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108,277 | <p>If Mochizuki's proof of abc is correct, why would this provide a new proof of FLT?</p>
<p>Edit: In proof of asymptotic FLT, does Mochizuki claim a specific value of n and if so what is this value?</p>
| José Hdz. Stgo. | 1,593 | <p>Let us suppose that <span class="math-container">$x^{n}+y^{n}= z^{n}$</span> with <span class="math-container">$x, y,$</span> and <span class="math-container">$z$</span> relatively prime. By the <em>abc</em> conjecture, <span class="math-container">$|x^{n}|\ll |xyz|^{1+\epsilon}$</span>, <span class="math-container">$|y^{n}|\ll |xyz|^{1+\epsilon}$</span>, and <span class="math-container">$|z^{n}|\ll |xyz|^{1+\epsilon}$</span>. Therefore, <span class="math-container">$|xyz|^{n}\ll |xyz|^{3+\epsilon}$</span> which implies that, for <span class="math-container">$|xyz|>1$</span>, <span class="math-container">$n$</span> is bounded. Unfortunately, this establishes only an <em>asymptotic</em> version of FLT. Nevertheless, if we had explicit information regarding the implied constant in the <em>abc</em> conjecture, we could in principle determine explicit upper bounds for those <span class="math-container">$n$</span>'s for which the <em>abc</em> conjecture doesn't settle FLT.</p>
<p><strong>UPDATE (October 26th, 2021)</strong>. In the abstract of "Explicit Estimates in Inter-universal Teichmuller Theory", Mochizuki et. al. we read this:</p>
<blockquote>
<p>In the final paper of a series of papers concerning <em>inter-universal Teichmüller theory</em>, Mochizuki verified various <em>numerically
non-effective versions of the Vojta, ABC, and Szpiro Conjectures</em> over
number fields. In the present paper, we obtain various <em>numerically effective versions</em> of Mochizuki’s results... These numerically effective versions imply <em>effective diophantine results</em> such as an <em>effective
version of the ABC inequality over mono-complex number fields</em> [i.e.,
the rational number field or an imaginary quadratic field] and <em>effective
versions of conjectures of Szpiro</em>. We also obtain an explicit estimate
concerning “Fermat’s Last Theorem” (FLT)—i.e., to the effect that
FLT holds for prime exponents greater than <span class="math-container">$1.615\cdot 10^{14}$</span>—which is sufficient to give an alternative proof of the first case of FLT. In the second case of FLT, if one combines the techniques of the present paper with a recent estimate due to <a href="https://www.sciencedirect.com/science/article/abs/pii/S0022314X2100055X" rel="nofollow noreferrer">Mihăilescu</a>, then the lower bound <span class="math-container">$1.615 \cdot 10^{14}$</span> can be improved to <span class="math-container">$3.35\cdot 10^{9}$</span>.</p>
</blockquote>
<p>If I understand correctly, <a href="https://www.kurims.kyoto-u.ac.jp/%7Emotizuki/papers-english.html" rel="nofollow noreferrer">this paper was uploaded</a> to Mochizuki's homepage four months ago. I dared to update my reply because I consider that the lines above give a definite answer to the OP's second question (regardless of what the generalized opinion about IUTT is).</p>
|
3,657,751 | <p>Consider the series <span class="math-container">$$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$</span> This is clearly not absolutely convergent. On the other hand, obvious choice, alternating series does not work here. Seems like the partial sum sequence is bounded but it is not monotone.</p>
<p>How can we prove that this series converges? and, where does it converge to? </p>
| Gary | 83,800 | <p>Observe that
<span class="math-container">$$
\left| {\sum\limits_{n = 1}^N {( - 1)^{T_n + 1} } } \right| \le 2
$$</span>
for any <span class="math-container">$N\geq 1$</span>. Also <span class="math-container">$\frac{1}{n}\to 0$</span> monotonically. Hence, by Dirichlet's test, the series converges. The limit is
<span class="math-container">\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{T_n + 1} }}{n}} & = \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^{2N} {\frac{{( - 1)^{T_n + 1} }}{n}} = \mathop {\lim }\limits_{N \to + \infty } \left( {\sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n - 1}}} + \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n}}} } \right)
\\ &
= \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n - 1}}} + \mathop {\lim }\limits_{N \to + \infty } \sum\limits_{n = 1}^N {\frac{{( - 1)^{n + 1} }}{{2n}}} = \frac{\pi }{4} + \frac{1}{2}\log 2.
\end{align*}</span></p>
|
23,942 | <p>I have tried to resolve the problem of the following link <a href="https://mathematica.stackexchange.com/questions/23931/how-can-i-solve-precision-problem">How can I solve precision problem</a></p>
<p>I can tell the problem described in that link shortly here, It's no mater how many precision is there after decimal(.) the result should be 2 or 3 precision after decimal as per user requirement.e.g</p>
<pre><code>N[10/3]
3
</code></pre>
<p>Output should be as per user wish precision as 2.</p>
<pre><code>3.33
3.00
</code></pre>
<p>There I found <code>NumberForm[]</code> function as a solution, since the <code>NumberForm[]</code> function is wrap up the result, I can't get the result as number out of it.</p>
<p>I tried like this </p>
<pre><code>NumberForm[N[10/3],{4,2}]
NumberForm[3, {4, 2}]
</code></pre>
<p>Output am getting is </p>
<pre><code>3.33
3.00
</code></pre>
<p>But when I try to do further operation, it's not computing</p>
<pre><code>NumberForm[N[10/3], {4, 2}] + 1
NumberForm[3, {4, 2}] + 1
</code></pre>
<p>Output</p>
<pre><code>1+3.33
1+3.00
</code></pre>
<p>I tried other way also which follows</p>
<pre><code>ToExpression[ToString[NumberForm[3, {4, 2}]]]
</code></pre>
<p>OutPut</p>
<pre><code>3.
</code></pre>
<p>but I need output as 3.00. Help me to find the solution.</p>
| Silvia | 17 | <p>Is <a href="http://reference.wolfram.com/mathematica/ref/SetAccuracy.html" rel="nofollow"><code>SetAccuracy</code></a> what you want?</p>
<pre><code>a = N[10/3, {∞, 3}]
</code></pre>
<blockquote>
<p>3.33</p>
</blockquote>
<pre><code>b = 3``3
</code></pre>
<blockquote>
<p>3.00</p>
</blockquote>
<pre><code>a + 1
</code></pre>
<blockquote>
<p>4.33</p>
</blockquote>
<pre><code>b + 1
</code></pre>
<blockquote>
<p>4.00</p>
</blockquote>
<pre><code>394.985674``3
</code></pre>
<blockquote>
<p>394.99</p>
</blockquote>
<p>Please note <code>Accuracy</code> is a different concept from <code>Precision</code>.</p>
|
32,150 | <p>I want to test if expressions (mix of variables, functions and numbers) are zero valued, as fast as possible, and <code>PossibleZeroQ</code> is sometimes very slow. One solution I found was to substitute the variables for random reals and test if the value of the substituted expression is less than, say, $0.0001$.</p>
<p>It works good, but maybe there are other solutions out there.</p>
<p>I know it can cause some wrong answers, but what is most important is the speed, since I can check the false positive later with <code>PossibleZeroQ</code>.</p>
<p>Can you think of an algorithm that can perform fast zero value tests in detriment of some loss of accuracy?</p>
<p>Edit:</p>
<p>I'll post my algorithm here:</p>
<pre><code>TestZeroValuedExpression[expression_,symbolslist_]:=Module[{numericvalue},
Quiet[TimeConstrained[If[Check[
numericvalue=N[Expand[expression/.Table[symbolslist[[i]]->RandomReal[{1,2}],{i,Length[symbolslist]}]]];
,False]=!=False,
If[Abs[numericvalue]>0.00001,False,PossibleZeroQ[expression]],
PossibleZeroQ[expression]
],3,False]]
];
</code></pre>
| Giovanni F. | 4,769 | <p>Ok, posting my version here. It has 0.1 seconds to test, otherwise returns false. I am assuming that if it takes longer than that, then the expression being tested is too complex (for my problem) and is not useful.</p>
<pre><code>TestZeroValuedExpression[expression_,symbolslist_]:=Module[{expressionrandomvalue},
Quiet[TimeConstrained[If[Check[
expressionrandomvalue=Chop[N[expression/.Table[symbolslist[[i]]->RandomInteger[{-3,3}],{i,Length[symbolslist]}]]];
,False]=!=False,
If[expressionrandomvalue!=0,False,PossibleZeroQ[expression]],
PossibleZeroQ[expression]
],0.1,False]]
];
</code></pre>
|
20,802 | <p>Look at the following example:</p>
<p>Which picture has four apples?</p>
<p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p>
<hr />
<p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p>
<hr />
<p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p>
<hr />
<p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p>
<p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p>
<blockquote>
<p>"Do you have three dollars?"
"Yes."</p>
</blockquote>
<p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p>
<p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as "which picture has <strong>exactly</strong> four apples"?</p>
| Barmar | 10,551 | <p>When we describe counts in natural language, there's almost always an implicit "exactly" when phrasing like this. We use phrases like "at least 4" when we want a more general description. Most children who have reached a development level where this quiz would be reasonable will probably already have learned this.</p>
<p>In fact, this is why there's a common joke:</p>
<blockquote>
<p>Q: How many months have 28 days?<br />
A: Just one, February.<br />
Q: No, they <em>all</em> do.</p>
</blockquote>
<p>The punch line works because we normally don't treat "have 28 days" as meaning "have 28 days, and possibly others", but when someone points it out we can see the potential ambiguity. But some people probably still won't get the joke, because the implicit "exactly" is so pervasive.</p>
|
2,530,298 | <p>I tried putting y alone and got y=(-6x-5)/5. Which I then put into the distance formula sqrt((x-1)^2+(y+5) and substitute the number above in for y but my answer never comes out correct.. Wondering if I could get some help.</p>
| Nosrati | 108,128 | <p>Another method. Take a circle centered in $(1,-5)$ that is
$$(x-1)^2+(y+5)^2=d^2$$
The slope of tangent line is
$$y'=-\dfrac{f_x}{f_y}=-\dfrac{x-1}{y+5}$$
will be $-\dfrac65$, the slope of given line. Then it's sufficient to solve the system
\begin{cases}
6x+5y+5=0,\\
5x-6y=35.
