X + Y = 148 Let X = the number of adults and Y= the number of kids 12X + 9Y = 1,410 The 2nd equation will be the lbs of food ordered -9(X + Y = 148) Multiply by a number to get a negative X or Y 12X + 9Y = 1,410 Line the equation up and Y cancels out -9X + -9Y = -1,332 12X = 1,410 Add the equations -9X = -1,332 3X = 78 Divide 3 from 3 and 78 __ __ 3 3 X = 26 Place the answer in for one of the equations above X(26) + Y = 148 Multiply 26 + Y = 148 Subtract 26 from 26 and 148 Y = 122 Finish writing the equation (26,122) X on the left Y on the right", null, "Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©azrefs.org 2016\nrəhbərliyinə müraciət" ]	[ null, "http://azrefs.org/22435_html_2f09230b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.935933,"math_prob":0.9994081,"size":534,"snap":"2019-13-2019-22","text_gpt3_token_len":155,"char_repetition_ratio":0.1490566,"word_repetition_ratio":0.034782607,"special_character_ratio":0.32397005,"punctuation_ratio":0.06557377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99953926,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-23T21:54:54Z\",\"WARC-Record-ID\":\"<urn:uuid:a86e8b84-13e5-40c3-bd44-d2be91b1654c>\",\"Content-Length\":\"9292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a434465f-bc07-4477-8981-0a80c5b09069>\",\"WARC-Concurrent-To\":\"<urn:uuid:980b27c1-9d8f-401c-86bd-9f42544b1e13>\",\"WARC-IP-Address\":\"176.9.99.84\",\"WARC-Target-URI\":\"http://azrefs.org/equation-will-be-the-lbs-of-food-ordered.html\",\"WARC-Payload-Digest\":\"sha1:SFROLGZ4I37V2AWPT7QQUUJRWMB3TQIL\",\"WARC-Block-Digest\":\"sha1:QTU5ZLXUVEURHC56AURCY7RHCMM6SOEC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257396.96_warc_CC-MAIN-20190523204120-20190523230120-00130.warc.gz\"}"}
https://www.studypool.com/discuss/3835191/Time-Value-of-Money	[ "", null, "Time Value of Money\n\nUser Generated\n\nbuuuuuu\n\nDescription\n\nAnswer all the questions in the file\n\nUnformatted Attachment Preview\n\n. ST-3 TIME VALUE OF MONEY It is now January 1, 2014; and you will need \\$1,000 on January 1, 2018, in 4 years. Your bank compounds interest at an 8% annual rate. a. How much must you deposit today to have a balance of \\$1,000 on January 1, 2018? b. If you want to make four equal payments on each January 1 from 2015 through 2018 to accumulate the \\$1,000, how large must each payment be? (Note that the payments begin a year from today.) c. If your father offers to make the payments calculated in Part b (\\$221 92) or to give you \\$750 on January 1, 2015 (a year from today), which would you choose? Explain. d. If you have only \\$750 on January 1, 2015, what interest rate, compounded annually for 3 years, must you earn to have \\$1,000 on January 1, 2018? e. Suppose you can deposit only \\$200 each January 1 from 2015 through 2018 (4 years). What interest rate, with annual compounding, must you earn to end up with \\$1,000 on January 1, 2018? f. Your father offers to give you \\$400 on January 1, 2015. You will then make six additional equal payments each 6 months from July 2015 through January 2018. If your bank pays 8% compounded semiannually, how large must each payment be for you to end up with \\$1,000 on January 1, 2018? g. What is the EAR, or EFF%, earned on the bank account in Part f? What is the APR earned on the account? . 5-1 What is an opportunity cost? How is this concept used in TVM analysis, and where is it shown on a time line? Is a single number used in all situations? Explain. . 5-3 If a firm’s earnings per share grew from \\$1 to \\$2 over a 10-year period, the total growth would be 100%, but the annual growth rate would be less than 10%. True or false? Explain. (Hint: If you aren’t sure, plug in some numbers and check it out.) . 5-5 To find the present value of an uneven series of cash flows, you must find the PVs of the individual cash flows and then sum them. Annuity procedures can never be of use, even when some of the cash flows constitute an annuity, because the entire series is not an annuity. True or false? Explain. . 5-7 Banks and other lenders are required to disclose a rate called the APR. What is this rate? Why did Congress require that it be disclosed? Is it the same as the effective annual rate? If you were comparing the costs of loans from different lenders, could you use their APRs to determine the loan with the lowest effective interest rate? Explain.\nPurchase answer to see full attachment", null, "User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.", null, "Hey, kindly check this out and let me know if you need any corrections/questions. I'm working on the other\n\nST-3 TIME VALUE OF MONEY\nAmount\n\n=\\$1000\n\nAnnual interest rate\n\n=8%\n\nPeriod\n\n=4 years\n\na) Find principle\n𝐴 = 𝑃(1 + 𝑟)𝑛\n1000 = 𝑃(1.08)4\n1000\n\n𝑃 = (1.08)4\n= \\$735.03\n\nb)\nAmount\n\n=\\$1000\n\nAnnual interest rate\n\n=8%\n\nPeriod\n\n=4 years\n\nCompounding periods/yr= 4\n\n1000 = 𝑃 (1 +\n\n0.08 4∗4\n)\n4\n\n= \\$728.45\n\nc)\n\nIf my father offered \\$221.92 quarterly, or give me \\$750 on January 1, 2015, I would choose\n\\$221.92 because it would earn more interest.\n\nd)\nPrincipal\nAmount\nYears\nAnnual interest rate =?\n𝐴 = 𝑃(1 + 𝑟)𝑛\n1000 = 750(1 + 𝑟)3\n\n=\\$750\n=\\$1,000\n=3\n\n1000\n= (1 + 𝑟)3\n750\n1.3333=(1 + 𝑟)3\n3\n\n3\n\n√1.3333= √(1 + 𝑟)3\n1.1006 = 1 + 𝑟\nR= 1.1006-1\n=0.1006\n= 10.06%\n\ne)\nprincipal\namount\nyears\nannual interest rate\n\n\\$200\n\\$1,000\n4\n?\n\n𝐴 = 𝑃(1 + 𝑟)𝑛\n1000 = 200(1 + 𝑟)4\n5 = (1 + 𝑟)4\n4\n\n√5 = 1 + 𝑟\n\n1.4953= 1+r\nR=0.4953\n= 49.53%\nf)\n\nFather’s offer\nAnnual interest rate\nAmount\nCompounding periods per year\n𝑟 2𝑛\n\n𝐴 = 𝑃 (1 + 2)\n\n1000 = 𝑝 (1 +\n\n0.08 2∗2.5\n)\n2\n\n1000 = 1.21665𝑃\nP = \\$821.9271\n\n=\\$400\n= 8%\n=\\$1000\n=2\n\ng) EAR= 𝐸𝐹𝐹% = (1 +\n\n𝐼𝑁𝑂𝑀 𝑁\n)\n𝑁\n2\n\n−1\n\n0.08\n) −1\n2\n𝐸𝐹𝐹% = 8.16%\n𝐸𝐹𝐹% = (1 +\n\nINOM=Periodic rate * number of periods per year\n0.08*2= 16%\n\n5-1 Opportunity cost\nOpportunity cost is the rate of interest a person could earn on an alternative investment with a risk\nsame as the risk of investment in question. It is the value of r in the Time Value of Money equations\nwhich is displayed at the top of a time line amid the 1st and 2nd tick marks. 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http://spiriteca.com/trading-withdraw-ldvhx/e38273-excel-best-practices-formulas	[ "Totals as intermediate results. In Excel 2007 and later versions, array formulas can handle whole-column references, but this forces calculation for all the cells in the column, including empty cells. The 8th best practice in Excel is to split up calculations into multiple smaller calculations. Excel VBA development best practices [closed] Ask Question Asked 10 years, 8 months ago. For more information, see Use faster VBA user-defined functions. This might not appear to reduce the number of calculations for a single array formula; however, most of the time it enables the smart recalculation process to recalculate only the formulas in the helper column that need to be recalculated. Applies to: Excel \| Excel 2013 \| Office 2016 \| VBA. To look up the table to use in addition to the row and the column, you can use the following techniques, focusing on how to make Excel look up or choose the table. To help you get the most out of Excel, we’ve put together a batch of the best Excel tips for beginners. Keep Formulas Readable. Application.DisplayStatusBar Turn off the status bar. It is generally faster than an exact match lookup. This is because there is a small overhead for each user-defined function call and significant overhead transferring information from Excel to the user-defined function. If there's no way to make it fast, make use of the manual calculation: -- Set calculation to manual as mentioned above. Most Excel users have difficulty understanding their own Excel formulas if it has not been looked at for a few months. Try to use simple direct cell references that work on closed workbooks. Avoid constants in formulas. 2-F9 A large collection of useful Excel formulas, beginner to advanced, with detailed explanations. Calendars, invoices, trackers and much more. @K Mac We’re amazed every day by the ways in which you, our customers, use Excel to make better decisions, leveraging the flexibility of the 2D grid and formulas to capture, analyze and collaborate on data. Up to this point, Excel has only had a … It's all here. Note: You can use Excel lists in 2003 version, which are similar to tables. In the text box, type %temp%, and then click OK. Tracking changes in a shared workbook causes your workbook file-size to increase rapidly. Thank you so much for this Jan. Excel will add the { and } to show you that this is an array formula). These formulas are re-calculated whenever there is a change in the workbook. Harnessing the power of functions 67. File size is almost the same as earlier versions (some additional information may be stored), and performance is slightly slower than earlier versions. I've written complex formulas in Excel for staff in other departments on the regular for years, but again my employer is a non profit in the SMB size category. note: When typing the syntax of a formula, Ctrl + A shows the function arguments. If your user-defined function is called early in the calculation chain, it can be passed as uncalculated arguments. Optimize your code by explicitly reducing the number of times data is transferred between Excel and your code. The beginning cell of the SUM is anchored in A1, but because the finishing cell has a relative row reference, it automatically increases for each row. Viewed 2k times 10. Unused cells are not examined, so whole column references are relatively efficient, but it is better to ensure that you do not include more used cells than you need. PivotTables are a great way to produce summary reports, but try to avoid creating formulas that use PivotTable results as intermediate totals and subtotals in your calculation chain unless you can ensure the following conditions: The PivotTable has been refreshed correctly during the calculation. Alot of my workbooks are distributed to people who have an old version of excel and do not need to see the workings - because of this I have written a very simple piece of code that loops through all of the calculation worksheets, and pastes as values - which resolves any compatibility issues (using formulas that arent supported in 03, and has a dramatic improvement in reopening time). -- Put the formula in the first data row, and make sure it's working before copying it to the rest of the cells in the column. that's all! If you cannot use two cells, use COUNTIF. Using the INDEX formula for a dynamic range is generally preferable to the OFFSET formula because OFFSET has the disadvantage of being a volatile function that will be calculated at every recalculation. Excel spreadsheet formulas usually work with numeric data; you can take advantage of data validation to specify the type of data that should be accepted by a cell i.e. This causes 7 calls to the UDF on each recalc. Try to move the circular calculations onto a single worksheet or optimize the worksheet calculation sequence to avoid unnecessary calculations. 3. Keep formulas in a separate sheet. However, it is slightly more flexible in that the range to sum may have, for example, multiple columns when the conditions have only one column. Avoid formulas that contain references to multiple worksheets. As follows: 1. Be sure that your Windows swap file is located on a disk that has a lot of space and that you defragment the disk periodically. However, the computer that is running Excel also requires memory resources. Starting in Excel 2007, you can use structured table references, which automatically expand and contract as the size of the referenced table increases or decreases. These reports fail for at least one of three reasons: bad strategy, bad structure, or bad technique. We outline the salary, skills, personality, and training you need for FP&A jobs and a successful finance career. One of the reasons for sluggish performance is that you are searching for something in a lot of un-sorted data. All free, fun and fantastic. TyPe oF FunCTion Common use FoRmulA To insert a function in a cell, use either Alt m F or shift + F3 (\"Function Wizard\"). To look up several contiguous columns, you can use the INDEX function in an array formula to return multiple columns at once (use 0 as the column number). If you put a formula in a cell of an empty table column, it will automatically fill and calculate even if you have manual calculation set. We recommend that you use the Microsoft Internet Information Services (IIS) setting to set the … It is important to reduce the number of cells in the circular calculation and the calculation time that is taken by these cells. The following code examples compare the two methods. CHAPTER 4. Use the first result second time too. I've started to use Names to do something similar, but changing arguments is the rub. If Excel stops responding for any reason, you might need to delete these files. Most Excel reports, forecasts, and analyses I’ve seen in my career could serve as excellent examples of what NOT to do in Excel. For example, instead of using OFFSET to construct a dynamic range, you can use INDEX. Many times we inherit un-sorted data thru data imports. Important: The calculated results of formulas and some Excel worksheet functions may differ slightly between a Windows PC using x86 or x86-64 architecture and a Windows RT PC using ARM architecture. Add an extra column for the MATCH to store the result (stored_row), and for each result column use the following: Alternatively, you can use VLOOKUP in an array formula. This page will direct you to all of our best content. As an example, if you have table called cs, then the formula sum(cs[column_name]) refers to sum of all values in the column_name of table cs. Starting in Excel 2007, you should use SUMIFS, COUNTIFS, and AVERAGEIFS functions instead of the DFunctions. 09/25/2017; 4 minutes to read; s; V; O; d; In this article. This article is the cheat sheet of formulas available in Microsoft Excel. With 1 million rows available starting in Excel 2007, an array formula that references a whole column is extremely slow to calculate. Learning Center Excel Tutorials and Practice Tests Welcome to Automate Excel! i hope that this tip would help. excel take time to calculate, when there are offsheet references (majorly used in sumif, sumproduct, index, match). These cells do not contain formulas. CONCATENATE. In large worksheets, you may frequently need to look up by using multiple indexes, such as looking up product volumes in a country. You can often reuse a stored exact MATCH many times. Other performance characteristics of Visual Basic 6 user-defined functions in Automation add-ins are similar to VBA functions. Well you are at the right place. But well-designed and called user-defined functions can be much faster than complex array formulas. This article covered ways to optimize Excel functionality such as links, lookups, formulas, functions, and VBA code to avoid common obstructions and improve performance. This method also avoids the problem of changed names in precedent workbooks. Mastering Excel formulas 1 CHAPTER 1. Essay on Excel Best Practices for Business The container for Excel spreadsheets — the grid where numbers, text, and formulas reside and calculations are performed — is a file called a workbook with If the lookup array is not sorted ascending, MATCH will return an incorrect answer. Forward referencing usually does not affect calculation performance, except in extreme cases for the first calculation of a workbook, where it might take longer to establish a sensible calculation sequence if there are many formulas that need to have their calculation deferred. (Array formulas must be entered by using Ctrl+-Shift+Enter. Built-in formulas tend to better than your own version – for example SUMIFS is easier to write and just as fast as SUMPRODUCT. vValue(lRow, lCol) = 2 * vValue(lRow, lCol) (any single character) and * (no character or any number of characters) in the criteria for alphabetical ranges as part of the SUMIF, COUNTIF, SUMIFS, COUNTIFS, and other IFS functions. Introduction to Excel Tables – what are they and how to use them? End Function, Name: test The following functionality can usually be turned off while your VBA macro executes: Application.ScreenUpdating Turn off screen updating. That said, for earlier Excel versions, Lookups continue to be frequently significant calculation obstructions. Forecast function available in excel is the simplest ever forecasting function that we could have. For example, create a defined name using one of the following formulas: When you use the dynamic range name in a formula, it automatically expands to include new entries. For #5. lookup data, compare results, calculations, etc. Pivot tables are an excellent way to calculate a lot of summary values with few clicks. Function name. Check the size of your toolbar file. These functions evaluate each of the conditions from left to right in turn. Excel Training Templates and Practice Workbooks. Function DemoFunction(r As Range) One of the reasons for sluggish performance is that you are searching for something … The calculation engine in Excel is optimized to exploit array formulas and functions that reference ranges. There are two methods of doing period-to-date or cumulative SUMs. We have free excel practice tests where you can sharpen your skill. Back on track… Investment bankers, including younger managing directors as most new MDs cut their teeth during a time when Excel best practices were already firmly enshrined, are very serious about Excel skills. But whether it be a project at work or just a personal budget, you need to know basic Excel to get things done right and done quickly. Use these tips & ideas to super-charge your sluggish workbook. For example, in cash flow and interest calculations, try to calculate the cash flow before interest, calculate the interest, and then calculate the cash flow including the interest. Forecast in Excel. If you need to produce totals and subtotals as part of the final results of your workbook, try using PivotTables. Building basic formulas 3. You know the functions and formulas but need to practive your Excel skills?Or do you need Excel Practice Tests online? Lookup time using the approximate match options of VLOOKUP, HLOOKUP, and MATCH on sorted data is fast and is not significantly increased by the length of the range you are looking up. Using many dynamic ranges can decrease performance. It is often more efficient to calculate a subset range for the lookup (for example, by finding the first and last row for the country, and then looking up the product within that subset range). However, OFFSET is a fast function and can often be used in creative ways that give fast calculation. using formulas – KPI Dashboard example, Image Lookup – How-to show dynamic picture in a cell [Excel Trick], 9 Box grid for talent mapping – HR for Excel – Template & Explanation, 6 Must Know Line Chart variations for Data Analysis, Excel formula to convert calendar format to table, Project Plan – Gantt Chart with drill-down capability [Templates], These Pivot Table tricks massively save your time, » Calculation Effeciency: Array Formulas Spreadsheet Budget and Consulting, http://professor-excel.com/performance-excel-study/. Sort your data. And since tables grow & shrink as you add / remove data, none of your formulas need to be dynamic. Make better worksheets and impress everyone (including your … Quite surprising: Some things have a great effect as switching the language to English (if you got Excel in German region setting, -81% of calcultion time) or avoiding SUMIFS as lookups (-53%). In your very 1st tip you state that \"And since tables grow & shrink as you add / remove data, none of your formulas need to be dynamic.\" Solution? Matchtype=0 requests an exact match and assumes that the data is not sorted. Go ahead and spend few minutes to be AWESOME. Using names that refer to other worksheets adds an additional level of complexity to the calculation process. For more information, see the \"Large data sets and the 64-bit version of Excel\" section in Excel performance: Performance and limit improvements. Although VLOOKUP is slightly faster (approximately 5 percent faster), simpler, and uses less memory than a combination of MATCH and INDEX, or OFFSET, the additional flexibility that MATCH and INDEX offer often enables you to significantly save time. If you switch the calculation mode to manual, you can wait until all the cells associated with the formula are updated before recalculating the workbook. One of the reasons for slow workbooks is lot of data. Great article on INDEX in Excel Hero (never seen so many responses in any forum as that one!) You can avoid the double exact lookup if you use exact MATCH once, store the result in a cell, and then test the result before doing an INDEX. ... Also, as you mentioned VLOOKUP: the most time consuming part of a VLOOKUP formula is the looking up - returning the relevant value is normally relatively quick once Excel has determined the location of that value. If you have a complex calculation that depends on cells in the circular reference, it can be faster to isolate this into a separate closed workbook and open it for recalculation after the circular calculation has converged. Caution it could be fatal as the model could change. -- Use a countif to see if you'll get a match at all in a lookup table, and only do the big, complicated multi-criteria array-formula index lookup for the records that have at least a basic match. Here are a few whacky ideas that I try in such cases: Optimization is not a one-shot exercise. [...] had some posts on speeding up Excel worksheets, one of the posts focuses on formulas and another he let the general readers make their suggestions. Data for Excel formula list in this guide. Most formula examples shown here are self-explanatory. The PivotTable has not been changed so that the information is still visible. You should have only one formula per row, meaning that whatever formula is used in the first cell of any given row should be the same formula uniformly applied across the entire row. Each user has a unique XLB file. 4. This guide gives you the answer. You can minimize this performance decrease by storing the COUNTA part of the formula in a separate cell or defined name, and then referring to the cell or name in the dynamic range: You can also use functions such as INDIRECT to construct dynamic ranges, but INDIRECT is volatile and always calculates single-threaded. The XLS format is the same format as earlier versions. Please see Office VBA support and feedback for guidance about the ways you can receive support and provide feedback. By doing this, you can avoid recalculating all your linked workbooks when you recalculate any workbook. Formulas!:!Functions!! \" Example: Customer Service Dashboard – Data & Calculations, Examples, Details & Downloads on Excel Pivot Tables, Remove duplicates & sort a list using Pivot Tables, Sorting numbers etc. Get ready to be initiated into the inner circles of Excel Ninjadom today I’m talking about array formulas also known as Crt Shift Enter formulas (CSE formula for short). Excel Nested IF statement: examples, best practices and alternatives. Enter =NOT() where is the Integrity Check’s Passed cell’s address for this alert. 1. Note that the size and shape of the ranges or arrays that are used in the conditional expressions and range to sum must be the same, and they cannot contain entire columns. This week, Speed up your Spreadsheets – Your Action Required, Speeding up & Optimizing Excel – Tips for Charting & Formatting [Speedy Spreadsheet Week]. So how do you optimize & speed-up your Excel formulas? The 64-bit version of Excel does not have a 2 GB or up to 4 GB limit. The \"imposing index\" article you mention is a must read, it truly will revolutionise the way you use excel. Create a backup copy first. I am working on a P&L constructed in a form of pivot table with a # of calculated items. Dynamic ranges have the following advantages and disadvantages: Dynamic ranges work well to limit the number of calculations performed by array formulas. Of INDIRECT SUMIFS is easier to write and just as fast as SUMPRODUCT do! Off sheet cell references in excel best practices formulas circular references by using IFERROR, or more web service input values d! Caution it could be fatal as the model could change the model could change, as it! Down calculation and training you need just a few months Comments are posted via e-mail decision-making tool evaluate. 09/25/2017 ; 4 minutes to be dynamic, cash flow projections and more pivots in one go to your... 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Will find the best free Excel practice Tests where you can pre-calculate a lower-range limit and upper-range limit the... In thinking thru various calculations and planning the formulas formulas but need to practive your Excel skills level... ( renaming it `` toolbar.OLD '' is safer ) data is transferred between Excel and your code,! 10-20 rows. ), by using concatenated lookup values on save set excel best practices formulas..., covers most of my formulas are re-calculated whenever there is a calculation-intensive operation 65,536 rows )... And their dependents strings is a fast function and TableLookup_Value to choose range! End you said `` please note that excel best practices formulas runs a compatibility check works smoothly you... Things manually this sounds interesting, but you must handle them carefully use two lookups range... Majorly used in formulas with structural references, can be much faster ways of the! T you many ways of improving lookup calculation time many binary files everything one ask! Of whole column references inefficiently remove any named ranges that result in error or missing links and columns than need... Restrict the range that you can also return many cells from one section to another may only be a of..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89865005,"math_prob":0.8544152,"size":26538,"snap":"2021-21-2021-25","text_gpt3_token_len":5372,"char_repetition_ratio":0.14909174,"word_repetition_ratio":0.04738041,"special_character_ratio":0.20340644,"punctuation_ratio":0.11942388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96543473,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T06:37:01Z\",\"WARC-Record-ID\":\"<urn:uuid:9cd68f3e-82a1-41c6-8ee4-4dd7a8d3f0bc>\",\"Content-Length\":\"38311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:924bb004-fba8-4e13-bb0b-4db132a155b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b00d7dd0-b591-424a-9da1-37efb1ad1a4d>\",\"WARC-IP-Address\":\"184.106.210.181\",\"WARC-Target-URI\":\"http://spiriteca.com/trading-withdraw-ldvhx/e38273-excel-best-practices-formulas\",\"WARC-Payload-Digest\":\"sha1:3H2AVDQBS7RWYS3NN4DQROS2QXI22UAU\",\"WARC-Block-Digest\":\"sha1:YWB3FFMASXB4KP7VPRIPBCVDQLS3XSG2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487643703.56_warc_CC-MAIN-20210619051239-20210619081239-00265.warc.gz\"}"}
