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https://www.iucr.org/news/newsletter/etc/legacy-articles?issue=2495&result_138864_result_page=9	[ "", null, "", null, "# Two-wavelength MAD phasing: in search of the optimal choice of wavelength\n\nGonzalez et al. [Acta Cryst. D55 (1999), 1449-1458] reported in their recent article that interpretable electron-density maps, similar in quality to those calculated with data collected at three wavelengths, could be obtained using only two wavelengths if the wavelengths were chosen so as to give a large contrast in the dispersive component of the scattering factor. Four different crystals, which contain iron, gold, iridium, or selenium atoms, were used in their study.\n\nThe multiwavelength anomalous dispersion (MAD) method exploits the structure-factor variation with wavelengths around the absorption edges of heavy atoms within the protein crystals. The variation consists of the real part (or the dispersive component, f') and the imaginary part (or the anomalous component, f''). The importance of using wavelengths that can provide the largest f' was explained as follows: “Firstly, the refinement of the anomalous scatterer positions and occupancies is highly dependent on the dispersive differences measured from centric reflections. Secondly, a MAD data collection is usually performed in a way which partially cancels out systematic errors in the structure factors at different wavelengths, leading to dispersive differences which are less affected by errors in the data.”\n\nTwo approaches have been used in calculating phase information from a MAD experiment. One approach uses an analytical method to solve the phase problem and calls for three or more wavelengths to optimize the results. The other approach treats MAD as a traditional isomorphous replacement phasing. The importance of selecting a large difference in the real part of the scattering factor in the two-wavelength experiment is interesting, as it resembles what is known from the single isomorphous replacement and anomalous scattering (SIRAS) experiment, i.e., phasing power increases with increased differences in the real part of the structure factors, which is achievable by the incorporation of heavier heavy atoms.", null, "(a) Detail of the 2.1 Å MAD map of cytochrome c553 calculated with phases from the three-wavelength data set, showing an a-helix and therefined model. (b) As (a), calculated with phases from a two-wavelength data set.\nMost structures determined by the MAD method thus farwere obtained from data collected at three or more wavelengths. Doing the MAD experiment using fewer data sets will reduce crystal exposure to X-rays and is a good option for structure determination using a high-flux synchrotron source or as an onbeamline map calculation whereby a two-wavelength map is calculated to evaluate the need, on the fly, for collecting data at more wavelengths. More interesting still, is the one-wavelength approach for on-the-fly map calculation using data from the veryfirst wavelength, where radiation damage is the least. Using a numerical method for resolving phase ambiguity, we have recently determined the structure of an Fe-containing protein (86kDa per asymmetric unit) using single-wavelength iron anomalous data collected in-house (P12.02.023 IUCr 99 program abstracts). For single-wavelength anomalous scattering data, collection at the absorption edge is preferable. However, the use of other wavelengths, in-house or at synchrotron, is an option provided that the measurements are accurately made.\n\nBi-Cheng Wang\nU. of Georgia, Athens, GA, USA" ]	[ null, "https://www.iucr.org/__data/assets/image/0008/136673/left_arrow.png", null, "https://www.iucr.org/__data/assets/image/0007/136672/right_arrow.png", null, "https://www.iucr.org/__data/assets/image/0004/2695/7_3_4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9319898,"math_prob":0.9465698,"size":2139,"snap":"2020-10-2020-16","text_gpt3_token_len":422,"char_repetition_ratio":0.12693208,"word_repetition_ratio":0.012698413,"special_character_ratio":0.18373072,"punctuation_ratio":0.101928376,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9533151,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-28T13:06:45Z\",\"WARC-Record-ID\":\"<urn:uuid:45df818d-8f9a-4580-9bf0-10680030a8f3>\",\"Content-Length\":\"113229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:159265af-0cda-4a83-a4d0-4b33abcb24e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d571887-d5c1-492c-8930-9024e2f0e744>\",\"WARC-IP-Address\":\"192.70.242.15\",\"WARC-Target-URI\":\"https://www.iucr.org/news/newsletter/etc/legacy-articles?issue=2495&result_138864_result_page=9\",\"WARC-Payload-Digest\":\"sha1:EWIFOLPD7LUP22JLX7EKYMQLPKVMST3C\",\"WARC-Block-Digest\":\"sha1:QXPPF6H7LDY7CGPY2NFQRB6SCV3VWIIW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875147154.70_warc_CC-MAIN-20200228104413-20200228134413-00438.warc.gz\"}"}
https://docs.gpytorch.ai/en/stable/_modules/gpytorch/settings.html	[ "# Source code for gpytorch.settings\n\n#!/usr/bin/env python3\n\nimport logging\nimport warnings\n\nimport torch\n\nclass _feature_flag:\nr\"\"\"Base class for feature flag settings with global scope.\nThe default is set via the _default class attribute.\n\"\"\"\n\n_default = False\n_state = None\n\n@classmethod\ndef is_default(cls):\nreturn cls._state is None\n\n@classmethod\ndef on(cls):\nif cls.is_default():\nreturn cls._default\nreturn cls._state\n\n@classmethod\ndef off(cls):\nreturn not cls.on()\n\n@classmethod\ndef _set_state(cls, state):\ncls._state = state\n\ndef __init__(self, state=True):\nself.prev = self.__class__._state\nself.state = state\n\ndef __enter__(self):\nself.__class__._set_state(self.state)\n\ndef __exit__(self, args):\nself.__class__._set_state(self.prev)\nreturn False\n\nclass _value_context:\n_global_value = None\n\n@classmethod\ndef value(cls):\nreturn cls._global_value\n\n@classmethod\ndef _set_value(cls, value):\ncls._global_value = value\n\ndef __init__(self, value):\nself._orig_value = self.__class__.value()\nself._instance_value = value\n\ndef __enter__(self,):\nself.__class__._set_value(self._instance_value)\n\ndef __exit__(self, args):\nself.__class__._set_value(self._orig_value)\nreturn False\n\nclass _dtype_value_context:\n_global_float_value = None\n_global_double_value = None\n_global_half_value = None\n\n@classmethod\ndef value(cls, dtype):\nif torch.is_tensor(dtype):\ndtype = dtype.dtype\nif dtype == torch.float:\nreturn cls._global_float_value\nelif dtype == torch.double:\nreturn cls._global_double_value\nelif dtype == torch.half:\nreturn cls._global_half_value\nelse:\nraise RuntimeError(f\"Unsupported dtype for {cls.__name__}.\")\n\n@classmethod\ndef _set_value(cls, float_value, double_value, half_value):\nif float_value is not None:\ncls._global_float_value = float_value\nif double_value is not None:\ncls._global_double_value = double_value\nif half_value is not None:\ncls._global_half_value = half_value\n\ndef __init__(self, float=None, double=None, half=None):\nself._orig_float_value = self.__class__.value()\nself._instance_float_value = float\nself._orig_double_value = self.__class__.value()\nself._instance_double_value = double\nself._orig_half_value = self.__class__.value()\nself._instance_half_value = half\n\ndef __enter__(self,):\nself.__class__._set_value(\nself._instance_float_value, self._instance_double_value, self._instance_half_value,\n)\n\ndef __exit__(self, args):\nself.__class__._set_value(self._orig_float_value, self._orig_double_value, self._orig_half_value)\nreturn False\n\nclass _fast_covar_root_decomposition(_feature_flag):\nr\"\"\"\nThis feature flag controls how matrix root decompositions (:math:K = L L^\\top) are computed\n(e.g. for sampling, computing caches, etc.).\n\nIf set to True, covariance matrices :math:K are decomposed with low-rank approximations :math:L L^\\top,\n(:math:L \\in \\mathbb R^{n \\times k}) using the Lanczos algorithm.\nThis is faster for large matrices and exploits structure in the covariance matrix if applicable.\n\nIf set to False, covariance matrices :math:K are decomposed using the Cholesky decomposition.\n\n.. warning ::\n\nSetting this to False will compute a complete Cholesky decomposition of covariance matrices.\nThis may be infeasible for GPs with structure covariance matrices.\n\nSee also: :class:gpytorch.settings.max_root_decomposition_size (to control the\nsize of the low rank decomposition used).\n\"\"\"\n\n_default = True\n\nclass _fast_log_prob(_feature_flag):\nr\"\"\"\nThis feature flag controls how to compute the marginal log likelihood of exact GPs\nand the log probability of multivariate normal distributions.\n\nIf set to True, log_prob is computed using a modified conjugate gradients algorithm (as\ndescribed in GPyTorch: Blackbox Matrix-Matrix Gaussian Process Inference with GPU Acceleration_.\nThis is a stochastic computation, but it is much faster for large matrices\nand exploits structure in the covariance matrix if applicable.\n\nIf set to False, log_prob is computed using the Cholesky decomposition.\n\n.. warning ::\n\nSetting this to False will compute a complete Cholesky decomposition of covariance matrices.\nThis may be infeasible for GPs with structure covariance matrices.\n\nSee also: :class:gpytorch.settings.num_trace_samples (to control the\nstochasticity of the fast log_prob estimates).\n\n.. _GPyTorch: Blackbox Matrix-Matrix Gaussian Process Inference with GPU Acceleration:\nhttps://arxiv.org/pdf/1809.11165.pdf\n\"\"\"\n\n_default = True\n\nclass _fast_solves(_feature_flag):\nr\"\"\"\nThis feature flag controls how to compute solves with positive definite matrices.\nIf set to True, solves are computed using preconditioned conjugate gradients.\nIf set to False, log_prob is computed using the Cholesky decomposition.\n\n.. warning ::\n\nSetting this to False will compute a complete Cholesky decomposition of covariance matrices.\nThis may be infeasible for GPs with structure covariance matrices.\n\"\"\"\n\n_default = True\n\n[docs]class skip_posterior_variances(_feature_flag):\n\"\"\"\nWhether or not to skip the posterior covariance matrix when doing an ExactGP\nforward pass. If this is on, the returned gpytorch MultivariateNormal will have a\nZeroLazyTensor as its covariance matrix. This allows gpytorch to not compute\nthe covariance matrix when it is not needed, speeding up computations.\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n[docs]class detach_test_caches(_feature_flag):\n\"\"\"\nWhether or not to detach caches computed for making predictions. In most cases, you will want this,\nas this will speed up derivative computations of the predictions with respect to test inputs. However,\nif you also need derivatives with respect to training inputs (e.g., because you have fantasy observations),\nthen you must disable this.\n\n(Default: True)\n\"\"\"\n\n_default = True\n\n[docs]class deterministic_probes(_feature_flag):\n\"\"\"\nWhether or not to resample probe vectors every iteration of training. If True, we use the same set of probe vectors\nfor computing log determinants each iteration. This introduces small amounts of bias in to the MLL, but allows us\nto compute a deterministic estimate of it which makes optimizers like L-BFGS more viable choices.\n\nNOTE: Currently, probe vectors are cached in a global scope. Therefore, this setting cannot be used\nif multiple independent GP models are being trained in the same context (i.e., it works fine with a single GP model)\n\n(Default: False)\n\"\"\"\n\nprobe_vectors = None\n\n@classmethod\ndef _set_state(cls, state):\nsuper()._set_state(state)\ncls.probe_vectors = None\n\n[docs]class debug(_feature_flag):\n\"\"\"\nWhether or not to perform \"safety\" checks on the supplied data.\n(For example, that the correct training data is supplied in Exact GP training mode)\nPros: fewer data checks, fewer warning messages\nCons: possibility of supplying incorrect data, model accidentially in wrong mode\n\n(Default: True)\n\"\"\"\n\n_default = True\n\n[docs]class fast_pred_var(_feature_flag):\n\"\"\"\nFast predictive variances using Lanczos Variance Estimates (LOVE)\nUse this for improved performance when computing predictive variances.\n\nAs described in the paper:\n\nConstant-Time Predictive Distributions for Gaussian Processes_.\n\nSee also: :class:gpytorch.settings.max_root_decomposition_size (to control the\nsize of the low rank decomposition used for variance estimates).\n\n(Default: False)\n\n.. _Constant-Time Predictive Distributions for Gaussian Processes:\nhttps://arxiv.org/pdf/1803.06058.pdf\n\"\"\"\n\n_num_probe_vectors = 1\n\n@classmethod\ndef num_probe_vectors(cls):\nreturn cls._num_probe_vectors\n\n@classmethod\ndef _set_num_probe_vectors(cls, value):\ncls._num_probe_vectors = value\n\ndef __init__(self, state=True, num_probe_vectors=1):\nself.orig_value = self.__class__.num_probe_vectors()\nself.value = num_probe_vectors\nsuper().__init__(state)\n\ndef __enter__(self):\nself.__class__._set_num_probe_vectors(self.value)\nsuper().__enter__()\n\ndef __exit__(self, args):\nself.__class__._set_num_probe_vectors(self.orig_value)\nreturn super().__exit__()\n\n[docs]class fast_pred_samples(_feature_flag):\n\"\"\"\nFast predictive samples using Lanczos Variance Estimates (LOVE).\nUse this for improved performance when sampling from a predictive posterior matrix.\n\nAs described in the paper:\n\nConstant-Time Predictive Distributions for Gaussian Processes_.\n\nSee also: :class:gpytorch.settings.max_root_decomposition_size (to control the\nsize of the low rank decomposition used for samples).\n\n(Default: False)\n\n.. _Constant-Time Predictive Distributions for Gaussian Processes:\nhttps://arxiv.org/pdf/1803.06058.pdf\n\"\"\"\n\n_default = False\n\n[docs]class fast_computations:\nr\"\"\"\nThis feature flag controls whether or not to use fast approximations to various mathematical\nfunctions used in GP inference.\nThe functions that can be controlled are:\n\n* :attr:covar_root_decomposition\nThis feature flag controls how matrix root decompositions\n(:math:K = L L^\\top) are computed (e.g. for sampling, computing caches, etc.).\n\n* If set to True,\ncovariance matrices :math:K are decomposed with low-rank approximations :math:L L^\\top,\n(:math:L \\in \\mathbb R^{n \\times k}) using the Lanczos algorithm.\nThis is faster for large matrices and exploits structure in the covariance matrix if applicable.\n\n* If set to False,\ncovariance matrices :math:K are decomposed using the Cholesky decomposition.\n\n* :attr:log_prob\nThis feature flag controls how GPyTorch computes the marginal log likelihood for exact GPs\nand log_prob for multivariate normal distributions\n\n* If set to True,\nlog_prob is computed using a modified conjugate gradients algorithm (as\ndescribed in GPyTorch Blackbox Matrix-Matrix Gaussian Process Inference with GPU Acceleration_.\nThis is a stochastic computation, but it is much faster for large matrices\nand exploits structure in the covariance matrix if applicable.\n\n* If set to False,\nlog_prob is computed using the Cholesky decomposition.\n\n* :attr:fast_solves\nThis feature flag controls how GPyTorch computes the solves of positive-definite matrices.\n\n* If set to True,\nSolves are computed with preconditioned conjugate gradients.\n\n* If set to False,\nSolves are computed using the Cholesky decomposition.\n\n.. warning ::\n\nSetting this to False will compute a complete Cholesky decomposition of covariance matrices.\nThis may be infeasible for GPs with structure covariance matrices.\n\nBy default, approximations are used for all of these functions (except for solves).\nSetting any of them to False will use exact computations instead.\n\n* :class:gpytorch.settings.max_root_decomposition_size\n(to control the size of the low rank decomposition used)\n* :class:gpytorch.settings.num_trace_samples\n(to control the stochasticity of the fast log_prob estimates)\n\n.. _GPyTorch Blackbox Matrix-Matrix Gaussian Process Inference with GPU Acceleration:\nhttps://arxiv.org/pdf/1809.11165.pdf\n\"\"\"\ncovar_root_decomposition = _fast_covar_root_decomposition\nlog_prob = _fast_log_prob\nsolves = _fast_solves\n\ndef __init__(self, covar_root_decomposition=True, log_prob=True, solves=True):\nself.covar_root_decomposition = _fast_covar_root_decomposition(covar_root_decomposition)\nself.log_prob = _fast_log_prob(log_prob)\nself.solves = _fast_solves(solves)\n\ndef __enter__(self):\nself.covar_root_decomposition.__enter__()\nself.log_prob.__enter__()\nself.solves.__enter__()\n\ndef __exit__(self, args):\nself.covar_root_decomposition.__exit__()\nself.log_prob.__exit__()\nself.solves.__exit__()\nreturn False\n\n[docs]class lazily_evaluate_kernels(_feature_flag):\n\"\"\"\nLazily compute the entries of covariance matrices (set to True by default).\nThis can result in memory and speed savings - if say cross covariance terms are not needed\nor if you only need to compute variances (not covariances).\n\nIf set to False, gpytorch will always compute the entire covariance matrix between\ntraining and test data.\n\n(Default: True)\n\"\"\"\n\n_default = True\n\n[docs]class max_eager_kernel_size(_value_context):\n\"\"\"\nIf the joint train/test covariance matrix is less than this size, then we will avoid as\nmuch lazy evaluation of the kernel as possible.\n\n(Default: 512)\n\"\"\"\n\n_global_value = 512\n\n[docs]class max_cg_iterations(_value_context):\n\"\"\"\nThe maximum number of conjugate gradient iterations to perform (when computing\nmatrix solves). A higher value rarely results in more accurate solves -- instead, lower the CG tolerance.\n\n(Default: 1000)\n\"\"\"\n\n_global_value = 1000\n\n[docs]class min_variance(_dtype_value_context):\n\"\"\"\nThe minimum variance that can be returned from :obj:~gpytorch.distributions.MultivariateNormal#variance.\nIf variances are smaller than this, they are rounded up and a warning is raised.\n\n- Default for float: 1e-6\n- Default for double: 1e-10\n- Default for half: 1e-3\n\"\"\"\n\n_global_float_value = 1e-6\n_global_double_value = 1e-10\n_global_half_value = 1e-3\n\n[docs]class cholesky_jitter(_dtype_value_context):\n\"\"\"\nThe jitter value passed to psd_safe_cholesky when using cholesky solves.\n\n- Default for float: 1e-6\n- Default for double: 1e-8\n\"\"\"\n\n_global_float_value = 1e-6\n_global_double_value = 1e-8\n\n@classmethod\ndef value(cls, dtype=None):\nif dtype is None:\n# Deprecated in 1.4: remove in 1.5\nwarnings.warn(\n\"cholesky_jitter is now a _dtype_value_context and should be called with a dtype argument\",\nDeprecationWarning,\n)\nreturn cls._global_float_value\nreturn super().value(dtype=dtype)\n\n[docs]class cg_tolerance(_value_context):\n\"\"\"\nRelative residual tolerance to use for terminating CG.\n\n(Default: 1)\n\"\"\"\n\n_global_value = 1\n\n[docs]class ciq_samples(_feature_flag):\n\"\"\"\nWhether to draw samples using Contour Integral Quadrature or not.\nThis may be slower than standard sampling methods for N < 5000.\nHowever, it should be faster with larger matrices.\n\nAs described in the paper:\n\nFast Matrix Square Roots with Applications to Gaussian Processes and Bayesian Optimization_.\n\n(Default: False)\n\n.. _Fast Matrix Square Roots with Applications to Gaussian Processes and Bayesian Optimization:\nhttps://arxiv.org/abs/2006.11267\n\"\"\"\n\n_default = False\n\n[docs]class preconditioner_tolerance(_value_context):\n\"\"\"\nDiagonal trace tolerance to use for checking preconditioner convergence.\n\n(Default: 1e-3)\n\"\"\"\n\n_global_value = 1e-3\n\n[docs]class eval_cg_tolerance(_value_context):\n\"\"\"\nRelative residual tolerance to use for terminating CG when making predictions.\n\n(Default: 1e-2)\n\"\"\"\n\n_global_value = 0.01\n\nclass _use_eval_tolerance(_feature_flag):\n_default = False\n\n[docs]class max_cholesky_size(_value_context):\n\"\"\"\nIf the size of of a LazyTensor is less than max_cholesky_size,\nthen root_decomposition and inv_matmul of LazyTensor will use Cholesky rather than Lanczos/CG.\n\n(Default: 800)\n\"\"\"\n\n_global_value = 800\n\n[docs]class max_root_decomposition_size(_value_context):\n\"\"\"\nThe maximum number of Lanczos iterations to perform\nThis is used when 1) computing variance estiamtes 2) when drawing from MVNs,\nor 3) for kernel multiplication\nMore values results in higher accuracy\n\n(Default: 100)\n\"\"\"\n\n_global_value = 100\n\n[docs]class max_preconditioner_size(_value_context):\n\"\"\"\nThe maximum size of preconditioner to use. 0 corresponds to turning\npreconditioning off. When enabled, usually a value of around ~10 works fairly well.\n\n(Default: 15)\n\"\"\"\n\n_global_value = 15\n\nr\"\"\"\nThe maximum number of Lanczos iterations to perform when doing stochastic\nLanczos quadrature. This is ONLY used for log determinant calculations and\ncomputing Tr(K^{-1}dK/d\\theta)\n\n(Default: 20)\n\"\"\"\n\n_global_value = 20\n\n[docs]class memory_efficient(_feature_flag):\n\"\"\"\nWhether or not to use Toeplitz math with gridded data, grid inducing point modules\nPros: memory efficient, faster on CPU\nCons: slower on GPUs with < 10000 inducing points\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n[docs]class min_preconditioning_size(_value_context):\n\"\"\"\nIf the size of of a LazyTensor is less than min_preconditioning_size,\nthen we won't use pivoted Cholesky based preconditioning.\n\n(Default: 2000)\n\"\"\"\n\n_global_value = 2000\n\n[docs]class minres_tolerance(_value_context):\n\"\"\"\nRelative update term tolerance to use for terminating MINRES.\n\n(Default: 1e-4)\n\"\"\"\n\n_global_value = 1e-4\n\n\"\"\"\nThe number of quadrature points to compute CIQ.\n\n(Default: 15)\n\"\"\"\n\n_global_value = 15\n\n[docs]class num_likelihood_samples(_value_context):\n\"\"\"\nThe number of samples to draw from a latent GP when computing a likelihood\nThis is used in variational inference and training\n\n(Default: 10)\n\"\"\"\n\n_global_value = 10\n\n[docs]class num_gauss_hermite_locs(_value_context):\n\"\"\"\nThe number of samples to draw from a latent GP when computing a likelihood\nThis is used in variational inference and training\n\n(Default: 20)\n\"\"\"\n\n_global_value = 20\n\n[docs]class num_trace_samples(_value_context):\n\"\"\"\nThe number of samples to draw when stochastically computing the trace of a matrix\nMore values results in more accurate trace estimations\nIf the value is set to 0, then the trace will be deterministically computed\n\n(Default: 10)\n\"\"\"\n\n_global_value = 10\n\n[docs]class prior_mode(_feature_flag):\n\"\"\"\nIf set to true, GP models will be evaluated in prior mode.\nThis allows evaluating any Exact GP model in prior mode, even it if has training data / targets.\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n[docs]class sgpr_diagonal_correction(_feature_flag):\n\"\"\"\nIf set to true, during posterior prediction the variances of the InducingPointKernel\nwill be corrected to match the variances of the exact kernel.\n\nIf false then no such correction will be performed (this is the default in other libraries).\n\n(Default: True)\n\"\"\"\n\n_default = True\n\n[docs]class skip_logdet_forward(_feature_flag):\n\"\"\"\n.. warning:\n\nADVANCED FEATURE. Use this feature ONLY IF you're using\ngpytorch.mlls.MarginalLogLikelihood as loss functions for optimizing\nhyperparameters/variational parameters. DO NOT use this feature if you\nneed accurate estimates of the MLL (i.e. for model selection, MCMC,\nsecond order optimizaiton methods, etc.)\n\nThis feature does not affect the gradients returned by\n:meth:gpytorch.distributions.MultivariateNormal.log_prob\n(used by gpytorch.mlls.MarginalLogLikelihood).\nThe gradients remain unbiased estimates, and therefore can be used with SGD.\nHowever, the actual likelihood value returned by the forward\npass will skip certain computations (i.e. the logdet computation), and will therefore\nbe improper estimates.\n\nIf you're using SGD (or a variant) to optimize parameters, you probably\ndon't need an accurate MLL estimate; you only need accurate gradients. So\nthis setting may give your model a performance boost.\n\n(Default: False)\n\"\"\"\n\n_default = False\n\nclass _linalg_dtype_symeig(_value_context):\n_global_value = torch.double\n\nclass _linalg_dtype_cholesky(_value_context):\n_global_value = torch.double\n\n[docs]class linalg_dtypes:\n\"\"\"\nWhether to perform less stable linalg calls in double precision or in a lower precision.\nCurrently, the default is to apply all symeig calls and cholesky calls within variational\nmethods in double precision.\n\n(Default: torch.double)\n\"\"\"\n\ndef __init__(self, default=torch.double, symeig=None, cholesky=None):\nsymeig = default if symeig is None else symeig\ncholesky = default if cholesky is None else cholesky\n\nself.symeig = _linalg_dtype_symeig(symeig)\nself.cholesky = _linalg_dtype_cholesky(cholesky)\n\ndef __enter__(self):\nself.symeig.__enter__()\nself.cholesky.__enter__()\n\ndef __exit__(self, args):\nself.symeig.__exit__()\nself.cholesky.__exit__()\nreturn False\n\n[docs]class terminate_cg_by_size(_feature_flag):\n\"\"\"\nIf set to true, cg will terminate after n iterations for an n x n matrix.\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n[docs]class trace_mode(_feature_flag):\n\"\"\"\nIf set to True, we will generally try to avoid calling our built in PyTorch functions, because these cannot\nbe run through torch.jit.trace.\n\nNote that this will sometimes involve explicitly evaluating lazy tensors and various other slowdowns and\ninefficiencies. As a result, you really shouldn't use this feature context unless you are calling torch.jit.trace\non a GPyTorch model.\n\nOur hope is that this flag will not be necessary long term, once https://github.com/pytorch/pytorch/issues/22329\nis fixed.\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n[docs]class tridiagonal_jitter(_value_context):\n\"\"\"\nThe (relative) amount of noise to add to the diagonal of tridiagonal matrices before\neigendecomposing. root_decomposition becomes slightly more stable with this, as we need\nto take the square root of the eigenvalues. Any eigenvalues still negative after adding jitter\nwill be zeroed out.\n\n(Default: 1e-6)\n\"\"\"\n\n_global_value = 1e-6\n\n[docs]class use_toeplitz(_feature_flag):\n\"\"\"\nWhether or not to use Toeplitz math with gridded data, grid inducing point modules\nPros: memory efficient, faster on CPU\nCons: slower on GPUs with < 10000 inducing points\n\n(Default: True)\n\"\"\"\n\n_default = True\n\n[docs]class verbose_linalg(_feature_flag):\n\"\"\"\nPrint out information whenever running an expensive linear algebra routine (e.g. Cholesky, CG, Lanczos, CIQ, etc.)\n\n(Default: False)\n\"\"\"\n\n_default = False\n\n# Create a global logger\nlogger = logging.getLogger(\"LinAlg (Verbose)\")\nlogger.setLevel(logging.DEBUG)\n\n# Output logging results to the stdout stream\nch = logging.StreamHandler()\nch.setLevel(logging.DEBUG)\nformatter = logging.Formatter(\"%(name)s - %(levelname)s - %(message)s\")\nch.setFormatter(formatter)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5885665,"math_prob":0.9134856,"size":21082,"snap":"2021-43-2021-49","text_gpt3_token_len":5015,"char_repetition_ratio":0.15305057,"word_repetition_ratio":0.25714287,"special_character_ratio":0.24380988,"punctuation_ratio":0.178106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9873938,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T05:03:14Z\",\"WARC-Record-ID\":\"<urn:uuid:69d0c1b8-9e59-41e4-9ec5-657a67a5eedc>\",\"Content-Length\":\"153205\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d665b164-fde0-45c3-aeec-670c306c8de9>\",\"WARC-Concurrent-To\":\"<urn:uuid:a47d305f-0bce-49e4-b3ed-e9431e5065e9>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://docs.gpytorch.ai/en/stable/_modules/gpytorch/settings.html\",\"WARC-Payload-Digest\":\"sha1:EIF7N7A7VRM45JVOUAEMCCOBH5KHZWZA\",\"WARC-Block-Digest\":\"sha1:D5DP35BMHBQ3NXTLRZ44WWRITMKNQFBV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585381.88_warc_CC-MAIN-20211021040342-20211021070342-00068.warc.gz\"}"}
