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3,913,005 |
<p>So say there is 26 possible characters and you've got 1800 long random string. Now how would you go about finding the chance of there being a specific 5 letter word within?</p>
<p>I already found a related question to this but the given formula in the answer doesn't seem to make sense: <a href="https://math.stackexchange.com/questions/815741/probability-of-a-four-letter-word-from-a-sequence-of-n-random-letters">Probability of a four letter word from a sequence of n random letters</a></p>
<p>According to the formula in the accepted answer, the denominator would get really big 26^1800 with 1800 length string and mean very little probability. But shouldn't probability increase when the random string length is increased instead of decrease?</p>
|
user2661923
| 464,411 |
<p>This is a long-winded comment that is being posted as an answer only for legibility.</p>
<p>The only approach that I am aware of is <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">Inclusion-Exclusion</a> which results in very ugly math.</p>
<p>I will make the simplying assumption that the word is composed of 5 distinct characters. My approach will be</p>
<p><span class="math-container">$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$</span></p>
<p>where</p>
<p><span class="math-container">$$D = (26)^{(1800)}.$$</span></p>
<p>There are <span class="math-container">$(1801 - 5) = 1796$</span> slots that the word can begin in. For each slot, there are <span class="math-container">$(1800 - 5) = 1795$</span> unconstrained letters, which can be chosen in <span class="math-container">$(26)^{(1795)}$</span> ways.</p>
<p>Therefore, I will set</p>
<p><span class="math-container">$$N_1 = (1801 - 5) \times (26)^{(1800 - 5)}.$$</span></p>
<p>The eventual answer will be something like</p>
<p><span class="math-container">$$N = N_1 - N_2 + N_3 - N_4 + \cdots.$$</span></p>
<p>To compute <span class="math-container">$N_2$</span>, it would be <strong>wrong</strong> to presume that the computation is</p>
<p><span class="math-container">$$N_2 = (1801 - [1 \times 5]) \times (1801 - [2 \times 5]) \times (26)^{(1800 - [2 \times 5])}.$$</span></p>
<p>The reason that the above approach is wrong is that it presumes that regardless of the placement of one of the words, there will automatically be 1791 possible locations for the start of the 2nd word.</p>
<p>If the first word happens to start in position 1 or position 1796, then yes there will be 1791 possible locations for the 2nd word. However, as the first letter of the first word changes location from 1 to 2 to 3 to 4 to 5, the possible locations for the start of the second word decrease by 1.</p>
<p>For example, if the first word begins on position 5, then the second word can not begin before position 10. Note that by the simplifying assumption that the 5 letters in the word are distinct, the two words <strong>can not overlap</strong>.</p>
<p>If the first letter of the first word is located anywhere between positions 5 and 1792 inclusive, (1788 positions) then there will be (1796 - 9) = 1787 possible starting points for the second word.</p>
<p>If the first word starts in position <span class="math-container">$i$</span>, for <span class="math-container">$i \in \{1,2,3,4\}$</span> then there will be <span class="math-container">$(1792 - i)$</span> possible starting positions for the second word. By symmetry, similar considerations apply if the first word starts on or after position 1793.</p>
<p>Let</p>
<p><span class="math-container">$$T_2 = \left[2 \times (1788 + 1789 + 1790 + 1791)\right] ~+~
\left[1788 \times 1787 \right].$$</span></p>
<p>Then the correct computation is</p>
<p><span class="math-container">$$N_2 = T_2 \times (26)^{(1800 - [2 \times 5])}.$$</span></p>
<p>The math for <span class="math-container">$N_3$</span> will get <strong>uglier still</strong> than the math for <span class="math-container">$N_2$</span>. This is because in addition to being concerned about the starting position of the leftmost word and the rightmost word,
you also have to be concerned about the <strong>gap between</strong> two of the words. This will also affect how many positions the third word can start in.</p>
<p>Similarly when considering <span class="math-container">$N_4, N_5, \cdots$</span> you have to be concerned about <strong>each</strong> of the gaps between words.</p>
<p>Told you the math was ugly.</p>
<p>If somebody has a better way of doing this that in no way involves a computer algorithm, I'd like to know about it.</p>
|
398,371 |
<p>How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get</p>
<p>$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?</p>
|
egreg
| 62,967 |
<p>I'll show you a trick: since $t\to 1^+$ you know that $\pi t\to \pi^+$, so $(\pi t -\pi)\to 0^+$.</p>
<p>First of all do the change of variable $u=\pi t-\pi$, so you get $\pi t=u-\pi$. Then $\sin(\pi t)=\sin(u-\pi)=-\sin u$ and $\cos(\pi t)=\cos(u-\pi)=-\cos u$. Thus your limit becomes</p>
<p>$$
\lim_{u\to 0^+}\frac{-\sin u}{\sqrt{1-\cos u}}=
-\lim_{u\to 0^+}\sqrt{\frac{\sin^2 u}{1-\cos u}}=
-\lim_{u\to 0^+}\sqrt{1+\cos u}=-\sqrt{2}
$$</p>
<p>The first equality is justified because $\sin u>0$ for $0<u<\pi$ (and you're interested in a right neighborhood of $0$).</p>
<p>Limits at zero are "psychologically" better, aren't they? There's really no difference with doing the limit at $\pi$ or using $\pi t$, but the region around $0$ is better known.</p>
|
878,686 |
<p>How do I derive the $m$ in the formula:
$$I=\left(1+\frac{r}{m}\right)^{mn} -1$$</p>
<p>all the values of the variables in the formula except $m$ is given and the question is find $m$.
I just don't know how to derive the formula using the knowledge of Algebra I have.</p>
|
Community
| -1 |
<p>Using the answer by David, we set $t=\dfrac rm$, and</p>
<p>$$\sqrt[nr]{1+I}=(1+t)^{1/t}=s,$$
or
$$1+t=s^t=e^{t\ln(s)}.$$</p>
<p>Then
$$(1+t)s=e^{(1+t)\ln(s)},$$
$$-(1+t)\ln(s)\,e^{-(1+t)\ln(s)}=-\frac{\ln(s)}s,$$
and
$$-(1+t)\ln(s)=W(-\frac{\ln(s)}s).$$
Hence,
$$m=\frac r{-\dfrac{W(-\frac{\ln(s)}s)}{\ln(s)}-1}.$$</p>
|
64,096 |
<p>I'm trying to define a function that is evaluated differently depending on its
options. This code</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b_, OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b_, OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>works as expected. Evaluating </p>
<pre><code>{Foo[x,y,Bar->True],
Foo[x,y,Bar->False],
SetOptions[Foo,Bar->True];
Foo[x,y,Bar->True],
Foo[x,y,Bar->False],
SetOptions[Foo,Bar->False];
Foo[x,y,{Bar->True}],
Foo[x,y,{Bar->False}],
SetOptions[Foo,Bar->True];
Foo[x,y],
SetOptions[Foo,Bar->False];
Foo[x,y]}
</code></pre>
<p>I get </p>
<pre><code>{A, B, A, B, A, B, A, B}
</code></pre>
<p>which is precisely what I want. However, if I try to make <code>b</code> a blank sequence, i.e.</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__, OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b__, OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>weird things start happening. Evaluating my second code sample I get</p>
<pre><code>{A, B, A, A, A, B, A, B}
</code></pre>
<p>which is obviously <strong>wrong</strong>. A possible workaround is to write</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__/;FreeQ[{b},Rule], OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b__/;FreeQ[{b},Rule], OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>However this doesn't seem right to me, since it's actually the job of OptionsPattern to fish out the options from my functions. Of course
I could also write</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__, OptionsPattern[]]:= If[OptionValue[Bar],A,B]
</code></pre>
<p>which works as well, but this is still another workaround. Moreover, if I want
to have something like (in addition to the regular function definition)</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,a_,c___, OptionsPattern[]]:= 0 /; OptionValue[Bar];
</code></pre>
<p>then running</p>
<pre><code>{Foo[x,x,Bar->True],
Foo[x,x,Bar->False],
SetOptions[Foo,Bar->True];
Foo[x,x,Bar->True],
Foo[x,x,Bar->False],
SetOptions[Foo,Bar->False];
Foo[x,x,{Bar->True}],
Foo[x,x,{Bar->False}],
SetOptions[Foo,Bar->True];
Foo[x,x],
SetOptions[Foo,Bar->False];
Foo[x,x]}
</code></pre>
<p>again produces wrong results</p>
<pre><code>{0, Foo[x, x, Bar -> False], 0, 0, 0, Foo[x, x, {Bar -> False}], 0,
Foo[x, x]}
</code></pre>
<p>and my only solution is to use </p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,a_,c___/;FreeQ[{c},Rule], OptionsPattern[]]:= 0 /; OptionValue[Bar];
</code></pre>
<p>which is kind of a hack.</p>
<p><strong>So my question is, why using conditions with OptionValue and OptionsPattern
works if the last pattern before OptionsPattern is a blank, but fails if it is
a blank sequence. Is it a bug, or am I missing something essential about OptionsPattern?</strong></p>
<p>I'm observing this behavior both on Mathematca 9.0.1 and Mathematica 10.0.1 under Linux.</p>
|
rcollyer
| 52 |
<p><strong>Note</strong>: This is an incomplete analysis and leads to the wrong conclusion about the cause of the difficulty. Mr.W's <a href="https://mathematica.stackexchange.com/a/64193/52">answer</a> below correctly identifies the culprit as <code>Condition</code>.</p>
<hr>
<p>The problem you are facing has nothing to do with <code>OptionValue</code>, <code>OptionsPattern</code>, or <code>Condition</code>. It is simply because <code>b__</code> is under specified and <code>SlotSequence</code> is greedy. Effectively, you have specified</p>
<pre><code>Foo[a_, b__, c___]
</code></pre>
<p>so that <code>b__</code> will pick up <em>everything</em>, including the options, because it hasn't been told not to. The simplest fix is to use <code>Except</code>, e.g.</p>
<pre><code>Foo[a_, b:Except[_?OptionQ].., OptionsPattern[]]
</code></pre>
<p>and similarly,</p>
<pre><code>Foo[a_, b_, c:Except[_?OptionQ]..., OptionsPattern[]]
</code></pre>
<p>Note the use of <code>...</code> in the second one.</p>
|
64,096 |
<p>I'm trying to define a function that is evaluated differently depending on its
options. This code</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b_, OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b_, OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>works as expected. Evaluating </p>
<pre><code>{Foo[x,y,Bar->True],
Foo[x,y,Bar->False],
SetOptions[Foo,Bar->True];
Foo[x,y,Bar->True],
Foo[x,y,Bar->False],
SetOptions[Foo,Bar->False];
Foo[x,y,{Bar->True}],
Foo[x,y,{Bar->False}],
SetOptions[Foo,Bar->True];
Foo[x,y],
SetOptions[Foo,Bar->False];
Foo[x,y]}
</code></pre>
<p>I get </p>
<pre><code>{A, B, A, B, A, B, A, B}
</code></pre>
<p>which is precisely what I want. However, if I try to make <code>b</code> a blank sequence, i.e.</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__, OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b__, OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>weird things start happening. Evaluating my second code sample I get</p>
<pre><code>{A, B, A, A, A, B, A, B}
</code></pre>
<p>which is obviously <strong>wrong</strong>. A possible workaround is to write</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__/;FreeQ[{b},Rule], OptionsPattern[]]:= A /; OptionValue[Bar];
Foo[a_,b__/;FreeQ[{b},Rule], OptionsPattern[]]:= B /; !OptionValue[Bar];
</code></pre>
<p>However this doesn't seem right to me, since it's actually the job of OptionsPattern to fish out the options from my functions. Of course
I could also write</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,b__, OptionsPattern[]]:= If[OptionValue[Bar],A,B]
</code></pre>
<p>which works as well, but this is still another workaround. Moreover, if I want
to have something like (in addition to the regular function definition)</p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,a_,c___, OptionsPattern[]]:= 0 /; OptionValue[Bar];
</code></pre>
<p>then running</p>
<pre><code>{Foo[x,x,Bar->True],
Foo[x,x,Bar->False],
SetOptions[Foo,Bar->True];
Foo[x,x,Bar->True],
Foo[x,x,Bar->False],
SetOptions[Foo,Bar->False];
Foo[x,x,{Bar->True}],
Foo[x,x,{Bar->False}],
SetOptions[Foo,Bar->True];
Foo[x,x],
SetOptions[Foo,Bar->False];
Foo[x,x]}
</code></pre>
<p>again produces wrong results</p>
<pre><code>{0, Foo[x, x, Bar -> False], 0, 0, 0, Foo[x, x, {Bar -> False}], 0,
Foo[x, x]}
</code></pre>
<p>and my only solution is to use </p>
<pre><code>Options[Foo]={Bar->True};
Foo[a_,a_,c___/;FreeQ[{c},Rule], OptionsPattern[]]:= 0 /; OptionValue[Bar];
</code></pre>
<p>which is kind of a hack.</p>
<p><strong>So my question is, why using conditions with OptionValue and OptionsPattern
works if the last pattern before OptionsPattern is a blank, but fails if it is
a blank sequence. Is it a bug, or am I missing something essential about OptionsPattern?</strong></p>
<p>I'm observing this behavior both on Mathematca 9.0.1 and Mathematica 10.0.1 under Linux.</p>
|
Mr.Wizard
| 121 |
<p>Alright, I took another look at this issue and I do <em>not</em> believe this is a duplicate of:</p>
<ul>
<li><a href="https://mathematica.stackexchange.com/q/1567/121">How can I create a function with "positional" or "named" optional arguments?</a></li>
</ul>
<p>However I also do not believe that rcollyer's analysis is <em>entirely</em> correct. Please consider this example:</p>
<pre><code>ClearAll[Foo]
Options[Foo] = {Bar -> True};
Foo[a_, b__, OptionsPattern[]] := If[OptionValue[Bar], A, B]
{Foo[x, y, Bar -> True], Foo[x, y, Bar -> False], SetOptions[Foo, Bar -> True];
Foo[x, y, Bar -> True], Foo[x, y, Bar -> False], SetOptions[Foo, Bar -> False];
Foo[x, y, {Bar -> True}], Foo[x, y, {Bar -> False}], SetOptions[Foo, Bar -> True];
Foo[x, y], SetOptions[Foo, Bar -> False];
Foo[x, y]}
</code></pre>
<blockquote>
<pre><code>{A, B, A, B, A, B, A, B}
</code></pre>
</blockquote>
<p>Observe that <code>OptionValue[Bar]</code> resolves correctly to <code>True</code> or <code>False</code> in each case. One can use a more verbose RHS definition to show that every <code>a_</code> and <code>b__</code> match is correct and that <code>b__</code> <em>does not include the option</em> as rcollyer stated:</p>
<pre><code>ClearAll[Foo]
Options[Foo] = {Bar -> True};
Foo[a_, b__, OptionsPattern[]] := {{a}, {b}, OptionValue[Bar]}
{Foo[x, y, Bar -> True], Foo[x, y, Bar -> False], SetOptions[Foo, Bar -> True];
Foo[x, y, Bar -> True], Foo[x, y, Bar -> False], SetOptions[Foo, Bar -> False];
Foo[x, y, {Bar -> True}], Foo[x, y, {Bar -> False}], SetOptions[Foo, Bar -> True];
Foo[x, y], SetOptions[Foo, Bar -> False];
Foo[x, y]} // MatrixForm
</code></pre>
<blockquote>
<p>$\left(
\begin{array}{ccc}
\{x\} & \{y\} & \text{True} \\
\{x\} & \{y\} & \text{False} \\
\{x\} & \{y\} & \text{True} \\
\{x\} & \{y\} & \text{False} \\
\{x\} & \{y\} & \text{True} \\
\{x\} & \{y\} & \text{False} \\
\{x\} & \{y\} & \text{True} \\
\{x\} & \{y\} & \text{False} \\
\end{array}
\right)$</p>
</blockquote>
<p>Rather I believe the problem you experienced is due the behavior when a <a href="http://reference.wolfram.com/language/ref/Condition.html" rel="nofollow noreferrer"><code>Condition</code></a> fails. One can see that more than one possible match is checked in this example:</p>
<pre><code>ClearAll[Foo]
Options[Foo] = {Bar -> True};
Foo[a_, b__, OptionsPattern[]] := Null /; Print[{{a}, {b}, OptionValue[Bar]}]
Foo[x, y, Bar -> True];
Foo[x, y, {Bar -> False}];
</code></pre>
<blockquote>
<p>{{x},{y},True}</p>
<p>{{x},{y,Bar->True},True}</p>
<p>{{x},{y},False}</p>
<p>{{x},{y,{Bar->False}},True}</p>
</blockquote>
<p>Note that each line results in two different alignments being checked: first the correct one, then an <em>incorrect</em> one. Blocking the incorrect one is how rcollyer's method works, but it is not because <code>b__</code> is "greedy" but rather because the first, correct alignment did not pass the <code>Condition</code> and another <em>possible</em> alignment is sought. In the second case above the alignment <code>{{x},{y,{Bar->False}},True}</code> is the source of error. (This may be a semantic dispute but I think it is an important one.)</p>
<p>Although I usually favor separate definitions in this case I think using <code>If</code> is a more direct solution without unnecessary additional argument testing. </p>
|
3,140,194 |
<p>how to solve a system equation with radical</p>
<p><span class="math-container">$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$</span></p>
<p>And <span class="math-container">$$\sqrt{x+y}+\sqrt{x}=x+3$$</span></p>
<p>This system has <span class="math-container">$1$</span> root is <span class="math-container">$x=1;y=8$</span>,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.</p>
|
TheSilverDoe
| 594,484 |
<p><span class="math-container">$f(x)^{g(x)} = \exp(g(x) \ln(f(x))$</span>, so if <span class="math-container">$f(x)$</span> has a positive finite limit <span class="math-container">$a$</span>, and <span class="math-container">$g(x)$</span> has a finite limit <span class="math-container">$b$</span>, then by continuity of <span class="math-container">$\ln$</span> and <span class="math-container">$\exp$</span>, yes, <span class="math-container">$f(x)^{g(x)}$</span> tends to <span class="math-container">$a^{b}$</span></p>
|
3,140,194 |
<p>how to solve a system equation with radical</p>
<p><span class="math-container">$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$</span></p>
<p>And <span class="math-container">$$\sqrt{x+y}+\sqrt{x}=x+3$$</span></p>
<p>This system has <span class="math-container">$1$</span> root is <span class="math-container">$x=1;y=8$</span>,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.</p>
|
Hagen von Eitzen
| 39,174 |
<p>Not always.</p>
<p>We have <span class="math-container">$0^0=1$</span>, but <span class="math-container">$\lim_{x\to x_0} f(x)=\lim_{x\to x-0} g(x)=0$</span> does not imply <span class="math-container">$\lim_{x\to x_0}f(x)^{g(x)}=1$</span>. The problem is that the exponentiation map <span class="math-container">$(x,y)\mapsto x^y$</span> behaves somewhat weird on its maximal domain of definition. In particular, while <em>defined</em> for <span class="math-container">$(x,y)=(0,0)$</span> and a couple of other exceptional points, exponentiation is not <em>continuous</em> there (or even defined in a neighbourhood of <span class="math-container">$(0,0)$</span>). On the other hand, for <span class="math-container">$a>0$</span> and <span class="math-container">$b\in \Bbb R$</span> arbitrary, exponentiation is defined and continuous in a neighbourhood of <span class="math-container">$(a,b)$</span>. Therefore, in these cases, <span class="math-container">$f(x)\to a$</span> and <span class="math-container">$f(b)\to b$</span> implies <span class="math-container">$f(x)^{g(x)}\to a^b$</span>.</p>
|
2,775,087 |
<blockquote>
<p>Without using a calculator, what is the sum of digits of the numbers from $1$ to $10^n$?</p>
</blockquote>
<p>Now, I'm familiar with the idea of pairing the numbers as follows:</p>
<p>$$\langle 0, 10^n -1 \rangle,\, \langle 1, 10^n - 2\rangle , \dotsc$$</p>
<p>The sum of digits of each pair is $9n$. How can I prove this property retains for all pairs?</p>
|
deficiencyOn
| 532,135 |
<p>Okay so the truly simple explanation is merely the fact that all those pairs sum to $999$ (i.e. in the case 0f $n=3$) . There's nothing more than that. In more details, one can see that this property is true for the first pair. Inductively, the next pair is made by increasing the first item by $1$ and decreasing the second by $1$, and so maintaining this property.</p>
|
2,775,087 |
<blockquote>
<p>Without using a calculator, what is the sum of digits of the numbers from $1$ to $10^n$?</p>
</blockquote>
<p>Now, I'm familiar with the idea of pairing the numbers as follows:</p>
<p>$$\langle 0, 10^n -1 \rangle,\, \langle 1, 10^n - 2\rangle , \dotsc$$</p>
<p>The sum of digits of each pair is $9n$. How can I prove this property retains for all pairs?</p>
|
Love Invariants
| 551,019 |
<p>There is a constant pattern of nos. I will demonstrate it for 2 digit nos.<br>
00 01 02 03 04 05 06 07 08 09<br>
10 11 12 13 14 15 16 17 18 19<br>
20 21 22 23 24 25 26 27 28 29 and it goes so on. </p>
<p>You can make a pattern of numbers for any digit nos.<br>
Just add $0s$ before a number to make it of n-digit. Now due to this symmetrical pattern, all nos. 0,1,2,3,4. . . will be in equal amount. </p>
|
368,800 |
<p>Let <span class="math-container">$G$</span> be a finite group, let <span class="math-container">$X$</span> be a locally compact Hausdorff space, and let <span class="math-container">$G$</span> act freely on <span class="math-container">$X$</span>. It is well-known that the canonical quotient map <span class="math-container">$\pi\colon X\to X/G$</span> onto the orbit space <span class="math-container">$X/G$</span> admits local cross-sections. More precisely, for every <span class="math-container">$z\in X/G$</span> there are an open set <span class="math-container">$U$</span> in <span class="math-container">$X/G$</span> containing <span class="math-container">$z$</span>, and a continuous function <span class="math-container">$s\colon U\to X$</span> such that <span class="math-container">$\pi\circ s$</span> is the identity on <span class="math-container">$U$</span>. In particular, there is an open cover of <span class="math-container">$X/G$</span> consisting of sets where a local cross-section can be defined.</p>
<p><strong>Question</strong>: is there a <em>finite</em> open cover of <span class="math-container">$X/G$</span> consisting of sets where a local cross-section can be defined?</p>
<p>(This is the same as asking whether the Schwarz genus of the fiber map <span class="math-container">$X\to X/G$</span> is finite.)</p>
<p>The answer is "yes" if <span class="math-container">$X$</span> (or at least <span class="math-container">$X/G$</span>) is finitistic, so in particular whenever <span class="math-container">$X$</span> has finite covering dimension, and clearly also whenever <span class="math-container">$X$</span> is compact. I wonder if it is true in general.</p>
|
Hannes Thiel
| 24,916 |
<p>Let <span class="math-container">$X=[-1,1]^\infty\setminus\{0\}$</span>, which is a metrizable, locally compact space. Consider the two-element group <span class="math-container">$G$</span>, and the free <span class="math-container">$G$</span>-action on <span class="math-container">$X$</span> given by <span class="math-container">$(x_j)_{j=1}^\infty\mapsto (-x_j)_{j=1}^\infty$</span>. We show that the fibration <span class="math-container">$X\to X/G$</span> has infinite Schwarz genus.</p>
<p>Consider the <span class="math-container">$n$</span>-sphere <span class="math-container">$S^n$</span> with the antipodal <span class="math-container">$G$</span>-action. Then <span class="math-container">$S^n$</span> can be embedded equivariantly into <span class="math-container">$X$</span> for all <span class="math-container">$n$</span>. (Use an equivariant map <span class="math-container">$S^n\to [-1,1]^{n+1}\setminus\{0\}$</span>.) By the Lusternik–Schnirelmann theorem (a strengthening of the Borsuk-Ulam theorem), <span class="math-container">$S^n$</span> cannot be covered by <span class="math-container">$n+1$</span> closed sets that do not contain antipodal points. It follows that the Schwarz genus of <span class="math-container">$S^n\to S^n/G$</span> is at least <span class="math-container">$n+2$</span>. Since the Schwarz genus of <span class="math-container">$X\to X/G$</span> is an upper bound for the Schwarz genus of <span class="math-container">$S^n\to S^n/G$</span>, it follows that <span class="math-container">$X\to X/G$</span> has infinite Schwarz genus.</p>
|
1,752,506 |
<p>Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$</p>
<p>My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$</p>
<p>$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.</p>
<p>Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: <a href="https://math.stackexchange.com/questions/1751410/how-to-square-both-the-sides-of-an-equation#comment3574205_1751435">Why one should never divide by an expression that contains a variable.</a></p>
<p>So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?</p>
|
Merkh
| 141,708 |
<p>Yes it is correct. It would do harm dividing out $\sqrt{x^2 + 9}.$ Perhaps an easier example of why this is so... Consider
\begin{equation}
x^2 = x.