https://math.stackexchange.com/questions/2943851/finding-the-complex-square-roots-of-a-complex-number-without-a-calculator/2943857	[ "# Finding the complex square roots of a complex number without a calculator\n\nThe complex number $$z$$ is given by $$z = -1 + (4 \\sqrt{3})i$$\n\nThe question asks you to find the two complex roots of this number in the form $$z = a + bi$$ where $$a$$ and $$b$$ are real and exact without using a calculator.\n\nSo far I have attempted to use the pattern $$z = (a+bi)^2$$, and the subsequent expansion $$z = a^2 + 2abi - b^2$$. Equating $$a^2 - b^2 = -1$$, and $$2abi = (4\\sqrt{3})i$$, but have not been able to find $$a$$ and $$b$$ through simultaneous equations.\n\nHow can I find $$a$$ and $$b$$ without a calculator?\n\n• You could notice that $u=a^2$ and $v=-b^2$ satisfy $\\begin{cases}u+v=-1\\\\ uv=-12\\\\ u\\ge 0\\\\ v\\le0\\end{cases}$. – Saucy O'Path Oct 5 '18 at 22:41\n• Why is $uv=-12?$ – user376343 Oct 5 '18 at 22:53\n\nThe second equation can be written $$ab=2\\sqrt{3}$$ which gives $$b = \\frac{2\\sqrt{3}}{a}$$. If we substitute back into the first equation we get $$a^2 - \\frac{12}{a^2} = -1$$. Multiplying both sides by $$a^2$$ gives $$a^4 - 12 = - a^2$$. This can be written as $$a^4 + a^2 - 12 = 0$$ which is a quadratic equation solvable for $$a^2$$.\n\n• Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i \\text{ and } \\pm \\sqrt{3}$, thanks for the help. – Pegladon Oct 5 '18 at 22:56\n• I'm glad that I could help. Don't forget the plus/minus $a=\\pm 2i$ and checking your final answer(s). – sfmiller940 Oct 5 '18 at 23:23\n• @Pegladon, in this method $a,b$ are considered reals. Moreover, any non-zero complex number has exactly two square roots. Thus $\\pm 2i$ are not convenient for $a.$ – user376343 Oct 6 '18 at 7:49\n\nObserve that $$\\\|z\\\| = \\sqrt{(-1)^2+(4\\sqrt3)^2} = \\sqrt{49} = 7$$. Therefore the root of $$z$$ will have length $$\\sqrt 7$$, so $$a^2+b^2=7$$. Combine this with $$a^2-b^2=-1$$ to get $$a$$ and $$b$$.\n\nOne way is to write $$z=r^2 e^{2\\theta}$$ and roots will be $$re^\\theta$$ and $$re^{\\pi -\\theta}$$.\n\nFrom $$z =-1+4\\sqrt{3}i$$, we obtain $$r=7$$ and $$\\tan{2\\theta} = \\frac{2\\tan\\theta}{1-\\tan^2\\theta} = -4\\sqrt{3}$$. Second expression gives you a quadratic equation, $$2\\sqrt{3}\\tan^2\\theta -2\\sqrt{3} + \\tan\\theta =0$$.\n\nRoots of the above quadratic equation are $$\\tan\\theta= \\sqrt{3}/2,-2/\\sqrt{3}$$ which form $$\\tan\\theta$$ and $$\\tan(\\pi-\\theta)$$.\n\nHence, square roots of $$z$$ are $$(1+\\sqrt{3}/2i)\\frac{7}{\\sqrt{1+3/4}} = \\sqrt{7}(2+\\sqrt{3}i)$$ and $$(1-2/\\sqrt{3}i)\\frac{7}{\\sqrt{1+4/3}} = \\sqrt{7}(\\sqrt{3}-2i)$$.\n\nThat the square roots are $$\\pm(\\sqrt 3 + 2i)$$ can be seen by elementary algebra and trigonometry as follows. \\begin{align} & \\left\|-1 + i4\\sqrt 3\\right\| = \\sqrt{(-1)^2 + (4\\sqrt 3)^2 } = 7. \\\\[10pt] \\text{Therefore } & -1+i4\\sqrt 3 = 7(\\cos\\varphi + i\\sin\\varphi). \\\\[10pt] \\text{Therefore } & \\pm\\sqrt{-1+i4\\sqrt 3} = \\pm\\sqrt 7 \\left( \\cos \\frac \\varphi 2 + i \\sin\\frac\\varphi 2 \\right). \\end{align}\n\nNotice that $$\\sin \\varphi = \\frac{4\\sqrt 3} 7 \\quad \\text{and} \\quad \\cos\\varphi = \\frac{-1} 7$$ and recall that \\begin{align} \\tan\\frac\\varphi 2 & = \\frac{\\sin\\varphi}{1+\\cos\\varphi} \\\\[12pt] \\text{so we have }\\tan\\frac\\varphi 2 & = \\frac{4\\sqrt 3}{7-1} = \\frac 2 {\\sqrt 3}. \\\\[10pt] \\text{Therefore } \\sin\\frac\\varphi2 & = \\frac 2 {\\sqrt 7} \\quad \\text{and} \\quad \\cos\\frac\\varphi2 = \\frac{\\sqrt3}{\\sqrt7}. \\end{align} Thus the desired square roots are $$\\pm \\left( \\sqrt 3 + 2i \\right).$$\n\nNote that$$z=7\\left(-\\frac17+\\frac{4\\sqrt3}7i\\right).\\tag1$$Now, since $$\\left(-\\frac17\\right)^2+\\left(\\frac{4\\sqrt3}7\\right)^2=1$$, the expression $$(1)$$ expresses $$z$$ as $$7\\bigl(\\cos(\\alpha)+\\sin(\\alpha)i\\bigr)$$, for some $$\\alpha$$. So, a square root of $$z$$ is $$\\sqrt7\\left(\\cos\\left(\\frac\\alpha2\\right)+\\sin\\left(\\frac\\alpha2\\right)i\\right)$$. Now, note that if $$c=\\cos\\left(\\frac\\alpha2\\right)$$ and $$s=\\sin\\left(\\frac\\alpha2\\right)$$, then $$c^2+s^2=1$$ and $$c^2-s^2=\\cos(\\alpha)=-\\frac17$$. This allows you to compute the square roots of $$z$$.\n\nLet$$z^2=(x+yi)^2=−1+4\\sqrt3i,$$ i.e.$$(x^2-y^2)+2xyi=−1+4\\sqrt3i.$$ Compare real parts and imaginary parts, $$\\begin{cases} x^2 - y^2 = -1&\\qquad\\qquad(1)\\\\ 2xy = 4\\sqrt3&\\qquad\\qquad(2) \\end{cases}$$ Now, consider the modulus: $$\|z\|^2 =\|z^2\|$$, then $$x^2 + y^2 = \\sqrt{\\smash[b]{(-1)^2+(4\\sqrt3)^2}} = 7\\tag3$$ Solving $$(1)$$ and $$(3)$$, we get $$x^2 = 3\\Rightarrow x = \\pm\\sqrt3$$ and $$y^2 = 4\\Rightarrow y = \\pm2$$.\n\nFrom $$(2)$$, $$x$$ and $$y$$ are of same sign, $$\\begin{cases} x = \\sqrt3\\\\ y = 2 \\end{cases}\\text{ or } \\begin{cases} x = -\\sqrt3\\\\ y = -2 \\end{cases}$$ then$$z = \\pm(\\sqrt3 + 2i).$$\n\n• Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions. – Théophile Oct 5 '18 at 22:51" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9052116,"math_prob":1.0000099,"size":506,"snap":"2019-35-2019-39","text_gpt3_token_len":163,"char_repetition_ratio":0.12151395,"word_repetition_ratio":0.0,"special_character_ratio":0.35177866,"punctuation_ratio":0.06422018,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T07:39:51Z\",\"WARC-Record-ID\":\"<urn:uuid:f8694938-3de3-482f-b720-cd9376390c97>\",\"Content-Length\":\"176198\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2627d43d-e248-4315-a0c4-3c58ba1d478c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d014b33f-efc0-4d33-94e0-9b00886dfb9e>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2943851/finding-the-complex-square-roots-of-a-complex-number-without-a-calculator/2943857\",\"WARC-Payload-Digest\":\"sha1:L63JBOARKQMXBQ4SA3KP735IGXTHZL2R\",\"WARC-Block-Digest\":\"sha1:X25WE57SMICB4G5JX4AG7SYXVL32KOCQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573908.70_warc_CC-MAIN-20190920071824-20190920093824-00098.warc.gz\"}"}
https://scholars.ncu.edu.tw/zh/publications/finite-size-corrections-and-scaling-for-the-dimer-model-on-the-ch	[ "# Finite-size corrections and scaling for the dimer model on the checkerboard lattice\n\nNickolay Sh Izmailian, Ming Chya Wu, Chin Kun Hu\n\n6 引文斯高帕斯（Scopus）\n\n## 摘要\n\nLattice models are useful for understanding behaviors of interacting complex many-body systems. The lattice dimer model has been proposed to study the adsorption of diatomic molecules on a substrate. Here we analyze the partition function of the dimer model on a 2M×2N checkerboard lattice wrapped on a torus and derive the exact asymptotic expansion of the logarithm of the partition function. We find that the internal energy at the critical point is equal to zero. We also derive the exact finite-size corrections for the free energy, the internal energy, and the specific heat. Using the exact partition function and finite-size corrections for the dimer model on a finite checkerboard lattice, we obtain finite-size scaling functions for the free energy, the internal energy, and the specific heat of the dimer model. We investigate the properties of the specific heat near the critical point and find that the specific-heat pseudocritical point coincides with the critical point of the thermodynamic limit, which means that the specific-heat shift exponent λ is equal to. We have also considered the limit N→' for which we obtain the expansion of the free energy for the dimer model on the infinitely long cylinder. From a finite-size analysis we have found that two conformal field theories with the central charges c=1 for the height function description and c=-2 for the construction using a mapping of spanning trees can be used to describe the dimer model on the checkerboard lattice.\n\n原文 ???core.languages.en_GB??? 052141 Physical Review E - Statistical, Nonlinear, and Soft Matter Physics 94 5 https://doi.org/10.1103/PhysRevE.94.052141 已出版 - 23 11月 2016" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8186868,"math_prob":0.9325797,"size":3628,"snap":"2023-14-2023-23","text_gpt3_token_len":830,"char_repetition_ratio":0.14486755,"word_repetition_ratio":0.8465517,"special_character_ratio":0.22133407,"punctuation_ratio":0.07762557,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9702847,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T22:49:49Z\",\"WARC-Record-ID\":\"<urn:uuid:1ba08f77-94fc-400e-873c-720e6eab3d6b>\",\"Content-Length\":\"55382\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db4b1c34-4fa8-4410-a9b7-7f643181cef5>\",\"WARC-Concurrent-To\":\"<urn:uuid:c529e76e-fda2-4a80-9d46-5a8603facb5f>\",\"WARC-IP-Address\":\"13.228.199.194\",\"WARC-Target-URI\":\"https://scholars.ncu.edu.tw/zh/publications/finite-size-corrections-and-scaling-for-the-dimer-model-on-the-ch\",\"WARC-Payload-Digest\":\"sha1:JDOLIM5HRC3GUW3SSQSCHTZTENCD7NMH\",\"WARC-Block-Digest\":\"sha1:W2MTVTSDSP2AYZ3D7QVKXJBKDDGOSVMU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00341.warc.gz\"}"}
https://puzzling.stackexchange.com/questions/50113/turn-lead-into-gold/50115	[ "Can you turn the word LEAD into GOLD in five steps or less? You can change or replace one letter at a time only. The newly formed word must be a legit word in a recognized dictionary. Of course you cannot add or subtract letters, so all new words must have 4 letters. I am sure there are several solutions here.\n\nThree steps/Four words:\n\n• WOW great. Didnt know GOAD as a word. – DrD Mar 18 '17 at 16:04\n• I only remembered it because of a word play ladder I created a little bit ago haha – n_plum Mar 18 '17 at 16:04\n• This is actually the optimal solution: it's impossible to do better. – Rand al'Thor Mar 18 '17 at 16:08\n• I verified that this is the unique optimal solution - there's nothing else with three steps (in my word list) – isaacg Mar 20 '17 at 6:43\n• Five steps:\n\nLEAD $\\rightarrow$ LEND $\\rightarrow$ FEND $\\rightarrow$ FOND $\\rightarrow$ FOLD $\\rightarrow$ GOLD\n\n• Four steps:\n\nLEAD $\\rightarrow$ READ $\\rightarrow$ ROAD $\\rightarrow$ GOAD $\\rightarrow$ GOLD\n\n• @n_palum's solution in three steps:\n\nLEAD $\\rightarrow$ LOAD $\\rightarrow$ GOAD $\\rightarrow$ GOLD\n\nThis is optimal because\n\nthree letters are different between LEAD and GOLD, and we can only change one letter at a time, so we need at least three steps.\n\n• Damn, I had that exact three steps solution in less than one minute from reading the question, but I'm 21 to 22 hours late >_> – Zachiel Mar 19 '17 at 13:28\n\nHere's one in four steps:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8494131,"math_prob":0.92236435,"size":1643,"snap":"2021-31-2021-39","text_gpt3_token_len":461,"char_repetition_ratio":0.17144601,"word_repetition_ratio":0.014084507,"special_character_ratio":0.28301886,"punctuation_ratio":0.08387097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9822395,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T03:44:11Z\",\"WARC-Record-ID\":\"<urn:uuid:990cb305-b229-42f3-a04e-4e376ecd3dcf>\",\"Content-Length\":\"186114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b0684af-c9ba-4380-9d21-d9307f1bfc1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d14fb456-b105-42b0-a87b-17710a1122fc>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://puzzling.stackexchange.com/questions/50113/turn-lead-into-gold/50115\",\"WARC-Payload-Digest\":\"sha1:ZBG5LX7V3UJRR6UY5M4GGKKO5KBJXCS4\",\"WARC-Block-Digest\":\"sha1:E6KYELX5XTMDRWABDRDJSX7ZNFNQFH4D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150129.50_warc_CC-MAIN-20210724032221-20210724062221-00314.warc.gz\"}"}
https://noncommutativeanalysis.com/2012/10/11/william-arveson/?replytocom=494	[ "William Arveson\n\nWilliam B. Arveson was born in 1934 and died last year on November 15, 2011. He was my mathematical hero; his written mathematics has influenced me more than anybody else’s. Of course, he has been much more than just my hero, his work has had deep and wide influence on the entire operator theory and operator algebras communities. Let me quickly give an example that everyone can appreciate: Arveson proved what may be considered as the “Hahn-Banach Theorem” appropriate for operator algebras. He did much more than that, and I will expand below on some of his early contributions, but I want to say something before that on what he was to me.\n\nWhen I was a PhD student I worked in noncommutative dynamics. Briefly, this is the study of actions of (one-parameter) semigroups of -endomorphisms on von Neumann algebras (in short E-semigroups). The definitive book on this subject is Arveson’s monograph “Noncommutative Dynamics and E-Semigroups”. So, naturally, I would carry this book around with me, and I would read it forwards and backwards. The wonderful thing about this book was that it made me feel as if all my dreams have come true! I mean my dreams about mathematics: as a graduate student you dream of working on something grand, something important, something beautiful, something elegant, brilliant and deep. You want your problem to be a focal point where different ideas, different fields, different techniques, in short, all things, meet.\n\nWhen reading Arveson there was no doubt in my heart that, e.g., the problem classifying E-semigroups of type I was a grand problem. And I was blown away by the fact that the solution was so beautiful. He introduced product systems with such elegance and completeness that one would think that this subject has been studied for the last 50 years. These product systems were measurable bundles of operator spaces – which turn out to be Hilbert spaces! – that have a group like structure with respect to tensor multiplication. And they turn out to be complete invariants of E-semigroups on", null, "$B(H)$. The theory set down used ideas and techniques from Hilbert space theory, operator space theory, C-algebras, group representation theory, measure theory, functional equations, and many new ideas – what more could you ask for? Well, you could ask that the new theory also contribute to the solution of the original problem.\n\nIt turned out that the introduction of product systems immensely advanced the understanding of E-semigroups, and in particular it led to the full classification of type I ones.\n\nSo Arveson became my hero because he has made my dreams come true. And more than once: when reading another book by him, or one of his great papers, I always had a very strong feeling: this is what I want to do. And when I felt that I gave a certain problem all I thought I had in me, and decided to move on to a new problem, it happened that he was waiting for me there too.\n\nI wish to bring here below a little piece that I wrote after he passed away, which explains from my point of view what was one of his greatest ideas.\n\nFor a (by far) more authoritative and complete review of Arveson’s contributions, see the two recent surveys by Davidson (link) and Izumi (link).\n\nSubalgebras of C-algebras I and II\n\nArveson was one of the first to realize the importance of non-selfadjoint operator algebras as well as the role of operator spaces, and he developed a theory which had both immediate applications (to problems operator theory, including “single operator theory”), as well as long term, far reaching implications. In his paper [“Analyticity in Operator Algebras”, Amer. J. Math. 89, 1967] he wrote:\n\nIn the study of families of operators on Hilbert space, the self-adjoint algebras have occupied a preeminent position.\n\nBy “self-adjoint” he means, of course, C-algebras and von Neumann algebras. He continues:\n\nNevertheless, many problems in operator theory lead obstinately toward questions about algebras that are not necessarily self-adjoint. Indeed, the general theory of a single operator on a finite-dimensional space rests on an analysis of the polynomials in that operator, and has nothing at all to do with the -operation.\n\nArveson’s great insight was that if C-algebras are a non-commutative analogue of algebras of continuous functions, then the theory of non-selfadjoint algebras should be considered as a non-commutative version of the theory of of function algebras. His vision was expounded in two ground breaking papers ([“Subalgebras of C-algebras”, Acta Math. 123, 1969], and [“Subalgebras of C-algebras II”, Acta Math. 128, 1972]), where he laid the foundations to the theory of (not necessarily self-adjoint) operator algebras and of operator spaces.\n\nIt was probably in these papers where the notions of complete positivity, complete boundedness and complete isometry were first shown to play such a fundamental role in operator theory.\n\nDigression: You may wonder: “what is complete this and complete that“. Let me briefly explain what is a complete isometry, and the rest will be clear. An operator algebra", null, "$A$ is just a subalgebra of", null, "$B(H)$, and it inherits the operator norm from", null, "$B(H)$. A linear map", null, "$\\phi : A \\rightarrow B$ (where", null, "$B$ is another operator algebra) is said to be an isometry if", null, "$\\\|\\phi(a)\\\| = \\\|a\\\|$ for all", null, "$a \\in A$. This is a very familiar notion, the existence of such a", null, "$\\phi$ means that", null, "$A$ and", null, "$B$ have the same Banach space structure. It turns out that the Banach space structure of an operator algebra is far from determining its other properties. Arveson somehow understood that the complete version of isometry is what one is required to look at. Here is what this means.\n\nOne can form the algebra", null, "$M_n(A)$ of", null, "$n \\times n$ matrices over", null, "$A$. This is a subalgebra of", null, "$M_n(B(H)) = B(\\underbrace{H \\oplus H \\oplus \\cdots \\oplus H}_{n \\textrm{ times}})$,\n\nso it is again an operator algebra. If", null, "$\\phi : A \\rightarrow B$ is a linear map, we define", null, "$\\phi_n : M_n(A) \\rightarrow M_n(B)$ by letting", null, "$\\phi$ act entrywise, that is, if", null, "$[a_{ij}] \\in M_n(A)$ then", null, "$\\phi_n ([a_{ij}])$ is the matrix", null, "$[\\phi(a_{ij})]$. Finally, the map", null, "$\\phi$ is said to be a complete isometry if", null, "$\\phi_n$ is an isometry for all", null, "$n$. The same notions can be defined when", null, "$A$ and", null, "$B$ are operator spaces, that is, simply subpaces of", null, "$B(H)$End of digression.\n\nThe main issue confronted in the two “Subalgebras” papers mentioned above is this: given an algebra", null, "$A$ of operators on a Hilbert space, one can form the C-algebra", null, "$C^(A)$ which it generates. However, if", null, "$\\tilde{A}$ is a completely isometrically isomorphic (i.e., indistinguishable) copy of", null, "$A$, it might be the case that", null, "$C^(\\tilde{A})$ is different than", null, "$C^(A)$. What is the precise relationship between an operator algebra and the C-algebra that it generates? What is the “correct” C-algebra in which one should study", null, "$A$? How can one study operator algebras abstractly?\n\nArveson’s approach was the following: if", null, "$I$ is an ideal in", null, "$C^(A)$ such that the quotient map", null, "$C^(A) \\rightarrow C^(A)/I$ is completely isometric when restricted to", null, "$A$, then the quotient", null, "$C^(A)/I$ is another C-algebra that contains", null, "$A$ completely isometrically isomorphically, and this C-algebra is better than", null, "$C^(A)$ because it is smaller. Such an ideal", null, "$I$ is said to be a boundary ideal. Now suppose that there exists an ideal", null, "$J$ which is the largest boundary ideal —", null, "$J$ is said to be the Shilov boundary ideal and in the commutative case (when", null, "$A$ is a function algebra) it is precisely the ideal of functions which vanish on the Shilov boundary of", null, "$A$. Then", null, "$C^(A)/J$ is the smallest such quotient that can be formed. Arveson proved that", null, "$C^(A)/J$ is a canonical invariant of", null, "$A$ and that it is in fact the smallest C-algebra that contains", null, "$A$ completely isometrically isomorphically. This canonical C-algebra later became to be known as the C-envelope of", null, "$A$.\n\nTo be precise, it was shown that if", null, "$J$ is the Shilov boundary ideal for", null, "$A$ in", null, "$C^(A)$, and if", null, "$\\tilde{J}$ is the Shilov boundary ideal for", null, "$\\tilde{A}$ in", null, "$C^(\\tilde{A})$, and if", null, "$\\phi : A \\rightarrow \\tilde{A}$ is a complete isometry, then", null, "$\\phi$ extends to a -isomorphism between", null, "$C^(A)/I$ and", null, "$C^(\\tilde{A})/J$.\n\nAs a beautiful and very simple example of the kind of result that flowed from this theory let me bring the following:\n\nTheorem: Let", null, "$A$ and", null, "$B$ be two irreducible compact operators on", null, "$H$. Then", null, "$A$ and", null, "$B$ are unitarily equivalent if and only if the two dimensional operator spaces", null, "$\\textrm{span}\\{ I, A\\}$ and", null, "$\\textrm{span}\\{ I, B\\}$ are completely isometric via a unital map that takes", null, "$A$ to", null, "$B$\n\nThe urgent question now becomes: does the Shilov boundary ideal exist?.\n\nTo prove the existence of the Shilov boundary ideal, Arveson introduced boundary representations. Although boundary representations were central to his investigations, and although he proved that they exist in some special cases, the question of whether boundary representations exist in general was left open (consequently, the existence of the C-envelope was also left open). The existence of boundary representations in general (for separable operator spaces) was established by Arveson four decades later (!) in [“The noncommutative Choquet boundary”, J. Amer. Math. Soc. 21, 2008] following important works of Dritschel and McCullough and of Muhly and Solel (the existence of the C-envelope was established by Hamana in 1979 by a different approach).\n\nOne of the immediate applications of this theory that Arveson presented was a dilation theorem that unified and clarified the various bits and pieces of dilation theory (initiated by Sz.-Nagy) that were present at the time:\n\nArveson’s Dilation Theorem (1972). Let", null, "$T$ be a", null, "$k$-tuple of commuting operators and let", null, "$X \\subset \\mathbb{C}^k$ be a compact set. Then", null, "$T$ has a normal dilation", null, "$N$ with", null, "$\\textrm{sp}(N) \\subseteq \\partial X$ if and only if", null, "$X$ is a complete spectral set for", null, "$T$\n\nI do not want to go into the definitions, the reader is referred to this exposition by Arveson for details.\n\nMany of Arveson’s results in the two “Subalgebras” papers mentioned above relied on a Hahn–Banach type extension theorem for completely positive maps which he proved in the first section of the first paper. Let me state a version for completely contractive maps which is perhaps clearer to the newcomer:\n\nArveson’s Extension Theorem (1969). Let", null, "$B$ be a C-algebra, let", null, "$S \\subseteq B$ be an linear subspace containing the identity and let", null, "$\\phi : S \\rightarrow B(H)$ be a unital, completely contractive map. Then", null, "$\\phi$ has an extension to a unital, completely contractive and completely positive map", null, "$\\Phi : B \\rightarrow B(H)$.\n\nAnyone with some experience in functional analysis can appreciate the utility of this result. Indeed, this theorem is used by operator algebraists on a daily basis, to this day.\n\nThe ideas and results of these two papers of Arveson which we mentioned have had a profound impact on operator algebras and operator theory. Without them, the study of abstract operator algebras, the solution of the Sz.-Nagy–Halmos problem (is every polynomially bounded operator similar to a contraction?) and the numerous applications of operator space techniques to operator algebras, would have all been unthinkable." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, 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https://www.physicsforums.com/search/4748291/	[ "# Search results\n\n1. ### A Time Dilation Paradox?\n\n(Quibble) Almost correct. Mathematically that should be: If ##Δx_A=0## then ##Δt_B>=Δt_A## and If ##Δx_B=0## then ##Δt_A>=Δt_B##. And then, if they insist on having them both in one scenario, then clearly: If ##Δx_A=Δx_B=0## then ##Δt_B>=Δt_A## and ##Δt_A>=Δt_B##, and thus...\n2. ### Maxwell's distribution\n\nGood question. See Maxwell–Jüttner distribution.\n3. ### Perspective on Relativity and Length Contraction\n\nFwiw--not much apparently--I think you missed the point, but feel free not to lose any sleep over it :cool:\n4. ### Perspective on Relativity and Length Contraction\n\nNice diagram! Fwiw, note that : (1) in the top figure, where Red touches the green car simultaneously with both hands, Green protests and says that Red is cheating because he does not touch his green car simultaneously. According to Green, Red touches the green car's front too late and the...\n5. ### Derivation of Schwarzschild radius from escape velocity\n\nThere's a nice comment about that on page 2-22 of http://www.eftaylor.com/pub/chapter2.pdf\n\n7. ### How does Maxwell equation suggest that the speed of light is the same\n\n(Quibble) ... which will or will not falsify the postulates. I think that \"validate\" is too strong, as it might be interpreted as a synonym of \"prove\".\n8. ### SR question\n\nNo, more like, from a distance you look smaller to me, and I look smaller to you.\n9. ### Problem with relativity\n\nMeasured time differences increase by a factor gamma, but c.t' = c.t.gamma only for x=0 (see Wikipedia Time dilation) Measured distances decrease by a factor gamma, but x' = x/gamma only for t'=0 (see Wikipedia Length contraction) Combining these two clearly distinct situations with two events...\n10. ### Four amateur questions about gravity and general relativity\n\nNah. Gravity can be modeled with geometry. The former is physical and the latter is mathematical. The Sun has a LOT of mass.\n11. ### Spacetime diagram drawing problem\n\nLots of diagrams at https://en.wikipedia.org/wiki/Slope\n12. ### Spacetime diagram drawing problem\n\nTo get a slope 1/2 between the horizontal and a line, you do not* bisect the 45 degrees angle. Instead you draw a line with slope 1/2, which means climbing 1 unit for every 2 units of horizontal displacement. That gives an angle of 26.6 degrees, not one of 22.5 degrees. See...\n13. ### Spacetime diagram drawing problem\n\nThen you need Arctan(1/2) between the ct-axis and the ct'-axis, and thus 45 degrees -Arctan(1/2) between the ct'-axis and the light diagonal.\n14. ### Spacetime diagram drawing problem\n\nOops, I meant 45 degrees - Arctan(1/2) = 18.4 degrees.\n15. ### Spacetime diagram drawing problem\n\nTry using an Arctan(1/2) = 26.6 degrees angle.\n\nFrom the travelers point of view they don't accelerate the same way. The one in front moves away from the one behind. The thread breaks.\n17. ### Trying to reconcile Lorentz Transformation and Length Contraction\n\nIt means making a measurement that is simultaneous with the (x',t') = (0,0) event in my coordinates. You and the lamp post are moving in my frame, so I must make sure that I measure the distances to you and to the lamp post simultaneously. You are at distance x'=0 at my time t'=0, so I must...\n18. ### Time moves forward, what if you move UP or DOWN? Does time stop?\n\n\"Moving with the arrow of time\" is just fancy speak for \"growing older\". As far as we know, we can't grow younger, hence the \"arrow\". It points \"where\" we go.\n19. ### Ratio of proper clock rates?\n\nWorth a little texercise: \\begin{align} \\Delta x' &= x'_2 - x'_1 &\\quad \\Delta t' &= t'_2 - t'_1 \\\\ &= \\gamma \\ ( x_2 - v \\ t_2 ) - \\gamma \\ ( x_1 - v \\ t_1 ) & &= \\gamma \\ ( t_2 - \\tfrac{v}{c^2} x_2 ) - \\gamma \\ ( t_1 -...\n20. ### Ratio of proper clock rates?\n\nThanks to its linearity, the Lorentz transformation works perfectly on coordinate differences between pairs of events." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.888128,"math_prob":0.9184069,"size":1443,"snap":"2021-31-2021-39","text_gpt3_token_len":397,"char_repetition_ratio":0.09451008,"word_repetition_ratio":0.008333334,"special_character_ratio":0.27720028,"punctuation_ratio":0.15483871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935213,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T04:29:19Z\",\"WARC-Record-ID\":\"<urn:uuid:f2123a00-d75a-495b-bb81-3a415212c8b6>\",\"Content-Length\":\"53574\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3e3f493-f0f9-4388-8be5-69339fd9cd93>\",\"WARC-Concurrent-To\":\"<urn:uuid:ead9689f-bdd7-4bf3-8ad4-9a76aed6f7cd>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/search/4748291/\",\"WARC-Payload-Digest\":\"sha1:4UJL33FBVPKCSW47NOLHPGGMZTBGATHS\",\"WARC-Block-Digest\":\"sha1:TRYRX23NGTXLHDNRZQEYJKI2HGJQRCXW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057158.19_warc_CC-MAIN-20210921041059-20210921071059-00553.warc.gz\"}"}