\end{cases}</p>
|
3,580,258 | <p>Hi: The definition I'll use is this: Let <span class="math-container">$F$</span> be an abelian group and <span class="math-container">$X$</span> a subset of <span class="math-container">$F$</span>. Then <span class="math-container">$F$</span> is a free abelian group on <span class="math-container">$X$</span> if for every abelian group <span class="math-container">$G$</span> and every function <span class="math-container">$f$</span> from <span class="math-container">$X$</span> to <span class="math-container">$G$</span> there is a homomorphism <span class="math-container">$\phi$</span> from <span class="math-container">$F$</span> to <span class="math-container">$G$</span> that extends <span class="math-container">$f$</span>.</p>
<p>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$X$</span> a subset of <span class="math-container">$G$</span>. Let <span class="math-container">$F$</span> be the free abelian group on <span class="math-container">$X$</span>. Then <span class="math-container">$F=\langle X\rangle$</span> and so <span class="math-container">$F\subseteq G$</span>. That is, every finite group has an infinite subgroup. What am I doing wrong?</p>
<p>EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.</p>
<p><a href="https://i.stack.imgur.com/BJioC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BJioC.png" alt=" "></a></p>
<p>From this a free group is not only always free on a subset but additionally that subset generates it. If <span class="math-container">$G$</span> is a group and <span class="math-container">$X$</span> is a subset, however, indeed there will exist a free group on <span class="math-container">$X$</span> but I am unable to show it will be generated by <span class="math-container">$X$</span> based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now, throws light in the paradox (paradox for me, of course).</p>
| user3482749 | 226,174 | <p>You are using <span class="math-container">$\langle X \rangle$</span> to mean two different things, and conflating them: </p>
<ol>
<li>You are using it to mean the free abelian group on <span class="math-container">$X$</span>. </li>
<li>You are using it to mean the subgroup of <span class="math-container">$G$</span> generated by <span class="math-container">$X$</span>. </li>
</ol>
<p>These are not the same thing, but you assume that they are. </p>
|
128,695 | <p>Is there any good guide on covering space for idiots? Like a really dumped down approach to it . As I have an exam on this, but don't understand it and it's like 1/6th of the exam. </p>
<p>So I'm doing Hatcher problem and stuck on 4.</p>
<ol>
<li>Construct a simply-connected covering space of the space $X \subset \mathbb{R}^3$ that is a union of a sphere and diameter. </li>
</ol>
<p>All I can think of is just connecting a bunch of spheres in a line.</p>
<p>But, yeah pretty scared will fail my degree because of this. So I need a good guide of covering spaces that isn't Hatcher. The only other uses heavy category theory which is even worse to read. </p>
| Ronnie Brown | 28,586 | <p>Books on algebraic topology are usually good on giving invariants to show that spaces are <strong>not</strong> homotopy equivalent, but not so good at showing why spaces <strong>are</strong> homotopy equivalent. In my book "Topology and groupoids" (2006) there is a chapter on cofibrations, which discusses the homotopy type of adjunction spaces $B \cup_f X$ where $A$ is a closed subspace of $X$ and $f: A \to B$. It is shown that if $f\simeq g$, and $(X,A)$ has the HEP, then $B \cup_f X $ is homotopy equivalent to $B \cup_g X$. You can use this to show your example is homotopy equivalent to $S^2 \vee S^1$, (see p. 293), and one knows the universal cover of this from other examples on this site. There is also a useful <strong>gluing theorem</strong> for homotopy equivalences, which is quite non trivial to prove, but easy to use. </p>
<p>That book has a different view on covering spaces than other books, by using groupoids systematically. Whether in the time available this can help your exam is another matter! Good luck!</p>
<p>(All these were in the 1968, 1988 editions.) </p>
|
2,439,863 | <p>I was working on the series </p>
<p>$\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}z^{n(n + 1)}}$ and I was to consider when $z = i$. I have that $$\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}i^{n(n + 1)}} = \sum_{n=1}^{\infty}{\frac{(-1)^{\frac{3}{2}n+\frac{1}{2}n^2}}{n}} = 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + ... \cong 0.43882457311697565541$$</p>
<p>I believe it converges. Does anyone have any suggestions to find an exact value for the infinite series?</p>
| mercio | 17,445 | <p>Let $f(z) = z - \frac 12 z^2 - \frac 13 z^3 + \frac 14 z^4 + \frac 15z^5 - \frac 16 z^6 - \frac 17 z^7 + \frac 18z^8 + \cdots$</p>
<p>Then $(1+z^2)f'(z) = (1+z^2)(1-z-z^2+z^3+z^4-z^5-z^6+z^7+\cdots) = 1-z$</p>
<p>So $f'(z) = \frac {1-z}{1+z^2} = -\frac 12 \frac {2z}{1+z^2} + \frac 1 {1+z^2}$</p>
<p>And $f(z) = - \frac 12 \log(1+z^2) + \arctan(z)$</p>
<p>And finally $f(1) = - \frac 12 \log(2) + \frac \pi 4$.</p>
<p>You can also recognize this if you know beforehand the two corresponding infinite series </p>
<p>$\frac \pi 4 = \int_0^1 \frac 1{1+z^2} dz = 1 - \frac 13 + \frac 15 - \frac 17 + \cdots$ </p>
<p>$\log 2 = \int_0^1 \frac 1{1+z} dz = 1 - \frac 12 + \frac 13 - \frac 14 + \cdots = 2(\frac 12 - \frac 14 + \frac 16 - \frac 18 + \cdots)$ </p>
|
106,775 | <p>I don't get this, need some help, examples and information</p>
<blockquote>
<p>The linear function $f$ is given by
$$f(x) = 3x - 2 ,\quad -2 \leq x \leq 4.$$</p>
<ol>
<li><p>Enter the independent variable and the dependent variable.</p></li>
<li><p>Determine the function values $f (-2)$, $f (-1)$, $f (0)$ and $f (4)$.</p></li>
</ol>
<p>Enter the definitions and values crowd. </p>
</blockquote>
<p>I know what a function is, but how can you find the independent variable and the dependent variable?</p>
<p>How one can determines the function values and how you specify the definitions and values crowd?</p>
<p>What I know about functions:</p>
<p>Variables can have any name, $x$, $y$, $z$, or "maria", "girl", "young"; at a specific value and function can be called anything.</p>
<p>My own example of functions:</p>
<p>Age = 18</p>
<p>year = 6</p>
<p>Maria (age, years) = age + years = 24</p>
<p>in 6 years Maria is going be 24 years</p>
| Yeujer | 207,161 | <p>f(2)= 3 times 2 − 2</p>
<p>f(-1)= 3 times -1 - 2</p>
<p>and so on because the -1 in f(-1) is the x-value and if that is the x-value then the the equation 3x-2 it would end up as 3 times -1 subtracted by 2. Which will end up with:
3 times -1=-3
-3 subtracted by 2=-5
So the answer for f(-1) is equal to -5. I hope that this helps even though this was a post form 2 years ago...</p>
|
3,327,094 | <p>Give an example of a non abelian group of order <span class="math-container">$55$</span>.</p>
<p>To find non abelian group the simplest way is to find one non abelian group whose order divides the order of given group and then we take the group which is the external direct product of the non abelian group and some other abelian group.</p>
<p>For example to find a non abelian group of order <span class="math-container">$36$</span> we take the permutation group <span class="math-container">$S_3$</span> and take the group <span class="math-container">$S_3\otimes \Bbb Z_6$</span>. But using this way we can not have a group of order <span class="math-container">$55$</span> since any group of order <span class="math-container">$5$</span> or <span class="math-container">$11$</span> will be abelian.</p>
<p>So how do we proceed?</p>
| Chinnapparaj R | 378,881 | <p>Consider the field <span class="math-container">$G=\Bbb Z_{11}$</span>. Now its multiplicative group <span class="math-container">$\Bbb Z_{11}^*$</span> is a group of order <span class="math-container">$10$</span>. Now <span class="math-container">$5$</span> divides <span class="math-container">$10$</span>, so By Cauchy's theorem, there exist a subgroup <span class="math-container">$H$</span> of order <span class="math-container">$5$</span> in <span class="math-container">$\Bbb Z_{11}^*$</span>. Now consider <span class="math-container">$$S=\left\{ \begin{pmatrix} 1& x \\ 0& y \end{pmatrix} : x \in \Bbb Z_{11}, y \in H\right\}$$</span> Then <span class="math-container">$S$</span> is a non abelian group under matrix multiplication of order <span class="math-container">$55$</span></p>
|
306,212 | <p>The only statement I'm sure of is that any hyperbolic or Euclidean manifold is a $K(G,1)$ (i.e. its higher homotopy groups vanish), since its universal cover must be $\mathbb H^n$ or $\mathbb E^n$. But for example, if a complete Riemannian manifold $M$ satisfies one of the following, can I conclude that $M$ is a $K(G,1)$?</p>
<ol>
<li><p>$M$ has sectional curvature bounded above by some negative number.</p></li>
<li><p>$M$ has negative sectional curvature.</p></li>
<li><p>$M$ has nonpositive sectional curvature.</p></li>
<li><p>$M$ has sectional curvature bounded above by $f(\operatorname{vol}(M))$ (where $f: \mathbb R \to \mathbb R$ is some function depending only on the dimension of $M$ that I don't know).</p></li>
<li><p>$M$ has scalar curvature bounded above by some negative number.</p></li>
<li><p>$M$ has negative scalar curvature.</p></li>
<li><p>$M$ has nonpositive scalar curvature.</p></li>
<li><p>$M$ has scalar curvature bounded above by $f(\operatorname{vol}(M))$.</p></li>
</ol>
<p>Do the answers change if I assume that $M$ is compact? Have I left out a relevant condition of some kind?</p>
| Igor Rivin | 11,142 | <p>For 1-3, yes, by the Cartan-Hadamard Theorem.5-... No. For example, every 3-manifold admits a metric of negative scalar curvature (I think this is actually true for any manifold, due to Lohkamp).</p>
|
316,016 | <p>Could you recommend any approachable books/papers/texts about matroids (maybe a chapter from somewhere)? The ideal reference would contain multiple examples, present some intuitions and keep formalism to a necessary minimum.</p>
<p>I would appreciate any hints or appropriate sources.</p>
| Nemo | 519,978 | <p>I have read chapter 1 of Gordon and
McNultey's book and think it provides an
Excellent clear intro. It will get you to first
Base quickly. </p>
|
1,506,763 | <p>This is not true for infinite measures (<a href="https://math.stackexchange.com/questions/342039/pointwise-convergence-but-not-in-measure">Pointwise convergence, but not in measure</a>). Is it true for a finite measure? Namely, let a finite (probability) measure $\mu(\cdot)$. Does a point-wise convergence of $\mu-$measurable functions, $\{f_n(x)\}$, imply convergence in measure $\mu(\cdot)$?</p>
<p>Equivalently, does
$$
\forall x~\lim_{n\to \infty} f_n(x) \to f(x)\,,
$$
imply
$$
\forall \varepsilon ~\lim_{n\to \infty} \mu\big{(}\left\{x~\big{|}~|(f_n(x) - f(x)|>\varepsilon\right\}\big{)} \to 0\,~?