\end{equation}
Obviously the two roots are 0,1, but do you see what happens when you divide by $x?$ You reduce the order of the polynomial, hence throwing away a root. </p>
|
2,482,564 |
<p>Let $(\Omega,\Sigma)$ be a given measurable space and let $f$ be a $\Sigma$-measurable function. If $h:[-\infty,\infty]\rightarrow[-\infty,\infty]$ is a continuous function, then the composite function $hf$ is measurable.</p>
<p>This can be proven easily if every continuous function is measurable as I just need to show that the inverse image of $hf$ is measurable for a given measurable set say $E$.</p>
<p>i.e. $(hf)^{-1}(E)=f^{-1}(h^{-1}(E))\in\Sigma$</p>
<p>But is a continuous function measurable? Or this can be proven in another way?</p>
|
Gono
| 384,471 |
<p>Usually with "measurable" on the real line "Borel measurable" is meant... if you do also every continuous function is ofc measurable because the preimage of open sets under continuous functions are open sets as well and open sets are a generator of the Borel Algebra</p>
|
11,922 |
<p>Which is correct terminology: "A Cartesian plane" or "The Cartesian plane"? (As in the directions for a section of homework being, "Plot a point on ______ Cartesian plane." In that context, I feel that one of the two should be used consistently, but find there's little agreement on Google for which is appropriate.</p>
|
Gerald Edgar
| 127 |
<p>Why do you think they should be used consistently? I think there are situations appropriate for each of these choices.</p>
|
653,449 |
<p>According to <a href="http://en.wikibooks.org/wiki/Haskell/Category_theory" rel="noreferrer">the Haskell wikibook on Category Theory</a>, the category below is not a valid category due to the addition of the morphism <em>h</em>. The hint says to " think about associativity of the composition operation." But I don't don't see why it fails.</p>
<p><span class="math-container">$$
f \circ (g \circ h) = (f \circ g) \circ h\\
f \circ (\mathit{id}_B) = (\mathit{id}_A) \circ h\\
$$</span></p>
<p>Does this then reduce to <span class="math-container">$f = h$</span> ?</p>
<p>And is that not true because <em>f</em> and <em>h</em>, despite both being from B to A, are not equivalent?</p>
<p><a href="https://i.stack.imgur.com/gllAK.png" rel="noreferrer"><img src="https://i.stack.imgur.com/gllAK.png" alt="" /></a><br />
<sub>(source: <a href="https://upload.wikimedia.org/wikibooks/en/6/65/Not-a-cat.png" rel="noreferrer">wikimedia.org</a>)</sub></p>
|
Martin Brandenburg
| 1,650 |
<p>From the graph we see that $f$ <em>and</em> $h$ are inverse to $g$. Now it is a general fact about categories that inverses of morphisms are unique. The proof is the same as the one for groups.</p>
|
2,066,501 |
<p>If Q = \begin{bmatrix}1&1/√6&1/3√3\\1&-1/√6&-1/3√3\\0 & √2/√3 & -1/3√3\end{bmatrix}</p>
<p>and R = \begin{bmatrix}2√2&1/√2&1/√2\\0&4/√6&1/√6\\0&0&1/3√3\end{bmatrix}.</p>
<p>Is A invertible? (no computation required) Is the system Ax=b solvable for each b in $R^3$ (give the formula for its solutions).</p>
|
Dave
| 334,366 |
<p>Since $A=QR$ we have: $\det(A)=\det(QR)=\det(Q)\det(R)$</p>
<p>Since $Q$ is orthogonal (i.e. $Q^{-1}=Q^T$), we have: $1=\det(I)=\det(QQ^T)=\det(Q)\det(Q^T)=(\det(Q))^2$ so $\det(Q)=\pm 1\neq 0$.</p>
<p>Since $R$ is upper triangular, and has all nonzero elements on its diagonal, $\det(R)\neq 0$.</p>
<p>Thus $\det(A)\neq 0\implies A$ is invertible $\implies Ax=b$ has a unique solution for every $b\in\Bbb R^3$. I'm not sure what you mean by a formula for the solution to $Ax=b$. You could find $x$ as $x=A^{-1}b=(QR)^{-1}b=R^{-1}Q^Tb$ since $A$ is invertible, but I'm not sure if this is the "formula" you're looking for.</p>
|
662,744 |
<h2>Question Statement</h2>
<blockquote>
<p>Let $K$ be a finite field of $q$ elements. Let $U$, $V$ be vector spaces over $K$ with
$\dim(U) = k$, $\dim(V) = l$. How many linear maps $U \rightarrow V$ are there?</p>
</blockquote>
<p>I'm struggling to answer this question. Given that a linear map is uniquely determined by its action on a basis we may as well consider the mappings of the standard basis. Even with this restriction, each of the $k$ basis vectors can map to not only $q$ distinct elements but also into $l$ different positions (or none, or into multiple positions).</p>
<p>What easy thing am I missing?</p>
|
Sugata Adhya
| 36,242 |
<p><strong>Hint:</strong> The space of all linear maps from $U\to V$ is isomorphic to $\text{Mat}_{l\times k}(K).$</p>
|
410,663 |
<p>Can anyone in detail explain the procedure for finding the harmonic conjugate of the function $$u(x,y) = x^{2} - y\cdot (y+1)$$</p>
<p>I am new to this and I would like to know. </p>
|
ˈjuː.zɚ79365
| 79,365 |
<p>Following Peter Tamaroff's solution (integration of $v_y=u_x=2x$ and $v_x=-u_y=2y+1$) one obtains $v(x,y)=2xy+f(x)$ and $v(x,y)=2xy+x+g(y)$, from where $v(x,y)=2xy+x+c$ where $c$ is a constant. This is the general form of harmonic conjugate. </p>
<p>With some fluency in complex arithmetics, you can solve the problem with less effort by recognizing $x^2-y^2$ as $\operatorname{Re}(z^2)$ and $-y$ as $\operatorname{Re}(iz)$. Since $u=\operatorname{Re}(z^2+iz)$, we get a harmonic conjugate as $\operatorname{Im}(z^2+iz) = 2xy+x$, up to an additive constant. </p>
|
268,461 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a> </p>
</blockquote>
<blockquote>
<p>Let $f\colon\mathbb C \to \mathbb C$ be entire. Show that if
$|\operatorname{Im}f(z)|\geqslant |\operatorname{Re}f(z)|$ for all $z \in \mathbb C$, then $f$ is constant on $\mathbb C$. </p>
</blockquote>
<p>Can I answer this by considering the distance between $f(z)$ and $i$ like in this problem <a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a>?</p>
|
Robert Israel
| 8,508 |
<p>Every member of $K$ is either in $F$ or in $F^c$. If it's in $F$, it's covered by the open cover of $F$. If it's in $F^c$, it's covered by $F^c$. So in either case, it's covered. </p>
|
2,361,275 |
<p>$$\neg(A\land B)\land(B\lor\neg C)\Leftrightarrow(\neg A\land B)\lor(\neg B\land\neg C)$$
Checked their truth tables are the same, but can you show the steps how to transform one to the other?</p>
|
Peter
| 463,556 |
<p>We have:
$$\overline{AB} (B \vee \bar{C}) \Leftrightarrow (\bar{A} \vee \bar{B})(B \vee \bar{C}) \Leftrightarrow \bar{A}B \vee \bar{A} \bar{C} \vee \bar{B} B \vee \bar{B} \bar{C} \Leftrightarrow \bar{A}B \vee \bar{A} \bar{C} \vee \bar{B} \bar{C}$$
Now if $\bar{A} \bar{C}$ turns out to be true, either $\bar{A} B$ or $\bar{B} \bar{C}$ must be true, because $B$ is either true or false. That's why you can leave $\bar{A} \bar{C}$ out and you get your result:
$$ \overline{AB} (B \vee \bar{C}) \Leftrightarrow \bar{A} B \vee \bar{B} \bar{C}$$</p>
<p>You can also use some boolean algebra to eliminate $\bar{A} \bar{C}$:</p>
<p>$$ \bar{A}B \vee \bar{A} \bar{C} \vee \bar{B} \bar{C} \Leftrightarrow \bar{A}B \vee \bar{A} \bar{C} (B \vee \bar{B}) \vee \bar{B} \bar{C} \Leftrightarrow \bar{A} B \vee \bar{A} \bar{C} B \vee \bar{A} \bar{C} \bar{B} \vee \bar{B}\bar{C} \\ \Leftrightarrow \bar{A} B (1 \vee \bar{C}) \vee \bar{B} \bar{C}(\bar{A} \vee 1) \Leftrightarrow \bar{A} B \vee \bar{B} \bar{C}$$</p>
|
1,655,884 |
<blockquote>
<p>How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?</p>
</blockquote>
<p><strong>Attempt:</strong></p>
<p>This seems like a hard question, since I can't even think of one example to this. Mathematically we have,</p>
<p>$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$ </p>
<p>where we have to find all positive <em>integer</em> solutions $(x,y,z)$. </p>
<p>I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,</p>
<p>$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$ </p>
<p>then we are done.</p>
|
poetasis
| 546,655 |
<p>For your equation, solutions include
<span class="math-container">$$(1,1,1)\quad (1,2,2)\quad (1,3,3)\quad (1,3,3)\quad
(1,4,4)\quad (1,5,5)\quad \cdots $$</span>
Any triple with zero or one for one of <span class="math-container">$\space x,y\space $</span> will work. All others would probably require a brute force search.</p>
|
182,301 |
<p>I have asked a question <a href="https://mathematica.stackexchange.com/questions/182247/constructing-possibilities-of-a-generating-function">here</a>.
I want to reproduce the coefficients of a generating function of the form:
<span class="math-container">$$(1 + x)^2 (1 + x + x^2 + x^3+\cdots+x^n)^{n-1}$$</span>
It is important that <span class="math-container">$x_0$</span> and <span class="math-container">$x_n$</span> to be strictly 0 or 1 and <span class="math-container">$x_1$</span> to <span class="math-container">$x_2$</span> can be any number within 0 and n (nothing higher). Here are two examples:
For <span class="math-container">$n=3$</span> we have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
</code></pre>
<p>We can produce the coefficients as: </p>
<pre><code>n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 8}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>which gives: </p>
<pre><code>{1, 4, 8, 12, 14, 12, 8, 4, 1}
</code></pre>
<p>as desired. For <span class="math-container">$n=4$</span> one had to add extra constraint and change the <span class="math-container">$t$</span> range.
We have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
</code></pre>
<p>thus <span class="math-container">$t$</span> should be from 0 to 14,
so we have: </p>
<pre><code>n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 14}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.</p>
<p><strong>Note: I want the output <code>table</code> to be in a format so that I can see all of the possibilities of the sums.</strong>
For instance, for n=3, <code>table[[2]]</code> should give all of the possibilities such that the total is equal to 1.</p>
<pre><code>{{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}
</code></pre>
|
Carl Woll
| 45,431 |
<p><strong>Update</strong></p>
<p>An even faster method (I also modified the order of each possibility as requested in the comments):</p>
<pre><code>tups[n_] := Values @ GroupBy[
Tuples[Join[{Range[2]}, ConstantArray[Range[n+1], n-1], {Range[2]}] - 1],
Total
]
</code></pre>
<p>A comparison with the accepted answer:</p>
<pre><code>r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
</code></pre>
<blockquote>
<p>{0.000123, Null}</p>
<p>{0.001075, Null}</p>
<p>True</p>
</blockquote>
<p>Almost an order of magnitude faster for <span class="math-container">$n=3$</span>. For <span class="math-container">$n=7$</span>:</p>
<pre><code>r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
</code></pre>
<blockquote>
<p>{0.195056, Null}</p>
<p>{29.7257, Null}</p>
<p>True</p>
</blockquote>
<p><strong>Original method</strong></p>
<p>Here's a modification of @kglr's answer to your linked question:</p>
<pre><code>sums[n_]:= Last @ Reap[
Array[Sow[{##}, Plus[##]]&, Join[{2}, ConstantArray[n+1, n-1], {2}], 0],
_,
#2&
]
</code></pre>
<p>For <span class="math-container">$n=3$</span>:</p>
<pre><code>sums[3][[2]]
</code></pre>
<blockquote>
<p>{{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}</p>
</blockquote>
<p>And a couple checks:</p>
<pre><code>Length /@ sums[3]
Length /@ sums[4]
</code></pre>
<blockquote>
<p>{1, 4, 8, 12, 14, 12, 8, 4, 1}</p>
<p>{1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1}</p>
</blockquote>
|
627,575 |
<p>There is a whiskey made up of 64% corn, 32% rye, and 4% barley that was made by blending other whiskies together. I am trying to figure out if there is a chance the ratio of this whiskey could be the result of blending two, maybe three whiskies of different ratios.</p>
<p>The possible whiskies:</p>
<p>Whiskey A is 60% corn, 36% rye and 4% barley.</p>
<p>Whiskey B is 81% corn, 15% rye, and 4% barley. </p>
<p>Whiskey C is 75% corn, 21% rye, and 4% barley</p>
<p>I have a feeling there is a possibility because these whiskies all have 4% barley, but I can't figure out if the other percentages match up in any 1:2:3 ratio in the blend. Any help will be greatly appreciated. Thank you.</p>
|
MJD
| 25,554 |
<p><strong>Hint</strong> Consider the three base whiskies as three unit vectors over the three-dimensional vector space of corn, rye, and barley. Mixing whiskeys then corresponds to simple operations on vectors, and you should be able to express this mixing operation in terms of operations you have already studied. Then see if the vector $(64, 32, 4)$ can be expressed in terms of these operations on $A,B,$ and $C$.</p>
|
4,594,043 |
<p>For <span class="math-container">$|x|<1,$</span> we have
<span class="math-container">$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$</span></p>
<hr />
<p><span class="math-container">$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$</span></p>
<p><span class="math-container">$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$</span></p>
<hr />
<p><span class="math-container">$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$</span></p>
<hr />
<p>In a similar way, I dare guess that</p>
<p><span class="math-container">$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$</span></p>
<p>where <span class="math-container">$n$</span> is a non-negative <strong>real</strong> number.</p>
<p>Proof:
<span class="math-container">$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$</span>
Letting <span class="math-container">$y=-(k+1)\ln x $</span> transforms the last integral into a Gamma function as</p>
<p><span class="math-container">$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$</span></p>
<p>Now we can conclude that
<span class="math-container">$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$</span></p>
<p><strong>Can we</strong> evaluate <span class="math-container">$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$</span> without expanding <span class="math-container">$\ln (1-x)$</span>?</p>
<p>Your comments and alternative methods are highly appreciated?</p>
|
Leucippus
| 148,155 |
<p>Starting with the integral
<span class="math-container">$$ \int_{0}^{1} (1-x)^m \, x^{-t} \, dx = B(m+1, 1-t), $$</span>
a case of the Beta function, then
<span class="math-container">\begin{align}
\partial_{m} \int_{0}^{1} (1-x)^m \, x^{-t} \, dx &= \partial_{m} \, B(m+1, 1-t) \\
\int_{0}^{1} \ln(1-x) \, (1-x)^m \, x^{-t} \, dx &= B(m+1, 1-t) \, \left(\psi(m+1) - \psi(m-t+2) \right),
\end{align}</span>
where <span class="math-container">$\psi(x)$</span> is the digamma function. Setting <span class="math-container">$m=0$</span> leads to
<span class="math-container">$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \frac{\psi(1) - \psi(1-t)}{1-t}. $$</span><br />
Now using
<span class="math-container">$$ \psi(x+1) = - \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \, \zeta(k+1) \, x^{k} $$</span>
then
<span class="math-container">$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \sum_{k=1}^{\infty} (-1)^k \, \zeta(k+1) \, (1-t)^{k-1}. $$</span>
Using the operator <span class="math-container">$(-1)^n \, \partial_{t}^{n}$</span> on both sides of this last expression the following is obtained.
<span class="math-container">\begin{align}
I_{n} &= \int_{0}^{1} \ln(1-x) \, \ln^{n}(x) \, x^{-t} \, dx \\
&= \sum_{k=1}^{\infty} (-1)^{k+n} \, \zeta(k+1) \, \partial_{t}^{n} \, (1-t)^{k-1} \\
&= \sum_{k=1}^{\infty} (-1)^{k} \, \zeta(k+1) \, \frac{(k-1)!}{(k-n-1)!} \, (1-t)^{k-n-1} \\
&= \sum_{k=n}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n} \\
&= (-1)^{n+1} \, \zeta(n+2) \, n! + \sum_{k=n+1}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n}.
\end{align}</span>
Setting <span class="math-container">$t = 1$</span> gives the desired result
<span class="math-container">$$ \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx = (-1)^{n+1} \, n! \, \zeta(n+2). $$</span></p>
<p>or
<span class="math-container">\begin{align}
(-1)^{n+1} \, n! \, \zeta(n+2) &= \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx \\
&= (-1)^n \, \partial_{t}^{n} \, \left. \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx \right|_{t=1} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. \int_{0}^{1} (1-x)^{m} \, x^{-t} \, dx \right|_{t=1}^{m=0} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. B(m+1, 1-t) \right|_{t=1}^{m=0}.