http://downloads.hindawi.com/journals/mpe/2009/716104.xml	[ "MPEMathematical Problems in Engineering1563-51471024-123XHindawi Publishing Corporation71610410.1155/2009/716104716104Research ArticleMacro- and Microsimulations for a Sublimation Growth of SiC Single CrystalsGeiserJürgen1IrleStephan2PiqueiraJosé Roberto Castilho1Humboldt University of BerlinUnter den Linden 610099 BerlinGermanyhu-berlin.de2Nagoya UniversityFuro-cho, Chikusa-kuNagoya 464-8602Japannagoya-u.ac.jp20091201200920090309200802112008181120082009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nThe numerous technical applications in electronic and optoelectronic devices, such as lasers, diodes, and sensors, demand high-quality silicon carbide (SiC) bulk single crystal for industrial applications. We consider an SiC crystal growth process by physical vapor transport (PVT), called modified Lely method. We deal with a model for the micro- and macroscales of the sublimation processes within the growth apparatus. The macroscopic model is based on the heat equation with heat sources due to induction heating and nonlocal interface conditions, representing the heat transfer by radiation. The microscopic model is based on the quantum interatomic potential and is computed with molecular dynamics. We study the temperature evolution in the apparatus and reflect the growth behavior of the microscopic model. We present results of some numerical simulations of the micro- and macromodels of our growth apparatus.\n\n1. Introduction\n\nThe motivation for this study comes from the technical demand to simulate a crystal growth apparatus for SiC single crystals. The single crystals are used as a high-valued and expensive material for optoelectronics and electronics (cf. ). We concentrate on a deterministic model for simulating crystal growth; alternative models are discussed with comprehensive probabilistic modeling (see ).\n\nThe silicon carbide (SiC) bulk single crystals are produced by a growth process through physical vapor transport (PVT), called modified Lely method. The modeling for the thermal processes within the growth apparatus is done in [3, 4]. In this paper, we propose one step more in the modeling of the macroscopic and microscopic parts. The idea is to exchange results from the macroscopic to the microscopic scale to obtain a feedback to control the growth process of the SiC bulk. Here the benefits are an acceleration of solving interactive growth processes of the crystal with their underlying temperature in the apparatus. Using only standard codes, which are decoupled, a simple parameter exchange of temperature and pressure in the deposition region cannot resolve the growth problem accurately. We propose a first framework of a combined model, which is based on the authors’ knowledge of a novel work and a first approach to a coupled solver method.\n\n2. Macroscopic Model: Heat-Flux\n\nIn the following, we discuss the macroscopic model, which is based on continuum equations for the heat-flux.\n\n2.1. Mathematical Model\n\nThe underlying equations of the model are given as follows.\n\n(a) In this work, we assume that the temperature evolution inside the gas region Ωg can be approximated by considering the gas as pure argon (Ar). The reduced heat equation isρgtUg(κgT)=0,Ug=zArRArT,where T is the temperature, t is the time, and Ug is the internal energy of the argon gas. The parameters are given as ρg being the density of the argon gas, κg being the thermal conductivity, zAr being the configuration number, and RAr being the gas constant for argon.\n\n(b) The temperature evolution inside the region of solid materials Ωs (e.g., inside the silicon carbide crystal, silicon carbide powder, graphite, and graphite insulation) is described by the heat equationρstUs(κsT)=f,Us=0Tcs(S)dS,where ρs is the density of the solid material, Us is the internal energy, κs is the thermal conductivity, and cs is the specific heat.\n\nThe equations hold in the domains of the respective materials and are coupled by interface conditions, for example, requiring the continuity for the temperature and for the normal components of the heat flux on the interfaces between opaque solid materials. On the boundary of the gas domain, that is, on the interface between the solid material and the gas domain, we consider the interface conditionκgTng+RJ=κsTng,where ng is the normal vector of the gas domain, R is the radiosity, and J is the irradiosity. The irradiosity is determined by integrating R along the whole boundary of the gas domain (cf. ). Moreover, we haveR=E+Jref,E=σϵT4(Stefan-Boltzmannequation),Jref=(1ϵ)J,where E is the radiation, Jref is the reflexed radiation, ϵ is the emissivity, and σ is the Boltzmann radiation constant.\n\nThe density of the heat source induced by the induction heating is determined by solving Maxwell’s equations. We deal with these equations under the simplifying assumption of an axisymmetric geometry, axisymmetric electromagnetic fields, and a sinusoidal time dependence of the involved electromagnetic quantities, following . The considered system and its derivation can be found in [3, 4, 7].\n\nIn this paper, we focus on the discretization and material properties, which are important for realistic simulations. Our underlying software tool WIAS-HiTNIHS (cf. ) allows us a flexibility in the grid generation and for the material parameters.\n\nIn the next section, we describe the used discretization.\n\n2.2. Discretization\n\nFor the discretization of the heat equation (diffusion equation), we apply the implicit Euler method in time and the finite volume method for the space discretization (cf. [3, 4, 8]). We consider a partition 𝒯=(ωi)iI of Ω such that, for m{s,g} (with s solid, g gas) and iI,ωm,i:=ωiΩm defines either a void subset or a nonvoid, connected, and open polyhedral subset of Ω. By integrating the corresponding heat equation (2.1) or (2.3) over ωm,i, we derive the following nonlinear equations for the temperature variables,ρmωm,i(Um(Tn+1)Um(Tn))rdxΔtn+1ωm,iκm(Tn+1)Tn+1nωm,irds=Δtn+1ωm,ifmrdx,where the time interval is Δtn+1=tn+1tn. The temperature is given as Tn+1=T(tn+1,x), where x represents cylindrical coordinates. For the right-hand sides, we demand fs:=f0 and fg=0.\n\nMore details of the discretization and of dealing with the interface conditions are presented in [3, 4, 9, 10].\n\nIn the next section, the properties of the materials in the crystal growth apparatus are described.\n\n2.3. Material Properties\n\nFor the technical realization of the apparatus, we implement the axisymmetric geometry given in , which is presented in Figure 1. Furthermore, the properties of the materials are specified in [3, 9, 12].\n\nThe growth apparatus’ dimensions: rmin=0,rmax=8.4cm,zmin=0,zmax=25.0cm, the coil rings’ dimensions: rmin=4.2cm,rmax=5.2cm,zmin=0,zmax=14.0cm.\n\nWithin the following specific material functions and parameters for the processes, the thermal conductivity κ is given in W/(m K), the electrical conductivity σc is given in 1/(Ohm m), the mass density ρ is given in kg/m3, the specific heat csp is given in J/(K kg), the temperature T is given in K, and the relative gas constant RAr is given in J/(K kg). Further, the emissivity ϵ and relative magnetic permeability μ are given dimensionless.\n\n(i) For the gas phase (argon), we haveκAr(T)={1.83914104T0.800404T500,7.128738+6.610288102T2.440839104T2+4.497633107T34.1325171010T4+1.5144631013T5500T600,4.1944104T0.671118600T, where σc,Ar=0.0,ρAr=3.73103,μAr=1.0,zAr=3/2,RAr=2.081308102.\n\n(ii) For graphite felt insulation, we haveκIns(T)={8.175102+2.485104TT1473,1.1902102+0.346838T3.9971104T2+2.2830107T36.460471011T4+7.25491015T51473T1873,0.7447+7.5104T1873T, where ϵIns=0.2,σc,Ins(T)=2.45102+9.82102T,ρIns=170.00,μIns=1.00,csp,Ins=2100.00.\n\n(iii) For the graphite, we haveκGraphite(T)=37.715exp(1.96104T),ϵGraphite(T)={0.67T1200,3.7527.436103T+6.4163106T22.3366109T33.08331013T41200T2200,0.792200T, where σc,Graphite=104,ρGraphite=1750.0,μGraphite=1.0,csp,Graphite(T)=1/(4.411102T2.306+7.97104T0.0665).\n\n(iv) For the SiC crystal, we have κSiC-C(T)=exp(9.892+(2.498102)/T0.844ln(T)),ϵSiC-C=0.85,σc,SiC-C=105,ρSiC-C=3140.0,μSiC-C=1.0,csp,SiC-C(T)=1/(3.91104T3.173+1.835103T0.117).\n\n(v) For the SiC powder, we have κSiC-P(T)=1.452102+5.471012T3,ϵSiC-P=0.85,σc,SiC-P=100.0,ρSiC-P=1700.0,μSiC-P=1.0,csp,SiC-P=1000.0.\n\nThe functions are programmed in our flexible software package WIAS-HiTNIHS.\n\nIn the next section, we discuss the microscopic model.\n\n2.4. Coupling Method for Macroscopic and Microscopic Models: Operator Splitting\n\nOften simple coupling via the parameters (e.g., target-temperature and growth velocity of the bulk) is enough for the problem.\n\nHere we propose a new idea of coupling the model equations together, on the one hand the heat equations and on the other hand the kinetic equations for molecules.\n\nFor a first idea, we deal with abstract operators, which include the heat- and the kinetics equations.\n\nUsing our two standard codes of the macro- and micromodels, we could implement a coupled model, by a so-called iterative operator-splitting method. Such a proposed method couples the two physical processes of the thermal situation in the growth apparatus and their important geometrical differences at the deposition layer with the kinetic molecular model. The benefits are a numerical algorithm, that exchanged the underlying operators of the thermal situation and the kinetic molecular situation, which are computed by each software code independently and coupled via an iterative solver step; see a detailed coupling analysis in .\n\nIn the following algorithm, an iteration method is proposed, with fixed splitting discretization step-size τ.\n\nDue to the underlying multiscale problem of kinetics and heat processes, we have to solve fine time scales of kinetic equations and coarse time scales for heat equations. On a time interval [tn,tn+1] that is sufficiently small to yield accurate kinetics, we solve the following subproblems consecutively for i=0,2,,2m (cf. [14, 15]):ci(t)t=Aci(t)+Bci1(t),withci(tn)=cn,i=1,2,,j,ci(t)t=Aci1(t)+Bci(t),withci+1(tn)=cn,i=j+1,j+2,,m, where we assume that the operator A has a large time scale (macroscopic model) and B has a small time scale (microscopic model). Further c0(tn)=cn,c1=0, and cn are initialization and starting conditions.\n\nIn the following, we give an overview to the accuracy of the method, which is given in the convergence and the rate of the convergence.\n\nTheorem 2.1.\n\nLet us consider the abstract Cauchy problem in a Banach space X:tc(t)=Ac(t)+Bc(t),0<tT,c(0)=c0, where A,B,A+B:XX are given linear operators being generators of the C0-semigroup and c0X is a given element. Then the iteration process (2.11)-(2.12) is convergent. The rate of convergence is of higher order and given as 𝒪(τn2m), where the iterations are i=1,3,,2m+1.\n\nThe proof is given in .\n\nIn the next subsection, we present the methods for the microscopic model.\n\n3. Microscopic Model: Quantum Chemical Molecular Dynamics (QM/MD) of SiC Condensation (Methodology)\n\nThe density-functional tight-binding (DFTB) method is employed as the quantum interatomic potential in our molecular dynamics (MD) simulations, using atomic and diatomic parameters obtained from density functional theory; see . DFTB is an approximate density functional theory method based on the tight binding approach, and utilizes an optimized minimal LCAO Slater-type all-valence basis set in combination with a two-center approximation for Hamiltonian matrix elements. Parameter sets for Si-Si and Si-C were taken from . Energies and gradients are evaluated direct (on the fly) during the dynamic simulation. As in our previous simulations of carbon cap and subsequent nanotube formation on the C- and Si-faces of SiC(000-1) surfaces during sublimation evaporation, we have not included charge- or spin-polarization in the present work. Further, we will consider in a next model electrokinetic effect on heat transfer in parallel-plate microchannels, hydrodynamic focusing effects, and nanoeffect as done in .\n\nFor time propagation we employed a velocity Verlet integrator with a time step of 1.209 fs (50 atomic units) and used a Nose-Hoover chain thermostat to generate a canonical ensemble for target temperature Tt; see . The thermostat was employed uniformly in the reaction system.\n\nRegarding the atomistic structure of the employed surface model systems, we have chosen the C-face of the same square SiC(000-1) slab unit cell as in our previous study, consisting of two SiC layers terminated by hydrogen atoms to mimic bulk effect in the direction away from the surface. Periodic boundary conditions were employed with a unit cell size of 1000 Å in the direction perpendicular to the surface and 16.0 Å and 15.4 Å in the other two surface-directions to achieve two-dimensional slab periodicity. The geometry optimized structure of this surface model is shown in Figure 2.\n\nOptimized geometry of the C-face of the (000-1) SiC surface as initial starting point for QM/MD simulations. Blue spheres correspond to silicon atoms, purple spheres correspond to carbon atoms, and white spheres correspond to hydrogen atoms terminating the slab model in bulk direction. The model is the unit cell used in periodic boundary calculations with infinite surface extension.\n\nDuring MD simulations, the movements of hydrogen terminating atoms were frozen. Using such an approach, we have effectively introduced a steep temperature gradient from the deepest bulk-side SiC layer to the atoms lying above on the surface. The slab model was then annealed at Tt=2000K for 1.20 picoseconds, and 3 structures were selected as initial starting geometries at t=0.60 picosecond (trajectory A50), t=0.72 picosecond (trajectory A60), and t=0.86 picosecond (trajectory A80). In the vicinity of the surface, 10 SiC molecules were randomly distributed in the gas phase. Since these molecules are nonbonded to the surface, they are subsequently thermostated at Tt. Gas phase molecules approaching the surface will experience immediate cooling, which will drive the condensation process during these simulations.\n\nIn the microscopic model, we can derive the growth rate v of the seed surface in dependence on temperature and pressure. Based on these growth rates, we can adapt the geometry for the macroscopic model. Such modification helps to give more accurate temperature differences in the macroscopic model and understand the growth process.\n\nIn the next section, we present results of our numerical experiments.\n\n4. Numerical Experiments\n\nWe present in the following our macro- and microscopic simulations, where the microscopic simulations take into account the target temperature of the macroscopic model.\n\n4.1. Macroscopic Model: Simulation of the Temperature Field in the Apparatus\n\nFor the numerical results, we apply the parameter functions in Section 2.3. We consider the geometry shown in Figure 1, using a constant total input power of 10kW (cf. ). The numerical experiments are performed using the software WIAS-HiTNIHS (cf. ) based on the software package pdelib (cf. ) which uses the sparse matrix solver PARDISO (cf. ). We compute the coupled system consisting of the heat equations and Maxwell’s equations. For the growth process, the temperature difference Tss=T(rsource,zsource)T(rseed,zseed) (with the coordinates (rsource,zsource)=(0,0.143) and (rseed,zseed)=(0,0.158), corresponding to the points Tsource and Tseed in Figure 1) is crucial. On the other hand, in the physical growth experiments, usually only the temperatures T(rbottom,zbottom) and T(rtop,ztop) (with the coordinates (rbottom,zbottom)=(0,0.028) and (rtop,ztop)=(0,0.173), corresponding to the points Tbottom and Ttop in Figure 1) are measurable and their difference Tbt=T(rbottom,zbottom)T(rtop,ztop) is often used as an indicator for Tss. In Figure 3, we present the temperature differences Tss and Tbt. As a result of our computations, the temperature difference Tbt can only restrictively be used as an indicator for the temperature difference Tss (cf. the discussions in [5, 9]).\n\nTransient results for the temperature differences Tbt and Tss.\n\nThe further computations are based on the stationary case, dealing with (2.1) by discarding the terms with a time derivative. For this case, the results are virtually equal to the one in the transient case with t>15000 seconds. For the stationary results, we focus on the error analysis for the space dimension by applying the grid refinement. The solutions for the heat equation are computed at the points T(rbottom,zbottom) and T(rtop,ztop) for successive grids. For the error analysis, we apply the following error differences:ϵabs=\|T˜j+1(r,z)T˜j(r,z)\|,where T˜j(r,z) and T˜j+1(r,z) are solutions evaluated at the point (r,z) which has been computed using the grids j and j+1, respectively. The elements of the grid j+1 are approximately 1/4 of the elements of the grid j. The results are presented in Table 1.\n\nComputations on different grids for the errors analysis with absolute differences (cf. (4.1)).\n\nGridGrid point (0,0.028) (Tbottom)Grid point (0,0.173) (Ttop)\nLevelNumber of nodesSolution T[K]Absolute difference T[K]Solution T[K]Absolute difference T[K]\n015322408.112813.29\n1230172409.781.672812.781.01\n2912902410.350.572811.790.49\n33642252410.460.112811.600.19\n\nThe result of the refinement indicates the reduction of the absolute difference as it is demanded for the convergence of the discretization method. The method is stabilized in the presented refinement by reducing the differences.\n\nIn Figure 4, the temperature field is presented for the stationary case. The temperature increases from the bottom to the middle of the graphite pot, and decreases from the middle to the top of the graphite pot.\n\nTemperature field for the apparatus simulated for the stationary case with 23017 nodes.\n\n4.2. Microscopic Model: Atomistic QM/MD Simulations of SiC Condensation on the C-Face of Si(000-1)\n\nThe total time of the three condensation simulations was 24.02 picoseconds. This is admittedly a time too short for the study of crystal growth, which would ideally require annealing simulations on the order of several 100 nanoseconds, but this study is focusing on the initial stages of SiC aggregation and tries to identify key features in the condensation process. As such, this is at present rather a preliminary study exploring the applicability of QM/MD simulations for SiC crystal growth. We have first concentrated on the polar C-surface of SiC (0001) since it has a maximum of dangling bonds with highest reactivity. The Si-face and other nonpolar surfaces are much less reactive; see . Since our seed crystal surface slab model contains only two SiC layers, we are also unable to address the issue of polymorphism at present, although it should be noted that our model system rather resembles the cubic 3C than the hexagonal polytypes.\n\nTt was chosen as 2000 K for all simulations, and representative snapshots from trajectory A50 are given in Figure 5. We find that under the present conditions with a relatively high density of SiC gas molecules, several of them attach very quickly to the surface (2 after 0.10 picosecond). Also, SiC molecules can react with each other to form dimers, preferably with C-C bonds. Eventually, an average of 5.3 SiC molecules become attached to the surface in the three simulations, with the other molecules being lost to the vacuum layer.\n\nSimulation of the addition of 10 SiC atoms on the C-face of the (000-1) SiC surface from Figure 2. Blue spheres correspond to silicon atoms, purple spheres correspond to carbon atoms, and white spheres correspond to hydrogen atoms terminating the slab model in bulk direction. Times are given in picoseconds (ps), indicating that the moment the snapshots were taken during the dynamics simulations.\n\nOnce attached, the Si atoms on the surface prove to be highly mobile, as their bond radius is larger than the case of carbon, and the binding energies are lower . The carbon atoms on the surface tend to form C2 units, and behave similar to “wobbling C2” entities that we had observed for high-temperature simulations of pure carbon; see . It seems from our simulations at this stage that the system tries to reach an equilibrium with a constant number of C-C in the new layers, and that the Si atoms are more isolated, becoming occasionally attacked by a C2 dimer. In particular, C2 units are oriented mainly perpendicular to the surface, while the more visible Si2 dimers do not show such an alignment preference. The surface itself retains the structure of alternating Si–C units. A new layer of Si–C units is being deposited with a somewhat inhomogeneous structure containing C2 and Si2 units at first, and gradually becoming more homogeneous due to annealing.\n\n5. Summary\n\nWe have presented a model for the heat transport inside a technical apparatus for crystal growth of SiC single crystals. We introduce the heat equation and the radiation of the apparatus and the coupled situation of the different materials. The equations are discretized by the finite volume method and the complex material functions are embedded in this method. Transient and stationary results are presented leading to some information about the processes within the technical apparatus. We present numerical results for the stationary case to support the accuracy of our solutions. We also presented atomistic quantum chemical molecular dynamics (QM/MD) simulations based on the density-functional tight-binding (DFTB) method for initial reactions of gaseous SiC on the polar C-face of SiC(000-1). In our future work, we concentrate on further implementations and numerical methods for a crystal growth model and use kinetic data obtained from more accurate microscopic model simulations in the simulation of the heat transport. Once longer and a larger number of trajectories are obtained in our microsimulations, it will be possible to deduct an accurate QM/MD-based estimate for the bulk growth, in dependence on the temperature to our macrosimulations. This data will then enter the iterative solution of the heat and kinetics equations of the coupled macroscopic and microscopic models.\n\nMüllerSt.stephan_mueller@cree.comGlassR. C.HobgoodH. M.Progress in the industrial production of SiC substrates for semiconductor devicesMaterials Science and Engineering B2001801–332733110.1016/S0921-5107(00)00658-9YaoX.HeB.WangH.ZhouX.Numerical simulation of dendrite growth during solidificationInternational Journal of Nonlinear Sciences and Numerical Simulation200672171176KleinO.PhilipP.philip@wias-berlin.deTransient numerical investigation of induction heating during sublimation growth of silicon carbide single crystalsJournal of Crystal Growth20032471-221923510.1016/S0022-0248(02)01903-6PhilipP.Transient numerical simulation of sublimation growth of SiC bulk single crystals. Modeling, finite volume method, results, Ph.D. thesis2003Berlin, GermanyWeierstrass-Institute for Applied Analysis and StochasticsReport No. 22KleinO.PhilipP.SprekelsJ.Modeling and simulation of sublimation growth of SiC bulk single crystalsInterfaces and Free Boundaries200463295314MR2095334ZBL1081.35117RappazJ.SwierkoszM.Modelling in numerical simulation of electromagnetic heatingModelling and Optimization of Distributed Parameter Systems1996New York, NY, USAChapman & Hall313320MR1388547ZBL0879.65091KleinO.PhilipP.Correct voltage distribution for axisymmetric sinusoidal modeling of induction heating with prescribed current, voltage, or powerIEEE Transactions on Magnetics20023831519152310.1109/20.999125GeiserJ.R3T: radioactive-retardation-reaction-transport-program for the simulation of radioactive waste disposalsApril 2004College Station, Tex, USAInstitute for Scientific Computation, Texas A&M UniversityKleinO.PhilipP.philip@wias-berlin.deTransient temperature phenomena during sublimation growth of silicon carbide single crystalsJournal of Crystal Growth20032493-451452210.1016/S0022-0248(02)02320-5KleinO.PhilipP.Transient conductive-radiative heat transfer: discrete existence and uniqueness for a finite volume schemeMathematical Models & Methods in Applied Sciences2005152227258MR2119998ZBL1070.65075PonsM.AnikinM.ChourouK.State of the art in the modelling of SiC sublimation growthMaterials Science and Engineering B199961-62182810.1016/S0921-5107(98)00439-5NilssonO.MehlingH.HornR.Determination of the thermal diffusivity and conductivity of monocrystalline silicon carbide (300–2300 K)High Temperatures-High Pressures1997291737910.1068/htec142GeiserJ.Iterative operator-splitting methods with higher-order time integration methods and applications for parabolic partial differential equationsJournal of Computational and Applied Mathematics20082171227242MR242744310.1016/j.cam.2007.06.028ZBL1144.65062KanneyJ. F.MillerC. T.caseyller@unc.eduKelleyC. T.Convergence of iterative split-operator approaches for approximating nonlinear reactive problemsAdvances in Water Resources200326324726110.1016/S0309-1708(02)00162-8FaragóI.faragois@cs.elte.huGeiserJ.geiser@mathematik.hu-berlin.deIterative operator-splitting methods for linear problemsInternational Journal of Computational Science and Engineering20073425526310.1504/IJCSE.2007.018264PorezagD.FrauenheimTh.KöhlerTh.SeifertG.KaschnerR.Construction of tight-binding-like potentials on the basis of density-functional theory: application to carbonPhysical Review B19955119129471295710.1103/PhysRevB.51.12947RaulsE.HajnalZ.DeákP.FrauenheimTh.Theoretical study of the nonpolar surfaces and their oxygen passivation in 4H- and 6H-SiCPhysical Review B20016424724532310.1103/PhysRevB.64.245323IrleS.WangZ.ZhengG.MorokumaK.KusunokiM.Theory and experiment agree: single-walled carbon nanotube caps grow catalyst-free with chirality preference on a SiC surfaceThe Journal of Chemical Physics20061254504470210.1063/1.2212402WangZ.IrleS.sirle@iar.nagoya-u.ac.jpZhengG.KusunokiM.MorokumaK.morokuma@emory.eduCarbon nanotubes grow on the C face of SiC (0001) during sublimation decomposition: quantum chemical molecular dynamics simulationsThe Journal of Physical Chemistry C200711135129601297210.1021/jp072208dChangC.-C.YangR.-J.rjyang@mail.ncku.edu.twHydrodynamic focusing effect on two-unmixed-fluid in microchannelsInternational Journal of Nonlinear Sciences and Numerical Simulation200893213220FanJ.WangL.lqwang@hku.hkChengL.Electrokinetic effects on flow and heat transfer in parallel-plate microchannelsInternational Journal of Nonlinear Sciences and Numerical Simulation200783335345FanJ.WangL.lqwang@hku.hkChengL.Forced convection in rectangular microchannels: electrokinetic effectsInternational Journal of Nonlinear Sciences and Numerical Simulation200783359374HeJ.-H.jhhe@dhu.edu.cnWanY.-Q.XuL.Nano-effects, quantum-like properties in electrospun nanofibersChaos, Solitons & Fractals2007331263710.1016/j.chaos.2006.09.023AllenM. P.TildesleyD. J.Computer Simulation of Liquids1989New York, NY, USAOxford University PressFuhrmannJ.KopruckiTh.LangmachH.HackbuschW.WittumG.pdelib: an open modular tool box for the numerical solution of partial differential equations. Design patternsProceeding of the 14th GAMM Seminar on Concepts of Numerical SoftwareJanuary 2001Kiel, GermanySchenkO.olaf.schenk@unibas.chGärtnerK.gaertner@wias-berlin.deSolving unsymmetric sparse systems of linear equations with PARDISOFuture Generation Computer Systems200420347548710.1016/j.future.2003.07.011PollmannJ.KrügerP.SabischM.Atomic and electronic structure of SiC surfaces from ab-initio calculationsPhysica Status Solidi (B)1997202142144510.1002/1521-3951(199707)202:1<421::AID-PSSB421>3.0.CO;2-DIrleS.ZhengG.ElstnerM.MorokumaK.morokuma@emory.eduFormation of fullerene molecules from carbon nanotubes: a quantum chemical molecular dynamics studyNano Letters20033446547010.1021/nl034023y" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87206566,"math_prob":0.9443012,"size":10046,"snap":"2020-34-2020-40","text_gpt3_token_len":2069,"char_repetition_ratio":0.1462856,"word_repetition_ratio":0.026282052,"special_character_ratio":0.19540116,"punctuation_ratio":0.08078477,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97785085,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T18:43:57Z\",\"WARC-Record-ID\":\"<urn:uuid:a9762f8e-5b13-4111-afa6-4ff59b496629>\",\"Content-Length\":\"122012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96cc5fa6-728c-482b-b779-c9ce03d27f7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:beba59d5-eac8-4480-af9b-4f984dac6efb>\",\"WARC-IP-Address\":\"52.217.46.204\",\"WARC-Target-URI\":\"http://downloads.hindawi.com/journals/mpe/2009/716104.xml\",\"WARC-Payload-Digest\":\"sha1:KWYBL2JVCYQMXS3SLLMNQXHACEYRGQKI\",\"WARC-Block-Digest\":\"sha1:ZKYR4O2IRD5ABHO2UBNBOCXHFMIYFTCV\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738562.5_warc_CC-MAIN-20200809162458-20200809192458-00286.warc.gz\"}"}