$$</p>
| Brian Ding | 197,478 | <p>By the reversed Fatou lemma, we have </p>
<p>$\lim \mu( |f_n-f|\geq \epsilon) = \lim \int 1_{\{ |f_n-f|\geq \epsilon \}}d\mu \leq \int \limsup 1_{\{|f_n-f|\geq \epsilon \}}d\mu = 0.$</p>
<p>$1_{\{\cdot\}}$ is the indicator function (or characteristic function for analysis people). Since indicator is bounded, it is integrable as the measure is finite. The use of Fatou lemma is then justified. </p>
|
4,489,898 | <p>After 18 months of studying an advanced junior high school mathematics course, I'm doing a review of the previous 6 months, starting with solving difficult quadratics that are not easily factored, for example:
<span class="math-container">$$x^2+6x+2=0$$</span>
This could be processed via the quadratic equation but the course I'm working through asks me to use the complete the square method. I can do it, and I appreciate the geometric illustration of what is happening.</p>
<p>But it's so powerful and elegant, I can't help but wonder where else this method of adding something into an expression only to take it away in another is employed in mathematics. And is there a name for the general case of this kind of operation?</p>
<p><a href="https://i.stack.imgur.com/I0fmi.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/I0fmi.jpg" alt="geometric complete the square" /></a></p>
| G Tony Jacobs | 92,129 | <p>We use a similar trick in teaching first semester calculus. To obtain the well-known product rule for derivatives, we have to deal with the expression:</p>
<p><span class="math-container">$$\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$$</span></p>
<p>This is inconvenient, as the two terms in the numerator have nothing in common to let us do any factoring. Thus, we add and subtract a term that shares something with each:</p>
<p><span class="math-container">$$=\frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}$$</span></p>
<p>Now the first two terms of the numerator have a common factor, and the second two terms have a common factor. Thus, we are able to complete the needed derivation.</p>
|
718,850 | <p>I see similar questions asked on here and obviously I did some research and read my book, but it seems like every explanation contradicts another in some way. There are basically infinite scenarios using these and every example problem/scenario I seem to convince myself it could be both!</p>
<p><strong>Here are some of my understandings of each:</strong></p>
<p><strong>Permutation:</strong> Every detail matters and ALL ways of doing something. "Think of permutations as a list."</p>
<p><strong>Combinations:</strong> Used for groups. Order and Position DOES NOT matter.</p>
<hr>
<p><strong>My Confusion:</strong> </p>
<p>a.) If permutations are ALL ways of doing something.. then why does order/position/type matter?</p>
<p>b.) If order does NOT matter with combinations.. why are "Locks" said to have a "combination" when clearly the order does matter with a lock? If the "combination" to unlock something is 1-2-3.. then clearly 1-3-2 would not work. Therefore it seems like order does matter..</p>
<p>c.) If permutations are ALL ways of doing something and if EVERY detail matters.. then why are the number of permutations larger than the number of combinations?</p>
<p>Sorry if I included too much. I'm really struggling with this and every time I think I understand a scenario/problem.. I look at another and have no idea how to do it! I'd greatly appreciate any help. Thank you!</p>
| Nilesh Ingle | 662,954 | <p>Thank you for question and explanations above. Below is the way I try to remember it.</p>
<p>'Does not matter' actually means 'no replacement' AND 'no repetition within arrangement'</p>
<p><em>[Note: In other words <span class="math-container">$[1,2]$</span> or <span class="math-container">$[2,1]$</span> does not matter, we can pick only one of those two arrangements.]</em></p>
<p>Briefly:</p>
<ol>
<li>Lock: All combinations allowed (with replacement AND repetition both)</li>
<li>Permutation: NO replacement AND repetition (or rearrangement of the same set) allowed</li>
<li>Combination: NO replacement AND NO repetition (or rearrangement of the same set)</li>
</ol>
<p><strong>Example:</strong></p>
<p><em>Code to a lock with 2-digits: -</em>- -<em>- using only three digits: 7,8 and 9</em></p>
<p>[Note: Here we have a total of 3 digits to work with: 7,8 and 9]</p>
<p><strong>1. Lock:</strong></p>
<p><em>Question:</em> </p>
<p>Find all possible 2-digit codes for the lock</p>
<p><em>Answer</em>: </p>
<p><span class="math-container">$3 \times 3 = 9$</span> </p>
<p><em>Possible codes for the lock:</em></p>
<ul>
<li>1,1 </li>
<li>1,2</li>
<li>1,3</li>
<li>2,1</li>
<li>2,2</li>
<li>2,3</li>
<li>3,1</li>
<li>3,2</li>
<li>3,3</li>
</ul>
<p><em>Explanation:</em></p>
<p>Because here we assume that our lock could have "replacement and repetition". Thus the possibilities of arranging the code could be anything.</p>
<p><strong>2. Permutation:</strong></p>
<p>Now let's assume that we have the same lock with 2-digit code, however now if a digit is used once in the 2-digit-code, then it cannot be used again i.e. NO replacement.</p>
<p><em>Question</em>: </p>
<p>Find 2-digit code with NO replacement AND with repetition.</p>
<p><em>Answer</em>: </p>
<p><span class="math-container">$3 \times 2 = 6$</span></p>
<p>or</p>
<p><span class="math-container">$^{n}\textrm{P}_k = \frac{n!}{(n-k)!} =\frac{3!}{(3-2)!} = 6$</span></p>
<p><em>Possible codes for the lock:</em></p>
<ul>
<li>1,2</li>
<li>1,3</li>
<li>2,1</li>
<li>2,3</li>
<li>3,1</li>
<li>3,2</li>
</ul>
<p>[Note: Removed 1,1; 2,2; and 3,3 because repetition is NOT allowed i.e. repetition of the same digit within a code]</p>
<p><em>Explanation:</em></p>
<p>Because let's say we use the digit '1' in the first place of the 2-digit code, then we have only 2 out of 3 digits remaining for the second place of the 2-digit code. Thus, one less number for each subsequent digit of the code. </p>
<p><strong>3. Combination</strong></p>
<p>Here, we add additional constraints on top of the Permutation i.e. NO replacement and NO repetition.</p>
<p><em>Question:</em> </p>
<p>Find 2-digit code with NO replacement and NO repetition</p>
<p><em>Answer:</em> </p>
<p><span class="math-container">$^{n}\textrm{C}_r = \frac{n!}{(r! * (n-r)!} =\frac{3!}{2!*(3-2)!} = 3$</span></p>
<p><em>Possible codes for the lock:</em></p>
<ul>
<li>1,2</li>
<li>1,3</li>
<li>2,3</li>
</ul>
<p>[Note-1: Removed 1,1; 2,2 and 3,3 because replacement is NOT allowed.]</p>
<p>[Note-2: Removed 2,1; 3,2 and 3,1 because repetition is NOT allowed. Because 2,1 or 1,2 is does not matter and we can pick only one of those two. The ORDER DOES NOT MATTER!]</p>
<p><em>Explanation:</em></p>
<p>Here, we start off with 3 total digits that we can use. However, we have two constraints. The first constraint is that we cannot replace or re-use a digit once it is used (same as permutation above). Additionally, now we cannot repeat an arrangement of digits if they ALL are the same digits.</p>
|
1,515,417 | <p>I understand the idea that some infinities are "bigger" than other infinities. The example I understand is that all real numbers between 0 and 1 would not be able to "fit" on an infinite list.</p>
<p>I have to show whether these sets are countable or uncountable. If countable, how would you enumerate the set? If uncountable, how would you prove using diagonalization? </p>
<p>Set 1. All real numbers represented only by 1's. EX) 1, .11, 111.11, 1.111...</p>
<p>Set 2. All real numbers represented only by 2's and 3's. EX) .2, 23.2, 22.2232...</p>
| Brian Tung | 224,454 | <p>For Set $2$: Just consider the subset of this set consisting of numbers in the interval $(0, 1)$. Assume you have a complete listing of such numbers, and write them out, padding them with trailing zeros as needed. (Remember that the defining characteristic of Set $2$ is that the numbers can be represented only with $2$s and $3$s—not that they <em>cannot</em> be represented otherwise.) Now each position in each number in the list is either a $0$, a $2$, or a $3$. If you can generate a number whose value in Set $2$ that nevertheless differs from the $n$th item in the list in the $n$th place value, then you have properly executed the diagonalization argument.</p>
<p>Since Set $2$ is a superset of its restriction to $(0, 1)$, if that restriction is uncountable, then Set $2$ itself must also be uncountable.</p>
|
335,258 | <p>Find the domain of the function:
$$f(x)= \sqrt{x^2 - 4x - 45}$$</p>
<p>I'm just guessing here; how about if I square everything and then put it in the graphing calculator?
Thanks,
Lauri</p>
| Sam | 66,646 | <p><em>Hint:</em> The domain of a function is the set of input values for which the function is defined. Do you know of any values for which the square root is not defined?</p>
|
2,493,481 | <p>I'm currently studying calculus of variations. I couldn't find a rigorous definition of a functional on this site.</p>
<ol>
<li>What is the general definition of a functional?</li>
<li>Why for calculus of variations in physics, I must to use for a functional <em>a convex function</em> for the space of the admissible functions?</li>
</ol>
| supinf | 168,859 | <p>Here is a rigorous definition:
Let $V$ be a vector space over a field $\mathbb K$.
Then every function $F:V\to\mathbb K$ is called a functional.</p>
<p>An important class of functionals are linear functionals.
If $\mathbb K\in \{\mathbb R, \mathbb C\}$ and $V$ is a normed vector space ( or even a topological vector space)
then the continuous linear functionals are important objects.</p>
|
33,646 | <h1>Preamble</h1>
<p>I am a novice SE user, a toddler. In this post I want to criticize some moderator actions, which seems risky and futile, regarding unwritten policies on the meta; however, just for the record, I write this post so that unbiased readers find it helpful.</p>
<p>Please note that I am not throwing a tantrum and that I have slept enough, so it is not needed to say to me, "It is time for you to go to bed, nighty night."</p>
<h1>what's up? Why are you crying again?</h1>
<p>I, as a one being interested in contributing to the community but lacking enough math knowledge to create high-quality math content, decided to suggest grammar-fixing edits. But, many community adults ordered me not to "tinker", so I had to follow their order and stop fixing the grammar of posts.</p>
<p>After that I decided to do some <em><strong>unpaid</strong></em> janitorial task, flagging unnecessary comments, regarding that there is an option for such comments when one flags a comment and that the following excerpt from the <a href="https://meta.stackexchange.com/a/17365/502725">comment guideline</a> confirms such flagging:</p>
<blockquote>
<h3>When should I flag a comment?</h3>
<p>...</p>
<ul>
<li>It's no longer needed. <br/><br/>This comment is outdated, conversational or not relevant to the post.</li>
</ul>
</blockquote>
<p>(Also, <a href="https://meta.stackexchange.com/a/279451/502725">this community manager's answer</a> may be helpful).</p>
<p>After cleaning up many threads, the moderator team made two actions against me, which are criticized in this post:</p>
<p><strong>First</strong>, they contacted me privately sending the following message:</p>
<blockquote>
<p>... these comments are inoffensive and, while they are not strictly needed any longer, they are not causing harm to the site. On the other hand, it takes time for us to handle your flags. In the future, please think about whether or not a comment's deletion is worth the time of the moderators before you flag it.</p>
</blockquote>
<p><strong>Second</strong>, they rejected a huge number of my flags, which seems to have been deleted by the moderator(s) rejecting such flags. Since deleted comments cannot be seen by users except moderators, I cannot prove the claim to you (I think the moderator team would not deny it); however, let me put a screenshot of my declined flags here:</p>
<p><a href="https://i.stack.imgur.com/Z0Bml.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z0Bml.jpg" alt="enter image description here" /></a></p>
<p>Now, let us consider the following points:</p>
<ul>
<li>It was said, "they are not causing harm to the site." This is wrong because of the following reasons:<br/><br/><strong>First</strong>, if they were not causing harm to the site, the Stack Exchange community team would have never put the mentioned option in the list of flagging reasons and explicitly mentioned their opinion about such comments in the guideline.<br/><br/><strong>Second</strong>, they are actually causing harm to the site because many users put their conversational and unnecessary comments under posts when they see others do that, so this site would contain an uncontrollable number of such comments.</li>
<li>It was said, "it takes time for us to handle your flags." According to <a href="https://math.meta.stackexchange.com/q/33035/656359">this information</a>, the number of comment flags handled by the moderators is on average about 25 per a day. Let us suppose that I want to raise 25 comment flags per a day, which needs to be handled by the moderators (Please note that many comments are handled by the system). Since there are currently 10 moderators on the site, each moderator needs to handle 5 flags per a day, each of which should not take more than several seconds. If some moderators are not active enough on the site, I think rejecting a toddler's unpaid janitorial task is not a good solution for such a problem.</li>
<li>It was said, "think about whether or not a comment's deletion is worth the time of the moderators before you flag it." It seems that the moderators presume that I am an idle person whose only job and concern is flagging unnecessary comments on the site (?). Do you know digging up unnecessary comments takes a considerable amount of time and energy? My time and energy are worthless?</li>
<li>As you know well, I did nothing wrong about flagging such comments; I just followed the <strong>written</strong> guideline, but was treated according to an <em>unwritten</em> norm(?). According to the guideline, rejection of many flags may result in some written consequences (for example, reducing my flag daily allowance or flag ban) or <em>unwritten</em> ones (For example, a community manager disliking me may want to take it as an excuse to suspend me for 99 years).</li>
<li>If my behavior was a misdeed, why did a moderator remove the flagged comments after rejecting my flags? This reminds me of my previous story saying that a user made the same edit, which had already been rejected by the user; however, there is a difference: in this case the one who did such a thing is a moderator, not a regular user. I do not know why such things happen to a toddler like me.</li>
</ul>
<p>Finally, let me answer your question, "Why are you crying?" I am crying because I was treated like the following:</p>
<p>Suppose that you live in a city in which there is a rule asking citizens to keep it clean. You decide to keep the city clean by taking trash into a trash can <strong>for free</strong>. After you being tired, a mayor shouts at you, "Why do you waste our time? Trash cans must be emptied." Then the mayor empties a trash can onto your head.</p>
<p><strong>P.S.</strong> Although this post may seem harsh to the moderators, let me <strong>appreciate</strong> their efforts for moderating this community and handling meta issues. Please note that if I disliked the moderators, I would have never criticized them to continue making mistakes so that they would not improve themselves.</p>
| Tryst with Freedom | 688,539 | <p>Point of this answer: I wish to discuss a more fundamental idea, that is , on why people comment such ways.</p>
<hr />
<p>I think the problem is that, at least culturally, mathematics stack exchange has evolved from a question and answer site into more of a community.</p>
<blockquote>
<p>A community is a social unit (a group of living things) with commonality such as norms, religion, values, customs, or identity. Communities may share a sense of place situated in a given geographical area (e.g. a country, village, town, or neighbourhood) or in virtual space through communication platforms.</p>
</blockquote>
<p>And you see, when you have such a community, it's not just the question and answers but also the askers and answerers themselves which are valued. Hence, comments appreciating or depreciating an answer are expected. Since, we are of course people.</p>
<p>This may seem to some that it compromises stack exchange's advantage over other platforms such as discord, reddit , Facebook etc; I.e. that of focused Q&A. However, I'll say that it doesn't because the vetting process on stack exchange is that much stricter, proactive and transparent than the ones on any of the mentioned sites. I genuinely believe that it is not easy for someone to use mathematics stack exchange because of the difficulties of understanding the culture here, learning mathjax and having and keeping a learning attitude.</p>
<p>Finally, I do agree with the sentiment that there is a bit of a disconnect between the written rules and how the rules are applied in practice but I suppose that kind of exists in any place with rules.</p>
<p>A personal comment I'd like to say is that I highly appreciate how passionate you are about the site and I wish that there will be more people like you in the future who are able to write thoughtful criticisms and contemplate policy deeply for mathematics stack exchange.</p>
<hr />
<p>Note: The above is simply how I view the situation, if you have criticisms on my views, I invite you to comment on it so that I can improve my views and answer here.</p>
|
30,402 | <p>The envelope of parabolic trajectories from a common launch point is itself a parabola.