https://math.stackexchange.com/questions/1872557/equivalence-between-ch-and-existence-of-certain-sets-in-mathbbr2	[ "# Equivalence between CH and existence of certain sets in $\\mathbb{R}^2$\n\nThe problem is to prove the following equivalence:\n\n$$2^{\\aleph_0}=\\aleph_1 \\iff \\exists\\ A,B\\in\\mathbb{R}^2:\\\\ \\textrm{a)}A\\cup B=\\mathbb{R}^2 \\\\\\textrm{b)}\\ \\forall\\ \\textrm{lines } l\\ \\textrm{parallel to x-axis, } \|A\\cap l\|\\leq\\aleph_0 \\\\\\textrm{c)}\\ \\forall\\ \\textrm{lines } l\\ \\textrm{parallel to y-axis, } \|B\\cap l\|\\leq\\aleph_0$$\n\nI think I know how to prove \"$\\impliedby$\":\n\nLet's assume the RHS and that $2^{\\aleph_0}\\geq\\aleph_2$. Now let's take a sequence $\\langle x_{\\alpha}:\\ \\alpha<\\omega_1\\rangle$. Since $\\bigcup_\\limits{\\alpha<\\omega_1}\|B\\cap x_{\\alpha}\|\\leq \\aleph_1\\cdot\\aleph_0=\\aleph_1$ there is such a $y$-coordinate that neither of $(x_{\\alpha},y)$ is in $B$. But hence we obtain $\\aleph_1$ point which all lie on the same line parallel to $x$-axis. Since $A\\cup B=\\mathbb{R}^2$ they all must be contained in $A$ $-$ a contradiction as $A$ has at most countably many of those points.\n\nBut what about the other implication? Do I contruct these sets by induction? Over all lines or something else? I haven't come up with anything yet, perhaps you could help me.\n\nHint. Let $\\sqsubseteq$ well-order $\\mathbb R$ of order type $\\omega_1$. Being a relation, $\\sqsubseteq$ is a subset of $\\mathbb R\\times\\mathbb R$. Perhaps that will work as either $A$ or $B$?\n• I still don't get it :( Maybe a hint to a hint? I mean: Every linear order contains a cofinal well-order. So what if $\\sqsubseteq$ is a cofinal well order on some line $l$ with natural order? Jul 27, 2016 at 9:58\n• @Jules: You're overcomplicating it :). Assuming CH, there is a bijection $\\varphi:\\mathbb R\\to\\omega_1$. Then define $x\\sqsubseteq y$ iff $\\varphi(x)\\le\\varphi(y)$. This gives you a well-ordering of all of $\\mathbb R$ of order type $\\omega_1$. In particular $\\{x\\in\\mathbb R\\mid x\\sqsubseteq a\\}$ is at most countable for every $a$. Jul 27, 2016 at 10:09\n• @Jules: Remember what an ordering is. An ordering such as $\\sqsubseteq$ is a particular case of a relation, which is, in and of itself, a set of pairs, that is, a subset of $\\mathbb R\\times\\mathbb R$. The set $\\sqsubseteq$ itself can be your $A$. And then $B$ becomes $\\mathbb R^2 \\setminus {\\sqsubseteq}$. Jul 27, 2016 at 10:33\n• But don't I need a set on which I impose this order? A set of elements that this relation pertains to? I guess it should be $\\mathbb{R}^2$ here. Jul 27, 2016 at 10:37" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78685683,"math_prob":0.9994376,"size":1077,"snap":"2023-40-2023-50","text_gpt3_token_len":388,"char_repetition_ratio":0.12581547,"word_repetition_ratio":0.028368793,"special_character_ratio":0.3249768,"punctuation_ratio":0.086124405,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999827,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T04:25:17Z\",\"WARC-Record-ID\":\"<urn:uuid:794f9858-a954-4c0f-8db2-4ea229a6039f>\",\"Content-Length\":\"140884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e0a9dce-d84a-4331-986d-eeaab8bf5ad0>\",\"WARC-Concurrent-To\":\"<urn:uuid:db6b7d75-d19e-4850-aa62-80156fb881a4>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1872557/equivalence-between-ch-and-existence-of-certain-sets-in-mathbbr2\",\"WARC-Payload-Digest\":\"sha1:AUMYP26A2RA6MEL5DJFHMCZYVSBC2ZKT\",\"WARC-Block-Digest\":\"sha1:2PJQED6OIVJ7BFVVSBP2FXNRQZGXNYUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506676.95_warc_CC-MAIN-20230925015430-20230925045430-00377.warc.gz\"}"}
https://www.rdocumentation.org/packages/ggraph/versions/2.0.5/topics/geom_edge_fan	[ "ggraph (version 2.0.5)\n\n# geom_edge_fan: Draw edges as curves of different curvature\n\n## Description\n\nThis geom draws edges as cubic beziers with the control point positioned half-way between the nodes and at an angle dependent on the presence of parallel edges. This results in parallel edges being drawn in a non-overlapping fashion resembling the standard approach used in `igraph::plot.igraph()`. Before calculating the curvature the edges are sorted by direction so that edges going the same way will be adjacent. This geom is currently the only choice for non-simple graphs if edges should not be overplotted.\n\n## Usage\n\n```geom_edge_fan(\nmapping = NULL,\ndata = get_edges(),\nposition = \"identity\",\narrow = NULL,\nstrength = 1,\nn = 100,\nlineend = \"butt\",\nlinejoin = \"round\",\nlinemitre = 1,\nlabel_colour = \"black\",\nlabel_alpha = 1,\nlabel_parse = FALSE,\ncheck_overlap = FALSE,\nangle_calc = \"rot\",\nforce_flip = TRUE,\nlabel_dodge = NULL,\nlabel_push = NULL,\nshow.legend = NA,\n...,\nspread\n)geom_edge_fan2(\nmapping = NULL,\ndata = get_edges(\"long\"),\nposition = \"identity\",\narrow = NULL,\nstrength = 1,\nn = 100,\nlineend = \"butt\",\nlinejoin = \"round\",\nlinemitre = 1,\nlabel_colour = \"black\",\nlabel_alpha = 1,\nlabel_parse = FALSE,\ncheck_overlap = FALSE,\nangle_calc = \"rot\",\nforce_flip = TRUE,\nlabel_dodge = NULL,\nlabel_push = NULL,\nshow.legend = NA,\n...,\nspread\n)geom_edge_fan0(\nmapping = NULL,\ndata = get_edges(),\nposition = \"identity\",\narrow = NULL,\nstrength = 1,\nlineend = \"butt\",\nshow.legend = NA,\n...,\nspread\n)```\n\n## Arguments\n\nmapping\n\nSet of aesthetic mappings created by `ggplot2::aes()` or `ggplot2::aes_()`. By default x, y, xend, yend, group and circular are mapped to x, y, xend, yend, edge.id and circular in the edge data.\n\ndata\n\nThe return of a call to `get_edges()` or a data.frame giving edges in correct format (see details for for guidance on the format). See `get_edges()` for more details on edge extraction.\n\nposition\n\nPosition adjustment, either as a string, or the result of a call to a position adjustment function.\n\narrow\n\nArrow specification, as created by `grid::arrow()`.\n\nstrength\n\nModify the width of the fans `strength > 1` will create wider fans while the reverse will make them more narrow.\n\nn\n\nThe number of points to create along the path.\n\nlineend\n\nLine end style (round, butt, square).\n\nlinejoin\n\nLine join style (round, mitre, bevel).\n\nlinemitre\n\nLine mitre limit (number greater than 1).\n\nlabel_colour\n\nThe colour of the edge label. If `NA` it will use the colour of the edge.\n\nlabel_alpha\n\nThe opacity of the edge label. If `NA` it will use the opacity of the edge.\n\nlabel_parse\n\nIf `TRUE`, the labels will be parsed into expressions and displayed as described in `grDevices::plotmath()`.\n\ncheck_overlap\n\nIf `TRUE`, text that overlaps previous text in the same layer will not be plotted. `check_overlap` happens at draw time and in the order of the data. Therefore data should be arranged by the label column before calling `geom_label()` or `geom_text()`.\n\nangle_calc\n\nEither 'none', 'along', or 'across'. If 'none' the label will use the angle aesthetic of the geom. If 'along' The label will be written along the edge direction. If 'across' the label will be written across the edge direction.\n\nforce_flip\n\nLogical. If `angle_calc` is either 'along' or 'across' should the label be flipped if it is on it's head. Default to `TRUE`.\n\nlabel_dodge\n\nA `grid::unit()` giving a fixed vertical shift to add to the label in case of `angle_calc` is either 'along' or 'across'\n\nlabel_push\n\nA `grid::unit()` giving a fixed horizontal shift to add to the label in case of `angle_calc` is either 'along' or 'across'\n\nshow.legend\n\nlogical. Should this layer be included in the legends? `NA`, the default, includes if any aesthetics are mapped. `FALSE` never includes, and `TRUE` always includes. It can also be a named logical vector to finely select the aesthetics to display.\n\n...\n\nOther arguments passed on to `layer()`. These are often aesthetics, used to set an aesthetic to a fixed value, like `colour = \"red\"` or `size = 3`. They may also be parameters to the paired geom/stat.\n\nspread\n\nDeprecated. Use `strength` instead.\n\n## Aesthetics\n\n`geom_edge_fan` and `geom_edge_fan0` understand the following aesthetics. Bold aesthetics are automatically set, but can be overridden.\n\n• x\n\n• y\n\n• xend\n\n• yend\n\n• from\n\n• to\n\n• edge_colour\n\n• edge_width\n\n• edge_linetype\n\n• edge_alpha\n\n• filter\n\n`geom_edge_fan2` understand the following aesthetics. Bold aesthetics are automatically set, but can be overridden.\n\n• x\n\n• y\n\n• group\n\n• from\n\n• to\n\n• edge_colour\n\n• edge_width\n\n• edge_linetype\n\n• edge_alpha\n\n• filter\n\n`geom_edge_fan` and `geom_edge_fan2` furthermore takes the following aesthetics.\n\n• start_cap\n\n• end_cap\n\n• label\n\n• label_pos\n\n• label_size\n\n• angle\n\n• hjust\n\n• vjust\n\n• family\n\n• fontface\n\n• lineheight\n\n## Computed variables\n\nindex\n\nThe position along the path (not computed for the 0 version)\n\n## Edge variants\n\nMany geom_edge_ layers comes in 3 flavors depending on the level of control needed over the drawing. The default (no numeric postfix) generate a number of points (`n`) along the edge and draws it as a path. Each point along the line has a numeric value associated with it giving the position along the path, and it is therefore possible to show the direction of the edge by mapping to this e.g. `colour = stat(index)`. The version postfixed with a \"2\" uses the \"long\" edge format (see `get_edges()`) and makes it possible to interpolate node parameter between the start and end node along the edge. It is considerable less performant so should only be used if this is needed. The version postfixed with a \"0\" draws the edge in the most performant way, often directly using an appropriate grob from the grid package, but does not allow for gradients along the edge.\n\nOften it is beneficial to stop the drawing of the edge before it reaches the node, for instance in cases where an arrow should be drawn and the arrowhead shouldn't lay on top or below the node point. geom_edge_* and geom_edge_2 supports this through the start_cap and end_cap aesthetics that takes a `geometry()` specification and dynamically caps the termini of the edges based on the given specifications. This means that if `end_cap = circle(1, 'cm')` the edges will end at a distance of 1cm even during resizing of the plot window.\n\nAll `geom_edge_` and `geom_edge_2` have the ability to draw a label along the edge. The reason this is not a separate geom is that in order for the label to know the location of the edge it needs to know the edge type etc. Labels are drawn by providing a label aesthetic. The label_pos can be used to specify where along the edge it should be drawn by supplying a number between 0 and 1. The label_size aesthetic can be used to control the size of the label. Often it is needed to have the label written along the direction of the edge, but since the actual angle is dependent on the plot dimensions this cannot be calculated beforehand. Using the angle_calc argument allows you to specify whether to use the supplied angle aesthetic or whether to draw the label along or across the edge.\n\n## Edge aesthetic name expansion\n\nIn order to avoid excessive typing edge aesthetic names are automatically expanded. Because of this it is not necessary to write `edge_colour` within the `aes()` call as `colour` will automatically be renamed appropriately.\n\n## See Also\n\nOther geom_edge_: `geom_edge_arc()`, `geom_edge_bend()`, `geom_edge_density()`, `geom_edge_diagonal()`, `geom_edge_elbow()`, `geom_edge_hive()`, `geom_edge_link()`, `geom_edge_loop()`, `geom_edge_parallel()`, `geom_edge_point()`, `geom_edge_span()`, `geom_edge_tile()`\n\n## Examples\n\n```# NOT RUN {\nrequire(tidygraph)\ngr <- create_notable('bull') %>%\nconvert(to_directed) %>%\nbind_edges(data.frame(from = c(1, 2, 2, 3), to = c(2, 1, 3, 2))) %E>%\nmutate(class = sample(letters[1:3], 9, TRUE)) %N>%\nmutate(class = sample(c('x', 'y'), 5, TRUE))\n\nggraph(gr, 'stress') +\ngeom_edge_fan(aes(alpha = stat(index)))\n\nggraph(gr, 'stress') +\ngeom_edge_fan2(aes(colour = node.class))\n\nggraph(gr, 'stress') +\ngeom_edge_fan0(aes(colour = class))\n# }\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76064235,"math_prob":0.9659521,"size":7473,"snap":"2021-21-2021-25","text_gpt3_token_len":1894,"char_repetition_ratio":0.14058107,"word_repetition_ratio":0.14708298,"special_character_ratio":0.250368,"punctuation_ratio":0.14637682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98411405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-13T14:06:37Z\",\"WARC-Record-ID\":\"<urn:uuid:b4eb490f-f3dc-457d-ac31-37e8b97ed3ff>\",\"Content-Length\":\"79412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8e140a6-4c0e-47e1-b0e7-1642fbb741a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:def3dc1b-d619-4a82-aca2-84e13dba5554>\",\"WARC-IP-Address\":\"99.84.109.63\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/ggraph/versions/2.0.5/topics/geom_edge_fan\",\"WARC-Payload-Digest\":\"sha1:TSFIIKI3SIGHX6VWURRRIYBHDWE463OV\",\"WARC-Block-Digest\":\"sha1:2KXOF4RP6AKALS3ZOBRTTRE5QORJZTTW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487608856.6_warc_CC-MAIN-20210613131257-20210613161257-00481.warc.gz\"}"}
https://www.geogebra.org/m/heqSg72K	[ "# The Derivative Curve\n\nAuthor:\nchisanga\nThis applet shows how to draw the derivative of a function. Point A can be moved along the curve. The red point has the same x coordinate as A and its y coordinate is the gradient of the function, f. Move the point A along the curve and the red point traces out the derivative curve.\n1. For what values are is the derivative positive?\n2. For what values are is the derivative negative?\n3. How do stationary points on translate to points on the derivative curve?\n4. How do stationary points on the derivative curve translate to points on the curve of ?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8881991,"math_prob":0.99367285,"size":611,"snap":"2021-31-2021-39","text_gpt3_token_len":133,"char_repetition_ratio":0.21746293,"word_repetition_ratio":0.10810811,"special_character_ratio":0.21767594,"punctuation_ratio":0.094017096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996719,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T01:55:46Z\",\"WARC-Record-ID\":\"<urn:uuid:8598a5a5-51c7-4013-99de-de92ba826c2d>\",\"Content-Length\":\"39353\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3c8f6a6-df08-4b99-8bac-d2333a8ed52c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3239c6f-b56e-48e6-b1e4-9cdc26fa05b1>\",\"WARC-IP-Address\":\"99.84.105.53\",\"WARC-Target-URI\":\"https://www.geogebra.org/m/heqSg72K\",\"WARC-Payload-Digest\":\"sha1:OQ4ZULUUMAMHHMZ5OLDNUPYE3E2VYWUN\",\"WARC-Block-Digest\":\"sha1:PTMKXMQO333YWLOEVSJJ2XM4IK7VNWYY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057131.88_warc_CC-MAIN-20210921011047-20210921041047-00010.warc.gz\"}"}
https://questions.llc/questions/1646435	[ "# home / study / engineering / computer science / computer science questions and answers / 4.7 problems in this exercise assume that the logic blocks used to implement a processor’s ...\n\nQuestion: 4.7 Problems in this exercise assume that the logic blocks used to implement a processor’s data...\n4.7 Problems in this exercise assume that the logic blocks used to implement a processor’s datapath have the following latencies:\n1-Mem/D-Mem\nRegister File\nMUX\nALU\nSingle gate\nRegister Setup\nSign extend\nControl\n250 ps\n150 ps\n25 ps\n200 ps\n150 ps\n5 ps\n30 ps\n20 ps\n50 ps\n50 ps\n“Register read” is the time needed after the rising clock edge for the new register value to appear on the output. This value applies to the PC only. “Register setup” is the amount of time a register’s data input must be stable before the rising edge of the clock. This value applies to both the PC and Register File.\n4.7.1 Although the control unit as a whole requires 50 ps, it so happens that we can extract the correct value of the Reg2Loc control wire directly from the instruction. Thus, the value of this control wire is available at the same time as the instruction. Explain how we can extract this value directly from the instruction. Hints: Carefully examine the opcodes shown in Figure 2.20. Also, remember that LSR and LSL do not use the Rm field. Finally, ignore STXR.\n4.7.2 What is the latency of an R-type instruction (i.e., how long must the clock period be to ensure that this instruction works correctly)?\n4.7.3 What is the latency of LDUR? (Check your answer carefully. Many students place extra muxes on the critical path.)\n4.7.4 What is the latency of STUR? (Check your answer carefully. Many students place extra muxes on the critical path.)\n4.7.5 What is the latency of CBZ?\n4.7.6 What is the latency of B?\n4.7.7 What is the latency of an I-type instruction?\n4.7.8 What is the minimum clock period for this CPU?\n\n## To answer the questions about the latencies and clock period for the CPU in this exercise, we need to refer to the given information about the latencies of various logic blocks used to implement the processor's datapath.\n\n4.7.1: The value of the Reg2Loc control wire can be directly extracted from the instruction because it is available at the same time as the instruction. This is possible because certain opcodes shown in Figure 2.20 allow us to extract this value directly from the instruction. The R-type instructions (except for LSR and LSL which do not use the Rm field) have the necessary opcode information to determine the value of Reg2Loc. We can therefore use this information to extract the value of Reg2Loc directly from the instruction.\n\n4.7.2: The latency of an R-type instruction is the time it takes for the instruction to complete its execution. In this case, we need to consider the latencies of the logic blocks involved in the execution of an R-type instruction. According to the given information, the latencies for the required logic blocks are as follows:\n- Register Setup: 20 ps\n- ALU: 200 ps\n\nTo ensure that the R-type instruction works correctly, the clock period should be equal to or greater than the sum of these latencies. Therefore, the latency of an R-type instruction is 30 ps + 20 ps + 200 ps = 250 ps.\n\n4.7.3: The latency of LDUR instruction is the time it takes for the instruction to complete its execution. To determine this, we need to consider the latencies of the logic blocks involved in the execution of the LDUR instruction. According to the given information, the latencies for the required logic blocks are as follows:\n- Register Setup: 20 ps\n- MUX: 25 ps\n- Mem/D-Mem: 250 ps\n\nTo check our answer carefully and avoid placing extra muxes on the critical path, we should consider the longest path among these latencies. Therefore, the latency of LDUR instruction is 30 ps + 20 ps + 250 ps = 300 ps.\n\n4.7.4: The latency of STUR instruction is the time it takes for the instruction to complete its execution. To determine this, we need to consider the latencies of the logic blocks involved in the execution of the STUR instruction. According to the given information, the latencies for the required logic blocks are as follows:\n- Register Setup: 20 ps\n- MUX: 25 ps\n- Mem/D-Mem: 250 ps\n\nTo check our answer carefully and avoid placing extra muxes on the critical path, we should consider the longest path among these latencies. Therefore, the latency of STUR instruction is 30 ps + 20 ps + 250 ps = 300 ps.\n\n4.7.5: The latency of CBZ instruction is the time it takes for the instruction to complete its execution. However, the given information does not provide specific latencies for the logic blocks involved in the execution of the CBZ instruction. Therefore, we cannot determine the exact latency of CBZ instruction based on the given information.\n\n4.7.6: The latency of B instruction is the time it takes for the instruction to complete its execution. However, the given information does not provide specific latencies for the logic blocks involved in the execution of the B instruction. Therefore, we cannot determine the exact latency of B instruction based on the given information.\n\n4.7.7: The latency of an I-type instruction is the time it takes for the instruction to complete its execution. However, the given information does not provide specific latencies for the logic blocks involved in the execution of an I-type instruction. Therefore, we cannot determine the exact latency of an I-type instruction based on the given information.\n\n4.7.8: The minimum clock period for this CPU should be equal to or greater than the sum of the latencies of all the logic blocks involved in the execution of the longest instruction. In this case, we can consider the LDUR instruction, which has the longest path among the given instructions. According to the given information, the latencies for the required logic blocks in LDUR instruction are as follows:\n- Register Setup: 20 ps\n- MUX: 25 ps\n- Mem/D-Mem: 250 ps\n\nTherefore, the minimum clock period for this CPU should be 30 ps + 20 ps + 25 ps + 250 ps = 325 ps.\n\n3170219\n\nCreated\n\nRating\n0\n\n## To find the answers to the given questions, we need to analyze the provided information about the latencies of the logic blocks used in the processor's datapath. Let's break down each question and explain how to find the answers.\n\n4.7.1 - To extract the value of the Reg2Loc control wire directly from the instruction, we need to examine the opcodes shown in Figure 2.20. By carefully examining the opcodes, we can determine if the instruction is an LSR or LSL type, as they do not use the Rm field. Finally, we can ignore STXR. The correct value of Reg2Loc control wire is available at the same time as the instruction.\n\n4.7.2 - The latency of an R-type instruction refers to how long the clock period should be to ensure that this instruction works correctly. To find the latency, we need to consider the slowest logic block in the datapath, which is the Register File with a latency of 150 ps. Therefore, the clock period should be at least 150 ps to ensure correct operation.\n\n4.7.3 - The latency of LDUR refers to how long it takes for an LDUR instruction to execute. To find the latency, we need to identify the critical path, i.e., the path with the longest latency. It is important to check for any extra muxes placed on the critical path, as they might introduce additional delay. By carefully examining the datapath, we can identify the critical path and determine the LDUR's latency.\n\n4.7.4 - Similar to question 4.7.3, the latency of STUR refers to the time it takes for an STUR instruction to execute. We need to identify the critical path and check for any extra muxes on it to find the correct latency of STUR.\n\n4.7.5 - The latency of CBZ refers to the time it takes for a CBZ instruction to execute. To find the latency, we need to identify the critical path and determine the delay introduced by the logic blocks on that path.\n\n4.7.6 - The latency of B refers to the time it takes for a B instruction to execute. Similar to previous questions, we need to identify the critical path and calculate the delay introduced by the logic blocks on that path.\n\n4.7.7 - The latency of an I-type instruction refers to the time it takes for an I-type instruction to execute. Again, we need to identify the critical path and calculate the delay introduced by the logic blocks on that path.\n\n4.7.8 - The minimum clock period for this CPU refers to the shortest time duration between clock cycles that allows correct operation. To find this period, we need to consider the longest latency of any logic block in the datapath and set the clock period to be at least that long.\n\nBy carefully analyzing the information provided about the latencies and following the instructions stated for each question, we can find the correct answers.\n\n3633184\n\nCreated\n\nRating\n0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9018421,"math_prob":0.8684431,"size":8959,"snap":"2023-40-2023-50","text_gpt3_token_len":2086,"char_repetition_ratio":0.18447794,"word_repetition_ratio":0.40761906,"special_character_ratio":0.23841947,"punctuation_ratio":0.118063115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97726405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T14:36:37Z\",\"WARC-Record-ID\":\"<urn:uuid:8cfdc414-225c-4240-bc12-56fd96be65cd>\",\"Content-Length\":\"47042\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eaffdcf3-6460-461f-9e5c-8793e1339463>\",\"WARC-Concurrent-To\":\"<urn:uuid:467b9d0c-e9d3-41d4-a48b-05b12a8f0bb9>\",\"WARC-IP-Address\":\"45.79.29.166\",\"WARC-Target-URI\":\"https://questions.llc/questions/1646435\",\"WARC-Payload-Digest\":\"sha1:H7G4N66RQLE2XZKIZJ7TOLFOZ4EPGERP\",\"WARC-Block-Digest\":\"sha1:SIQFNQLNSVOX4ZOKPGA52T6VQ7KE3LF3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100427.59_warc_CC-MAIN-20231202140407-20231202170407-00188.warc.gz\"}"}
https://www.colorhexa.com/e1e1ed	[ "# #e1e1ed Color Information\n\nIn a RGB color space, hex #e1e1ed is composed of 88.2% red, 88.2% green and 92.9% blue. Whereas in a CMYK color space, it is composed of 5.1% cyan, 5.1% magenta, 0% yellow and 7.1% black. It has a hue angle of 240 degrees, a saturation of 25% and a lightness of 90.6%. #e1e1ed color hex could be obtained by blending #ffffff with #c3c3db. Closest websafe color is: #ccccff.\n\n• R 88\n• G 88\n• B 93\nRGB color chart\n• C 5\n• M 5\n• Y 0\n• K 7\nCMYK color chart\n\n#e1e1ed color description : Light grayish blue.\n\n# #e1e1ed Color Conversion\n\nThe hexadecimal color #e1e1ed has RGB values of R:225, G:225, B:237 and CMYK values of C:0.05, M:0.05, Y:0, K:0.07. Its decimal value is 14803437.\n\nHex triplet RGB Decimal e1e1ed `#e1e1ed` 225, 225, 237 `rgb(225,225,237)` 88.2, 88.2, 92.9 `rgb(88.2%,88.2%,92.9%)` 5, 5, 0, 7 240°, 25, 90.6 `hsl(240,25%,90.6%)` 240°, 5.1, 92.9 ccccff `#ccccff`\nCIE-LAB 89.846, 2.202, -5.841 73.26, 75.972, 90.921 0.305, 0.316, 75.972 89.846, 6.242, 290.653 89.846, -0.686, -9.445 87.162, -2.504, -0.833 11100001, 11100001, 11101101\n\n# Color Schemes with #e1e1ed\n\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #edede1\n``#edede1` `rgb(237,237,225)``\nComplementary Color\n• #e1e7ed\n``#e1e7ed` `rgb(225,231,237)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #e7e1ed\n``#e7e1ed` `rgb(231,225,237)``\nAnalogous Color\n• #e7ede1\n``#e7ede1` `rgb(231,237,225)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #ede7e1\n``#ede7e1` `rgb(237,231,225)``\nSplit Complementary Color\n• #e1ede1\n``#e1ede1` `rgb(225,237,225)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #ede1e1\n``#ede1e1` `rgb(237,225,225)``\nTriadic Color\n• #e1eded\n``#e1eded` `rgb(225,237,237)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #ede1e1\n``#ede1e1` `rgb(237,225,225)``\n• #edede1\n``#edede1` `rgb(237,237,225)``\nTetradic Color\n• #b1b1d0\n``#b1b1d0` `rgb(177,177,208)``\n• #c1c1da\n``#c1c1da` `rgb(193,193,218)``\n• #d1d1e3\n``#d1d1e3` `rgb(209,209,227)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #f1f1f7\n``#f1f1f7` `rgb(241,241,247)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #e1e1ed\n\nBelow, you can see some colors close to #e1e1ed. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #e1e4ed\n``#e1e4ed` `rgb(225,228,237)``\n• #e1e3ed\n``#e1e3ed` `rgb(225,227,237)``\n• #e1e2ed\n``#e1e2ed` `rgb(225,226,237)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #e2e1ed\n``#e2e1ed` `rgb(226,225,237)``\n• #e3e1ed\n``#e3e1ed` `rgb(227,225,237)``\n• #e4e1ed\n``#e4e1ed` `rgb(228,225,237)``\nSimilar Colors\n\n# #e1e1ed Preview\n\nText with hexadecimal color #e1e1ed\n\nThis text has a font color of #e1e1ed.\n\n``<span style=\"color:#e1e1ed;\">Text here</span>``\n#e1e1ed background color\n\nThis paragraph has a background color of #e1e1ed.\n\n``<p style=\"background-color:#e1e1ed;\">Content here</p>``\n#e1e1ed border color\n\nThis element has a border color of #e1e1ed.\n\n``<div style=\"border:1px solid #e1e1ed;\">Content here</div>``\nCSS codes\n``.text {color:#e1e1ed;}``\n``.background {background-color:#e1e1ed;}``\n``.border {border:1px solid #e1e1ed;}``\n\n# Shades and Tints of #e1e1ed\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, #040407 is the darkest color, while #fafafc is the lightest one.\n\n• #040407\n``#040407` `rgb(4,4,7)``\n• #0b0b13\n``#0b0b13` `rgb(11,11,19)``\n• #13131f\n``#13131f` `rgb(19,19,31)``\n• #1a1a2c\n``#1a1a2c` `rgb(26,26,44)``\n• #212138\n``#212138` `rgb(33,33,56)``\n• #292944\n``#292944` `rgb(41,41,68)``\n• #303050\n``#303050` `rgb(48,48,80)``\n• #38385d\n``#38385d` `rgb(56,56,93)``\n• #3f3f69\n``#3f3f69` `rgb(63,63,105)``\n• #464675\n``#464675` `rgb(70,70,117)``\n• #4e4e81\n``#4e4e81` `rgb(78,78,129)``\n• #55558e\n``#55558e` `rgb(85,85,142)``\n• #5c5c9a\n``#5c5c9a` `rgb(92,92,154)``\nShade Color Variation\n• #6666a3\n``#6666a3` `rgb(102,102,163)``\n• #7373ab\n``#7373ab` `rgb(115,115,171)``\n• #7f7fb2\n``#7f7fb2` `rgb(127,127,178)``\n• #8b8bba\n``#8b8bba` `rgb(139,139,186)``\n• #9797c1\n``#9797c1` `rgb(151,151,193)``\n• #a4a4c8\n``#a4a4c8` `rgb(164,164,200)``\n• #b0b0d0\n``#b0b0d0` `rgb(176,176,208)``\n• #bcbcd7\n``#bcbcd7` `rgb(188,188,215)``\n• #c8c8de\n``#c8c8de` `rgb(200,200,222)``\n• #d5d5e6\n``#d5d5e6` `rgb(213,213,230)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #ededf4\n``#ededf4` `rgb(237,237,244)``\n• #fafafc\n``#fafafc` `rgb(250,250,252)``\nTint Color Variation\n\n# Tones of #e1e1ed\n\nA tone is produced by adding gray to any pure hue. In this case, #e7e7e7 is the less saturated color, while #d0d0fe is the most saturated one.\n\n• #e7e7e7\n``#e7e7e7` `rgb(231,231,231)``\n• #e5e5e9\n``#e5e5e9` `rgb(229,229,233)``\n• #e3e3eb\n``#e3e3eb` `rgb(227,227,235)``\n• #e1e1ed\n``#e1e1ed` `rgb(225,225,237)``\n• #dfdfef\n``#dfdfef` `rgb(223,223,239)``\n• #ddddf1\n``#ddddf1` `rgb(221,221,241)``\n• #dbdbf3\n``#dbdbf3` `rgb(219,219,243)``\n• #dadaf4\n``#dadaf4` `rgb(218,218,244)``\n• #d8d8f6\n``#d8d8f6` `rgb(216,216,246)``\n• #d6d6f8\n``#d6d6f8` `rgb(214,214,248)``\n• #d4d4fa\n``#d4d4fa` `rgb(212,212,250)``\n• #d2d2fc\n``#d2d2fc` `rgb(210,210,252)``\n• #d0d0fe\n``#d0d0fe` `rgb(208,208,254)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #e1e1ed 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":0.50826216,"math_prob":0.6718516,"size":3706,"snap":"2021-04-2021-17","text_gpt3_token_len":1726,"char_repetition_ratio":0.124257155,"word_repetition_ratio":0.011090573,"special_character_ratio":0.52320564,"punctuation_ratio":0.23370786,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97159946,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-20T05:13:10Z\",\"WARC-Record-ID\":\"<urn:uuid:4ca7fe34-2771-45c1-8851-afecdfdf0968>\",\"Content-Length\":\"36338\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9894cbad-e30b-4a6b-82a6-5af3d4dec3a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab849b2b-15aa-4f59-917a-13f35bad164a>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/e1e1ed\",\"WARC-Payload-Digest\":\"sha1:E3CYAAILK4PGZAK34ICRN6OZTAINAXFN\",\"WARC-Block-Digest\":\"sha1:4DUUXBZEMEAVE5OBHP3W5USXV26NEZ2E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039375537.73_warc_CC-MAIN-20210420025739-20210420055739-00145.warc.gz\"}"}