https://answers.ros.org/question/239417/cart-controller-help/	[ "# Cart Controller help!\n\nHi!\n\nI am making a cart controller. The idea is using a delta of time, iterate a given number of times to converge to the desired position. So the set parameters are dt (delta time) and N (iterations)\n\n1. Get the desired task positions\n2. Get the actual joint positions\n3. Calculate jacobian for actual joint position\n4. Calculate actual task position for the joint position\n5. Calculate the error (for position and orientation)\n6. Divide the error by delta time to get velocity in task space.\n7. Using the jacobian and the velocity in task space, calculate the velocity in joint space.\n8. Using the velocity in joint space and the delta of time, calculate the necessary (new) joint positions.\n9. Iterate from 3 to 8 a number of times.\n\nMy variables are:\n\nKDL::JntArray q_ (actual joint positions)\n\nKDL::Frame x_des (the desired task position)\n\nKDL::JntArray q_ (joint positions calculations)\n\nKDL::Jacobian J_ (jacobian)\n\nNow this is the algoritm.\n\n...\n\nq_s=q_;\n\nk=0;\n\nwhile(!k>iter){\n\nk++;\n\njnt_to_jac_solver_->JntToJac(q_s, J_); //Calculate jacobian for joint position\n\njnt_to_pose_solver_->JntToCart(q_s, x_s); //Calculate task position for the joint position\n\nxerr_.vel = x_des.p-x_s.p; //Calculate the error for position\n\nxerr_.rot = 0.5 * (x_des.M.UnitX() * x_s.M.UnitX() + x_des.M.UnitY() * x_s.M.UnitY() + x_des.M.UnitZ() * x_s.M.UnitZ()); //Calculate the error for orientation\n\nxdot_s.vel = xerr_.vel/dt; //Divide the error by dt to get xdot_s\n\nxdot_s.rot = xerr_.rot/dt; //Divide the error by dt to get xdot_s\n\nqdot_s=f(xdot_s,J_); //Then using xdot_s, and the jacobian, I calculate qdot_s\n\nq_s=q_s+dt*qdot_s; //New joint position\n\nThe problem is that using that it doesn't perform well. I have check that it works when I don't use the error of rotation to calculate xdot_s (that happens when I comment the Bold line in my code). So I'm pretty sure I am not calculating well the error in rotation (and so the velocity in task space)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73913306,"math_prob":0.99027115,"size":2302,"snap":"2021-31-2021-39","text_gpt3_token_len":626,"char_repetition_ratio":0.16710183,"word_repetition_ratio":0.03314917,"special_character_ratio":0.26020852,"punctuation_ratio":0.1663113,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99944586,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T02:44:15Z\",\"WARC-Record-ID\":\"<urn:uuid:d1b8a107-68b9-42cf-8178-1ba8b5662bf9>\",\"Content-Length\":\"53052\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75188faf-6093-4e6f-9e81-aeb198dcef29>\",\"WARC-Concurrent-To\":\"<urn:uuid:a03ce15b-41aa-4a05-802b-0d480330bb81>\",\"WARC-IP-Address\":\"140.211.15.248\",\"WARC-Target-URI\":\"https://answers.ros.org/question/239417/cart-controller-help/\",\"WARC-Payload-Digest\":\"sha1:CPOOKVNQJG3ULOAKQH6YVFKGJYKAVRN5\",\"WARC-Block-Digest\":\"sha1:POE3S552IVWFIJVLTDKAIV3DCA3NQ4J7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154042.23_warc_CC-MAIN-20210731011529-20210731041529-00186.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/hep-ph/0208177/	[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\nLAL 02-81\n\nhep-ph/0208177.v3\n\nMay 25, 2020\n\nConfronting Spectral Functions from Annihilation and Decays: Consequences for the Muon Magnetic Moment\n\nM. Davier, S. Eidelman, A. Höcker and Z. Zhang111 E-mail: , , ,\n\nLaboratoire de l’Accélérateur Linéaire,\n\nIN2P3-CNRS et Université de Paris-Sud, F-91898 Orsay, France\n\nBudker Institute of Nuclear Physics, Novosibirsk, 630090, Russia\n\nAbstract\n\nVacuum polarization integrals involve the vector spectral functions which can be experimentally determined from two sources: (i) annihilation cross sections and (ii) hadronic decays. Recently results with comparable precision have become available from CMD-2 on one side, and ALEPH, CLEO and OPAL on the other. The comparison of the respective spectral functions involves a correction from isospin-breaking effects, which is evaluated. After the correction it is found that the dominant spectral functions do not agree within experimental and theoretical uncertainties. Some disagreement is also found for the spectral functions. The consequences of these discrepancies for vacuum polarization calculations are presented, with the emphasis on the muon anomalous magnetic moment. The work includes a complete re-evaluation of all exclusive cross sections, taking into account the most recent data that became available in particular from the Novosibirsk experiments and applying corrections for the missing radiative corrections. The values found for the lowest-order hadronic vacuum polarization contributions are\n\nwhere the errors have been separated according to their sources: experimental, missing radiative corrections in data, and isospin breaking. The Standard Model predictions for the muon magnetic anomaly read\n\nwhere the errors account for the hadronic, light-by-light scattering and electroweak contributions. We observe deviations with the recent BNL measurement at the 3.0 () and 0.9 () level, when adding experimental and theoretical errors in quadrature.\n\n## 1 Introduction\n\nHadronic vacuum polarization in the photon propagator plays an important role in the precision tests of the Standard Model. This is the case for the evaluation of the electromagnetic coupling at the mass scale, , which receives a contribution of the order of that must be known to an accuracy of better than 1% so that it does not limit the accuracy on the indirect determination of the Higgs boson mass from the measurement of . Another example is provided by the anomalous magnetic moment of the muon where the hadronic vacuum polarization component is the leading contributor to the uncertainty of the theoretical prediction.\nStarting from Refs. [1, 2] there is a long history of calculating the contributions from hadronic vacuum polarization in these processes. As they cannot be obtained from first principles because of the low energy scale involved, the computation relies on analyticity and unitarity so that the relevant integrals can be expressed in terms of an experimentally determined spectral function which is proportional to the cross section for annihilation into hadrons. The accuracy of the calculations has therefore followed the progress in the quality of the corresponding data . Because the latter was not always suitable, it was deemed necessary to resort to other sources of information. One such possibility was the use of the vector spectral functions derived from the study of hadronic decays for the energy range less than 1.8 GeV. Another one occurred when it was realized in the study of decays that perturbative QCD could be applied to energy scales as low as 1-2 GeV, thus offering a way to replace poor data in some energy regions by a reliable and precise theoretical prescription [7, 8, 9]. Finally, without any further theoretical assumption, it was proposed to use QCD sum rules [10, 11] in order to improve the evaluation in energy regions dominated by resonances where one has to rely on experimental data. Using these improvements the lowest-order hadronic contribution to was found to be \n\nThe complete theoretical prediction includes in addition QED, weak and higher order hadronic contributions.\nThe anomalous magnetic moment of the muon is experimentally known to very high accuracy. Combined with the older less precise results from CERN , the measurements from the E821 experiment at BNL [13, 14, 15], including the most recent result , yield\n\n aexpμ=(11659203±8) 10−10 , (2)\n\nand are aiming at an ultimate precision of in the future. The previous experimental result was found to deviate from the theoretical prediction by 2.6 , but a large part of the discrepancy was actually originating from a sign mistake in the calculation of the small contribution from the so-called light-by-light (LBL) scattering diagrams [17, 18]. The new calculations of the LBL contribution [19, 20, 21] have reduced the discrepancy to a nonsignificant 1.6 level. At any rate it is clear that the presently achieved experimental accuracy already calls for a more precise evaluation of .\nIn this paper we critically review the available experimental input to vacuum polarization integrals. Such a re-evaluation is necessary because\n\n• new results have been obtained at Novosibirsk with the CMD-2 detector in the region dominated by the resonance with a much higher precision than before, and more accurate R measurements have been performed in Beijing with the BES detector in the 2-5 GeV energy range .\n\n• new preliminary results are available from the final analysis of decays with ALEPH using the full statistics accumulated at LEP1 ; also the information from the spectral functions measured by CLEO [25, 26] and OPAL was not used previously and can be incorporated in the analysis.\n\n• new results on the evaluation of isospin breaking have been produced [28, 29, 30], thus providing a better understanding of this critical area when relating vector and isovector spectral functions.\n\nSince we are mostly dealing with the low energy region, common to both and data, and because of the current interest in the muon magnetic moment prompted by the new experimental result, the emphasis in this paper is on rather than . It is true that the presently achieved accuracy on is meeting the goals for the LEP/SLD/FNAL global electroweak fit. However the situation will change in the long run when very precise determinations of , as could be available from the beam polarization asymmetry at the future Linear Collider, necessitate a significant increase of the accuracy on .\nDisclaimer: ’theoretical’ predictions using vacuum polarization integrals are based on experimental data as input. The data incorporated in this analysis are used as quoted by their authors. In particular, no attempt has been made to re-evaluate systematic uncertainties even if their size was deemed to be questionable in some cases. However, whenever significant incompatibilities between experiments occur, we apply an appropriate rescaling of the combined error. The analysis thus heavily relies on the quality of the work performed in the experiments.\n\n## 2 Muon Magnetic Anomaly\n\nIt is convenient to separate the Standard Model prediction for the anomalous magnetic moment of the muon into its different contributions,\n\nwith\n\nwhere is the pure electromagnetic contribution (see [32, 33] and references therein), is the lowest-order contribution from hadronic vacuum polarization, is the corresponding higher-order part [34, 4], and , where the first error is the hadronic uncertainty and the second is due to the Higgs mass range, accounts for corrections due to exchange of the weakly interacting bosons up to two loops . For the LBL part we add the values for the pion-pole contribution [19, 20, 21] and the other terms [20, 21] to obtain .\nBy virtue of the analyticity of the vacuum polarization correlator, the contribution of the hadronic vacuum polarization to can be calculated via the dispersion integral \n\nwhere is the QED kernel ,\n\n K(s)=x2(1−x22)+(1+x)2(1+1x2)(ln(1+x)−x+x22)+(1+x)(1−x)x2lnx , (6)\n\nwith and . In Eq. (5), denotes the ratio of the ’bare’ cross section for annihilation into hadrons to the pointlike muon-pair cross section. The ’bare’ cross section is defined as the measured cross section, corrected for initial state radiation, electron-vertex loop contributions and vacuum polarization effects in the photon propagator (see Section 4 for details). The reason for using the ’bare’ (i.e. lowest order) cross section is that a full treatment of higher orders is anyhow needed at the level of , so that the use of ’dressed’ cross sections would entail the risk of double-counting some of the higher-order contributions.\nThe function decreases monotonically with increasing . It gives a strong weight to the low energy part of the integral (5). About 91 of the total contribution to is accumulated at center-of-mass energies below 1.8 GeV and 73 of is covered by the two-pion final state which is dominated by the resonance.\n\n## 3 The Input Data\n\n### 3.1 e+e− Annihilation Data\n\nThe exclusive low energy cross sections have been mainly measured by experiments running at colliders in Novosibirsk and Orsay. Due to the high hadron multiplicity at energies above GeV, the exclusive measurement of the respective hadronic final states is not practicable. Consequently, the experiments at the high energy colliders ADONE, SPEAR, DORIS, PETRA, PEP, VEPP-4, CESR and BEPC have measured the total inclusive cross section ratio .\nWe give in the following a compilation of the data used in this analysis:\n\n• The measurements are taken from OLYA [38, 39], TOF , CMD , DM1 and DM2 .\n\nThe most precise data from CMD-2 are now available in their final form . They differ from the preliminary ones, released two years ago , mostly in the treatment of radiative corrections. Compared to the preliminary ones, the new results are corrected (see Section 4) for leptonic and hadronic vacuum polarization, and for photon radiation by the pions (final state radiation – FSR), so that the measured final state corresponds to including pion-radiated photons. The various changes resulted into a reduction of the cross section by about 1% below the peak and 5% above. The dominant contribution stemmed from vacuum polarization, while the (included) FSR correction increased the cross section by about 0.8% in the peak region. The overall systematic error of the final data is quoted to be 0.6% and is dominated by the uncertainties in the radiative corrections (0.4%).\n\nWe do not use the data from NA7 as they are known to suffer from a systematic bias in the energy scale . All the experiments agree with each other within their quoted errors, but the high precision claimed by CMD-2 makes this experiment unique and consequently not cross-checked by the others at that level.\n\nThe comparison between the cross section results from CMD-2 and from previous experiments (corrected for vacuum polarization and FSR, according to the procedure discussed in Section 4) is shown in Fig. 1. Note that the errors bars given contain both statistical and systematic errors, added in quadrature (this is the case for all figures in this paper). The agreement is good within the much larger uncertainties (2-10%) quoted by the older experiments.\n\n• The situation of the data on the and resonances has significantly improved recently [46, 47, 48]. The numerical procedure for integrating their cross sections is described in detail in Section 8.2.\n\n• The cross sections for and not originating from the decay of the and resonances are taken from SND and CMD-2 data in the continuum. They include contributions from and .\n\n• The reaction is dominated by the and intermediate resonances discussed above. The continuum data are taken from ND , DM1 , DM2 , SND and CMD .\n\n• The data are available from M3N , OLYA , ND , DM2 [58, 59, 60], CMD-2 and SND . It is fair to say that large discrepancies are observed between the different results, which are probably related to problems in the calculation of the detection efficiency (the cross sections can be seen in Fig. 9 shown in Section 6.1). The efficiencies are small in general (%) and are affected by uncertainties in the decay dynamics that is assumed in the Monte Carlo simulation. One could expect the more recent experiments (CMD-2 and SND) to be more reliable in this context because of specific studies performed in order to identify the major decay processes involved. Accordingly we do not include the ND data in the analysis.\n\n• The reaction is mainly reconstructed in the final state and is thus already accounted for. It was studied by the collaborations ND , DM2 , CMD-2 and SND [62, 63]. We use these cross section measurements to compute the contribution corresponding to the decay mode.\n\n• The final state was studied by the experiments OLYA , ND , CMD , DM1 [66, 67], DM2 [58, 59, 60], CMD-2 and SND . The experiments agree reasonably well within their quoted uncertainties (see Fig. 9 in Section 6.1).\n\n• The data are taken from M3N and CMD . It contains a contribution from the channel with which has to be treated separately because the decay violates isospin. The other five-pion mode is not measured, but can be accounted for using the isospin relation . The relation is used after subtracting the contribution in the rate. Then the contribution with is added to obtain the full rate.\n\n• For the reaction , measured by the groups DM1 , DM2 and CMD-2 , a contribution is calculated for decaying into . The dominant three-pion decay already appears in the five-pion final state.\n\n• Similarly, the contribution for , with , is taken by isospin symmetry to be half of .\n\n• The process was studied by ND , DM2 and CMD-2 . We subtract from its cross section the contributions which are already counted in the and final states.\n\n• The cross sections of the six-pion final states and were measured by DM1 , CMD and DM2 . For the missing channel one can rely on isospin relations in order to estimate its contribution. If only data are used, the isospin bound is weak, leading to a possibly large contribution with an equally large uncertainty. However, some information can be found in the isospin-rotated processes222Throughout this paper, charge conjugate states are implied for decays. and , where the hadronic system has been shown to be dominated by and , once the axial-vector and contributions are discarded. An isospin analysis then reveals the dominance of the final state. As a consequence the channel in annihilation only receives a very small contribution, determined by the cross section. We include a component for .\n\n• The and cross sections above the resonance are taken from OLYA , DM1 , DM2 , CMD and CMD-2 .\n\n• The reactions and were studied by DM1 [79, 80] and DM2 . Using isospin symmetry the cross section of the final state is obtained from the relation .\n\n• The inclusive reaction + was analyzed by DM1 . After subtracting from its cross section the separately measured contributions of the final states , and , it still includes the modes , and . With the assumption that the cross sections for the processes and are equal, one can summarize the total contribution as twice the above corrected + cross section. Implied by the assumption made, it is reasonable to quote as the systematic uncertainty one-half of the cross section for the channel measured by DM1 and DM2 .\n\n• Baryon-pair production is included using the cross sections for from DM1 and DM2 , and for from FENICE .\n\n• At energies larger than 2 GeV the total cross section ratio is measured inclusively. Data are provided by the experiments , MARK I , DELCO , DASP , PLUTO , LENA , Crystal Ball [92, 93], MD-1 , CELLO , JADE , MARK-J , TASSO , CLEO , CUSB , MAC , and BES . Due to their weak experimental precision, the data of are not used in this analysis. The measurements of the MARK I Collaboration are significantly higher than those from more recent and more precise experiments. In addition, the QCD prediction of , which should be reliable in this energy regime, favours lower values, in agreement with the other experiments. Consequently the MARK I results on have been discarded.\n\nAlthough small, the enhancement of the cross section due to interference is corrected for energies above the mass. We use a factorial ansatz according to Ref. [102, 3], yielding a negligible contribution to .\n\nThe data in the charm region are displayed in Fig. 2. Good agreement is found among the experiments.\n\n• The narrow and resonances are treated in Section 8.3.\n\n### 3.2 Data from Hadronic τ Decays\n\nData from decays into two- and four-pion final states , and , are available from ALEPH , CLEO [25, 26] and OPAL . Very recently, preliminary results on the full LEP1 statistics have been presented by ALEPH . They agree with the published results, but correspond to a complete reanalysis with refined systematic studies allowed by the 2.5 times larger data set. The branching fraction for the decay mode is of particular interest since it provides the normalization of the corresponding spectral function. The new value , , turns out to be larger than the previously published one based on the 1991-93 LEP1 statistics, .\nAssuming (for the moment) isospin invariance to hold, the corresponding isovector cross sections are calculated via the CVC relations\n\n σI=1e+e−→π+π− = 4πα2svπ−π0 , (7) σI=1e+e−→π+π−π+π− = 2⋅4πα2svπ−3π0 , (8) σI=1e+e−→π+π−π0π0 = 4πα2s[v2π−π+π0−vπ−3π0] . (9)\n\nThe spectral function for a given vector hadronic state is defined by \n\n vV(s)≡m2τ6\|Vud\|2SEWB(τ−→ντV−)B(τ−→ντe−¯νe)dNVNVds[(1−sm2τ)2(1+2sm2τ)]−1, (10)\n\nwhere is obtained from averaging333 Since the two determinations, and are not consistent, the final error has been enlarged correspondingly. the two independent determinations from nuclear decays and kaon decays (assuming unitarity of the CKM matrix) and accounts for electroweak radiative corrections as discussed in Section 5.1. The spectral functions are obtained from the corresponding invariant mass distributions, after subtracting out the non- background and the feedthrough from other decay channels, and after a final unfolding from detector effects such as energy and angular resolutions, acceptance, calibration and photon identification.\nIt is important to note that decay experiments measure decay rates that include the possibility of photon radiation in the decay final state. Depending on the experiment, the analysis may (ALEPH) or may not (CLEO) keep events with radiative photons in the final state, but all experiments rely on the TAUOLA decay library to compute their efficiencies. In TAUOLA charged particles are given a probability to produce bremsstrahlung using the PHOTOS procedure which is based on the leading logarithm approximation valid at low photon energy. Thus the measured spectral functions correspond to given final states inclusive with respect to radiative photons in the decay.\nIt should be pointed out that the experimental conditions at (ALEPH, OPAL) and (CLEO) energies are very different. On the one hand, at LEP, the events can be selected with high efficiency () and small non- background (), thus ensuring little bias in the efficiency determination. The situation is not as favorable at lower energy: because the dominant hadronic cross section has a smaller particle multiplicity, it is more likely to pollute the sample and strong cuts must be applied, hence resulting in smaller efficiencies. On the other hand, CLEO has an advantage for the reconstruction of the decay final state since particles are more separated in space. The LEP detectors have to cope with collimated decay products and the granularity of the detectors, particularly the calorimeters, plays a crucial role. One can therefore consider ALEPH/OPAL and CLEO data to be approximately uncorrelated as far as experimental procedures are concerned. The fact that their respective spectral functions for the and modes agree, as demonstrated in Fig. 3 for , is therefore a valuable experimental consistency test.\n\n## 4 Radiative Corrections for e+e− Data\n\nRadiative corrections applied to the measured cross sections are an important step in the experimental analyses. They involve the consideration of several physical processes and lead to large corrections. We stress again that the evaluation of the integral in Eq. (5) requires the use of the ’bare’ hadronic cross section, so that the input data must be analyzed with care in this respect.\nSeveral steps are to be considered in the radiative correction procedure:\n\n• Corrections are applied to the luminosity determination, based on large-angle Bhabha scattering and muon-pair production in the low-energy experiments, and small-angle Bhabha scattering at high energies. These processes are usually corrected for external radiation, vertex corrections and vacuum polarization from lepton loops.\n\n• The hadronic cross sections given by the experiments are always corrected for initial state radiation and the effect of loops at the electron vertex.\n\n• The vacuum polarization correction in the photon propagator is a more delicate point. The cross sections need to be fully corrected for our use, i.e.\n\n σbare=σdressed(α(0)α(s))2 , (11)\n\nwhere is the measured cross section already corrected for initial state radiation, and is obtained from resummation of the lowest-order evaluation\n\nWhereas can be analytically calculated (here given to leading order)\n\n Δαlep(s)=α(0)3π∑l(logsm2l−53) , (13)\n\nis related by analyticity and unitarity to a dispersion integral, akin to (5),\n\nwhich must also be evaluated using input data. Since the hadronic correction involves the knowledge of at all energies, including those where the measurements are made, the procedure has to be iterative, and requires experimental as well as theoretical information over a large energy range.\n\nThis may explain why the vacuum polarization correction is in general not applied by the experiments to their published cross sections. Here the main difficulty is even to find out whether the correction (and which one? leptonic at least? hadronic?) has actually been used, as unfortunately this is almost never clearly stated in the publications. The new data from CMD-2 are explicitly corrected for both leptonic and hadronic vacuum polarization effects, whereas the preliminary data from the same experiment were not.\n\nIn fact, what really matters is the correction to the ratio of the hadronic cross section to the cross section for the process used for the luminosity determination. In the simplest case (for example, DM2 for the channel) of the normalization to the process, the vacuum polarization effects cancel. However, generally the normalization is done with respect to large angle Bhabha scattering events or to both Bhabha and . In the latter case, Bhabha events dominate due to the -channel contribution. In the mode, all experiments before the latest CMD-2 results corrected their measured processes (, and ) for radiative effects using calculations which took only leptonic vacuum polarization into account [108, 109]. For the other channels, it is harder to find out as information about the luminosity determination and the detailed procedure for radiative corrections is in general not given in the publications.\n\nFor all experimental results, but the newest from CMD-2 and DM2, we apply a correction for the missing hadronic vacuum polarization given by \n\nwhere the correction in the denominator applies to the Bhabha cross section evaluated at a mean value of the squared momentum transfer , which depends on the angular acceptance in each experiment. A uncertainty is assigned to . For the and resonance cross sections, we were informed that the recent CMD-2 and SND results were not corrected for leptonic vacuum polarization, so in their case we applied a full correction taking into account both leptonic and hadronic components.\n\n• In Eq. (5) one must incorporate in the contributions of all hadronic states produced at the energy . In particular, radiative effects in the hadronic final state must be considered, i.e., final states such as have to be included.\n\nInvestigating the existing data in this respect is also a difficult task. In the data from CMD-2 most additional photons are experimentally rejected to reduce backgrounds from other channels and the fraction kept is subtracted using the Monte Carlo simulation which includes a model for FSR. Then the full FSR contribution is added back as a correction, using an analytical expression computed in scalar QED (point-like pions) . As this effect was not included in earlier analyses, we applied the same correction to older data.\n\nIn principle one must worry about FSR effects in other channels as well. For the inclusive measurements it is included by definition. When is evaluated from QCD at high energy, the prediction must be corrected for FSR from the quarks, but this is a negligible effect for . The situation for the exclusive channels is less clear because it depends on the experimental cuts and whether or not FSR is included in the simulation. Taking as an educated guess the effect in the channel, we correspondingly correct the contributions to from all remaining exclusive channels by the factor where is the charged particle multiplicity in the final state.\n\nIn summary, we correct each experimental result, but those from CMD-2 , by the factor . As an illustration of the orders of magnitude involved, the different corrections in the contribution amount to for the leptonic vacuum polarization, for the hadronic vacuum polarization, and for the FSR correction. The correction to the / ratio from the missing hadronic vacuum polarization is small, typically . Both the vacuum polarization and FSR corrections apply only to experiments other than CMD-2, therefore the overall correction to the channel is considerably reduced.