\end{align}</span></p>
|
2,287,878 |
<p>Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)</p>
|
florence
| 343,842 |
<p>Note that the roots of $x^2+x+1$ are
$$\zeta_3 =e^{2\pi i/3} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$$
and
$$\zeta_3^2 =e^{4\pi i/3} = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$$
Both of these are roots of $P(x)$ with multiplicity $100$. Therefore,
$$\sum_{k=1}^{200}\frac{1}{1+x_k} = 100\left( \frac{1}{1+\zeta_3}+\frac{1}{1+\zeta_3^2}\right)$$
$$ = 100\left(\frac{1+\zeta_3+1+\zeta_3^2}{1+\zeta_3+\zeta_3^2+\zeta_3^3} \right) = 100$$</p>
<p>As for the polynomial division, suppose
$$P(x) = g(x)(x^2+x)+r(x)$$
where $r$ is linear. We have $P(0) = 1$, which implies that $r(0) = 1$. Further, $P(-1) = 1$, and so $r(-1)=1$. Consequently, $r(x)=1$ for all $x$, so $a_1=0$.</p>
|
3,757,972 |
<p>If <span class="math-container">$\frac{ab} {a+b} = y$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are greater than zero, why is <span class="math-container">$y$</span> always smaller than the smallest number substituted?</p>
<p>Say <span class="math-container">$a=2$</span> , <span class="math-container">$b=4$</span> (smallest number here is <span class="math-container">$2$</span>. Thus, the answer would be smaller than <span class="math-container">$2$</span>)</p>
<p><span class="math-container">$\frac{2\cdot4}{ 2+4} = 1.\bar 3$</span></p>
<p>I got this equation from physics. It's for getting total resistance and the miss told us to not waste time in mcq on it because the answer will always be smaller than the smallest number. But I can't explain to myself in words or by intuition why this happens. Any help??</p>
|
Peter
| 82,961 |
<p>We have <span class="math-container">$$\frac{ab}{a+b}<\frac{ab}{a}=b$$</span> and <span class="math-container">$$\frac{ab}{a+b}<\frac{ab}{b}=a$$</span> if <span class="math-container">$a,b>0$</span></p>
|
3,757,972 |
<p>If <span class="math-container">$\frac{ab} {a+b} = y$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are greater than zero, why is <span class="math-container">$y$</span> always smaller than the smallest number substituted?</p>
<p>Say <span class="math-container">$a=2$</span> , <span class="math-container">$b=4$</span> (smallest number here is <span class="math-container">$2$</span>. Thus, the answer would be smaller than <span class="math-container">$2$</span>)</p>
<p><span class="math-container">$\frac{2\cdot4}{ 2+4} = 1.\bar 3$</span></p>
<p>I got this equation from physics. It's for getting total resistance and the miss told us to not waste time in mcq on it because the answer will always be smaller than the smallest number. But I can't explain to myself in words or by intuition why this happens. Any help??</p>
|
J. W. Tanner
| 615,567 |
<p>If <span class="math-container">$a,b>0$</span> then <span class="math-container">$a<a+b$</span> and <span class="math-container">$\dfrac a{a+b}<1$</span> so <span class="math-container">$\dfrac {ab}{a+b}<b$</span>.</p>
<p>A similar argument shows that <span class="math-container">$\dfrac{ab}{a+b}<a$</span>.</p>
|
3,757,972 |
<p>If <span class="math-container">$\frac{ab} {a+b} = y$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are greater than zero, why is <span class="math-container">$y$</span> always smaller than the smallest number substituted?</p>
<p>Say <span class="math-container">$a=2$</span> , <span class="math-container">$b=4$</span> (smallest number here is <span class="math-container">$2$</span>. Thus, the answer would be smaller than <span class="math-container">$2$</span>)</p>
<p><span class="math-container">$\frac{2\cdot4}{ 2+4} = 1.\bar 3$</span></p>
<p>I got this equation from physics. It's for getting total resistance and the miss told us to not waste time in mcq on it because the answer will always be smaller than the smallest number. But I can't explain to myself in words or by intuition why this happens. Any help??</p>
|
DMcMor
| 155,622 |
<p>Well, so long as <span class="math-container">$a,b > 0$</span> it's certainly true that <span class="math-container">$$ab < aa + ab,$$</span> but that gives us <span class="math-container">$$ab < a(a+b) \implies \frac{ab}{a+b} < a.$$</span></p>
<p>Similarly, <span class="math-container">$$ab < ab + bb$$</span> which gives us <span class="math-container">$$ab < b(a+b) \implies \frac{ab}{a+b} < b.$$</span></p>
|
3,007,785 |
<p><span class="math-container">$\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$</span> I know how to count this limit with the help of l'Hopital rule. But it is very awful, because I need 3 times derivate it. So, there is very difficult calculations. I have the answer <span class="math-container">$\frac{2}{5}$</span>. </p>
<p>I want to know if there is other ways to calculate it, without 3 times using l'Hopital rule? (I could write my steps, but they are very big. I just took third derivative of numerator and denominator)</p>
|
Community
| -1 |
<p>There is a simple way to address the denominator:</p>
<p><span class="math-container">$$(x+\tan x-\sin2x)'=1+\tan^2x+1-2\cos 2x=\tan^2x+4\sin^2x=\sin^2x\left(\frac1{\cos^2x}+4\right).$$</span></p>
<p>The second factor will tend to <span class="math-container">$5$</span> and the first can be replaced by <span class="math-container">$x^2$</span>.</p>
<p>Now using @trancelocation's trick and applying L'Hospital once, you need to find the limit of</p>
<p><span class="math-container">$$\frac{2e^{2x}-(2+4x)}{2\cdot5x^2}.$$</span></p>
|
2,874,840 |
<blockquote>
<p>If $P\left(A\right)=0.8\:$ and $P\left(B\right)=0.4$, find the maximum and minimum values of $\:P(A|B)$.</p>
</blockquote>
<p>My textbook says the answer is $0.5$ to $1$. But I think the answer should be $0$ to $1$.</p>
<p>The textbook claims $P(A∩B)$ is $0.2$ when $P(A'∩B')=0$</p>
<p>I think that the minimum value arises when $A$ and $B$ are mutually exclusive. So there isn't a chance of both happening and so you have $\frac{0}{0.4} = P(A|B) = 0$ (right?)</p>
<p>I agree with the textbook in saying the maximum value is one.</p>
|
StubbornAtom
| 321,264 |
<p>Since $P(A\cap B)$ is less than both $P(A)$ and $P(B)$,</p>
<p>we have $$P(A\cap B)\le \min(P(A),P(B))=0.4$$ And from Bonferroni's inequality it follows that $$P(A\cap B)\ge \max(0,P(A)+P(B)-1)=0.2$$</p>
<p>Now, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{0.4}$$</p>
<p>So, $$0.5\le P(A\mid B)\le 1$$</p>
|
11,290 |
<p>I have this code to produce an interactive visualization of a tangent plane to a function:</p>
<pre><code>Clear[f]
f[x_, y_] := x^3 + 2*y^3
Manipulate[
Show[
Plot3D[f[x, y], {x, -1, 1}, {y, -1, 1}, PlotStyle -> Opacity[0.8]],
Plot3D[f[point[[1]], point[[2]]] +
Limit[(f[point[[1]] + h, point[[2]]] - f[point[[1]], point[[2]]])/
h, h -> 0]*(x - point[[1]]) +
Limit[(f[point[[1]], point[[2]] + h] - f[point[[1]], point[[2]]])/
h, h -> 0]*(y - point[[2]]), {x, -1, 1}, {y, -1, 1},
PlotStyle -> Opacity[0.8], MeshStyle -> Gray]],
{{point, {0, 0}}, {-1, -1}, {1, 1}},
SaveDefinitions -> True]
</code></pre>
<p>It is, however, extremely slow. I suspect that the reason is that I unnecessarily compute the partial derivatives over and over again inside the second <code>Plot3D</code>, so my question is: how to change it?</p>
<p>Note: I am using the above code also for other functions as <a href="https://mathematica.stackexchange.com/questions/11278/how-to-find-numerical-value-of-a-derivative-at-point">discussed here</a>, and that is the reason for the <code>Limit</code> in computation of partial derivatives.</p>
|
Mark McClure
| 36 |
<p>Simply move the derivative computation outside the scope of the <code>Manipulate</code>, using either <code>D</code> or <code>Limit</code>.</p>
<pre><code>f[x_, y_] = x^3 + 2 y^3;
fx[x_, y_] = D[f[x, y], x];
fx[x_, y_] = Limit[(f[x + h, y] - f[x, y])/h, h -> 0];
fy[x_, y_] = D[f[x, y], y];
fy[x_, y_] = Limit[(f[x, y + h] - f[x, y])/h, h -> 0]
Manipulate[
x0 = p[[1]]; y0 = p[[2]];
Show[{
Plot3D[{f[x, y],
f[x0, y0] + fx[x0, y0] (x - x0) + fy[x0, y0] (y - y0)},
{x, -2, 2}, {y, -2, 2}, BoxRatios -> {1, 1, 1},
PlotRange -> {{-2, 2}, {-2, 2}, {-25, 25}},
PlotStyle -> {Directive[Opacity[0.6]],
Directive[Orange, Opacity[0.6]]},
ViewPoint -> {2.5, -2, 1}, ClippingStyle -> None],
Graphics3D[{PointSize[Large],
Point[{p[[1]], p[[2]], f[p[[1]], p[[2]]]}]}]
}],
{{p, {1, 1}}, {-1, -1}, {2, 2}}]
</code></pre>
|
338,099 |
<p>Are there general ways for given rational coefficients <span class="math-container">$a,b,c$</span> (I am particularly interested in <span class="math-container">$a=3,b=1,c=8076$</span>, but in general case too) to answer whether this equation has a rational solution or not?</p>
|
Xarles
| 24,442 |
<p>Your curve is a genus 1 curve, usually expressed as
<span class="math-container">$$ y^2= -abx^4+bc$$</span>
(just by multiplying everything by <span class="math-container">$b$</span> and changing <span class="math-container">$y$</span> by <span class="math-container">$by$</span>). </p>
<p>The curve has local points everywhere, but since it looks like has no rational points, you can try to do a 2-descent. This can be done easily with the TwoCoverDescent algorithm as explained in </p>
<p>Nils Bruin and Michael Stoll. Two-cover descent on hyperelliptic curves. Math. Comp., 78:2347--2370, 2009. </p>
<p>It is not guaranteed to succeed, but in the special case you are interested <span class="math-container">$a=3$</span>, <span class="math-container">$b=1$</span> and <span class="math-container">$c=8076$</span> it does and the answer is that <strong>it has no rational points</strong>. </p>
<p>All this has been implemented in MAGMA. You can use the following code</p>
<blockquote>
<p>P:=PolynomialRing(Rationals());
H:=HyperellipticCurve((-3)*P.1^4+8076);
Hk:=TwoCoverDescent(H);#Hk;</p>
</blockquote>
<p>It answers 0, so there is no 2-cover with rational points, so the original curve has no rational points. </p>
|
2,299,678 |
<p>Question:</p>
<p>Assume $x, y$ are elements of a field $F$. Prove that if $xy = 0$, then $x = 0$ or $y = 0$.</p>
<p>My thinking:</p>
<p>I am not sure how to prove this. <strong>I can only use basic field axioms.</strong> Should I assume that both x and y are not equal to 0 and then prove by contradiction or should I assume one of x and y is not 0 and then prove the other one has to equal 0?</p>
<p>Thanks </p>
|
CY Aries
| 268,334 |
<p>Hint:</p>
<p>If $x=0$, then we are done.</p>
<p>If $x\ne 0$, then $x^{-1}$ exists. Multiply $x^{-1}$ to both sides of $xy=0$.</p>
|
101,078 |
<p>While reading through several articles concerned with mathematical constants, I kept on finding things like this:</p>
<blockquote>
<p>The continued fraction for $\mu$(Soldner's Constant) is given by $\left[1, 2, 4, 1, 1, 1, 3, 1, 1, 1, 2, 47, 2, ...\right]$. </p>
<p>The <strong>high-water marks</strong> are 1, 2, 4, 47, 99, 294, 527, 616, 1152, ... ,
which occur at positions 1, 2, 3, 12, 70, 126, 202, 585, 1592, ... . </p>
</blockquote>
<p>(copied from <a href="http://mathworld.wolfram.com/SoldnersConstant.html" rel="nofollow">here</a>)</p>
<p>I didn't find a definition of <strong>high-water marks</strong> in the web, so I assume that it's a listing of increasing largest integers, while going through the continued fraction expansion.</p>
<p>Is this correct and is there special meaing behind them?</p>
|
Bill Dubuque
| 242 |
<p>Truncating a continued fraction right before a "high-water" (large) partial quotient $\rm\: a_{i+1}\:$ yields a particularly good rational approximation since </p>
<p>$$\rm r\ =\ [a_0;\ a_1;\ a_2;\ \cdots\ ]\ \ \Rightarrow\ \ \left|\ r\ -\ \frac{p_i}{q_i}\:\right|\ \le\: \frac{1}{a_{i+1}\:q_i^2}$$</p>
<p>For example</p>
<p>$$\rm \pi\ =\ [3;\ 7;\ 15;\ 1;\ 292;\ \cdots\ ]\ \ \Rightarrow \ \ \left|\ \pi - \frac{355}{113}\:\right|\ \le\: \frac{1}{292\cdot 113^2}\: =\ 2.68\cdot 10^{-7}$$ </p>
<p>In fact we have $\rm\quad \pi\ -\ \dfrac{355}{113}\ =\ {-}2.67\cdot 10^{-7}\:.$</p>
|
3,756,932 |
<p>Overall, I think I understand the Monty Hall problem, but there's one particular part that I either don't understand, or I don't agree with, and I'm hoping for an intuitive explanation why I'm wrong or confirmation that I'm right.</p>
<p>The variant that I'm working on is the one with 3 doors. Prize is uniformly located behind 1 of the doors. You choose Door 3. The host opens door 2 and reveals that it's empty. Do you switch to door 1 or stay with your original choice of door 3?</p>
<p>Okay so, if the problem was instead that the host opens either door 1 or door 2 (which ever is empty), then I would switch.</p>
<p>But I think this problem is different because the problem statement states explicitly that the host opens door 2 and reveals that it's empty. So therefore the prize is behind either door 1 or door 3, each with equal probability. So switching is not expected to be advantageous.</p>
<p>The author of the book I am studying from states this as a solution:</p>
<p><em>Suppose you play the game repeatedly and always
choose Door 3. If you look at all the times the host reveals Door 2 empty, you
will find that two-thirds of the time the prize lies behind Door 1, and one-third
of the time it is behind Door 3. Seeing Door 2 empty is thus a stronger signal
that Door 1 has the prize than it is that Door 3 has it. This argument is
more general, of course. Whichever door you choose, seeing the host reveal an
empty door is a signal that you should switch.</em></p>
<p>I disagree with this reasoning because of my intuition above. Am I wrong?</p>
<p>Edit: I agree with the last sentence, which doesn't constrain the host to only opening door 2, but I disagree with the first few sentences where the host is constrained to opening door 2.</p>
<p>Another edit: The original problem statement is (Verbatim):</p>
<p><em>You choose Door 3. He opens Door 2 and reveals that it is empty. You now know that the prize lies behind either Door 3 or Door 1. Should you switch your choice to Door 1?</em></p>
<p>I feel that this problem is equivalent to just eliminating the host and the second door, and asking the question "If the prize is uniformly distributed behind doors 1 and 3, and you choose door 3, should you switch to door 1."</p>
|
Barry Cipra
| 86,747 |
<p>If Monty tells you <em>in advance</em> that, after you've made your initial selection, he will open one of the other two doors, revealing it to be empty, and then give the option of switching to the door he didn't open, he is, in effect, telling you that you can either stick with what's behind your initial selection or else have what's behind <em>both</em> the other two doors. Viewed this way, it's obvious you have a <span class="math-container">$1/3$</span> probability of winning the prize by sticking and a <span class="math-container">$2/3$</span> probability by taking the sum of the other two doors, i.e., by switching after the reveal.</p>
<p>The key is that the rules are established in advance. If Monty is allowed the option of sometimes making the offer and sometimes not, all bets are off.</p>
|
3,756,932 |
<p>Overall, I think I understand the Monty Hall problem, but there's one particular part that I either don't understand, or I don't agree with, and I'm hoping for an intuitive explanation why I'm wrong or confirmation that I'm right.</p>
<p>The variant that I'm working on is the one with 3 doors. Prize is uniformly located behind 1 of the doors. You choose Door 3. The host opens door 2 and reveals that it's empty. Do you switch to door 1 or stay with your original choice of door 3?</p>
<p>Okay so, if the problem was instead that the host opens either door 1 or door 2 (which ever is empty), then I would switch.</p>
<p>But I think this problem is different because the problem statement states explicitly that the host opens door 2 and reveals that it's empty. So therefore the prize is behind either door 1 or door 3, each with equal probability. So switching is not expected to be advantageous.</p>
<p>The author of the book I am studying from states this as a solution:</p>
<p><em>Suppose you play the game repeatedly and always
choose Door 3. If you look at all the times the host reveals Door 2 empty, you
will find that two-thirds of the time the prize lies behind Door 1, and one-third
of the time it is behind Door 3. Seeing Door 2 empty is thus a stronger signal
that Door 1 has the prize than it is that Door 3 has it. This argument is
more general, of course. Whichever door you choose, seeing the host reveal an
empty door is a signal that you should switch.</em></p>
<p>I disagree with this reasoning because of my intuition above. Am I wrong?</p>
<p>Edit: I agree with the last sentence, which doesn't constrain the host to only opening door 2, but I disagree with the first few sentences where the host is constrained to opening door 2.</p>
<p>Another edit: The original problem statement is (Verbatim):</p>
<p><em>You choose Door 3. He opens Door 2 and reveals that it is empty. You now know that the prize lies behind either Door 3 or Door 1. Should you switch your choice to Door 1?</em></p>
<p>I feel that this problem is equivalent to just eliminating the host and the second door, and asking the question "If the prize is uniformly distributed behind doors 1 and 3, and you choose door 3, should you switch to door 1."</p>
|
zkutch
| 775,801 |
<p>I prefer following explanation:</p>
<p>From scratch I have probability <span class="math-container">$\frac{1}{3}$</span> to win and <span class="math-container">$\frac{2}{3}$</span> against me. Because Monty should open door with empty choice, then that <span class="math-container">$\frac{2}{3}$</span> which was against me now work for me, if I change selection, because in these cases I'll win now.</p>
|
220,946 |
<p>I am a newcomer to Mathematica and I am trying to solve this differential equation:</p>
<pre><code>s = NDSolve[{y'[x] == -4/3*y[x]/x + 4 a/(3 x^2) + 4/(9 x^3) a b -
8/(9 x^3) c, y[0.01] == 20}, y, {x, 0.01, 10}]
</code></pre>
<p>I am getting the error message "Encountered non-numerical value for a derivative at x == 0.01" and I don't know what I am doing wrong. Could someone help me?</p>
|
Cesareo
| 62,129 |
<p>A "brute force" solution with Genetic Algorithms </p>
<p>Given a symbolic matrix, first we convert to a zero-one's matrix in which the ones represent non null elements. This is done as follows. Given <strong>M</strong> we obtain <strong>M0</strong></p>
<pre><code>{n, n} = Dimensions[M]
M0 = Table[If[NumericQ[M[[i, j]]] && M[[i, j]] == 0, 0, 1], {i, 1, n}, {j, 1, n}]
</code></pre>
<p>After that the fitness is calculated as the diagonal sum for the resulting transformed matrix, after a change in rows followed for a change in columns. This can be observed in the module <strong>fitnessFunction</strong>. The crossover operation is implemented as a single point crossover as can be observed in the module <strong>doSingleCrossover</strong>. The script can be optimized but it was left as it is as a means to show easily the GA procedures. </p>
<pre><code>Clear[recover]
recover[M0_, bestIndividual_] := Module[{Mopt = {}, Mopt0, i},