https://sensing.konicaminolta.us/us/blog/identifying-color-differences-using-l-a-b-or-l-c-h-coordinates/	[ "Deprecated: strcasecmp(): Passing null to parameter #1 (\\$string1) of type string is deprecated in C:\\inetpub\\wwwroot\\km-sensing\\wp-content\\themes\\uncode\\core\\inc\\wp-bootstrap-navwalker.php on line 107 Deprecated: strcasecmp(): Passing null to parameter #1 (\\$string1) of type string is deprecated in C:\\inetpub\\wwwroot\\km-sensing\\wp-content\\themes\\uncode\\core\\inc\\wp-bootstrap-navwalker.php on line 107 Deprecated: strcasecmp(): Passing null to parameter #1 (\\$string1) of type string is deprecated in C:\\inetpub\\wwwroot\\km-sensing\\wp-content\\themes\\uncode\\core\\inc\\wp-bootstrap-navwalker.php on line 107\n\n# Identifying Color Differences Using Lab* or LCH* Coordinates\n\nHow Closely Does the Sample Match the Standard?\n\nEven if two colors look the same to one person, slight differences may be found when evaluated with a color measurement instrument. If the color of a sample does not match the standard, customer satisfaction is compromised and the amount of rework and costs increase. Because of this, identifying color differences between a sample and the standard as early in the production process as possible is important.\n\nColor difference can be defined as the numerical comparison of a sample’s color to the standard. It indicates the differences in absolute color coordinates and is referred to as Delta (Δ). These formulas calculate the difference between two colors to identify inconsistencies and help users control the color of their products more effectively.\n\nTo begin, the sample color and the standard color should be measured and the values for each measurement saved. The color differences between the sample and standard are calculated using the resulting colorimetric values.\n\nIdentifying Color Differences Using CIE Lab* Coordinates\n\nDefined by the Commission Internationale de l’Eclairage (CIE), the Lab* color space was modeled after a color-opponent theory stating that two colors cannot be red and green at the same time or yellow and blue at the same time. As shown below, L* indicates lightness, a* is the red/green coordinate, and b* is the yellow/blue coordinate. Deltas for L* (ΔL), a (Δa) and b (Δb) may be positive (+) or negative ( -). The total difference, Delta E (ΔE), however, is always positive.\n\nΔL* (L* sample minus L* standard) = difference in lightness and darkness (+ = lighter, – = darker)\nΔa* (a* sample minus a* standard) = difference in red and green (+ = redder, – = greener)\nΔb* (b* sample minus b* standard) = difference in yellow and blue (+ = yellower, – = bluer)", null, "Let’s compare Apple 1 to Apple 2 (see Figure 1).", null, "Figure 1\n\nLooking at the Lab* values for each apple in Figure 1, we can objectively determine that the apples don’t match in color. These values tell us that Apple 2 (sample) is lighter, less red, and more yellow in color than Apple 1 (standard). If we put the values of ΔL=+4.03, Δa=-3.05, and Δb=+1.04 into the color difference equation, it can be determined that the total color difference between the two apples is 5.16.\n\n5.16 = [4.03^2 + -3.05^2 + 1.04^2] ^1/2\n\nIdentifying Color Differences Using CIE LCH Coordinates\n\nThe LCh color space is similar to Lab, but it describes color differently using cylindrical coordinates instead of rectangular coordinates. In this color space, L indicates lightness, C* represents chroma, and h is the hue angle. Chroma and hue are calculated from the a* and b* coordinates in Lab. Deltas for lightness (ΔL), chroma (ΔC), and hue (ΔH) may be positive (+) or negative ( -). These are expressed as:\n\nΔL* (L* sample minus L* standard) = difference in lightness and darkness (+ = lighter, – = darker)\nΔC* (C* sample minus C* standard) = difference in chroma (+ = brighter, – = duller)\nΔH* (H* sample minus H* standard) = difference in hue\n\nLet’s compare Apple 1 to Apple 2 (see Figure 2).", null, "Figure 2\n\nLooking at the LCh values for each apple in Figure 2, we can objectively determine that the apples don’t match in color. Like the Lab* values, these values tell us that Apple 2 (sample) is lighter and duller in appearance than Apple 1 (standard). The positive ΔH* value of +1.92 indicates Apple 2 falls counterclockwise to Apple 1 in the LCh color space. This tells us that Apple 2 is less red than Apple 1.\n\nColor measurement instruments, such as colorimeters and spectrophotometers, can detect differences indiscernible to the human eye and then instantly display these differences in numerical terms. After identifying color differences using Lab* or LCh values, it should be decided whether the sample is acceptable or not using tolerance limits." ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2020502%202085'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20588%20272'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20588%20272'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5048553,"math_prob":0.9732572,"size":20943,"snap":"2023-40-2023-50","text_gpt3_token_len":4991,"char_repetition_ratio":0.1533502,"word_repetition_ratio":0.48715416,"special_character_ratio":0.19385952,"punctuation_ratio":0.14091435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9785497,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T14:43:00Z\",\"WARC-Record-ID\":\"<urn:uuid:f658d0ee-e5dc-47dc-8819-59692b36392e>\",\"Content-Length\":\"202528\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a80ce6e-4d6d-4c00-a8cc-aadf77ea24b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:67743557-4a40-46bb-aeab-77d3c199b383>\",\"WARC-IP-Address\":\"207.18.59.111\",\"WARC-Target-URI\":\"https://sensing.konicaminolta.us/us/blog/identifying-color-differences-using-l-a-b-or-l-c-h-coordinates/\",\"WARC-Payload-Digest\":\"sha1:6BTCMWLFJPR3AW2ZOGPPKBOC6OAA2FC6\",\"WARC-Block-Digest\":\"sha1:T6WSYNUGAO5Z2LQVXM42IDKFJ2AV2SAY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510300.41_warc_CC-MAIN-20230927135227-20230927165227-00206.warc.gz\"}"}
https://converter.ninja/velocity/meters-per-second-to-knots/980-mps-to-knot/	[ "# 980 meters per second in knots\n\n## Conversion\n\n980 meters per second is equivalent to 1904.96760259179 knots.", null, "## Conversion in the opposite direction\n\nThe inverse of the conversion factor is that 1 knot is equal to 0.000524943310657596 times 980 meters per second.\n\nIt can also be expressed as: 980 meters per second is equal to $\\frac{1}{\\mathrm{0.000524943310657596}}$ knots.\n\n## Approximation\n\nAn approximate numerical result would be: nine hundred and eighty meters per second is about one thousand, nine hundred and four point nine seven knots, or alternatively, a knot is about zero times nine hundred and eighty meters per second.\n\n## Footnotes\n\n The precision is 15 significant digits (fourteen digits to the right of the decimal point).\n\nResults may contain small errors due to the use of floating point arithmetic." ]	[ null, "https://converter.ninja/images/980_mps_in_knot.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84914196,"math_prob":0.99247915,"size":690,"snap":"2020-34-2020-40","text_gpt3_token_len":147,"char_repetition_ratio":0.14577259,"word_repetition_ratio":0.036036037,"special_character_ratio":0.23333333,"punctuation_ratio":0.08661418,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9878654,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-26T07:25:24Z\",\"WARC-Record-ID\":\"<urn:uuid:4474f60e-a049-4b87-b91c-2c07db4f67fe>\",\"Content-Length\":\"16135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dad9153f-0fc1-47f2-b928-90ee5ca9a4b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:004404bc-5c09-4701-b981-cf8eac4863e8>\",\"WARC-IP-Address\":\"172.67.167.118\",\"WARC-Target-URI\":\"https://converter.ninja/velocity/meters-per-second-to-knots/980-mps-to-knot/\",\"WARC-Payload-Digest\":\"sha1:44IKIK5TRL53BB7KHMZMPFR35RDZB5WY\",\"WARC-Block-Digest\":\"sha1:CIAC4V5OLZRWQFDU6YGVE3LFEVTZTTCF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400238038.76_warc_CC-MAIN-20200926071311-20200926101311-00143.warc.gz\"}"}
http://www.oilfieldknowledge.com/resistivity-logs-and-its-applications/	[ "# Blog Details\n\n• ##### Professional Services\n• Ask any technical question and we will get you answer from experienced technical person\n\n• Talk to our recruitment expert for recruitment services.\n\n• ##### Oilfield expertise at your desktop\nPetrophysics", null, "#### Resistivity logs and its applications\n\nResistivity of material is an intrinsic property of material which measures how strongly that material opposes the flow of electric current.Its unit is Ohm-meter.\n\nThe potential difference V between two ends of wire of length L,cross section area A and supplied by electric current I is given as :-\n\nV= Ir …………………………..eq 1\n\nWhere r is resistance of wire ,which is constant for that wire,unit of r is ohms,Unit of potential difference V is volts and unit of electric current I is ampere\n\nResistance offered by wire is directly proportional to length of wire and inversely proportional to cross sectional area A.\n\nr a L/A ………………………….eq 2\n\nIn the above equation constant of proportionality is resistivity R ,hence equation 2 can written as:-\n\nr = R (L/A) …………………………eq 3\n\nResistivity is a physical property of material ,its value is constant for same material at same temperature.\n\nRe-arranging equation 3 we get:-\n\nR = r (A/L)\n\nAs A is cross-section area with unit in meter2 , L is length with unit in meter and r is resistance in ohms,unit of resistivity R comes out as Ohm- m2 /m or Ohm-m.\n\nReciprocal of resistance is conductance and reciprocal of resistivity is conductivity.\n\nUnit of conductivity as used in logging for practical reasons is mili mho / meter\n\nR= 1000/conductivity\n\nApplications of resistivity log:-\n\n1.Water Saturation:- Resistivity is important input for water saturation estimation.Archie equation provides relation for resistivity based water saturation calculation as below:-\n\nSw = [ (a Rw)/(qm *Rt)]1/n\n\nWhere:-\n\nSw = Water Saturation\n\na = Cementation factor\n\nRw = Formation Water resistivity\n\nq = Porosity\n\nm = Cementation exponent\n\nRt = True formation resistivity\n\nn = Saturation exponent\n\nAs we can see in above equation true formation resistivity and formation water resistivity are important input for water saturation calculation and resistivity tools help us in getting it.\n\nThere are certain assumptions associated with Archie equation as given below :-\n\n1a. Only water in pores acts as electrical conductor,matrix and hydrocarbon acts as insulator.\n\n1b. There is no shale in sands.As most of the reservoirs around the world have shales in matrix so several shaly sand equations are developed such as Indonesian equation,Simandoux equation,Waxman-smits and Dual-water.Most of these equations are modified form of archie equation.\n\n1c. Diameter of invasion:- Diameter of mud filtrate invasion is determined during resistivity environmental correction during determination of true resistivity.Tornado chart is used for determination of true resistivity and diameter of invasion.\n\n2.Well to well Correlation:- Doing well to well correlation using resisistivity curves has distinct advantage as correlation of resistivity is correlation of fluid volume and salinity in formation.\n\n3.Pore pressure estimation:- Resistivity can also be used for pore pressure estimation as per below equation:-\n\nPp = Phyd Rn / Rlog\n\nWhere:-\n\n1. Pp =Pore Pressure\n2. Phyd = Normal hydrostatic pore pressure\n3. Rn = Normal resistivity\n4. Rlog = Measured resistivity" ]	[ null, "http://www.oilfieldknowledge.com/wp-content/uploads/2017/02/resistivitylogs.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91731423,"math_prob":0.98722786,"size":3061,"snap":"2022-40-2023-06","text_gpt3_token_len":673,"char_repetition_ratio":0.1573438,"word_repetition_ratio":0.004210526,"special_character_ratio":0.20581509,"punctuation_ratio":0.09074074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99550295,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T02:28:37Z\",\"WARC-Record-ID\":\"<urn:uuid:b1bb13fb-ec48-42c9-894f-c57ce52250ed>\",\"Content-Length\":\"36440\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5476304-3f9d-4b79-b23e-06efa88367d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:563fa39f-8092-415a-b043-908021c676ab>\",\"WARC-IP-Address\":\"148.66.137.16\",\"WARC-Target-URI\":\"http://www.oilfieldknowledge.com/resistivity-logs-and-its-applications/\",\"WARC-Payload-Digest\":\"sha1:AQMMXDHWINICEZIKWE45ABPX32EXNIX2\",\"WARC-Block-Digest\":\"sha1:LWBKSUIRY7D3NEFVQP7JQWKUZUMV4B4E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500080.82_warc_CC-MAIN-20230204012622-20230204042622-00695.warc.gz\"}"}
https://smartpy.dev/docs/manual/syntax/integers-and-mutez	[ "# Integers and mutez \n\nSmartPy has several types for integers: `sp.nat` for non-negative integers, `sp.int` for (all) integers, and `sp.mutez` for non-negative amounts of micro-Tez, the token of the Tezos blockchain.\n\n## `sp.int` and `sp.nat`\n\nA literal integer such as `2` is either of type `sp.nat` or `sp.int`, depending on how it is used. In order to be explicit about the type, you can simply write `sp.nat(2)` or `sp.int(2)`.\n\n### Basic arithmetic \n\nThe operators `+`, `-`, `` and `/` perform addition, subtraction, multiplication, and division, respectively. They are homogeneous, meaning that both arguments must be of the same type (either both `sp.nat` or both `sp.int`).\n\nThe type of the result is the same as the arguments, except for `-`, which always returns an `sp.int`.\n\nExamples:\n\nsmartpy\n``````assert sp.nat(2) + sp.nat(3) == sp.nat(5)\nassert sp.int(2) + sp.int(3) == sp.int(5)\nassert sp.nat(2) - sp.nat(3) == sp.int(-1)\nassert sp.int(2) - sp.int(3) == sp.int(-1)\n``````\n\nThe unary negation operator `-` can take either `sp.nat` or `sp.int` and always returns an `sp.int`. For example, `- sp.nat(2) == sp.int(-2)`.\n\nMixing different types yields an error: for example `sp.nat(2) + sp.int(3)` is invalid. For heterogeneous arguments there are `sp.add`, `sp.sub`, `sp.mul`.\n\nExample:\n\nsmartpy\n``````a = sp.nat(2)\nb = sp.int(3)\n``````\n\n### Division \n\n`sp.ediv` performs Euclidean division. If both of its arguments are of type `sp.nat`, its result type is `sp.option[sp.pair[sp.nat, sp.nat]]`, otherwise `sp.option[sp.pair[sp.int, sp.nat]]`.\n\nThe option catches division by zero: `sp.ediv(a, 0) == None`. If `b != 0`, then `sp.ediv(a, b) = sp.Some(q, r)` where `q` (the quotient) and `r` (the remainder) are the unique integers such that `a == q b + r` and both `0 <= r` and `r < b`.\n\nExample:\n\nsmartpy\n``````assert sp.ediv( 14, 3) == sp.Some(( 4, 2)) # 14 == 4 * 3 + 2\nassert sp.ediv(-14, 3) == sp.Some((-5, 1)) # -14 == -5 * 3 + 1\nassert sp.ediv( 14,-3) == sp.Some((-4, 2)) # 14 == -4 * -3 + 2\nassert sp.ediv(-14,-3) == sp.Some(( 5, 1)) # -14 == 5 * -3 + 1\n``````\n\nIf `a` and `b` are of the same type quotient and remainder can be obtained as `a / b` and `sp.mod(a, b)`, respectively. In both cases a `Division_by_zero` error is raised if `b == 0`.\n\nWARNING\n\nFor negative denominators the result of division differs between SmartPy and Python. SmartPy follows the Michelson semantics, whereas Python rounds the quotient towards negative infinity (yielding negative remainders!). To avoid confusion between the two, the SmartPy syntax uses `/` and `sp.mod` instead of `//` and `%`.\n\n### Comparison \n\nTwo integers of the same type (either `sp.nat` or `sp.int`) can be compared with the operators `==`, `!=`, `<`, `<=`, `>`, `>=`. The result is of type `sp.bool`. For example, `2 == 3` is `False`.\n\n### From `sp.nat` to `sp.int`\n\nTo convert from `sp.nat` to `sp.int`, there is `sp.to_int`. For example `sp.to_int(sp.nat(2)) == sp.to_int(2)`, or (thanks to type inference) `sp.to_int(2) == 2`.\n\n### From `sp.int` to `sp.nat`\n\n`sp.is_nat` takes an `sp.int` and returns a value of type `sp.option[sp.nat]`. If `a >= 0`, then `sp.is_nat(a) == sp.Some(b)`, where `b` is the unique `sp.nat` such that `sp.to_int(b) == a` (i.e. `a` and `b` represent the same integer). If `a < 0`, then `sp.is_nat(a) == None`. For example:\n\nsmartpy\n``````assert sp.is_nat( 2) == sp.Some(2)\nassert sp.is_nat(-2) == None\n``````\n\n`sp.as_nat` performs the same conversion, but instead of returning an option raises an error on negative numbers. Thus `sp.as_nat(2) == 2`, whereas `sp.as_nat(-2)` yields an error.\n\n### Bitwise arithmetic \n\nBitwise and and or operations are available as `\|` and `&` on `sp.nat`:\n\nsmartpy\n``````assert 42 & 1 == 0\nassert 42 \| 1 == 43\n``````\n\n## Token amounts \n\nMicro-Tez amounts are always prefixed by `sp.mutez`, so e.g. `sp.mutez(42)` denotes 42 micro-Tez, e.g. `0.000042` Tez. This expression is of type `sp.mutez`.\n\n`sp.tez(a)` is equivalent to `sp.mutez(a * 1_000_000)`.\n\nAddition and subtraction on `sp.mutez` can be performed with `+` and `-`. If subtraction were to yield a negative value, an error is raised instead. Thus `sp.mutez(5) - sp.mutez(2) == sp.mutez(3)`, whereas `sp.mutez(5) - sp.mutez(10)` yields an error.\n\nsmartpy\n``````sp.split_tokens(amount: sp.mutez, quantity: sp.nat, division: sp.nat)\n-> sp.mutez\n``````\n\nFor combined multiplication and division on `sp.mutez` values there is `sp.split_tokens(a, b, c)`, which calculates the value \"`a * b / c`\".\n\nThis enables the computation of mutez percentages, for example:\n\nsmartpy\n``````sp.split_tokens(sp.mutez(200), 5, 100) == sp.mutez(10) # 5% of 200 mutez\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63034016,"math_prob":0.9861737,"size":4261,"snap":"2023-14-2023-23","text_gpt3_token_len":1379,"char_repetition_ratio":0.14869626,"word_repetition_ratio":0.0028328612,"special_character_ratio":0.3264492,"punctuation_ratio":0.22447014,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972961,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T02:21:56Z\",\"WARC-Record-ID\":\"<urn:uuid:51c1be19-88cc-4c54-92d1-15e9d045e745>\",\"Content-Length\":\"131410\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7edba96e-b116-4475-8837-765f823d7450>\",\"WARC-Concurrent-To\":\"<urn:uuid:669bd929-d369-4217-836c-146b8d1a0a80>\",\"WARC-IP-Address\":\"104.26.0.198\",\"WARC-Target-URI\":\"https://smartpy.dev/docs/manual/syntax/integers-and-mutez\",\"WARC-Payload-Digest\":\"sha1:7WQ3VB3FWSE4OOD3ZHKAK7VIG3T3WAYU\",\"WARC-Block-Digest\":\"sha1:UEHQ3NX5B7QW24GN7KGSC3CS7JQA5PIE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296944606.5_warc_CC-MAIN-20230323003026-20230323033026-00132.warc.gz\"}"}
https://bookdown.org/ltupper/340f21_notes/an-optional-historical-side-note-gosset-and-the-t.html	[ "## 4.9 An optional historical side-note: Gosset and the t\n\nWilliam Sealy Gosset was the quality control chemist and mathematician for the Guinness company – which brewed beer, but also did a bunch of agriculture – in the early 1900’s. So what did that have to do with all this distribution theory?\n\nWell, Gosset worked on quality control, of the product and the ingredients that went into it. As a QC person, he spent a lot of time measuring various quantities – checking the chemical characteristics of barley, that kind of thing. He noticed that when he created confidence intervals using the normal distribution,\n\n$\\bar{y} \\pm z^* \\frac{s}{\\sqrt{n}}$\n\nthe confidence intervals were too narrow: he was sending back too many samples as abnormal (the old “false positive” problem). So he did some simulations with data whose mean and SD he knew. (Please note that this was before computers: he did his simulations by writing values on three thousand pieces of cardboard and manually shuffling them until they were in random order. Don’t you feel better about the tidyverse now.) He divided his data into many small groups of size 4, then calculated the mean and SD of each little sample. Then he found the “z-score” of each group mean, but pretended he didn’t know $$\\sigma$$ and used the group’s $$s$$ instead: $z = \\frac{\\bar{y} - \\mu}{s/\\sqrt{4}}$\n\nWhen he did a histogram of these purported z-scores, it turned out that the “tails” were way too long for a normal distribution (such a distribution is sometimes called heavy-tailed). Why would that happen?\n\nWell, as we’ve seen, $$s$$ is often smaller than the true $$\\sigma$$. This makes $$z$$ too big. And because these little samples were so small, $$s$$ and therefore $$z$$ were often way off. The effect would be less pronounced with larger samples, since $$s$$ would be more reliable and a better estimate of $$\\sigma$$.\n\nGosset figured out what the right distribution actually was, but Guinness wouldn’t let him publish his work under his own name for… reasons. (Some say they were afraid of revealing trade secrets.) So he published it under a pseudonym: Student.\n\nSee this surprisingly entertaining article: https://www.jstor.org/stable/2683142.\n\nAs a side note to the side note, Gosset had an extended correspondence with a couple of other pro statisticians about this. In one letter to them, he wrote (about reading other people’s mathematical writing): “It’s not so much the mathematics. I can often say ‘Well, of course, that’s beyond me, but we’ll take it as correct’ but when I come to ‘Evidently’ I know that means two hours hard work at least before I can see why.”\n\nSome things never change." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.97533154,"math_prob":0.95890653,"size":2654,"snap":"2023-40-2023-50","text_gpt3_token_len":611,"char_repetition_ratio":0.086792454,"word_repetition_ratio":0.0,"special_character_ratio":0.23737754,"punctuation_ratio":0.1023166,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9585466,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T21:05:28Z\",\"WARC-Record-ID\":\"<urn:uuid:e6d7eab9-72c2-4b66-95b5-85f73bbb12dc>\",\"Content-Length\":\"66308\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7719f06a-7bbc-461d-b7b9-44f97a1983e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:dcd92abb-f7cc-4698-9b65-a3e3885bf58b>\",\"WARC-IP-Address\":\"34.198.117.108\",\"WARC-Target-URI\":\"https://bookdown.org/ltupper/340f21_notes/an-optional-historical-side-note-gosset-and-the-t.html\",\"WARC-Payload-Digest\":\"sha1:BXBE7RTUPVDOYQ3TMVOOCYDV32K7VKOJ\",\"WARC-Block-Digest\":\"sha1:NVFWI55EPCFSEOP7AU2PEDR5SN54MKII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100146.5_warc_CC-MAIN-20231129204528-20231129234528-00358.warc.gz\"}"}
http://itur.info/elapsed-time-worksheets-3rd-grade/	[ "# Elapsed Time Worksheets 3rd Grade\n\ntelling time worksheets grade free printable elapsed 3rd.\n\ntelling time worksheets grade free printable on elapsed for third.\n\ntelling time worksheets grade elapsed 3rd free.\n\nprintable time worksheets grade 3 elapsed 3rd pdf.\n\nanalog elapsed time worksheets 3rd grade easy.\n\neasy elapsed time worksheets free printable 3rd grade.\n\nelapsed time third grade worksheets within the hour 3rd.\n\nelapsed time third grade worksheets 3rd free.\n\nelapsed time worksheets grade math teaching squared 3rd easy.\n\ntelling time worksheets free first grade printable for graders 2 elapsed within the hour 3rd.\n\nelapsed time worksheets math 3rd grade number line.\n\ntime worksheets grade 3 math elapsed 3rd easy.\n\nthird grade time worksheets math elapsed 3rd.\n\nelapsed time worksheets grade within the hour 3rd.\n\nelapsed time worksheets grade worksheet telling 3rd easy.\n\ntelling time worksheets for grade elapsed 3rd easy.\n\nclocks worksheets time worksheet grade 3 telling the in half hour elapsed within 3rd.\n\ngrade 2 math time worksheets free printable elapsed 3rd.\n\ntell grade telling time worksheet eureka math free printable worksheets on elapsed for third.\n\nsecond grade telling time worksheets free printable clock for elapsed within the hour 3rd.\n\nelapsed time worksheets 3rd grade word problems.\n\nelapsed time worksheets free printable 3rd grade.\n\nfirst grade time worksheets printable math elapsed 3rd word problems.\n\ngrade time worksheets elapsed 3rd word problems.\n\nanalog elapsed time start from quarter hours worksheets 3rd grade word problems.\n\ntelling time worksheets grade 3 elapsed 3rd number line.\n\nprintable time worksheets grade 3 free on elapsed for third.\n\nmath elapsed time worksheets with clocks 3rd grade free.\n\nelapsed time worksheets grade math 3rd.\n\nelapsed time word problems grade worksheets 3rd number line.\n\nelapsed time worksheets math with clocks grade printable free for 3rd.\n\ntelling time worksheets grade 3 hour elapsed to the clock half 3rd easy.\n\nfree printable time worksheets elapsed 3rd grade." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7877494,"math_prob":0.7913542,"size":2069,"snap":"2019-43-2019-47","text_gpt3_token_len":433,"char_repetition_ratio":0.35690072,"word_repetition_ratio":0.15806451,"special_character_ratio":0.18656355,"punctuation_ratio":0.09770115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9848009,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T23:39:42Z\",\"WARC-Record-ID\":\"<urn:uuid:b6c33ca3-bd01-4ef2-b13d-c146db60b0e9>\",\"Content-Length\":\"36614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b3e353b-13fa-42b6-8bcc-3a23eb7d8b7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc9d701b-98cd-405c-bfff-1ec2b8b308e7>\",\"WARC-IP-Address\":\"104.28.5.77\",\"WARC-Target-URI\":\"http://itur.info/elapsed-time-worksheets-3rd-grade/\",\"WARC-Payload-Digest\":\"sha1:LBRDQ6BVKYCXRUQ7BKK2M7S5JQDEAHDN\",\"WARC-Block-Digest\":\"sha1:4PXGXIBTAZVEZ7JZDTWBTFOSYXHIJSLR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670268.0_warc_CC-MAIN-20191119222644-20191120010644-00357.warc.gz\"}"}