\nThe uncertainties on the missing vacuum polarization () and the FSR corrections () are conservatively considered to be fully correlated between all channels to which the correction applies. The total error from these missing radiative corrections, taken as the quadratic sum of the two contributions, is given separately for the final results.\n\n## 5 Isospin Breaking in e+e− and τ Spectral Functions\n\n### 5.1 Sources of Isospin Symmetry Breaking\n\nThe relationships (7), (8) and (9) between and spectral functions only hold in the limit of exact isospin invariance. This is the Conserved Vector Current (CVC) property of weak decays. It follows from the factorization of strong interaction physics as produced through the and propagators out of the QCD vacuum. However, we know that we must expect symmetry breaking at some level from electromagnetic effects and even in QCD because of the up and down quark mass splitting. Since the normalization of the spectral functions is experimentally known at the 0.5% level, it is clear that isospin-breaking effects must be carefully examined if one wants this precision to be maintained in the vacuum polarization integrals. Various identified sources of isospin breaking are considered in this section and discussed in turn.\nBecause of the dominance of the contribution in the energy range of interest for data, we discuss mainly this channel, following our earlier analysis . The corrections on from isospin breaking are given in Table 1. A more complete evaluation is given in the next section. Finally, the 4-pion modes will be briefly discussed.\n\n• Electroweak radiative corrections must be taken into account. Their dominant contribution comes from the short distance correction to the effective four-fermion coupling enhancing the amplitude by the factor , where is the average charge of the final state partons . While this correction vanishes for leptonic decays, it contributes for quarks. All higher-order logarithms can be resummed using the renormalization group [112, 113], and the short distance correction can be absorbed into an overall multiplicative electroweak correction ,\n\n (16)\n\nwhich is equal to 1.0194 when using the current fermion and boson masses and for consistency the quark-level expressions for as given in Ref. . The difference between the resummed value and the lowest-order estimate (1.0188) can be taken as a conservative estimate of the uncertainty. QCD corrections to have been calculated [112, 116] and found to be small, reducing its value to 1.0189.\n\nSubleading non-logarithmic short distance corrections have been calculated to order at the quark level , , and for the leptonic width , . Summing up all the short distance corrections, one obtains the value for that must be used for the inclusive hadronic width\n\nOther uncertainties on the quark mass, the running of , and QCD corrections are at the level.\n\nLong distance corrections are expected to be final-state dependent in general. They have been computed for the decay leading to a total radiative correction of 2.03 , which is dominated by the leading logarithm from the short distance contribution. Although very encouraging, this result may not apply to all hadronic decays, in particular for the important mode. Therefore an uncertainty of 0.0040 was previously assigned to (see the following Section 5.2 for more) to cover the final-state dependence of the correction with respect to the calculation at the quark level.\n\n• A contribution [28, 4] for isospin breaking occurs because of the mass difference between charged and neutral pions, which is essentially of electromagnetic origin. The spectral function has a kinematic factor which is different in () and decay (). We write\n\n v0(s) = β30(s)12\|F0π(s)\|2 , (18) v−(s) = β3−(s)12\|F−π(s)\|2 , (19)\n\nwith obvious notations, being the electromagnetic and weak pion form factors, respectively, and defined by\n\n β0,−=β(s,mπ−,mπ0,−) , (20)\n\nwhere\n\n β(s,m1,m2)=[(1−(m1+m2)2s)(1−(m1−m2)2s)]1/2 . (21)\n\nHence, a correction equal to is applied to the spectral function.\n\n• Other corrections occur in the form factor itself. It turns out that it is affected by the pion mass difference because the same factor enters in the width. This effect partially compensates the corrections (18), (19) of the cross section, as seen in Table 1.\n\n• Similarly a possible mass difference between the charged and neutral meson affects the value of the corresponding width and shifts the resonance lineshape. Theoretical estimates and experimental determinations [5, 120] show that the mass difference is compatible with zero within about 1 MeV.\n\n• interference occurs in the mode and thus represents an obvious source of isospin symmetry breaking. Its contribution can be readily introduced into the spectral function using the parameters determined in the CMD-2 fit . The integral over the interference almost vanishes by itself since it changes sign at the mass, however the -dependent integration kernel produces a net effect (Table 1).\n\n• Electromagnetic decays explicitly break SU(2) symmetry. This is the case for the decays through an intermediate state because of identical ’s, , and . The decay deserves particular attention: calculations have been done with an effective model for both charged and neutral ’s. The different contributions are listed in Table 1.\n\n• A breakdown of CVC is due to quark mass effects: different from generates for a charge-changing hadronic current between and quarks. Expected deviations from CVC due to so-called second class currents such as, e.g., the decay where the corresponding final state (C=+1) is forbidden, lead to an estimated branching fraction of the order of , while the experimental upper limit amounts to .\n\n### 5.2 A More Elaborate Treatment of Isospin Breaking in the 2π Channel\n\nThe above analysis of isospin breaking leaves out the possibility of sizeable contributions from virtual loops. This problem was studied recently within a model based on Chiral Perturbation Theory. In this way the correct low-energy hadronic structure is implemented and a consistent framework can be set up to calculate electroweak and strong processes, such as the radiative corrections in the decay. One might worry that the mass is too large for such a low-energy approach. However a reasonable matching with the resonance region and even beyond is claimed to be achieved, providing a very useful tool to study radiative decays.\nA new analysis has been issued which is more suited to our purpose, in the sense that it applies to the inclusive radiative rate, , as measured by the experiments. A consistent calculation of radiative corrections is presented including real photon emission and the effect of virtual loops. All the contributions listed in the previous section are included and the isospin-breaking contributions in the pion form factor are now more complete. Following Ref. , the relation between the Born level spectral function and the spectral function (19) reads\n\n vπ+π−(s)=1GEM(s)β30β3−∣∣∣F0π(s)F−π(s)∣∣∣2vπ−π0(γ)(s) , (22)\n\nwhere is the long-distance radiative correction involving both real photon emission and virtual loops (the infrared divergence cancels in the sum). Note that the short-distance correction, discussed above, is already applied in the definition of (cf. Eq. (10)), but its value differs from Eq. (17) because subleading quark-level and hadron-level contributions should not be added, as double counting would occur. The correct expression for the mode therefore reads\n\nthe subleading hadronic corrections being now incorporated in the mass-dependent factor. The form factor correction is dominated by the effect of the pion mass difference in the width, but it also includes a small contribution at the level from the ’chiral’ form used for the lineshape. In practice, however, the correction is independent of the chosen parametrization of the form factor. The different contributions to the isospin-breaking corrections are shown in the second column of Table 1. The values slightly differ from those given in Ref. because the authors use a model for the pion form factor rather than integrating experimental data. The largest difference however stems from our re-evaluation of the short-distance electroweak correction, , including the subleading leptonic contribution. The sum amounts to to be compared with given in Ref. .\nThe dominant uncertainty in this method stems from the - mass difference. Indeed, in the chiral model used in Ref. the only parameter entering the pion form factor is the mass, since the width is given by . In the method previously used and recalled in Section 5.1, the width at the pole was taken as an independent parameter with , so that the effect of the mass difference approximately cancels after integration. This explains the large difference in the uncertainties quoted for the two evaluations in Table 1.\nSince the integral (5) requires as input the spectral function including FSR photon emission, a final correction is necessary. It is identical to that applied in the CMD-2 analysis [22, 111] (cf. Section 4). All the corrections are drawn versus in Fig. 4. The overall correction reduces the rate below the peak, but, somewhat unexpectedly, has the opposite effect above. This behavior is driven by the long-distance radiative corrections contained in .\n\nThe total correction to the result in this method, not including the FSR contribution, amounts to , where the main contribution to the error is due to the experimental limits on the mass difference. After including the FSR contribution, it becomes , a value consistent with the result in the first column of Table 1 which does not include the virtual corrections and uses a less sophisticated treatment of radiative decays. In the following we apply the correction functions from the more complete analysis (method (II) in Table 1) and keep the corresponding uncertainty separate from the purely experimental errors.\n\n### 5.3 Isospin Breaking in 4π Channels\n\nThere exists no comparable study of isospin breaking in the channels. Only kinematic corrections resulting from the pion mass difference have been considered so far , which we have applied in this analysis. It creates shifts of () and () for and , respectively. However, since the four-pion contribution to is relatively less important than the two-pion part (by a little more than an order of magnitude in the integration range up to 1.8 GeV) and the experimental uncertainties are much larger, we feel this is a justified procedure at the present level of accuracy of the data. Moreover, the entire correction has been attributed as systematic error which is kept separate from the experimental errors on from these channels.\n\nIt should also be pointed out that the systematic uncertainties from isospin breaking are essentially uncorrelated between the and modes: as Table 1 shows, the dominant sources of uncertainties are the - mass difference for and the threshold factors in where large errors have been given to cover uncertainties in the decay dynamics and the missing pieces.\n\n## 6 Comparison of e+e− and τ Spectral Functions\n\nThe and the isospin-breaking corrected spectral functions can be directly compared for the dominant and final states. For the channel, the -dominated form factor falls off very rapidly at high energy so that the comparison can be performed in practice over the full energy range of interest. The situation is different for the channels where the decay kinematics limits the exercise to energies less than 1.6 GeV, with only limited statistics beyond.\n\n### 6.1 Direct Comparison\n\nFig. 5 shows the comparison for the spectral functions. Visually, the agreement seems satisfactory, however the large dynamical range involved does not permit an accurate test. To do so, the data are plotted as a point-by-point ratio to the spectral function in Fig. 7, and enlarged in Fig. 7, to better emphasize the region of the peak444 The central bands in Figs. 7 and 7 give the quadratic sum of the statistical and systematic errors of the combined spectral functions. Local bumps in these bands stem from increased errors when combining different experiments having local inconsistencies. We use the procedure described in Section 7.1. . The data are significantly lower by 2-3% below the peak, the discrepancy increasing to about 10% in the 0.9-1.0 GeV region.\nThe comparison for the cross sections is given in Fig. 9 for the channel and in Fig. 9 for . As noted before, the latter suffers from large differences between the results from the different experiments. The data, combining two measured spectral functions according to Eq. (9) and corrected for isospin breaking as discussed in Section 5, lie somewhat in between with large uncertainties above 1.4 GeV because of the lack of statistics and a large feedthrough background in the mode. In spite of these difficulties the spectral function is in agreement with data as can be seen in Fig. 9. It is clear that intrinsic discrepancies exist among the experiments and that a quantitative test of CVC in the channel is premature.\n\n### 6.2 Branching Ratios in τ Decays and CVC\n\nA convenient way to assess the compatibility between and spectral functions proceeds with the evaluation of decay fractions using the relevant spectral functions as input. All the isospin-breaking corrections detailed in Section 5.2 are included. The advantage of this procedure is to allow a quantitative comparison using a single number. The weighting of the spectral function is however different from the vacuum polarization kernels. Using the branching fraction , obtained assuming leptonic universality in the charged weak current , the results for the main channels are given in Table 2. The errors quoted for the CVC values are split into uncertainties from (i) the experimental input (the annihilation cross sections) and the numerical integration procedure, (ii) the missing radiative corrections applied to the relevant data, and (iii) the isospin-breaking corrections when relating and spectral functions. The values for the branching ratios involve measurements [24, 124, 125] given without charged hadron identification, i.e. for the , and final states. The corresponding channels with charged kaons have been measured [126, 127] and their contributions can be subtracted out in order to obtain the pure pionic modes." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91505957,"math_prob":0.9413088,"size":37425,"snap":"2020-45-2020-50","text_gpt3_token_len":8086,"char_repetition_ratio":0.17054595,"word_repetition_ratio":0.009384951,"special_character_ratio":0.20785572,"punctuation_ratio":0.095569335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96155035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T12:01:25Z\",\"WARC-Record-ID\":\"<urn:uuid:1f739d10-3d38-4643-929b-84a018d7fa0a>\",\"Content-Length\":\"1049567\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36e69f6d-4088-4d05-9d0b-74207e60aeb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1aea95f3-1ae2-4b80-88e1-6af7dccaee85>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-ph/0208177/\",\"WARC-Payload-Digest\":\"sha1:JI25UNUP5GPDPAQI6EY7YTLBO4V536UC\",\"WARC-Block-Digest\":\"sha1:NHDJ7YJJ7FIMWK2ZBJAXG4SJX4JQVFXE\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107882581.13_warc_CC-MAIN-20201024110118-20201024140118-00110.warc.gz\"}"}
https://blogs.ams.org/visualinsight/2016/04/15/barth-sextic/	[ "# Barth Sextic\n\nA sextic surface is one defined by a polynomial equation of degree 6. The Barth sextic, drawn above by Craig Kaplan, is the sextic surface with the maximum possible number of ordinary double points: that is, points where it looks like the origin of the cone in 3-dimensional space defined by\n\n$$x^2 + y^2 = z^2 .$$\n\nIn 1946, Francesco Severi claimed that 52 is the maximum number of ordinary double points possible for a sextic surface. This was refuted in 1994 when Wolf Barth discovered what is now called the Barth sextic, which has 65 double points:\n\n• Wolf Barth, Two projective surfaces with many nodes, admitting the symmetries of the icosahedron, Journal of Algebraic Geometry 5 (1994), 173–186.\n\nIn 1997, David B. Jaffe and Daniel Ruberman proved that 65 is the maximum number of double points for a sextic surface that has no singularities other than nodes, which are a generalization of double points:\n\n• David B. Jaffe, and Daniel Ruberman, A sextic surface cannot have 66 nodes, Journal of Algebraic Geometry 6 (1997), 151–168.\n\nNot all sextics with 65 double points have icosahedral symmetry, but as Barth noted in the title of his paper, his does! 20 of the double points lie at the vertices of a regular dodcahedron, and 30 lie at the midpoints of the edges of another, concentric, regular dodecahedron. To see this more clearly, look at the beautiful animations here:\n\n• Hermann Serras, Barth’s sextic surface and two associated regular dodecahedra.\n\nThat accounts for 50 of the 65 double points. Where are the other 15? They are hiding at ‘points at infinity’. To understand this, and to bring them into view, we need to dig a bit deeper into the geometry.\n\nA sextic surface is defined by a homogeneous polynomial equation of degree 6. This equation determines a subset $S \\subset \\mathbb{C}^4$ with complex dimension 2. Note that if $(x_1, \\dots, x_4) \\in \\mathbb{C}^4$ is a solution, so is any multiple $(cx_0, \\dots , cx_4)$. We may thus projectivize $S$, treating any solution as ‘the same’ as any multiple of that solution. The result is an algebraic variety $X$ in the complex projective space $\\mathbb{C}\\mathrm{P}^3$. This is a smooth manifold of complex dimension 2, so it is called a smooth surface. To obtain an ordinary real 2-dimensional surface we may take its intersection with a copy of $\\mathbb{R}\\mathrm{P}^3$ in $\\mathbb{C}\\mathrm{P}^3$.\n\nSitting inside $\\mathbb{R}\\mathrm{P}^3$ we have ordinary 3-dimensional space, $\\mathbb{R}^3$. But we also have ‘points at infinity’. If you march off in either of two opposite directions in $\\mathbb{R}^3$, you will approach one of these points at infinity. The points at infinity form a projective plane, that is, a copy of $\\mathbb{R}\\mathrm{P}^2$.\n\nThe Barth sextic is usually described by the sextic equation\n\n$$4(\\Phi^2 x^2-y^2)(\\Phi^2 y^2 – z^2)(\\Phi^2 z^2-x^2) – (1+2\\Phi)(x^2+y^2+z^2-w^2)^2 w^2 = 0$$\n\nwhere\n\n$$\\Phi = \\frac{\\sqrt{5} + 1}{2}$$\n\nis the ‘large’ golden ratio.\n\nWith a standard choice of $\\mathbb{R}^3$ inside $\\mathbb{R}\\mathrm{P}^3$, 15 of the double points of this surface lie at infinity. You can reach them by taking either of the aforementioned dodecahedra, drawing lines between the midpoints of its 15 pairs of opposite edges, and following these lines to infinity!\n\nWe can bring these double points into view by rotating $\\mathbb{R}\\mathrm{P}^3$ slightly. Then we get a view like this, drawn by Abdelaziz Nait Merzouk:\n\nIn this view, the Barth sextic has also been sliced by a plane, to make its inside visible.\n\nMerzouk has also brought the double points at infinity into view using a transformation that compresses all of $\\mathbb{R}^3$ down to the unit ball. Then these double points lie on the unit sphere:\n\nHowever, we must count antipodal points on the sphere as the same point in $\\mathbb{R}\\mathrm{P}^2$. So, the 30 double points you see on the unit sphere above give 15 points at infinity.\n\nFor other views, see:\n\n• Abdelaziz Nait Merzouk, Barth sextic.\n\nJaffe and Ruberman actually proved that a sextic surface with only nodes as singularities cannot have more than 65 nodes. Their proof uses coding theory in an intriguing way. They show that an ‘even’ set of nodes (see their paper for the definition) gives rise to a binary linear code: that is, a linear subspace of $\\mathbb{F}_2^n$, where $n$ is the number of nodes. Their proof then uses a mixture of algebraic geometry and coding theory.\n\nThis is another picture of the Barth sextic, obtained here:\n\n• Oliver Labs, Gallery of surfaces, Imaginary: Open Mathematics.\n\nVisual Insight is a place to share striking images that help explain advanced topics in mathematics. I’m always looking for truly beautiful images, so if you know about one, please drop a comment here and let me know!\n\n## 4 thoughts on “Barth Sextic”\n\n1.", null, "Michael Nelson says:\n\nI think I can see how the dodecahedron is involved. First, dehomogenize by setting w = 1. Then if you assume that x,y,z lie on the the unit sphere in R^3, the second term in the equation vanishes, and what you are left with is six lines. I think you’ll get a picture that looks like this:\n\nhttp://myweb.tiscali.co.uk/oaktree/butterfly/backpages/pics/great-circles/icosa/XT_Icosa_F1_Sphere_PPT.jpg\n\nI got this picture from this website:\n\n2.", null, "Andreas Daniel says:\n\nWith the free software SURFER you can visualize/manipulate the Barth Sextic in real time (zoom in/out, rotate it, change parameters, change the equation, etc.).\n\nThe program is available at:\nhttp://imaginary.org/program/surfer\n\nYou will find the Barth Sextic in the “World record surfaces” gallery. There you will find other surfaces with maximum possible numbers of double points (also ones, which are not yet proven to be the “maximum” ones).\n\n•", null, "John Baez says:\n\nCool!\n\nI don’t see the “world record surfaces” gallery.\n\n•", null, "Andreas Daniel says:\n\nDear John,\n\njust click on the “Start” button and you will see three galleries (Tutorial, Fantasy and World Record Surfaces). There you will find the double point world record quadrics, cubics, quartics, sextics, septics, octics, etc.\n\nEnjoy! And thanks for the great article!" ]	[ null, "https://secure.gravatar.com/avatar/80214fc45aa81137288a6415e741c3bc", null, "https://secure.gravatar.com/avatar/4a3149a3377a5721bf7768229079df61", null, "https://secure.gravatar.com/avatar/34784534843022b3541c8ddd693718cb", null, "https://secure.gravatar.com/avatar/4a3149a3377a5721bf7768229079df61", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88671273,"math_prob":0.98825675,"size":6223,"snap":"2022-40-2023-06","text_gpt3_token_len":1659,"char_repetition_ratio":0.1312108,"word_repetition_ratio":0.004052685,"special_character_ratio":0.25389683,"punctuation_ratio":0.12228479,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952404,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T09:13:33Z\",\"WARC-Record-ID\":\"<urn:uuid:21b62e96-a4c5-41ed-ab02-0b8fe0901677>\",\"Content-Length\":\"67723\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2297433f-adea-4f92-b89c-21bf153857d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:399005d1-dcf0-4621-9581-387a70eadd13>\",\"WARC-IP-Address\":\"130.44.204.215\",\"WARC-Target-URI\":\"https://blogs.ams.org/visualinsight/2016/04/15/barth-sextic/\",\"WARC-Payload-Digest\":\"sha1:2PYBFRZVWUJY3WOTWZNRAKPYSGLYZYUG\",\"WARC-Block-Digest\":\"sha1:67YNFZ2Q4EPMT5RKWLQZK7RHSVY5ZMSW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499541.63_warc_CC-MAIN-20230128090359-20230128120359-00475.warc.gz\"}"}
https://blog.lcrun.com/step-by-step-write-algorithm-of-cycle-and-recursion/	[ "May 31, 2016\n\n# 一步一步写算法之循环和递归\n\n## 求和递归函数\n\n``````int calculate(int m)\n{\nint count = 0;\nif(m <0)\nreturn -1;\n\nfor(int index = 0; index <= m; index++)\ncount += index;\n\nreturn count;\n}\n``````\n\n``````int calculate(int m)\n{\nif(m == 0)\nreturn 0;\nelse\nreturn calculate(m -1) + m;\n}\n``````\n\n1. 第一段代码从0，开始计算，从0到m逐步计算；第二段代码是从10开始计算，逐步到0之后这回，这样同样可以达到计算的效果\n2. 第一段代码不需要重复的压栈操作，第二段需要重复的函数操作，当然这也是递归的本质\n3. 第一段代码比较长，第二段代码较短\n\n## 查找递归函数\n\n``````int find(int array[], int length, int value)\n{\nint index = 0;\nif(NULL == array \|\| 0 == length)\nreturn -1;\n\nfor(; index < length; index++)\n{\nif(value == array[index])\nreturn index;\n}\nreturn -1;\n}\n``````\n\n``````int _find(int index, int array[], int length, int value)\n{\nif(index == length)\nreturn -1;\n\nif(value == array[index])\nreturn index;\n\nreturn _find(index + 1, array, length, value);\n}\n\nint find(int array[], int length, int value)\n{\nif(NULL == array \|\| length == 0)\nreturn -1;\n\nreturn _find(0, array, length, value);\n}\n``````\n\n## 指针变量遍历\n\n``````typedef struct _NODE\n{\nint data;\nstruct _NODE* next;\n}NODE;\n``````\n\n``````void print(const NODE* pNode)\n{\nif(NULL == pNode)\nreturn;\n\nwhile(pNode){\nprintf(\"%dn\", pNode->data);\npNode = pNode->next;\n}\n}\n``````\n\n``````void print(const NODE* pNode)\n{\nif(NULL == pNode)\nreturn;\nelse\nprintf(\"%dn\", pNode->data);\n\nprint(pNode->next);\n}\n``````" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.7054432,"math_prob":0.99964786,"size":1654,"snap":"2021-43-2021-49","text_gpt3_token_len":927,"char_repetition_ratio":0.15757576,"word_repetition_ratio":0.11518325,"special_character_ratio":0.286578,"punctuation_ratio":0.16176471,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980677,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T10:16:19Z\",\"WARC-Record-ID\":\"<urn:uuid:dcdab371-e3f5-48c0-b2f4-3cd54ebeceae>\",\"Content-Length\":\"21295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cd57b5d9-d2e4-423d-8f92-fae71aa18892>\",\"WARC-Concurrent-To\":\"<urn:uuid:8caac7e5-aff9-46f8-8948-b0de39dc656a>\",\"WARC-IP-Address\":\"42.193.175.138\",\"WARC-Target-URI\":\"https://blog.lcrun.com/step-by-step-write-algorithm-of-cycle-and-recursion/\",\"WARC-Payload-Digest\":\"sha1:QITSMBPPRC34MQ35ZX3TFD5AF272RZG2\",\"WARC-Block-Digest\":\"sha1:GELWAYDXGH32GLYBZYUR2IN65MUEJR5M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00396.warc.gz\"}"}
https://www.opencascade.com/doc/occt-6.9.1/refman/html/class_b_rep_builder_a_p_i___find_plane.html	[ "BRepBuilderAPI_FindPlane Class Reference\n\nDescribes functions to find the plane in which the edges of a given shape are located. A FindPlane object provides a framework for: More...\n\n`#include <BRepBuilderAPI_FindPlane.hxx>`\n\n## Public Member Functions\n\nBRepBuilderAPI_FindPlane ()\nInitializes an empty algorithm. The function Init is then used to define the shape. More...\n\nBRepBuilderAPI_FindPlane (const TopoDS_Shape &S, const Standard_Real Tol=-1)\nConstructs the plane containing the edges of the shape S. A plane is built only if all the edges are within a distance of less than or equal to tolerance from a planar surface. This tolerance value is equal to the larger of the following two values: More...\n\nvoid Init (const TopoDS_Shape &S, const Standard_Real Tol=-1)\nConstructs the plane containing the edges of the shape S. A plane is built only if all the edges are within a distance of less than or equal to tolerance from a planar surface. This tolerance value is equal to the larger of the following two values: More...\n\nStandard_Boolean Found () const\nReturns true if a plane containing the edges of the shape is found and built. Use the function Plane to consult the result. More...\n\nHandle< Geom_PlanePlane () const\nReturns the plane containing the edges of the shape. Warning Use the function Found to verify that the plane is built. If a plane is not found, Plane returns a null handle. More...\n\n## Detailed Description\n\nDescribes functions to find the plane in which the edges of a given shape are located. A FindPlane object provides a framework for:\n\n• extracting the edges of a given shape,\n• implementing the construction algorithm, and\n• consulting the result.\n\n## Constructor & Destructor Documentation\n\n BRepBuilderAPI_FindPlane::BRepBuilderAPI_FindPlane ( )\n\nInitializes an empty algorithm. The function Init is then used to define the shape.\n\n BRepBuilderAPI_FindPlane::BRepBuilderAPI_FindPlane ( const TopoDS_Shape & S, const Standard_Real Tol = `-1` )\n\nConstructs the plane containing the edges of the shape S. A plane is built only if all the edges are within a distance of less than or equal to tolerance from a planar surface. This tolerance value is equal to the larger of the following two values:\n\n• Tol, where the default value is negative, or\n• the largest of the tolerance values assigned to the individual edges of S. Use the function Found to verify that a plane is built. The resulting plane is then retrieved using the function Plane.\n\n## Member Function Documentation\n\n Standard_Boolean BRepBuilderAPI_FindPlane::Found ( ) const\n\nReturns true if a plane containing the edges of the shape is found and built. Use the function Plane to consult the result.\n\n void BRepBuilderAPI_FindPlane::Init ( const TopoDS_Shape & S, const Standard_Real Tol = `-1` )\n\nConstructs the plane containing the edges of the shape S. A plane is built only if all the edges are within a distance of less than or equal to tolerance from a planar surface. This tolerance value is equal to the larger of the following two values:\n\n• Tol, where the default value is negative, or\n• the largest of the tolerance values assigned to the individual edges of S. Use the function Found to verify that a plane is built. The resulting plane is then retrieved using the function Plane.\n Handle< Geom_Plane > BRepBuilderAPI_FindPlane::Plane ( ) const\n\nReturns the plane containing the edges of the shape. Warning Use the function Found to verify that the plane is built. If a plane is not found, Plane returns a null handle.\n\nThe documentation for this class was generated from the following file:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80318034,"math_prob":0.8976546,"size":2822,"snap":"2019-51-2020-05","text_gpt3_token_len":667,"char_repetition_ratio":0.15294535,"word_repetition_ratio":0.80794704,"special_character_ratio":0.22324592,"punctuation_ratio":0.13085938,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9818908,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T06:48:28Z\",\"WARC-Record-ID\":\"<urn:uuid:2e0a81be-ebbb-4739-a716-058fc1dcfb56>\",\"Content-Length\":\"14953\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2f0b785-cf65-4cbc-8ae6-550e133c5fcc>\",\"WARC-Concurrent-To\":\"<urn:uuid:665203c3-533a-44e9-a50b-a3c740324d8f>\",\"WARC-IP-Address\":\"188.165.114.136\",\"WARC-Target-URI\":\"https://www.opencascade.com/doc/occt-6.9.1/refman/html/class_b_rep_builder_a_p_i___find_plane.html\",\"WARC-Payload-Digest\":\"sha1:CSY4OOFOVP2XNXFX4DQUPEOYLQIVRXIL\",\"WARC-Block-Digest\":\"sha1:2EBLXN3MYRFTSXGYDDBR5KWYQ3FHOMD3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592261.1_warc_CC-MAIN-20200118052321-20200118080321-00535.warc.gz\"}"}