For[i = 1, i <= length, i++, AppendTo[Mopt, M0[[bestIndividual[[1, i]]]]]];
Mopt0 = Transpose[Mopt];
Mopt = {};
For[i = 1, i <= length, i++, AppendTo[Mopt, M0[[bestIndividual[[2, i]]]]]];
Return[Mopt]
]
Clear[doMutation];
doMutation[{stringh_, stringv_}] := Module[{tempstring, i, ind1, ind2, atom, choice},
choice = RandomInteger[1];
If[choice == 1, tempstring = stringh, tempstring = stringv];
If[Random[] < mutationRate, ind1 = RandomInteger[{1, length}];
ind2 = RandomInteger[{1, length}];
atom = tempstring[[ind1]];
tempstring[[ind1]] = tempstring[[ind2]];
tempstring[[ind2]] = atom];
If[choice == 1, Return[{tempstring, stringv}], Return[{stringh, tempstring}]]
]
Clear[fitnessFunction];
fitnessFunction[{listh_, listv_}] := Module[{n = Length[M0], Mdum = {}, i, j, sum = 0, Mdum0, rowi},
For[i = 1, i <= n, i++, rowi = M0[[listh[[i]]]];
AppendTo[Mdum, rowi]];
Mdum0 = Transpose[Mdum];
Mdum = {};
For[i = 1, i <= n, i++, rowi = M0[[listv[[i]]]];
AppendTo[Mdum, rowi]];
Return[Total[Diagonal[Mdum]]]
]
Clear[doSingleCrossover];
doSingleCrossover[{stringh1_, stringv1_}, {stringh2_, stringv2_}] :=
Module[{cuth, cutv, temph1, temph2, tempv1, tempv2},
cuth = RandomInteger[{1, length}]; cutv = RandomInteger[{1, length}];
temph1 = Join[Take[stringh1, cuth], Drop[stringh2, cuth]];
temph2 = Join[Take[stringh2, cuth], Drop[stringh1, cuth]];
tempv1 = Join[Take[stringv1, cutv], Drop[stringv2, cutv]];
tempv2 = Join[Take[stringv2, cutv], Drop[stringv1, cutv]];
Return[{{temph1, tempv1}, {temph2, tempv2}}]
]
Clear[doCumSumOfFitness];
doCumSumOfFitness := Module[{temp}, temp = 0.0;Table[temp += popFitness[[i]], {i, popSize}]]
Clear[doSingleSelection];
doSingleSelection := Module[{rfitness, ind},
rfitness = RandomReal[{0, cumFitness[[popSize]]}];
ind = 1;
While[rfitness > cumFitness[[ind]], ind++];
Return[ind]
]
Clear[selectPair];
selectPair := Module[{ind1, ind2}, ind1 = doSingleSelection;
While[(ind2 = doSingleSelection) == ind1];
{ind1, ind2}
]
Clear[pickRandomPair];
pickRandomPair := Module[{ind1, ind2}, ind1 = RandomInteger[{1, popSize}];
While[(ind2 = RandomInteger[{1, popSize}]) == ind1];
{ind1, ind2}
]
Clear[exchangeString];
exchangeString[ind_, newstring_, newF_] := Module[{}, popStrings[[ind]] = newstring;
popFitness[[ind]] = newF
]
Clear[renormalizeFitness];
renormalizeFitness[fitness0_List] :=
Module[{minF, maxF, a, b, fitness = fitness0, i}, minF = Min[fitness];
maxF = Max[fitness];
a = 0.5*maxF/(maxF + minF);
b = (1 - a)*maxF;
Map[a # + b &, fitness]
]
Clear[bestDet]
bestDet := Module[{bestFitness = -1, i, ibest = 1},
For[i = 1, i <= popSize, i++,
If[popFitness[[i]] > bestFitness, bestFitness = popFitness[[i]];
ibest = i]];
If[bestFitness > bestOfAll, bestOfAll = bestFitness;
bestIndividual = popStrings[[ibest]]];
Return[popStrings[[ibest]]]
]
Clear[doInitialize];
doInitialize := Module[{i},
popFitness = Table[fitnessFunction[popStrings[[i]]], {i, popSize}];
popFitness = renormalizeFitness[popFitness];
cumFitness = doCumSumOfFitness;
listOfCumFitness = {cumFitness[[popSize]]};
historyOfPop = {bestDet}
]
Clear[updateGenerationSync];
updateGenerationSync := Module[{parentsid, children, ip}, parentsid = {};
Do[AppendTo[parentsid, selectPair], {popSize/2}];
children = {};
Do[AppendTo[children,
doSingleCrossover[popStrings[[parentsid[[ip, 1]]]],
popStrings[[parentsid[[ip, 2]]]]]], {ip, popSize/2}];
popStrings = Flatten[children, 1];
popStrings = Map[doMutation, popStrings];
popFitness = Map[fitnessFunction, popStrings];
popFitness = renormalizeFitness[popFitness];
cumFitness = doCumSumOfFitness
]
</code></pre>
<p>and now the main program</p>
<pre><code>SeedRandom[4];
bestOfAll = -1;
popSize = 600;(*should be even*)
numberOfEpochs = 300;
mutationRate = 0.007;
n = Length[M0];
length = n;
popStrings = Table[{RandomSample[Table[i, {i, 1, n}]],
RandomSample[Table[i, {i, 1, n}]]}, {popSize}];
doInitialize;
Do[updateGenerationSync;
AppendTo[historyOfPop, bestDet];
AppendTo[listOfCumFitness,
cumFitness[[popSize]]], {numberOfEpochs}
];
ListLinePlot[Map[fitnessFunction, historyOfPop], PlotRange -> All]
bestIndividual
fitnessFunction[bestIndividual]
recover[M, bestIndividual] // Diagonal
(* {1, m1m1, 1, m8m19, m5m21, m6m20, m15m7, m9m8, m9m19, m14m10, m14m21, m15m21, m18m13, m17m14, m12m26, m10m16, m10m17, 1, m8m19, m4m20, m14m21, m8m19, m9m20, m7m21, m16m25, m16m26, m4m27, 1, m10m26, m10m27}*)
</code></pre>
<p>NOTE</p>
<p>This matrix has null determinant. Follows the fitness evolution plot, and the best individual.</p>
<p><a href="https://i.stack.imgur.com/Z45l8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z45l8.jpg" alt="enter image description here"></a></p>
<pre><code>(* {{25, 22, 27, 15, 12, 12, 16, 18, 5, 8, 27, 7, 19, 22, 19, 27, 22, 14, 30, 26, 18, 19, 12, 10, 25, 13, 29, 28, 7, 6}, {28, 2, 27, 20, 27, 26, 15, 9, 25, 14, 27, 27, 18, 17, 29, 10, 10, 24, 8, 4, 14, 14, 15, 13, 16, 16, 4, 4, 16, 16}}} *)
</code></pre>
|
1,414,690 |
<p>This is a followup to my question <a href="https://math.stackexchange.com/questions/1414636/eigenvalues-of-matrix-with-all-1s">here</a>.</p>
<p>Let $A$ be the $n \times n$ matrix over a field of characteristic 0, all of whose entries are 1. Is $A$ diagonalizable?</p>
|
user1551
| 1,551 |
<p>$A$ is always diagonalisable when the field has characteristic 0 or characteristic $>n$.</p>
<p>Let $e_i$ denotes the vector with a $1$ at the $i$-th position and zeros elsewhere. Let $P$ be the matrix whose first and second columns are respectivelly $\sum_je_j$ and $e_1-e_2$ and whose $j$-th column is $e_2-e_j$ when $j\ge3$. Clearly, the columns of $P$ are eigenvectors of $A$. It is not hard to show that $\det P=(-1)^{n+1}n$. Therefore $P$ is invertible and the aforementioned eigenvectors form an eigenbasis when the characteristic of the field is either $0$ or $>n$. </p>
|
69,187 |
<p>Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of course be obtained if $A$, $B$ have a common eigenspace of multiplicity at least $m$.</p>
<p>My question is: is it the only possibility?</p>
<p>A way to proceed is the following: the characteristic polynomia of the generic matrix is $\det(xA+yB-tI)$, and its discriminant $\Delta$ (with respect to $t$) is a homogeneous polynomial in $x,y$ of degree $n^2-n$, where $n$ is the number of rows and columns in $A$ and $B$. Since every matrix in the family has some eigenvalue of multiplicity $>1$, the polynomial $\Delta$ vanishes identically, hance all the $n^2-n+1$ coefficients in $\Delta$ vanish. This gives $n^2-n+1$ polynomial conditions on the $n^2+n$ coefficients of $A$ and $B$, and this might help somehow.</p>
<p>Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?</p>
<p>Thanks in advance!</p>
|
Igor Rivin
| 11,142 |
<p>If you restrict to the postulated $m$-dimensional eigenspace $E$ of $C=xA + yB,$ the fact that $C+\epsilon A$ has the same multiplicity, means that $A$ restriced to $E$ is also a multiple of identity, as is $B$ (by the same argument). Since a small perturbation of a matrix with distinct eigenvalues still has distinct eigenvalues, the situation in the previous sentence must occur for your condition to hold.</p>
<p>I strongly suggest you read the first couple of chapters of Kato's "Perturbation theory of linear operators" (the first two chapters deal with the finite-dimensional case). </p>
|
69,187 |
<p>Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of course be obtained if $A$, $B$ have a common eigenspace of multiplicity at least $m$.</p>
<p>My question is: is it the only possibility?</p>
<p>A way to proceed is the following: the characteristic polynomia of the generic matrix is $\det(xA+yB-tI)$, and its discriminant $\Delta$ (with respect to $t$) is a homogeneous polynomial in $x,y$ of degree $n^2-n$, where $n$ is the number of rows and columns in $A$ and $B$. Since every matrix in the family has some eigenvalue of multiplicity $>1$, the polynomial $\Delta$ vanishes identically, hance all the $n^2-n+1$ coefficients in $\Delta$ vanish. This gives $n^2-n+1$ polynomial conditions on the $n^2+n$ coefficients of $A$ and $B$, and this might help somehow.</p>
<p>Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?</p>
<p>Thanks in advance!</p>
|
David E Speyer
| 297 |
<p>This is false. Let $A_0$ and $B_0$ be $k \times k$ symmetric matrices with no common eigenspace. Let $A$ and $B$ be the $2k \times 2k$ matrices with block forms $\left( \begin{smallmatrix} A_0 & 0 \\ 0 & A_0 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} B_0 & 0 \\ 0 & B_0 \end{smallmatrix} \right)$. Then $A$ and $B$ have no common eigenspace, but every eigenvalue of $Ax+By$ has multiplicity $\geq 2$. And, of course, you can replace $2$ by any $m$.</p>
<p>Here is a different example. Let $A$ and $B$ be symmetric matrices of the form
$$\begin{pmatrix}
\ast & \ast & \ast & \ast & \ast & \ast \\
\ast & \ast & \ast & \ast & \ast & \ast \\
\ast & \ast & 0 & 0 & 0 & 0 \\
\ast & \ast & 0 & 0 & 0 & 0 \\
\ast & \ast & 0 & 0 & 0 & 0 \\
\ast & \ast & 0 & 0 & 0 & 0 \\
\end{pmatrix}$$</p>
<p>Then $Ax+By$ is also of this form, and you can check that any matrix of this form has a zero eigenspace of multiplicity $\geq 2$. However, there is no reason for the $0$-eigenspaces of $A$ and $B$ to meet at all!</p>
<hr>
<p>More generally, regarding "a better way": For any matrices $A$ and $B$, let $F(x,y,z) = \det(z \mathrm{Id} + x A + y B)$. This defines a curve in $\mathbb{P}^2$. Your condition is every line through $(0:0:1)$ intersects this curve in a point of multiplicity $\geq m$. This turns out to imply that $F$ must have a component of multiplicity $\geq m$: i.e. $F$ factors as $G^m H$ for some homogenous polynomials $G$ and $H$. </p>
<p>If $A$ and $B$ had a common eigenspace, then $G$ would be linear, and $A$ and $B$ would have direct sum decompositions $A = A_1^{\oplus m} \oplus A_2$ and $B = B_1^{\oplus m} \oplus B_2$ where $(A_1, B_1)$ and $(A_2, B_2)$ give rise to the plane curves $G$ and $H$.</p>
<p>The preceeding examples show that life can be more complicated. In the first example, $F=G^2$, for some nonlinear $G$. In the second, the factorization of $F$ does not correspond to a direct sum decomposition of $(A,B)$. </p>
<p>There is a fair amount of literature on expressing plane curves as determinants, so there is probably someone who has looked specifically in the case of a multiple factor, but I don't know who.</p>
|
69,187 |
<p>Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of course be obtained if $A$, $B$ have a common eigenspace of multiplicity at least $m$.</p>
<p>My question is: is it the only possibility?</p>
<p>A way to proceed is the following: the characteristic polynomia of the generic matrix is $\det(xA+yB-tI)$, and its discriminant $\Delta$ (with respect to $t$) is a homogeneous polynomial in $x,y$ of degree $n^2-n$, where $n$ is the number of rows and columns in $A$ and $B$. Since every matrix in the family has some eigenvalue of multiplicity $>1$, the polynomial $\Delta$ vanishes identically, hance all the $n^2-n+1$ coefficients in $\Delta$ vanish. This gives $n^2-n+1$ polynomial conditions on the $n^2+n$ coefficients of $A$ and $B$, and this might help somehow.</p>
<p>Still, both finding these polynomial conditions and solving them, seems to be painful and extremely computational. Maybe there are better ways to proceed $\ldots$?</p>
<p>Thanks in advance!</p>
|
Denis Serre
| 8,799 |
<p>I think that the most interesting examples come from systems of conservation laws in physics. These are systems of PDEs (partial differential equations). When they are first-order and linear (linearity can be achived by linearizing about a constant state), you have a system
$$\partial_tU+\sum_{\alpha=1}^dA^\alpha U=f,$$
where $A^\alpha\in{\bf M}_n({\mathbb R})$. The space dimension is usually $d=3$ but may be smaller for waves propagating in specific directions. Exemples cover the Euler equations for gas dynamics, the Maxwell's equation for electro-magnetism, Magneto-hydrodynamics, elasticity and so on.</p>
<p>The symbol $A(\xi)=\sum_\alpha A^\alpha$ plays a fundamental role. Its eigenvalues are the wave velocities, times the modulus $|\xi|$. In several exemples, especially when the system has a group invariance, the eigenvalues have constant multiplicities, yet the eigenspace do vary with $\xi$. Let us take the example of Maxwell's equations in vacuum. Here $d=3$ and $n=6$. We have
$$A(\xi)=\begin{pmatrix} 0_3 & J(\xi) \\\\ -J(\xi) & 0_3 \end{pmatrix},$$
where $J(\xi)$ is the matrix of the cross product by $\xi$: $J(\xi)X=\xi\times X$.
The eigenvalues of $A(\xi)$ are $0$ and $\pm|\xi|$. None of the eigenvectors is constant. In particular, two matrices $A^\alpha$ don't have a common eigenspace.</p>
|
2,291,175 |
<p>I know that the domain is: $(-\infty,0) \cup (0,\infty)$, or all Reals except Zero.</p>
<p>But if a take the "nearest negative number to zero" and then, the "nearest positive number to zero", my function will hugely increase (Y will approach +infinity).</p>
<p>And my domain have all numbers that approach Zero in both sides, except Zero.</p>
<p>So, why the function 1/x does not increase between $(-\infty,0)$ and $(0,\infty)$? (Sorry for any english error)</p>
|
James S. Cook
| 36,530 |
<p>The question is how you define <strong>increasing function</strong> ? Suppose we define increasing function as one for which $a,b \in \text{dom}(f)$ with $a\leq b$ implies $f(a)\leq f(b)$. Then, for $f(x) = 1/x$ for $x \in \mathbb{R} - \{ 0 \} = (-\infty,0) \cup (0, \infty)$ we certainly <strong>cannot</strong> claim $f$ is increasing. Consider, $-1<1$ and $-1,1 \in \text{dom}(f)$ yet $f(-1) = -1 \nleq 1 = f(1)$.</p>
<p>On the other hand, we can also <strong>define</strong> $f$ to be <strong>increasing</strong> on $U \subseteq \text{dom}(f)$ if whenever $a,b \in U$ and $a \leq b$ we obtain $f(a) \leq f(b)$. In this terminology, a function is increasing iff it is increasing on its domain.</p>
<p>I think where students get confused is in calculus where we talk about derivatives and their connection to increase and decrease of a function. That local idea is really played out on a small neighborhood of the point. If we discard the role of the subset in the discussion of increase and decrease then it leads to the sort of confusion that is seen in your question.</p>
<p>To avoid pathological things from a geometric perspective, focus your attention to intervals. An interval is a connected subset of real numbers.</p>
|
2,716,036 |
<p>In reviewing some old homework assignments, I found two problems that I really do not understand, despite the fact that I have the answers.</p>
<p>The first is: R(x, y) if y = 2^d * x for some nonnegative integer d. What I do not understand about this relation is how it can possibly be transitive (according to my notes it is). My understanding is that if the relation were transitive, the following would apply: if y = 2^d * x and x = 2^d * z, then y = 2^d * z. That seems impossible unless x = z. Am I missing something?</p>
<p>The second is: R(x, y) if x and y are both divisible by 17. What I do not understand about this relation is why it is not reflexive. My understanding is that if the relation is reflexive, if x is divisible by 17 then both x and x are divisible by 17. I think that I am possibly applying the quality of reflexiveness incorrectly to this relation, but I am not quite sure.</p>
<p>Thank you for any help in correcting these misunderstandings!</p>
|
José Carlos Santos
| 446,262 |
<ol>
<li>If $x\mathop Ry$ and $y\mathop Rz$, then there are non-negative integers $d$ and $d'$ such that $y=2^dx$ and $z=2^{d'}y$. Therefore, $z=2^{d+d'}x$ and, since $d+d'$ is a non-negative integer, $x\mathop Rz$.</li>
<li>No, it is not reflexive. For instance, $1\not\mathop R1$.</li>
</ol>
|
2,900,283 |
<p>I'm having trouble proving this theorem from Rudin: that if $x \neq 0$ then $\frac{1}{\frac{1}{x}} = x$. </p>
<p>Rudin seems to solve this by referring to an earlier result that $xy = xz$ for $x \neq 0$ implies that $y = z$. I haven't quite been able to grasp this approach, as we don't seem to have access to this assumption. Another, perhaps more intuitive approach, is to deduce it from the field axioms:
\begin{align*}
\frac{1}{\frac{1}{x}} & = 1 \cdot \frac{1}{\frac{1}{x}} & & \text{Mult Identity} \\
& = \left(x \cdot \frac{1}{x}\right) \cdot \frac{1}{\frac{1}{x}} & & \text{Mult Inverse} \\
& = x \left(\frac{1}{x \cdot \frac{1}{x}} \right) & & \text{Associativity, simplification} \\
& = x \cdot 1 & & \text{Mult Inverse} \\
& = x & & \text{Mult Identity}
\end{align*}
I'm particularly unsure on whether we can perform the third line of this proof, wherein we write that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = \frac{1}{x \cdot \frac{1}{x}}$. This is surely the multiplication law for rational numbers, but our only assumption is that $x$ is an element of some field. Considering all of the possible fields -- reals, rationals, complex, finite fields, etc. -- it seems that this law would work, but I can't think of an axiom by which it would other than the fact that multiplication is defined and behaves as we would expect in fields, so this seems like a standard result. I'm unsure on whether I ought to prove such a result prior to using it, or if it simply follows from the definition.</p>
<p>Any helpful insights or hints would be greatly appreciated. </p>
|
Peter Szilas
| 408,605 |
<p>Suggestion:</p>
<p>Let $x\not =0.$</p>
<p>$x^{-1}= 1/x.$</p>
<p>Starting with the second line:</p>
<p>$ (xx^{-1})(x^{-1})^{-1}= $</p>
<p>$x(x^{-1}(x^{-1})^{-1})=x \cdot 1=x$.</p>
|
1,248,329 |
<p>Let $R$ be an integral domain, and let $r \in R$ be a non-zero non-unit. Prove that $r$ is irreducible if and only if every divisor of $r$ is either a unit or an associate of $r$.</p>
<p>Proof. ($\leftarrow$) Suppose $r$ is reducible then $r$ can be expressed as $r = ab$ where $a$, $b$ are not units. This contradicts the fact that every divisor of $r$ is either a unit or an associate of $r$.</p>
<p>($\rightarrow$) Suppose every divisor of $r$ is neither a unit nor an associate of $r$. Then $r$ is reducible.</p>
<p>Can someone verify if my proof is correct. However, I prefer a direct proof. This proof doesn't look nice. Can someone show me how I can do this directly?</p>
|
Stefan Hamcke
| 41,672 |
<p>You are right about the first part. If $p\omega$ is nullhomotopic in $A$, then the homotopy to the constant loop lifts to a homotopy between $\omega$ and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map $p:\tilde X\to X$ restricts to a covering map $p^{-1}(A)\to A$ for every subset $A$ of $X$.</p>
<p>Regarding the second part, let every path component of $p^{-1}(A)$ be simply-connected. If $\omega$ is a loop in $A$, nullhomotopic as a loop in $X$, then what can you say about its lift $\tilde\omega$. Recall that every homotopy of paths lifts to a homotopy between their unique lifts at a given point $\tilde x$ lifting the starting point of the paths.</p>
|
510,488 |
<p>Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...</p>
<p>"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?</p>
<p>It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction?