In the U.S. soon many will have a chance to observe this fact directly, as the 4th of July is traditionally celebrated with fireworks.</p>
<p>If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows that the parabola
$$
y = x \tan \theta - [x^2 /(2 v^2)] (1 + \tan^2 \theta)
$$
and the envelope of all such trajectories, is another parabola:
$$
y = v^2 /2 - x^2 / (2v^2)
$$
<hr>
<img src="https://i.stack.imgur.com/tn5pg.jpg" alt="envelope"></p>
<hr>
<p>These equations are not difficult to derive.
I have two questions.
First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations?
Is there a purely geometric argument?
Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.</p>
<p>Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity.
Is the envelope of these elliptical trajectories also an ellipse?
(I didn't try to work out the equations.)
Perhaps the same geometric viewpoint (if it exists) could apply, e.g. by slightly tilting the sections.</p>
| Arseniy Akopyan | 2,158 | <p>It is easy to see that all these parabolas have the same directrix. Height of a directirix correspond to energy of the body. So you have the family of parabolas with the common point $P$ and the directrix $l$. It is easy to prove, (using just definition of parabola as a locus of points...) that all of the touched the parabola with the focus at $P$ and the directrix $l_1$, which parallel $l$ (actually $l$ is midline of $P$ and $l_1$).</p>
<p>The same holds for sun-earth set. If Earth decides to fly in other direction (but with the same speed) its path will be always touch the fixed ellipse with foci in Sun and this position of Earth.</p>
|
400,715 | <p>Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$. Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$</p>
<p>Is $E$ closed and bounded in $\mathbb{Q}?$ Is it compact? Justify your answers.</p>
| Community | -1 | <p>It is not closed. You can always find a sequence of $x \in E$ such that converge to a real number.</p>
|
758,950 | <p>I have a pretty straightforward combinatorical problem which is an exercise to one paper about generating functions.</p>
<ol>
<li>How many ways are there to get a sum of 14 when 4 distinguishable dice are rolled? </li>
</ol>
<p>So, one die has numbers 1..6 and as dice are distinguishable then we should use exponential generating functions (we count sequences of rolled dice), because $3,4,3,4$ differs from $3,3,4,4$. So, we end up with answer
$$[\frac{x^{14}}{14!}](x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!})^4$$
How can we nicely calculate the coefficient of $\frac{x^{14}}{14!}$? I don't want to do this brutally, because next task is </p>
<p>2 Find generating function for the number of ways a sum of n can occur when rolling a die an infinite (or at least n) number of times.</p>
<p>I'd appreciate some help on such problems and how to "wrap" such limited exponential series.</p>
| epi163sqrt | 132,007 | <p>As <a href="https://math.stackexchange.com/users/205/shreevatsar">ShreevatsaR</a> pointed out it's sufficient to consider ordinary generating functions, since they already take into account that $3,4,3,4$ and $3,3,4,4$ are different. The first is coded as the coefficient of $x^3x^4x^3x^4$, while the second as the coefficient of $x^3x^3x^4x^4$ when considering the ogf $(x^1+\cdots+x^6)^4$.</p>
<p>Therefore we get for the first part</p>
<p>\begin{align*}
[x^{14}]&(x^1+\cdots+x^6)^4\\
&=[x^{14}]x^4(1+\cdots+x^5)^4\\
&=[x^{10}](1+\cdots+x^5)^4\\
&=[x^{10}]\left(\frac{1-x^6}{1-x}\right)^4\\
&=[x^{10}](1-4x^6+6x^{12}-4x^{18}+x^{24})\sum_{k\geq0}\binom{-4}{k}(-x)^k\\
&=([x^{10}]-4[x^{4}])\sum_{k\geq0}\binom{k+3}{k}x^k\\
&=\binom{13}{10}-4\binom{7}{4}\\
&=146
\end{align*}</p>
<p>For the second part we observe that each roll contributes at least $1$ to the value $n$. We can therefore restrict ourselves to the bracketed formulation: A die will be rolled <em>at least $n$ number of times</em> since all further rolls will not contribute to $n$.</p>
<p>An ordinary generating function in this case is</p>
<p>$$x^n\left(\frac{1-x^6}{1-x}\right)^n$$</p>
<blockquote>
<p><strong>Added 2014-04-19:</strong> Supplement - Using exponential generating functions instead.</p>
</blockquote>
<p>This is admittedly a <em>rather simple minded attempt</em> to answer the question of <a href="https://math.stackexchange.com/users/10394/chris">Chris</a> from the comment below: What do we count, if <em>we would have used exponential generating functions here?</em> Please feel free, to provide better examples with four dice, if you like.</p>
<p>If we use <em>exponential</em> instead of ordinary generating functions, we could imagine that we also have the <em>pips</em> of the faces of the dies <em>distinguishable</em>. Let's assume we have four <em>magic dice</em> $(M_1,M_2,M_3,M_4)$ (<em>M</em> for <em>magic</em>) and we are asking what's the contribution of a roll $(3,4,3,4) \rightarrow 14$ with respect to</p>
<p>$$\left[\frac{x^{14}}{14!}\right](x+\frac{x^2}{2}+\cdots+\frac{x^6}{6!})^4$$</p>
<p>The contribution is the multinomial coefficient</p>
<p>$$\left[\frac{x^{14}}{14!}\right]\frac{x^3}{3!}\frac{x^4}{4!}\frac{x^3}{3!}\frac{x^4}{4!}=\binom{14}{3,4,3,4}=\frac{14!}{3!^{2}4!^{2}}=4204200$$</p>
<p>and the <em>explanation</em>: In this case (namely roll with $14$ pips) the <em>magic dice can choose for the resulting pips from</em> $14$ <em>different colors</em>. $M_1$ chooses $3$ colors, $M_2$ chooses $4$ from the remaining $11$ colors, $M_3$ takes $3$ from the remaining $7$ and the rest of $4$ colors is used by $M_4$. So, we have</p>
<p>$$\binom{14}{3}\binom{11}{4}\binom{7}{3}\binom{4}{4} = \binom{14}{3,4,3,4}$$</p>
<p>different possibilities to see the <em>magic dice colorized with</em> $14$ <em>different colors</em>.</p>
|
2,078,737 | <p>I will gladly appreciate explanation on how to do so on this matrix:</p>
<p>$$
\begin{pmatrix}
i & 0 \\
0 & i \\
\end{pmatrix}
$$</p>
<p>I got as far as calculating the eigenvalues and came up with $λ = i$.