https://www.convertunits.com/from/half+US+gallon/to/litre	[ "## ››Convert half US gallon to liter\n\n half US gallon litre\n\nHow many half US gallon in 1 litre? The answer is 0.52834410527459.\nWe assume you are converting between half US gallon and liter.\nYou can view more details on each measurement unit:\nhalf US gallon or litre\nThe SI derived unit for volume is the cubic meter.\n1 cubic meter is equal to 528.34410527459 half US gallon, or 1000 litre.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between half US gallons and liters.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of half US gallon to litre\n\n1 half US gallon to litre = 1.89271 litre\n\n5 half US gallon to litre = 9.46353 litre\n\n10 half US gallon to litre = 18.92706 litre\n\n15 half US gallon to litre = 28.39059 litre\n\n20 half US gallon to litre = 37.85412 litre\n\n25 half US gallon to litre = 47.31765 litre\n\n30 half US gallon to litre = 56.78118 litre\n\n40 half US gallon to litre = 75.70824 litre\n\n50 half US gallon to litre = 94.63529 litre\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from litre to half US gallon, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Litre\n\nThe litre (spelled liter in American English and German) is a metric unit of volume. The litre is not an SI unit, but (along with units such as hours and days) is listed as one of the \"units outside the SI that are accepted for use with the SI.\" The SI unit of volume is the cubic metre (m³).\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8239935,"math_prob":0.8045273,"size":2062,"snap":"2021-21-2021-25","text_gpt3_token_len":524,"char_repetition_ratio":0.27065113,"word_repetition_ratio":0.057591625,"special_character_ratio":0.2803104,"punctuation_ratio":0.1199095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9693232,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-10T06:54:30Z\",\"WARC-Record-ID\":\"<urn:uuid:3dcde571-1013-47a0-8bd6-bad8e6640912>\",\"Content-Length\":\"50204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff22071d-bac1-4530-85f5-6ed4bc832c39>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa420f4c-7080-462e-877e-1b963adaabdc>\",\"WARC-IP-Address\":\"54.80.109.124\",\"WARC-Target-URI\":\"https://www.convertunits.com/from/half+US+gallon/to/litre\",\"WARC-Payload-Digest\":\"sha1:ANUJVCSMFRY2E6HTS6KTBRYRLUFCZHHK\",\"WARC-Block-Digest\":\"sha1:KXPN3VPVOELFY2YTPJMWK2FUZQ5XYLBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989115.2_warc_CC-MAIN-20210510064318-20210510094318-00395.warc.gz\"}"}
https://www.rdocumentation.org/packages/MASS/versions/7.3-22/topics/qda	[ "# qda\n\n0th\n\nPercentile\n\nKeywords\nmultivariate\n##### Usage\nqda(x, ...)## S3 method for class 'formula':\nqda(formula, data, \\dots, subset, na.action)## S3 method for class 'default':\nqda(x, grouping, prior = proportions,\nmethod, CV = FALSE, nu, \\dots)## S3 method for class 'data.frame':\nqda(x, \\dots)## S3 method for class 'matrix':\nqda(x, grouping, \\dots, subset, na.action)\n##### Arguments\nformula\nA formula of the form groups ~ x1 + x2 + ... That is, the response is the grouping factor and the right hand side specifies the (non-factor) discriminators.\ndata\nData frame from which variables specified in formula are preferentially to be taken.\nx\n(required if no formula is given as the principal argument.) a matrix or data frame or Matrix containing the explanatory variables.\ngrouping\n(required if no formula principal argument is given.) a factor specifying the class for each observation.\nprior\nthe prior probabilities of class membership. If unspecified, the class proportions for the training set are used. If specified, the probabilities should be specified in the order of the factor levels.\nsubset\nAn index vector specifying the cases to be used in the training sample. (NOTE: If given, this argument must be named.)\nna.action\nA function to specify the action to be taken if NAs are found. The default action is for the procedure to fail. An alternative is na.omit, which leads to rejection of cases with missing values on any required variable. (NOTE: If given, this\nmethod\n\"moment\" for standard estimators of the mean and variance, \"mle\" for MLEs, \"mve\" to use cov.mve, or \"t\" for robust estimates based on a t distribution.\nCV\nIf true, returns results (classes and posterior probabilities) for leave-out-out cross-validation. Note that if the prior is estimated, the proportions in the whole dataset are used.\nnu\ndegrees of freedom for method = \"t\".\n...\narguments passed to or from other methods.\n##### Details\n\nUses a QR decomposition which will give an error message if the within-group variance is singular for any group.\n\n##### Value\n\n• an object of class \"qda\" containing the following components:\n• priorthe prior probabilities used.\n• meansthe group means.\n• scalingfor each group i, scaling[,,i] is an array which transforms observations so that within-groups covariance matrix is spherical.\n• ldeta vector of half log determinants of the dispersion matrix.\n• levthe levels of the grouping factor.\n• terms(if formula is a formula) an object of mode expression and class term summarizing the formula.\n• callthe (matched) function call.\n• unless CV=TRUE, when the return value is a list with components:\n• classThe MAP classification (a factor)\n• posteriorposterior probabilities for the classes\n\n##### References\n\nVenables, W. N. and Ripley, B. D. (2002) Modern Applied Statistics with S. Fourth edition. Springer.\n\nRipley, B. D. (1996) Pattern Recognition and Neural Networks. Cambridge University Press.\n\npredict.qda, lda\n\n##### Aliases\n• qda\n• qda.data.frame\n• qda.default\n• qda.formula\n• qda.matrix\n• model.frame.qda\n• print.qda\n##### Examples\ntr <- sample(1:50, 25)\ntrain <- rbind(iris3[tr,,1], iris3[tr,,2], iris3[tr,,3])\ntest <- rbind(iris3[-tr,,1], iris3[-tr,,2], iris3[-tr,,3])\ncl <- factor(c(rep(\"s\",25), rep(\"c\",25), rep(\"v\",25)))\nz <- qda(train, cl)\npredict(z,test)\\$class\nDocumentation reproduced from package MASS, version 7.3-22, License: GPL-2 \| GPL-3\n\n### Community examples\n\nLooks like there are no examples yet." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67669797,"math_prob":0.91457933,"size":3284,"snap":"2019-51-2020-05","text_gpt3_token_len":804,"char_repetition_ratio":0.108841464,"word_repetition_ratio":0.013592233,"special_character_ratio":0.25334957,"punctuation_ratio":0.18827161,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98956555,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T07:07:36Z\",\"WARC-Record-ID\":\"<urn:uuid:66255c7d-a1af-420a-bb87-b13ca6910e6a>\",\"Content-Length\":\"18144\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ef1b384-1788-4e00-bc24-7ae9c8525742>\",\"WARC-Concurrent-To\":\"<urn:uuid:df2afc4b-0ea6-4e65-a592-953dbb6c5090>\",\"WARC-IP-Address\":\"3.226.88.111\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/MASS/versions/7.3-22/topics/qda\",\"WARC-Payload-Digest\":\"sha1:K4774DPC7B3FH5BILGUIGGHIM4ZPRRQC\",\"WARC-Block-Digest\":\"sha1:HIQPWB7Y5KFRJWPYWPB5J2FIDK4UNQKL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592261.1_warc_CC-MAIN-20200118052321-20200118080321-00096.warc.gz\"}"}
https://www.webqc.org/molecularweightcalculated-190211-198.html	[ "", null, "#### Chemical Equations Balanced on 02/11/19\n\n Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19\nCalculate molecular weight\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232\nMolar mass of h2o is 18,01528\nMolar mass of H2O is 18.01528\nMolar mass of MgSiO3 is 100.3887\nMolar mass of TiO2 is 79.8658\nMolar mass of Ni2O3 is 165,385\nMolar mass of C4H4S is 84.13956\nMolar mass of Pb(NO3)2 is 331.2098\nMolar mass of O is 15.9994\nMolar mass of AgNO3 is 169.8731\nMolar mass of CH3OH is 32.04186\nMolar mass of C2H14O2 is 70.13136\nMolar mass of ZnSO47H2O is 287.54956\nMolar mass of H2O2 is 34.01468\nMolar mass of K2CuC4O85H2O is 407.857\nMolar mass of h3po4 is 838,9535416\nMolar mass of MgCl2 is 95.211\nMolar mass of MgCl2 is 95.211\nMolar mass of C7H16 is 100.20194\nMolar mass of H2 is 2,01588\nMolar mass of K2CuC4O82H2O is 353.81116\nMolar mass of MgSO4 is 120.3676\nMolar mass of (ch3)2co acetone is 48.10746\nMolar mass of Na2CO3 is 105.98843856\nMolar mass of O2 is 31,9988\nMolar mass of NH4po3 is 644.9857512\nMolar mass of s04 is 128.26\nMolar mass of (NH4)3PO4 is 149.086742\nMolar mass of C2 is 24,0214\nMolar mass of Ni2O3 is 165.385\nMolar mass of c3h8o3 is 92.09382\nMolar mass of SO4 is 96.0626\nMolar mass of f2ca is 78,0748064\nMolar mass of CuSO4Fe is 215.4536\nMolar mass of H2O is 18,01528\nMolar mass of O2 is 31,9988\nMolar mass of H3PO4 is 97.995182\nMolar mass of CH4 is 16,04246\nMolar mass of Sc2(SO4)3 is 378.099624\nMolar mass of C6H12O6 is 180,15588\nMolar mass of SO2 is 64.0638\nMolar mass of B4H10 is 53.3234\nMolar mass of CO2 is 44.0095\nMolar mass of C2 is 24,0214\nMolar mass of k2(cu(c2o4)2) is 317.7806\nMolar mass of NH3 is 17,03052\nMolar mass of glucose is 180.15588\nMolar mass of H2SO4 is 98,07848\nMolar mass of FeCl3 is 162.204\nMolar mass of H2o is 18,01528\nMolar mass of O2 is 31.9988\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232\nCalculate molecular weight\n Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19\nMolecular masses on 02/04/19\nMolecular masses on 01/12/19\nMolecular masses on 02/11/18" ]	[ null, "https://www.webqc.org/images/logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8534067,"math_prob":0.93103266,"size":1822,"snap":"2020-34-2020-40","text_gpt3_token_len":803,"char_repetition_ratio":0.3910891,"word_repetition_ratio":0.119266056,"special_character_ratio":0.4720088,"punctuation_ratio":0.11581292,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96657056,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T06:09:36Z\",\"WARC-Record-ID\":\"<urn:uuid:92c28523-ed66-488e-be2f-d4098389bc53>\",\"Content-Length\":\"57253\",\"Content-Type\":\"application/http; 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https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-63242	[ "# Answer to Question #63242 in Microeconomics for kamna\n\nQuestion #63242\nThe Lumins Lamp Company, a producer of old-style oil lamps, estimated the following demand function for its product:\n\nQ = 120,000 – 10,000P\n\nwhere Q is the quantity demanded per year and P is the price per lamp.\n\nThe firm’s fixed costs are $12,000 and variable costs are$1.50 per lamp.\n\na. Write an equation for the total revenue (TR) function in terms of Q.\nb. Specify the marginal revenue function.\nc. Write an equation for the total cost (TC) function in terms of Q.\nd. Specify the marginal cost function.\ne. Write an equation for total profits (π) in terms of Q. At what level of output (Q) are total profits maximized? What price will be charged? What are total profits at this output level?\nf. Check your answers in Part (e) by equating the marginal revenue and marginal cost functions, determined in Parts (b) and (d), and solving for Q.\ng. What model of market pricing behavior has been assumed in this problem?\n1\n2016-11-10T09:42:11-0500\n\nNeed a fast expert's response?\n\nSubmit order\n\nand get a quick answer at the best price\n\nfor any assignment or question with DETAILED EXPLANATIONS!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8865718,"math_prob":0.92788285,"size":941,"snap":"2022-27-2022-33","text_gpt3_token_len":226,"char_repetition_ratio":0.13020277,"word_repetition_ratio":0.036363635,"special_character_ratio":0.25079703,"punctuation_ratio":0.13930348,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954991,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T19:13:37Z\",\"WARC-Record-ID\":\"<urn:uuid:624051fb-be56-47de-b39c-d275b5d3a6b0>\",\"Content-Length\":\"308231\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d1d430e-4534-493f-b30a-6532cc50a3a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0c2530a-c1ec-4404-a683-11a28f6fe724>\",\"WARC-IP-Address\":\"44.240.111.95\",\"WARC-Target-URI\":\"https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-63242\",\"WARC-Payload-Digest\":\"sha1:23TFP24YADBSVO4V2QVPUGFYY3XJ3TPN\",\"WARC-Block-Digest\":\"sha1:COCI52WW5CAPH2VBLDFGOJUDKC2BSMJY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103642979.38_warc_CC-MAIN-20220629180939-20220629210939-00119.warc.gz\"}"}
https://factorization.info/prime-factors/9/prime-factors-of-98370.html	[ "Prime Factors of 98370", null, "Here we have a collection of all the information you may need about the Prime Factors of 98370. We will give you the definition of Prime Factors of 98370, show you how to find the Prime Factors of 98370 (Prime Factorization of 98370) by creating a Prime Factor Tree of 98370, tell you how many Prime Factors of 98370 there are, and we will show you the Product of Prime Factors of 98370.\n\nPrime Factors of 98370 definition\nFirst note that prime numbers are all positive integers that can only be evenly divided by 1 and itself. Prime Factors of 98370 are all the prime numbers that when multiplied together equal 98370.\n\nHow to find the Prime Factors of 98370\nThe process of finding the Prime Factors of 98370 is called Prime Factorization of 98370. To get the Prime Factors of 98370, you divide 98370 by the smallest prime number possible. Then you take the result from that and divide that by the smallest prime number. Repeat this process until you end up with 1.\n\nThis Prime Factorization process creates what we call the Prime Factor Tree of 98370. See illustration below.", null, "All the prime numbers that are used to divide in the Prime Factor Tree are the Prime Factors of 98370. Here is the math to illustrate:\n\n98370 ÷ 2 = 49185\n49185 ÷ 3 = 16395\n16395 ÷ 3 = 5465\n5465 ÷ 5 = 1093\n1093 ÷ 1093 = 1\n\nAgain, all the prime numbers you used to divide above are the Prime Factors of 98370. Thus, the Prime Factors of 98370 are:\n\n2, 3, 3, 5, 1093.\n\nHow many Prime Factors of 98370?\nWhen we count the number of prime numbers above, we find that 98370 has a total of 5 Prime Factors.\n\nProduct of Prime Factors of 98370\nThe Prime Factors of 98370 are unique to 98370. When you multiply all the Prime Factors of 98370 together it will result in 98370. This is called the Product of Prime Factors of 98370. The Product of Prime Factors of 98370 is:\n\n2 × 3 × 3 × 5 × 1093 = 98370\n\nPrime Factor Calculator\nDo you need the Prime Factors for a particular number? You can submit a number below to find the Prime Factors of that number with detailed explanations like we did with Prime Factors of 98370 above.\n\nPrime Factors of 98371\nWe hope this step-by-step tutorial to teach you about Prime Factors of 98370 was helpful. Do you want a test? If so, try to find the Prime Factors of the next number on our list and then check your answer here." ]	[ null, "https://factorization.info/images/prime-factors-of.png", null, "https://factorization.info/images/factor-trees/9/factor-tree-of-98370.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88623846,"math_prob":0.9543392,"size":2390,"snap":"2023-40-2023-50","text_gpt3_token_len":613,"char_repetition_ratio":0.32648784,"word_repetition_ratio":0.071428575,"special_character_ratio":0.31171548,"punctuation_ratio":0.080168776,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997284,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T06:06:07Z\",\"WARC-Record-ID\":\"<urn:uuid:491c2e8d-5e82-41ef-8039-348524bdf114>\",\"Content-Length\":\"7668\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eb495142-45ff-4d8d-96b1-bba55a0684a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:e331d80c-3977-452a-85d3-ec6a5e345a77>\",\"WARC-IP-Address\":\"18.67.65.2\",\"WARC-Target-URI\":\"https://factorization.info/prime-factors/9/prime-factors-of-98370.html\",\"WARC-Payload-Digest\":\"sha1:EJOV2XCI3KPCEGCJCDFAFY5JLNDZ4675\",\"WARC-Block-Digest\":\"sha1:O3UTXKPOMH32A2IYXY3FOLPMOHEU33VN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100650.21_warc_CC-MAIN-20231207054219-20231207084219-00356.warc.gz\"}"}
https://pixel-druid.com/articles/motivation-for-the-compact-open-topology.html	[ "## § Motivation for the compact-open topology\n\n• If $X$ is a compact space and $Y$ is a metric space, consider two functions $f, g: X \\to Y$.\n• We can define a distance $d(f, g) \\equiv \\min_{x \\in X} d(f(x), g(x))$.\n• The $\\min_{x \\in X}$ has a maximum because $X$ is compact.\n• Thus this is a real metric on the function space $Map(X, Y)$.\n• Now suppose $Y$ is no longer a metric space, but is Haussdorf. Can we still define a topology on $Map(X, Y)$?\n• Let $K \\subseteq X$ be compact, and let $U \\subseteq Y$ be open such that $f(K) \\subseteq U$.\n• Since $Y$ is Hausdorff, $K \\subseteq X$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.837763,"math_prob":1.000001,"size":484,"snap":"2023-40-2023-50","text_gpt3_token_len":156,"char_repetition_ratio":0.12083333,"word_repetition_ratio":0.0,"special_character_ratio":0.2768595,"punctuation_ratio":0.13821138,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000083,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T04:14:09Z\",\"WARC-Record-ID\":\"<urn:uuid:3cfc6bf4-13b7-43c4-b076-056a5b589120>\",\"Content-Length\":\"17567\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dfbc7f6-7594-4643-a567-6748c2a18692>\",\"WARC-Concurrent-To\":\"<urn:uuid:21b88164-8440-4706-b8f1-d16effd63287>\",\"WARC-IP-Address\":\"159.65.151.13\",\"WARC-Target-URI\":\"https://pixel-druid.com/articles/motivation-for-the-compact-open-topology.html\",\"WARC-Payload-Digest\":\"sha1:6JDLQD4YWZJSGY4CW42IZU4D24UNRTOK\",\"WARC-Block-Digest\":\"sha1:4VHIAXNRAGC6VR4X5E2J43JTJHKWCNBW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510358.68_warc_CC-MAIN-20230928031105-20230928061105-00539.warc.gz\"}"}
https://origin.geeksforgeeks.org/looping-techniques-python/?ref=lbp	[ "# Looping Techniques in Python\n\n• Difficulty Level : Easy\n• Last Updated : 23 Nov, 2021\n\nPython supports various looping techniques by certain inbuilt functions, in various sequential containers. These methods are primarily very useful in competitive programming and also in various projects which require a specific technique with loops maintaining the overall structure of code. A lot of time and memory space is been saved as there is no need to declare the extra variables which we declare in the traditional approach of loops.\n\nWhere they are used?\nDifferent looping techniques are primarily useful in the places where we don’t need to actually manipulate the structure and order of the overall containers, rather only print the elements for a single-use instance, no in-place change occurs in the container. This can also be used in instances to save time.\n\nAttention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.\n\nTo begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning - Basic Level Course\n\nDifferent looping techniques using Python data structures are:\n\n• Using enumerate(): enumerate() is used to loop through the containers printing the index number along with the value present in that particular index.\n\n## Python3\n\n `# python code to demonstrate working of enumerate()` `for` `key, value ``in` `enumerate``([``'The'``, ``'Big'``, ``'Bang'``, ``'Theory'``]):` ` ``print``(key, value)`\n\nOutput:\n\n```0 The\n1 Big\n2 Bang\n3 Theory```\n\n## Python3\n\n `# python code to demonstrate working of enumerate()` `for` `key, value ``in` `enumerate``([``'Geeks'``, ``'for'``, ``'Geeks'``, ``'is'``, ``'the'``, ``'Best'``, ``'Coding'``, ``'Platform'``]):` ` ``print``(value, end``=``' '``)`\n\nOutput:\n\n`Geeks for Geeks is the Best Coding Platform `\n• Using zip(): zip() is used to combine 2 similar containers(list-list or dict-dict) printing the values sequentially. The loop exists only till the smaller container ends. A detailed explanation of zip() and enumerate() can be found here.\n\n## Python3\n\n `# python code to demonstrate working of zip()` `# initializing list` `questions ``=` `[``'name'``, ``'colour'``, ``'shape'``]` `answers ``=` `[``'apple'``, ``'red'``, ``'a circle'``]` `# using zip() to combine two containers ` `# and print values` `for` `question, answer ``in` `zip``(questions, answers):` ` ``print``(``'What is your {0}? I am {1}.'``.``format``(question, answer))`\n\nOutput:\n\n```What is your name? I am apple.\nWhat is your color? I am red.\nWhat is your shape? I am a circle.```\n• Using iteritem(): iteritems() is used to loop through the dictionary printing the dictionary key-value pair sequentially.\n• Using items(): items() performs the similar task on dictionary as iteritems() but have certain disadvantages when compared with iteritems().\n• It is very time-consuming. Calling it on large dictionaries consumes quite a lot of time.\n• It takes a lot of memory. Sometimes takes double the memory when called on a dictionary.\n\nExample 1:\n\n## Python3\n\n `# python code to demonstrate working of iteritems(),items()` `d ``=` `{ ``\"geeks\"` `: ``\"for\"``, ``\"only\"` `: ``\"geeks\"` `}` `# using iteritems to print the dictionary key-value pair` `print` `(``\"The key value pair using iteritems is : \"``)` `for` `i,j ``in` `d.items():` ` ``print``( i,j )` ` ` `# using items to print the dictionary key-value pair` `print` `(``\"The key value pair using items is : \"``)` `for` `i,j ``in` `d.items():` ` ``print``( i,j )`\n\nOutput:\n\n```The key value pair using iteritems is :\ngeeks for\nonly geeks\nThe key value pair using items is :\ngeeks for\nonly geeks```\n\nExample 2:\n\n## Python3\n\n `# python code to demonstrate working of items()` `king ``=` `{``'Akbar'``: ``'The Great'``, ``'Chandragupta'``: ``'The Maurya'``, ` ` ``'Modi'` `: ``'The Changer'``}` `# using items to print the dictionary key-value pair` `for` `key, value ``in` `king.items():` ` ``print``(key, value)`\n\nOutput:\n\n```Akbar The Great\nChandragupta The Maurya\nModi The Changer```\n• Using sorted(): sorted() is used to print the container is sorted order. It doesn’t sort the container but just prints the container in sorted order for 1 instance. The use of set() can be combined to remove duplicate occurrences.\n\nExample 1:\n\n## Python3\n\n `# python code to demonstrate working of sorted()` `# initializing list` `lis ``=` `[ ``1` `, ``3``, ``5``, ``6``, ``2``, ``1``, ``3` `]` `# using sorted() to print the list in sorted order` `print` `(``\"The list in sorted order is : \"``)` `for` `i ``in` `sorted``(lis) :` ` ``print` `(i,end``=``\" \"``)` ` ` `print` `(``\"\\r\"``)` ` ` `# using sorted() and set() to print the list in sorted order` `# use of set() removes duplicates.` `print` `(``\"The list in sorted order (without duplicates) is : \"``)` `for` `i ``in` `sorted``(``set``(lis)) :` ` ``print` `(i,end``=``\" \"``)`\n\nOutput:\n\n```The list in sorted order is :\n1 1 2 3 3 5 6\nThe list in sorted order (without duplicates) is :\n1 2 3 5 6 ```\n\nExample 2:\n\n## Python3\n\n `# python code to demonstrate working of sorted()` `# initializing list` `basket ``=` `[``'guave'``, ``'orange'``, ``'apple'``, ``'pear'``, ` ` ``'guava'``, ``'banana'``, ``'grape'``]` `# using sorted() and set() to print the list` `# in sorted order` `for` `fruit ``in` `sorted``(``set``(basket)):` ` ``print``(fruit)`\n\nOutput:\n\n```apple\nbanana\ngrape\nguava\nguave\norange\npear```\n• Using reversed(): reversed() is used to print the values of the container in the reversed order. It does not reflect any changes to the original list\n\nExample 1:\n\n## Python3\n\n `# python code to demonstrate working of reversed()` `# initializing list` `lis ``=` `[ ``1` `, ``3``, ``5``, ``6``, ``2``, ``1``, ``3` `]` `# using revered() to print the list in reversed order` `print` `(``\"The list in reversed order is : \"``)` `for` `i ``in` `reversed``(lis) :` ` ``print` `(i,end``=``\" \"``)`\n\nOutput:\n\n```The list in reversed order is :\n3 1 2 6 5 3 1 ```\n\nExample 2:\n\n## Python3\n\n `# python code to demonstrate working of reversed()` `# using reversed() to print in reverse order` `for` `i ``in` `reversed``(``range``(``1``, ``10``, ``3``)):` ` ``print` `(i)`\n\nOutput:\n\n```7\n4\n1```\n• These techniques are quick to use and reduce coding effort. for, while loops need the entire structure of the container to be changed.\n• These Looping techniques do not require any structural changes to the container. They have keywords that present the exact purpose of usage. Whereas, no pre-predictions or guesses can be made in for, while loop i.e not easily understand the purpose at a glance.\n• Looping technique makes the code more concise than using for & while looping.\n\nMy Personal Notes arrow_drop_up\nRecommended Articles\nPage :" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7287738,"math_prob":0.60460126,"size":6083,"snap":"2021-43-2021-49","text_gpt3_token_len":1564,"char_repetition_ratio":0.12847508,"word_repetition_ratio":0.23408072,"special_character_ratio":0.27305606,"punctuation_ratio":0.13551816,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9820238,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T03:12:23Z\",\"WARC-Record-ID\":\"<urn:uuid:9dd00cc8-f38c-4c2b-b0a7-6f13aa1661b3>\",\"Content-Length\":\"253512\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96bd245d-c10b-45fc-9ed3-12fdc2e6a151>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9237a26-384a-4271-93de-e106341c805d>\",\"WARC-IP-Address\":\"44.228.100.190\",\"WARC-Target-URI\":\"https://origin.geeksforgeeks.org/looping-techniques-python/?ref=lbp\",\"WARC-Payload-Digest\":\"sha1:2C56ORF57UBP3C2Y5AQ36G3C55UN5W7B\",\"WARC-Block-Digest\":\"sha1:MECKQT4X3PMWBEFWTK5GZEMH6DTCY6ER\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358443.87_warc_CC-MAIN-20211128013650-20211128043650-00581.warc.gz\"}"}