I think I'm missing some key point. </p>
|
bof
| 97,206 |
<p>Averaging the inequalities $(x+y)^2\ge0$ and $x^2+y^2\ge0$ we get $\frac12(x+y)^2+\frac12(x^2+y^2)\ge0$, that is $x^2+xy+y^2\ge0$, contradicting the assumption that $x^2+xy+y^2\lt0$.</p>
|
2,071,828 |
<p>Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.</p>
|
GRE
| 378,788 |
<p>My factorization was incorrect, I have understood my error. Now to calculate the Galois group of the extension (which is actually a Galois extension, as $\mathbb {F}_9$ is the splitting field of a separable polynomial, having first derivative different from zero, with coefficients in $\mathbb {F}_3$) I just note that the degree of the extension is $[\mathbb {F}_9:\mathbb {F}_3]=2$, so the group has 2 elements and it is isomorphic to $\mathbb {Z}_2$. Is it right?</p>
|
1,619,679 |
<p>If the series $\sum\limits_{n=1}^{\infty} u_n $ is convergent then the sequence $u_n \rightarrow 0$ as $n \rightarrow \infty$. Therefore if the ratio test $R=\frac{u_{n+1}}{u_n}$ gives $R<1$ then we can conclude that $(u_{n})_{n\in \mathbb{N}}$ is convergent, right?</p>
<p>Generally, if we can find that a series is convergent then can we always conclude that the sequence that is summed up is convergent, provided that it is not an alternating series?</p>
|
Ben Grossmann
| 81,360 |
<p>Yes, you have the right idea. Whenever
the series $\sum_{n=1}^\infty u_n$ converges, the sequence $u_n$ must converge to zero as $n \to \infty$. This is true <em>even if</em> the sum is alternating. Certainly, then: if $\sum u_n$ converges by the ratio test, then we must have $u_n \to 0$ as $n \to \infty$.</p>
<p>Note, however, that the converse does not hold. That is, it is possible that $u_n \to 0$ as $n \to \infty$, but $\sum_{n=1}^\infty u_n$ diverges.</p>
|
2,928,806 |
<p><span class="math-container">$m(t)=\frac{t^2}{1-t^2},t<1$</span></p>
<p>The suggested answer provided by our teacher states that m(0)=0, so there's no real value with this function as mgf; hence mgf doesn't exist.</p>
<p>Intuitively I understand that <span class="math-container">$e^{xt}>0$</span> should always be true, so the expectation should always be positive under this transformation. But I still don't quite get what does "m(0)=0 so there's no real value to this function" and how this statement backed the conclusion "mgf does not exist".</p>
|
J.G.
| 56,861 |
<p>For any mgf <span class="math-container">$M$</span>, <span class="math-container">$M(0)=1$</span>. Since <span class="math-container">$m(0)=0\ne 1=M(0)$</span>, <span class="math-container">$m$</span> is not an mgf.</p>
|
1,004,056 |
<p>Pick a piece of paper and a pen. Put the pen on a starting point and begin to draw an arbitrary curve and don't withdraw your hand until you reached the starting point. You can meet your curve during drawing in some <em>one point</em> intersections but not more (like a line intersection). For example the left curve is allowed but the right one is not.</p>
<p><img src="https://i.stack.imgur.com/ItJT2.png" alt="enter image description here"></p>
<p>Now your paper is divided to some areas. We call two areas "neighbors" if their <em>common borderline</em> has more than one point. A "coloring" for the areas of the paper is such that two neighbor areas have different colors. Recall the chess board coloring.</p>
<blockquote>
<p><strong>Question:</strong> Is it true that there is a $2$-coloring (Black and White) for the areas of a paper when we draw an arbitrary curve on it as described above? If yes, why?</p>
</blockquote>
<p><strong>An Example:</strong> Note to the coloring in the following shape. You can try on more complicated shapes by your own. It seems to be always true that we can do this by $2$ colors.</p>
<p><img src="https://i.stack.imgur.com/fhlXZ.png" alt="enter image description here"></p>
|
John Hughes
| 114,036 |
<p>If the curve is smooth enough, and there are finitely many self-intersections, the answer is yes. For this answer, I'm going to limit myself to a particular class of curves, though: those where at most <em>two</em> bits of curve intersect at any crossing, so you can have something that locally looks like $\times$, but not something that looks locally like an asterisk (*). </p>
<p>Why the smoothness and finiteness requirements? </p>
<p>If you think of, say, the graph of $x^2 \sin \frac{1}{x}$, between $-1$ and $1$, together with the segment of the $x$-axis between $-1$ and $1$, and connect the two ends with vertical segments, you get something with infinitely many regions, and it gets tough to say whether it's really chess-board-like. </p>
<p>For the case I mentioned, here's a "proof": pick a direction $v$ with the properties that</p>
<ol>
<li><p>the tangent to your curve is in direction $v$ only finitely often, and </p></li>
<li><p>at these tangencies, the curve lies on one side of $v$ or the other (i.e., no "inflections" at points where the tangent is $v$). </p></li>
<li><p>Furthermore, at each crossing, neither tangent vector should be $v$. </p></li>
<li><p>All lines in direction $v$ intersect the curve in at most finitely many points. </p></li>
<li><p>Finally, the ray between any two crossing points or any two tangents should not be parallel to $v$. </p></li>
</ol>
<p>I'm using smoothness in a big way here. [The existence of $v$ is guaranteed by Sard's Theorem, which is not an easy theorem at all, and I'm not going to prove it here.] </p>
<p>Rotate things so that $v$ is the positive $x$ direction. </p>
<p>From each point in your curve-complement, draw a ray in the positive-$x$ direction. Count the number of intersections of this ray with your curve. If it's odd, color the square black; otherwise color it white. Don't count "tangent" intersections, or crossings.
That gives a coloring scheme; the only remaining question is "is it checkerboard-like?". </p>
<p>A "sweep-line" argument takes care of this: consider the coloring of two very nearby points in a region. Are they the same? Compare their ray-curve intersections: if the points are close enough, the ray-curve intersections are pairwise close to each other, and you're good...except in two cases: </p>
<ol>
<li><p>One ray is tangent to the curve and the other is not. Then the intersection count for the second ray is either two greater than for the first, or equal to it, as the second ray either intersection the curve near the tangency twice or not at all (by the requirement that the curve lie entirely on one side of its tangent line, i.e, condition 2). (I'm using condition 5 to ensure that the ray is tangent to the curve at only ONE point so that this analysis makes sense) </p></li>
<li><p>The two rays have a crossing between them. This doesn't actually cause a problem, as the intersection of ray1 with the crossing parts can be paired up with the intersections of ray2 with the crossing parts. (The details for this part use property 3 and property 5)</p></li>
</ol>
<p>So evidently a connected region has locally-constant color, hence constant color. </p>
<p>Clearly if two regions share a boundary (they're "neighboring countries") then their intersection-counts will differ in parity: A right-pointing ray from a point near the boundary between the left country and the right country will pass through that boundary; for a point just on the other side of that boundary (such a point exists by condition 4), the right-pointing ray will lack that intersection. </p>
<p>In short: even stating the claim really clearly isn't all that easy, and the proof isn't trivial either. It's somewhat simpler if you're willing to say that your curve is a (possibly self-intersecting) polygon with finitely many vertices (i.e., a "connect the dots drawing"); in that case, the conditions I wrote above are all easy to verify. </p>
<p>I believe that there's a one-liner proof using Alexander duality and mod-2 homology/cohomology, but that's not very illuminating, and there's a ton of machinery behind it. </p>
|
33,743 |
<p>I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!</p>
<p>P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...</p>
<p>Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?</p>
<p>(edit)</p>
<p>Thanks guys! This got me the following result:</p>
<p>$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$</p>
<p>summing over all primes $p$. (And RHS is Riemann zeta function.)</p>
<p>First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (<a href="http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series" rel="nofollow">reference: wikipedia</a>. I know.). Now we have (with rearranging):</p>
<p>$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$</p>
<p>By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have</p>
<p>$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$</p>
<p>I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....</p>
|
Steven Stadnicki
| 785 |
<p>It's also easy to show that this series converges through more elementary means (though without getting a good estimate of the value), by a version of the same elementary technique often used to show that the harmonic series diverges. Start by fixing $k=2$; since the terms for any other $k$ are smaller than this, then clearly if this series converges the others will. Next, break it into chunks of length $2^n$: $S = {1\over 1^2} + ({1\over 2^2}+{1\over 3^2}) + ({1\over 4^2}+{1\over 5^2}+{1\over 6^2}+{1\over 7^2}) + \ldots$ Now, replace each of the terms in the sets of parentheses by the first term there; since we know that e.g. ${1\over 7^2} \lt {1\over 4^2}$, our sum will be bounded by the result; $S\lt S'$, where $S' = {1\over 1^2} + ({1\over 2^2}+{1\over 2^2}) + ({1\over 4^2}+{1\over 4^2}+{1\over 4^2}+{1\over 4^2}) + \ldots$ - and here it's easy to see that, for instance, the $4$ terms of $1\over 4^2$ will add up to $1\over 4$, the (unshown) $8$ terms of $1\over 8^2$ will add up to $1\over 8$, etc; so $S'$ is just $1+{1\over 2}+{1\over 4}+{1\over 8}+\ldots = 2$, and so we know that the original series $S$ converges (and in fact that its value is less than $2$).</p>
|
854,671 |
<p>So I'm a bit confused with calculating a double integral when a circle isn't centered on $(0,0)$. </p>
<p>For example: Calculating $\iint(x+4y)\,dx\,dy$ of the area $D: x^2-6x+y^2-4y\le12$.
So I kind of understand how to center the circle and solve this with polar coordinates. Since the circle equation is $(x-3)^2+(y-2)^2=25$, I can translate it to $(u+3)^2+(v+2)^2=25$ and go on from there.</p>
<p>However I would like to know if I could solve this without translating the circle to the origin. I thought I could, so I simply tried solving $\iint(x+4y)\,dx\,dy$ by doing this:
$\int_0^{2\pi}\,d\phi\int_0^5(r\cos\phi + 4r\sin\phi)r\,dr$ but this doesn't work. I'm sure I'm missing something, but why should it be different? the radius is between 0 and 5 in the original circle as well, etc.</p>
<p>So my questions are:</p>
<ol>
<li><p>How can I calculate something like the above integral without translating the circle to the origin? What am I doing wrong?</p></li>
<li><p>I would appreciate a good explanation of what are the steps exactly when translating the circle. I kind of "winged it" with just saying "OK, I have to move the $X$ back by 3, so I'll call it $X+3$, the same with the $Y$ etc. If someone could give a clear breakdown of the steps that would be very nice :)</p></li>
</ol>
<p>Thanks!</p>
|
Michael Hardy
| 11,667 |
<p>Just brute force:</p>
<p>There is a calendar for a common year (i.e. a non-leap year) beginning on Sunday.</p>
<p>There is a calendar for a common year beginning on Monday.</p>
<p>There is a calendar for a common year beginning on Tuesday.</p>
<p>. . . and so on. Seven calendars. Then seven more for leap years.</p>
<p>Go through all 14 of them and observe that each has at least one Friday the 13th. And some have two, and some have three, and none have more than three.</p>
<p>There may be no way to reach this conclusion except this kind of brute force, because the structure of the calendar (how many months, how many days in each month) is not defined by orderly rules.</p>
|
3,429,350 |
<p>If we a sample of <span class="math-container">$n$</span> values from a given population and if <span class="math-container">$X$</span> is the variable of the sample, then the mean of <span class="math-container">$X$</span> is just <span class="math-container">$\dfrac{ \sum x }{n}$</span></p>
<p>Now, suppose <span class="math-container">$X$</span> is random variable. For concreteness, let us take <span class="math-container">$X$</span> to be the number of heads in a toss of a coin. Now, <span class="math-container">$X$</span> can be 0 or 1. Therefore, the mean in this case is <span class="math-container">$\dfrac{0+1}{2} = \dfrac{1}{2}$</span> and the formula coincides with the above, the one for samples. </p>
<p>In general, for random variable <span class="math-container">$X$</span>, we know the mean is <span class="math-container">$\sum x P(X=x)$</span>.</p>
<p>But, how is this different from the mean for samples? What is the motivation for this defition? </p>
|
Randy Marsh
| 391,136 |
<p>It is equal to one because it contains <span class="math-container">$K_5$</span> and we can exhibit an embedding in genus 1. In general the problem of determining the genus of a graph is NP-hard.</p>
<p>Here is an approach for finding an embedding in orientable genus 1:</p>
<p>Cut out two open disks from the sphere, and straighten it out into a tube, so that what you get is a tube with boundary. Put three vertices <span class="math-container">$1$</span>, <span class="math-container">$2$</span>, <span class="math-container">$3$</span> on the bottom boundary, and three vertices <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span> on the top. You can immediately get the following graph without any edge intersections:</p>
<p><span class="math-container">$$1A, 1B, 1C, 12, 2C, 23, 3C,AB,BC$$</span></p>
<p>We can also go around the tube, so in addition to the edges above we also have</p>
<p><span class="math-container">$$13, 3A, AC.$$</span></p>
<p>Now we have a graph with 12 edges, two vertices <span class="math-container">$(1, C)$</span> of degree 5, two vertices (3, A) of degree 4 and two vertices (2, B) of degree 3.</p>
<p>Take now a sphere, trace out two circles on it with the same radius and centers on the north and south pole. Mark three points on the first circle with <span class="math-container">$123$</span>, and three points on the second circle with <span class="math-container">$ABC$</span> so that <span class="math-container">$1$</span> and <span class="math-container">$A$</span>, <span class="math-container">$2$</span> and <span class="math-container">$B$</span>, and <span class="math-container">$3$</span> and <span class="math-container">$C$</span> are mirror images with respect to the equator. Now glue the tube along the two circles so that the points and the vertices match-up (and remove the interiors of the two circles that contain the poles). It is easy to see that the three missing edges <span class="math-container">$2A$</span>, <span class="math-container">$2B$</span> and <span class="math-container">$3B$</span> can be drawn without any interstections.</p>
|
2,990,642 |
<p><span class="math-container">$\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$</span></p>
<p>Using Binomial theorem:</p>
<p><span class="math-container">$(0.9999+\frac{1}{n})^n={n \choose 0}*0.9999^n+{n \choose 1}*0.9999^{n-1}*\frac{1}{n}+{n \choose 2}*0.9999^{n-2}*(\frac{1}{n})^2+...+{n \choose n-1}*0.9999*(\frac{1}{n})^{n-1}+{n \choose n}*(\frac{1}{n})^n=0.9999^n+0.9999^{n-1}+\frac{n-1}{2n}*0.9999^{n-2}+...+n*0.9999*(\frac{1}{n})^{n-1}+(\frac{1}{n})^n$</span></p>
<p>A limit of each element presented above is 0. How should I prove that limit of "invisible" elements (I mean elements in "+..+") is also 0?</p>
|
William M.
| 396,761 |
<p><strong>Hint</strong>. Use that <span class="math-container">$\lim (1 + \frac{x}{n})^n \to e^x$</span> for any real number <span class="math-container">$x.$</span></p>
|
708,633 |
<p>My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?</p>
<p>What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then $R^n/N \simeq R^n$ implies $N=0$. It seems like $N\neq 0$ implies $R^n/N$ has nontrivial relations, so $R^n/N$ can't be free.</p>
<p>If $R^n/N \simeq R^n$, we'd have an exact sequence</p>
<p>$0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0$</p>
<p>which splits since $R^n$ is free, so $R^n \simeq R^n \oplus N$. Does this imply $N=0$? What if we assume $R$ is commutative, or even local? Maybe Nakayama can come in handy.</p>
<p>I'm interested in noncommutative examples too. Thanks!</p>
|
egreg
| 62,967 |
<p>A classical example is the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$ ($p$ a prime); for each proper subgroup $H$ of it, it holds $\mathbb{Z}(p^{\infty})\cong \mathbb{Z}(p^{\infty})/H$ (and there's plenty of subgroups).</p>
<p>If $R$ is the endomorphism ring of an infinite dimensional vector space, then, as (left) modules, $R\oplus R\cong R$, so
$$
\frac{R\oplus R}{0\oplus R}\cong R\cong R\oplus R
$$</p>
|
932,430 |
<p>I am studying about how a real number is defined by its properties. The three type of properties that make the real numbers what they are.</p>
<ol>
<li><p>Algebraic properties i.e, the axioms of addition, subtraction multiplication and division.</p></li>
<li><p>Order properties i.e., the inequality properties</p></li>
<li><p>Completeness property </p></li>
</ol>
<p>Here is the question : I am not able to understand the completeness property and please explain it to me in detail(it says upper bound lower bound ......)as I am a self learner.</p>
|
Ben
| 33,679 |
<p>Suppose I have a set $X \subseteq \mathbb{R}$ such that for all $x \in X$ we have $x \le t$ for some number $t$. Then the completeness axiom guarantees the existence of a smallest number, call it $s$, such that $x \le s$.</p>
<p>You write this down more formally by saying that if $X$ is bounded, <em>ie.</em> if all $x$ are such that $x \le t$ for some $t$, then there exists a number $s \in \mathbb{R}$ satisfying the following:</p>
<ul>
<li><p>$x \le s$</p></li>
<li><p>If $x \le t'$ then $s \le t'$</p></li>
</ul>
<p>we call $s$ the <strong>supremum</strong> of $X$ and denote it by $\sup(X)$. </p>
<p>The completeness axiom is declaring the existence of a supremum on a bounded set.</p>
|
348,674 |
<p>I am a beginner at matrices and I am trying to find out whether or not the linear transformation defined by the matrix $A$ is onto, and also whether it is 1-1.</p>
<p>Here is the matrix $A$: </p>
<p>$$\begin{bmatrix}
1 & 3 & 4 & -1 & 2\\
2 & 6 & 6 & 0 & -3\\
3 & 9 & 3 & 6 & -3\\
3 & 9 & 0 & 9 & 0
\end{bmatrix}$$</p>
<p>I reduced it to echeleon row form but I am not sure what to do from there, thank you for any help.</p>
|
Dimitri
| 62,474 |
<p>If the reduced matrix have a zero column, then it is not 1-1 because for example $A(00001)^{T}$ will be zero, this will be the case, beacuse the matrix $A$ that a vector in $\mathbb{R}^5$ to $\mathbb{R}^4$, so it can´t be 1-1.