https://www.physicsforums.com/threads/gyromagnetic-ratio.90678/	[ "# Gyromagnetic ratio\n\nHomework Helper\nGold Member\n\n## Main Question or Discussion Point\n\nHi,\n\nThe gyromagnetic ratio is the ratio of the magnetic dipole moment to the angular momentum.\n\nI really don't get how the fact that the gyromagnetic ratio of a rotating circular loop of mass M and charge Q is g = Q/2M implies that the gyromagnetic ratio of a uniform rotating spere is also g = Q/2M!\n\nRelated Classical Physics News on Phys.org\nTide\nHomework Helper\nPoints to ponder:\n\n(a) Does the gyromagnetic ratio of the rotating loop depend on the size of the loop?\n\n(b) Can you determine the dipole moment of a rotating spherically symmetric charge distribution by summing over loops?\n\nHomework Helper\nGold Member\nTide said:\nPoints to ponder:\n\n(a) Does the gyromagnetic ratio of the rotating loop depend on the size of the loop?\nNo.\n\nTide said:\n(b) Can you determine the dipole moment of a rotating spherically symmetric charge distribution by summing over loops?\nThis is exactly what's throwing me off! In principle, since the sphere could be considered as an infinity of circular loops of different sizes, its dipole moment is the sum of all those. But since the sphere is uniform, the ratio of its mass to its charge is the same at every point of it. Hence, the dipole moment of each loops is the same at Q/2M, making the total dipole of the sphere infinite!\n\nDr Transport\nGold Member\nThe dipole moment for a charged rotating sphere is a problem in Griffths, Wangsness and Jackson amongst other texts. It is not infinite, look in some these texts to get an idea on how to calculate it. I think that the easiest manner to go about it is to start with the vector potential for a rotating sphere and go on from there.\n\nHomework Helper\nGold Member\nGriffiths says:\n\nDavid J. said:\nWhat is the gyromagnetic ratio for a uniform spinnin sphere? [This requires no new calculation; simply decompose the sphere into infinitesimal rings, and apply the result of part (a)\nThe result of part (a) was of course that g = Q/2M for a ring.\n\nSo he seems to be saying that just by reasoning, we can determine that it is the same. I exposed my reasoning to you; what is flawed in it?\n\nHomework Helper\nGold Member\nI did it by integration.\n\nDoes anyone know what is the argument that allows one to conclude that the gyromagnetic ratio of any solid of revolution is Q/2M ?\n\nTide\nHomework Helper\nYou should have gotten a clue from the work you did for the sphere. The integrals for the magnetic moment and angular momentum cancel when you form the ratio! :)\n\nGokul43201\nStaff Emeritus\nGold Member\nquasar987 said:\nNo.\n\nThis is exactly what's throwing me off! .... Hence, the dipole moment of each loops is the same at Q/2M, making the total dipole of the sphere infinite!\nNo, each loop only has some infinitesimal charge dQ and infinitesimal mass dM.\n\nHomework Helper\nGold Member\nGokul43201 said:\nNo, each loop only has some infinitesimal charge dQ and infinitesimal mass dM.\nSure but I'm convinced that nevertheless, their ratio is the same:\n\n$$\\frac{dQ}{dM} = \\frac{dQ/dV}{dM/dV} = \\frac{\\rho_q}{\\rho_m} = \\frac{Q}{M}$$\n\nHomework Helper\nGold Member\nTide said:\nYou should have gotten a clue from the work you did for the sphere. The integrals for the magnetic moment and angular momentum cancel when you form the ratio! :)\nFor the sphere ok. But that seemed almost like magic you know. There are all these constant that come in from integration and they end up being 1/2 at the end. How do I know it wasn't a big coincidence and that for another random revolution figure, it will be different?\n\nEdit: Ok, now I see it from the integral, by generalizing to an arbitrary volume of revolution, that it will always give Q/2M. Thanks for the hint Tide.\n\nLast edited:\nTide" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89799047,"math_prob":0.9548465,"size":706,"snap":"2019-51-2020-05","text_gpt3_token_len":183,"char_repetition_ratio":0.18518518,"word_repetition_ratio":0.8,"special_character_ratio":0.23937677,"punctuation_ratio":0.044444446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9909166,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T05:29:40Z\",\"WARC-Record-ID\":\"<urn:uuid:c145feeb-1f34-432a-9a4d-513bd37cd65b>\",\"Content-Length\":\"106350\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec0cc854-415d-4499-866b-a25a3198c6b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:420f60eb-427a-4b50-9852-e3c74a7657c7>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/gyromagnetic-ratio.90678/\",\"WARC-Payload-Digest\":\"sha1:VMM2TUB627P6B65CXOSKH2XZTBBH35IG\",\"WARC-Block-Digest\":\"sha1:OLSTK5O3NCD2KKA6GYCASCNAJZXQUGQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601615.66_warc_CC-MAIN-20200121044233-20200121073233-00380.warc.gz\"}"}
https://waseda.pure.elsevier.com/en/publications/impossibility-of-weak-convergence-of-kernel-density-estimators-to	[ "# Impossibility of weak convergence of kernel density estimators to a non-degenerate law in L2(ℝd)\n\nResearch output: Contribution to journalArticle\n\n3 Citations (Scopus)\n\n### Abstract\n\nIt is well known that the kernel estimator, for the probability density f on ℝd has pointwise asymptotic normality and that its weak convergence in a function space, especially with the uniform topology, is a difficult problem. One may conjecture that the weak convergence in L2(ℝd) could be possible. In this paper, we deny this conjecture. That is, letting, we prove that for any sequence {rn} of positive constants such that rn = o(√n), if the rescaled residual, converges weakly to a Borel limit in L2(ℝd), then the limit is necessarily degenerate.\n\nOriginal language English 129-135 7 Journal of Nonparametric Statistics 23 1 https://doi.org/10.1080/10485251003678507 Published - 2011 Mar Yes\n\n### Fingerprint\n\nKernel Density Estimator\nWeak Convergence\nKernel Estimator\nProbability Density\nAsymptotic Normality\nFunction Space\nTopology\nConverge\nEstimator\nKernel density\nWeak convergence\nImpossibility\nKernel estimator\nAsymptotic normality\n\n### Keywords\n\n• Kernel estimator\n• Weak convergence in L Space\n\n### ASJC Scopus subject areas\n\n• Statistics and Probability\n• Statistics, Probability and Uncertainty\n\n### Cite this\n\nIn: Journal of Nonparametric Statistics, Vol. 23, No. 1, 03.2011, p. 129-135.\n\nResearch output: Contribution to journalArticle\n\n@article{23de33109ef644f38caafeb4e1f0e25c,\ntitle = \"Impossibility of weak convergence of kernel density estimators to a non-degenerate law in L2(ℝd)\",\nabstract = \"It is well known that the kernel estimator, for the probability density f on ℝd has pointwise asymptotic normality and that its weak convergence in a function space, especially with the uniform topology, is a difficult problem. One may conjecture that the weak convergence in L2(ℝd) could be possible. In this paper, we deny this conjecture. That is, letting, we prove that for any sequence {rn} of positive constants such that rn = o(√n), if the rescaled residual, converges weakly to a Borel limit in L2(ℝd), then the limit is necessarily degenerate.\",\nkeywords = \"Kernel estimator, Weak convergence in L Space\",\nauthor = \"Yoichi Nishiyama\",\nyear = \"2011\",\nmonth = \"3\",\ndoi = \"10.1080/10485251003678507\",\nlanguage = \"English\",\nvolume = \"23\",\npages = \"129--135\",\njournal = \"Journal of Nonparametric Statistics\",\nissn = \"1048-5252\",\npublisher = \"Taylor and Francis Ltd.\",\nnumber = \"1\",\n\n}\n\nTY - JOUR\n\nT1 - Impossibility of weak convergence of kernel density estimators to a non-degenerate law in L2(ℝd)\n\nAU - Nishiyama, Yoichi\n\nPY - 2011/3\n\nY1 - 2011/3\n\nN2 - It is well known that the kernel estimator, for the probability density f on ℝd has pointwise asymptotic normality and that its weak convergence in a function space, especially with the uniform topology, is a difficult problem. One may conjecture that the weak convergence in L2(ℝd) could be possible. In this paper, we deny this conjecture. That is, letting, we prove that for any sequence {rn} of positive constants such that rn = o(√n), if the rescaled residual, converges weakly to a Borel limit in L2(ℝd), then the limit is necessarily degenerate.\n\nAB - It is well known that the kernel estimator, for the probability density f on ℝd has pointwise asymptotic normality and that its weak convergence in a function space, especially with the uniform topology, is a difficult problem. One may conjecture that the weak convergence in L2(ℝd) could be possible. In this paper, we deny this conjecture. That is, letting, we prove that for any sequence {rn} of positive constants such that rn = o(√n), if the rescaled residual, converges weakly to a Borel limit in L2(ℝd), then the limit is necessarily degenerate.\n\nKW - Kernel estimator\n\nKW - Weak convergence in L Space\n\nUR - http://www.scopus.com/inward/record.url?scp=79952460890&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=79952460890&partnerID=8YFLogxK\n\nU2 - 10.1080/10485251003678507\n\nDO - 10.1080/10485251003678507\n\nM3 - Article\n\nAN - SCOPUS:79952460890\n\nVL - 23\n\nSP - 129\n\nEP - 135\n\nJO - Journal of Nonparametric Statistics\n\nJF - Journal of Nonparametric Statistics\n\nSN - 1048-5252\n\nIS - 1\n\nER -" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8051411,"math_prob":0.93796563,"size":2936,"snap":"2019-51-2020-05","text_gpt3_token_len":816,"char_repetition_ratio":0.11732606,"word_repetition_ratio":0.63596493,"special_character_ratio":0.26737058,"punctuation_ratio":0.12592593,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9823167,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T06:44:50Z\",\"WARC-Record-ID\":\"<urn:uuid:c4112c13-e771-4ee0-a232-882df90ceae6>\",\"Content-Length\":\"38278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19614d01-1dc7-4e36-829f-c98187b7fa0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6b3da27-f175-4f57-830d-689c214024a7>\",\"WARC-IP-Address\":\"52.220.215.79\",\"WARC-Target-URI\":\"https://waseda.pure.elsevier.com/en/publications/impossibility-of-weak-convergence-of-kernel-density-estimators-to\",\"WARC-Payload-Digest\":\"sha1:UH5OXHFXF4GX2MKOWT6HS6JQVYJ7X5U2\",\"WARC-Block-Digest\":\"sha1:VHWURTRGWY3YZIXTGSV5RRT3RF7QOEGS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250615407.46_warc_CC-MAIN-20200124040939-20200124065939-00185.warc.gz\"}"}
http://narrabio.com/ebooks/advanced-visual-quantum-mechanics	[ "# Download Advanced visual quantum mechanics by Bernd Thaller PDF", null, "By Bernd Thaller\n\nVisual Quantum Mechanics is a scientific attempt to enquire and to coach quantum mechanics simply by computer-generated animations. even though it is self-contained, this e-book is a part of a two-volume set on visible Quantum Mechanics. the 1st booklet seemed in 2000, and earned the eu educational software program Award in 2001 for oustanding innovation in its box. whereas themes in publication One quite often involved quantum mechanics in a single- and two-dimensions, e-book units out to give third-dimensional structures, the hydrogen atom, debris with spin, and relativistic particles. It additionally includes a uncomplicated path on quantum info conception, introducing issues like quantum teleportation, the EPR paradox, and quantum desktops. jointly the 2 volumes represent an entire direction in quantum mechanics that areas an emphasis on rules and ideas, with a good to average quantity of mathematical rigor. The reader is anticipated to be conversant in calculus and common linear algebra. to any extent further mathematical suggestions can be illustrated within the textual content.\n\nTh CD-ROM encompasses a huge variety of Quick-Time video clips provided in a multimedia-like setting. the films illustrate and upload colour to the text, and let the reader to view time-dependent examples with a degree of interactivity. The point-and-click interface isn't any tougher than utilizing the web.\n\nRead or Download Advanced visual quantum mechanics PDF\n\nBest quantum physics books\n\nPath integrals and their applications in quantum, statistical, and solid state physics\n\nThe complicated research Institute on \"Path Integrals and Their functions in Quantum, Statistical, and reliable kingdom Physics\" used to be held on the college of Antwerpen (R. U. C. A. ), July 17-30, 1977. The Institute used to be subsidized by means of NATO. Co-sponsors have been: A. C. E. C. (Belgium), Agfa-Gevaert (Belgium), l'Air Li uide BeIge (Belgium), Be1gonucleaire (Belgium), Bell cell Mfg.\n\nQuantum theory and measurement\n\nThe forty-nine papers gathered the following remove darkness from the that means of quantum thought because it is disclosed within the size procedure. including an advent and a supplemental annotated bibliography, they speak about matters that make quantum idea, overarching precept of twentieth-century physics, seem to many to prefigure a brand new revolution in technology.\n\nQuantum statistical mechanics and Lie group harmonic analysis\n\nHarm N. , Hermann R. Quantum statistical mechanics and lie staff harmonic research (Math Sci Press, 1980)(ISBN 0915692309)(260s)\n\nExtra info for Advanced visual quantum mechanics\n\nSample text\n\n156) where Jν and Nν are the Bessel function and Neumann functions of order ν. ˆ is singular for > 0. 15 allows investigation of the dependence on . 2. Special Topic: Properties of the Riccati-Bessel functions The real-valued functions jˆ (z) and n ˆ (z) are for z > 0 solutions of the equation − ( + 1) d2 y(z) + y(z) − y(z) = 0. 158) 46 1. 159) jˆ (z) = −(−z) +1 jˆ0 (z) , z dz z 1 d 1 n ˆ 0 (z) . 160) z dz z Their limiting behavior for small z is given by 2 ! 161) (2 + 1)! (2 )! 162) z 1 + O(z 2 ) , as z → 0.\n\nDuring one period, the trajectory goes twice through ϑ, hence we multiply the time by 2 to obtain the total time spent in dϑ during one period. The time needed to complete a period is T = 2π/L. Hence, the probability of ﬁnding the particle in dϑ is L dt(ϑ) L 2 dt(ϑ) = dϑ. 134) 2π π dϑ Hence, one needs to invert the function ϑ(t) and diﬀerentiate with respect to ϑ. It is suﬃcient to invert the function ϑ(t) on a part of the time interval during which the trajectory goes through the point ϑ under consideration.\n\nIn a classical picture, the angular momentum vector (if measured with respect to a certain direction) thus lies on certain cones. Eigenvalues of the orbital angular momentum: The eigenvalues of the operator L2 are precisely the numbers 2 ( + 1), where is a non-negative integer. For each , the operator L3 has the eigenvalues m, where m = − , − + 1, . . , . 7. In quantum mechanics, this picture should not be taken seriously because the same result would be obtained for the possible values of L1 and L2 (or the component of L in an arbitrary direction).\n\nDownload PDF sample\n\nRated 4.44 of 5 – based on 29 votes" ]	[ null, "https://images-na.ssl-images-amazon.com/images/I/41OWXLJedRL._SX332_BO1,204,203,200_.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8668677,"math_prob":0.9290829,"size":4596,"snap":"2021-21-2021-25","text_gpt3_token_len":1124,"char_repetition_ratio":0.1119338,"word_repetition_ratio":0.0026075619,"special_character_ratio":0.23781548,"punctuation_ratio":0.114716105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96925366,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-22T14:34:10Z\",\"WARC-Record-ID\":\"<urn:uuid:bd659226-86dc-4c4d-97ed-b680955da63f>\",\"Content-Length\":\"32187\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e8b3bdd-d98b-409f-b4c0-c032a1557549>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1cde7e8-70cd-443e-a41c-2cc92b277db6>\",\"WARC-IP-Address\":\"173.201.92.128\",\"WARC-Target-URI\":\"http://narrabio.com/ebooks/advanced-visual-quantum-mechanics\",\"WARC-Payload-Digest\":\"sha1:UN62PUCPM63PBR73KWSSUMEQE6VY67FO\",\"WARC-Block-Digest\":\"sha1:7T5F65HYEBTGDD46Z4RZI2V56SC6MHG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488517820.68_warc_CC-MAIN-20210622124548-20210622154548-00066.warc.gz\"}"}
https://www.vocal.com/noise-reduction/a-basic-voice-activity-detection-vad/	[ "The voice activity detection (VAD) is a sensitive component in spectrum subtraction. Its performance can dramatically influence the noise reduction level and the speech distortion severity.\n\nThe power spectrum of a speech segment has several characteristics that can be used to decide on the presence or absence of speech. The commonly used measures include short-time averaged energy, zero-crossing rate, and linear prediction coefficients etc.\n\nFigure 1 shows a basic VAD system. It has three main components. Feature extraction unit analyzes and calculates a set of decision metrics, feature vectors, to be used for VAD threshold computation and VAD decision making. Various implementations may use application specific definitions of feature vectors.\n\nEnergy based approaches are the most often used due to their simplicity. The design rationale is that speech bursts have higher transient power than the background noise. A long-term energy averaging can be used to measure the background noise level while a short-term averaging can be used to measure the transient speech power. If the short time averaging is higher than the long time averaging by a certain threshold, VAD declares speech presence. The decision mechanism may be implemented by the following logic,", null, "$VAD\\ =\\ 1,ifEs>El0, otherwise$\n\nwhere", null, "$E_s$ and", null, "$E_l$ are the short- and long-term energy values. They can be calculated frame by frame in time domain.", null, "$E_s=\\frac{1}{N}\\sum_{t\\ =\\ 1}^{N}{x^2\\left(t\\right)}$\n\nwhere N is the frame length.\n\nThe long term can be obtained from a frame by frame moving average with an adjustable decay parameter as below,", null, "$E_l=\\left(1-\\gamma\\right)E_s+\\gamma\\ E_l^{OLD}$\n\nwhere", null, "$E_l^{OLD}$ is from previous frame and", null, "$\\gamma$ is the decay parameter.\n\nThe performance of the VAD can obviously be improved if a feedback approach is adopted to evaluate the long term energy smoothing." ]	[ 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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9234314,"math_prob":0.99000925,"size":1706,"snap":"2020-34-2020-40","text_gpt3_token_len":310,"char_repetition_ratio":0.113396004,"word_repetition_ratio":0.02255639,"special_character_ratio":0.17702228,"punctuation_ratio":0.080267556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99319077,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T11:44:43Z\",\"WARC-Record-ID\":\"<urn:uuid:05586406-215d-44c2-a9f1-f08b3eb260b4>\",\"Content-Length\":\"93161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0994e3f-5ae3-4af0-86cb-c0299df35989>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6936171-c870-4ce2-afcd-5610226ec8cc>\",\"WARC-IP-Address\":\"72.236.255.161\",\"WARC-Target-URI\":\"https://www.vocal.com/noise-reduction/a-basic-voice-activity-detection-vad/\",\"WARC-Payload-Digest\":\"sha1:2XMEKBFVB3SPU66E3IP6NCUJRQDXLOMA\",\"WARC-Block-Digest\":\"sha1:JYZR7BVMJLRS3ATDN4J6KY4PJMHKBPMG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401641638.83_warc_CC-MAIN-20200929091913-20200929121913-00665.warc.gz\"}"}
http://www.51wenwen.com/question-12353.html	[ "", null, "", null, "分享\n\n1个回答\n\n• 热门排序\n• 时间排序\n\n0个人赞同了Ta的回答", null, "0", null, "评论0\n\n• 关注数\n1\n• 浏览数\n109", null, "相关问题\n\n8个回答\n\n1个回答\n\n1个回答\n\n• 我邀请的列表\n• 擅长该话题的人\n• 关注该话题的人\n• 我关注的人\n•", null, "雙禧\n\n该用户一共参与了0 个回答\n\n邀请回答\n•", null, "孟令轩\n\n该用户一共参与了1 个回答\n\n邀请回答\n•", null, "戴伟\n\n该用户一共参与了0 个回答\n\n邀请回答\n•", null, "问问管家\n\n该用户一共参与了 14 个回答\n\n邀请回答\n•", null, "任勇洙\n\n该用户一共参与了 84 个回答\n\n邀请回答\n•", null, "Kaiser\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "张朗朗\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "村口小猎人\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "张大东\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "popo\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "陈查理\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "云淡风轻\n\n该用户一共参与了 1 个回答\n\n邀请回答\n•", null, "陈晓畅\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "Alives\n\n该用户一共参与了 1 个回答\n\n邀请回答\n•", null, "范思瑶\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "王自飞\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "燕子翩然\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "心灵\n\n该用户一共参与了 0 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null, "云川\n\n该用户一共参与了 1 个回答\n\n邀请回答\n•", null, "KimRin\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "阳光\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "海陵居士\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "风烈\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "木人小\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "者也\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "风之羽\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "王东来\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "锄禾\n\n该用户一共参与了 0 个回答\n\n邀请回答\n•", null, "此去经年\n\n该用户一共参与了 6 个回答\n\n邀请回答\n•", null, "问问旅行\n\n该用户一共参与了 0 个回答\n\n邀请回答\n\n• 移民\n• 留学\n• 海外投资\n• 新加坡移民\n• 出国旅游\n• 德国投资\n• 新加坡投资\n• 香港投资\n• 泰国投资\n• 塞浦路斯投资\n• 日本投资\n• 马来西亚投资\n• 菲律宾投资\n• 英国投资\n• 荷兰投资\n• 比利时投资\n• 爱尔兰投资\n• 意大利投资\n• 希腊投资\n• 西班牙投资\n• 葡萄牙投资\n• 美国投资\n• 加拿大投资\n• 新西兰投资\n• 斐济投资\n• 澳大利亚投资\n• 瑞典投资\n• 瑞士留学\n• 德国留学\n• 波兰留学\n• 新加坡留学\n• 香港留学\n• 泰国留学\n• 日本留学\n• 马来西亚留学\n• 菲律宾留学\n• 英国留学\n• 荷兰留学\n• 比利时留学\n• 爱尔兰留学\n• 意大利留学\n• 希腊留学\n• 西班牙留学\n• 葡萄牙留学\n• 马耳他留学\n• 美国留学\n• 加拿大留学\n• 新西兰留学\n• 澳大利亚留学\n• 瑞典留学\n• 挪威留学\n• 芬兰留学\n• 匈牙利移民\n• 瑞士移民\n• 德国移民\n• 波兰移民\n• 奥地利移民\n• 香港移民\n• 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耶路撒冷音乐和舞蹈学院\n• 牙买加留学\n• 摩纳哥留学\n• 西印度大学\n• 牙买加理工大学\n• 加勒比海事大学\n• 摩纳哥国际大学\n• 格拉茨大学\n• 因斯布鲁克大学\n• 维也纳医科大学\n• 格拉茨医科大学\n• 因斯布鲁克医科大学\n• 萨尔茨堡大学\n• 莱奥本矿业大学\n• 维也纳农业大学\n• 维也纳兽医大学\n• 维也纳经济大学\n• 林茨大学\n• 克拉根福大学\n• 克莱姆斯多瑙大学\n• 维也纳美术学院\n• 维也纳应用艺术大学\n• 维也纳音乐与表演艺术大学\n• 萨尔茨堡莫扎特音乐大学\n• 格拉茨音乐与表演艺术大学\n• 林茨艺术与工业设计大学\n• 维也纳模都尔大学\n• 维也纳音乐与艺术市立大学\n• 维也纳爵士流行音乐私立大学\n• 西伯格堡私立大学\n• 海法大学\n• 奥博学术大学\n• 奥卢大学\n• 诺维阿应用科学大学\n• 城市应用科学大学\n• 罗约阿应用科学大学\n• 拉普兰应用科学大学\n• 拉彭兰塔工业大学\n• 拉赫蒂应用科学大学\n• 汉肯经济学院\n• 卡莱利亚应用科学大学\n• 卡亚尼应用科学大学\n• 芬兰东南应用科学大学\n• 于韦斯屈莱应用科学大学\n• 海门应用科学大学\n• 胡马克应用科学大学\n• 哈格哈里亚应用科学大学\n• 迪亚科尼亚应用科学大学\n• 芬兰中央应用科技大学\n• 图尔库大学\n• 阿卡达应用科学大学\n• 瓦萨大学\n• 拉普兰大学\n• 赫尔辛基艺术大学\n• 东芬兰大学\n• 赫尔辛基大学\n• 奥卢应用科学大学\n• 塞马应用科学大学\n• 萨塔昆塔应用科学大学\n• 萨沃尼亚应用科学大学\n• 塞伊奈约基应用科学大学\n• 坦佩雷应用科学大学\n• 图尔库应用科学大学\n• 瓦萨应用科学大学\n• 于韦斯屈莱大学\n• 埃及留学\n• 开罗大学\n• 亚历山大大学\n• 艾因夏姆斯大学\n• 艾斯尤特大学\n• 坦塔大学\n• 曼苏尔大学\n• 扎加齐克大学\n• 哈勒旺大学\n• 米尼亚大学\n• 姆努菲亚大学\n• 苏伊士运河大学\n• 南河谷大学\n• 法尤姆大学\n• 班尼苏维夫大学\n• 本哈大学\n• 谢赫村大学\n• 苏哈贾大学\n• 塞得港大学\n• 德曼胡尔大学\n• 阿斯旺大学\n• 杜姆亚特大学\n• 苏伊士大学\n• 萨达特城大学\n• 埃及科技大学\n• 埃及国际大学\n• 文学及现代科学十月大学\n• 十月六号大学\n• 现代技术与信息大学\n• 埃及未来大学\n• 法鲁斯大学\n• 尼罗大学\n• 复兴大学\n• 西奈大学\n• 三角洲科技大学\n• 斋月十号大学\n• 班达尔大学\n• 拉科鲁尼亚大学\n• 阿尔卡拉大学\n• 阿利坎特大学\n• 阿尔梅利亚大学\n• 马德里自治大学\n• 布尔戈斯大学\n• 卡的斯大学\n• 坎塔布里亚大学\n• 马德里卡洛斯三世大学\n• 卡斯蒂利亚拉曼却大学\n• 马德里康普顿斯大学\n• 科尔多瓦大学\n• 埃斯特雷马杜拉大学\n• 赫罗纳大学\n• 格拉纳达大学\n• 维尔瓦大学\n• 巴利阿里群岛大学\n• 安达卢西亚国际大学\n• 梅南德斯·佩拉尤国际大学\n• 哈恩大学\n• 海梅一世大学\n• 拉古纳大学\n• 里奥哈大学\n• 加那利群岛拉斯帕尔马斯大学\n• 莱昂大学\n• 莱里达大学\n• 马拉加大学\n• 米格尔·埃尔南德斯·德埃尔切大学\n• 穆尔西亚大学\n• 奥维耶多大学\n• 帕布罗·德奥拉韦德大学\n• 巴斯克大学\n• 卡塔赫纳理工大学\n• 加泰罗尼亚理工大学\n• 马德里理工大学\n• 瓦伦西亚理工大学\n• 朋培法普拉大学\n• 纳瓦拉公立大学\n• 胡安卡洛斯国王大学\n• 罗维拉·依维尔基里大学\n• 萨拉曼卡大学\n• 圣地亚哥德孔波斯特拉大学\n• 塞维利亚大学\n• 瓦伦西亚大学\n• 巴里亚多利德大学\n• 维戈大学\n• 萨拉戈萨大学\n• 卡米亚斯大主教大学\n• 德乌斯托大学\n• 拉蒙尤以大学\n• 圣帕布洛大学\n• 安东尼奥·德·内布里哈大学\n• 欧洲塞万提斯大学\n• 萨拉曼卡天主教大学\n• 维克大学——加泰罗尼亚中央大学\n• IE大学\n• 马德里欧洲大学\n• 蒙德拉贡大学\n• 智者阿方索十世大学\n• 圣安东尼奥天主教大学\n• 捷克留学\n• 查理大学\n• 捷克布杰约维采南波西米亚大学\n• 拉贝河畔乌斯季扬•埃万盖利斯塔•普尔基涅大学\n• 马萨里克大学\n• 帕拉茨基大学\n• 奥斯特拉发大学\n• 赫拉德茨－克拉洛维大学\n• 奥帕瓦西里西亚大学\n• 捷克技术大学\n• 布拉格化工大学\n• 皮尔森西波西米亚大学\n• 利贝雷茨技术大学\n• 帕尔杜比采大学\n• 布尔诺技术大学\n• 奥斯特拉发技术大学\n• 托马斯拔佳大学\n• 布拉格经济大学\n• 捷克生命科学大学\n• 布尔诺孟德尔大学\n• 布拉格表演艺术学院\n• 布拉格美术学院\n• 布拉格艺术、建筑与设计学院\n• 布尔诺亚纳切克音乐与表演艺术学院\n• 伊赫拉瓦理工学院\n• 捷克布杰约维采技术与商务学院\n• 区域发展与银行学院\n• 欧洲理工学院\n• 布拉格酒店管理学院\n• 弗兰克-戴森教育学院\n• 高级法律研究大学\n• 经济与管理大学\n• 纽约大学布拉格分校\n• 布拉格国际和公共关系学院\n• 政治和社会科学学院\n• 欧洲和区域研究学院\n• 皮塞克国际电影研究学院\n• 体育学院\n• 牛顿学院\n• 物流学院\n• 护理医学院\n• 布尔诺国际商学院\n• 私立经济研究学院\n• 布拉格商业大学\n• 斯丁学院\n• 布拉格大都会大学\n• 夸美纽斯大学\n• 卡雷尔英语学院\n• 英美大学\n• 布拉格心理学研究学院\n• 西摩拉维亚学院\n• 奥洛穆茨摩拉维亚商学院\n• Cervo学院\n• 麒麟学院\n• 商业与酒店管理学院\n• 社会与行政学院\n• Akcent学院\n• 布拉格建筑学院\n• 应用心理学学院\n• 斯柯达汽车大学\n• 艺术与设计学院\n• 创业与法律学院\n• 创意传播大学\n• 布拉格金融管理大学\n• 捷克共和国警察学院\n• 捷克国防大学\n• 土耳其移民\n• 亚洲大学\n• 安东国立大学\n• 安养大学\n• 白石大学\n• 釜山教育大学\n• 釜山外国语大学\n• 加尔文大学\n• 大邱加图立大学\n• 釜山加图立大学\n• 昌原国立大学\n• 清州国立教育大学\n• 清州大学\n• 晋州教育大学\n• 草堂大学\n• 全北国立大学\n• 总神大学\n• 全南大学\n• 朝鲜大学\n• 秋溪艺术大学\n• 春川教育大学\n• 中央大学\n• 忠北大学\n• 忠南大学\n• 青云大学\n• 车医科学大学\n• 昌信大学\n• 加图立关东大学\n• 大邱庆北科学技术院\n• 大邱韩医大学\n• 大邱教育大学\n• 大邱大学\n• 大田加图立大学\n• 大田大学\n• 大真大学\n• 檀国大学\n• 东亚大学\n• 同德女子大学\n• 东义大学\n• 东国大学\n• 东西大学\n• 东新大学\n• 东洋大学\n• 德成女子大学\n• 大邱艺术大学\n• 梨花女子大学\n• 乙支大学\n• 极东大学\n• 嘉泉大学\n• 金刚大学\n• 金泉大学\n• 公州教育大学\n• 光州加图立大学\n• 光州科学技术院\n• 光州教育大学\n• 光州大学\n• 京仁教育大学\n• 庆州大学\n• 庆南科学技术大学\n• 庆尚大学\n• 江陵原州大学\n• 汉拿大学\n• 翰林大学\n• 韩博大学\n• 韩东大学\n• 韩国外国语大学\n• 韩京国立大学\n• 汉丽大学\n• 韩南大学\n• 韩世大学\n• 韩瑞大学\n• 韩信大学\n• 汉城大学\n• 汉阳大学\n• 湖南大学\n• 弘益大学\n• 湖西大学\n• 湖原大学\n• 协成大学\n• 仁川加图立大学\n• 仁荷大学\n• 仁济大学\n• 韩国国际大学\n• 仁川大学\n• 济州国立大学\n• 全州教育大学\n• 全州大学\n• 中部大学\n• 中源大学\n• 济州国际大学\n• 韩国交通大学\n• 韩国科学技术院\n• 江南大学\n• 江原大学\n• 加耶大学\n• 启明大学\n• 公州国立大学\n• 建国大学\n• 建阳大学\n• 国民大学\n• 韩国航空大学\n• 韩国海洋大学\n• 韩国放送通信大学\n• 韩国体育大学\n• 韩国艺术综合大学\n• 韩国教员大学\n• 拿撒勒大学\n• 韩国产业技术大学\n• 高丽大学\n• 韩国技术教育大学\n• 金乌工科大学\n• 群山大学\n• 光州女子大学\n• 光云大学\n• 京畿大学\n• 庆熙大学\n• 庆北大学\n• 京东大学\n• 庆一大学\n• 庆南大学\n• 庆星大学\n• 庆云大学\n• 韩国传统文化大学\n• 木浦加图立大学\n• 木浦海洋大学\n• 木浦大学\n• 牧园大学\n• 明知大学\n• 南部大学\n• 南首尔大学\n• 培材大学\n• 浦项工科大学\n• 釜庆大学\n• 釜山大学\n• 平泽大学\n• 世翰大学\n• 信韩大学\n• 三育大学\n• 尚志大学\n• 祥明大学\n• 世宗大学\n• 世明大学\n• 西京大学\n• 首尔大学\n• 首尔教育大学\n• 首尔科学技术大学\n• 首尔女子大学\n• 西原大学\n• 新京大学\n• 新罗大学\n• 西江大学\n• 松源大学\n• 淑明女子大学\n• 顺天乡大学\n• 崇实大学\n• 顺天大学\n• 成均馆大学\n• 诚信女子大学\n• 水原加图立大学\n• 加图立大学\n• 首尔市立大学\n• 水原大学\n• 东明大学\n• 威德大学\n• 蔚山科学技术院\n• 蔚山大学\n• 圆光大学\n• 又松大学\n• 又石大学\n• 岭南大学\n• 艺苑艺术大学\n• 龙仁大学\n• 延世大学\n• 唯元大学\n• 灵山大学\n• 根特大学国际校区\n• 韩国纽约州立大学\n• 犹他大学亚洲校区\n• 韩国乔治梅森大学\n• 丹麦媒体与新闻学院\n• 西兰大学学院\n• 南丹麦大学学院\n• 达尼亚高等教育学院\n• IBA国际商业学院\n• MidtVest商业学院\n• 西兰商务与技术学院\n• EA西南商业学院\n• 奥胡斯商业学院\n• 哥本哈根商业学院\n• 小贝尔特学院\n• 哥本哈根设计与技术学院\n• 奥胡斯建筑学院\n• 丹麦皇家美术学院-建筑、设计与修复学院\n• 科尔丁设计学院\n• 丹麦皇家音乐学院\n• 皇家音乐学院-奥胡斯/奥尔堡\n• 韵律音乐学院\n• 丹麦国家表演艺术学院\n• 丹麦皇家美术学院-视觉艺术学院\n• 丹麦国家电影学院\n• 丹麦国家音乐学院\n• 奥胡斯海洋与技术工程学院\n• 哥本哈根海洋工程与技术管理学院\n• 斯文堡国家海洋学院\n• 腓特烈西亚海洋与技术工程学院\n• 爱尔兰国立梅努斯大学\n• 利莫瑞克大学\n• 都柏林城市大学\n• 国立艺术设计学院\n• St.Angela’s College of E\n• 香侬酒店管理学院\n• 公共管理学院\n• 都柏林理工大学\n• 卡罗理工学院\n• 阿斯隆理工学院\n• 科克理工学院\n• 敦达克理工学院\n• 高威理工学院\n• 莱特肯尼理工学院\n• 利莫瑞克理工学院\n• 斯莱戈理工学院\n• 特拉利理工学院\n• 沃特福德理工学院\n• 邓莱里文艺理工学院\n• 格里菲斯学院\n• 爱尔兰国家学院\n• 都柏林商学院\n• 福拉尔贝格高等专业学院\n• 布尔根兰高等专业学院教学点\n• 约阿内高等专业学院\n• 格拉茨高等专业学院\n• 因斯布鲁克管理中心\n• 克雷姆斯高等专业学院\n• 蒂罗尔州库夫施泰因高等专业学院\n• 萨尔茨堡高等专业学院\n• 圣•帕尔藤高等专业学院\n• 克恩藤技术高等专业学院\n• 上奥地利州高等专业学院\n• 维也纳经济高等专业学院\n• 维也纳职业促进高等专业学院\n• 维也纳康普斯高等专业学院\n• 维也纳技术高等专业学院\n• 劳德尔国际商务学校\n• 维也纳新城经济与技术高等专业学院\n• 特雷西娅军事指挥学院\n• 蒂罗尔卫生保健职业中心高等专业学院教学点\n• 马来西亚国际伊斯兰大学\n• 马来西亚国民大学\n• 马来西亚吉兰丹大学\n• 马来西亚彭亨大学\n• 马来西亚玻璃市大学\n• 马来西亚沙巴大学\n• 马来西亚沙捞越大学\n• 马来西亚丁加奴大学\n• 苏丹依德理斯教育大学\n• 马来西亚国防大学\n• 马来西亚博特拉大学\n• 马来西亚伊斯兰理科大学\n• 苏丹再纳阿比汀大学\n• 马六甲马来西亚技术大学\n• 玛拉工艺大学\n• 马来西亚敦胡先翁大学\n• 马来西亚北方大学\n• 科学及科技大学\n• 亚洲城市大学\n• 亚太科技大学\n• 百纳利管理与创业大学\n• 城市大学\n• DRB-HICOM马来西亚汽车工业大学\n• 环球大学\n• 精英大学\n• 吉隆坡建设大学\n• 国际伊斯兰金融教育中心\n• 国际医药大学\n• 英迪国际大学\n• 马来亚-威尔士国际大学\n• 林国荣创意科技大学\n• 玛莎大学\n• 马来西亚供应链创新学院\n• 马来西亚理科与工艺大学\n• 管理与科学大学\n• 玛尼帕尔国际大学\n• 多媒体大学\n• 汝来大学\n• 首要大学\n• 博特拉商学院\n• 霹雳州奎斯特国际大学\n• 依斯干达莱佛士大学\n• 世纪大学\n• 双威大学\n• 泰莱大学\n• 思特雅大学\n• UNITAR国际大学\n• 吉隆坡大学\n• 雪兰莪大学\n• Universiti Sultan Azlan Shah\n• 国油科技大学\n• 国家能源大学\n• 敦阿都拉萨大学\n• 马来西亚计算机科学与工程大学\n• 成功大学学院\n• 赛柏再也医药理科大学学院\n• 万达大学学院\n• 英沙尼亚大学学院\n• 雪兰莪国际伊斯兰大学学院\n• 双德国际科技大学学院\n• 伯乐大学学院\n• KPJ国际大学学院\n• 吉隆坡首都大学学院\n• 林肯大学学院\n• 林登大学学院\n• 南方大学学院\n• TATI大学学院\n• 拉曼大学学院\n• 北斯达利大学学院\n• 马六甲伊斯兰大学学院\n• 沙捞越科技大学学院\n• 国际艺术学院\n• 师范学院\n• 密德萨斯大学\n• 曼彻斯特城市大学\n• 曼彻斯特大学\n• 拉夫堡大学\n• 伦敦大学学院\n• 伦敦南岸大学\n• 伦敦政治经济学院\n• 伦敦卫生与热带医药学院\n• 伦敦城市大学\n• 伦敦商学院\n• 伦敦大学\n• 利物浦约翰摩尔大学\n• 利物浦赫普大学\n• 利物浦大学\n• 林肯大学\n• 莱斯特大学\n• 利兹艺术大学\n• 利兹三一大学\n• 利兹贝克特大学\n• 利兹大学\n• 英国法学大学\n• 兰卡斯特大学\n• 金斯顿大学\n• 伦敦国王学院\n• 肯特大学\n• 基尔大学\n• 伦敦大学学院教育学院\n• 帝国理工学院\n• 伦敦银行与金融学院\n• 赫尔大学\n• 哈德斯菲尔德大学\n• 高地与群岛大学\n• 伦敦大学海斯洛普学院\n• 赫特福德大学\n• 赫瑞•瓦特大学\n• 哈珀亚当斯大学\n• 吉尔德霍尔音乐与戏剧学院\n• 格林威治大学\n• 伦敦大学金史密斯学院\n• 格林多大学\n• 格鲁斯特大学\n• 格拉斯哥卡利多尼亚大学\n• 格拉斯哥大学\n• 法尔茅斯大学\n• 地产管理大学学院\n• 埃克塞特大学\n• 埃塞克斯大学\n• 爱丁堡龙比亚大学\n• 爱丁堡大学\n• 知山大学\n• 东伦敦大学\n• 东英吉利亚大学\n• 杜伦大学\n• 邓迪大学\n• 德比大学\n• 德蒙福特大学\n• 坎布里亚大学\n• 创意艺术大学\n• 克兰菲尔德大学\n• 考文垂大学\n• 伦敦大学城市学院\n• 奇切斯特大学\n• 切斯特大学\n• 中央兰开夏大学\n• 卡迪夫大学\n• 卡迪夫城市大学\n• 坎特伯雷大学\n• 剑桥大学\n• 白金汉郡新大学\n• 骨科大学学院\n• 白金汉大学\n• 伦敦布鲁内尔大学\n• 布里斯托大学\n• 布莱顿大学\n• 布拉德福德大学\n• 英博夏尔大学\n• 伯恩茅斯大学\n• 博尔顿大学\n• 格罗斯泰斯特主教大学\n• 伯明翰大学学院\n• 伯明翰城市大学\n• 伯明翰大学\n• 伦敦大学伯贝克学院\n• 贝德福特大学\n• 巴斯斯巴大学\n• 巴斯大学\n• 班戈大学\n• 雅顿大学\n• 英欧脊椎推拿疗法学院\n• 阿斯顿大学\n• 阿什里奇商学院\n• 伦敦艺术大学\n• 伯恩茅斯艺术大学\n• 坎特伯雷大教主大学\n• 安格利亚鲁斯金大学\n• 亚伯大学\n• 阿伯泰大学\n• 阿伯丁大学\n• 纽卡斯尔大学\n• 纽曼大学\n• 北安普顿大学\n• 诺森比亚大学\n• 诺里奇艺术大学\n• 诺丁汉大学\n• 诺丁汉特伦特大学\n• NCG\n• 皇家护理学院\n• 开放大学\n• 牛津大学\n• 牛津布鲁克斯大学\n• 普利茅斯大学\n• 朴次茅斯大学\n• 爱丁堡玛格丽特女王大学\n• 伦敦玛丽女王大学\n• 贝尔法斯特女王大学\n• 雷丁大学\n• 伦敦摄政大学\n• 罗伯特戈登大学\n• 罗汉普顿大学\n• 英国皇家音乐学院\n• 英国皇家农业大学\n• 中央演讲与戏剧学院\n• 英国皇家艺术学院\n• 皇家音乐学院\n• 苏格兰皇家音乐戏剧学院\n• 伦敦大学皇家霍洛威学院\n• 皇家北方音乐学院\n• 皇家兽医学院\n• 伦敦里士满美国国际大学\n• 索尔福德大学\n• 伦敦大学亚非学院\n• 谢菲尔德大学\n• 谢菲尔德哈勒姆大学\n• 南威尔士大学\n• 南安普顿大学\n• 南安普顿索伦特大学\n• 圣安德鲁斯大学\n• 伦敦大学圣乔治医学院\n• 圣马克与圣约翰大学\n• 斯泰福厦大学\n• 斯特灵大学\n• 思克莱德大学\n• 桑德兰大学\n• 萨里大学\n• 萨赛克斯大学\n• 斯旺西大学\n• 萨福克大学\n• 特威克南圣玛丽大学\n• 提赛德大学\n• 圣三一拉邦音乐舞蹈学院\n• 奥斯特大学\n• 威尔士大学\n• 威尔士三一圣大卫大学\n• 华威大学\n• 西英格兰大学\n• 西伦敦大学\n• 西苏格兰大学\n• 威斯敏斯特大学\n• 温切斯特大学\n• 胡弗汉顿大学\n• 伍斯特大学\n• 瑞特大学学院\n• 约克大学\n• 约克圣约翰大学\n• 林肯大学\n• 梅西大学\n• 惠灵顿维多利亚大学\n• Ara坎特伯雷理工学院\n• 新西兰东部理工学院\n• 马努卡理工学院\n• 尼尔森马尔伯勒理工学院\n• 北方理工学院\n• 奥塔哥理工学院\n• 南方理工学院\n• 泰普迪尼理工学院\n• 新西兰理工学院 (Unitec)\n• 国立联合学院\n• 新西兰国立中部理工学院\n• 怀卡托理工学院\n• 惠灵顿理工学院\n• 塔拉纳基西部理工学院\n• 维特利亚理工学院\n• 奥克兰商学院\n• 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https://math.stackexchange.com/questions/2109192/simplifying-fraction-with-factorials-frac3n13n/2109210	[ "# Simplifying fraction with factorials: $\\frac{(3(n+1))!}{(3n)!}$\n\nI was trying to solve the limit:\n\n$$\\lim_{n \\to \\infty} \\sqrt[n]{\\frac{(3n)!}{(5n)^{3n}}}$$\n\nBy using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):\n\n$$L= \\lim_{n \\to \\infty} \\frac{a_n}{b_n} = \\lim_{n \\to \\infty} \\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$\n\nNow I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.\n\nJust so you can correct me, my reasoning was:\n\n$$\\frac{(3(n+1))!}{(3n)!} = \\frac{3\\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 4 \\cdot (...) \\cdot 3 \\cdot (n+1)}{3 \\cdot 1 \\cdot 3 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot (...) \\cdot 3 \\cdot n } =$$ $$= \\frac{3^n(n+1)!}{3^{n}n!} = \\frac{(n+1)!}{n!} = n+1$$\n\nWhich apparently isn't correct. I must have failed at something very silly. Thanks in advance!\n\n• $(3(n+1)!)$ is the product of all the natural numbers from $1$ to $3(n+1)$, and $3\\cdot1\\cdot 3\\cdot 2\\ldots\\cdot 3(n+1)$ is the product of the numbers that are multiples of $3$. – ajotatxe Jan 22 '17 at 19:08\n• @Manuel Also you can apply Stirling's approximation. – Behrouz Maleki Jan 22 '17 at 19:11\n• @ajotatxe I don't have (3(n+1)!), but (3(n+1))! (the factorial includes everything between the parentheses). – mar Jan 22 '17 at 19:17\n• This question is somewhat similar: Need to simplify ratio test: $\\frac{(2(k+1))!}{(2k)!}$ - it might also be useful for you. I found it using Approach0. – Martin Sleziak Jan 23 '17 at 6:49\n\nBy Stirling's inequality the answer is clearly $\\left(\\frac{3}{5e}\\right)^3$. To prove it, you may notice that by setting $$a_n = \\frac{(3n)!