https://jumk.de/math-physics-formulary/exponential-growth.php	[ "## Formulary\n\n### Math and Physics Calculators\n\n\| Start: Formulary \| Imprint & Privacy\n\nAnzeige\n\n# Calculate Exponential Growth\n\nCalculates the exponential growth for a doubling with each cycle. The number of cycles can be calculated from cycle length and run time.\n\na * 2n = b\na = start value, b = end value, n = number of cycles\n\n a: b: n:\nPlease enter two values, the third will be calculated.\n\n cycle length: d, h, m, s run time: d, h, m, s number of cycles:\nPlease enter times for cycle length and run time.\nd = days, h = hours, m = minutes, s = seconds.\n\nAlso see exponential decay.", null, "The function y = 2x, made with the Function Graphs Plotter.\n\nFormulary and calculators math and physics.\n© jumk.de Webprojects \| German: Formelsammlung Mathe & Physik\n\nAnzeige" ]	[ null, "https://jumk.de/math-physics-formulary/exp.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8171635,"math_prob":0.99846804,"size":479,"snap":"2021-43-2021-49","text_gpt3_token_len":119,"char_repetition_ratio":0.13263159,"word_repetition_ratio":0.022988506,"special_character_ratio":0.24425887,"punctuation_ratio":0.14893617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99952877,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T09:47:53Z\",\"WARC-Record-ID\":\"<urn:uuid:58720b2a-209b-480d-b219-fc2c0563102a>\",\"Content-Length\":\"19763\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5598749a-06fd-48ba-8bbf-bdbf3495b3dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d152cdf-c3fb-40b0-affc-3597dbebef84>\",\"WARC-IP-Address\":\"92.204.58.21\",\"WARC-Target-URI\":\"https://jumk.de/math-physics-formulary/exponential-growth.php\",\"WARC-Payload-Digest\":\"sha1:ILLP7LXCWX6KOT5ZZ5FWH3W4QFJB42E2\",\"WARC-Block-Digest\":\"sha1:SDEXKEDLPGEBKPLWATPSRJFXLDZDLNW3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584554.98_warc_CC-MAIN-20211016074500-20211016104500-00573.warc.gz\"}"}
https://intl.siyavula.com/read/maths/grade-10/finance-and-growth/09-finance-and-growth-03	[ "We think you are located in United States. Is this correct?\n\n# 9.4 Calculations using simple and compound interest\n\n## 9.4 Calculations using simple and compound interest (EMA6Q)\n\n### Hire purchase (EMA6R)\n\nAs a general rule, it is not wise to buy items on credit. When buying on credit you have to borrow money to pay for the object, meaning you will have to pay more for it due to the interest on the loan. That being said, occasionally there are appliances, such as a fridge, that are very difficult to live without. Most people don't have the cash up front to purchase such items, so they buy it on a hire purchase agreement.\n\nA hire purchase agreement is a financial agreement between the shop and the customer about how the customer will pay for the desired product. The interest on a hire purchase loan is always charged at a simple interest rate and only charged on the amount owing. Most agreements require that a deposit is paid before the product can be taken by the customer. The principal amount of the loan is therefore the cash price minus the deposit. The accumulated loan will be worked out using the number of years the loan is needed for. The total loan amount is then divided into monthly payments over the period of the loan.\n\nHire purchase is charged at a simple interest rate. When you are asked a hire purchase question, don't forget to always use the simple interest formula.\n\nThis video explains hire purchase and shows some examples of hire purchase calculations.\n\nVideo: 2GHF\n\n## Worked example 7: Hire purchase\n\nTroy wants to buy an additional screen for his computer which he saw advertised for $$\\text{R}\\,\\text{2 500}$$ on the internet. There is an option of paying a $$\\text{10}\\%$$ deposit and then making $$\\text{24}$$ monthly payments using a hire purchase agreement, where interest is calculated at $$\\text{7,5}\\%$$ p.a. simple interest. Calculate what Troy's monthly payments will be.\n\n### Write down the known variables\n\nA new opening balance is required, as the $$\\text{10}\\%$$ deposit is paid in cash.\n\n\\begin{align} \\text{10}\\% \\text{ of } \\text{2 500} & = 250 \\\\ \\therefore P & = \\text{2 500} - 250 = \\text{2 250} \\\\ i & = \\text{0,075} \\\\ n & = \\frac{24}{12} = 2 \\end{align}\n\n### Write down the formula\n\n$A = P\\left(1 + in\\right)$\n\n### Substitute the values\n\n\\begin{align} A & = \\text{2 250}\\left(1 + \\text{0,075}\\times 2\\right) \\\\ & = \\text{2 587,50} \\end{align}\n\n### Calculate the monthly repayments on the hire purchase agreement\n\n\\begin{align} \\text{Monthly payment } & = \\frac{\\text{2 587,50}}{24} \\\\ & = \\text{107,81} \\end{align}\n\nTroy's monthly payment is $$\\text{R}\\,\\text{107,81}$$.\n\ntemp text\n\nA shop can also add a monthly insurance premium to the monthly instalments. This insurance premium will be an amount of money paid monthly and gives the customer more time between a missed payment and possible repossession of the product.\n\nThe monthly payment is also called the monthly instalment.\n\n## Worked example 8: Hire purchase with extra conditions\n\nCassidy wants to buy a TV and decides to buy one on a hire purchase agreement. The TV's cash price is $$\\text{R}\\,\\text{5 500}$$. She will pay it off over $$\\text{54}$$ months at an interest rate of $$\\text{21}\\%$$ p.a. An insurance premium of $$\\text{R}\\,\\text{12,50}$$ is added to every monthly payment. How much are her monthly payments?\n\n### Write down the known variables\n\n\\begin{align} P & = \\text{5 500} \\\\ i & = \\text{0,21} \\\\ n & = \\frac{54}{12} = \\text{4,5} \\end{align}\n\nThe question does not mention a deposit, therefore we assume that Cassidy did not pay one.\n\n### Write down the formula\n\n$A = P\\left(1 + in\\right)$\n\n### Substitute the values\n\n\\begin{align} A & = \\text{5 500}\\left(1 + \\text{0,21}\\times \\text{4,5}\\right) \\\\ & = \\text{10 697,50} \\end{align}\n\n### Calculate the monthly repayments on the hire purchase agreement\n\n\\begin{align} \\text{Monthly payment } & = \\frac{\\text{10 697,50}}{54} \\\\ & = \\text{198,10} \\end{align}\n\n$$\\text{198,10} + \\text{12,50} = \\text{210,60}$$\n\nCassidy will pay $$\\text{R}\\,\\text{210,60}$$ per month for $$\\text{54}$$ months until her TV is paid off.\n\nTextbook Exercise 9.3\n\nAngelique wants to buy a microwave on a hire purchase agreement. The cash price of the microwave is $$\\text{R}\\,\\text{4 400}$$. She is required to pay a deposit of $$\\text{10}\\%$$ and pay the remaining loan amount off over $$\\text{12}$$ months at an interest rate of $$\\text{9}\\%$$ p.a.\n\nWhat is the principal loan amount?\n\nFirst calculate the amount for the deposit:\n\n\\begin{align} \\text{deposit} &= \\text{4 400} \\times \\frac{\\text{10}}{\\text{100}}\\\\ &= \\text{440} \\end{align}\n\nTo determine the principal loan amount, we must subtract the deposit amount from the cash price:\n\n\\begin{align} P &= \\text{cash price} - \\text{deposit}\\\\ &= \\text{4 400} - \\text{440}\\\\ &= \\text{R}\\,\\text{3 960,00} \\end{align}\n\nWhat is the accumulated loan amount?\n\nRead the question carefully and write down the given information:\n\n\\begin{align} A & = ? \\\\ P & = \\text{R}\\,\\text{3 960,00} \\\\ i & = \\frac{9}{100} = \\text{0,09} \\\\ n & = 1 \\end{align}\n\nTo determine the accumulated loan amount, we use the simple interest formula:\n\n\\begin{align} A &= P(1+in)\\\\ &= \\text{R}\\,\\text{3 960,00}\\left(\\text{1} + \\text{0,09} \\times 1\\right)\\\\ &= \\text{R}\\,\\text{4 316,40} \\end{align}\n\nWhat are Angelique's monthly repayments?\n\nTo determine the monthly payment amount, we divide the accumulated amount $$A$$ by the total number of months: \\begin{align} \\text{Monthly repayment} &= \\frac{A}{\\text{no. of months}}\\\\ &= \\frac{\\text{R}\\,\\text{4 316,40}}{\\text{12}}\\\\ &= \\text{R}\\,\\text{359,70} \\end{align}\n\nWhat is the total amount she has paid for the microwave?\n\nTo determine the total amount paid, we add the accumulated loan amount and the deposit: \\begin{align} \\text{Total amount} &= A + \\text{deposit amount}\\\\ &= \\text{R}\\,\\text{4 316,40} + \\text{440}\\\\ &= \\text{R}\\,\\text{4 756,40} \\end{align}\n\nNyakallo wants to buy a television on a hire purchase agreement. The cash price of the television is $$\\text{R}\\,\\text{5 600}$$. She is required to pay a deposit of $$\\text{15}\\%$$ and pay the remaining loan amount off over $$\\text{24}$$ months at an interest rate of $$\\text{14}\\%$$ p.a.\n\nWhat is the principal loan amount?\n\nFirst calculate the amount for the deposit:\n\n\\begin{align} \\text{deposit} &= \\text{5 600} \\times \\frac{\\text{15}}{\\text{100}}\\\\ &= \\text{840} \\end{align}\n\nTo determine the principal loan amount, we must subtract the deposit amount from the cash price:\n\n\\begin{align} P &= \\text{cash price} - \\text{deposit}\\\\ &= \\text{5 600} - \\text{840}\\\\ &= \\text{R}\\,\\text{4 760,00} \\end{align}\n\nWhat is the accumulated loan amount?\n\nRead the question carefully and write down the given information:\n\n\\begin{align} A & = ? \\\\ P & = \\text{R}\\,\\text{4 760,00} \\\\ i & = \\frac{14}{100} = \\text{0,14} \\\\ n & = 2 \\end{align}\n\nTo determine the accumulated loan amount, we use the simple interest formula:\n\n\\begin{align} A &= P(1+in)\\\\ &= \\text{R}\\,\\text{4 760,00}\\left(\\text{1} + \\text{0,14} \\times 2\\right)\\\\ &= \\text{R}\\,\\text{6 092,80} \\end{align}\n\nWhat are Nyakallo's monthly repayments?\n\nTo determine the monthly payment amount, we divide the accumulated amount $$A$$ by the total number of months: \\begin{align} \\text{Monthly repayment} &= \\frac{A}{\\text{no. of months}}\\\\ &= \\frac{\\text{R}\\,\\text{6 092,80}}{\\text{24}}\\\\ &= \\text{R}\\,\\text{253,87} \\end{align}\n\nWhat is the total amount she has paid for the television?\n\nTo determine the total amount paid we add the accumulated loan amount and the deposit: \\begin{align} \\text{Total amount} &= A + \\text{deposit amount}\\\\ &= \\text{R}\\,\\text{6 092,80} + \\text{840}\\\\ &= \\text{R}\\,\\text{6 932,80} \\end{align}\n\nA company wants to purchase a printer. The cash price of the printer is $$\\text{R}\\,\\text{4 500}$$. A deposit of $$\\text{15}\\%$$ is required on the printer. The remaining loan amount will be paid off over $$\\text{24}$$ months at an interest rate of $$\\text{12}\\%$$ p.a.\n\nWhat is the principal loan amount?\n\nTo calculate the principal loan amount, we first calculate the amount for the deposit and then subtract the deposit amount from the cash price:\n\n\\begin{align} P & = \\text{4 500} - (\\text{4 500} \\times \\text{0,15}) \\\\ &= \\text{4 500} - 675 \\\\ &= \\text{R}\\,\\text{3 825} \\end{align}\n\nWhat is the accumulated loan amount?\n\nRemember that hire purchase uses simple interest. We write down the given information and then substitute these values into the simple interest formula.\n\n\\begin{align} P &= \\text{R}\\,\\text{3 825} \\\\ i & = \\text{0,12}\\\\ n & = \\frac{24}{12} = 2 \\\\\\\\ A = & P(1 + in) \\\\ A &= \\text{3 825}(1 + (\\text{0,12})(2))\\\\ A & = \\text{R}\\,\\text{4 743} \\end{align}\n\nHow much will the company pay each month?\n\nTo determine the monthly payment amount (how much the company pays each month), we divide the accumulated amount $$A$$ by the total number of months:\n\n$$\\dfrac{\\text{4 743}}{24} = \\text{R}\\,\\text{197,63}$$\n\nWhat is the total amount the company paid for the printer?\n\nTo determine the total amount paid we add the accumulated loan amount and the deposit:\n\n$$675 + \\text{4 743} = \\text{R}\\,\\text{5 418}$$\n\nSandile buys a dining room table costing $$\\text{R}\\,\\text{8 500}$$ on a hire purchase agreement. He is charged an interest rate of $$\\text{17,5}\\%$$ p.a. over $$\\text{3}$$ years.\n\nHow much will Sandile pay in total?\n\nThe question does not mention a deposit so we assume Sandile did not pay one. We write down the given information and then use the simple interest formula to calculate the accumulated amount.\n\n\\begin{align} A & = ? \\\\ P & = \\text{8 500}\\\\ i & = \\text{0,175}\\\\ n & = 3\\\\\\\\ A & = P(1 + in)\\\\ A & = \\text{8 500}(1 + (\\text{0,175})(3))\\\\ A & = \\text{R}\\,\\text{12 962,50} \\end{align}\n\nHow much interest does he pay?\n\nTo calculate the total interest paid we subtract the cash price from the accumulated amount.\n\n$$\\text{12 962,50} - \\text{8 500} = \\text{R}\\,\\text{4 462,50}$$\n\nWhat is his monthly instalment?\n\nTo determine the monthly payment amount, we divide the accumulated amount $$A$$ by the total number of months:\n\n$$\\dfrac{\\text{12 962,50}}{36} = \\text{R}\\,\\text{360,07}$$\n\nMike buys a table costing $$\\text{R}\\,\\text{6 400}$$ on a hire purchase agreement. He is charged an interest rate of $$\\text{15}\\%$$ p.a. over $$\\text{4}$$ years.\n\nHow much will Mike pay in total?\n\nRead the question carefully and write down the given information:\n\n\\begin{align} A & = ? \\\\ P & = \\text{R}\\,\\text{6 400} \\\\ i & = \\frac{15}{100} = \\text{0,15} \\\\ n & = 4 \\end{align}\n\nTo determine the accumulated loan amount, we use the simple interest formula:\n\n\\begin{align} A &= P(1+in)\\\\ &= \\text{6 400} (\\text{1} + \\text{0,15} \\times \\text{4})\\\\ &= \\text{R}\\,\\text{10 240} \\end{align}\n\nHow much interest does he pay?\n\nTo determine the interest amount, we subtract the principal amount from the accumulated amount: \\begin{align} \\text{Interest amount} &= A-P\\\\ &= \\text{10 240} - \\text{6 400}\\\\ &= \\text{R}\\,\\text{3 840} \\end{align}\n\nWhat is his monthly instalment?\n\nTo determine the monthly instalment amount, we divide the accumulated amount $$A$$ by the total number of months: \\begin{align} \\text{Monthly instalment} &= \\frac{A}{\\text{no. of months}}\\\\ &= \\frac{\\text{10 240}}{\\text{4} \\times \\text{12}}\\\\ &= \\text{R}\\,\\text{213,33} \\end{align}\n\nTalwar buys a cupboard costing $$\\text{R}\\,\\text{5 100}$$ on a hire purchase agreement. He is charged an interest rate of $$\\text{12}\\%$$ p.a. over $$\\text{2}$$ years.\n\nHow much will Talwar pay in total?\n\nRead the question carefully and write down the given information:\n\n\\begin{align} A & = ? \\\\ P & = \\text{R}\\,\\text{5 100} \\\\ i & = \\frac{12}{100} = \\text{0,12} \\\\ n & = 2 \\end{align}\n\nTo determine the accumulated loan amount, we use the simple interest formula: \\begin{align} A &= P(1+in)\\\\ &= \\text{5 100} (\\text{1} + \\text{0,12} \\times \\text{2})\\\\ &= \\text{R}\\,\\text{6 324} \\end{align}\n\nHow much interest does he pay?\n\nTo determine the interest amount, we subtract the principal amount from the accumulated amount: \\begin{align} \\text{Interest amount} &= A-P\\\\ &= \\text{6 324} - \\text{5 100}\\\\ &= \\text{R}\\,\\text{1 224} \\end{align}\n\nWhat is his monthly instalment?\n\nTo determine the monthly instalment amount, we divide the accumulated amount $$A$$ by the total number of months: \\begin{align} \\text{Monthly instalment} &= \\frac{A}{\\text{no. of months}}\\\\ &= \\frac{\\text{6 324}}{\\text{2} \\times \\text{12}}\\\\ &= \\text{R}\\,\\text{263,50} \\end{align}\n\nA lounge suite is advertised for sale on TV, to be paid off over $$\\text{36}$$ months at $$\\text{R}\\,\\text{150}$$ per month.\n\nAssuming that no deposit is needed, how much will the buyer pay for the lounge suite once it has been paid off?\n\n$$36 \\times 150 = \\text{R}\\,\\text{5 400}$$\n\nIf the interest rate is $$\\text{9}\\%$$ p.a., what is the cash price of the suite?\n\n\\begin{align} A & = \\text{5 400}\\\\ P & = ? \\\\ i & = \\text{0,09} \\\\ n & = 3 \\\\\\\\ A & = P(1 + in) \\\\ \\text{5 400} & = P(1 + (\\text{0,09})(3)) \\\\ \\frac{\\text{5 400}}{\\text{1,27}} & = P\\\\ P & = \\text{R}\\,\\text{4 251,97} \\end{align}\n\nTwo stores are offering a fridge and washing machine combo package. Store A offers a monthly payment of $$\\text{R}\\,\\text{350}$$ over $$\\text{24}$$ months. Store B offers a monthly payment of $$\\text{R}\\,\\text{175}$$ over $$\\text{48}$$ months.\n\nIf both stores offer $$\\text{7,5}\\%$$ interest, which store should you purchase the fridge and washing machine from if you want to pay the least amount of interest?\n\nTo calculate the interest paid at each store we need to first find the cash price of the fridge and washing machine.\n\nStore A:\n\n\\begin{align} A & = \\text{350} \\times \\text{24} = \\text{8 400}\\\\ P & = ? \\\\ i & = \\text{0,075} \\\\ n & = 2 \\\\\\\\ A & = P(1 + in) \\\\ \\text{8 400} & = P(1 + (\\text{0,075})(2)) \\\\ \\frac{\\text{8 400}}{\\text{1,15}} & = P\\\\ P & = \\text{R}\\,\\text{7 304,45} \\end{align}\n\nTherefore the interest is $$\\text{R}\\,\\text{8 400} - \\text{R}\\,\\text{7 304,45} = \\text{R}\\,\\text{1 095,65}$$\n\nStore B:\n\n\\begin{align} A & = \\text{175} \\times \\text{48} = \\text{8 400}\\\\ P & = ? \\\\ i & = \\text{0,075} \\\\ n & = 4 \\\\\\\\ A & = P(1 + in) \\\\ \\text{8 400} & = P(1 + (\\text{0,075})(4)) \\\\ \\frac{\\text{8 400}}{\\text{1,3}} & = P\\\\ P & = \\text{R}\\,\\text{6 461,54} \\end{align}\n\nTherefore the interest is $$\\text{R}\\,\\text{8 400} - \\text{R}\\,\\text{6 461,54} = \\text{R}\\,\\text{1 938,46}$$\n\nIf you want to pay the least amount in interest you should purchase the fridge and washing machine from store A.\n\nTlali wants to buy a new computer and decides to buy one on a hire purchase agreement. The computers cash price is $$\\text{R}\\,\\text{4 250}$$. He will pay it off over $$\\text{30}$$ months at an interest rate of $$\\text{9,5}\\%$$ p.a. An insurance premium of $$\\text{R}\\,\\text{10,75}$$ is added to every monthly payment. How much are his monthly payments?\n\n\\begin{align} P & = \\text{4 250} \\\\ i & = \\text{0,095} \\\\ n & = \\frac{30}{12} = \\text{2,5} \\end{align}\n\nThe question does not mention a deposit, therefore we assume that Tlali did not pay one.\n\n\\begin{align} A & = P\\left(1 + in\\right) \\\\ A & = \\text{4 250}\\left(1 + \\text{0,095}\\times \\text{2,5}\\right) \\\\ & = \\text{5 259,38} \\end{align}\n\nThe monthly payment is:\n\n\\begin{align} \\text{Monthly payment } & = \\frac{\\text{5 259,38}}{36} \\\\ & = \\text{146,09} \\end{align}\n\nAdd the insurance premium: $$\\text{R}\\,\\text{146,09} + \\text{R}\\,\\text{10,75} = \\text{R}\\,\\text{156,84}$$\n\nRichard is planning to buy a new stove on hire purchase. The cash price of the stove is $$\\text{R}\\,\\text{6 420}$$. He has to pay a $$\\text{10}\\%$$ deposit and then pay the remaining amount off over $$\\text{36}$$ months at an interest rate of $$\\text{8}\\%$$ p.a. An insurance premium of $$\\text{R}\\,\\text{11,20}$$ is added to every monthly payment. Calculate Richard's monthly payments.\n\n\\begin{align} P & = \\text{6 420} - (\\text{0,10})(\\text{6 420}) = \\text{5 778} \\\\ i & = \\text{0,08} \\\\ n & = \\frac{36}{12} = \\text{3} \\end{align}\n\nCalculate the accumulated amount:\n\n\\begin{align} A & = P\\left(1 + in\\right) \\\\ A & = \\text{5 778}\\left(1 + \\text{0,08}\\times \\text{3}\\right) \\\\ & = \\text{7 164,72} \\end{align}\n\nCalculate the monthly repayments on the hire purchase agreement:\n\n\\begin{align} \\text{Monthly payment } & = \\frac{\\text{7 164,72}}{36} \\\\ & = \\text{199,02} \\end{align}\n\nAdd the insurance premium: $$\\text{R}\\,\\text{199,02} + \\text{R}\\,\\text{11,20} = \\text{R}\\,\\text{210,22}$$\n\n### Inflation (EMA6S)\n\nThere are many factors that influence the change in price of an item, one of them is inflation. Inflation is the average increase in the price of goods each year and is given as a percentage. Since the rate of inflation increases year on year, it is calculated using the compound interest formula.\n\n## Worked example 9: Calculating future cost based on inflation\n\nMilk costs $$\\text{R}\\,\\text{14}$$ for two litres. How much will it cost in $$\\text{4}$$ years time if the inflation rate is $$\\text{9}\\%$$ p.a.?\n\n### Write down the known variables\n\n\\begin{align} P & = 14 \\\\ i & = \\text{0,09} \\\\ n & = 4 \\end{align}\n\n### Write down the formula\n\n$A = P{\\left(1 + i\\right)}^{n}$\n\n### Substitute the values\n\n\\begin{align} A & = 14{\\left(1 + \\text{0,09}\\right)}^{4} \\\\ & = \\text{19,76} \\end{align}\n\nIn four years time, two litres of milk will cost $$\\text{R}\\,\\text{19,76}$$.\n\n## Worked example 10: Calculating past cost based on inflation\n\nA box of chocolates costs $$\\text{R}\\,\\text{55}$$ today. How much did it cost $$\\text{3}$$ years ago if the average rate of inflation was $$\\text{11}\\%$$ p.a.?\n\n### Write down the known variables\n\n\\begin{align} A & = 55 \\\\ i & = \\text{0,11} \\\\ n & = 3 \\end{align}\n\n### Write down the formula\n\n$A = P{\\left(1 + i\\right)}^{n}$\n\n### Substitute the values and solve for $$P$$\n\n\\begin{align} 55 & = P{\\left(1 + \\text{0,11}\\right)}^{3} \\\\ \\frac{55}{{\\left(1 + \\text{0,11}\\right)}^{3}} & = P \\\\ \\therefore P & = \\text{40,22} \\end{align}\n\nThree years ago, the box of chocolates would have cost $$\\text{R}\\,\\text{40,22}$$.\n\ntemp text\nTextbook Exercise 9.4\n\nThe price of a bag of apples is $$\\text{R}\\,\\text{12}$$. How much will it cost in $$\\text{9}$$ years time if the inflation rate is $$\\text{12}\\%$$ p.a.?\n\nRead the question carefully and write down the given information:\n\n• $$A = ?$$\n• $$P = \\text{R}\\,\\text{12}$$\n• $$n = \\text{9}$$\n• $$i = \\frac{12}{100}$$\n\nTo determine the future cost, we use the compound interest formula: \\begin{align} A &= P\\left(1+i\\right)^n \\\\ &= \\text{12} \\times \\left(\\text{1} + \\frac{\\text{12}}{\\text{100}}\\right)^{\\text{9}}\\\\ &= \\text{R}\\,\\text{33,28} \\end{align}\n\nThe price of a bag of potatoes is $$\\text{R}\\,\\text{15}$$.\n\nHow much will it cost in $$\\text{6}$$ years time if the inflation rate is $$\\text{12}\\%$$ p.a.?\n\nRead the question carefully and write down the given information:\n\n• $$A = ?$$\n• $$P = \\text{R}\\,\\text{15}$$\n• $$n = \\text{6}$$\n• $$i = \\frac{12}{100}$$\n\nTo determine the future cost, we use the compound interest formula: \\begin{align} A &= P\\left(1+i\\right)^n \\\\ &= \\text{15} \\times \\left(\\text{1} + \\frac{\\text{12}}{\\text{100}}\\right)^{\\text{6}}\\\\ &= \\text{R}\\,\\text{29,61} \\end{align}\n\nThe price of a box of popcorn is $$\\text{R}\\,\\text{15}$$. How much will it cost in $$\\text{4}$$ years time if the inflation rate is $$\\text{11}\\%$$ p.a.?\n\nRead the question carefully and write down the given information:\n\n• $$A = ?$$\n• $$P = \\text{R}\\,\\text{15}$$\n• $$n = \\text{4}$$\n• $$i = \\frac{11}{100}$$\n\nTo determine the future cost, we use the compound interest formula: \\begin{align} A &= P\\left(1+i\\right)^n \\\\ &= \\text{15} \\times \\left(\\text{1} + \\frac{\\text{11}}{\\text{100}}\\right)^{\\text{4}}\\\\ &= \\text{R}\\,\\text{22,77} \\end{align}\n\nA box of raisins costs $$\\text{R}\\,\\text{24}$$ today. How much did it cost $$\\text{4}$$ years ago if the average rate of inflation was $$\\text{13}\\%$$ p.a.? Round your answer to 2 decimal places.\n\nRead the question carefully and write down the given information:\n\n• $$A = \\text{R}\\,\\text{24}$$\n• $$P = ?$$\n• $$i = \\frac{13}{100}$$\n• $$n = \\text{4}$$\n\nWe use the compound interest formula and make $$P$$ the subject: \\begin{align} A &= P\\left(1+i\\right)^n \\\\ P &= \\frac{A}{\\left(1+i\\right)^n} \\\\ &= \\frac{\\text{24}}{\\left(\\text{1} + \\frac{\\text{13}}{\\text{100}}\\right)^{\\text{4}}} \\\\ &= \\text{R}\\,\\text{14,72} \\end{align}\n\nA box of biscuits costs $$\\text{R}\\,\\text{24}$$ today. How much did it cost $$\\text{5}$$ years ago if the average rate of inflation was $$\\text{11}\\%$$ p.a.? Round your answer to 2 decimal places.\n\nRead the question carefully and write down the given information:\n\n• $$A = \\text{R}\\,\\text{24}$$\n• $$P = ?$$\n• $$i = \\frac{11}{100}$$\n• $$n = \\text{5}$$\n\nWe use the compound interest formula and make $$P$$ the subject: \\begin{align} A &= P\\left(1+i\\right)^n \\\\ P &= \\frac{A}{\\left(1+i\\right)^n} \\\\ &= \\frac{\\text{24}}{\\left(\\text{1} + \\frac{\\text{11}}{\\text{100}}\\right)^{\\text{5}}} \\\\ &= \\text{R}\\,\\text{14,24} \\end{align}\n\nIf the average rate of inflation for the past few years was $$\\text{7,3}\\%$$ p.a. and your water and electricity account is $$\\text{R}\\,\\text{1 425}$$ on average, what would you expect to pay in $$\\text{6}$$ years time?\n\n\\begin{align} A & = ? \\\\ P & = \\text{1 425} \\\\ i & = \\text{0,073} \\\\ n & = 6 \\\\\\\\ A & = P(1 + i)^{n} \\\\ A & = \\text{1 425}(1 + \\text{0,073})^{6} \\\\ A & = \\text{R}\\,\\text{2 174,77} \\end{align}\n\nThe price of popcorn and a cooldrink at the movies is now $$\\text{R}\\,\\text{60}$$. If the average rate of inflation is $$\\text{9,2}\\%$$ p.a. what was the price of popcorn and cooldrink $$\\text{5}$$ years ago?\n\n\\begin{align} A & = \\text{R}\\,\\text{60}\\\\ P & = ?\\\\ i & = \\text{0,092}\\\\ n & = 5\\\\ \\\\ A & = P(1 + i)^{n}\\\\ \\\\ 60 &= P(1 + \\text{0,092})^{5}\\\\ \\frac{60}{(\\text{1,092})^5} & = P \\\\ P &= \\text{R}\\,\\text{38,64} \\end{align}\n\n### Population growth (EMA6T)\n\nFamily trees increase exponentially as every person born has the ability to start another family. For this reason we calculate population growth using the compound interest formula.\n\n## Worked example 11: Population growth\n\nIf the current population of Johannesburg is $$\\text{3 888 180}$$, and the average rate of population growth in South Africa is $$\\text{2,1}\\%$$ p.a., what can city planners expect the population of Johannesburg to be in $$\\text{10}$$ years?\n\n### Write down the known variables\n\n\\begin{align} P & = \\text{3 888 180} \\\\ i & = \\text{0,021} \\\\ n & = 10 \\end{align}\n\n### Write down the formula\n\n$A = P{\\left(1 + i\\right)}^{n}$\n\n### Substitute the values\n\n\\begin{align} A & = \\text{3 888 180}{\\left(1 + \\text{0,021}\\right)}^{10} \\\\ & = \\text{4 786 343} \\end{align}\n\nCity planners can expect Johannesburg's population to be $$\\text{4 786 343}$$ in ten years time.\n\ntemp text\nTextbook Exercise 9.5\n\nThe current population of Durban is $$\\text{3 879 090}$$ and the average rate of population growth in South Africa is $$\\text{1,1}\\%$$ p.a.\n\nWhat can city planners expect the population of Durban to be in $$\\text{6}$$ years time? Round your answer to the nearest integer.\n\nRead the question carefully and write down the given information:\n\n• $$A = ?$$\n• $$P = \\text{3 879 090}$$\n• $$i = \\frac{1.1}{100}$$\n• $$n = \\text{6}$$\n\nWe use the following formula to determine the expected population for Durban: \\begin{align} A &= P\\left(1+i\\right)^n\\\\ &= \\text{3 879 090} \\left(\\text{1} + \\frac{\\text{1,1}}{\\text{100}}\\right)^{\\text{6}} \\\\ &= \\text{4 142 255} \\end{align}\n\nThe current population of Polokwane is $$\\text{3 878 970}$$ and the average rate of population growth in South Africa is $$\\text{0,7}\\%$$ p.a.\n\nWhat can city planners expect the population of Polokwane to be in $$\\text{12}$$ years time? Round your answer to the nearest integer.\n\nRead the question carefully and write down the given information:\n\n• $$A = ?$$\n• $$P = \\text{3 878 970}$$\n• $$i = \\frac{0.7}{100}$$\n• $$n = \\text{12}$$\n\nWe use the following formula to determine the expected population for Polokwane: \\begin{align} A &= P\\left(1+i\\right)^n\\\\ &= \\text{3 878 970} \\left(\\text{1} + \\frac{\\text{0,7}}{\\text{100}}\\right)^{\\text{12}} \\\\ &= \\text{4 217 645} \\end{align}\n\nA small town in Ohio, USA is experiencing a huge increase in births. If the average growth rate of the population is $$\\text{16}\\%$$ p.a., how many babies will be born to the $$\\text{1 600}$$ residents in the next $$\\text{2}$$ years?\n\n\\begin{align} A & = ? \\\\ P & = \\text{1 600} \\\\ i & = \\text{0,16} \\\\ n & = 2 \\\\\\\\ A & = P(1 + i)^{n} \\\\ A &= \\text{1 600}(1 + \\text{0,16})^{2}\\\\ A & = \\text{2 152,96}\\\\ \\text{2 153} - \\text{1 600} & = 553 \\end{align}\n\nThere will be roughly $$\\text{553}$$ babies born in the next two years." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77006716,"math_prob":0.9999014,"size":3311,"snap":"2021-31-2021-39","text_gpt3_token_len":1223,"char_repetition_ratio":0.21983671,"word_repetition_ratio":0.19646366,"special_character_ratio":0.45303532,"punctuation_ratio":0.12481645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T00:07:56Z\",\"WARC-Record-ID\":\"<urn:uuid:beb8f2cd-eac7-4a22-bd04-f1d1014fd578>\",\"Content-Length\":\"88639\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92c00725-a1bd-4698-b57e-01e9d27e8a37>\",\"WARC-Concurrent-To\":\"<urn:uuid:b51b99fc-8d97-4cd2-8fdb-1ccb3787c016>\",\"WARC-IP-Address\":\"104.26.8.190\",\"WARC-Target-URI\":\"https://intl.siyavula.com/read/maths/grade-10/finance-and-growth/09-finance-and-growth-03\",\"WARC-Payload-Digest\":\"sha1:WZK36P7NJCCNO6SQTACEOCA6V4PF63XX\",\"WARC-Block-Digest\":\"sha1:MG4XIRRA4JOWWLD5NGCCT22PTTDCJ673\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057787.63_warc_CC-MAIN-20210925232725-20210926022725-00613.warc.gz\"}"}