when trying to find the eigenvectors I came up with the $0$ matrix.</p>
<p>what am I doing wrong? </p>
<p>Much appreciation and thanks in advance.</p>
| B. Goddard | 362,009 | <p>I think most people would say that limits that go to infinity do not exist. But it's a special case of non-existence. Some limit's don't exist because the function bounces around too much. But this limit doesn't exist because it increases without bound. </p>
<p>Some folks will say "What about the extended reals?" which contains $\pm \infty.$
In this case, it would be fair to say the limit exists, but that's not going to be the most common usage.</p>
<p>Response to edit: In that case you don't need to know this limit. $1$ is not i your domain. The fact that the derivative is unbounded (in the open interval) is sufficient to show that the function is not uniformly continuous. </p>
|
26,192 | <p>I have a list of rules that represents a list of parameters to be applied to a circuit model (Wolfram SystemModeler model): </p>
<pre><code>sk = {
{R1 -> 10080., R2 -> 10080., C1 -> 1.*10^-7, C2 -> 9.8419*10^-8},
{R1 -> 10820., R2 -> 4984.51, R3 -> 10000., R4 -> 10000., C1 -> 1.*10^-7,
C2 -> 1.85417*10^-7},
{R1 -> 12600., R2 -> 12600., C1 -> 1.*10^-7, C2 -> 6.29882*10^-8},
{R1 -> 16420., R2 -> 16420., C1 -> 1.*10^-7, C2 -> 3.70897*10^-8},
{R1 -> 26120., R2 -> 26120., C1 -> 1.*10^-7, C2 -> 1.46573*10^-8},
{R1 -> 76600., R2 -> 1283.61, R3 -> 10000., R4 -> 10000., C1 -> 1.*10^-7,
C2 -> 1.01704*10^-7}};
</code></pre>
<p>Before I can apply these values to the model parameters I have to rename them. The list above consists of six lists - four lists of four rules and two lists of six rules. Those that have 4 rules should be named "sallenKeyUnityGain" and those that have 6 rules should be named "sallenKey". This is what I have so far: </p>
<pre><code>Table[If[Length[sk[[i]]] > 4, cirname = "sallenKey", cirname = "sallenKeyUnityGain"];
(cirname <> ToString[i] <> "." <> ToString[sk[[i]][[All, 1]][[j]]]) -> sk[[i]][[All, 2]][[j]]
, {i, Length[sk]}, {j, Length[sk[[i]]]}]
</code></pre>
<p>And this is the output: </p>
<pre><code>{{"sallenKeyUnityGain1.R1" -> 10080., "sallenKeyUnityGain1.R2" -> 10080.,
"sallenKeyUnityGain1.C1" -> 1.*10^-7, "sallenKeyUnityGain1.C2" -> 9.8419*10^-8},
{"sallenKey2.R1" -> 10820., "sallenKey2.R2" -> 4984.51, "sallenKey2.R3" -> 10000.,
"sallenKey2.R4" -> 10000., "sallenKey2.C1" -> 1.*10^-7,
"sallenKey2.C2" -> 1.85417*10^-7},
{"sallenKeyUnityGain3.R1" -> 12600., "sallenKeyUnityGain3.R2" -> 12600.,
"sallenKeyUnityGain3.C1" -> 1.*10^-7, "sallenKeyUnityGain3.C2" -> 6.29882*10^-8},
{"sallenKeyUnityGain4.R1" -> 16420., "sallenKeyUnityGain4.R2" -> 16420.,
"sallenKeyUnityGain4.C1" -> 1.*10^-7, "sallenKeyUnityGain4.C2" -> 3.70897*10^-8},
{"sallenKeyUnityGain5.R1" -> 26120., "sallenKeyUnityGain5.R2" -> 26120.,
"sallenKeyUnityGain5.C1" -> 1.*10^-7, "sallenKeyUnityGain5.C2" -> 1.46573*10^-8},
{"sallenKey6.R1" -> 76600., "sallenKey6.R2" -> 1283.61, "sallenKey6.R3" -> 10000.,
"sallenKey6.R4" -> 10000., "sallenKey6.C1" -> 1.*10^-7,
"sallenKey6.C2" -> 1.01704*10^-7}}
</code></pre>
<p>This would work fine if all were sallenKey or if all were sallenKeyUnityGain. However, I would like the output to look like this:</p>
<pre><code> {{"sallenKeyUnityGain1.R1" -> 10080., "sallenKeyUnityGain1.R2" -> 10080.,
"sallenKeyUnityGain1.C1" -> 1.*10^-7, "sallenKeyUnityGain1.C2" -> 9.8419*10^-8},
{"sallenKey1.R1" -> 10820., "sallenKey1.R2" -> 4984.51, "sallenKey1.R3" -> 10000.,
"sallenKey1.R4" -> 10000., "sallenKey1.C1" -> 1.*10^-7,
"sallenKey1.C2" -> 1.85417*10^-7},
{"sallenKeyUnityGain2.R1" -> 12600., "sallenKeyUnityGain2.R2" -> 12600.,
"sallenKeyUnityGain2.C1" -> 1.*10^-7, "sallenKeyUnityGain2.C2" -> 6.29882*10^-8},
{"sallenKeyUnityGain3.R1" -> 16420., "sallenKeyUnityGain3.R2" -> 16420.,
"sallenKeyUnityGain3.C1" -> 1.*10^-7, "sallenKeyUnityGain3.C2" -> 3.70897*10^-8},
{"sallenKeyUnityGain4.R1" -> 26120., "sallenKeyUnityGain4.R2" -> 26120.,
"sallenKeyUnityGain4.C1" -> 1.*10^-7, "sallenKeyUnityGain4.C2" -> 1.46573*10^-8},
{"sallenKey2.R1" -> 76600., "sallenKey2.R2" -> 1283.61, "sallenKey2.R3" -> 10000.,
"sallenKey2.R4" -> 10000., "sallenKey2.C1" -> 1.*10^-7,
"sallenKey2.C2" -> 1.01704*10^-7}}
</code></pre>
<p>In other words, if I have 4 apples and 2 oranges, instead of having:<br>
apple1, orange2, apple3, apple4, apple5, orange6, I would like:<br>
apple1, orange1, apple2, apple3, apple4, orange2. How would I do this?<br>
Thank you<br>
Tatjana</p>
| Chris Degnen | 363 | <p>This version implements a pair of counters, <code>a</code> & <code>b</code>, while retaining much of the original code :-</p>
<pre><code>a = b = 0;
Table[
If[Length[sk[[i]]] > 4,
cirname = "sallenKey";
(cirname <> ToString[If[a + b != i, ++a, a]] <> "." <>
ToString[sk[[i]][[All, 1]][[j]]]) -> sk[[i]][[All, 2]][[j]],
cirname = "sallenKeyUnityGain";
(cirname <> ToString[If[a + b != i, ++b, b]] <> "." <>
ToString[sk[[i]][[All, 1]][[j]]]) -> sk[[i]][[All, 2]][[j]]
], {i, Length[sk]}, {j, Length[sk[[i]]]}]
</code></pre>
|
3,664,717 | <p>Let's say there is a function <span class="math-container">$g: B \rightarrow B$</span> and <span class="math-container">$B$</span> is some set.</p>
<p>A relation <span class="math-container">$Rx$</span> over set <span class="math-container">$B$</span> is when </p>
<p><span class="math-container">$a Rx b$</span> </p>
<p>if <span class="math-container">$g(a) = b$</span>.</p>
<p>In this case, what kind of function <span class="math-container">$g: N \rightarrow N$</span> makes <span class="math-container">$Rx$</span> symmetric but not transitive nor reflexive?</p>
<p>I don't really understand this question. Does <span class="math-container">$g: N \rightarrow N$</span> mean the input is a natural number and output is also?</p>
<p>I know symmetry is when for elements of <span class="math-container">$B, b_1$</span> and<span class="math-container">$ b_2$</span>, if <span class="math-container">$b_1$</span> is related to <span class="math-container">$b_2$</span> then <span class="math-container">$b_2$</span> is related to b1 but I was wondering how this would be applied here. </p>
<p>Thanks!</p>
| SagarM | 142,677 | <p>As far as I understand you are looking for the following kind of example.</p>
<p>let <span class="math-container">$f: N \rightarrow N $</span> be defined as follows
<span class="math-container">$$f(x) = \text{smallest prime larger than }x -x$$</span>
Then,
if f(x)=y, then the relation <span class="math-container">$xRy$</span> is equivalently given as follows
<span class="math-container">$$xRy \Leftrightarrow x+y = \text{prime number} $$</span>
The above relation is symmetric but not reflexive or transitive.</p>
|
3,664,717 | <p>Let's say there is a function <span class="math-container">$g: B \rightarrow B$</span> and <span class="math-container">$B$</span> is some set.</p>
<p>A relation <span class="math-container">$Rx$</span> over set <span class="math-container">$B$</span> is when </p>
<p><span class="math-container">$a Rx b$</span> </p>
<p>if <span class="math-container">$g(a) = b$</span>.</p>
<p>In this case, what kind of function <span class="math-container">$g: N \rightarrow N$</span> makes <span class="math-container">$Rx$</span> symmetric but not transitive nor reflexive?</p>
<p>I don't really understand this question. Does <span class="math-container">$g: N \rightarrow N$</span> mean the input is a natural number and output is also?</p>
<p>I know symmetry is when for elements of <span class="math-container">$B, b_1$</span> and<span class="math-container">$ b_2$</span>, if <span class="math-container">$b_1$</span> is related to <span class="math-container">$b_2$</span> then <span class="math-container">$b_2$</span> is related to b1 but I was wondering how this would be applied here. </p>
<p>Thanks!</p>
| J. C. | 388,924 | <p>Consider the following function (here I'm considering <span class="math-container">$0 \notin \mathbb{N}$</span>, but it can be easily adapted to include <span class="math-container">$0$</span>):
<span class="math-container">$$f: \mathbb{N} \to \mathbb{N} $$</span>
<span class="math-container">$$ f(n)=
\begin{cases}
n + 1&\text{if}\, n \text{ is odd}\\
n - 1&\text{if}\, n \text{ is even}
\end{cases} $$</span></p>
<p>Then <span class="math-container">$f$</span>, seen as a relation, is symmetric, but not reflexive or transitive.</p>
|
262,500 | <p>What is the Green’s function of the boundary value problem
$$
\frac{\mathrm d^2 y}{\mathrm d x^2}-\frac{1}{x}\frac{\mathrm dy}{\mathrm dx}=1,\quad y(0)=y(1)=0,
$$</p>
<p>this boundary problem is not self adjoint, so please help me how to solve it.</p>
| user26872 | 26,872 | <p>First note that the solution to the homogeneous problem is
$y(x) = a + b x^2.$</p>
<p>We wish to solve
$$\begin{equation*}
\frac{d^2}{dx^2}G(x,t) - \frac{1}{x} \frac{d}{dx} G(x,t) = \delta(x-t),\tag{1}
\end{equation*}$$
where $G$ satisfies the boundary conditions
$G(0,t) = G(1,t) = 0$.
Therefore,
$$G(x,t) =
\begin{cases}
a(t)x^2, & x<t \\
b(t)(1-x^2), & x>t.
\end{cases}$$</p>
<p>Integrating (1) from $t-\epsilon$ to $t+\epsilon$ we find the jump condition<br>
$$\left.\frac{d}{dx}G(x,t)\right|_{x=t+\epsilon}
- \left.\frac{d}{dx}G(x,t)\right|_{x=t-\epsilon} = 1.$$
Integrating once more we find the continuity condition<br>
$$\left.G(x,t)\right|_{x=t+\epsilon}
- \left.G(x,t)\right|_{x=t-\epsilon} = 0.$$
These two conditions determine $a(t)$ and $b(t)$.
After a little work one finds
$$G(x,t) =
\begin{cases}
\displaystyle\frac{(t^2-1)x^2}{2t}, & x<t \\
\displaystyle\frac{(x^2-1)t}{2}, & x>t.
\end{cases}$$
Finally, we can use $G$ to solve the inhomogeneous problem,
$$y(x) = \int_0^1 dt\, G(x,t) = \frac{1}{2}x^2\log x.$$
This is the solution found by @Marvis. </p>
<p>Although it is often used, I don't remember ever seeing the above method given a proper name.
I call it the <em>jump method.</em>
There are at least two other common approaches to getting Green's functions.