}{(5n)^{3n}}$$ you have: $$\\frac{a_{n+1}}{a_n} = \\frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \\frac{\\frac{3n+3}{5n+5}\\cdot\\frac{3n+2}{5n+5}\\cdot\\frac{3n+1}{5n+5}}{\\left(1+\\frac{1}{n}\\right)^{3n}}\\to\\frac{\\left(\\frac{3}{5}\\right)^3}{e^3}$$ as $n\\to +\\infty$.\n\n• I don't want to know the answer to the limit, but how I can go from a reason of factorials to the polynomial Mathematica gave me. But thanks, I'll use your answer to check if I got it right, haha. – mar Jan 22 '17 at 19:15\n• @Manuel: the reason is simple: if you have a positive sequence such that $\\frac{a_{n+1}}{a_n}\\to c$, then $\\sqrt[n]{a_n}\\to c$, too. – Jack D'Aurizio Jan 22 '17 at 19:16\n• And $$\\frac{(3(n+1))!}{(3n)!}=\\frac{(3n+3)!}{(3n)!}=(3n+3)(3n+2)(3n+1).$$ – Jack D'Aurizio Jan 22 '17 at 19:18\n• Oh, that makes sense. Why did I not see it before? Thanks a lot, Jack. – mar Jan 22 '17 at 19:19\n\nI thought it might be useful to present an approach that relies on elementary tools only. To that end, we proceed.\n\nFirst, we write\n\n\\begin{align} \\frac{1}{n}\\log((3n!))&=\\frac1n\\sum_{k=1}^{3n}\\log(k)\\\\\\\\ &=\\left(\\frac1n\\sum_{k=1}^{3n}\\log(k/n)\\right)+3\\log(n) \\tag 1 \\end{align}\n\nNow, note that the parenthetical term on the right-hand side of $(1)$ is the Riemann sum for $\\int_0^3 \\log(x)\\,dx=3\\log(3)-3$.\n\nUsing $(1)$, we have\n\n\\begin{align} \\lim_{n\\to \\infty}\\sqrt[n]{\\frac{(3n)!}{(5n)^{3n}}}&=\\lim_{n\\to \\infty}e^{\\frac1n \\log((3n)!)-3\\log(5n)}\\\\\\\\ &=\\lim_{n\\to \\infty}e^{\\left(\\frac1n\\sum_{k=1}^{3n}\\log(k/n)\\right)+3\\log(n)-3\\log(5n)}\\\\\\\\ &=\\lim_{n\\to \\infty}e^{\\left(\\frac1n\\sum_{k=1}^{3n}\\log(k/n)\\right)-3\\log(5)}\\\\\\\\ &= e^{3\\log(3)-3-3\\log(5)}\\\\\\\\ &=\\left(\\frac{3}{5e}\\right)^3 \\end{align}\n\nas expected!\n\nTools used: Riemann sums and the continuity of the exponential function.\n\n$$(3(n+1))! \\neq 3 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 4 \\cdots 3 \\cdot (n+1)$$ To make it clear what the problem is, let's write the right-hand side with brackets: $$(3 \\cdot 2) \\cdot (3 \\cdot 3) \\cdot (3 \\cdot 4) \\cdots (3 \\cdot (n+1))$$ That's just multiplying all the positive multiples of 3 less than $3(n+1)$ together; the factorial is defined as multiplying together all positive integers less than $3(n+1)$. So the correct expansions are \\begin{align} (3(n+1))! &= 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdots (3n-2) \\cdot (3n-1) \\cdot 3n \\cdot (3n+1) \\cdot (3n+2) \\cdot 3(n+1) \\\\ (3n)! &= 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdots (3n-2) \\cdot (3n-1) \\cdot 3n \\end{align} which clearly have the quotient Mathematica gave you.\n\n• Thanks a lot, I noticed what you just said reading Jack's response. I haven't used the factorial function a lot and now I see I didn't completely understand it (or maybe I'm especially slow today). – mar Jan 22 '17 at 19:35\n\nMy favorite elementary inequality for $n!$: $\\def\\lfrac#1#2{{\\large\\frac{#1}{#2}}}$ $(n/e)^n < n! < (n/e)^{n+1}$.\n\nTaking n-th roots, $n/e < (n!)^{1/n} < (n/e)^{1+1/n}$.\n\nTherefore $\\sqrt[n]{\\lfrac{(3n)!}{(5n)^{3n}}} =\\lfrac{((3n)!)^{1/n}}{(5n)^{3}} >\\lfrac{((3n/e)^{3n})^{1/n}}{(5n)^{3}} =\\lfrac{(3n/e)^{3}}{(5n)^{3}} =\\lfrac{27n^3}{125e^3n^{3}} =\\lfrac{27}{125e^3}$.\n\nArguing the other way, $\\sqrt[n]{\\lfrac{(3n)!}{(5n)^{3n}}} =\\lfrac{((3n)!)^{1/n}}{(5n)^{3}} <\\lfrac{((3n/e)^{3n+1})^{1/n}}{(5n)^{3}} =\\lfrac{(3n/e)^{3}(3n/e)^{1/n}}{(5n)^{3}} =\\lfrac{27n^3(3n/e)^{1/n}}{125e^3n^{3}} =\\lfrac{27}{125e^3}(3n/e)^{1/n}$. Since, as $n \\to \\infty$, $n^{1/n} \\to 1$ and $a^{1/n} \\to 1$ for any $a > 0$, $(3n/e)^{1/n} \\to 1$ so the limit is $\\lfrac{27}{125e^3}$.\n\nA more general result:\n\n$\\sqrt[n]{\\lfrac{(an)!}{(bn)^{an}}} =\\lfrac{((an)!)^{1/n}}{(bn)^a} \\gt \\lfrac{((an/e)^{an})^{1/n}}{(bn)^a} =\\lfrac{(an/e)^a}{(bn)^a} =\\lfrac{a^a n^a}{b^ae^a} =\\lfrac{a^a}{b^ae^a} =\\left(\\lfrac{a}{be}\\right)^a$.\n\nThe other way goes exactly as above, so that $\\lim_{n \\to \\infty} \\sqrt[n]{\\lfrac{(an)!}{(bn)^{an}}} =\\left(\\lfrac{a}{be}\\right)^a$.\n\n• I get the part where you say this: $\\frac{((an)!)^{1/n}}{(bn)^a} \\gt \\frac{((an/e)^{an})^{1/n}}{(bn)^a}$ but why would that imply that $(a/be)^a$ is the result of the limit? – mar Jan 22 '17 at 20:25\n• The $n^a$ terms cancel out. – marty cohen Jan 22 '17 at 23:13\n• I made your fractions a bit bigger so that they are easier to read. Hope you don't mind, but feel free to revert if you don't like it. =) – user21820 Jan 23 '17 at 3:49\n• I don't mind editing as long as the meaning isn't changed. – marty cohen Jan 23 '17 at 4:12" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8811432,"math_prob":0.999884,"size":1104,"snap":"2020-10-2020-16","text_gpt3_token_len":388,"char_repetition_ratio":0.15181819,"word_repetition_ratio":0.04117647,"special_character_ratio":0.3777174,"punctuation_ratio":0.117154814,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99992764,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-20T14:48:28Z\",\"WARC-Record-ID\":\"<urn:uuid:fe412a15-d4d7-4669-b85a-d9dc810bab9b>\",\"Content-Length\":\"177841\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba8e0eae-05d7-4585-9717-80ce6ea32bc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:d72b5015-94a8-4ff0-a6b5-d4a3c1e181fb>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2109192/simplifying-fraction-with-factorials-frac3n13n/2109210\",\"WARC-Payload-Digest\":\"sha1:2EIDC5ZBFLHBQTAZ6ZOAK4JRXXSNQHDX\",\"WARC-Block-Digest\":\"sha1:LZN4A2AJHHQCX4N5SZ66JOII7XPD45IF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144979.91_warc_CC-MAIN-20200220131529-20200220161529-00116.warc.gz\"}"}
https://electronicsdesk.com/cyclic-code.html	[ "# Cyclic Code\n\nCyclic Code is known to be a subclass of linear block codes where cyclic shift in the bits of the codeword results in another codeword. It is quite important as it offers easy implementation and thus finds applications in various systems.\n\nCyclic codes are widely used in satellite communication as the information sent digitally is encoded and decoded using cyclic coding. These are error-correcting codes where the actual information is sent over the channel by combining with the parity bits.\n\n## Content: Cyclic Code\n\n### Introduction\n\nCyclic codes are known to be a crucial subcategory of linear coding technique because these offers efficient encoding and decoding schemes using a shift register. These are used in error correction as they can check for double or burst errors. Various other important codes like, Reed Solomon, Golay, Hamming, BCH, etc. can be represented using cyclic codes.\n\nBasically, a shift register and a modulo-2 adder are the two crucial elements considered as building blocks of cyclic encoding. Using a shift register, encoding can be efficiently performed. The fundamental elements of shift registers are flip flops (that acts as a storage unit) and input-output. While the other i.e., a binary adder has two inputs and one output.\n\n### Properties of Cyclic Code\n\nWe have mentioned at the beginning itself that cyclic codes fall under the category of linear block codes. Let us now understand on following which properties a code is said to be of cyclic nature.\n\nProperty 1: Property of Linearity\n\nAccording to this property, a linear combination of two codewords must be another codeword.\n\nSuppose we have two codewords Ci and Cj. So, on adding", null, "where this Cp must also be a codeword.\n\nFor example, suppose we have given 3 codewords (110, 101, 011).\n\nSo, according to linearity property, the addition of any of the two given codewords must produce the third codeword. Let’s check", null, "Hence, the given code is linear.\n\nProperty 2: Property of Cyclic Shifting\n\nAccording to this property, after a right or left shift in the bits of codewords the resultant code generated must be another codeword.\n\nSuppose, C is a codeword given as:", null, "Then after cyclic shifts", null, "Here we have performed the right cyclic shift that has produced these codewords.\n\nFor example, consider again those 3 codewords (110, 101, 011) which we considered for linearity property.\n\nSo, according to cyclic shifting property, an either right or left shift in the bits of a codeword must generate another codeword.\n\n110: shifting the bits towards the right will provide 011.\n\n101: a right shift in the bits of this codeword will give 110.\n\n011: right shifting of these bits will provide 101.\n\nHence, it is clear that shifting the bits of the codewords gives rise to another codeword thus cyclic shifting property is verified.\n\nCodes that follow both linearity, as well as cyclic shifting, are called cyclic codes.\n\nIt is to be noted here that if a codeword has all 0’s then it is called a valid codeword. But at the same time, 0 is regarded as a necessary condition and not a sufficient condition.\n\n## Example of Cyclic Code\n\nWe have discussed in linear block codes as well that a linear codeword is generally given as c(n,k). Here n represents the total bits in the codeword and k denotes the message bits. Thus, the parity bits are (n-k). Suppose we are given a code (7,4) then on comparing with general format the codeword will have 7 bits and the actual message bits are 4 while rest 3 are parity bits.\n\nA cyclic codeword is given as:", null, "Then the codeword polynomial will be represented as:", null, "For a given codeword C = , the codeword polynomial will be given as:\n\nC(X) = 1X0 + 0X1 + 1X2 + 1X3\n\nC(X) = X3 + X2 + 1\n\nIn a similar way, for any message codeword m, the message polynomial", null, "And generator polynomial", null, "Codewords are classified as systematic and non-systematic codewords.\n\nA systematic codeword is one in which the parity bits and message bits are present in separated forms.\n\nC = [parity bit, message bits]\n\nBut a non-systematic codeword is the one in which the message and parity bits exist in intermixed format and cannot be separated just by noticing the initial and final bits.\n\nLet us now understand the coding and decoding of codewords using cyclic code.\n\n### Encoding\n\nNon-Systematic Cyclic Encoding: Consider the message signal given as:\n\nm = \n\nThus,\n\nM(X) = 1X0 + 1X1 + 1X2 + 0X3\n\nM(X) = X2 + X + 1\n\nwith generator polynomial G(X) = X3 + X + 1\n\nFor non-systematic code, the codeword is given as:", null, "C(X) = (X2 + X + 1) (X3 + X + 1)\n\nC(X) = X5 + X3 + X2 + X4 + X2 + X + X3 + X + 1\n\nHere modulo 2 addition will be performed and in modulo 2 addition, the sum of 2 similar bits results in 0.\n\nC(X) = X5 + X3 + X2 + X4 + X2 + X + X3 + X + 1\n\nThus, X3, X2 and X will get cancelled. So,\n\nC(X) = 1 + X4 + X5\n\nHence, from the above codeword polynomial, the codeword will be:\n\nC = \n\nFrom the codeword bits, we can clearly interpret that the encoded codeword contains the message and parity bits in an intermixed pattern. Thus, is a non-systematic codeword and direct determination of message bits is not possible.\n\nSystematic Cyclic Encoding: Consider another message signal:\n\nm = \n\nSo, message polynomial will be:\n\nM(X) = 1 + X2 + X3\n\nWhile the generator polynomial G(X) = X3 + X + 1\n\nThe equation for determining 7 bits codeword for systematic code is given as:", null, ": P(X) represents the parity polynomial and is given by:", null, "So, to construct the systematic codeword first we have to determined P(X).\n\nSince n= 7 and k = 4", null, "Therefore,", null, "Hence, the obtained value of P(X) = 1\n\nNow, substituting the values in codeword polynomial equation,", null, "C(X) = X3 (X3 + X2 + 1) + 1\n\nC(X) = 1 + X3 + X5 + X6\n\nHence, the codeword for the above code polynomial will be:\n\nC = \n\nSo, here the first 3 bits are parity bits and the last four bits are message bits. And we can cross-check that we have considered as the message bits and parity polynomial remainder was 1 i.e., code .\n\nHence, in this way encoding of non-systematic and systematic codewords is performed.\n\n### Decoding\n\nTo understand how the detection of cyclic code takes place. Consider R(X) as received polynomial, C(X) as the encoded polynomial, G(X) to be generator polynomial, and E(X) as error polynomial.\n\nSyndrome or error involved during transmission is given as:", null, "Since R(X) is a combination of encoded polynomial and error polynomial thus above equation can be written as:", null, "Here, if the obtained remainder is 0 then there will be no error and if the obtained remainder is not 0 then there will be an error that needs to be corrected.", null, "This is so because actually coded bit does not contain error thus syndrome is not required to be calculated.\n\nNow, let us check for error polynomial. So, consider the error table given below where we have assumed a single error in each of the bits of the code:", null, "So, now using the syndrome for error polynomial formula:", null, "We will determine, the erroneous bit. Consider the generator polynomial G(X) = X3 + X + 1 and dividing each error polynomial with it.", null, "Now, tabulating it,", null, "Consider an example where transmitted codeword is received code, r(n,k) = . Hence,\n\nR(X) = 1 + X2 + X5\n\nNow, let us check for the syndrome, by dividing the received polynomial R(X) by the generator polynomial G(X).", null, "We will get,", null, "S = X ≠ 0\n\nThis means the received codeword contains an error. From the tabular representation, it is clear that X has an error code . Thus, this represents an error in the second bit of the received code.\n\nHence, by using the same approach we can perform encoding and decoding using cyclic code." ]	[ null, "https://electronicsdesk.com/wp-content/uploads/2021/06/property-of-linearity.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/linearity-property.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/codeword-representation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/cyclic-shift.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/codeword-representation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/codeword-polynomial-representation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/message-polynomial-representation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/generator-polynomial-representation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/non-systematic-codeword.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/codeword-equation-for-systematic-code.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/parity-polynomial-equation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/parity-polynomial-equation-1.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/division-1.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/codeword-equation-for-systematic-code-1.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/syndrome-equation.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/syndrome-equation-1.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/syndrome-equation-2.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/Table-1-for-detection-of-error-of-cyclic-code.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/syndrome-equation-3.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/division-2.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/Table-2-for-detection-of-error-of-cyclic-code.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/syndrome-equation-4.jpg", null, "https://electronicsdesk.com/wp-content/uploads/2021/06/division-3.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8698733,"math_prob":0.97855556,"size":7555,"snap":"2023-40-2023-50","text_gpt3_token_len":1895,"char_repetition_ratio":0.16567342,"word_repetition_ratio":0.052748885,"special_character_ratio":0.2485771,"punctuation_ratio":0.10872483,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994727,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46],"im_url_duplicate_count":[null,2,null,2,null,4,null,2,null,4,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T20:17:10Z\",\"WARC-Record-ID\":\"<urn:uuid:8275524c-3db2-40a7-97f7-eb4d3ce79033>\",\"Content-Length\":\"112120\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:312b4e88-390d-41f3-ba02-e365b95e1dd1>\",\"WARC-Concurrent-To\":\"<urn:uuid:3970ff41-84f9-4849-b867-de03839ca9dd>\",\"WARC-IP-Address\":\"67.43.12.246\",\"WARC-Target-URI\":\"https://electronicsdesk.com/cyclic-code.html\",\"WARC-Payload-Digest\":\"sha1:5HLHZ5JNKMYOVTG2CLDNAZBFO4CLZXV3\",\"WARC-Block-Digest\":\"sha1:MUJNRJB3VS47HHR7GWLTD4GSPH4NXGPC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100232.63_warc_CC-MAIN-20231130193829-20231130223829-00352.warc.gz\"}"}
https://www.flightpedia.org/convert/1199-pounds-to-decagram.html	[ "# Convert 1,199.0 Pounds to Decagrams\n\n 1,199.0 Pounds (lb) = 54,385.73 Decagrams (dag) 1 lb = 45.3592 dag 1 dag = 0.022046 lb\n\n• Q: How many Pounds in a Decagram?\n\n• Q: How do you convert 1199 Pound (lb) to Decagram (dag)?\n\n1199 Pound is equal to 54,385.73 Decagram. Formula to convert 1199 lb to dag is 1199 * 45.359237\n\n• Q: How many Pounds in 1199 Decagrams?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6456379,"math_prob":0.9710339,"size":1401,"snap":"2021-43-2021-49","text_gpt3_token_len":447,"char_repetition_ratio":0.19470294,"word_repetition_ratio":0.00754717,"special_character_ratio":0.38329765,"punctuation_ratio":0.077220075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.972919,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T16:32:39Z\",\"WARC-Record-ID\":\"<urn:uuid:0ce37056-57fa-4d6b-a23c-b373aed07cc1>\",\"Content-Length\":\"21676\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0434cdb6-636a-40fa-b552-0d24ad6305aa>\",\"WARC-Concurrent-To\":\"<urn:uuid:229e2f2a-1348-41f3-99bf-ad306b54fa91>\",\"WARC-IP-Address\":\"35.202.210.90\",\"WARC-Target-URI\":\"https://www.flightpedia.org/convert/1199-pounds-to-decagram.html\",\"WARC-Payload-Digest\":\"sha1:SSN7L2V7BV22WN7Z6VPIXSNLISEW6UOI\",\"WARC-Block-Digest\":\"sha1:HJ4UBB2OCMBWKYL25JCBQFJQ5OCCN55X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362999.66_warc_CC-MAIN-20211204154554-20211204184554-00111.warc.gz\"}"}
https://nforum.ncatlab.org/discussion/2501/	[ "## Not signed in\n\nWant to take part in these discussions? Sign in if you have an account, or apply for one below\n\n## Site Tag Cloud\n\nVanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.\n\n• CommentRowNumber1.\n• CommentAuthorUrs\n• CommentTimeFeb 18th 2011\n• CommentRowNumber2.\n• CommentAuthorUrs\n• CommentTimeFeb 18th 2011\n• (edited Feb 18th 2011)\n\nDoes fat geometic realization send global Kan fibrations to topological fibrations?\n\nFor $G$ a simplicial topological group, is $\\Vert W G\\Vert \\to \\Vert \\bar W G\\Vert$ a fibration?\n\n• CommentRowNumber3.\n• CommentAuthorUrs\n• CommentTimeFeb 18th 2011\n• (edited Feb 19th 2011)\n\nI am thinking about the following:\n\nClaim For $G$ a well-sectioned simplicial topological group, $X$ a good simplicial topological space, every simplicial $G$-principal bundle $P$ over $X$ is also proper as a simplicial topological space.\n\nIdea of the proof\n\nEvery such $G$-bundle $P \\to X$ arises as the pullback\n\n$\\array{ P &\\to& W G \\\\ \\downarrow && \\downarrow \\\\ X &\\stackrel{\\tau}{\\to}& \\bar W G }$\n\nfor some morphism $\\tau$ of simplicial topological spaces.\n\nTherefore for each $n \\in \\mathbb{N}$ we have that $P_n$ is given by the pullback\n\n$\\array{ P_n &\\to& (W G)_n \\\\ \\downarrow && \\downarrow \\\\ X_n &\\stackrel{\\tau_n}{\\to}& (\\bar W G)_n }$\n\nin $Top$.\n\nNow, $W$ and $\\bar W G$ are both proper simplicial topological spaces. And $(W G)_n \\to (\\bar W )_n$ is a Hurewicz fibration for all $n$. Therefore by the theorem now included at closed cofibration, we have that the degeneracy maps of $P$ are induced by morphisms of pullback diagrams one of whose legs is a fibration along a degreewise closed cofibration. hence are themselves closed cofibrations. Hence $P$ is proper.\n\nRight?\n\n• CommentRowNumber4.\n• CommentAuthorDavidRoberts\n• CommentTimeFeb 19th 2011\n\nUrs, have you seen the paper that Danny and I are working on? We address very similar points.\n\n• CommentRowNumber5.\n• CommentAuthorUrs\n• CommentTimeFeb 19th 2011\n• (edited Feb 19th 2011)\n\nHi David,\n\nso apparently you didn’t get the email that I sent to you. What’s your current working email address?\n\nYes, I am looking at your paper with Danny as we speak! That’s where this question originates from: I think you provide almost all the statements to show that the left derived functor of geometric realization of simplicial topological spaces preserves homotopy fibers of morphisms of the form $X \\to W G$, hence sends topological $\\infty$-bundles to their underlying topological bundles.\n\nI need that statement for differential cohomology in a cohesive topos (schreiber). But yesterday I realized that in my proof that the intrinsic $\\Pi : ETop\\infty Grpd \\to Top$ preserves homotopy fibers (see Euclidean topological infinity-groupoid) I had been silently asssuming that with $X$ and $W G$ proper simplicial spaces, also the simplicial bundle $P$ that is defined by $\\tau : X \\to W G$ has the propety that its geometric realization coincides with its homotopy colimit.\n\nThis seems to be an important statement also for the main statement in your article with Danny, and so I tried to check with you and Danny if my idea how to show this is correct. Danny has meanwhile confirmed that this argument is indeed true.\n\nLet me just point out again what that has to do with homotopy fibers:\n\nsince for a simplicial topological group $G$ we have\n\n• $\\bar W G$ is globally fibrant (the maps $(\\bar W G)_n \\to [\\Lambda^n_k, \\bar W G]$ have global continuous sections) ;\n\n• $W G$ is globally fibrant\n\n• $W G \\to \\bar W G$ is a global fibration\n\nwe have that $W G \\to \\bar W G$ is a presentation by a fibration of the point inclusion $* \\to \\mathbf{B}G$ in the projective model structure for Euclidean-topological $\\infty$-groupoids $[CartSp_{top}^{op}, sSet]_{proj}$.\n\nThis means that for all simplicial spaces $X$ and morphisms $\\tau : X \\to \\bar W G$ the ordinary pullback\n\n$\\array{ P &\\to& W G \\\\ \\downarrow && \\downarrow \\\\ X &\\stackrel{\\tau}{\\to}& \\bar W G }$\n\nis a model for the homotopy pullback of $* \\to \\mathbf{B}G$ along $\\tau$, hence a model for its homotopy fiber, hence that $P$ is indeed a presentation of the correct topological principal $\\infty$-bundle given by $\\tau$.\n\nNow assume all topological spaces here are degreewise paracompact and admit good open covers. Then one can prove that the intrinsic fundamental $\\infty$-groupoid functor $\\Pi : ETop\\infty Grpd \\to \\infty Grpd \\simeq Top$ is presented on these by fat geometric realization. But now, since $X$ and $\\bar W G$ and $W G$ are assumed to be proper and since we find that then $P$ is implied to be proper, it follows that on our situation here $\\Pi$ is already presented by ordinary geometric realization.\n\nNow moreover, by your statement with Danny we have that $\|W G\| \\to \|\\bar W G\|$ is again a fibration resolution of the point inclusion $* \\to \\mathbf{B}\|G\|$. This shows (and I would think you could mention this as a nice immediate corollary in your article) that\n\n$\\array{ \|P\| &\\to& \|W G\| \\\\ \\downarrow && \\downarrow \\\\ \|X\| &\\stackrel{\\tau}{\\to}& \|\\bar W G\| }$\n\nwhich is an ordinary pullback diagram by the general fact that geometric realization preserves pullbacks, is again even a homotopy pullback, hence itself exhibits $\|P\|$ as the homotopy fiber of $\|\\tau\|$.\n\nSo in total this corollary of your work with Danny shows:\n\nTheorem On morphisms of Euclidean-topological $\\infty$-groupoids $X \\to \\mathbf{B}G$ that are presented by morphisms of degreewise paracompact simplicial topological spaces, we have that $\\Pi : ETop\\infty Grpd \\to \\infty Grpd$ preserves homotopy fibers.\n\nThat’s a very powerful statement for doing refinements of Whitehead towers from Top to cohesive $\\infty$-groupoids. For instance there is a refinement of the first fractional Pontryagin class $\\frac{1}{2}p_1 : B Spin \\to B^4 \\mathbb{Z}$ to a morphism of simplicial topological (even Lie) groups $\\frac{1}{2}\\mathbf{p}_1 : W Spin \\to W \\Xi U(1)$. Its homotopy fiber will be some topological 2-group: the string 2-group. By the above theorem it is immediate that its geometric realization is the ordinary topological string group. And so on for higher steps in the Whitehead tower.\n\nThat’s what I am after. Only that yesterday I reallized that I had missed to argue that $P$ in the above is proper. So therefore I was fishing here for sanity checks that indeed it is.\n\n• CommentRowNumber6.\n• CommentAuthorDavidRoberts\n• CommentTimeFeb 19th 2011\n\nAh, I did get that email - after I checked here. So you do have the right email for me. Then I’ve been out all day, so couldn’t fix my error. I’ll have to think about this - I need an early night tonight!\n\n• CommentRowNumber7.\n• CommentAuthorUrs\n• CommentTimeFeb 19th 2011\n• (edited Feb 19th 2011)\n\nI have written out the argument at geometric realization of simp top spaces – Applications – realization of topological principal oo-bundles.\n\nHave to dash off now. Can try to expand more later.\n\n• CommentRowNumber8.\n• CommentAuthorUrs\n• CommentTimeSep 17th 2021\n\nI have adjusted the referencing for the proposition (here) that the topological realization of a well-pointed topological group is well-pointed" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94639295,"math_prob":0.95629364,"size":5134,"snap":"2023-40-2023-50","text_gpt3_token_len":1132,"char_repetition_ratio":0.15419103,"word_repetition_ratio":0.0023584906,"special_character_ratio":0.19068952,"punctuation_ratio":0.08458244,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948023,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T04:21:38Z\",\"WARC-Record-ID\":\"<urn:uuid:218fa7aa-8aff-414c-84a8-293fc2b03e50>\",\"Content-Length\":\"74295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8caf0352-e6ff-450e-b075-7d5352e6d41a>\",\"WARC-Concurrent-To\":\"<urn:uuid:cafe7424-7041-467c-bd5c-c46b0e597693>\",\"WARC-IP-Address\":\"128.2.25.48\",\"WARC-Target-URI\":\"https://nforum.ncatlab.org/discussion/2501/\",\"WARC-Payload-Digest\":\"sha1:B47EFY4IIGBXT5MII2RUPUDTRMUV5ZFG\",\"WARC-Block-Digest\":\"sha1:YMS3MFIDXZQ4KDXT6FZ74JJEXPF5B6E4\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101195.85_warc_CC-MAIN-20231210025335-20231210055335-00519.warc.gz\"}"}
https://learnindex.com/order-of-operations/	[ "# Order of Operations\n\nThe order of operations is a set of rules used to determine the sequence in which mathematical operations should be performed in an equation. It is important to follow these rules to ensure that the calculations are performed correctly and consistently. Failure to follow the order of operations can result in incorrect solutions to mathematical problems.\n\n## The Order of Operations\n\nThe order of operations is commonly remembered by the acronym PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).\n\nParentheses: Perform operations inside parentheses first. If there are nested parentheses, start with the innermost set of parentheses and work outward.\n\nExponents: Perform any calculations involving exponents or powers.\n\nMultiplication and Division: Perform multiplication and division in order from left to right.\n\nAddition and Subtraction: Perform addition and subtraction in order from left to right.\n\n## Example:\n\nLet’s take the following expression: 6 ÷ 2(1 + 2)\n\nTo solve this equation using the order of operations, we would follow the steps as follows:\n\nNext, perform multiplication and division in order from left to right: 6 ÷ 2(3)\nNow, we have to use PEMDAS again since we have multiplication and division at the same level: 2(3) = 6\nFinally, perform the division: 6 ÷ 6 = 1\nSo the solution to the equation is 1.\n\n## Practice Exercise\n\nNow that you have learned about the order of operations in mathematics, let’s try a practice exercise.\n\n2 + 3 × 4\n(5 + 2) × 3\n10 − 2 × 3²\n6 ÷ 2 + 1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92913705,"math_prob":0.9737684,"size":2004,"snap":"2023-40-2023-50","text_gpt3_token_len":443,"char_repetition_ratio":0.1555,"word_repetition_ratio":0.055384614,"special_character_ratio":0.21756487,"punctuation_ratio":0.11420613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99805653,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T10:55:24Z\",\"WARC-Record-ID\":\"<urn:uuid:4417b6a7-2903-4cba-83ba-9ef39f622cd4>\",\"Content-Length\":\"102215\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:883be31a-fe13-4378-b817-b83dfeccac1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:3bb86e2b-32a9-4f80-8ced-80523944a0fc>\",\"WARC-IP-Address\":\"141.193.213.10\",\"WARC-Target-URI\":\"https://learnindex.com/order-of-operations/\",\"WARC-Payload-Digest\":\"sha1:CXFBCO4T4LEX2AM32OVYYBSZKJPGCA43\",\"WARC-Block-Digest\":\"sha1:ABG7POGMLP3V7V4HQZSZM62LIV4QSGVU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100651.34_warc_CC-MAIN-20231207090036-20231207120036-00402.warc.gz\"}"}