To see if it is into, you have to see that the columns generate $\mathbb{R}^4$, so if the reduced matrix have $4$ non-zero rows, that will be the case.</p>
|
1,672,882 |
<p>Why we always see extended operations, like arbitrary unions, products, etc. in different parts of mathematics in the form of extensions of finite ones upon arbitrary <em>sets</em> (called index set)? Why never use an index-<em>class</em> for a proper class?</p>
<p>EDIT:
Of course I think this a reason, for example, that in the context of category theory, some theorems only hold for small/locally small categories: we are not allowed to operate upon arbitrary proper classes.</p>
|
hmakholm left over Monica
| 14,366 |
<p>The restrictions on what you can use proper classes for are there to make sure that everything you write down can actually be proved to exist using the restricted set-building axioms of ZF set theory. We can prove once and for all that a union or product (or whatever) indexed by a <em>set</em> will always describe something that has an existence proof, but this proof generally doesn't go through if the indices are pulled from a proper class.</p>
<p>It's more of a rule of thumb than an absolute truth, though -- for example, an <em>intersection</em> indexed by a proper class is actually unproblematic from a ZF point of view. But it is much easier to <em>remember</em> simply not using proper classes for indexing, and the cases where you need the exceptions are sufficiently rare that it is cost-effective (in mental effort) simply to say not to index by proper classes, and in the few cases where that is insufficient argue directly from the axioms instead of relying on the slick index notation.</p>
<p>The reason why ZF restricts set-building in the first place is to avoid allowing the reasoning that leads to set-theoretical paradoxes such as <em>Russel's</em> or <em>Buralli-Forti's</em> paradox. There are a few competing suggestions for other restrictions that aim to achieve that by other means, but the general consensus is that the ZF restrictions are the most convenient to stay within for most ordinary mathematics.</p>
|
108,297 |
<p>I've been told that strong induction and weak induction are equivalent. However, in all of the proofs I've seen, I've only seen the proof done with the easier method in that case. I've never seen a proof (in the context of teaching mathematical induction), that does the same proof in both ways, and I can't seem to figure out how to do them myself. It would put my mind at ease if I could see with my own eyes that a proof done with strong induction can be completed with weak induction. Does anyone have a link to proofs proved with both, or could anyone show me a simple proof here? I'm more interested in proofs were strong induction is the easier method.</p>
|
Patrick Da Silva
| 10,704 |
<p>The idea is that if something is proved with "strong" induction, i.e. by assuming all preceding cases, then you can use "weak" induction on the hypothesis "all preceding cases hold". Let me explain with mathematical notation, perhaps it'll be a little clearer.</p>
<p>Suppose you want to prove a proposition for all $n \ge 1$, i.e you want to show that for all $n \ge 1$, $P(n)$ is true, where $P(n)$ is some proposition. Define the proposition $Q(n)$ by "$P(k)$ is true for all $k$ with $1 \le k \le n$". Then showing that $P(n)$ is true using "strong" induction is equivalent to showing that $Q(n)$ is true using "weak" induction. But $P(n)$ is true for all $n$ if and only if $Q(n)$ is true for all $n$, hence the proof techniques are completely equivalent (in the sense that using one technique or the other has the same veracity ; it doesn't mean that one is more or less complicated to use than the other).</p>
<p>At some point in the study of mathematics you stop making the distinction between "strong" and "weak". You just say that you're using "induction". I wouldn't be sure that you stop distinguishing this if you study logic though, but let's just leave those kind of problems to logicians, shall we.</p>
<p>Hope that helps,</p>
|
3,902,501 |
<p>Over <span class="math-container">$\mathbb R$</span>, the only linear maps are those of the form <span class="math-container">$ax$</span>.</p>
<p>If we discuss rational functions over <span class="math-container">$\mathbb R$</span>, this extra structure would allow us to describe a wider variety of linear maps.</p>
<p>But the obvious maps such as limits, differentiation, summation, doing <span class="math-container">$f(x)\mapsto f(x+k)$</span> seem too easy. Is there a linear map which specifically takes advantage of the fact that we have a rational function?</p>
<p>For instance, something like "double the coefficient of <span class="math-container">$x$</span> of the numerator and triple that of <span class="math-container">$x$</span> in the denominator". I know this is not a good example because it isn't linear, but it illustrates what I mean by "taking advantage of the fact that we have a rational function", i.e., something which doesn't generalise easily to a wider class of functions (say <span class="math-container">$C^1$</span>).</p>
|
Robert Israel
| 8,508 |
<p>Any linear map <span class="math-container">$T$</span> of vector spaces <span class="math-container">$X \to Y$</span> can be "generalized" to a map of
<span class="math-container">$Z \to Y$</span>, where <span class="math-container">$Z$</span> is any vector space that contains <span class="math-container">$X$</span>. Namely, let <span class="math-container">$P$</span> be
a projection from <span class="math-container">$Z$</span> onto <span class="math-container">$X$</span>, and take the linear map <span class="math-container">$T \circ P$</span>. Of course,
you might not be able to obtain <span class="math-container">$P$</span> explicitly: you might need the Axiom of Choice.</p>
|
2,574,962 |
<p>An even graph is a graph all of whose vertices have even degree. </p>
<p>A spanning subgraph $H$ of $G$, denoted by $H \subseteq_{sp} G$, is a graph obtained by $G$ by deleting <em>only</em> edges of $G$.</p>
<p>I want to show that if $G$ is a connected graph, then
$\big|\{H \subseteq_{sp} G | H$ $is$ $even\}\big| = 2^{e-n+1}$, where $e$ is the number of edges and $n$ the number of vertices of $G$.</p>
<p>Can anyone give me a solution or a hint? Thanks in advance!</p>
|
Yly
| 253,140 |
<p>Pick a base vertex $v$. For any other vertex $w$, pick a path $P(v,w)$ from $v$ to $w$. $P(v,w)$ exists because $G$ is connected. </p>
<p>For a spanning subgraph $H\subseteq_{sp} G$, we speak of "flipping an edge $e$" to mean removing $e$ from $H$ if $e\in H$, and adding $e$ to $H$ if $e$ is not in $H$. This transformation yields another spanning subgraph. Note that flipping all edges along a path changes the degree of the end vertices of the path by one (provided the path is not a loop), but not any other vertices.</p>
<p>Define an equivalence relation $\sim$ on the set of all spanning subgraphs of $G$ as follows: $H_1\sim H_2$ if $H_2$ is obtained from $H_1$ by flipping all the edges along the path $P(v,w)$ for some $w$. Thus each equivalence class contains $2^{n-1}$ spanning subgraphs, as we have $n-1$ paths $P(v,w)$ to flip. </p>
<p>For any spanning subgraph $H$, there is a unique equivalent graph with even degree for all vertices $w\neq v$ (just flip all paths $P(v,w)$ for which $w$ has odd degree). Call this equivalent subgraph $H'$. In $H'$, $v$ must also have even degree, because the sum of degrees for all vertices is even (being equal to $2\times\text{number of edges}$), and the sum of degrees of all vertices $w\neq v$ is also even by hypothesis. Hence, each equivalence class has a unique even spanning subgraph.</p>
<p>There are $2^e$ spanning subgraphs, and thus $2^{e-n+1}$ equivalence classes, and thus $2^{e-n+1}$ even spanning subgraphs. </p>
|
713,626 |
<p>If $X$ is $Beta\left(\dfrac{ \alpha_1}{ 2 }, \dfrac{\alpha_2}{2}\right)$ then $\dfrac{\alpha_2 X}{\alpha_1(1-X)}$ is $F(\alpha_1, \alpha_2)$? </p>
<p>Any help is appreciated I don't know where to start. I'm assuming I need the pdf's of each distribution?</p>
|
Frazer
| 125,581 |
<p>Yes, I use the pdf.</p>
<p>$f(x)=\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}x^{\frac{m}{2}-1}(1-x)^{\frac{n}{2}-1}$</p>
<p>and the pdf of function $g(x)$ is $l(y)=f(h(y))|h'(y)|$, where $h$ is the inverse of $g$. </p>
<p>We get $h(y)=\frac{my}{my+n}$ and $|h'(y)|=\frac{mn}{(my+n)^2}$, then after patient insertion and simplification we get the result of
$m^{m/2}n^{n/2}\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}y^{m/2-1}(my+n)^{-(m+n)/2}$ , which is indeed $F(m,n)$.</p>
|
1,615,732 |
<p>Consider $CW$-complex $X$ obtained from $S^1\vee S^1$ by glueing two $2$-cells by the loops $a^5b^{-3}$ and $b^3(ab)^{-2}$. As we can see in Hatcher (p. 142), abelianisation of $\pi_1(X)$ is trivial, so we have $\widetilde H_i(X)=0$ for all $i$. And if $\pi_2(X)=0$, we have that $X$ is $K(\pi,1)$.</p>
<p>My question is: <em>how can one compute</em> $\pi_2(X)$? Computing homotopy groups is hard, what methods may i use?</p>
|
Hari Rau-Murthy
| 142,843 |
<p>$\newcommand{\Z}{\mathbb{Z}}$
The statement $\pi_2(X)=0$ would imply that $X$ is a $K(\pi,1)$, is not true. I will construct for you now an acyclic space,(the space $X$ below) with $\pi_2=0$, that is not an eilenberg maclane space.</p>
<p>Let $2I$ be the $\Z/2$ extension of the icosahedral group, $I$, given by the preimage of $I$ under $spin(3) \to SO(3)$(FYI spin(3) is the connected double cover of $SO(3)$). Since $I$ is perfect,i.e. $H_1(I)=0$, the hoschild-serre spectral sequence of this extension implies that $H_1(2I)$ is perfect. Therefore by the Quillen plus construction there is a space with $X$ with $\tilde H_*(X)=0$, and $\pi_1(X)=2I$. Since $2I$ is not an acyclic group, $K(2I,1) \neq X$ in the homotopy category. But if $X$ were an eilenberg maclane space it would have to be $K(2I,1)$. Therefore $X$ is not an eilenberg maclane space. Furthermore, since $\pi_2(\text{quillen plus construction on } A_5)=\Z/2$, $\pi_2(X)=0$.</p>
|
119,686 |
<p>I have a very dumb question. Let $X = \mathbb{P}^2_k = Proj(k[x,y,z])$ where $k$ is algebraically closed. We have an invertible sheaf $\mathcal{O}(2)$ on $X$. Its space of global sections contains the elements $x^2, y^2, z^2, xy, yz, xz$. </p>
<p>It seems to me (by my calculations), however, that $\mathcal{O}(2)$ is generated by $x^2, y^2$, and $z^2$. Meaning, these 3 global sections generate the stalks at each point of $X$. I'm suspicious, though. Is this true?</p>
<p>David</p>
|
rfauffar
| 12,158 |
<p>The thing is, with $x^2$, $y^2$ and $z^2$, there's no way of generating $xy$, $xz$ and $yz$; that's the problem. That's why these necessarily have to be part of the generating set.</p>
|
1,874,159 |
<blockquote>
<p>Given real numbers $a_0, a_1, ..., a_n$ such that $\dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1}=0,$ prove that $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n=0$ has at least one real solution.</p>
</blockquote>
<p>My solution:</p>
<hr>
<p>Let $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$</p>
<p>$$\int f(x) = \dfrac {a_0}{1} x + \dfrac {a_1}{2}x^2 + \cdots + \dfrac {a_n}{n+1} x^{n+1} + C$$</p>
<p>$$\int_0^1 f(x) = \left[ \dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1} \right]-0$$</p>
<p>$$\int_0^1 f(x) = 0$$</p>
<p>Since $f$ is continuous, by the area interpretation of integration, it must have at least one zero.</p>
<hr>
<p>My question is, is this rigorous enough? Do I need to prove the last statement, perhaps by contradiction using Riemann sums? Is this a theorem I can/should quote?</p>
|
Marco Cantarini
| 171,547 |
<p>You can also prove it using the mean value theorem. You showed that $$\int_{0}^{1}f\left(x\right)dx=0
$$ now since $f$ is continuous by the <a href="https://en.wikipedia.org/wiki/Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals">mean value theorem for integrals</a> we have that exists some $c\in\left(0,1\right)
$ such that $$f\left(c\right)=\int_{0}^{1}f\left(x\right)dx
$$ so $$f\left(c\right)=0$$ as wanted.</p>
|
3,702,309 |
<p>The question reads,
Find all differentiable functions <span class="math-container">$f$</span> such that <span class="math-container">$$f(x+y)=f(x)+f(y)+x^2y$$</span> for all <span class="math-container">$x,y \in \mathbb{R} $</span>. The function <span class="math-container">$f$</span> also satisfies <span class="math-container">$$\lim_{x \rightarrow 0}\frac {f(x)}x = 0$$</span></p>
<p>To solve the problem I wrote the expression for <span class="math-container">$\frac{df}{dx}$</span> using first principle and found that <span class="math-container">$\frac{df}{dx} = x^2.$</span>(Due to the given conditions.)</p>
<p>Using this and calculating <span class="math-container">$f(0)$</span> as zero I got the implication that <span class="math-container">$f(x)$</span> must be <span class="math-container">$\frac{x^3}{3}.$</span></p>
<p>But,
Clearly the calculated <span class="math-container">$f(x)$</span> does not satisfy the required condition for the problem. Thinking about where I had committed the mistake, I realized that <span class="math-container">$f(x+y)=f(x)+f(y)+x^2y$</span> cannot be true for all real <span class="math-container">$x,y$</span>. </p>
<p>So how is it that these operations that implied <span class="math-container">$f(x)=\frac{x^3}{3}$</span> went wrong?</p>
<p>And why is it that while doing these operations I could not 'see' that the conditions cannot be satisfied for all real <span class="math-container">$x$</span>?</p>
<p>EDIT:Reading the comments, I want to make the clarification that I realise the fact that there can not be any <span class="math-container">$ f(x) $</span> that satisfies these conditions using a different method. However I fail to understand why the method elaborated in the question does not reflect this fact to me??</p>
|
nonuser
| 463,553 |
<p>First notice by letting <span class="math-container">$x=y=0$</span> we get <span class="math-container">$f(0)=0$</span>. Further we have <span class="math-container">$$f(x+y)-f(x) = f(y) +x^2y$$</span> now if <span class="math-container">$y\ne 0$</span> we have <span class="math-container">$$\lim _{y\to 0} {f(x+y)-f(x)\over y} = \lim _{y\to 0} \big({f(y)\over y} +x^2\big)$$</span></p>
<p>so we have <span class="math-container">$f'(x) =x^2$</span> and thus, for some real <span class="math-container">$c$</span> we have <span class="math-container">$$f(x) ={x^3\over 3}+c$$</span>
Since <span class="math-container">$f(0)=0$</span> we get <span class="math-container">$c=0$</span>. As always we have to chek it buy pluging it in starting equation:
<span class="math-container">$${(x+y)^3\over 3} = {x^3\over 3}+{y^3\over 3} +x^2y\implies y^2x=0$$</span>
for all <span class="math-container">$x$</span> and <span class="math-container">$y$</span> which is clearly nosene.</p>
|
471,561 |
<p><img src="https://ukpstq.bn1.livefilestore.com/y2p5St1yRZbdxBvzMbBTYjqjqtwDvBaoWtc7YRGZXCwBTax0XseUIh_l_O92NO6XAbLeGaqU67bkBI4lroIlcD2ade_rxfDast52B_7ECcMd68/question3.png?psid=1" alt="question"></p>
<p>I have attempted these few simple questions, can someone let me know if this is correct please? If not please provide the answer as I learn better that way and if possible explain.</p>
<p>i) FA1 Start = -X1 and FA1 End = +X2</p>
<p>FA2 Start = -Y1 and FA2 End = +Y3</p>
<p>ii) FA1 NOT ACCEPTED and FA2 ACCEPTED</p>
<p>iii) In FA1 the string must begin with 'a' and this is in a loop, the next string is a 'b' followed by another 'b' in a loop. it can then go back round through 'a' and start again but must end with a 'b'</p>
<p>In FA2 it must start with a 'b' and is in a loop, then it is followed by an 'a' and go back through 'b' or continue onto 'a' and then end or go through 'a' in a loop or start the cycle again through 'b' back to the start but must end with an 'a' at y3.</p>
|
dtldarek
| 26,306 |
<p>That's not quite right. Here's a <strong>hint</strong>:</p>
<ul>
<li>What are the necessary suffixes for $L_1$ and $L_2$?</li>
<li>Suppose $w_1 \in L_1$. Does $\mathtt{a}w_1$ or $\mathtt{b}w_1$ belong to $L_1$?</li>
<li>Suppose $w_2 \in L_2$. Does $\mathtt{a}w_2$ or $\mathtt{b}w_2$ belong to $L_2$?</li>
</ul>
<p>I hope this helps $\ddot\smile$</p>
|
4,623,748 |
<p>Let <span class="math-container">$X,Y$</span> be smooth vector fields on the unit circle <span class="math-container">$M = S^1$</span> such that <span class="math-container">$[X,Y] = 0$</span>, i.e. (treating tangent vectors as derivations) <span class="math-container">$X(x)(Yf) = Y(x)(Xf)$</span> for all <span class="math-container">$x\in M$</span>. Assuming that <span class="math-container">$X(x)\ne 0$</span> for all <span class="math-container">$x\in M$</span> I want to show that there is a <span class="math-container">$c\in \Bbb R$</span> such that <span class="math-container">$Y = cX$</span>.</p>
<p>My attempt: for all <span class="math-container">$x\in M$</span> we have <span class="math-container">$X(x), Y(x) \in T_x M$</span> which is one dimensional, and since <span class="math-container">$X(x) \ne 0$</span> there is a <span class="math-container">$c(x)\in \Bbb R$</span> such that <span class="math-container">$Y(x) = c(x) X(x)$</span>. Assuming this <span class="math-container">$c: M\to \Bbb R$</span> is smooth (I don't know how to show this) we get that <span class="math-container">$Y(x)(Xf) = X(x)(Yf) = X(x)(cXf) = X(x)(c) (Xf)(x) + c(x) X(x)(Xf) = X(x)(c)X(x)(f) + Y(x)(Xf)$</span>, so <span class="math-container">$X(x)(c) X(x)(f) = 0$</span>. This holds for all smooth <span class="math-container">$f: M\to \Bbb R$</span>, so since <span class="math-container">$X(x)\ne 0$</span> we get <span class="math-container">$X(x)(c) = 0$</span>. I'm not sure how to proceed or if this is even correct.</p>
|
José Carlos Santos
| 446,262 |
<p>If <span class="math-container">$z^2+|z|=0$</span>, then <span class="math-container">$z^2=-|z|\leqslant0$</span>. The only complex numbers whose square is a real number smaller than or equal to <span class="math-container">$0$</span> are those numbers <span class="math-container">$z$</span> of the form <span class="math-container">$\lambda i$</span>, for some <span class="math-container">$\lambda\in\Bbb R$</span>. But then<span class="math-container">$$z^2+|z|=0\iff-\lambda^2+|\lambda|=0,$$</span>whose only solutions are <span class="math-container">$-1$</span>, <span class="math-container">$0$</span>, and <span class="math-container">$1$</span>. Therefore, the solutions of your equation are <span class="math-container">$-i$</span>, <span class="math-container">$0$</span>, and <span class="math-container">$i$</span>.</p>
|
1,851,698 |
<p>While playing with the results of defining a new operation, I came across a number of interesting properties with little literature surrounding it; the link to my original post is here: <a href="https://math.stackexchange.com/questions/1785715/finding-properties-of-operation-defined-by-x%E2%8A%95y-frac1-frac1x-frac1y">Finding properties of operation defined by $x⊕y=\frac{1}{\frac{1}{x}+\frac{1}{y}}$? ("Reciprocal addition" common for parallel resistors)</a></p>
<p>and as you can see, the operation of interest is $x⊕y = \frac{1}{\frac{1}{x}+\frac{1}{y}} = \frac{xy}{x+y}$. </p>
<p>In wanting to find a condition such that $x⊕y = x-y$, I found that the ratio between x and y mus be φ=1.618... the golden ratio, for this to work!</p>
<p>$x⊕y=x-y$</p>
<p>$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x-y$</p>
<p>$\frac{xy}{x+y} = x-y$</p>
<p>$xy = x^2-y^2$</p>
<p>$0 = x^2-xy-y^2$</p>
<p>and, using the quadratic formula,</p>
<p>$x = \frac{y±\sqrt{y^2+4y^2}}{2}$</p>
<p>$x = y\frac{1±\sqrt{5}}{2}$</p>
<p>$x = φy$</p>
<p>This result is amazing in and of itself. Yet through the same basic setup, we find a new ratio pops out if we try $x⊕y = x+y$ and it is complex. </p>
<p>$x⊕y = x+y$</p>
<p>$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x+y$</p>
<p>$\frac{xy}{x+y} = x+y$</p>
<p>$xy = x^2+2xy+y^2$</p>
<p>$0 = x^2+xy+y^2$</p>
<p>$x = \frac{-y±\sqrt{y^2-4y^2}}{2}$</p>
<p>$x = y\frac{1±\sqrt{-3}}{2}$</p>
<p>$x = y\frac{1±\sqrt{3}i}{2}$</p>
<p>and this is the "imaginary golden ratio"!</p>
<p>$φ_i = \frac{1+\sqrt{3}i}{2}$</p>
<p>It has many properties of the golden ratio, mirrored. This forum from 2011 is the only literature I could dig up on it, and it explains most of the properties I also found and more. <a href="http://mymathforum.com/number-theory/17605-imaginary-golden-ratio.html" rel="noreferrer">http://mymathforum.com/number-theory/17605-imaginary-golden-ratio.html</a></p>
<p>This number is extremely cool, because its mathematical properties mirror φ but also have their own coolness. </p>
<p>$φ_i = 1-\frac{1}{φ_i}$</p>
<p>$φ_i^2 = φ_i - 1$</p>
<p>and generally </p>
<p>$φ_i^n = φ_i^{n-1} - φ_i^{n-2}$</p>
<p>This complex ratio also lies on the unit circle in the complex plane, and has a representation as a power of e! </p>
<p>$φ_i = cos(π/3)+ isin(π/3) = e^{iπ/3}$</p>
<p>$|φ_i|=1$</p>
<p>It is also a nested radical, because of the identity $φ_i^2 + 1 = φ_i$</p>
<p>$φ_i=\sqrt{-1+\sqrt{-1+\sqrt{-1+\sqrt{-1+...}}}}$</p>
<p>Since the only other forum which I could find that has acknowledged the existence of the imaginary golden ratio (other than the context of it as a special case imaginary power of e) I'd like to share my findings and ask if anybody has heard of this ratio before, and if anybody could offer more fine tuned ideas or explorations into the properties of this number. One specific qustion I have involves its supposed connection (according to the 2011 forum) to the sequence </p>
<p>$f_n = f_{n-1} - f_{n-2}$</p>
<p>$f_0=0$</p>
<p>$f_1=1$</p>
<p>$0,1,1,0,-1,-1,0,1,1,...$</p>
<p>could somebody explain to me how this sequence is connected to φ_i? The forum states there is a connection, but I can't figure out what it is based on the wording. What am I missing?</p>
<p>Thanks for your help with my question/exploration. </p>
|
Mitchell Spector
| 350,214 |
<p>$$f_n = (\phi_i^{\;-n}-\phi_i^{\;n})\frac{i}{\sqrt{3}}$$</p>
<p>You can prove this by induction on $n,$ using the fact that $\phi_i$ and $\phi_i^{\;-1}$ are the two solutions to the quadratic equation $x^2-x+1=0.$ (Alternatively, it's easy to see that the three sequences $\langle \phi_i^{\;n}\;\vert\;n \in \mathbb{N}\rangle,$ $\langle \phi_i^{\;-n}\;\vert\;n \in \mathbb{N}\rangle,$ and $\langle f_n \;\vert\;n \in \mathbb{N}\rangle$ are all periodic with period 6, so it's actually sufficient to check the formula above for $n=\;$0, 1, 2, 3, 4, and 5.)</p>
<p>However, a little linear algebra shows what's really going on. The set of all complex-valued sequences satisfying the same recurrence relation as your $f_n$ is closed under pointwise addition and multiplication by a constant, so it forms a vector space $V$ over the field of complex numbers. The two sequences $u_1=\langle \phi_i^{\;n}\;\vert\;n \in \mathbb{N}\rangle$ and $u_2=\langle \phi_i^{\;-n}\;\vert\;n \in \mathbb{N}\rangle$ satisfy the recurrence relation (because $\phi_i$ and $\phi_i^{\;-1}$ are roots of the quadratic equation above), so $u_1$ and $u_2$ belong to $V.$</p>
<p>You can check that $\lbrace u_1,u_2 \rbrace$ is a basis for $V.$ The formula I gave for $f_n$ is just the particular linear combination of $u_1$ and $u_2$ that happens to yield the sequence $\langle f_n \;\vert\;n \in \mathbb{N}\rangle.$</p>
|
1,099,214 |
<p>I'm having trouble calculating some probabilities for a simple card game I'm into </p>
<p>1) So say we have a deck of 13 cards total. You draw 3 cards so there are 10 cards left in the deck. Say there is 1 special card in the deck that we want to figure out the probability of getting it in the hand. How do we figure it out? Would it be:</p>
<p>$$ \dbinom{1}{1} * \dbinom{12}{2} / \dbinom{13}{3} $$</p>
<p>1 choose 1 for the special card. 12 choose 2 for the other 2 cards which we don't care what they are. I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand but that would make the probability greater than 1.</p>
<p>2) Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card? </p>
<p>3) Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc).</p>
<p>I remember doing math like this in high school and it being really basic but don't quite remember exactly how to solve them. Thanks!</p>
|
drhab
| 75,923 |
<p>Hint on 2)</p>
<p>$P(\text{the special AND}\geq1 \text{ copies of...})=P(\text{the special})-P(\text{the special AND 0 copies of...})$</p>
|
238,673 |
<p><a href="http://vihart.com/doodling/" rel="nofollow">Vi Hart's doodling videos</a> and a 4 year old son interested in mazes has made me wonder:</p>
<p>What are some interesting mathematical "doodling" diversions/games that satisfy the following criteria:</p>
<p>1) They are "solitaire" games, i.e. require only one player;
2) They require only a pencil and blank sheet of paper;
3) They don't rely on abstract mathematical language/substitutions. More precisely, I'm interested in "picture" type games with very simple rules, and not the cheeky answer: "mathematics".</p>
<p>What I'd really like to see is a game with simple rules that, out of the planarity of a sheet of paper, somehow "generates" a maze for the player from the "automaton" type rules...A good starting point may be a solitaire version of dots-and-boxes or something like this...</p>
<p>If such a thing is impossible, under certain simple desirable assumptions, I'd like to see proofs of such a fact, too.</p>
|
Rosie F
| 344,044 |
<p>Or run a <a href="https://en.wikipedia.org/wiki/Turmite" rel="nofollow">turmite</a> which moves on a grid of square cells, and never "decreases" the colour of a cell (where this colour is viewed as a number). Represent cell-colours as e.g. $\Box, \boxminus, \boxplus, \blacksquare$ (where every cell starts as $\Box$), so you never need to rub anything out.</p>
<p>For example ($r=90^\circ $ turn clockwise, etc.)</p>
<p>$\begin{array}{c|cc}\text{old} & \text{new} & \text{turn} \\
\Box & \boxminus & r \\ \boxminus & \boxplus & f \\ \boxplus & \blacksquare & u \\ \blacksquare & \blacksquare & f \end{array}$</p>
<p>If we call this $rfuf$ for short (from the above table's right-hand column) then the following are some more 4-colour turmites whose behaviour is not trivial: $rflf, rrlf, rurf, rulf, rlrf$. Unfortunately the most interesting turmites can both increase and decrease the number for a cell's colour, so running such a turmite on paper would need a pencil and a lot of rubbing out. For simplicity, I've kept to 1-state turmites so you don't have to keep track of the turmite's state as you doodle.</p>
|
19,478 |
<p>Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?</p>
|
Bjorn Poonen
| 2,757 |
<p>No, not even for genus $0$ curves, if "<strong>$X$ is defined over $K$</strong>" means that the variety $X$, initially defined over an extension $F$ of $K$, is to be isomorphic to the base extension of some variety over $K$. (Pete in his answer was implicitly assuming that he was allowed to base extend to an algebraic closure of $F$ before going down to $K$.)</p>
<p>For finite extensions $F \supseteq E$ of $\mathbb{Q}_p$, if $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $E$ if and only if $[F:E]$ is odd (because $\operatorname{Br}(E)[2] \to \operatorname{Br}(F)[2]$ is multiplication by $[F:E]$ from $\frac{1}{2} \mathbb{Z}/\mathbb{Z}$ to itself).</p>
<p>So if $F$ is an $A_4$-extension of $k:=\mathbb{Q}_2$ (such an extension exists), and $K$ and $L$ are the subfields of $F$ fixed by two $3$-cycles in $A_4$, and $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $K$, and similarly over $L$, but not over $K \cap L = \mathbb{Q}_2$.</p>
|
973,985 |
<p>I don't know where to start. </p>
<p>I know that the sum of the angles is less than or equal to 180.