https://mathoverflow.net/questions/148925/fantastic-properties-of-z-2z/148942	[ "Fantastic properties of Z/2Z\n\nRecently I gave a lecture to master's students about some nice properties of the group with two elements $\\mathbb{Z}/2\\mathbb{Z}$. Typically, I wanted to present simple, natural situations where the only group satisfying the given constraints is $\\mathbb{Z}/2\\mathbb{Z}$ (also $\\mathbb{Z}/2\\mathbb{Z}$ as a ring or as a field could qualify, but I'd prefer to stick to the group if possible). Here are some examples of theorems that I proved to the students :\n\n1. Let $G$ be a nontrivial group with trivial automorphism group. Then $G$ is isomorphic to $\\mathbb{Z}/2\\mathbb{Z}$.\n2. Let $G$ be a nontrivial quotient of the symmetric group on $n>4$ letters (nontrivial meaning here different from 1 and the symmetric group itself). Then $G$ is isomorphic to $\\mathbb{Z}/2\\mathbb{Z}$.\n3. Let $k$ be an algebraically closed field and let $k_0$ be a subfield such that $k/k_0$ is finite. Then $k/k_0$ is Galois and $G=\\text{Gal}(k/k_0)$ is isomorphic to $\\mathbb{Z}/2\\mathbb{Z}$. (Moreover $k$ has characteristic $0$ and $k=k_0(i)$ where $i^2=-1$.) This is a theorem of Emil Artin and I actually did not prove it because my students did not have enough background in field theory.\n4. Let $k$ be a field with the following property: there exists a $k$-vector space $E$ of finite dimension $n>1$ and an isomorphism $E\\simeq E^$ between the space and its linear dual which does not depend on the choice of a basis, i.e. is invariant under $\\text{GL}(E)$. Then $k=\\mathbb{Z}/2\\mathbb{Z}$, $n=2$ and the isomorphism $E\\simeq E^$ corresponds to the nondegenerate bilinear form given by the determinant.\n\nI am looking for some more fantastic apparitions of $\\mathbb{Z}/2\\mathbb{Z}$. Do you know some?\n\n• Why do you use ${\\bf Z}/2{\\bf Z} =\\{\\{2k+1\\}_{k \\in {\\bf Z}},\\{2k\\}_{k \\in {\\bf Z}}\\}$ and not $G = \\{-1,+1\\}$ ? – Patrick I-Z Nov 14 '13 at 22:15\n• @Patrick: You mean, in order to focus on the group rather than the field? You're right. But as my fourth example shows, I still have some interest for the field $\\mathbb{Z}/2\\mathbb{Z}$, which I certainly write $\\mathbb{F}_2$ in this context, and that's maybe why I wrote things the way I did. – Matthieu Romagny Nov 14 '13 at 22:20\n• For the symmetric group $S_n$, the outer automorphism group $\\textrm{Out}(S_n)=\\textrm{Aut}(S_n) /\\textrm{Inn}(S_n)$ is trivial (i.e. any automorphism is an inner one) unless $n=6$. In fact, $\\textrm{Out}(S_6)=\\mathbf{Z}/2 \\mathbf{Z}$. – Francesco Polizzi Nov 14 '13 at 23:00\n• @MatthieuRomagny: For a proof that without AC you can have such spaces, see this math.SE post. If you have a vector space that has a basis (and hence dimension) with more than one element, then swapping two basis vectors gives you a nontrivial automorphism. If you know that such a basis exists, then you don't need AC to define this automorphism; you need AC to guarantee the existence of bases. However, I believe that existence of nontrivial automorphisms does not imply existence of bases. – Arturo Magidin Nov 15 '13 at 18:32\n• youtube.com/watch?v=UTby_e4-Rhg – Noam D. Elkies Jul 3 '14 at 2:19\n\nA nice theorem is: $\\{\\pm 1\\}$ is the only group that can act freely on a sphere of even dimension. In contrast: There are infinitely many groups acting freely on every odd-dimensional sphere.\n\n• I would say that in the examples above it's the objects that are nice, not the group of two elements. – Lev Borisov Nov 14 '13 at 23:48\n• This is very nice; do you have a reference? – Matthieu Romagny Nov 15 '13 at 10:55\n• Do you mean: the only finite group? – Matthieu Romagny Nov 15 '13 at 11:02\n• @MatthieuRomagny: No, the theorem holds for arbitrary groups. A can be found in Hatcher's book on algebraic topology. It is very easy, once you have the machinery: the mapping degree induces a group morphism $d: G\\to \\{\\pm 1\\}$. Since a fixed point free map $S^n \\to S^n$ has degree $(-1)^{n+1}$ and $n$ is even, the kernel of $d$ is trivial and hence $G\\leq\\{\\pm 1\\}$. – Johannes Hahn Nov 15 '13 at 14:12\n• @JohannesHahn Very nice answer. What other compact manifold $M$ satisfies this property:$Z_{2}$ is the only group with free action on $M$? – Ali Taghavi Nov 9 '14 at 15:39\n\nThis is more of a joke than a serious example. Let $K$ be a field, $K_+$ its additive group, and $K_$ its multiplicative group. Thus $\\mathbb{R}_\\cong \\mathbb{R}_+\\times (\\mathbb{Z}/2\\mathbb{Z})$. What fields have the \"opposite\" property, that is, $K_+\\cong K_\\times (\\mathbb{Z}/2\\mathbb{Z})$? Answer: only $\\mathbb{Z}/2\\mathbb{Z}$.\n\n• Which fields $K$ have $K_{} \\simeq K_{+} \\times (\\mathbb{Z}/2\\mathbb{Z})$? – Sam Hopkins Nov 15 '13 at 2:27\n• Sam: for starters, $K$ is an exponential field. Take $i$ to be the inclusion of $K_+$ into the product; then $E = \\iota \\circ i_{K_+}$ (where $\\iota$ is the above isomorphism) is an exponential function. According to wikipedia $E$ has trivial image if the characteristic of $K$ is nonzero. $E$ is also injective in this case, but $K_+$ can't be trivial. Therefore $K$ has characteristic zero. Can anyone take it farther than that? – Ibrahim Tencer Nov 15 '13 at 5:50\n• I have asked this question separately: mathoverflow.net/questions/148949 – Sam Hopkins Nov 15 '13 at 6:10\n\nIt is the only non-trivial group whose free square ($GG$) satisfies a non-trivial identity (or is solvable, or is amenable...)\n\nEdit (Nov 9, 2014), suggested by Sam Nead\n\n... or is virtually cyclic, or is two-ended, or contains no nonabelian free subgroups...\n\n• Let me try, the identity $x^2=e$ ? – Denis Serre Nov 3 '14 at 7:40\n• @Denis, of course not. The infinite dihedral group $Z/2ZZ/2Z$ has elements of infinite order. – Anton Klyachko Nov 3 '14 at 7:46\n• There are several nice identities - the simplest after a few moments thought is $[x^2, y^2]$. – Sam Nead Nov 9 '14 at 10:51\n• @Sam, I bet your single identity forms a basis of identities of the group. (This means that all other identities are consequences of this one.) – Anton Klyachko Nov 9 '14 at 11:12\n• Oh, that looks nice - I'll have to think about that. By the way, I sort of want to edit your answer: $Z_2 * Z_2$ is the only non-trivial free product that is two-ended (thus is virtually cyclic) while all others have infinitely many ends (so contain rank two free groups, etc). – Sam Nead Nov 9 '14 at 13:44\n\nIt is the only finite group with exactly two conjugacy classes and it is the only non-trivial group with trivial automorphism group.\n\n• That's interesting; I managed to extend this property to all groups whose elements have finite order, and when nontrivial elements have infinite order there are counterexamples built by Higman, Neumann and Neumann in the paper Embedding Theorems for Groups in J. London Math. Soc. 1949. – Matthieu Romagny Nov 17 '13 at 17:03\n\n(In the density model) a random group is either $\\mathbb{Z} / 2\\mathbb{Z}$ or the trivial group. (I learnt this from Danny Calegari but I believe this is origionally due to Gromov.)\n\nMore precisely, fix $k \\geq 2$ and $D \\geq 1/2$. For each $n$ let $X_n$ be the set of reduced words in $x_1^{\\pm 1}, \\ldots, x_k^{\\pm 1}$.\n\nNow consider the group: $$G_n = \\langle x_1, \\ldots, x_k \| r_1, \\ldots, r_l \\rangle$$ where each $r_i$ is chosen randomly (independently) from $X_n$ and $l = \|X_n\|^{D} \\approx (2k - 1)^{nD}$. Then $\\mathbb{P}(G_n \\cong \\mathbb{Z / 2Z} \\textrm{ or } \\mathbb{1}) \\to 1$ as $n \\to \\infty$.\n\nThe proof of this is quite slick and relies on the \"Random Pigeon Hole Principle\": That because of the number of pair of relators are so large you are likely to get a number of almost equal relators, that differ at a single letter, which kills off a generator.\n\n• That is terrific! Do you know a reference where I can read more on this? – Matthieu Romagny Nov 4 '14 at 23:57\n• So I guess that the correct reference should be Gromov's \"Asymptotic invariants of infinite groups\" but it is ~300 pages and I haven't read it. Danny gave a talk at Cornell and in the first ~15 mins he covers the density model and gives a sketch of this result. It's available here: cornell.edu/video/danny-calegari-random-groups-diamonds-glass – Mark Bell Nov 5 '14 at 7:50\n\nJust a comment. You are probably looking for \"fantastic properties\" stated in terms of group theory, \"if an (abstract) group has such and such properties then it is $Z/2Z$\". However various representations of this group also have \"fantastic properties\". I mean first of all the group of automorphisms $C/R$ which consists of complex conjugation and identity. Several important subjects, like \"real algebraic geometry\" are based on these properties. Once I was strongly tempted to call my paper \"Some applications of representation theory of the group of 2 elements\", but my co-author convinced me that this title would be too radical.\n\nThe paper I mentioned is: Wronski map and Grassmannians of real codimension 2 subspaces, Computational Methods and Function Theory, 1 (2001) 1-25.\n\nAn example from differential geometry: If M is a compact, even-dimensional Riemannian manifold of positive sectional curvature, then its fundamental group is either 1 or Z/2Z.\n\nequivalently: a group acting freely on a compact, even-dimensional Riemannian manifold of positive sectional curvature is 1 or Z/2Z.\n\nThis is a variant of Synge's Theorem.\n\nThe automorphism group of the category of categories is $\\mathbb{Z}/2$.\n\nThat is, the group of invertible functors $F: \\mathrm{Cat} \\to \\mathrm{Cat}$ is $\\mathbb{Z}/2$.\n\n• Do you have a reference for this? – Ivan Di Liberti Aug 6 '18 at 23:14\n• Sorry, I can't find one now. It's pretty easy to prove. – John Baez Aug 8 '18 at 3:39\n• In higher category theory there are results as described here: golem.ph.utexas.edu/category/2011/11/… – David Corfield Aug 8 '18 at 8:36\n• Can you tell what is the generator of this group? If you're thinking about the functor $F$ that takes a category to its opposite, then I'm not sure it's a functor $F:\\mathrm{Cat}\\to \\mathrm{Cat}$. The problem is that $F$ maps $\\mathrm{Fun}(C,D)$ into $\\mathrm{Fun}(C^\\circ,D^\\circ)^\\circ$ rather than into $\\mathrm{Fun}(C^\\circ,D^\\circ)$. – Matthieu Romagny Sep 28 '18 at 13:09\n• Ah, maybe you mean $\\mathrm{Cat}$ as a 1-category? In this case $\\mathrm{Fun}(C,D)$ is just a set and the problem does not arise. – Matthieu Romagny Sep 28 '18 at 19:10\n\n$Z/2Z$ is abelian group which is canonically isomorphic to its dual abelian group. (Since dual group is again $Z/2Z$ and there is only one automorphism of $Z/2Z$).\n\nActually it is not the only group with this property, and another one is again related to $Z/2Z$. It is the group $Z/2Z \\oplus Z/2Z$: the dual group is group of characters, the kernel of each character contains two elements - identity and another one , so we can set a bijection: character <-> non-identity element in the kernel. This example is the same as item 4, in the original question.\n\nIsn't it remarkable that the fundamental group of the special orthogonal group $SO_n$ is $\\mathbb Z/2\\mathbb Z$ for $n\\ge3$ ?\n\n• Why is that remarkable? – Todd Trimble Nov 3 '14 at 7:52\n\nThe largest group which has embeddings into every nonabelian finite simple group is the Cartesian square of this group.\n\nLet $G$ be a group. The holomorph $Hol(G) = G \\rtimes Aut(G)$ can be regarded as a subgroup of the symmetric group $S_{\|G\|}$, by considering the functions $f: G \\to G$ sending $x \\in G$ to $x^{\\alpha} \\cdot g$, for $g \\in G$ and $\\alpha \\in Aut(G)$.\nThis is never a self-normalizing subgroup of $S_{\|G\|}$ when $G$ is nonabelian, because any anti-automorphism of $G$, including $x \\to x^{-1}$, normalizes it but is not in it. So $N_{ S_{ \|G\|} } (Hol(G))/Hol(G)$ always has a subgroup of order 2 in this case.\n\nThe only nontrivial groups lacking proper subgroups are the cyclic groups of prime order. Among these, only a cyclic group of order 2 can be isomorphic to the unique minimal subgroup of two nonisomorphic finite groups of the same order (a cyclic group and a quaternion group, when the order is a power of 2).\n\nLet $G$ be a nonabelian group in which all subgroups are normal. Then $G$ is isomorphic to the Cartesian product of a abelian torsion group in which all elements have odd order, the quaternion group $Q_{8}$, and a group of exponent 2 (by the Axiom of Choice, this last factor is a vector space over the field $\\mathbb{Z}/(2)$, as mentioned before).\n\nNotice how the group of exponent 2 above did not need to be required to be abelian? If $G$ is a group and all cyclic subgroups of $G$ are of order 2, it immediately follows that $G$ is abelian.\n\n• Instead of \"by the Axiom of Choice, this last factor is a vector space over the field Z/(2), as mentioned before\" I would rather write : \"this last factor is a vector space over the field Z/(2), and by the Axiom of Choice it has a basis, as mentioned before\". – Matthieu Romagny Aug 28 '18 at 13:09\n\nRH holds if and only if the group of isometries of the complex plane that preserve globally the multiset of non-trivial zeroes of the Riemann Zeta function is isomorphic to $Z/2Z$ (otherwise, it would be isomorphic to $Z/2Z\\times Z/2Z$).\n\n• Can you elaborate on this example or maybe give some reference where this result is explained or implied? – bonif Jan 8 '18 at 11:00\n• From the functional equation of $\\zeta$, a hypothetical non trivial zero off the critical line gives rise to a second one which is its image under the map $s\\mapsto 1-\\bar{s}$. This map coincides with identity for zeroes on this line. As the Dirichlet coefficients of $\\zeta$ are real, the map $s\\mapsto\\bar{s}$ maps a non trivial zero of zeta to a non trivial zero of $\\zeta$. Hence the two maps $s\\mapsto s$ and $s\\mapsto\\bar{s}$ are the symmetries of the multiset of non trivial zeroes of $\\zeta$. Adding a hypothetical zero off the line gives rise to the additional... – Sylvain JULIEN Jan 8 '18 at 12:03\n• ...symmetry $s\\mapsto 1-\\bar{s}$ and hence generates a global symmetry group isomorphic to $\\Z/2\\Z\\times\\Z/2\\Z$. As I said, the absence of zero off the line forces the maps $s\\mapsto s$ and $s\\mapsto 1-\\bar{s}$ to coincide, reducing the global isometry group to a single copy of $\\Z/2\\Z$ . – Sylvain JULIEN Jan 8 '18 at 12:06\n• Has this been published somewhere? Coming from the theoretical physics camp I can see off the top of my head there are some possible immediate physical analogies with this. – bonif Jan 8 '18 at 12:33\n• I strongly doubt it. As far as I know, I'm probably the only person to consider RH might be true investigating the symmetries of $\\zeta$ or equivalently of the multiset of non trivial zeroes thereof. I'd be glad to be proven wrong though. – Sylvain JULIEN Jan 8 '18 at 13:47\n\nEdit after Emil's comment (so my answer is not really good then): It's a group (or any product of it) where addition and substraction seen as a binary (that is, $- = + \\circ (Id, i)$ where $i$ is the inverse map) actually coincides. Note that this also means that it's a group where $-$ is actually associative.\n\n• Every group of exponent 2 has this property. These are not just products of $\\mathbb Z/2\\mathbb Z$, you also have to close them under subgroups. Again, this doesn't make the two-element group special, as every nontrivial group of exponent 2 generates the class of all of them using products and subgroups. – Emil Jeřábek Nov 2 '14 at 12:39\n• @Emil, every group of exponent two is a direct product of several two-element groups (if you believe the Axiom of Choice). – Anton Klyachko Nov 3 '14 at 2:13\n• @AntonKlyachko Every group of exponent two is a direct sum of two-element groups. (Finite direct sums coincide with finite direct products, but infinite direct sums need not be infinite direct products.) – Todd Trimble Nov 3 '14 at 7:21\n• @Todd, I understand you but I prefer to use slightly different terminology. See this discussion in comments: mathoverflow.net/q/92972/24165 – Anton Klyachko Nov 3 '14 at 7:42\n• @AntonKlyachko Okay, thanks for clarifying. The boldfaced \"is\" in your response to Emil sounded to me like a correction, when in fact what he wrote was already correct according to his word usage (and resonates in other ways as well). – Todd Trimble Nov 3 '14 at 12:13\n\nThe central multiplicative action of $(\\mathbb{Z}/2\\mathbb{Z})^r$ on the cohomology ring $H^∗(\\mathfrak{X}_r(SU(2)))$ is trivial, where $\\mathfrak{X}_r(SU(2))$ is the character variety $SU(2)^r/SU(2)$. This ultimately allows for the computation of the Poincaré polynomial of the real moduli spaces $\\mathfrak{X}_r(SL(3,\\mathbb{R}))$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85462064,"math_prob":0.99292386,"size":1683,"snap":"2019-43-2019-47","text_gpt3_token_len":500,"char_repetition_ratio":0.16557474,"word_repetition_ratio":0.038910504,"special_character_ratio":0.2774807,"punctuation_ratio":0.0694864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99941146,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T08:01:49Z\",\"WARC-Record-ID\":\"<urn:uuid:f8d3743c-b94b-4a2a-8e53-fc751d3a2eb8>\",\"Content-Length\":\"236863\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42cbab71-f388-43d5-a7cd-8199d81c9e31>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9689457-44d6-418e-bf9e-0f9ec2fad0a3>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/148925/fantastic-properties-of-z-2z/148942\",\"WARC-Payload-Digest\":\"sha1:B6UR3RS4UGZRAROSAILUT7VIVFVSRWF3\",\"WARC-Block-Digest\":\"sha1:2HWN2TLKQOJWH5JVG3GS7L7ELDCB7AZW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986677964.40_warc_CC-MAIN-20191018055014-20191018082514-00456.warc.gz\"}"}
https://www.colorhexa.com/18f02e	[ "# #18f02e Color Information\n\nIn a RGB color space, hex #18f02e is composed of 9.4% red, 94.1% green and 18% blue. Whereas in a CMYK color space, it is composed of 90% cyan, 0% magenta, 80.8% yellow and 5.9% black. It has a hue angle of 126.1 degrees, a saturation of 87.8% and a lightness of 51.8%. #18f02e color hex could be obtained by blending #30ff5c with #00e100. Closest websafe color is: #00ff33.\n\n• R 9\n• G 94\n• B 18\nRGB color chart\n• C 90\n• M 0\n• Y 81\n• K 6\nCMYK color chart\n\n#18f02e color description : Vivid lime green.\n\n# #18f02e Color Conversion\n\nThe hexadecimal color #18f02e has RGB values of R:24, G:240, B:46 and CMYK values of C:0.9, M:0, Y:0.81, K:0.06. Its decimal value is 1634350.\n\nHex triplet RGB Decimal 18f02e `#18f02e` 24, 240, 46 `rgb(24,240,46)` 9.4, 94.1, 18 `rgb(9.4%,94.1%,18%)` 90, 0, 81, 6 126.1°, 87.8, 51.8 `hsl(126.1,87.8%,51.8%)` 126.1°, 90, 94.1 00ff33 `#00ff33`\nCIE-LAB 83.288, -80.031, 72.704 32.028, 62.708, 13 0.297, 0.582, 62.708 83.288, 108.124, 137.746 83.288, -77.095, 96.946 79.188, -66.385, 45.698 00011000, 11110000, 00101110\n\n# Color Schemes with #18f02e\n\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #f018da\n``#f018da` `rgb(240,24,218)``\nComplementary Color\n• #6ef018\n``#6ef018` `rgb(110,240,24)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #18f09a\n``#18f09a` `rgb(24,240,154)``\nAnalogous Color\n• #f0186e\n``#f0186e` `rgb(240,24,110)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #9a18f0\n``#9a18f0` `rgb(154,24,240)``\nSplit Complementary Color\n• #f02e18\n``#f02e18` `rgb(240,46,24)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #2e18f0\n``#2e18f0` `rgb(46,24,240)``\n• #daf018\n``#daf018` `rgb(218,240,24)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #2e18f0\n``#2e18f0` `rgb(46,24,240)``\n• #f018da\n``#f018da` `rgb(240,24,218)``\n• #0bb01c\n``#0bb01c` `rgb(11,176,28)``\n• #0dc820\n``#0dc820` `rgb(13,200,32)``\n• #0fe024\n``#0fe024` `rgb(15,224,36)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #30f244\n``#30f244` `rgb(48,242,68)``\n• #48f359\n``#48f359` `rgb(72,243,89)``\n• #60f56f\n``#60f56f` `rgb(96,245,111)``\nMonochromatic Color\n\n# Alternatives to #18f02e\n\nBelow, you can see some colors close to #18f02e. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #38f018\n``#38f018` `rgb(56,240,24)``\n• #26f018\n``#26f018` `rgb(38,240,24)``\n• #18f01c\n``#18f01c` `rgb(24,240,28)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #18f040\n``#18f040` `rgb(24,240,64)``\n• #18f052\n``#18f052` `rgb(24,240,82)``\n• #18f064\n``#18f064` `rgb(24,240,100)``\nSimilar Colors\n\n# #18f02e Preview\n\nThis text has a font color of #18f02e.\n\n``<span style=\"color:#18f02e;\">Text here</span>``\n#18f02e background color\n\nThis paragraph has a background color of #18f02e.\n\n``<p style=\"background-color:#18f02e;\">Content here</p>``\n#18f02e border color\n\nThis element has a border color of #18f02e.\n\n``<div style=\"border:1px solid #18f02e;\">Content here</div>``\nCSS codes\n``.text {color:#18f02e;}``\n``.background {background-color:#18f02e;}``\n``.border {border:1px solid #18f02e;}``\n\n# Shades and Tints of #18f02e\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010801 is the darkest color, while #f5fef6 is the lightest one.\n\n• #010801\n``#010801` `rgb(1,8,1)``\n• #021b04\n``#021b04` `rgb(2,27,4)``\n• #032d07\n``#032d07` `rgb(3,45,7)``\n• #04400a\n``#04400a` `rgb(4,64,10)``\n• #05520d\n``#05520d` `rgb(5,82,13)``\n• #076510\n``#076510` `rgb(7,101,16)``\n• #087713\n``#087713` `rgb(8,119,19)``\n• #098916\n``#098916` `rgb(9,137,22)``\n• #0a9c19\n``#0a9c19` `rgb(10,156,25)``\n• #0bae1c\n``#0bae1c` `rgb(11,174,28)``\n• #0dc11f\n``#0dc11f` `rgb(13,193,31)``\n• #0ed322\n``#0ed322` `rgb(14,211,34)``\n• #0fe525\n``#0fe525` `rgb(15,229,37)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #2af13f\n``#2af13f` `rgb(42,241,63)``\n• #3df24f\n``#3df24f` `rgb(61,242,79)``\n• #4ff460\n``#4ff460` `rgb(79,244,96)``\n• #62f571\n``#62f571` `rgb(98,245,113)``\n• #74f681\n``#74f681` `rgb(116,246,129)``\n• #87f792\n``#87f792` `rgb(135,247,146)``\n• #99f8a3\n``#99f8a3` `rgb(153,248,163)``\n• #abfab3\n``#abfab3` `rgb(171,250,179)``\n• #befbc4\n``#befbc4` `rgb(190,251,196)``\n• #d0fcd5\n``#d0fcd5` `rgb(208,252,213)``\n• #e3fde5\n``#e3fde5` `rgb(227,253,229)``\n• #f5fef6\n``#f5fef6` `rgb(245,254,246)``\nTint Color Variation\n\n# Tones of #18f02e\n\nA tone is produced by adding gray to any pure hue. In this case, #808881 is the less saturated color, while #0ff926 is the most saturated one.\n\n• #808881\n``#808881` `rgb(128,136,129)``\n• #779179\n``#779179` `rgb(119,145,121)``\n• #6d9b72\n``#6d9b72` `rgb(109,155,114)``\n• #64a46a\n``#64a46a` `rgb(100,164,106)``\n• #5aae63\n``#5aae63` `rgb(90,174,99)``\n• #51b75b\n``#51b75b` `rgb(81,183,91)``\n• #47c154\n``#47c154` `rgb(71,193,84)``\n• #3eca4c\n``#3eca4c` `rgb(62,202,76)``\n• #34d445\n``#34d445` `rgb(52,212,69)``\n• #2bdd3d\n``#2bdd3d` `rgb(43,221,61)``\n• #21e736\n``#21e736` `rgb(33,231,54)``\n• #18f02e\n``#18f02e` `rgb(24,240,46)``\n• #0ff926\n``#0ff926` `rgb(15,249,38)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #18f02e is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50598323,"math_prob":0.6940066,"size":3673,"snap":"2020-24-2020-29","text_gpt3_token_len":1642,"char_repetition_ratio":0.12591986,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5578546,"punctuation_ratio":0.23430493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98540115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-12T19:23:26Z\",\"WARC-Record-ID\":\"<urn:uuid:6849c422-e22a-4114-bd5f-ae5ffea3c723>\",\"Content-Length\":\"36257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddf29c44-5c45-4b04-a18d-6933e8c51a23>\",\"WARC-Concurrent-To\":\"<urn:uuid:4715bf4a-5808-4da9-afab-33975cd0828a>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/18f02e\",\"WARC-Payload-Digest\":\"sha1:ZIOFSOEQE7VIYKEWN72FMOKCVJEO42LO\",\"WARC-Block-Digest\":\"sha1:MZO5QURWROGCNLJV4DFUOVQZYYAXRRVA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657139167.74_warc_CC-MAIN-20200712175843-20200712205843-00070.warc.gz\"}"}
https://www.electronics-tutorial.net/Digital-CMOS-Design/CMOS-Inverter/Noise-Margin/	[ "Home > Digital CMOS Design > CMOS Inverter > Noise Margin\n\nCMOS-Inverter\n\nNoise Margin :\n\nIn digital integrated circuits, to minimize the noise it is necessary to keep \"0\" and \"1\" intervals broader. Hence noise margin is the measure of the sensitivity of a gate to noise and expressed by, NML (noise margin Low) and NMH (noise margin High).\n\nNML and NMH are defined as,\n\nNML = VIL  VOL and NMH = VOH  VIH\n\nIn order to define the terms VIL, VOL, VOH and VIH again consider the VTC of Inverter as shown in Figure below. The VOH is the maximum output voltage at which the output is \"logic high\". The VOL is the minimum output voltage at which the output is \"logic low\". The regions of acceptable high and low voltages are defined by VIH and VIL respectively. VIH and VIL represents the points where the gain dVoutdVin of VTC is equals to  1 as shown in above Figure. The region between VIH and VIL is called as the undefined region or transition width.", null, "Fig2-Noise-Margin\n\nFigure below shows the NMH and NML levels of two cascaded inverters. The noise margin shows the levels of noise when the gates are connected together. For the digital integrated circuits the noise margin is larger than '0' and ideally it is high.", null, "Fig2-Noise-Margin" ]	[ null, "https://www.electronics-tutorial.net/Digital-CMOS-Design/CMOS-Inverter/Noise-Margin/Fig1-Noise-Margin.png", null, "https://www.electronics-tutorial.net/Digital-CMOS-Design/CMOS-Inverter/Noise-Margin/Fig2-Noise-Margin.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90029603,"math_prob":0.9698782,"size":1165,"snap":"2023-40-2023-50","text_gpt3_token_len":284,"char_repetition_ratio":0.13953489,"word_repetition_ratio":0.03902439,"special_character_ratio":0.21630901,"punctuation_ratio":0.0745614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9677044,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T20:18:20Z\",\"WARC-Record-ID\":\"<urn:uuid:7511283d-7c53-44db-9a4b-11d8dd6a83ea>\",\"Content-Length\":\"109227\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab3f6669-e23a-44c4-a1c2-44ecaac481d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:078df1b1-d547-4393-bb43-8125fd8567ef>\",\"WARC-IP-Address\":\"172.67.134.138\",\"WARC-Target-URI\":\"https://www.electronics-tutorial.net/Digital-CMOS-Design/CMOS-Inverter/Noise-Margin/\",\"WARC-Payload-Digest\":\"sha1:3AMFHD7IYAXQ2YWVRQBJH2COXVYXDBXN\",\"WARC-Block-Digest\":\"sha1:TBYAUK2UXZYUYEHEIVDTQ4HNGYWBLPFA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510707.90_warc_CC-MAIN-20230930181852-20230930211852-00333.warc.gz\"}"}
https://lms.aotawhiti.school.nz/?q=node/208160	[ "# 2018 Term 4 HB Hawk Stage 6 Maths Group\n\n• Posted on: 10 October 2018\n• By: libbyboyd\n\nStage 6 maths group will attend workshops 2-3x per week and complete 3-4 Maths buddy sessions as follow up Must Dos..\n\nThis term we will be focusing on FRACTIONS.\nWe will be learning to order fractions, make and convert improper fractions, convert fractions to decimals and percentages, all using materials, imaging and number properties.\n1. Know fractions and percentages and decimals in everyday use.\n\nWe will also be practicing our time tables for speedy recall with regular speed tests each session.\n\nWe will continue to use and revise Number Strategies and Knowledge:\n- Use a range of multiplicative strategies when operating on whole numbers.\n- Understand addition and subtraction of fractions, decimals, and integers.\n- Apply simple linear proportions, including ordering fractions.\n- Know the relative size and place value structure of positive and negative integers and decimals to three decimals.\n\nLearning Area:\nTo Do List:\npractice my times tables\nhave a good attitude when learning new and maybe difficult concepts\nCome to my maths group and participate positively\nComplete must dos on Maths Buddy on time.\nLA Code:\nLevels:\n2\n3\nYear:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.759583,"math_prob":0.88462526,"size":943,"snap":"2019-13-2019-22","text_gpt3_token_len":183,"char_repetition_ratio":0.14483494,"word_repetition_ratio":0.0,"special_character_ratio":0.19724284,"punctuation_ratio":0.11377245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9611071,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T10:24:28Z\",\"WARC-Record-ID\":\"<urn:uuid:d601f57f-819c-4638-80c2-d844eb2abe2a>\",\"Content-Length\":\"15235\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2d1658e-7dba-4444-b023-5f3a24687dfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:2566c768-5db1-45de-90bb-ea6877c1b929>\",\"WARC-IP-Address\":\"111.69.22.43\",\"WARC-Target-URI\":\"https://lms.aotawhiti.school.nz/?q=node/208160\",\"WARC-Payload-Digest\":\"sha1:KK5PPQTB2DNYNFH7TFJ54ETVBULEQ6JK\",\"WARC-Block-Digest\":\"sha1:EXNSTDRFHKS4BBB23Y3N5COQWJAQWHC5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201953.19_warc_CC-MAIN-20190319093341-20190319115341-00516.warc.gz\"}"}