One involves an eigenfunction expansion and the other the fundamental solution.</p>
|
2,941,311 | <p>Given diophantine equation <span class="math-container">$11x+17y +19z =2561$</span> , which <span class="math-container">$x,y,z \geq 1$</span></p>
<p>Find minimum and maximum value of <span class="math-container">$x+y+z$</span></p>
<p>I'm start with reduces equation to <span class="math-container">$11x+17y +19z =2514$</span> , which help us guarantee that <span class="math-container">$x,y,z \geq 1$</span> then set each variable to <span class="math-container">$0$</span> to find bound of <span class="math-container">$x+y+z$</span> </p>
<p>that comes to this<br>
<a href="https://i.stack.imgur.com/N3fwl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N3fwl.png" alt="enter image description here"></a></p>
<p>I'm not sure that the solution I did was correct. So, minimum <span class="math-container">$x+y+z = 137$</span> and maximum <span class="math-container">$x+y+z = 231$</span> ???. sorry for my english and thank you in advance.</p>
| hmakholm left over Monica | 14,366 | <p>There are cases where <span class="math-container">$\lim\frac{f(x)}{g(x)}$</span> exists but <span class="math-container">$\lim\frac{f'(x)}{g'(x)}$</span> does <em>not</em> exist. For example, with <span class="math-container">$a=0$</span>:</p>
<p><span class="math-container">$$ f(x) = x^2\sin(1/x) \qquad\qquad g(x) = x $$</span></p>
<p>Here <span class="math-container">$\lim\limits_{x\to 0}\frac{f(x)}{g(x)}=0$</span>, but <span class="math-container">$\frac{f'(x)}{g'(x)}$</span> does not have a limit for <span class="math-container">$x\to 0$</span>.</p>
<p><strong>Therefore</strong> L'Hospital's rule can only go in one direction: If <span class="math-container">${f'(x)}/{g'(x)}$</span> happens to have a limit (<em>and</em> <span class="math-container">$f(x), g(x)$</span> both tend to <span class="math-container">$0$</span> or <span class="math-container">$\infty$</span>), <em>then</em> this is also the limit of <span class="math-container">${f(x)}/{g(x)}$</span>.</p>
<p>But if <span class="math-container">${f'(x)}/{g'(x)}$</span> does <em>not</em> exist, then this is not enough to conclude anything about whether <span class="math-container">${f(x)}/{g(x)}$</span> has a limit or not.</p>
|
3,921 | <p>Say I have a triangle with vertices $(0,0), (2,4), (4,0)$ that I want to rotate along the origin. Rotation by multiples of $90^{\circ}$ is simple. However, I want to rotate by something a bit more complicated, such as $54^{\circ}$. How do I figure out where the vertices would be then?</p>
| Isaac | 72 | <p>One way is to use complex numbers. Multiplying by $\cos\theta+i\sin\theta$ rotates by $\theta$ about 0, so you could multiply $(2+4i)(\cos 54^\circ+i\sin 54^\circ)$ to get the rotation image of (2,4).</p>
|
3,921 | <p>Say I have a triangle with vertices $(0,0), (2,4), (4,0)$ that I want to rotate along the origin. Rotation by multiples of $90^{\circ}$ is simple. However, I want to rotate by something a bit more complicated, such as $54^{\circ}$. How do I figure out where the vertices would be then?</p>
| J. M. ain't a mathematician | 498 | <p>In the answer to <a href="https://math.stackexchange.com/questions/2429">this question</a>, I mentioned the formula for the rotation matrix; one merely takes the product of the rotation matrix with the coordinates (treated as 2-vectors) to get the new rotated coordinates. Note that I gave the matrix for clockwise rotation; for anticlockwise rotation, negate the angle (thus switching the sign of the two sine components).</p>
|
1,749,284 | <p>As part of my homework i've the following question:</p>
<p>The tangent line $ L $ is crossing the graph of $ y = ax^3 + bx $ at point $ x = x_0 $, find another point where the tangent-line $L$ is crossing the graph. Define $ a = 1$ and $b = 0$.</p>
<p>Second part of the question is to graph $y = x^3$ and show the corssing points.</p>
<p>I was able to find find the equation of $L$ using the derivative of the function and the point: $$ y - x_0^3 = 3x_0^2(x - x_0) \\ y = 3x_0^2x - 2x_0^3 $$</p>
<p>I'm unable to find another point the tangent $L$ cross the graph (which is assume is $y = x^3$).
Should i resolve the value of $x_0$?</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Quote from <a href="http://jtra.cz/stuff/essays/math-self-reference/index.html" rel="nofollow">Self Referential Formula in Math</a>:</p>
<blockquote>
Its graph contains all possible bitmaps that fit in region of 17 * 106 grid. So it is not much of wonder that one of those many bitmaps contains meaningful representation of the formula itself.
</blockquote>
<p>So, the formula actually generates a long mosaic of all the possible monochromatic (1 bit per pixel) 17 * 106 images. The image of the formula is one (actually many) of the subimages.</p>
<p>The formula can be very aptly described as a <a href="http://hyperdiscordia.crywalt.com/library_of_babel.html" rel="nofollow">Library of Babel</a> generator.</p>
<p>A 5K screenshot is too big to fit, but the formula can be generalized to bigger sizes.</p>
|
3,207,453 | <p>studying the series <span class="math-container">$\sum_\limits{n=2}^\infty \frac{1}{n(\log n)^ {2}}$</span>.</p>
<p>I've tried with the root criterion</p>
<p><span class="math-container">$\lim_{n \to \infty} \sqrt[n]{\frac{1}{n(\log n)^ {2}}}>1$</span>
and the series should diverge.</p>
<p>But I'm not sure
Can someone help me to understand?</p>
| Wojowu | 127,263 | <p>Hint: you can also use the <a href="https://en.wikipedia.org/wiki/Cauchy_condensation_test" rel="nofollow noreferrer">Cauchy condensation test</a>.</p>
|
7,247 | <p>Let $f(x) \in L^p(\mathbb{R})$ and $K \in C^m(\mathbb{R})$. Can I then say that $(f \ast K) (x) = \int_{\mathbb{R}} f(t) K(x-t) dt$ is in $C^m$? </p>
<p>I know that this is true if $K$ has compact support, but I was wondering if it is possible to have a stronger result (perhaps $K$ vanishing at $\infty$?). </p>
| Amitesh Datta | 10,467 | <p>I wrote an article on analysis recently and I included the following relevant result (with proof) in the article; I hope it is helpful:</p>
<p><strong>Theorem</strong> Let $f\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ for some $1\leq p \leq \infty$. Also, let $g\in L^1(\mathbb{R}^n)$ be a function all of whose partial derivatives of the first order exist and are such that $\frac{\partial g}{\partial x_i}$ is bounded on $\mathbb{R}^n$ for all $1\leq i\leq n$. We conclude that the partial derivatives of the convolution $f\ast g$ of the first order exist on $\mathbb{R}^n$. In fact, $\frac{\partial (f\ast g)}{\partial x_i}=f\ast (\frac{\partial g}{\partial x_i})$ for all $1\leq i\leq n$. </p>
<p><em>Proof</em>. First note that the convolution $f\ast g\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ by Minkowski's inquality
and is therefore finite (and well-defined) a.e. Let us fix $1\leq i\leq n$.
Note that </p>
<p>$\frac{\left(f\ast g\right)\left(x+he_i\right)-\left(f\ast g\right)(x)}{h} - \left(f\ast \left(\frac{\partial g}{\partial x_i}\right)\right)\left(x\right)$</p>
<p>$= \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right) \right]dy$ </p>
<p>In particular, </p>
<p>$\frac{\partial \left(f\ast g\right)}{\partial x_i}\left(x\right)$</p>
<p>$=\lim_{h\to 0} \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]dy$</p>
<p>$= \int_{\mathbb{R}^{n}} \left[\lim_{h\to 0} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]\right]dy$</p>
<p>$= \int_{\mathbb{R}^n} f\left(y\right)\frac{\partial g}{\partial x_i}\left(x-y\right) dy$</p>
<p>$= \left(f\ast \frac{\partial g}{\partial x_i}\right)\left(x\right)$</p>
<p>We will justify this computation using the
Lebesgue dominated convergence theorem. In particular, we will show that if $x\in \mathbb{R}^n$ is fixed,
the expression $\left|\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$ is bounded by
an $L^1(f)$ function in $y$ for all $h>0$ sufficiently small. (Let us recall that $L^1(f)$ is the $L^1$ space
associated to the complex measure $\mu_f$ defined by $\mu_f(E)=\int_{E} f$ for every measurable $E\subseteq \mathbb{R}^n$.
Clearly, every constant function is in $L^1(f)$.)
However, this is an easy consequence of the mean value theorem: we know that there exists
$\delta>0$ such that $0<h<\delta$ implies </p>
<p>$\left|\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right] -
\left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$</p>
<p>$\leq 2\sup_{c\in\mathbb{R}^n}\left|\frac{\partial g}{\partial x_i}\left(c\right)\right|$</p>
<p>and the result now follows from the hypotheses. <strong>Q.E.D.</strong></p>
|
4,215,824 | <p>Is there a formal definition of "almost always less than" or "almost always greater than"? I think one could define it using probabilities but not sure how to go about it. If one could show the following, then I think you could say <span class="math-container">$X$</span> is almost always less than a value <span class="math-container">$x$</span>. Is there other ways of going about this?</p>
<p><span class="math-container">$$
P(X<x)=1
$$</span></p>
| tommik | 791,458 | <p>This means that</p>
<p><span class="math-container">$$\mathbb{P}[X\geq x]=0$$</span></p>
<p>Thus there can be some sets where <span class="math-container">$X\geq x$</span> but they are all sets with probability zero</p>
|
4,215,824 | <p>Is there a formal definition of "almost always less than" or "almost always greater than"? I think one could define it using probabilities but not sure how to go about it. If one could show the following, then I think you could say <span class="math-container">$X$</span> is almost always less than a value <span class="math-container">$x$</span>. Is there other ways of going about this?</p>
<p><span class="math-container">$$
P(X<x)=1
$$</span></p>
| Flitwick | 602,312 | <p>If something holds almost surely, it means it is true on a set of probability measure 1. Let <span class="math-container">$X$</span> be a r.v. on some probability space <span class="math-container">$(\Omega,\mathcal{F},\mathbb{P})$</span> and <span class="math-container">$c\in\mathbb{R}$</span>. Then the following are by definition the same:</p>
<p>i) <span class="math-container">$X<c$</span> <span class="math-container">$\mathbb{P}$</span>-almost surely</p>
<p>ii) <span class="math-container">$\{\omega\in\Omega: X(\omega)<c\}$</span> occurs with probability 1</p>
<p>iii) <span class="math-container">$\mathbb{P}(X<c)=1$</span></p>
|
2,354,383 | <p>Why doesn't a previous event affect the probability of (say) a coin showing tails?</p>
<p>Let's say I have a <strong>fair</strong> and <strong>unbiased</strong> coin with two sides, <em>heads</em> and <em>tails</em>.</p>
<p>For the first time I toss it up the probabilities of both events are equal to $\frac{1}{2}$.
This much is <em>intuitive and clear</em> to me.</p>
<p>Now suppose that I toss it up $1000000000$ times and the scores are,</p>
<p>$501000000$ Heads</p>
<p>$499000000$ Tails</p>
<p>Now, for the $1001000000^{th}$ toss, shouldn't the probability of a tail coming up be <strong>greater</strong> than that of heads showing up?</p>
<p>I have seen many books which say that even for the $1001000000^{th}$ toss, the probabilities of both events are equal to $\frac{1}{2}$.</p>
<p>This <strong>seems wrong to me</strong> since the same books affirm that <strong>if a coin is tossed a large number of times, the quantity $\frac{heads}{tails}$ will approach $1$.</strong></p>
<p>I know this is very elementary and naive, yet I had only superficially studied probability and I hope you all will bear with me.</p>
<p><strong>My Objections with some of the top-voted answers</strong></p>
<blockquote>
<p>It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter. </p>
</blockquote>
<p>I don't get this statement. What exactly does the second sentence mean? Moreover, if what you said is true then, the following comment by a user should be wrong,</p>
<blockquote>
<p>Law of large numbers</p>
</blockquote>
<p>So these are contradicting each other I feel. Please bear with my lack of knowledge.</p>
| D_S | 28,556 | <p>The ratio of heads to tails will approach $1$ as the number of times it's tossed tends towards infinity. If you've tossed the coin $1000$ times, you have hardly tossed it at all.</p>
<p>Another way to think about it: we agree that the first time you toss the coin, the probability of heads is $1/2$. Suppose you toss the coin $1000$ times, and you have $501$ heads and $499$ tails. You ask your friend to toss the coin for you. He does not know that you have already been tossing the coin. What is the probability that he will get heads?</p>
|
2,354,383 | <p>Why doesn't a previous event affect the probability of (say) a coin showing tails?</p>
<p>Let's say I have a <strong>fair</strong> and <strong>unbiased</strong> coin with two sides, <em>heads</em> and <em>tails</em>.</p>
<p>For the first time I toss it up the probabilities of both events are equal to $\frac{1}{2}$.