but how do i prove this.</p>
|
Michael Hardy
| 11,667 |
<p>It seems what must have been intended is this: Find <em>one</em> triangle in which it is easy to prove that the sum is $180^\circ$.</p>
<p>Then deduce that if it's the same in all other triangles, then it must be $180^\circ$ in all triangles.</p>
<p><b>PS:</b> OK, let's look at the isosceles right triangle. The <b>conventional</b> way to view it would say that since one angle is $90^\circ$ and the other two are equal to each other, and the sum of all three must be $180^\circ$, it follows that the sum of those two must be $90^\circ$, so each must be $45^\circ$. <b>But</b> we cannot do that here because the fact that the sum is $180^\circ$ is what we must <b>prove</b> rather than a fact that we can rely on in this context. So here's what I propose: Draw the diagonal of a square, dividing it into two triangles of the shape just described. The diagonal splits the $90^\circ$ angle at one corner of the square into two parts. Argue that those two parts are congruent, so each must be half of $90^\circ$. Then you have the two $45^\circ$ angles, so the sum is proved to be $180^\circ$.</p>
<p>Finally, if the sum must be the same in all triangles, then $180^\circ$ is it.</p>
|
973,985 |
<p>I don't know where to start. </p>
<p>I know that the sum of the angles is less than or equal to 180.
but how do i prove this.</p>
|
Sherlock Holmes
| 173,514 |
<p>Draw a generic triangle. Produce a line through a vertex parallel to the opposite side. Applying alternate angles, the result is seen. </p>
|
1,038,060 |
<p>Can anyone help me with this question: I know it before, but I have tried to solve it myself and didnt succeed.
what is the regular expression for this language: L=all words that have 00 or 11 but not both.</p>
<p>Thank you!</p>
|
Aaron Maroja
| 143,413 |
<p>As suggested, I am writing the complete answer.</p>
<p>First we use the not so simple result: </p>
<p><strong>Theorem:</strong> Let $n = 3$ or $n\geq 5$. Then $A_n$ is simple. </p>
<p><strong>Statement:</strong> Now for $n=3$ or $n \geq 5$ let's show that $\{id\}, A_n$ and $S_n$ are the only normal subgroups of $S_n$.In particular, $A_n$ is the <strong>only</strong> sugbroup of $S_n$ of index $2$. </p>
<p>Proof: It is clear that $\{id\}, A_n$ and $S_n$ are normal subgroups of $S_n$. Now let $H$ be a normal subgroup of $S_n$ and consider the group homomorphism $$\psi: H \to \{-1,+1\}$$</p>
<p>defined by $$\psi(\alpha)= \begin{cases}1&, \text {if}\ \ \alpha\ \ \text{is even}\\-1&, \text {if}\ \ \alpha\ \ \text{is odd}\end{cases} $$</p>
<p>Naturally, $\ker \psi = H \cap A_n$ and $(H: \ker \psi) = |\psi(H)| = 1$ or $2$. Thus, $(H: H \cap A_n) = 1$ or $2$.</p>
<ul>
<li><strong>$1^{st}$ case:</strong> </li>
</ul>
<p>$(H: H\cap A_n) = 1$, that is, $H \subset A_n$. As $H \lhd S_n$, then a fortiori $H \lhd A_n$, then it follows from the theorem that $H=\{id\}$ or $H=A_n$. </p>
<ul>
<li><strong>$2^{nd}$ case:</strong></li>
</ul>
<p>$(H:H\cap A_n) = 2$. As $H \lhd S_n$, then $H \cap A_n \lhd A_n$, then from the theorem, $H \cap A_n = \{id\}$ or $H \cap A_n = A_n$. Therefore $|H| =2$ or $|H| = S_n$. </p>
<p>Let's suppose $|H|=2$. As $H \cap A_n = \{id\}$, then $H$ contains an odd permutation $\tau$ of order $2$. Such permutation $\tau$ is necessarily a product of disjoint transpositions, say $\tau = (12)\rho_{2}\ldots\rho_{s}$. Then </p>
<p>$$\tau ' = (13)\tau(13)^{-1} \in H , \text{because}\ \ \tau \in H \lhd S_n$$ </p>
<p>As $$\tau ' (2) = [(13)\tau(13)](2) = [(13)\tau](2) = [(13)](1) =3$$</p>
<p>and $\tau(2) =1$, we obtain $\tau \neq \tau '$ and it follows that $|H| \geq 3$. So assuming that $|H| = 2$ leads us to a contradiction and then $H = S_n$. </p>
|
1,038,060 |
<p>Can anyone help me with this question: I know it before, but I have tried to solve it myself and didnt succeed.
what is the regular expression for this language: L=all words that have 00 or 11 but not both.</p>
<p>Thank you!</p>
|
user1729
| 10,513 |
<p>The following idea is super-simple. However, it only works if $n$ is odd (so it works precisely half the time!). But as it is so simple I thought it would be nice to record it here.</p>
<p>Suppose $n$ is odd. Note that $S_n$ can be generated by the elements $\alpha:=(1, 2)$ and $\beta:=(1, 2, \ldots, n)$. Now, every subgroup $H$ of index 2 is the kernel of some homomorphism $S_n\rightarrow \mathbb{Z}_2$. The key point is the following:</p>
<blockquote>
<p>Homomorphisms are defined by the images of the generators. As $\beta:=(1, 2, \ldots, n)$ has odd order, it is killed by every homomorphism $\phi:S_n\rightarrow \mathbb{Z}_2$, so $\beta\in\ker\phi$ for all such $\phi$.</p>
</blockquote>
<p>Therefore, if $\phi$ has non-trivial image it must be <em>precisely</em> the map defined by $\alpha\mapsto 1$, $\beta\mapsto0$. Hence, there is a unique homomorphism $\phi:S_n\rightarrow \mathbb{Z}_2$, and hence $S_n$ contains a unique subgroup of index 2 (for $n$ odd).</p>
|
3,672,839 |
<p>What is the derivative of <span class="math-container">$f(x)=|x|^\frac{3}{2},\forall x\in \mathbb{R}$</span>.</p>
<p>When <span class="math-container">$x>0$</span> it is fine but the problem is when <span class="math-container">$x\leq 0$</span>.I was trying by the defintion of derivative but what is <span class="math-container">$|x|^\frac{3}{2}$</span> when <span class="math-container">$x<0$</span>?</p>
|
Community
| -1 |
<p>We were hit with Zorn's lemma in the first group theory course before any other (this is in India). I never had the pleasure of taking courses in logic and set theory, and ring theory naturally came only later, so this answer is technically different from Asaf's.</p>
<blockquote>
<p><em>Exercise.</em> Let <span class="math-container">$G$</span> be a non-trivial finitely generated group, and <span class="math-container">$H$</span> a proper subgroup of <span class="math-container">$G$</span>. Use Zorn's lemma to show that there is a maximal proper subgroup of <span class="math-container">$G$</span> that contains <span class="math-container">$H$</span>.</p>
</blockquote>
<p>This was our first experience getting punched in the gut by the damn thing (assignment 3 - week 5), and by the end of the course we were all bent over crying for mercy. It took several years, including courses in ring theory and topology, before we were acclimatized to it.</p>
|
136,808 |
<p>Let $A$ be a nonempty subset of $\omega$, the set of natural numbers.
I want to prove this statement:</p>
<blockquote>
<p>If $\bigcup A=A$ then $n\in A \implies n^+\in A$.</p>
</blockquote>
<p>Help...</p>
|
hmakholm left over Monica
| 14,366 |
<p>$\bigcup A\subseteq A$ says that $A$ is transitive and is therefore an ordinal. Now if $A$ were a successor ordinal $\alpha+1 = \alpha\cup\{\alpha\}$, then $\alpha\in A$ but $\bigcup A = (\bigcup\alpha)\cup\alpha \not\ni \alpha$. Thus $A$ must be either $0$ or $\omega$.</p>
<hr>
<p>Or more directly: Assume $n\in A$. Then $n\in\bigcup A$, that is, there exits $y$ such that $n\in y \in A$. Then $n^+ \le y$. In the case $n^+=y$ we have $n\in A$ directly. Otherwise $n^+\in y\in A$ so $n^+\in\bigcup A$.</p>
|
2,867,718 |
<p>How do I integrate $\sin\theta + \sin\theta \tan^2\theta$ ?</p>
<p>First thing,
I have been studying maths for business for approximately 3 months now. Since then, I studied algebra and then I started studying calculus. Yet, my friend stopped me there, and asked me to study Fourier series as we'll need it for our incoming projects. So I feel that I am missing a lot of things as I haven't studied integrals yet. </p>
<p>Today, I encountered this solution that I couldn't understand at all. </p>
<p>Apparently, $\int(\sin\theta + \sin\theta \tan^2\theta)d\theta = \int(\sin\theta (1 + \tan^2\theta))d\theta$. </p>
<p>Below, a link to the solution at 5:08.
<a href="https://youtu.be/aw_VM_ZDeIo" rel="nofollow noreferrer">https://youtu.be/aw_VM_ZDeIo</a></p>
<p>He stated the we have to learn the integral identities. So, I started searching the whole internet looking for them. But, I think I couldn't find them. The only thing that I found was something called Magic hexagon. I thought of reading about $\theta$ as it might mean something. But, after all I learned that it is just a regular greek letter used as a variable.</p>
|
PM.
| 416,252 |
<p>$$
\sin\theta+\sin\theta\tan^2\theta=\sin\theta(1+\tan^2\theta)=\sin\theta\sec^2\theta=\sec\theta\tan\theta
$$
because $(1+\tan^2\theta)=\sec^2\theta$ is a 'standard' identity.</p>
<p>Also
$$
\frac{d\sec\theta}{d\theta}=\sec\theta\tan\theta
$$
is a 'standard' derivative.</p>
<p>Therefore
$$
\int(\sin\theta+\sin\theta\tan^2\theta)\, d\theta = \int \sec\theta\tan\theta\, d\theta=\sec\theta+C
$$
where $C$ is an arbitrary constant</p>
|
1,905,308 |
<p>In the book "A Course in Metric Geometry" (By Dmitri Burago, Yuri Burago Sergei Ivanov), there is a short proof of lower semicontinuity of length induced by a metric: (Prop 2.3.4, <a href="https://books.google.com/books?id=dRmIAwAAQBAJ&pg=PA35" rel="nofollow noreferrer">pg 35</a>)</p>
<p>Let $\gamma_j:[a,b] \to (X,d)$ converge pointwise to $\gamma$. Fix $\epsilon >0$.Let $Y=\{y_1,...y_N \}$ be a partition of $\gamma$ such that $\Sigma(Y) \ge L(\gamma)-\epsilon$, where $$\Sigma(Y)=\sum_i d(\gamma(y_i),\gamma(y_{i+1})).$$
Define $\Sigma_j(Y)=\sum_i d(\gamma_j(y_i),\gamma_j(y_{i+1}))$, and choose $j$ large enough so that the inequality $d(\gamma(y_i),\gamma_j(y_i))< \epsilon$ hols for all $y_i \in Y$. Then:</p>
<p>$$ L(\gamma) \le \Sigma(Y) + \epsilon \le \Sigma_j(Y) + 2N\epsilon +\epsilon \le L(\gamma_j)+(2N+1)\epsilon$$ where the $2N\epsilon$ factor comes from going through $\gamma(y_i) \to \gamma_j(y_i) \to \gamma_j(y_{i+1}) \to \gamma_j(y_i) \to \gamma_(y_{i+1})$.</p>
<p>The authors then say we take $\epsilon \to 0$ and finish. </p>
<p><strong>My problem:</strong></p>
<p>$N$ is in fact a function of $\epsilon$. How can I be sure that $N(\epsilon)\epsilon \to 0$ when $\epsilon \to 0$.</p>
<p>More formally, we can think of $N(\epsilon)$ as the <strong>minimal</strong> size of a partition which induces an $\epsilon$-approximation of $\gamma$. clearly this quantity increases with $\epsilon$, and without any knowledge on how "wild" is $\gamma$ I see no way how to gain control over it.</p>
|
David C. Ullrich
| 248,223 |
<p>This is not an answer to your specific question about the proof in the book. It seems worthwhile to point out that the lower semicontinuity of arc length is much more trivial than that proof makes it appear - the proof you sketch seems to me to be sort of missing the point.</p>
<p>Since $\sum_j(Y)$ is just a finite sum it is clear that $\sum(Y)=\lim_j\sum_j(Y)$. And $\sum_j(Y)\le L(\gamma_j)$ by definition. So $$\sum(Y)=\liminf_j\sum_j(Y)\le\liminf_j L(\gamma_j).$$Hence the definition of $L$ shows that $$L(\gamma)=\sup_Y\sum(Y)\le\liminf_j L(\gamma_j).$$</p>
|
765,738 |
<p>I am trying to prove the following inequality:</p>
<p>$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$</p>
<p>for all $a>0, b>0$.</p>
<p>Does anyone know how to prove it?</p>
<p>Thanks a lot in advance!</p>
|
egreg
| 62,967 |
<p>It's not restictive to assume $a>b$, so we can write $a=e^{2s}$, $b=e^{2t}$, with $s>t$ and the inequality to prove becomes
$$
4(e^s-e^t)^2\le(e^s-e^t)(e^s+e^t)(2s-2t)
$$
or
$$
\frac{e^s-e^t}{s-t}\le\frac{e^s+e^t}{2}
$$
By Lagrange's theorem, we know that
$$
\frac{e^s-e^t}{s-t}=e^u
$$
for some $u$, $t<u<s$, so we want
$$
e^u\le \frac{e^s+e^t}{2}
$$
which is true for all $u\in(t,s)$ by the convexity of the exponential function.</p>
|
2,359,408 |
<p>Question:</p>
<p>Let $\\f: \ \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ via $ f(x) = (x+2, x-3)$. Is $f$ injective? Is $f$ surjective?</p>
<p>I was able to prove that $f$ is injective. However, I am not quite sure if $f$ is surjective. If it is surjective could someone please tell me how to prove that. If not, could someone provide a counter example.</p>
<p>Thanks </p>
|
Siong Thye Goh
| 306,553 |
<p>Hint:</p>
<p>Let $y=f(x)=( y_1 , y_2 ),$ notice that $y_1-y_2=5$</p>
<p>Now consider point $(6,0)$, does it have a preimage?</p>
|
2,359,408 |
<p>Question:</p>
<p>Let $\\f: \ \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ via $ f(x) = (x+2, x-3)$. Is $f$ injective? Is $f$ surjective?</p>
<p>I was able to prove that $f$ is injective. However, I am not quite sure if $f$ is surjective. If it is surjective could someone please tell me how to prove that. If not, could someone provide a counter example.</p>
<p>Thanks </p>
|
freakish
| 340,986 |
<p>Well, you need to show that for any $(a,b)\in\mathbb{R}^2$ there exists $x\in\mathbb{R}$ such that $f(x)=(a,b)$. Since $f(x)=(x+2, x-3)$ then $a=x+2$ and $b=x-3$. In particular $x=a-2$ and $x=b+3$ and therefore $a-2=b+3$ and finally $a=b+5$. So that's the constraint on $(a,b)$ in order for it to be in the image of $f$. Finding a counterexample is easy now, e.g. $(a,b)=(0,0)$.</p>
|
548,563 |
<p>Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$</p>
<p>I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$</p>
<p>I have no idea about it. :|</p>
|
Bruno Joyal
| 12,507 |
<p>Recall the Euler Beta integral</p>
<p>$$\beta(a, b) = \int_0^1 x^{a-1} (1-x)^{b-1} dx$$</p>
<p>and Euler's formula for it in terms of the Gamma function,</p>
<p>$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$</p>
<p>In particular, since ${2n \choose n} = (2n!)/(n!)^2$, we have</p>
<p>$$\beta(n+1, n+1) = \frac{1}{(2n+1){2n \choose n}}.$$</p>
<p>Thus, </p>
<p>$$\sum_{n=0}^\infty \frac{1}{2n \choose n} = \int_0^1 \sum_{n=0}^\infty {(2n+1) (x(1-x))^{n}} dx$$</p>
<p>and we have the series $\sum_{n\geq 0} (2n+1)y^n = (y+1)/(y-1)^2$. Therefore,</p>
<p>$$\sum_{n=0}^\infty \frac{1}{2n \choose n} = \int_0^1 \frac{x(1-x)+1}{(x(1-x)-1)^2} dx$$</p>
<p>and by routine integration (partial fractions or your favorite standard method), this <a href="http://www.wolframalpha.com/input/?i=integrate%20%28x%281-x%29%2b1%29/%28x%281-x%29-1%29%5E2%20dx%20from%20x=0%20to%201" rel="nofollow">equals</a> $\frac{2}{27}(18+\sqrt 3 \pi)$.</p>
|
1,867,352 |
<p>Find all the angles $v$ between $-\pi$ and $\pi$ such that</p>
<p>$$-\sin(v)+ \sqrt3 \cos(v) = \sqrt2$$</p>
<p>The answer has to be in the form of: $\pi/2$ (it must include $\pi$)</p>
<p>I have tried squaring but I get nowhere.</p>
|
bthmas
| 222,365 |
<p>As noted by another user, it's useful to combine a sum of sines and cosines into a single function.