https://math.stackexchange.com/questions/482726/would-proof-of-legendres-conjecture-also-prove-riemanns-hypothesis	[ "# Would proof of Legendre's conjecture also prove Riemann's hypothesis?\n\nLegendre's conjecture is that there exists a prime number between $n^2$ and $(n+1)^2$. This has been shown to be very likely using computers, but this is merely a heuristic. I have read that if this conjecture is true, the biggest gap between two consecutive primes is $O(\\sqrt{p})$; the Riemann Hypothesis, on the other hand, implies that this gap is $O(\\sqrt{p}\\log{p})$, which is a wider gap for sufficiently large inputs. This leaves me asking the following question, which I am aware may be a bit naive, but I want to be sure:\n\nWould a proof of Legendre's conjecture also be considered a proof of Riemann's hypothesis?\n\nEDIT: I am aware that the way the asymptotic above were expressed was in terms of prime numbers, and that the Riemann Hypothesis is concerned about everything in between as well; would a proof of this sort need to show this upper bound for all inputs, or would the prime numbers be sufficient?\n\nEDIT 2: It seems to me that, if a proof of Legendre's conjecture could result in a proof of the Riemann hypothesis, that this would come from the $\\log{x}$ term of the asymptotic, showing that this term is never smaller than what results from the distance between primes. In other words, this would have to be an inductive proof, showing an initial case and, as a result of that initial case and the fact that the gap is $O(\\sqrt{p})$ for that case, all future cases must therefore be $O(\\sqrt{p} \\log{p})$. I hate to add to my question once again, but please tell me if this is completely off.\n\nThe Riemann hypothesis (RH) implies that $p_{k+1}-p_k=O(\\sqrt{p_k}\\log p_k)$, which was shown by Cramer in $1919$. However, this corollary is much weaker than RH. A proof of Legendre's conjecture would imply that $p_{k+1}-p_k=O(\\sqrt{p_k})$, which is indeed better than Cramer's result. But this does not necessarily mean that it implies RH. It only means that Legendre is better than a consequence of RH. In fact, it is conjectured that the gap $p_{k+1}-p_k$ is even better than $O(\\sqrt{p_k})$, i.e., $O(\\log^2 p_k)$ which is much smaller, and that between $n^2$ and $(n+1)^2$ there lies not only one prime but quite a lot of primes.\n• Good answer; along the same lines, would showing that $p_{k+1} - p_k = O(\\log^2{p_k})$ prove the Riemann hypothesis? In other words, is it possible to prove the Riemann hypothesis by finding any upper bound on the distance between consecutive primes? – Adam Sep 3 '13 at 13:54\n• No, I do not think that a proof of $p_{k+1}-p_k=O(\\log^2 p_k)$ would give a proof of RH, because the relation between RH and the gaps is not sufficiently strong. – Dietrich Burde Sep 3 '13 at 14:36\n• I have since read that the Riemann hypothesis has been proven to be equivalent to stating that the absolute value of the difference between the pi function and the logarithmic integral is smaller than $c(\\sqrt{p} \\log{p})$ for some constant $c$. When I read \"equivalent\", my immediate conclusion is that proving one proves the other, both ways. Is this not the case? Is there anything about the distribution of the primes that, if proven, would prove the Riemann Hypothesis? – Adam Sep 5 '13 at 6:03" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9753218,"math_prob":0.9951874,"size":1511,"snap":"2019-26-2019-30","text_gpt3_token_len":369,"char_repetition_ratio":0.12276045,"word_repetition_ratio":0.007633588,"special_character_ratio":0.2322965,"punctuation_ratio":0.096463025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99981505,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-20T22:40:55Z\",\"WARC-Record-ID\":\"<urn:uuid:1df8be83-04a4-47cc-aa4d-31f8f62d6a8f>\",\"Content-Length\":\"150136\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b5f465a-12e6-4e25-bc0b-f625ae9363fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:f58b90ad-c669-4d1e-affa-cf8ca6409182>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/482726/would-proof-of-legendres-conjecture-also-prove-riemanns-hypothesis\",\"WARC-Payload-Digest\":\"sha1:27GX5COPQQUZJQ63XUSBWKWDQD3QWA3G\",\"WARC-Block-Digest\":\"sha1:UU3XFAPPPEK66PHKFIGQ4GWJF7JSKCAK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526714.15_warc_CC-MAIN-20190720214645-20190721000645-00430.warc.gz\"}"}
https://quics.umd.edu/publications/computational-notions-quantum-min-entropy	[ "# Computational Notions of Quantum Min-Entropy\n\n Title Computational Notions of Quantum Min-Entropy Publication Type Journal Article Year of Publication 2017 Authors Chen, Y-H, Chung, K-M, Lai, C-Y, Vadhan, SP, Wu, X Date Published 2017/09/09 Abstract We initiate the study of computational entropy in the quantum setting. We investigate to what extent the classical notions of computational entropy generalize to the quantum setting, and whether quantum analogues of classical theorems hold. Our main results are as follows. (1) The classical Leakage Chain Rule for pseudoentropy can be extended to the case that the leakage information is quantum (while the source remains classical). Specifically, if the source has pseudoentropy at least k, then it has pseudoentropy at least k − ℓ conditioned on an ℓ- qubit leakage. (2) As an application of the Leakage Chain Rule, we construct the first quantum leakage-resilient stream-cipher in the bounded-quantum-storage model, assuming the existence of a quantum-secure pseudorandom generator. (3) We show that the general form of the classical Dense Model Theorem (interpreted as the equivalence between two definitions of pseudo-relativemin-entropy) does not extend to quantum states. Along the way, we develop quantum analogues of some classical techniques (e.g., the Leakage Simulation Lemma, which is proven by a Nonuniform Min-Max Theorem or Boosting). On the other hand, we also identify some classical techniques (e.g., Gap Amplification) that do not work in the quantum setting. Moreover, we introduce a variety of notions that combine quantum information and quantum complexity, and this raises several directions for future work. URL https://arxiv.org/abs/1704.07309" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8322555,"math_prob":0.5735357,"size":1688,"snap":"2022-05-2022-21","text_gpt3_token_len":381,"char_repetition_ratio":0.14370546,"word_repetition_ratio":0.0,"special_character_ratio":0.21090047,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9697324,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T18:15:15Z\",\"WARC-Record-ID\":\"<urn:uuid:423fd01d-b84c-4f82-af09-ce97e154314b>\",\"Content-Length\":\"21478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1e645bb-eb8f-40aa-bee4-a11c8f7024a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:67133b09-feaf-44cf-850b-21ab6046d992>\",\"WARC-IP-Address\":\"128.8.120.41\",\"WARC-Target-URI\":\"https://quics.umd.edu/publications/computational-notions-quantum-min-entropy\",\"WARC-Payload-Digest\":\"sha1:RV3DK63COZ777TF4ZWDNQQVZOVEWTES3\",\"WARC-Block-Digest\":\"sha1:G526JP2TLVPR5CY6PPP7UYATJ4SAFOSE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662675072.99_warc_CC-MAIN-20220527174336-20220527204336-00726.warc.gz\"}"}
https://number.rocks/squared/1638/1624	[ "1638/1624 squared as a fraction\n\nCalculate how much is 1638 square divided by 1624 squared\n\nThe fraction 1638/1624 square is equal to 2683044/2637376 in fraction\n\n(1638/1624)2 =\n2683044/2637376\n\nEvaluate 1638/1624 square\n\n= (1638/1624)2\n\n= 16382/16242\n\n= 2683044/2637376\n\nNext Fraction:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8396079,"math_prob":0.99875265,"size":174,"snap":"2019-43-2019-47","text_gpt3_token_len":65,"char_repetition_ratio":0.15882353,"word_repetition_ratio":0.0,"special_character_ratio":0.5632184,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997762,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T15:53:52Z\",\"WARC-Record-ID\":\"<urn:uuid:f9621415-024c-44b7-b5fc-ee9f55e27517>\",\"Content-Length\":\"6278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44d37fc7-5a00-4f10-889a-ef8e8407e5ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:c63a8d0c-4a81-484e-9956-0f62723e9aae>\",\"WARC-IP-Address\":\"166.62.6.39\",\"WARC-Target-URI\":\"https://number.rocks/squared/1638/1624\",\"WARC-Payload-Digest\":\"sha1:X2FZ3QUIJ7VQPHEP3UWF7ZNEWITPE2IN\",\"WARC-Block-Digest\":\"sha1:BAY53ZDEBJ4LY46MIVDJJWGUGS2LLRTA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986653876.31_warc_CC-MAIN-20191014150930-20191014174430-00222.warc.gz\"}"}
https://www.fernuni-hagen.de/analysis/en/research/zacher.shtml	[ "# Talk by Rico Zacher\n\nOn March 21st, 2022, Prof. Dr. Rico Zacher (University Ulm) gave a talk about \"Li-Yau inequalities for general non-local diffusion equations via reduction to the heat kernel\" as part of the research seminar Analysis of the FernUniversität in Hagen.\n\n## Abstract\n\nI will present a reduction principle to derive Li-Yau inequalities for non-local diffusion problems in a very general framework, which covers both the discrete and continuous setting. The approach is not based on curvature-dimension inequalities but on heat kernel representations of the solutions and consists in reducing the problem to the heat kernel. As an important application we obtain a Li-Yau inequality for positive solutions $u$ to the fractional (in space) heat equation of the form $(-\\Delta)^{\\beta/2}(\\log u)\\le C/t$, where $\\beta\\in (0,2)$. I will also show that this Li-Yau inequality allows to derive a Harnack inequality. The general result is further illustrated with an example in the discrete setting by proving a sharp Li-Yau inequality for diffusion on a complete graph. This is joint work with Frederic Weber (Münster).\n\nSlides of the talk (PDF 299 KB)\n\n## Video of the talk\n\nMarvin Plümer \| 04.05.2022" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9146587,"math_prob":0.9424723,"size":1154,"snap":"2023-40-2023-50","text_gpt3_token_len":257,"char_repetition_ratio":0.12782608,"word_repetition_ratio":0.0,"special_character_ratio":0.21403813,"punctuation_ratio":0.066985644,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97651696,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T11:11:36Z\",\"WARC-Record-ID\":\"<urn:uuid:f3280419-e538-40ee-a65b-07d0b8f518db>\",\"Content-Length\":\"19265\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2e1fdb8-ea26-4bf4-92b9-c8def873a3a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:54f5efa1-a9f7-42cf-bebf-f07656390a8a>\",\"WARC-IP-Address\":\"132.176.184.160\",\"WARC-Target-URI\":\"https://www.fernuni-hagen.de/analysis/en/research/zacher.shtml\",\"WARC-Payload-Digest\":\"sha1:5VHNDMZ4XMNKXUE2CLQLYXPCYXYRJNKC\",\"WARC-Block-Digest\":\"sha1:5QNIGV2UWEI4JKWFPGH2775FICUVFMPJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506399.24_warc_CC-MAIN-20230922102329-20230922132329-00766.warc.gz\"}"}
https://k4lra.org/and-pdf/205-torque-and-equilibrium-problems-and-solutions-pdf-562-464.php	[ "", null, "# Torque And Equilibrium Problems And Solutions Pdf\n\nFile Name: torque and equilibrium problems and solutions .zip\nSize: 19532Kb\nPublished: 07.12.2020", null, "", null, "Refer to the following information for the next five questions.\n\nIf your institution subscribes to this resource, and you don't have a MyAccess Profile, please contact your library's reference desk for information on how to gain access to this resource from off-campus. Please consult the latest official manual style if you have any questions regarding the format accuracy.\n\nAll examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation We introduced a problem-solving strategy in Example Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies.\n\n## Section Summary\n\nAll examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation We introduced a problem-solving strategy in Example Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.\n\nSkip to main content. Search form Search. Static equilibrium problems and solutions. Those forc Static Equilibrium and Elasticity. Solving Statics problems. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution proce When both 3. An object in equilibrium which is also at rest is said to be in static equilibrium: its velocity and acceleration are both zero.", null, "## Equilibrium and Statics\n\nIf you're seeing this message, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos. Science High school physics Torque and angular momentum Torque and equilibrium. Finding torque for angled forces. Practice: Calculating torque. Practice: Equilibrium and applied force.", null, "## 12.3: Examples of Static Equilibrium\n\nThe second condition necessary to achieve equilibrium involves avoiding accelerated rotation maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges.\n\nTorque and equilibrium. Rotational inertia and angular second law. There is no resultant force and therefore zero acceleration. This is the currently selected item. So, if you watched the presentation on the center of mass, which you should have, you might have gotten a little bit of a glancing view of what torque is.", null, "#### INTRODUCTION\n\nСьюзан была настолько ошеломлена, что отказывалась понимать слова коммандера. - О чем вы говорите. Стратмор вздохнул. - У Танкадо наверняка была при себе копия ключа в тот момент, когда его настигла смерть. И я меньше всего хотел, чтобы кто-нибудь в севильском морге завладел ею. - И вы послали туда Дэвида Беккера? - Сьюзан все еще не могла прийти в .\n\nУ него случился инфаркт. Я сам. Никакой крови. Никакой пули. Беккер снисходительно покачал головой: - Иногда все выглядит не так, как есть на самом деле.\n\nБринкерхофф опрокинул директорский стул и бросился к двери. Он сразу же узнал этот голос.\n\n## Melancholy death of oyster boy and other stories pdf\n\n0 comments\n\n### Leave a comment\n\nit’s easy to post a comment\n\nYou may use these HTML tags and attributes: ```<a href=\"\" title=\"\"> <abbr title=\"\"> <acronym title=\"\"> <b> <blockquote cite=\"\"> <cite> <code> <del datetime=\"\"> <em> <i> <q cite=\"\"> <strike> <strong> ```" ]	[ null, "https://k4lra.org/img/31dfc900361a321c0c1c0ea1a282608c.jpg", null, "https://k4lra.org/1.jpg", null, "https://k4lra.org/2.png", null, "https://k4lra.org/1.jpg", null, "https://k4lra.org/img/367bb5024af510b55f40e28121d184f8.jpg", null, "https://k4lra.org/img/782777.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55815715,"math_prob":0.6705195,"size":3618,"snap":"2021-21-2021-25","text_gpt3_token_len":872,"char_repetition_ratio":0.12617598,"word_repetition_ratio":0.18021202,"special_character_ratio":0.18656716,"punctuation_ratio":0.111455105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97279423,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T00:18:33Z\",\"WARC-Record-ID\":\"<urn:uuid:37b831d1-239d-4ec7-86f3-730c0c2ded95>\",\"Content-Length\":\"22303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:527a2964-3806-4ebc-a079-bc8e819e61c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e445c79-3e46-4bea-8565-3c17e3af15a3>\",\"WARC-IP-Address\":\"172.67.168.110\",\"WARC-Target-URI\":\"https://k4lra.org/and-pdf/205-torque-and-equilibrium-problems-and-solutions-pdf-562-464.php\",\"WARC-Payload-Digest\":\"sha1:KHJJMKWQRCJTAMYP2RPPXMEOAXPEJTA3\",\"WARC-Block-Digest\":\"sha1:TD5TNBXHFN7KMFCXIQMCAK5YJCPJEQJ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487611089.19_warc_CC-MAIN-20210613222907-20210614012907-00565.warc.gz\"}"}
https://physics.stackexchange.com/questions/61000/phase-shift-of-resonance?noredirect=1	[ "# Phase shift of resonance [duplicate]\n\nThis question already has an answer here:\n\nFor resonance to occur, is it true that the force lags behind the motion by $\\pi/2$? I saw some notes written that the motion lags behind the force by $\\pi/2$ which makes no sense to me. As I watched many videos and I worked out the motion, it always happens before the force pushes. E.g. if the force is $F=\\cos\\omega t$ and $x = A\\cos(\\omega t -\\pi/2)$, is it true that force lags behind the motion?\n\nIf it's $\\pi/2$ , does there happen anything special?\n\nAlso, I realised that if there is no damping, we can only get 180 or 0 phase difference, which is quite counterintuitive to me. Can anyone give any example to help me feel better?\n\n## marked as duplicate by Emilio Pisanty, Waffle's Crazy Peanut, user10851, Alan Rominger, Brandon EnrightMay 31 '13 at 1:57\n\nThis question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.\n\n## 1 Answer\n\nImagine that the oscillator is a swing and you are the force pushing it. The phase shift is nothing more than the statement that you have to act differently than the swing. Obviously, you shouldn't push in the exact opposite direction (which rules out a phase shift of $\\pi$).", null, "Imagine the red line being the amplitude of the swing, and the green line is your push strength.\n\nWhat the optimal phase shift of $\\pi/2$ (which is equivalent to switching $\\sin$ with $\\cos$) tells you is that you change your pushing direction every time the swing is at its maximum amplitude. So, instead of pushing the strongest when the swing amplitude is the biggest, you push the strongest when the amplitude is 0 and don't push at all when the amplitude is at its maximum." ]	[ null, "https://i.stack.imgur.com/SDC3G.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9638654,"math_prob":0.89927286,"size":677,"snap":"2019-35-2019-39","text_gpt3_token_len":178,"char_repetition_ratio":0.13818721,"word_repetition_ratio":0.0,"special_character_ratio":0.26292467,"punctuation_ratio":0.11333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9674684,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T01:21:45Z\",\"WARC-Record-ID\":\"<urn:uuid:8427e0b0-1163-42a3-8d9e-1179cdbf18e9>\",\"Content-Length\":\"118628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75e6aef8-ae09-4c80-9d0a-987801b6a603>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d897213-a990-4a4f-80af-17188c4012f2>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/61000/phase-shift-of-resonance?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:I2QGBYZQFOBBFQYF5WQPSPY3XGB3HDSA\",\"WARC-Block-Digest\":\"sha1:JTB6MMUE3QA7FOFPNM5UW2GWL72RAJUF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027314638.49_warc_CC-MAIN-20190819011034-20190819033034-00263.warc.gz\"}"}
http://www.carsten-koenig.de/2020/09/12/matplotlib-performance/	[ "# Matplotlib Performance\n\nWhen generating images for ever larger growing data sets, the performance in creating and writing them to the disk is no longer negligible. Imagine a data set with 10,000 astronomical sources, each observed in 10 bands. Just writing out an image for each band would result in 100,000 images – leaving apart any analysis plots. Now if each image takes just 100 milliseconds to be generated and saved, just the generation of the figures would already take more than 2.5 hours – and that’s without any data loading, processing, normalization or leave alone analysis. It would therefore be the goal to increase the performance and reduce the computing workload to the absolute minimum necessary, in order not to end up with too long runtime for the data reduction just to get a quick view on the data.\n\nIn this article I will quickly go over a few things you might want to keep in mind when generating a large number of figures using Matplotlib for Python. There are a couple obvious and not so obvious decisions you can make, in order to accelerate creation of your plots.\n\n### The test setup\n\nThis is a simplified version of the test setup I used, in order to investigate the creation times. In order to avoid any cached objects to be reused and flaw the benchmark, it was run for each setup individually in a fresh iPython session. To reduce the parameter space, I only compared PNG (Portable Network Graphics) and PDF (Portable Document Format) files, likely the two most widely used formats for screen and print media, respectively. Similarly only 100 and 300 DPI were used for resolution tests.\n\n```from matplotlib import pyplot as plt\nimport numpy as np\n\ndef testSavefig(format, method, dpi, data):\n'''\nCreate a test figure.\n\n@param format: The file format (png or pdf).\n@param method: The method to be used (either 'fig' or 'plt')\n@param dpi: The resolution to use in DPI.\n@param data: The data to be plotted.\n\n'''\n# Backends to use\nbe = {'png': 'agg',\n'pdf': 'pdf'}\n\n# Create the figure\nfig = plt.Figure()\nax.plot(data)\n\n# Save the figure to a file\nif method == 'fig':\nfig.savefig('test.%s'%format,\ndpi=dpi,\nbackend=be[format])\nelif method == 'plt':\nplt.savefig('test.%s'%format,\ndpi=dpi,\nbackend=be[format])\n\n> format = 'png' # or 'pdf'\n> method = 'fig' # or 'plt'\n> dpi = 100 # resolution in DPI (here 100 or 300)\n> n = 2 # data size order of magnitude\n> data = np.arange(10**n) # simple slope\n> %timeit testSavefig(format, data, method)\n```\n\n### 1.) How to save a figure? (pyplot vs. figure)\n\nThere are two major ways how to create a figure in Matplotlib. When working with pyplot you can use `plt.savefig()` in order to save your active figure. Alternatively you can use the figures own `fig.savefig()` method to save them. As the time to save a figure also depends on the number of objects in it, the number of data points is another parameter to keep in mind.\n\nWe see here a couple of interesting trends:\n\n• The pyplot.savefig() method is significantly slower than the figures own savefig method.\n• For the pyplot.savefig() method it does not matter if we save the figure as either PNG or PDF.\n• PDF creation is faster than PNG creation when created with figure.savefig().\n\nThis means that if you want to create a large amount of figures, use the `figure.savefig()` method, not the `pyplot.savefig()` one, as the former outperforms the latter significantly.\n\nInteresting side note: the bump in the pyplot curves at 105 data points is real and likely arises from some form of memory-object-size mismatch (simply speaking: imagine what happens if you have to put a a 9 byte object through a 8 byte register – you would have to do the operation twice, wasting almost 50% of the bandwidth)\n\n### 2.) The resolution and format (PNG vs PDF vs DPI)\n\nAs we have already seen, saving a figure as PDF is faster than saving it as a PNG. So we will now compare PNG and PDF creation, and will also take into accound two different resolutions: 100 DPI and 300 DPI.\n\nAgain, we see some interesting trends:\n\n• Creation of PNGs is slower for higher resolutions, which is expected for a bitmap format.\n• As seen before, PDFs are created faster than PNGs.\n• For PDFs the resolution has no effect on the creation time (which is expected from a vector format).\n• No matter which format or resolution is used, for extremely high number of data points, the differences become insignificant, indicating some other mechanism dominating the file generation process.\n• On the other hand, below 100,000 data points the differences are approximately constant.\n\nIn conclusion this means that if your plot has less than 100,000 data points, the method you choose can make a significant difference on the runtime. If you create figures for screen inspection, choosing a lower resolution is advisable, although you should still be able to see the details you want to identify in your plots. A hybrid approach of creating PNG thumbnails and linking the PDFs for detailed inspection might be favorable over creating high resolution PNGs.\n\nAlthough not included in the test, it is noteworthy that creating an Encapsulated Post Script (EPS) file, rather than a PDF file, results in another speed-up gain of up to 10 percent over PDF.\n\n### 3.) The use of Latex\n\nIn Matplotlib you have three options when annotating you figures with mathematical expressions:\n\n• use plain text and keep it simple (i.e. use no \\$[…]\\$ notation at all),\n• use Matplotlibs own mathtext capabilities (usetex=False),\n• or use an external Latex package (usetex=True).\n\nAlthough it seems obvious, it is worth mentioning that the choice you make has a significant impact on the creation time of your figures. The following table lists the runtimes for the three methods, when only the x- and y-axis labels contain Latex expressions (or don’t contain it for the plain text example):\n\nFrom these results it is clear, that you should only use an external Latex package, when it is absolutely necessary, as it is about three times slower than the plain text generation and still almost a factor of two slower than using Matplotlibs own mathtext extension. You would probably only use an external Latex package for a small amount of figures, or even just for publication, when you do not iterate any more.\n\n## Closing remarks\n\nNow, PNG is a screen format, whereas PDF is made for print media. So you might actually not have too much of a choice. For print media (E)PS might be an option, although these formats are widely considered outdated. For screen presentations you could maybe switch to SVG (Scalable Vector Graphics), when e.g. working with Jupyter Notebooks in a browser. This should result in similar results like PDF, as it is also a vector format, but as SVG can not be used in all kind of software and displaying it might even in these days still be slightly inconsistent, you would have to check on a case-by-case basis if this is an option.\n\nNote that all the discussion in this article are only relevant, if you have to create a really large number of plots. If you only have to create a few figures for a journal publication or a presentation at a conference – do not bother. In these cases just use the most convenient approach." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.872042,"math_prob":0.7507506,"size":7650,"snap":"2021-31-2021-39","text_gpt3_token_len":1831,"char_repetition_ratio":0.10881507,"word_repetition_ratio":0.015117158,"special_character_ratio":0.26496732,"punctuation_ratio":0.12244898,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95232594,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T04:24:25Z\",\"WARC-Record-ID\":\"<urn:uuid:cae32b53-6100-4241-8e6b-5a7c9befd0b1>\",\"Content-Length\":\"67091\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c28c0d3d-6e9c-4f91-aaec-66f9a70d8008>\",\"WARC-Concurrent-To\":\"<urn:uuid:59c5f2e9-68a6-447a-9702-f615e435dd47>\",\"WARC-IP-Address\":\"134.119.225.67\",\"WARC-Target-URI\":\"http://www.carsten-koenig.de/2020/09/12/matplotlib-performance/\",\"WARC-Payload-Digest\":\"sha1:YEERAGHNAS6PMBTW7AVMZUAZ5OD27W7C\",\"WARC-Block-Digest\":\"sha1:NS4CDNHIAE76SIQDXZOK77DBM5XJ53ZR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155322.12_warc_CC-MAIN-20210805032134-20210805062134-00367.warc.gz\"}"}