This much is <em>intuitive and clear</em> to me.</p>
<p>Now suppose that I toss it up $1000000000$ times and the scores are,</p>
<p>$501000000$ Heads</p>
<p>$499000000$ Tails</p>
<p>Now, for the $1001000000^{th}$ toss, shouldn't the probability of a tail coming up be <strong>greater</strong> than that of heads showing up?</p>
<p>I have seen many books which say that even for the $1001000000^{th}$ toss, the probabilities of both events are equal to $\frac{1}{2}$.</p>
<p>This <strong>seems wrong to me</strong> since the same books affirm that <strong>if a coin is tossed a large number of times, the quantity $\frac{heads}{tails}$ will approach $1$.</strong></p>
<p>I know this is very elementary and naive, yet I had only superficially studied probability and I hope you all will bear with me.</p>
<p><strong>My Objections with some of the top-voted answers</strong></p>
<blockquote>
<p>It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter. </p>
</blockquote>
<p>I don't get this statement. What exactly does the second sentence mean? Moreover, if what you said is true then, the following comment by a user should be wrong,</p>
<blockquote>
<p>Law of large numbers</p>
</blockquote>
<p>So these are contradicting each other I feel. Please bear with my lack of knowledge.</p>
| Ross Millikan | 1,827 | <p>The <em>assumption</em> for a coin is that is has no memory. That means that the chance of heads is the same on every toss. For a fair coin, that chance is $\frac 12$ <strong>regardless of the history</strong>. If you toss $100$ times and get heads every time (very unlikely, but it could happen) the most probable event after a million tosses (including the $100$ you already did) is $500050$ heads and $499950$ tails. It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter. Look how close the head/tail ratio would be to $1$ at that point.</p>
|
3,758,536 | <p>Theorem 3.29: If <span class="math-container">$p>1$</span>,<br />
<span class="math-container">$$
\sum_{n=2}^{\infty}\frac{1}{n(\log\ n)^p}
$$</span>
converges; if <span class="math-container">$p\leq1$</span>, the series diverges.</p>
<p>Proof: The monotonicity of the logarithmic function implies that <span class="math-container">$\{log\ n\}$</span> increases. Hence <span class="math-container">$\{1/n\ \log\ n\}$</span> decreases, and we can apply Theorem 3.27 to the series above; this leads us to the series
<span class="math-container">$$
\sum_{k=1}^{\infty}2^k\cdot \frac{1}{2^k(\log\ 2^k)^p}=\sum_{k=1}^{\infty}\frac{1}{(k\log\ 2)^p}=\frac{1}{(\log\ 2)^p}\sum_{k=1}^{\infty}\frac{1}{k^p}
$$</span>
and Theorem 3.29 follows from Theorem 3.28.</p>
<p>I have two questions:<br />
(1) How can we get the decrease of <span class="math-container">$\{1/n\ \log\ n\}$</span> from the increase of <span class="math-container">$\{\log\ n\}$</span>? I didn't see any conncection between them.<br />
(2) The author said that we can apply Theorem 3.27 to the series. However, in order to apply Theorem 3.27, I think we need to show <span class="math-container">$\{1/n\ (\log\ n)^p\}$</span> is decreasing. But I don't know how to do that.</p>
<p>Theorem 3.27: Suppose <span class="math-container">$a_1\geq a_2\geq\cdots\geq0$</span>. Then the series <span class="math-container">$\sum_{n=1}^{\infty} a_n$</span> converges if and only if the series
<span class="math-container">$$
\sum_{k=0}^{\infty}2^ka_{2^k}=a_1+2a_2+4a_4+8a_8+\cdots
$$</span>
converges.</p>
| Oliver Díaz | 121,671 | <p>The function <span class="math-container">$g(x)=x\log^p(x)$</span> increases and is positive in the interval <span class="math-container">$(1,\infty)$</span>. From that, it follows that <span class="math-container">$f(x)=\frac{1}{x\log^px}$</span> decreases on <span class="math-container">$(1,\infty)$</span>.</p>
<p>The convergence of the series can then be analyzed either by the integral test or Cauchy's condensation theorem, as you proposed in your problem.</p>
|
3,758,536 | <p>Theorem 3.29: If <span class="math-container">$p>1$</span>,<br />
<span class="math-container">$$
\sum_{n=2}^{\infty}\frac{1}{n(\log\ n)^p}
$$</span>
converges; if <span class="math-container">$p\leq1$</span>, the series diverges.</p>
<p>Proof: The monotonicity of the logarithmic function implies that <span class="math-container">$\{log\ n\}$</span> increases. Hence <span class="math-container">$\{1/n\ \log\ n\}$</span> decreases, and we can apply Theorem 3.27 to the series above; this leads us to the series
<span class="math-container">$$
\sum_{k=1}^{\infty}2^k\cdot \frac{1}{2^k(\log\ 2^k)^p}=\sum_{k=1}^{\infty}\frac{1}{(k\log\ 2)^p}=\frac{1}{(\log\ 2)^p}\sum_{k=1}^{\infty}\frac{1}{k^p}
$$</span>
and Theorem 3.29 follows from Theorem 3.28.</p>
<p>I have two questions:<br />
(1) How can we get the decrease of <span class="math-container">$\{1/n\ \log\ n\}$</span> from the increase of <span class="math-container">$\{\log\ n\}$</span>? I didn't see any conncection between them.<br />
(2) The author said that we can apply Theorem 3.27 to the series. However, in order to apply Theorem 3.27, I think we need to show <span class="math-container">$\{1/n\ (\log\ n)^p\}$</span> is decreasing. But I don't know how to do that.</p>
<p>Theorem 3.27: Suppose <span class="math-container">$a_1\geq a_2\geq\cdots\geq0$</span>. Then the series <span class="math-container">$\sum_{n=1}^{\infty} a_n$</span> converges if and only if the series
<span class="math-container">$$
\sum_{k=0}^{\infty}2^ka_{2^k}=a_1+2a_2+4a_4+8a_8+\cdots
$$</span>
converges.</p>
| Oliver Díaz | 121,671 | <ul>
<li><p>When <span class="math-container">$p=0$</span> divergence is direct for in such case you get the harmonic series <span class="math-container">$\sum_n\frac1n$</span>.</p>
</li>
<li><p>For <span class="math-container">$p<0$</span> notice that <span class="math-container">$\frac{\log^{-p}}{n}\geq \log^{-p}2\frac{1}{n}$</span> for all <span class="math-container">$n\geq2$</span> and so, the series diverges in such cases.</p>
</li>
<li><p>For <span class="math-container">$p>0$</span>, as explained <a href="https://math.stackexchange.com/a/3758547/121671">here</a>, your summand is increasing in which case you can apply Cauchy's condensation test as you suggested.</p>
</li>
</ul>
|
73,039 | <p>I have a huge data file which I can't ListPlot.</p>
<p>This code generates similar kind of data:</p>
<pre><code>datatest =RandomSample[Join[RandomReal[{0.5, 15}, 20], RandomReal[.1, 10000]]];
datatest2 = 5 + Riffle[datatest, -datatest];
</code></pre>
<p>I want to filter (delete) the part of the data that is not necessary as follows:</p>
<pre><code>peaks = FindPeaks[datatest2, 0, 0, 5.2];
ListPlot[datatest2, PlotRange -> All, Joined -> True,
Epilog -> {Red, PointSize[0.01], Point[peaks]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/RpKLb.jpg" alt="enter image description here">
Currently I am using some kind of long way to do the task.</p>
<p>Is there any signal processing functionality in MMA that can do this easily?</p>
<p>Thank you</p>
| bill s | 1,783 | <p>A common way to remove outliers is with the Median filter. What you want to do is the opposite: to keep the outliers and remove the inliers. Subtracting the data from the median, then clipping the result and selecting all those larger than a threshold is one way to proceed. </p>
<pre><code>short = Select[Chop[datatest2 - MedianFilter[datatest2, 5], 0.5], Abs[#] > 0.1 &];
ListPlot[short, PlotRange -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/5oX8a.png" alt="enter image description here"></p>
<p>One downside is that this removes the mean value (about five in your data). You may wish to add this back in, and also to fiddle with the parameters to get the width you are looking for.</p>
|
73,039 | <p>I have a huge data file which I can't ListPlot.</p>
<p>This code generates similar kind of data:</p>
<pre><code>datatest =RandomSample[Join[RandomReal[{0.5, 15}, 20], RandomReal[.1, 10000]]];
datatest2 = 5 + Riffle[datatest, -datatest];
</code></pre>
<p>I want to filter (delete) the part of the data that is not necessary as follows:</p>
<pre><code>peaks = FindPeaks[datatest2, 0, 0, 5.2];
ListPlot[datatest2, PlotRange -> All, Joined -> True,
Epilog -> {Red, PointSize[0.01], Point[peaks]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/RpKLb.jpg" alt="enter image description here">
Currently I am using some kind of long way to do the task.</p>
<p>Is there any signal processing functionality in MMA that can do this easily?</p>
<p>Thank you</p>
| gpap | 1,079 | <p>I am not an expert (understatemnt of the year!) in signal processing but you can use the band function and create a sparse matrix that has as many 1s as you want around the positions of the peaks. I am not sure the following is the best way to do this but it works:</p>
<pre><code>With[{width = 200},
spArray = SparseArray[
Thread[
(Band[# - width, # + width] & /@ peaks[[All, 1]]) -> 1],
Length[datatest2]]
];
</code></pre>
<p>so this means that I have made a sparse array with 1s in 400 samples ($\pm 200$) around each of your peaks and if I multiply this with your original array <code>dataset2</code> these are gonna be the only elements that will survive:</p>
<pre><code>ListPlot[spArray datatest2, PlotRange -> All, Joined -> True,
Epilog -> {Red, PointSize[0.01], Point[peaks]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/TrR5U.png" alt="enter image description here"></p>
|
5,927 | <p>I have a problem with the binomial coefficient $\binom{5}{7}$. I know that the solution is zero, but I have problems to reproduce that:</p>