$$
\sqrt{3}\cos(v) - \sin(v) = \sqrt{2}
$$
In order to turn the LHS into a single function we have a bit of a choice here we can make the single function a sine or a cosine, and we can also choose if we want an addition or subtraction in the angle.</p>
<p>For this I'll show how to do it with a subtraction in the angle and a cosine function.</p>
<p>Recall from the angle difference formula for cosine
$$
\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)
$$
Now we set the equality
$$
\begin{align}
\sqrt{3}\cos(v) - \sin(v) &= A\cos(v -\theta)\\
\sqrt{3}\cos(v) - \sin(v) &= A \left [ \cos(v)\cos(\theta) + \sin(v)\sin(\theta) \right ]\\
\sqrt{3}\cos(v) - \sin(v) &= A\cos(\theta)\cos(v) + A\sin(\theta)\sin(v)
\end{align}
$$
From here we equate the coefficients
$$
\begin{align}
A\cos(\theta) &= \sqrt{3}\\
A\sin(\theta) &= -1
\end{align}
$$
Solving for $\theta$ gives
$$
\tan(\theta) = \frac{-1}{\sqrt{3}} \implies \theta = \frac{-\pi}{6}
$$
Solving for $A$ gives
$$
A\cos \left ( \frac{-\pi}{6} \right ) = \sqrt{3} \implies A\frac{\sqrt{3}}{2} = \sqrt{3} \implies A = 2
$$</p>
<p>Therefore,
$$
\sqrt{3}\cos(v) - \sin(v) = 2\cos(v +\frac{\pi}{6})
$$</p>
<p>Just make this substitution, and I'm sure you can do the rest.</p>
|
1,910,085 |
<p>For all integers $n \ge 0$, prove that the value $4^n + 1$ is not divisible by 3.</p>
<p>I need to use Proof by Induction to solve this problem. The base case is obviously 0, so I solved $4^0 + 1 = 2$. 2 is not divisible by 3.</p>
<p>I just need help proving the inductive step. I was trying to use proof by contradiction by saying that $4^n + 1 = 4m - 1$ for some integer $m$ and then disproving it. But I'd rather use proof by induction to solve this question. Thanks so much.</p>
|
BenPen
| 365,278 |
<p>I can see a superset proof: If you transform it into $2^{2n} + 1$, it's a well known proof that shows it's not divisible by 3. (maybe the original answer gave it away too much...)</p>
|
245,623 |
<p><em>For the following vectors $v_1 = (3,2,0)$ and $v_2 = (3,2,1)$, find a third vector $v_3 = (x,y,z)$ which together build a base for $\mathbb{R}^3$.</em></p>
<p>My thoughts:</p>
<p>So the following must hold:</p>
<p>$$\left(\begin{matrix}
3 & 3 & x \\
2 & 2 & y \\
0 & 1 & z
\end{matrix}\right)
\left(\begin{matrix}
{\lambda}_1 \\
{\lambda}_2 \\
{\lambda}_3
\end{matrix}\right) =
\left(\begin{matrix}
0 \\
0 \\
0
\end{matrix}\right)
$$</p>
<p>The gauss reduction gives</p>
<p>$$
\left(\begin{matrix}
3 & 3 & x \\
0 & 1 & z \\
0 & 0 & -\frac{2}{3}x+y
\end{matrix}\right)
$$</p>
<p>(but here I'm not sure if I'm allowed to swap the $y$ and $z$ axes)</p>
<p>For ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$, this gives me</p>
<p>$$
x = 0 \\
y = 0 \\
z = 0
$$</p>
<p>Is this third vector $v_3$ building a base of $\mathbb{R}^3$ together with the other two vectors? If not, where are my mistakes?</p>
|
Gerry Myerson
| 8,269 |
<p>The big mistake is at the very beginning --- there is no reason at all why you should want that equation to hold. </p>
<p>There are infinitely many correct choices for $v_3$. One simple one is the cross product of $v_1$ and $v_2$ (warning --- this choice won't be available in other vector spaces). </p>
|
245,623 |
<p><em>For the following vectors $v_1 = (3,2,0)$ and $v_2 = (3,2,1)$, find a third vector $v_3 = (x,y,z)$ which together build a base for $\mathbb{R}^3$.</em></p>
<p>My thoughts:</p>
<p>So the following must hold:</p>
<p>$$\left(\begin{matrix}
3 & 3 & x \\
2 & 2 & y \\
0 & 1 & z
\end{matrix}\right)
\left(\begin{matrix}
{\lambda}_1 \\
{\lambda}_2 \\
{\lambda}_3
\end{matrix}\right) =
\left(\begin{matrix}
0 \\
0 \\
0
\end{matrix}\right)
$$</p>
<p>The gauss reduction gives</p>
<p>$$
\left(\begin{matrix}
3 & 3 & x \\
0 & 1 & z \\
0 & 0 & -\frac{2}{3}x+y
\end{matrix}\right)
$$</p>
<p>(but here I'm not sure if I'm allowed to swap the $y$ and $z$ axes)</p>
<p>For ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$, this gives me</p>
<p>$$
x = 0 \\
y = 0 \\
z = 0
$$</p>
<p>Is this third vector $v_3$ building a base of $\mathbb{R}^3$ together with the other two vectors? If not, where are my mistakes?</p>
|
Adam Rubinson
| 29,156 |
<p>Well (3,2,0) and (3,2,1) give you (0,0,1). So for your third one (0,1,0) would work.</p>
<p>Then (1,0,0) = 1/3 [(3,2,0) - 2(0,1,0)]</p>
|
1,207,247 |
<blockquote>
<p>Prove that if $\gcd(a,b)=1$, then $\gcd(a^2+b^2, a^2b^2)=1$.</p>
</blockquote>
<p>My attempt:</p>
<p>If $a$ is prime to $b$ the $gcd(a,b)=1$. Assume that $a^2+b^2$ and $a^2b^2$ are not prime to each other. Let $d=gcd(a^2+b^2, a^2b^2)$. Then $d|a^2+b^2, ~~ \&~~d|a^2b^2$. We shall have to prove that $d=1$.</p>
<p>How to prove that $d=1$?</p>
|
Awais Chishti
| 223,303 |
<p>We have $d|a^2(a^2+b^2)-a^2b^2$, $d|a^4$ and by symmetry, $d|b^4$. So $d|gcd(a^4,b^4)=gcd(a,b)^4=1$</p>
|
577,163 |
<p>Let $A=(a_{ij})_{n\times n}$ such
$a_{ij}>0$ and $\det(A)>0$.</p>
<p>Defining the matrix $B:=(a_{ij}^{\frac{1}{n}})$, show that $\det(B)>0?$.</p>
<p>This problem is from my friend, and I have considered sometimes, but I can't. Thank you </p>
|
user1551
| 1,551 |
<p>The statement is false. Counterexample:
$$
B=\pmatrix{2&2&2\\ 3&4&2\\ 3&1&4},\ A=B\circ B\circ B=\pmatrix{8&8&8\\ 27&64&8\\ 27&1&64},\ \det(B)=-2,\ \det(A)=7000.
$$</p>
<p><strong>Edit.</strong> However, if $B$ is positive definite, the <em>converse</em> of your statement is true. More specifically, suppose $B$ is positive definite (whether it is entrywise positive is unimportant) and $A$ is the $n$-fold Hadamard product of $B$ (i.e. $A=(b_{ij}^n)$). Then $\det(A)>0$. This is because for positive semidefinite matrices $X$ and $Y$, we have $\det(X\circ Y)\ge\det(X)\det(Y)$.</p>
|
1,545,045 |
<p>Is it possible to cut a pentagon into two equal pentagons?</p>
<p>Pentagons are not neccesarily convex as otherwise it would be trivially impossible.</p>
<p>I was given this problem at a contest but cannot figure the solution out, can somebody help?</p>
<p>Edit: Buy "cut" I mean "describe as a union of two pentagons which intersect only on the boudaries".</p>
|
Graham Kemp
| 135,106 |
<p>Since gender doesn't restrict anything, and people are unique individuals, we just need to count ways to divide twenty-eight people into fourteen pairs.</p>
<p>There are $28!$ ways to line everyone up, then split into $14$ pairs. However, each pair can be formed in $2!$ ways, and we don't care about the $14!$ ways to arrange the teams either.</p>
<blockquote class="spoiler">
<p> $$\dfrac{28!}{{2!}^{14}\,14!}$$</p>
</blockquote>
|
1,668,487 |
<p>$$2^{-1} \equiv 6\mod{11}$$</p>
<p>Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?</p>
<p>$$6^{-1} \equiv 2\mod11$$</p>
|
5xum
| 112,884 |
<p>It's because $2\cdot 6 \equiv 1 \mod 11$</p>
|
2,683,788 |
<blockquote>
<p>An urn contains $nr$ balls numbered $1,2..,n$ in such a way that $r$
balls bear the same number $i$ for each $i=1,2,...n$. $N$ balls are drawn at random without replacement. Find the probability that exactly $m$ of the numbers will appear in the sample.</p>
</blockquote>
<p>Any hints would be great, I tried solving it, finally relented and checked the solution given in the text, I can't seem to understand the working though I get the idea that inclusion-exclusion is the key to solving the problem.</p>
|
drhab
| 75,923 |
<p>Think of it as if the socks are picked up one by one.$$P(bbb)+P(bbg)+P(bgb)+P(gbb)=$$$$\frac{18}{34}\frac{17}{33}\frac{16}{32}+\frac{18}{34}\frac{17}{33}\frac{16}{32}+\frac{18}{34}\frac{16}{33}\frac{17}{32}+\frac{16}{34}\frac{18}{33}\frac{17}{32}=$$$$\frac{18}{34}\frac{17}{33}\frac{16}{32}+3\cdot\frac{18}{34}\frac{17}{33}\frac{16}{32}=4\cdot\frac{18}{34}\frac{17}{33}\frac{16}{32}=\frac6{11}$$</p>
|
2,294,585 |
<p>I can't figure out why $((p → q) ∧ (r → ¬q)) → (p ∧ r)$ isn’t a tautology. </p>
<p>I tried solving it like this: </p>
<p>$$((p ∧ ¬p) ∨ (r ∧ q)) ∨ (p ∧ r)$$
resulting in $(T) ∨ (p ∧ r)$ in the end that should result in $T$. What am I doing wrong?</p>
|
DanielV
| 97,045 |
<p>$$((p \Rightarrow q) \land (r \Rightarrow ¬q)) \Rightarrow (p ∧ r)$$</p>
<p>If it is January, then I am cold.</p>
<p>If it is July, then I am not cold.</p>
<p>Therefore, it is January and July.</p>
<p>Would you accept that?</p>
|
2,473,132 |
<p>Compute $$ \int_{0}^{ \infty }\frac{\ln xdx}{x^{2}+ax+b}$$
I tried to compute the indefinite integral by factorisation and partial fraction decomposition but it became nasty pretty soon. There must be another way to directly evaluate it without actually computing the indefinite integral which I don't know!</p>
|
pisco
| 257,943 |
<p>In order for your integral to converge, we must have $b>0$.</p>
<p>Hence
$$\begin{aligned}\int_0^\infty {\frac{{\ln x}}{{{x^2} + ax + b}}dx} &= \sqrt b \int_0^\infty {\frac{{\ln (\sqrt b x)}}{{b{x^2} + a\sqrt b x + b}}dx} \\ &= \sqrt b \ln \sqrt b \int_0^\infty {\frac{1}{{b{x^2} + a\sqrt b x + b}}dx} + \sqrt b \underbrace{\int_0^\infty {\frac{{\ln x}}{{b{x^2} + a\sqrt b x + b}}dx}}_{=0} \end{aligned}$$</p>
<p>The last integral is zero by making the subsitution $x\mapsto 1/x$. I think you will have no difficulty in evaluating the remaining integral. Although there are a lot of cases to discuss (depending on $a,b$, the integral might diverge, involve logarithm, or involve arc-tangent).</p>
|
4,013,638 |
<p>a) <span class="math-container">$F(x)>0$</span> for every <span class="math-container">$x>0$</span>.<br>
b) <span class="math-container">$F(x)>F(e)$</span> for every <span class="math-container">$x>1$</span>.<br>
c) <span class="math-container">$x=e^{-2}$</span> is an inflection point. <br>
d) None of the answers are correct. <br>
<br>
Note: This was a question in my exam and I wrote it from my memory so if there's something that doesn't make sense it might be that I messed up. <br>
My Work was to take second derivative of <span class="math-container">$F$</span> and saw that <span class="math-container">$F''(e^{-2})\ne 0$</span>, so I took out that option, and didn't know how to decide between (a) or (b), but I noticed that for (a) I might have to flip the integral, so I just went with (b), I would appreciate any explanation on how to really approach this question and not just with my intuition. (and know if I got it right too).</p>
|
VIVID
| 752,069 |
<ul>
<li>(<span class="math-container">$a$</span>) is wrong because <span class="math-container">$F(1) = 0$</span>.</li>
<li>(<span class="math-container">$b$</span>) is wrong because <span class="math-container">$F'(x)=x^{\sqrt {x}} > 0$</span> so the function is increasing.</li>
<li>(<span class="math-container">$c$</span>) is wrong as you have checked.</li>
<li>(<span class="math-container">$d$</span>) therefore, is the correct option.</li>
</ul>
<p><strong>Edit:</strong> (<span class="math-container">$c$</span>) option appears to be correct as @Bernard has shown. I was just convinced by the claim in the question.</p>
|
86,657 |
<p>Throughout my upbringing, I encountered the following annotations on Gauss's diary in several so-called accounts of the history of mathematics:</p>
<blockquote>
<p>"... A few of the entries indicate that the diary was a strictly private affair of its author's (sic). Thus for July 10, 1796, there is the entry</p>
<p>ΕΥΡΗΚΑ! num = Δ + Δ + Δ.</p>
<p>Translated , this echoes Archimedes' exultant "Eureka!" and states that every positive integer is the sum of three triangular numbers—such a number is one of the sequence 0, 1, 3, 6, 10, 15, ... where each (after 0) is of the form $\frac{1}{2}n(n+1)$, $n$ being a positive integer. Another way of saying the same thing is that every number of the form $8n+3$ is a sum of three odd squares... It is not easy to prove this from scratch.</p>
<p>Less intelligible is the cryptic entry for October 11, 1796, "Vicimus GEGAN." What dragon had Gauss conquered this time? Or what giant had he overcome on April 8, 1799, when he boxes REV. GALEN up in a neat rectangle? Although the meaning of these is lost forever the remaining 144 are for the most part clear enough."
"</p>
</blockquote>
<p>The preceding paragraphs have been quoted <i>verbatim</i> from J. Newman's The World of MATHEMATICS (Vol. I, pages 304-305) and the questions that I pose today were motivated from my recent spotting of [2]:</p>
<ul>
<li><p>Why is there no mention whatsoever to the REV. GALEN inscription in either Klein's or Gray's work?</p></li>
<li><p>What is the reason that E. T. Bell expressed that Gauss had written the Vicimus GEGAN entry on October 11, 1796? According to Klein, Gray, and (even) the Wikipedians it was written on October 21, 1796. As far as I understand, Klein and Gray are just reporting the dates that appear on the original manuscript. Did Bell actually go over it?</p></li>
<li><p>Last but not least, is there a compendium out there of all known potential explanations to the Vicimus GEGAN enigma? The only ones whereof I have notice can be found on page 112 of [1]:</p></li>
</ul>
<blockquote>
<p>"... Following a suggestion of Schlesinger [Gauss, Werke, X.1, part 2, 29], Biermann ... proposed that GA stood for Geometricas, Arithmeticas, so reading GEGAN in reverse as Vicimus N[exum] A[rithmetico] G[eometrici cum] E[xspectationibus] G[eneralibus]. Schumann has since proposed other variants; including, for GA, (La) G(rangianae) A(nalysis)..."</p>
</blockquote>
<p>Heartfelt thanks for your comments, reading suggestions, and replies.</p>
<p><strong>References</strong></p>
<ol>
<li>J. J. Gray. " A commentary on Gauss's mathematical diary, 1796-1814, with an English translation". <em>Expo. Math.</em> 2 (1984), 97-130.</li>
<li>F. Klein. "Gauß' wissenschaftliches Tagebuch 1796–1814". <em>Math. Ann.</em> 57 (1903), 1–34.</li>
<li>M. Perero. Historia e Historias de Matemáticas. Grupo Editorial Iberoamérica, 1994, pág. 40.</li>
</ol>
|
Kristal Cantwell
| 1,098 |
<p>I believe I have a reference for Schuhmann:</p>
<p>Schuhmann E. 1976 Vicimus GEGAN, Interpretationsvarianten zu einer Tagebuchnotiz von C.F. Gauss, Naturwiss. Tech. Medezin. 13.2, 17-20.</p>
|
1,139,723 |
<p>Given the differential equation $\frac{dy}{dt}+a(t)y=f(t)$ </p>
<p>with a(t) and f(t) continuous for:</p>
<p>$-\infty<t<\infty$</p>
<p>$a(t) \ge c>0$</p>
<p>$lim_{t_->0}f(t)=0$</p>
<p>Show that every solution tends to 0 as t approaches infinity.</p>
<p>$$\frac{dy}{dt}+a(t)y=f(t)$$</p>
<p>$$μ(t)=e^{\int a(t)dt}$$
$$μ(t)=e^{\frac{1}{2}a^2(t)}$$</p>
<p>Multiplying both sides of the equation by $μ(t)$:</p>
<p>$$\frac{dy}{dt}+a(t)y=f(t)$$
$$μ(t)[\frac{dy}{dt}+a(t)y]=μ(t)[f(t)]$$
$$e^{\frac{1}{2}a^2(t)}[\frac{dy}{dt}+a(t)y]=e^{\frac{1}{2}a^2(t)}[f(t)]$$
$$\frac{d}{dt}e^{\frac{1}{2}a^2(t)}y=e^{\frac{1}{2}a^2(t)}[f(t)]$$
$$e^{\frac{1}{2}a^2(t)}y=\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$ </p>
<p>Did I make a mistake in the integrating factor step? I am not sure how to proceed..</p>
|
Victor
| 213,782 |
<p>After your last step:</p>
<p>$$e^{\frac{1}{2}a^2(t)}y=\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$</p>
<p>You want to divide both sides by $e^{\frac{1}{2}a^2(t)}$ to solve for $y$</p>
<p>Giving us:</p>
<p>$$y=e^{-\frac{1}{2}a^2(t)}\int e^{\frac{1}{2}a^2(t)}[f(t)]dt$$</p>
|
3,129,852 |
<p>My question is pretty basic.</p>
<p>Here it goes:</p>
<blockquote>
<p>Is it always true that <span class="math-container">$$\prod_{j=1}^{w}{(2s_j + 1)} \equiv 1 \pmod 4$$</span>
where the <span class="math-container">$s_j$</span>'s are positive integers, and may be odd or even?</p>
</blockquote>
<p>We can perhaps assume that <span class="math-container">$w \geq 10$</span>.</p>
<p>I am thinking that it might be possible to disprove this, but I cannot think of a counterexample at this moment.</p>
<p><strong>MY ATTEMPT</strong></p>
<p>How about
<span class="math-container">$$w=10$$</span>
<span class="math-container">$$s_1 = s_2 = s_3 = s_4 = s_5 = s_6 = s_7 = s_8 = s_9 = 2$$</span>
and
<span class="math-container">$$s_{10} = 1$$</span>
for a counterexample?</p>
|
Carl Schildkraut
| 253,966 |
<p>Each factor on the left hand side may be either <span class="math-container">$\equiv 1$</span> or <span class="math-container">$\equiv 3\bmod 4$</span>. If there are an odd number of the latter (i.e. an odd number of odd <span class="math-container">$s_j$</span>), the product will be <span class="math-container">$\equiv 3\bmod 4$</span>. </p>
|
2,055,621 |
<p>Prove or disprove the following proposition: </p>
<blockquote>
<p>There are no positive integers $x$ and $y$ such that:
$x^2−3xy+2y^2=10$</p>
</blockquote>
<p>Thanks!</p>
|
Erik M
| 42,176 |
<p>Notice that $x^2−3xy+2y^2=(x-2y)(x-y)=10$.</p>
|
2,055,621 |
<p>Prove or disprove the following proposition: </p>
<blockquote>
<p>There are no positive integers $x$ and $y$ such that:
$x^2−3xy+2y^2=10$</p>
</blockquote>
<p>Thanks!</p>
|
Behrouz Maleki
| 343,616 |
<p>$$(x-2y)(x-y)=10=2\times 5=5\times 2=-2\times (-5)=-5\times (-2)$$
or
$$(x-2y)(x-y)=10=1\times 10=10\times 1=-1\times (-10)=-10\times (-1)$$
For example
\begin{cases}
x-2y=2\\
x-y=5
\end{cases}
we have $x=8$ and $y=3$</p>
|
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