https://googology.fandom.com/wiki/-illion	[ "The -illion system is a suffix used to represent the powers of one thousand. It is the only large number system that is widely in use except for systems in eastern Asia which are based on a myriad.\n\n## Short and long scale\n\nThe -illion system has two distinct forms, the long scale and the short scale. Accoring to long and short scale#Current usage, Most English-speaking and Arabic-speaking countries and regions use the short scale, while most continental European countries such as French and German use the long scale, and some countries use both. The long scale was also used in UK English until 1974, when the British usage and the American usage became identical.\n\nThe short scale goes as follows:\n\nThe long scale goes as follows:\n\nSbiis Saibian replaces the long-scale -illion with -illiad.\n\n## Naming systems of zillion numbers\n\n### Conway-Wechsler system\n\nConway and Guy developed a system for extending the zillion numbers according to the rule in this table, which determines n-illion numbers for n<1000. For n<10, standard English name is accepted.\n\nN units tens hundreds\n1 un N deci NX centi\n2 duo MS viginti N ducenti\n3 tre () NS triginta NS trecenti\n5 quin () NS quinquaginta NS quingenti\n6 se () N sexaginta N sescenti\n7 septe () N septuaginta N septingenti\n8 octo MX octoginta MX octingenti\n9 nove () nonaginta nongenti\n• () Note: when it is immediately before a component marked with S or X, \"tre\" increases to \"tres\" and \"se\" to \"ses\" or \"sex\" as appropriate. Similarly \"septe\" and \"nove\" increase to \"septem\" and \"novem\" or \"septen\" and \"noven\" immediately before components marked with M or N.\n• (*) quin is changed from original quinqua, as explained below.\n\nAny zillion from the 10th to the 999th is made by combining parts from the appropriate columns of this table, and then replacing the final vowel by \"illion\". For example,\n\n• 234-th zillion number is named as \"quattuor\" + \"triginta\" + \"ducenti\" + \"llion\" = quattuortrigintaducentillion.\n• 34-th zillion number is named as \"quattuor\" + \"triginta\" = quattuortriginta, replace the final \"a\" with \"illion\" = quattuortrigintillion\n\nMiakinen proposed that as quindecillion is a common name and Latin name of 15 is quindecim, not quinquadecim, quinqua should be replaced to quin, and a similar change to all the Conway-Wechsler names involving the quinqua- prefix should be updated. Robert adopted the suggestion, and call this modified version as the above table as the Conway-Wechsler system.\n\nConway and Guy wrote that this complete system of -illion words first appears in their book, and this system has been widely accepted, with some modifications described below.\n\n### Extending the system\n\nConway and Guy extended this system with Allan Wechsler for N≥1000 as follows.\n\nWith Allan Wechsler we propose to extend this system indefinitely by combining these according to the convention that \"XilliYilliZillion\" (say) denotes the (1000000X + 1000Y + Z)th zillion, using \"nillion\" for the zeroth \"zillion\" when this is needed as a placeholder. So for example the million-and-third zillion is a \"millinillitrillion.\"\n\n### Modified Conway-Wechsler system\n\nEven after changing quinqua to quin, Conway-Wechsler names have some difference from the common names, for example, sedecillion and novendecillion are better known as sexdecillion and novemdecillion respectively. Many people simplified the Conway-Wechsler system, where Cookiefonster summarized as follows. Cookiefonster says that this modified system is faithful to the dictionary -illions and it is accepted so much.\n\nHowever, this system is invalid because the name trecentillion means 2 different values for N=103 and N=300. In googology wiki, the modified system was used for the title of -illion numbers up to 999-illion, but because of this failure of the system, the title is now being changed to the original Conway-Wechsler system, except for the standard dictionary names of sexdecillion and novemdecillion.\n\nN units tens hundreds\n1 un deci centi\n2 duo viginti ducenti\n3 tre triginta trecenti\n5 quin quinquaginta quingenti\n6 sex sexaginta sescenti\n7 septen septuaginta septingenti\n8 octo octoginta octingenti\n9 novem nonaginta nongenti\n\nFish made a list of the -illion numbers of the Conway-Wechsler system and the modified system and a program which created the table.\n\n## In googological notations\n\nFor the short scale:\n\nNotation Expression\nFast-growing hierarchy between $$f_2(f_1^3(n))$$ and $$f_2(f_1^4(n))$$\nHardy hierarchy between $$H_{\\omega^2+\\omega 3}(n)$$ and $$H_{\\omega^2+\\omega 4}(n)$$\nSlow-growing hierarchy $$g_{\\omega^{n+1}}(1,000)$$; just below $$g_{\\omega^\\omega}(n)$$\n\nFor the long scale:\n\nNotation Expression\nFast-growing hierarchy between $$f_2(f_1^4(n))$$ and $$f_2(f_1^5(n))$$\nHardy hierarchy between $$H_{\\omega^2+\\omega 4}(n)$$ and $$H_{\\omega^2+\\omega 5}(n)$$\nSlow-growing hierarchy $$g_{\\omega^{2n}}(1,000)$$; just below $$g_{\\omega^\\omega}(n)$$\n\nFor -illiard:\n\nNotation Expression\nFast-growing hierarchy between $$f_2(f_1^4(n))$$ and $$f_2(f_1^5(n))$$\nHardy hierarchy between $$H_{\\omega^2+\\omega 4}(n)$$ and $$H_{\\omega^2+\\omega 5}(n)$$\nSlow-growing hierarchy $$g_{\\omega^{2n+1}}(1,000)$$; just below $$g_{\\omega^\\omega}(n)$$\n\n## Etymology\n\nAll \"-illion\" terms derive from \"million\", the word for which came from the Italian millione; the augmentative suffix -one applied to the word for \"thousand\", mille. This gives \"million\" the literal definition of roughly \"super thousand\". The pattern of extracting \"illion\" from it was established in 1484 by Nicolas Chuquet in his unpublished book Triparty en la science des nombres, which defined the long-scale senses of Byllion up to Nonyllion.\n\n## Quote\n\n• Although we value [your] donations, we were somewhat surprised to note that none of them ended in \"-illion.\"" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6223637,"math_prob":0.9179569,"size":10290,"snap":"2022-05-2022-21","text_gpt3_token_len":3420,"char_repetition_ratio":0.1404822,"word_repetition_ratio":0.08104814,"special_character_ratio":0.3325559,"punctuation_ratio":0.080684595,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9810951,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T08:52:58Z\",\"WARC-Record-ID\":\"<urn:uuid:531a459f-9b28-40b2-a0b1-7bdd7f74205b>\",\"Content-Length\":\"339469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:268be7b0-f880-42a5-a8b9-50c0ebbd2bc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:561887f5-3576-4d42-addc-963afc662472>\",\"WARC-IP-Address\":\"151.101.192.194\",\"WARC-Target-URI\":\"https://googology.fandom.com/wiki/-illion\",\"WARC-Payload-Digest\":\"sha1:A2LMKGRALSLYCGJK2J5UHVDS4Q265MJN\",\"WARC-Block-Digest\":\"sha1:4IAOXEVQWFTGS2YQS7BDRTTYK7M7QL5C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662604495.84_warc_CC-MAIN-20220526065603-20220526095603-00512.warc.gz\"}"}
https://zbmath.org/?q=an%3A0868.65025	[ "## Numerical stability of methods for solving augmented systems.(English)Zbl 0868.65025\n\nCensor, Yair (ed.) et al., Recent developments in optimization theory and nonlinear analysis. AMS/IMU special session on optimization and nonlinear analysis, May 24–26, 1995, Jerusalem, Israel. Providence, RI: American Mathematical Society. Contemp. Math. 204, 51-60 (1997).\nSummary: We consider questions of scaling and accuracy of methods for solving the two coupled symmetric linear systems $$Hy+Ax=b$$, $$A^Ty=c$$, where $$A\\in\\mathbb{R}^{m\\times n}$$, $$m\\geq n$$, and $$H\\in \\mathbb{R}^{m\\times m}$$ is symmetric positive definite. Such systems, known in optimization as Karush-Kuhn-Tucker (KKT) systems, are symmetric indefinite and form the kernel in many optimization algorithms. An often used method for solving KKT systems is to compute an $$LDL^T$$ factorization of the system matrix, where $$L$$ is lower triangular and $$D$$ is block-diagonal with $$1\\times 1$$ and symmetric $$2\\times 2$$ pivots. The pivots are chosen according to a scheme by J. R. Bunch and L. Kaufman [Math. Comput. 31, 163-179 (1977; Zbl 0355.65023)].\nFor the standard case, when $$H=I$$, we review results from Å. Björck [Pitman Res. Notes Math. Ser. 260, 1-16 (1992; Zbl 0796.65056)] which show that for the above method to be numerically stable a scaling of the (1,1) block is needed. We give a closed expression for the optimal scaling factor, and show that for this scaling the errors in $$x$$ and $$y$$ are of the same size as from a norm-wise backward stable method. For the case when $$H=D^{-2}$$ is diagonal we recommend a diagonal scaling such that rows of $$A$$ are equilibrated before a block scaling is performed as in the standard case.\nFor sparse problems the scaling has to be a compromise between stability and sparsity, since the optimal scaling usually leads to unacceptable fill-in in the factorization. Accuracy of the computed solution can then usually be restored by fixed precision iterative refinement. A different scheme, due to C. C. Paige and M. A. Saunders [SIAM J. Numer. Anal. 12, 617-629 (1975; Zbl 0319.65025)], which uses SYMMLQ with a preconditioner derived from the $$LDL^T$$ factorization is also described.\nFor the entire collection see [Zbl 0861.00014].\n\n### MSC:\n\n 65F35 Numerical computation of matrix norms, conditioning, scaling 65F50 Computational methods for sparse matrices 65F05 Direct numerical methods for linear systems and matrix inversion 65K05 Numerical mathematical programming methods 90C20 Quadratic programming\n\n### Citations:\n\nZbl 0355.65023; Zbl 0796.65056; Zbl 0319.65025\n\n### Software:\n\nMA47; Harwell-Boeing sparse matrix collection" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84703255,"math_prob":0.9931343,"size":2943,"snap":"2022-40-2023-06","text_gpt3_token_len":794,"char_repetition_ratio":0.111942835,"word_repetition_ratio":0.0,"special_character_ratio":0.28576282,"punctuation_ratio":0.17102967,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987669,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T14:47:33Z\",\"WARC-Record-ID\":\"<urn:uuid:284ae3c2-237c-4209-b365-b1e9bd0cc23b>\",\"Content-Length\":\"53179\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:307b736b-6663-4559-949f-5595187c240a>\",\"WARC-Concurrent-To\":\"<urn:uuid:9086a0c5-6311-4e20-9de7-cdd09cf9ef92>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A0868.65025\",\"WARC-Payload-Digest\":\"sha1:ILRAZLQBSIZPYMPMZJ3ITICGKPFANNR6\",\"WARC-Block-Digest\":\"sha1:VQHY7S45J7IKTG6O3G3S7UBPWSKG65KP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337421.33_warc_CC-MAIN-20221003133425-20221003163425-00708.warc.gz\"}"}
https://metanumbers.com/53933	[ "## 53933\n\n53,933 (fifty-three thousand nine hundred thirty-three) is an odd five-digits composite number following 53932 and preceding 53934. In scientific notation, it is written as 5.3933 × 104. The sum of its digits is 23. It has a total of 2 prime factors and 4 positive divisors. There are 49,020 positive integers (up to 53933) that are relatively prime to 53933.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 23\n• Digital Root 5\n\n## Name\n\nShort name 53 thousand 933 fifty-three thousand nine hundred thirty-three\n\n## Notation\n\nScientific notation 5.3933 × 104 53.933 × 103\n\n## Prime Factorization of 53933\n\nPrime Factorization 11 × 4903\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 53933 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 53,933 is 11 × 4903. Since it has a total of 2 prime factors, 53,933 is a composite number.\n\n## Divisors of 53933\n\n1, 11, 4903, 53933\n\n4 divisors\n\n Even divisors 0 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 58848 Sum of all the positive divisors of n s(n) 4915 Sum of the proper positive divisors of n A(n) 14712 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 232.235 Returns the nth root of the product of n divisors H(n) 3.66592 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 53,933 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 53,933) is 58,848, the average is 14,712.\n\n## Other Arithmetic Functions (n = 53933)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 49020 Total number of positive integers not greater than n that are coprime to n λ(n) 24510 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5491 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 49,020 positive integers (less than 53,933) that are coprime with 53,933. And there are approximately 5,491 prime numbers less than or equal to 53,933.\n\n## Divisibility of 53933\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 1 3 5 5 5 5\n\n53,933 is not divisible by any number less than or equal to 9.\n\n## Classification of 53933\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (53933)\n\nBase System Value\n2 Binary 1101001010101101\n3 Ternary 2201222112\n4 Quaternary 31022231\n5 Quinary 3211213\n6 Senary 1053405\n8 Octal 151255\n10 Decimal 53933\n12 Duodecimal 27265\n20 Vigesimal 6egd\n36 Base36 15m5\n\n## Basic calculations (n = 53933)\n\n### Multiplication\n\nn×i\n n×2 107866 161799 215732 269665\n\n### Division\n\nni\n n⁄2 26966.5 17977.7 13483.2 10786.6\n\n### Exponentiation\n\nni\n n2 2908768489 156878610917237 8460934122599343121 456323560034150372544893\n\n### Nth Root\n\ni√n\n 2√n 232.235 37.782 15.2393 8.83834\n\n## 53933 as geometric shapes\n\n### Circle\n\n Diameter 107866 338871 9.13817e+09\n\n### Sphere\n\n Volume 6.57132e+14 3.65527e+10 338871\n\n### Square\n\nLength = n\n Perimeter 215732 2.90877e+09 76272.8\n\n### Cube\n\nLength = n\n Surface area 1.74526e+10 1.56879e+14 93414.7\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 161799 1.25953e+09 46707.3\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.03813e+09 1.84883e+13 44036.1\n\n## Cryptographic Hash Functions\n\nmd5 566cf8f56fd4d12cc7a11cbb152b1dfa de31055250dc542bbdf27a89443915d3004f6b1b 526b2d74dc013365ec9d4fc37362deb532ead9d94df9726f60cc8e979d08ce57 f405f9730cc09673acb026b7c7cc4a9c9a9f6d2e540ac8ab62025ee88919e405bf7a3f4c785ca4e1a357b013c732204a27bae5979e9e4bcfb9ae809c0b005c18 aa783c69b0faa185d465e1580697770a31182e64" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6235927,"math_prob":0.983557,"size":4592,"snap":"2020-34-2020-40","text_gpt3_token_len":1615,"char_repetition_ratio":0.11726242,"word_repetition_ratio":0.029411765,"special_character_ratio":0.4483885,"punctuation_ratio":0.07435898,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99554455,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T16:28:04Z\",\"WARC-Record-ID\":\"<urn:uuid:e01ca37a-4e7a-45d3-87e9-74905b2ca534>\",\"Content-Length\":\"48306\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ea2e135-7d88-4538-a463-e6a4defafaf9>\",\"WARC-Concurrent-To\":\"<urn:uuid:144c5d59-3ccf-42bb-b71f-16e8f034dd16>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/53933\",\"WARC-Payload-Digest\":\"sha1:UOLINMLOZTNJMSL7XLXJWGFDI2DJUOP6\",\"WARC-Block-Digest\":\"sha1:O472NMZVHQKSKFO5TKON4Y56LTSZCK67\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739048.46_warc_CC-MAIN-20200813161908-20200813191908-00260.warc.gz\"}"}
https://robotics.stackexchange.com/questions/6859/how-to-implement-tracking-problem-with-pid-controller	[ "# how to implement tracking problem with PID controller\n\nI'm trying to implement the tracking problem for this example using PID controller. The dynamic equation is\n\n$$I \\ddot{\\theta} + d \\dot{\\theta} + mgL \\sin(\\theta) = u$$\n\nwhere\n\n$\\theta$ : joint variable.\n\n$u$ : joint torque\n\n$m$ : mass.\n\n$L$ : distance between centre mass and joint.\n\n$d$ : viscous friction coefficient\n\n$I$ : inertia seen at the rotation axis.\n\n$\\textbf{Regulation Problem:}$\n\nIn this problem, the desired angle $\\theta_{d}$ is constant and $\\theta(t)$ $\\rightarrow \\theta_{d}$ and $\\dot{\\theta}(t)$ $\\rightarrow 0$ as $t$ $\\rightarrow \\infty$. For PID controller, the input $u$ is determined as follows\n\n$$u = K_{p} (\\theta_{d} - \\theta(t)) + K_{d}( \\underbrace{0}_{\\dot{\\theta}_{d}} - \\dot{\\theta}(t) ) + \\int^{t}_{0} (\\theta_{d} - \\theta(\\tau)) d\\tau$$\n\nThe result is", null, "and this is my code main.m\n\nclear all\nclc\n\nglobal error;\nerror = 0;\n\nt = 0:0.1:5;\n\nx0 = [0; 0];\n\n[t, x] = ode45('ODESolver', t, x0);\n\ne = x(:,1) - (pi/2); % Error theta\n\nplot(t, e, 'r', 'LineWidth', 2);\ntitle('Regulation Problem','Interpreter','LaTex');\nxlabel('time (sec)');\nylabel('$\\theta_{d} - \\theta(t)$', 'Interpreter','LaTex');\ngrid on\n\n\nand ODESolver.m is\n\nfunction dx = ODESolver(t, x)\n\nglobal error; % for PID controller\n\ndx = zeros(2,1);\n\n%Parameters:\nm = 0.5; % mass (Kg)\nd = 0.0023e-6; % viscous friction coefficient\nL = 1; % arm length (m)\nI = 1/3mL^2; % inertia seen at the rotation axis. (Kg.m^2)\ng = 9.81; % acceleration due to gravity m/s^2\n\n% PID tuning\nKp = 5;\nKd = 1.9;\nKi = 0.02;\n\n% u: joint torque\nu = Kp(pi/2 - x(1)) + Kd(-x(2)) + Kierror;\nerror = error + (pi/2 - x(1));\n\ndx(1) = x(2);\ndx(2) = 1/I(u - dx(2) - mgLsin(x(1)));\n\nend\n\n\n$\\textbf{Tracking Problem:}$\n\nNow I would like to implement the tracking problem in which the desired angle $\\theta_{d}$ is not constant (i.e. $\\theta_{d}(t)$); therefore, $\\theta(t)$ $\\rightarrow \\theta_{d}(t)$ and $\\dot{\\theta}(t)$ $\\rightarrow \\dot{\\theta}_{d}(t)$ as $t$ $\\rightarrow \\infty$. The input is\n\n$$u = K_{p} (\\theta_{d} - \\theta(t)) + K_{d}( \\dot{\\theta}_{d}(t) - \\dot{\\theta}(t) ) + \\int^{t}_{0} (\\theta_{d}(t) - \\theta(\\tau)) d\\tau$$\n\nNow I have two problems namely to compute $\\dot{\\theta}_{d}(t)$ sufficiently and how to read from txt file since the step size of ode45 is not fixed. For the first problem, if I use the naive approach which is\n\n$$\\dot{f}(x) = \\frac{f(x+h)-f(x)}{h}$$\n\nthe error is getting bigger if the step size is not small enough. The second problem is that the desired trajectory is stored in txt file which means I have to read the data with fixed step size but I'v read about ode45 which its step size is not fixed. Any suggestions!\n\nEdit:\n\nFor tracking problem, this is my code\n\nmain.m\n\nclear all\nclc\n\nglobal error theta_d dt;\nerror = 0;\n\ni = 1;\nt(i) = 0;\ndt = 0.1;\nnumel(theta_d)\nwhile ( i < numel(theta_d) )\ni = i + 1;\nt(i) = t(i-1) + dt;\nend\n\nx0 = [0; 0];\noptions= odeset('Reltol',dt,'Stats','on');\n[t, x] = ode45(@ODESolver, t, x0, options);\n\ne = x(:,1) - theta_d; % Error theta\n\nplot(t, x(:,2), 'r', 'LineWidth', 2);\ntitle('Tracking Problem','Interpreter','LaTex');\nxlabel('time (sec)');\nylabel('$\\dot{\\theta}(t)$', 'Interpreter','LaTex');\ngrid on\n\n\nODESolver.m\n\nfunction dx = ODESolver(t, x)\n\npersistent i theta_dPrev\n\nif isempty(i)\ni = 1;\ntheta_dPrev = 0;\nend\n\nglobal error theta_d dt ;\n\ndx = zeros(2,1);\n\n%Parameters:\nm = 0.5; % mass (Kg)\nd = 0.0023e-6; % viscous friction coefficient\nL = 1; % arm length (m)\nI = 1/3mL^2; % inertia seen at the rotation axis. (Kg.m^2)\ng = 9.81; % acceleration due to gravity m/s^2\n\n% PID tuning\nKp = 35.5;\nKd = 12.9;\nKi = 1.5;\n\nif ( i == 49 )\ni = 48;\nend\n% theta_d first derivative\ntheta_dDot = ( theta_d(i) - theta_dPrev ) / dt;\ntheta_dPrev = theta_d(i);\n\n% u: joint torque\nu = Kp(theta_d(i) - x(1)) + Kd( theta_dDot - x(2)) + Kierror;\nerror = error + (theta_dDot - x(1));\n\ndx(1) = x(2);\ndx(2) = 1/I(u - dx(2) - mgLsin(x(1)));\n\ni = i + 1;\nend\n\n\ntrajectory's code is\n\nclear all\nclc\n\na = 0:0.1:(3pi)/2;\n\nfile = fopen('trajectory.txt','w');\n\nfor i = 1:length(a)\nfprintf(file,'%4f \\n',a(i));\nend\n\nfclose(file);\n\n\nThe result of the velocity is", null, "Is this correct approach to solve the tracking problem?\n\n1. The time step requested by ODE45 is not going to match what you have in your file. In your ODESolver() desired thetas are read one after the other and then the last one is repeated. As a result the desired theta was not a function of $t$. I used interp1() to fix that.\n\n2. The time difference between two iteration of ODE45 is not constant so you can't use dt to calculate the $\\Delta{\\theta}$. This is fixed by storing the t in a variable for future reference.\n\n3. Same goes for calculating the integral part of PID.\n\nHere is the new ODESolver():\n\nfunction dx = ODESolver(t, x)\npersistent theta_dPrev time_prev\n\nif isempty(theta_dPrev)\ntheta_dPrev = 0;\ntime_prev = 0;\nend\n\nglobal error theta_d dt time_array;\n\ndx = zeros(2,1);\n\n%Parameters:\nm = 0.5; % mass (Kg)\nd = 0.0023e-6; % viscous friction coefficient\nL = 1; % arm length (m)\nI = 1/3mL^2; % inertia seen at the rotation axis. (Kg.m^2)\ng = 9.81; % acceleration due to gravity m/s^2\n\n% PID tuning\nKp = 40;%35.5;\nKd = 10;%12.9;\nKi = 20;%1.5;\n\n% theta_d first derivative\ntheta_desired_current = interp1(time_array,theta_d,t);\nif t==time_prev\ntheta_dDot = 0;\nelse\ntheta_dDot = ( theta_desired_current - theta_dPrev ) / (t-time_prev);\nend\ntheta_dPrev = theta_desired_current;\n\n% u: joint torque\nu = Kp(theta_desired_current - x(1)) + Kd( theta_dDot - x(2)) + Kierror;\nerror = error + (theta_desired_current - x(1))(t-time_prev);\n\ndx(1) = x(2);\ndx(2) = 1/I(u - dx(2) - mgLsin(x(1)));\n\ntime_prev = t;\nend\n\n\nI also changed the main.m to the following:\n\nclear all; close all; clc;\n\nglobal error theta_d dt time_array;\nerror = 0;\n\n% theta_d = sin([1:.1:20]/3);\n\ntime_array=0:dt:dt(numel(theta_d)-1);\n\nx0 = [0; 0];\noptions= odeset('Reltol',dt,'Stats','on');\n[t_ode, x] = ode45(@ODESolver, [time_array(1),time_array(end)], x0, options);\n\ntheta_desired_ode = interp1(time_array,theta_d,t_ode);\ne = x(:,1) - theta_desired_ode; % Error theta\n\nfigure(1)\nplot(t_ode, x(:,2), 'r', 'LineWidth', 2);\ntitle('Velocity','Interpreter','LaTex');\nxlabel('time (sec)');\nylabel('$\\dot{\\theta}(t)$', 'Interpreter','LaTex');\ngrid on\n\nfigure(2)\nplot(t_ode,x(:,1))\nhold on\nplot(t_ode,theta_desired_ode)\ntitle('Theta','Interpreter','LaTex');\nxlabel('time (sec)');\nylabel('$\\dot{\\theta}(t)$', 'Interpreter','LaTex');\nlegend('Actual Theta', 'Desired Theta')\ngrid on\n\n• wow this is really nice work. Thank you so much. The results are perfect. Mar 24, 2015 at 19:31\n• I'm glad you found it useful. If you plan to implement it on hardware or want to take your control system one step further, try using anti-windup PID instead of the regular one. Mar 24, 2015 at 19:36\n• Any good reference for anti-windup PID? Mar 24, 2015 at 19:41\n• Mar 24, 2015 at 19:44\n• I'm not a big fan of global variables but since you are already using them, you can also set u as global. A better way would be using object-oriented programming. I also want to suggest not using ODE45. Make a loop with fixed time step specially since you seem to be interested in hardware in future. A loop would simulate how your controller and sensors work much much better. Also, it is going to be easier to implement your already developed codes on hardware. Mar 25, 2015 at 16:29\n\nSince both $\\theta_d\\left(t\\right)$ and $\\dot\\theta_d\\left(t\\right)$ are references at your disposal, i.e. you have to provide them in some way, why don't you simply play with $\\dot\\theta_d\\left(t\\right)$ and then compute $\\theta_d\\left(t\\right)$ accordingly by means of integration? As you might know, integration is a well posed operation compared with the derivative.\n\nOn the other hand, $\\dot\\theta\\left(t\\right)$ is a different \"beast\". In physical systems you barely have access to it (you'd need ad-hoc sensors for measuring it). Valid alternatives are: Kalman estimation, Savitzky-Golay filtering, use of proper D term in the controller equipped with a high-frequency pole.\n\n• position reference is more intuitive than velocity reference even though what you are saying is perfect and a good alternative. Since I'm doing simulation, I am able to access to the velocity, therefore data is available every instant. Having said that, your answer is alternative but not a solution to my problem. Thank you so much though. Mar 24, 2015 at 19:17\n• Are you sure that you have a too generic $\\theta_d\\left(t\\right)$ so that you cannot come up with a analytical derivative of it? How do you produce your position reference profile? I'm insisting because these details are very important in practice. What is the purpose of your simulation then? Not to go to a real implementation? Does it have only a didactic purpose? I understand you can solve the specific problem differently, but this peculiarity is too narrow and there's the risk that it hides other issues, as I said. Mar 24, 2015 at 19:54\n• I'm focusing now on simulation to enhance my skills. The problem with hardware is costly first and it needs some space and tools. A lot of academic papers are based on pure simulation. I will stick with it until I got some money and space. :) Mar 24, 2015 at 20:10\n• I see, you've started tackling the problem. \"A lot of academic papers are based on pure simulation\", that's right why our field (I'm in academic, tough I'll never forget I'm an engineer too) is being populated with questionable works and everything's becoming messy. Remember: exercise always your critics and never trust them :-) Good luck for the future. Mar 24, 2015 at 20:21\n\nYou can treat your differential equation as a third order ODE, because of the integral in the PID controller. For this I will denote the integral of $\\theta$ as $\\Theta$, such that the ODE becomes,\n\n$$I\\ddot{\\theta}+d\\dot{\\theta}+mgL\\sin(\\theta)+K_p\\left(\\theta-\\theta_d(t)\\right)+K_i\\left(\\Theta-\\int_0^t\\theta_d(\\tau)d\\tau\\right)+K_d\\left(\\dot{\\theta}-\\dot{\\theta}_d(t)\\right)=0$$\n\nOne way you can write this into MATLAB code would be:\n\nclear all\nclc\n\n% Parameters:\nm = 0.5; % mass (Kg)\nd = 0.0023e-6; % viscous friction coefficient\nL = 1; % arm length (m)\nI = 1/3mL^2; % inertia seen at the rotation axis. (Kg.m^2)\ng = 9.81; % acceleration due to gravity m/s^2\n\n% PID tuning\nKp = 5000;\nKi = 166667;\nKd = 50;\n\n% Define reference signal\nsyms t\na = 3 pi / 10;\ntemp = a * t;\n% temp = pi/2 + 0t;\nr = matlabFunction(temp, 'Vars', t);\nR = matlabFunction(int(temp,t), 'Vars', t);\ndr = matlabFunction(diff(temp), 'Vars', t);\nclear temp t\n\nfun = @(t,x) [x(2); x(3); -mgL/Isin(x(2))-d/Ix(3)] - [0; 0; ...\nKp / I (x(2) - r(t)) + ...\nKi / I * (x(1) - R(t)) + ...\nKd / I * (x(3) - dr(t))];\n\nT = linspace(0, 5, 1e3);\nx0 = [0; 0; 0];\n[t, Y] = ode45(@(t,x) fun(t,x), T, x0);\n\ny = zeros(size(t));\nfor i = 1 : length(t)\ny(i) = r(t(i));\nend\n\nfigure\nsubplot(2,1,1)\nplot(t, y, t, Y(:,2))\nxlabel('Time [s]')" ]	[ null, "https://i.stack.imgur.com/3EFpL.png", null, "https://i.stack.imgur.com/Lr1EO.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5871743,"math_prob":0.9978928,"size":4119,"snap":"2023-40-2023-50","text_gpt3_token_len":1465,"char_repetition_ratio":0.117375456,"word_repetition_ratio":0.21502209,"special_character_ratio":0.40058267,"punctuation_ratio":0.1799569,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988997,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T06:18:16Z\",\"WARC-Record-ID\":\"<urn:uuid:f2930e21-b421-4bff-9d52-d1a12f48269f>\",\"Content-Length\":\"213938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c8210f4-70ac-4f5f-9cd7-b4693c4cb9c5>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c9497a8-eaaa-4814-ab1e-f4198e4d12a8>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://robotics.stackexchange.com/questions/6859/how-to-implement-tracking-problem-with-pid-controller\",\"WARC-Payload-Digest\":\"sha1:X6TZYB57FASCNQYX4OFKKTZP53TMMDWU\",\"WARC-Block-Digest\":\"sha1:NOLHUMR53F7TJGXPEETMAX6KSX2BCQ5K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100525.55_warc_CC-MAIN-20231204052342-20231204082342-00021.warc.gz\"}"}
https://www.practiceaptitudetests.com/resources/numerical-reasoning-test-practice-currency-conversion/	[ "Currency conversions can sometimes seem daunting, but if you know how to approach them they become a lot easier.\n\nImagine you are told that the exchange rate for pounds to euros was £1 to €1.175 and you are asked to calculate how many pounds you could exchange for one euro.\n\nSaying £1 to €1.175 euros is the same as saying £1 equals €1.175.\n\nWhen applying mathematics to one side of a ratio the same must be done to the other. SO, if we wanted to find how many pounds we could exchange for one euro we need to do a calculation that changes this ratio to one that looks like this:\n\nIf you divide any number by itself you will always be left with 1, which is what we are trying to turn 1.175 into.\n\nSo we need to divide 1.175 by itself, and if we are dividing one side of the ratio by 1.175 we must do the same to the other. This leaves us with 0.851 pounds.\n\nSo, the answer is 0.851 pounds to 1 dollar." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.97751987,"math_prob":0.97956383,"size":896,"snap":"2020-45-2020-50","text_gpt3_token_len":222,"char_repetition_ratio":0.117713004,"word_repetition_ratio":0.0,"special_character_ratio":0.2700893,"punctuation_ratio":0.10576923,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9846655,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T09:45:00Z\",\"WARC-Record-ID\":\"<urn:uuid:adf52d28-5e5e-4a39-84c0-df18d44b3816>\",\"Content-Length\":\"123681\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31dd27c3-69aa-47c9-a363-38bf1f4ec1a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e7259e2-43a1-482e-885f-0f727502f92c>\",\"WARC-IP-Address\":\"104.248.63.231\",\"WARC-Target-URI\":\"https://www.practiceaptitudetests.com/resources/numerical-reasoning-test-practice-currency-conversion/\",\"WARC-Payload-Digest\":\"sha1:GLDNABHK5LI3JGYWDHSDZVPUYN26FI3F\",\"WARC-Block-Digest\":\"sha1:J2VLE76KW2UTJ5SFED3A6JRMFOPPY5T4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197593.33_warc_CC-MAIN-20201129093434-20201129123434-00389.warc.gz\"}"}
http://kamerynjw.net/2019/12/04/omegath-hod.html	[ "McAloon proved in the 70s that $\\mathrm{HOD}^\\omega$—i.e. the intersection of $\\mathrm{HOD}$, the $\\mathrm{HOD}$ of $\\mathrm{HOD}$, and so on iterated out finitely far—may fail to satisfy choice. I decided to write up an exposition of his result because (1) it seems to be underappreciated and (2) I wanted to understand it and writing up an exposition of it was a good way to make that happen. (Added after writing: In this regard this was successful! There were some details I convinced myself I understood on the first read-through but had to revisit to really understand to write this post.)\n\nMy main reference here is section 6 of Zadrożny’s 1983 paper “Iterated ordinal definability”. References for the below-cited results can be found therein.\n\nLet me start with the background context. It is well-known that there is a class forcing which forces the ground model to be the $\\mathrm{HOD}$ of the extension. Thus, the only axioms which ZFC proves must be satisfied by $\\mathrm{HOD}$ are ZFC itself. In particular, $\\mathrm{HOD}$ may fail to satisfy $V = \\mathrm{HOD}$ and it is possible that $\\mathrm{HOD}^\\mathrm{HOD} \\ne \\mathrm{HOD}$.\n\nAs such it is sensible to iterated $\\mathrm{HOD}$. Let $\\mathrm{HOD}^0 = V$, $\\mathrm{HOD}^{\\alpha+1} = \\mathrm{HOD}^{\\mathrm{HOD}^\\alpha}$ and $\\mathrm{HOD}^\\gamma = \\bigcap_{\\alpha < \\gamma} \\mathrm{HOD}^\\alpha$ for limit ordinals $\\gamma$. As this definition takes place in the metatheory, it is natural to inquire as to whether $\\mathrm{HOD}^\\alpha$ is definable. Of course, the sticking point is that you need uniform access to the sequence of iterated $\\mathrm{HOD}$s below a limit ordinal, because the successor stage preserves definability.\n\nTheorem (Grigorieff): If the universe is constructible from a set then $\\mathrm{HOD}^\\alpha$ is uniformly definable for all $\\alpha$.\n\nIn general, however, even the $\\omega$-th $\\mathrm{HOD}$ may fail to be definable. This is a consequence of the following proposition and theorem.\n\nProposition: If the sequence $\\langle \\mathrm{HOD}^\\alpha : \\alpha < \\lambda \\rangle$ is definable, then for each $\\alpha < \\lambda$ we have $\\mathrm{HOD}^\\alpha$ satisfies ZF.\n\nProof: We have to see that the iterated $\\mathrm{HOD}$s are transitive, closed under the Gödel operations, and almost universal. The first two of these are obvious, so let us check the third. The only substantive case is if $\\gamma < \\lambda$ is limit. Suppose that $X$ is a subset of $\\mathrm{HOD}^\\gamma$. We want to find a set $Y \\in \\mathrm{HOD}^\\gamma$ which covers $X$. Let $\\beta$ be large enough so that $X \\subseteq V_\\beta$. Then $Y = V_\\beta \\cap \\mathrm{HOD}^\\gamma = V_\\beta \\cap \\bigcap_{\\alpha < \\gamma} \\mathrm{HOD}^\\alpha$ is in $V_{\\beta+1} \\cap \\mathrm{HOD}^\\gamma$ and covers $X$, as desired. QED\n\nTheorem (Harrington): There is a (transitive) model of ZFC whose $\\omega$-th $\\mathrm{HOD}$ does not satisfy ZF.\n\nPutting these together, in Harrington’s model the $\\omega$-th $\\mathrm{HOD}$ is not definable.\n\nThe background now made clear, let me move to the result I want to talk about.\n\nTheorem (McAloon): There is a set-forcing extension of $L$ whose $\\omega$-th $\\mathrm{HOD}$ lacks a well-ordering of its reals.\n\nNote that since McAloon’s model is constructible from a set that the sequence of iterated $\\mathrm{HOD}$s over his model is definable and thus $\\mathrm{HOD}^\\omega$ satisfies ZF.\n\nProof (Following Zadrożny): Let $\\mathbb P$ be the poset to add $\\omega_1$ many Cohen reals, as defined in $L$. Since I will be needing the implementation details, let me be clear just what I mean by this. Conditions in $\\mathbb P$ are finite partial functions from $\\omega \\times \\omega_1$ to $2$, ordered by reverse inclusion. Let $G \\subseteq \\mathbb P$ be generic over $L$ and set $A_k = { (\\ell,\\alpha) \\in G: \\ell \\ge k }$ to be the chunk of $G$ from the $k$-th column on.\n\nWe next force over $L[G]$ to code the $A_k$ in some ordinal definable way. There are multiple ways to do this. Since we are starting over $L$, I will code by the (non)existence of Cohen-generic subsets of a cardinal. Namely, if $X$ is a set of ordinals then we can code $X$ starting at a cardinal $\\kappa$ by the product of $\\mathrm{Add}(\\kappa^{+i},1)$ for $i \\in X$. In $L[G]$, the only cardinal with subsets which are Cohen-generic over $L$ is $\\omega$ itself. So in the further extension by the coding forcing, so long as we choose $\\kappa$ to be uncountable, $X$ is definable using $\\kappa$ as a parameter. And this coding forcing preserves cardinals and preserves the $\\mathrm{GCH}$.\n\nLet me be a bit more precise. Fix a set of ordinals $X$ and a cardinal $\\kappa$. Let $\\mathbb C(X,\\kappa)$ be the full-support product of $\\mathrm{Add}(\\kappa^{+i},1)$ for $i \\in X$. Then $\\mathbb C(X,\\kappa)$ is $\\mathord<\\kappa$-closed and doesn’t add any Cohen-generic subsets to cardinals $>\\kappa^{+\\sup X}$. In particular, if we put enough space between coding points, then we can code multiple sets without one coding interfering with the other.\n\nWe can iterate this procedure. Start with $X = X^{(0)}$ and $\\kappa = \\kappa_0$ and let $X^{(1)} \\subseteq \\mathbb C(X^{(0)},\\kappa_0)$ be generic over the ground model $M$. We can think of $X^{(1)}$ as a set of ordinals; literally it is a function from a set of ordinals to sets of ordinals, but since the sets in the range are disjoint we can just think of $X^{(1)}$ as being the union of the range. We can then take $\\kappa_1$ large enough and force over $M[X^{(1)}]$ with $\\mathbb C(X^{(1)},\\kappa_1)$. And we can do this out to $\\omega$, coding the $k$-th generic $X^{(k)}$ above $\\kappa_k$, amounting to a full-support $\\omega$-length iteration. I will be using this later, so let’s call this $\\omega$-length iteration $\\mathbb S(X, \\kappa)$. (Earlier, $\\mathbb C$ was for “coding”, here $\\mathbb S$ is for “self-encoding”, as the generic codes itself into the pattern of which cardinals have Cohen-generic subsets.)\n\nAs we can space out $\\mathbb C(X,\\kappa)$ codings to not interfere with each other, it is clear that we can space out $\\mathbb S(X,\\kappa)$ codings to not interfere with each other. We know in advance how much space we need to code $X^{(0)} = X$, from which we can then calculate how much space we need to code $X^{(1)}$, and then how much to code $X^{(2)}$, and so on. So given a bunch of sets of ordinals $X_i$, $i \\in I$, to code, we can pick $\\kappa_i$ so that the $\\mathbb S(X_i,\\kappa_i)$ codings don’t interfere with each other. That is, if we force with the full-support product of the $\\mathbb S(X_i, \\kappa_i)$ then all of the $X_i^{(k)}$ are ordinal definable by looking at a contiguous block of cardinals and querying which of them have subsets which are Cohen-generic over $L$. (Observe that we are using full-support to ensure we don’t add any Cohen reals.)\n\nLet me summarize the situation and explain how we will apply these coding forcings. We forced over $L$ to get $L[G]$ where we added $\\omega_1$ many Cohen reals. We then take the chunks $A_k$ of $G$, $k < \\omega$, and we force with $\\prod_{k < \\omega} \\mathbb S(A_k, \\kappa_k)$ to code them in an ordinal-definable way. We will require that $\\kappa_0 > \\omega_1$ to make a later closure argument work. (The $A_k$ are not sets of ordinals, but we can consider them as such via our favorite pairing function.) Let $H$ be generic over $L[G]$ for this forcing. So in $L[G][H]$ not only are each $A_k$ ordinal definable, but so are the sets witnessing their ordinal definability, the sets witnessing the ordinal definability of those sets, and so on. In fact, we don’t even need ordinal parameters to define them; since we are only coding countably many sets we can code them at sufficiently spaced-out definable without parameters coding points to get that they are themselves definable without parameters.\n\nWe don’t actually want to go all the way to $L[G][H]$. It has too much information and so its $\\mathrm{HOD}$ is too big, being all of $L[G][H]$. We have to thin it out. Inside $L[G][H]$, let $B = \\bigcup_{k < \\omega} A_k^{(k)}$ and for $\\ell \\in \\omega$ let $B_\\ell = \\bigcup_{k < \\omega} A_{\\ell + k}^{(k)} = \\bigcup_{\\ell \\le k < \\omega} A_k^{(k-\\ell)}$. In particular, $B = B_0$. Our desired model is $L[B]$.\n\nClaim: In $L[B]$, we have $\\mathrm{HOD}^\\ell = L[B_\\ell]$ for $\\ell < \\omega$.\n\nThis is seen by induction. Work inside $(\\mathrm{HOD}^{\\ell})^{L[B]} = L[B_\\ell]$. Observe that $B_{\\ell+1}$ is ordinal definable, because an ordinal $i$ is in $B_{\\ell+1}$ if and only if $i$ is in one of the coding blocks, say the $k$-th one, and there is a subset of $(\\kappa_k)_k$ which is Cohen-generic over $L$. Thus, $L[B_{\\ell+1}] \\subseteq \\mathrm{HOD}^{L[B_\\ell]}$. For the other inclusion, note that $L[B_\\ell]$ is generic over $L[B_{\\ell+1}]$ for a forcing that adds Cohen-generic subsets to certain cardinals. So by a standard homogeneity argument we get that $\\mathrm{HOD}^{L[B_\\ell]} \\subseteq L[B_{\\ell+1}]$.\n\nThus, we get that in $L[B]$ that the $\\ell$-th iterated $\\mathrm{HOD}$s are all different, for $k < \\omega$. Since $L[B]$ is constructible from a set $(\\mathrm{HOD}^\\omega)^{L[B]} = \\bigcap_{\\ell < \\omega} L[B_\\ell]$ is a definable class over $L[B]$ and by the above proposition must satisfy ZF. It remains only to see that this model, call it $N$ so as to not have to write out $(\\mathrm{HOD}^\\omega)^{L[B]}$ ever again, does not have a well-order of its reals.\n\nEach of the models $L[B_\\ell]$ satisfies the $\\mathrm{GCH}$ and agrees with $L$ about which ordinals are cardinals. So if $N$ did have a well-order of its reals then it would have a well-order of ordertype ${\\omega_1}^L$. Thus there would be $X \\subseteq {\\omega_1}^L$ in $N$ so that every real in $N$ is in $L[X]$. As such, to show that $N$ does not have a well-order of its reals it suffices to prove there is no such $X$. Toward this, fix $X \\subseteq \\omega_1$ in $N$.\n\nObserve that $L[G][B] = L[B]$ because $A_0 = G \\in L[B]$. But the forcing to add $B$ over $L[G]$ doesn’t add new subsets of $\\omega_1$, because we did all the coding above $\\omega_1$. So it must be that $X \\in L[G]$. Because $G$ is generic for the forcing to add $\\omega_1$ many Cohen reals, a standard ccc argument gives that there is some countable ordinal $\\alpha$ so that $X \\in L[G \\cap (\\omega \\times \\alpha)]$. That is, we only needed a countable piece of $G$ to add $X$. Now let $c$ be the $\\alpha$-th row of $G$, i.e. $c = { n \\in \\omega : (n,\\alpha) \\in G }$. Then $c$ is a Cohen real over $L[G \\cap (\\omega \\times \\alpha)]$ and so in particular $c \\not \\in L[X]$. Notice, however, that $c \\in L[B_\\ell]$ for each $\\ell$, because a tail of $c$ is the $\\alpha$-th row of $A_\\ell \\in L[B_\\ell]$. So $c \\in N$, witnessing that there is a real in $N$ which is not constructible from $X$. This completes the argument that $N$ does not have a well-order of its reals. QED" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86482847,"math_prob":0.99989295,"size":10846,"snap":"2021-43-2021-49","text_gpt3_token_len":3211,"char_repetition_ratio":0.14951116,"word_repetition_ratio":0.026598755,"special_character_ratio":0.29651484,"punctuation_ratio":0.091533184,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999943,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T14:47:19Z\",\"WARC-Record-ID\":\"<urn:uuid:4a1c6513-8393-4dcb-8646-a2a59f0dda77>\",\"Content-Length\":\"16255\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9562fe61-1053-4952-905e-082a3c5c7d5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:8991e439-1f76-4c54-a4e0-b6e941c89472>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://kamerynjw.net/2019/12/04/omegath-hod.html\",\"WARC-Payload-Digest\":\"sha1:5SF4UR5PE5VWCR533YHFF734Q3TIVPBT\",\"WARC-Block-Digest\":\"sha1:2GBYZAVBS7YSKTMSTSBIM44BTOTFHQMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585270.40_warc_CC-MAIN-20211019140046-20211019170046-00687.warc.gz\"}"}
https://ptmts.org.pl/jtam/index.php/jtam/article/view/4158	[ "Journal of Theoretical\nand Applied Mechanics\n\n55, 2, pp. 407-420, Warsaw 2017\nDOI: 10.15632/jtam-pl.55.2.407\n\n### Parameter estimation of a discrete model of a reinforced concrete slab\n\nMałgorzata Abramowicz, Stefan Berczyński, Tomasz Wróblewski\nThis paper presents parameter estimation of a mathematical model regarding natural vibrations\nof a reinforced concrete slab. Parameter estimation is based on experiments conducted\non a real reinforced concrete slab. Estimated parameters include: substitute longitudinal\nmodulus of elasticity of the reinforced concrete slab, which takes into account longitudinal\nreinforcement, effective thickness of the reinforced concrete slab and coefficient of damping.\nUsing appropriate criteria during, the process of parameter estimation of the reinforcement\nconcrete slab models has a great impact on obtaining precise results. The estimation criteria\nare selected in order to achieve consistency of natural vibration frequencies along with the\nFrequency Response Function measured during experiments with those calculated with the\nmathematical model. The model and all the calculations have been made using MATLAB\nprogramming environment.\nKeywords: rigid finite element (RFE) model, reinforced concrete slab, estimation criteria, modal parameters, vibrations" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7960281,"math_prob":0.88635135,"size":2153,"snap":"2021-21-2021-25","text_gpt3_token_len":404,"char_repetition_ratio":0.15774779,"word_repetition_ratio":0.8181818,"special_character_ratio":0.18207152,"punctuation_ratio":0.12391931,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9760658,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-15T20:04:12Z\",\"WARC-Record-ID\":\"<urn:uuid:df6f5621-312b-4db7-a5f2-531b52f73dea>\",\"Content-Length\":\"18776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea935767-f0a4-4ec4-aabb-e55af03183a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:89971e27-da43-4b80-85b9-11ec5f05af3b>\",\"WARC-IP-Address\":\"212.85.123.71\",\"WARC-Target-URI\":\"https://ptmts.org.pl/jtam/index.php/jtam/article/view/4158\",\"WARC-Payload-Digest\":\"sha1:5Y4ISO3OGQ3FP7KW5U7QNHAOXRP7XRAS\",\"WARC-Block-Digest\":\"sha1:GU2QHM2OWFC2J2KI7VWRFFUG4UXLXSAR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991378.52_warc_CC-MAIN-20210515192444-20210515222444-00177.warc.gz\"}"}
https://matholympiad.org.bd/forum/viewtopic.php?f=41&t=4088&p=20739	[ "## Number theory\n\nProblem for Secondary Group from Divisional Mathematical Olympiad will be solved here.\nForum rules\nPlease don't post problems (by starting a topic) in the \"Secondary: Solved\" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.\nDrake\nPosts: 1\nJoined: Tue Dec 26, 2017 12:00 pm\n\n### Number theory\n\nThe energy of an ordered triple(a,b,c) formed by 3 positive integers a,b,c is said to be n and a<=b<=c and GCD(a,b,c)=1. There are some possible triples for which a^n+b^n+c^n is divisible by a+b+c for all n>0. Find the maximum possible value of a+b+c.\n\nsamiul_samin\nPosts: 1007\nJoined: Sat Dec 09, 2017 1:32 pm\n\n### Re: Number theory\n\nI'll Latex it for you:\nThe energy of a ordered triple ($a,b,c$) formed by $3$ positive integers $a,b,c$ is said to be $n$ and $a\\leq b\\leq c$ & GCD($a,b,c$)$=1$.There are some possible triples for $a^n+b^n+c^n$ is divisible by $a+b+c$ for all $n>0$.Find the maximum possible value of $a+b+c$.\n\nNote\n\nNABILA\nPosts: 28\nJoined: Sat Dec 15, 2018 5:19 pm\nLocation: Munshigonj, Dhaka\n\n### Re: Number theory\n\nsamiul_samin wrote:\nSat Mar 03, 2018 9:53 pm\nThere are some possible triples for $a^n+b^n+c^n$ is divisible by $a+b+c$ for all $n>0$.\n@Drake\nAre you sure about this?", null, "", null, "", null, "Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè samiul_samin Posts: 1007 Joined: Sat Dec 09, 2017 1:32 pm ### Re: Number theory NABILA wrote: Wed Jan 16, 2019 1:07 pm samiul_samin wrote: Sat Mar 03, 2018 9:53 pm There are some possible triples for$a^n+b^n+c^n$is divisible by$a+b+c$for all$n>0\\$.\n@Drake\nAre you sure about this?", null, "", null, "", null, "This is from dhaka regional 2017." ]	[ null, "https://matholympiad.org.bd/forum/images/smilies/icon_exclaim.gif", null, "https://matholympiad.org.bd/forum/images/smilies/icon_question.gif", null, "https://matholympiad.org.bd/forum/images/smilies/icon_e_ugeek.gif", null, "https://matholympiad.org.bd/forum/images/smilies/icon_exclaim.gif", null, "https://matholympiad.org.bd/forum/images/smilies/icon_question.gif", null, "https://matholympiad.org.bd/forum/images/smilies/icon_e_ugeek.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84198517,"math_prob":0.9906683,"size":2146,"snap":"2019-51-2020-05","text_gpt3_token_len":684,"char_repetition_ratio":0.11111111,"word_repetition_ratio":0.7025496,"special_character_ratio":0.30102515,"punctuation_ratio":0.13293651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921168,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T16:20:20Z\",\"WARC-Record-ID\":\"<urn:uuid:ed3f8448-34c5-43e2-bdcd-cdaee9fbd068>\",\"Content-Length\":\"38248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e8f8668-3dd3-4549-bf4b-6dc33dec12c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:790a80d6-7b87-4221-a15b-9398b15cfcbd>\",\"WARC-IP-Address\":\"167.71.232.37\",\"WARC-Target-URI\":\"https://matholympiad.org.bd/forum/viewtopic.php?f=41&t=4088&p=20739\",\"WARC-Payload-Digest\":\"sha1:CQ6EVIH4DLKQHI5BCA5OEINPXE3WRC5T\",\"WARC-Block-Digest\":\"sha1:TKFHL4N47DH7G4LWM73VHGC4CWJ7UHTE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540511946.30_warc_CC-MAIN-20191208150734-20191208174734-00080.warc.gz\"}"}
https://testbook.com/question-answer/which-of-the-following-is-not-one-of-the-methods-t--615dd1f8876baf0766602bd9	[ "Which of the following is NOT one of the methods to improve string efficiency of insulators?\n\nThis question was previously asked in\nUPPCL JE Electrical 7 Sept 2021 Official Paper (Shift 2)\nView all UPPCL JE Papers >\n1. Selection of k\n2. Disc porosity\n3. Guard ring\n\nOption 2 : Disc porosity\n\nDetailed Solution\n\nThe ratio of voltage across the whole string to the product of number of discs and the voltage across the disc nearest to the conductor is known as string efficiency.\n\nString efficiency = (conductor voltage)/(number of discs × voltage across the disc nearest to the conductor)\n\nString efficiency of a string of disc insulators can be improved by using following methods.\n\n• By using longer cross-arms: The value of string efficiency depends upon the value of K i.e., ratio of shunt capacitance to mutual capacitance. The lesser the value of K, the greater is the string efficiency and more uniform is the voltage distribution. The value of K can be decreased by reducing the shunt capacitance. In order to reduce shunt capacitance, the distance of conductor from tower must be increased i.e., longer cross-arms should be used. However, limitations of cost and strength of tower do not allow the use of very long cross-arms. In practice, K = 0·1 is the limit that can be achieved by this method.\n• By grading the insulators: In this method, insulators of different dimensions are so chosen that each has a different capacitance. The insulators are capacitance graded i.e. they are assembled in the string in such a way that the top unit has the minimum capacitance, increasing progressively as the bottom unit (i.e., nearest to conductor) is reached. Since voltage is inversely proportional to capacitance, this method tends to equalise the potential distribution across the units in the string. This method has the disadvantage that a large number of different-sized insulators are required. However, good results can be obtained by using standard insulators for most of the string and larger units for that near to the line conductor.\n• By using a guard ring (Static Shielding): The potential across each unit in a string can be equalised by using a guard ring which is a metal ring electrically connected to the conductor and surrounding the bottom insulator. The guard ring introduces capacitance between metal fittings and the line conductor. The guard ring is contoured in such a way that shunt capacitance currents i1, i2 etc. are equal to metal fitting line capacitance currents i′1, i′2 etc. The result is that same charging current I flows through each unit of string. Consequently, there will be uniform potential distribution across the units.\n\nSolution:\n\nDisc porosity is NOT one of the methods to improve string efficiency of insulators." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9253817,"math_prob":0.9391496,"size":2760,"snap":"2022-05-2022-21","text_gpt3_token_len":568,"char_repetition_ratio":0.14114659,"word_repetition_ratio":0.049217,"special_character_ratio":0.19456522,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9723859,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T20:22:44Z\",\"WARC-Record-ID\":\"<urn:uuid:5815b19f-e0d8-4c13-822f-61cb2207ea94>\",\"Content-Length\":\"123854\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02652072-8511-45f7-ab7b-d8355787d0fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a83b2f7-1426-49a7-a094-ffe6749d26bc>\",\"WARC-IP-Address\":\"104.22.45.238\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/which-of-the-following-is-not-one-of-the-methods-t--615dd1f8876baf0766602bd9\",\"WARC-Payload-Digest\":\"sha1:GIPFFPQWVU4DPLOLOFPKVVSDLKW7B5E4\",\"WARC-Block-Digest\":\"sha1:KJ74I3ONU36XNVNCQXZAA3P5EPPMI42Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301488.71_warc_CC-MAIN-20220119185232-20220119215232-00295.warc.gz\"}"}
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Time_Dependent_Quantum_Mechanics_and_Spectroscopy_(Tokmakoff)/02%3A_Introduction_to_Time-Dependent_Quantum_Mechanics/2.01%3A_Time-Evolution_with_a_Time-Independent_Hamiltonian	[ "# 2.1: Time-Evolution with a Time-Independent Hamiltonian\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\nThe time evolution of the state of a quantum system is described by the time-dependent Schrödinger equation (TDSE):\n\n$i \\hbar \\frac {\\partial} {\\partial t} \\psi ( \\overline {r} , t ) = \\hat {H} ( \\overline {r} , t ) \\psi ( \\overline {r} , t ) \\label{1.1}$\n\n$$\\hat{H}$$ is the Hamiltonian operator which describes all interactions between particles and fields, and determines the state of the system in time and space. $$\\hat{H}$$ is the sum of the kinetic and potential energy. For one particle under the influence of a potential\n\n$\\hat {H} = - \\frac {\\hbar^{2}} {2 m} \\hat {\\nabla}^{2} + \\hat {V} ( \\overline {r} , t ) \\label{1.2}$\n\nThe state of the system is expressed through the wavefunction $$\\psi ( \\overline {r} , t )$$. The wavefunction is complex and cannot be observed itself, but through it we obtain the probability density\n\n$P = \| \\psi ( \\overline {r} , t ) \|^{2},$\n\nwhich characterizes the spatial probability distribution for the particles described by $$\\hat{H}$$ at time $$t$$. Also, it is used to calculate the expectation value of an operator $$\\hat{A}$$\n\n\\begin{align} \\langle \\hat {A} (t) \\rangle &= \\int \\psi^{} ( \\overline {r} , t ) \\hat {A} \\psi ( \\overline {r} , t ) d \\overline {r} \\\\[4pt] &= \\langle \\psi (t) \| \\hat {A} \| \\psi (t) \\rangle \\label{1.3} \\end{align}\n\nPhysical observables must be real, and therefore will correspond to the expectation values of Hermitian operators ($$\\hat {A} = \\hat {A}^{\\dagger}$$).\n\nOur first exposure to time-dependence in quantum mechanics is often for the specific case in which the Hamiltonian $$\\hat{H}$$ is assumed to be independent of time: $$\\hat {H} = \\hat {H} ( \\overline {r} )$$. We then assume a solution with a form in which the spatial and temporal variables in the wavefunction are separable:\n\n$\\psi ( \\overline {r} , t ) = \\varphi ( \\overline {r} ) T (t) \\label{1.4}$\n\n$i \\hbar \\frac {1} {T (t)} \\frac {\\partial} {\\partial t} T (t) = \\frac {\\hat {H} ( \\overline {r} ) \\varphi ( \\overline {r} )} {\\varphi ( \\overline {r} )} \\label{1.5}$\n\nHere the left-hand side is a function only of time, and the right-hand side is a function of space only ($$\\overline {r}$$, or rather position and momentum). Equation \\ref{1.5} can only be satisfied if both sides are equal to the same constant, $$E$$. Taking the right-hand side we have\n\n$\\frac {\\hat {H} ( \\overline {r} ) \\varphi ( \\overline {r} )} {\\varphi ( \\overline {r} )} = E \\quad \\Rightarrow \\quad \\hat {H} ( \\overline {r} ) \\varphi ( \\overline {r} ) = E \\varphi ( \\overline {r} ) \\label{1.6}$\n\nThis is the Time-Independent Schrödinger Equation (TISE), an eigenvalue equation, for which $$\\varphi ( \\overline {r} )$$ are the eigenstates and $$E$$ are the eigenvalues. Here we note that\n\n$\\langle \\hat {H} \\rangle = \\langle \\psi \| \\hat {H} \| \\psi \\rangle = E,$\n\nso $$\\hat{H}$$ is the operator corresponding to $$E$$ and drawing on classical mechanics we associate $$\\hat{H}$$ with the expectation value of the energy of the system. Now taking the left-hand side of Equation \\ref{1.5} and integrating:\n\n\\begin{align} i \\hbar \\frac {1} {T (t)} \\frac {\\partial T} {\\partial t} &= E \\\\[4pt] \\left( \\frac {\\partial} {\\partial t} + \\frac {i E} {\\hbar} \\right) T (t) &= 0 \\label{1.7} \\end{align}\n\nwhich has solutions like this:\n\n$T (t) = \\exp ( - i E t / \\hbar ) \\label{1.8}$\n\nSo, in the case of a bound potential we will have a discrete set of eigenfunctions $$\\varphi _ {n} ( \\overline {r} )$$ with corresponding energy eigenvalues $$E_n$$ from the TISE, and there are a set of corresponding solutions to the TDSE.\n\n$\\psi _ {n} ( \\overline {r} , t ) = \\varphi _ {n} ( \\overline {r} ) \\underbrace{\\exp \\left( - i E _ {n} t / \\hbar \\right)}_{\\text{phase factor}} \\label{1.9}$\n\nPhase Factor\n\nFor any complex number written in polar form (such as $$re^{iθ}$$), the phase factor is the complex exponential factor ($$e^{iθ}$$). The phase factor does not have any physical meaning, since the introduction of a phase factor does not change the expectation values of a Hermitian operator. That is\n\n$\\langle \\phi \|A\|\\phi \\rangle = \\langle \\phi \|e^{-i\\theta}Ae^{i\\theta}\|\\phi \\rangle$\n\nSince the only time-dependence in $$\\psi _ {n}$$ is a phase factor, the probability density for an eigenstate is independent of time:\n\n$P = \\left\| \\psi _ {n} (t) \\right\|^{2} = \\text {constant}.$\n\nTherefore, the eigenstates $$\\varphi ( \\overline {r} )$$ do not change with time and are called stationary states.\n\nHowever, more generally, a system may exist as a linear combination of eigenstates:\n\n\\begin{align} \\psi ( \\overline {r} , t ) &= \\sum _ {n} c _ {n} \\psi _ {n} ( \\overline {r} , t ) \\\\[4pt] &= \\sum _ {n} c _ {n} e^{- i E _ {n} t h} \\varphi _ {n} ( \\overline {r} ) \\label{1.10} \\end{align}\n\nwhere $$c_n$$ are complex amplitudes, with\n\n$\\sum _ {n} \\left\| c _ {n} \\right\|^{2} = 1. \\nonumber$\n\nFor such a case, the probability density will oscillate with time. As an example, consider two eigenstates\n\n\\begin{align} \\psi ( \\overline {r} , t ) &= \\psi _ {1} + \\psi _ {2} \\nonumber \\\\[4pt] &= c _ {1} \\varphi _ {1} e^{- i E _ {1} t / h} + c _ {2} \\varphi _ {2} e^{- i E _ {2} t / h} \\label{1.11} \\end{align}\n\nFor this state the probability density oscillates in time as\n\n\\begin{align} P (t) & = \| \\psi \|^{2} \\nonumber \\\\[4pt] &= \\left\| \\psi _ {1} + \\psi _ {2} \\right\|^{2} \\nonumber \\\\[4pt] & = \\left\| c _ {1} \\varphi _ {1} \\right\|^{2} + \\left\| c _ {2} \\varphi _ {2} \\right\|^{2} + c _ {1}^{} c _ {2} \\varphi _ {1}^{} \\varphi _ {2} e^{- i \\left( \\alpha _ {2} - \\omega _ {1} \\right) t} + c _ {2}^{} c _ {1} \\varphi _ {2}^{*} \\varphi _ {1} e^{+ i \\left( a _ {2} - \\omega _ {1} \\right) t} \\nonumber \\\\[4pt] & = \\left\| \\psi _ {1} \\right\|^{2} + \\left\| \\psi _ {2} \\right\|^{2} + 2 \\left\| \\psi _ {1} \\psi _ {2} \\right\| \\cos \\left( \\omega _ {2} - \\omega _ {1} \\right) t \\label{1.12} \\end{align}\n\nwhere $$\\omega _ {n} = E _ {n} / \\hbar$$. We refer to this state of the system that gives rise to this time-dependent oscillation in probability density as a coherent superposition state, or coherence. More generally, the oscillation term in Equation \\ref{1.12} may also include a time-independent phase factor $$\\phi$$ that arises from the complex expansion coefficients.\n\nAs an example, consider the superposition of the ground and first excited states of the quantum harmonic oscillator. The basis wavefunctions, $$\\psi _ {0} (x)$$ and $$\\psi _ {1} (x)$$, and their stationary probability densities $$P _ {i} = \\left\\langle \\psi _ {i} (x) \| \\psi _ {i} (x) \\right\\rangle$$ are", null, "If we create a superposition of these states with Equation \\ref{1.11}, the time-dependent probability density oscillates, with $$\\langle x (t) \\rangle$$ bearing similarity to the classical motion. (Here $$c_0 = 0.5$$ and $$c_1 = 0.87$$.)", null, "" ]	[ null, "https://chem.libretexts.org/@api/deki/files/142104/Figure_1.png", null, "https://chem.libretexts.org/@api/deki/files/142105/Figure_2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71384543,"math_prob":0.9999523,"size":7110,"snap":"2022-27-2022-33","text_gpt3_token_len":2391,"char_repetition_ratio":0.1636645,"word_repetition_ratio":0.107344635,"special_character_ratio":0.3724332,"punctuation_ratio":0.09931766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999297,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T07:07:01Z\",\"WARC-Record-ID\":\"<urn:uuid:8af6b7c3-bc80-41c2-b6d1-96a342f3305b>\",\"Content-Length\":\"112167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3ca1314-6b6f-4edc-8aed-fe16c352d40d>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0af0356-5ab7-4ff3-a0bc-4b426bb2772a>\",\"WARC-IP-Address\":\"18.67.76.102\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Time_Dependent_Quantum_Mechanics_and_Spectroscopy_(Tokmakoff)/02%3A_Introduction_to_Time-Dependent_Quantum_Mechanics/2.01%3A_Time-Evolution_with_a_Time-Independent_Hamiltonian\",\"WARC-Payload-Digest\":\"sha1:TX73JVQKBZ74RKD7TDL6XNMY3DSY4HFM\",\"WARC-Block-Digest\":\"sha1:D62D5OVA7UFQ2V2GNSSKC5ILHJ6G5TUU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103034877.9_warc_CC-MAIN-20220625065404-20220625095404-00796.warc.gz\"}"}