<p>${\displaystyle \binom{5}{7}=\frac{5!}{7!\times(5-7)!}=\frac{5!}{7!\times(-2)!}=\frac{120}{5040\times-2}=\frac{120}{-10080}=-\frac{1}{84}}$</p>
<p>Where is my mistake?</p>
| svenwltr | 2,280 | <p>Yes, now I see the problem.</p>
<p>First, (-2)! really isn't defined. And I can't use the factorial method if $n\notin\mathbb{N}$. So I have to go these way:</p>
<p>${\displaystyle \binom{5}{7}=\frac{5\times4\times3\times2\times1\times0\times-1}{7!}=\frac{0}{7!}=0}$</p>
<p>Thus, if $k>n$ the solution will always be zero, because the numerator has always the factor zero.</p>
|
5,927 | <p>I have a problem with the binomial coefficient $\binom{5}{7}$. I know that the solution is zero, but I have problems to reproduce that:</p>
<p>${\displaystyle \binom{5}{7}=\frac{5!}{7!\times(5-7)!}=\frac{5!}{7!\times(-2)!}=\frac{120}{5040\times-2}=\frac{120}{-10080}=-\frac{1}{84}}$</p>
<p>Where is my mistake?</p>
| Darsh Ranjan | 2,032 | <p>$(-2)!$ is actually infinite. A more palatable way to phrase that, perhaps, is in terms of the reciprocal factorial: $1/(-2)! = 0$. We only need the recurrence relation $n! = n(n-1)!$, or in terms of reciprocal factorials: $$\frac{1}{(n-1)!} = n\cdot\frac{1}{n!}.$$ That means $\frac{1}{(-2)!} = \frac{0\cdot (-1)}{0!} = 0$. Then $\binom{5}{7} = \frac{5!}{7!}\cdot \frac{1}{(-2)!} = 0$, QED. </p>
|
2,356,593 | <blockquote>
<p>Quoting:" Prove: if $f$ and $g$ are continuous on $(a,b)$ and $f(x)=g(x)$ for every $x$ in a dense subset of $(a,b)$, then $f(x)=g(x)$ for all $x$ in $(a,b)$."</p>
</blockquote>
<p>Let $S \subset (a,b)$ be a dense subset such that every point $x \in (a,b)$ either belongs to S or is a limit point of S.</p>
<p>There exists $x_n \in S$ such that $\lim\limits_{n \rightarrow \infty} x_n =x$.</p>
<p>As $f$ and $g$ are continuous on (a,b),
$$f(x)= \lim\limits_{n \rightarrow \infty} f(x_n)= \lim\limits_{n \rightarrow \infty} g(x_n)= g(x)$$</p>
<ol>
<li>How can I conclude that "$f(x)=g(x)$ for all $x$ in $(a,b)$"?</li>
<li>Is there a proof with $\epsilon$, $\delta$ definition of limits?</li>
</ol>
<p>Much appreciated</p>
| tattwamasi amrutam | 90,328 | <p>Let $S$ be the dense subset of $(a,b)$. Like you wrote, for any $x \in (a,b)$, there exists $x_n \in S$ such that $x_n \to x$. Then $$f(x)=f(\lim_n x_n)=\lim_nf(x_n)=\lim_ng(x_n)=g(x)$$</p>
|
2,839,554 | <p>I would like to formalise some operations I am doing, however it is unclear how I should deal with categorical variables. </p>
<p>Imagine a dataset with 15 distinct couples (<code>ID</code>). Each couple was observed 3 times (<code>time</code>). </p>
<p>Each partner has responded to two questions: <code>p</code> and <code>act</code>. <code>p</code> is a dummy vector {0, 1} while <code>act</code> is a categorical variable with 4 levels {a,b,c,d}. <code>_m</code> refers to partner male and <code>_w</code> to partner female. </p>
<pre><code> ID time p_m p_w act_m act_w
1 A 1 1 1 c b
2 A 2 1 1 b c
3 A 3 1 1 c d
4 B 1 1 1 b b
5 B 2 0 1 a a
6 B 3 1 1 b b
7 C 1 1 1 b b
8 C 2 1 1 c c
9 C 3 1 1 c b
10 D 1 1 1 c b
11 D 2 1 0 b a
12 D 3 1 1 c b
13 E 1 1 1 d d
14 E 2 1 1 b c
15 E 3 1 1 c c
</code></pre>
<p>First, I am interested in formalising the matches on <code>p</code>. Because <code>p</code> is a dummy variable, it seems that I can simply write: </p>
<p>$joint_{jt} = (p_{jt}^{m} \times p_{jt}^{w})$</p>
<p>where $t$ denotes <em>time</em>, $p_{jt}^{m}$ denotes partner $m$ response at time $t$ belonging to couple number $j$ (and vice versa for partner $w$). </p>
<pre><code> ID time p_m p_w act_m act_w joint_j
1 A 1 1 1 c b 1
2 A 2 1 1 b c 1
3 A 3 1 1 c d 1
4 B 1 1 1 b b 1
5 B 2 0 1 a a 0
6 B 3 1 1 b b 1
7 C 1 1 1 b b 1
8 C 2 1 1 c c 1
9 C 3 1 1 c b 1
10 D 1 1 1 c b 1
11 D 2 1 0 b a 0
12 D 3 1 1 c b 1
13 E 1 1 1 d d 1
14 E 2 1 1 b c 1
15 E 3 1 1 c c 1
</code></pre>
<p>However, it is unclear to me if I can use the $\times$ operator for categorical variables. Basically, what I am interested in is when $joint_{jt} = 1$ and $p_m == p_w$, then 1. </p>
<p>My question is, how do you formalise $p_m == p_w$, when $p$ is categorical?
I have been told that I could not use <em>logical operators</em> in my papers (econ, sociology field), but that I should use the arithmetic operators. So, how do you express <code>TRUE/FALSE</code> mathematically? </p>
<p>Now I have </p>
<p>$jointact_{jt} = (joint_{jt} \times act_{jt}^{m} \times act_{jt}^{w})$</p>
<p>But this seems wrong to me. </p>
<p>Could I for instance do, first define a vector $act$ if $act_{jt}^{m} = act_{jt}^{j}$ is true</p>
<p>$$
act_{jt} = \left\{\begin{array}{ll}
1 & act_{jt}^{m} = act_{jt}^{w}\\
0 & otherwise
\end{array}\right.
$$</p>
<p>and then combine $act_{jt}$, with $join_{jt}$, like </p>
<p>$jointact_{jt} = (joint_{jt} \times act_{jt})$</p>
<pre><code> ID time p_m p_w act_m act_w join_j joinact_j
1 A 1 1 1 c b 1 0
2 A 2 1 1 b c 1 0
3 A 3 1 1 c d 1 0
4 B 1 1 1 b b 1 1
5 B 2 0 1 a a 0 0
6 B 3 1 1 b b 1 1
7 C 1 1 1 b b 1 1
8 C 2 1 1 c c 1 1
9 C 3 1 1 c b 1 0
10 D 1 1 1 c b 1 0
11 D 2 1 0 b a 0 0
12 D 3 1 1 c b 1 0
13 E 1 1 1 d d 1 1
14 E 2 1 1 b c 1 0
15 E 3 1 1 c c 1 1
</code></pre>
| Chinmaya mishra | 445,597 | <p>A system is linear if $A(t),B(t),C(t),D(t)$ are linear i.e $dx/dt=2x+3u$ . If these depend on time as the questions looks like , then (5) and (6) are correct.</p>
<p>If you look at (5) and (6) $A,B,C,D,x,y$ are independent .To decide to what dimension it goes to , just recall the definition of vector field .
For example , $h(x)=f(u(x),h(x))$ here h goes to $R^2$ . Similarly calculate all possible depencies of your $f.f$ goes from $R$ to $R^{n^2+nk+2}$.($n^2$ total no. of tuples in $A$ $nk$ is total no. of tuples in $B$ and others are $x(t)$ and $u(t)$) .</p>
|
2,900,014 | <p>How would solve for $a$ in this equation without using an approximation ?
is it possible?</p>
<p>where $x>0$ and $0<a<\infty$</p>
<p>$x=\Sigma _{i=1}^{n} i^a$</p>
<p>for example $120=\Sigma _{i=1}^{6} i^a$ what is $a$ in this equation?</p>
| David G. Stork | 210,401 | <p>The sum you give yields the <em><a href="https://en.wikipedia.org/wiki/Harmonic_number" rel="nofollow noreferrer">Harmonic number</a></em>, $H_n^{(-a)}$. This can be solved, in <em>Mathematica</em> for instance:</p>
<pre><code>Solve[k == HarmonicNumber[n, -a], a][[1]]
</code></pre>
<p>When you plug in $k=120$ and $n=6$ as required by the problem example one finds:</p>
<p>$a = 2.17926794327541855469750380268$</p>
<p>As a check:</p>
<p>$\sum\limits_{i=1}^6 i^a = 119.9999$ (close enough!).</p>
|
2,588,408 | <p>A question from <em>Introduction to Analysis</em> by Arthur Mattuck:</p>
<p>Suppose $f(x)$ is continuous for all $x$ and $f(a+b)=f(a)+f(b)$ for all $a$ and $b$. Prove that $f(x)=Cx$, where $C=f(1)$, as follows:</p>
<p>(a)prove, in order, that it is true when $x=n, {1\over n}$ and $m\over n$, where $m, n$ are integers, $n\ne 0$;</p>
<p>(b)use the continuity of $f$ to show it is true for all $x$.</p>
<p>I can show the statement is true when $x=n$. As for $x={1\over n},{m\over n}$, I don't know how.</p>
| Angina Seng | 436,618 | <p>$f(1/2)+f(1/2)=f(1)=C$: what is $f(1/2)$?</p>
<p>$f(1/3)+f(1/3)+f(1/3)=f(2/3)+f(1/3)=f(1)=C$: what is $f(1/3)$?</p>
<p>etc.</p>
|
690,569 | <p>Suppose a function is given by:
$$
f(x)=
\begin{cases}
\cos\left(\dfrac{1}{x}\right) & x\neq 0 \\
0 & x=0
\end{cases}
$$</p>
<p>Show that this function is not continuous. Please help - I don't know how to proceed with formally using the limits.</p>
| Brian Fitzpatrick | 56,960 | <p>There are lots of ways to do this. One way is to use the following result.</p>
<p><strong>Proposition.</strong> A function $f:\mathbb R\to\mathbb R$ is continuous at $p\in\mathbb R$ if and only if for every sequence $\{t_n\}$ with $t_n\to p$ we have $f(t_n)\to f(p)$.</p>
<p>Now, let $t_n=\dfrac{1}{2n\pi}$. Then $t_n\to 0$ but
$$
f(t_n)=f\left(\dfrac{1}{2n\pi}\right)=\cos\left(2n\pi\right)=1\to 1\neq 0=f(0)
$$
so $f$ is not continuous at $0$.</p>
|
690,569 | <p>Suppose a function is given by:
$$
f(x)=
\begin{cases}
\cos\left(\dfrac{1}{x}\right) & x\neq 0 \\
0 & x=0
\end{cases}
$$</p>
<p>Show that this function is not continuous. Please help - I don't know how to proceed with formally using the limits.</p>
| kevin | 131,542 | <p>The problem is at the point $x=0$. Consider one of the two sequences which converge to $0$; $x_n=\frac{1}{2n\pi }$ or $y_n=\frac{1}{(2n+1)\pi }$ for instance, then both sequences go to zero when $n$ goes to $+\infty$ but $f(x_n)=1$ and $f(y_n)=-1$ do not go to zero... thus the function cannot be continuous at $x=0$.</p>
|
620,370 | <p>I am tackling a problem which asks:</p>
<p>Find the sum of all the multiples of 3 or 5 below 1000.</p>
<p>My reasoning is that since
Since $\left\lfloor\frac{1000}{3}\right\rfloor = 333$ and $\left\lfloor\frac{1000}{5}\right\rfloor = 200$</p>
<p>This sum can be denoted as:
\begin{equation}
\sum\limits_{n=1}^{333}3n + \sum\limits_{m=1}^{200} 5m = 3\sum\limits_{n=1}^{333}n + 5\sum\limits_{m=1}^{200} m
\end{equation}
Using the identity
$$
\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}
$$
It can be solved analytically as
$$
3\sum\limits_{n=1}^{333}n + 5\sum\limits_{m=1}^{200} m = \frac{3}{2}(333(333+1)) + \frac{5}{2}(200(200+1)) = 267333
$$</p>
<p>But apparently this is incorrect? Where did I go wrong in my reasoning?</p>
<h2>Edit</h2>
<p>I tried to correct the double multiples with:
$$3\sum\limits_{i=1}^{333}i + 5\sum\limits_{j=1}^{200}j - 15\sum\limits_{k=1}^{66} k = 234168$$</p>
<p>But this seems to be wrong as well...</p>
<h2>Edit 2:</h2>
<p>GOT IT!</p>
<p>Up to and <strong>not</strong> including.</p>
<p>$$3\sum\limits_{i=1}^{333}i + 5\sum\limits_{j=1}^{199}j - 15\sum\limits_{k=1}^{66} k = 233168$$</p>
<p>Or, more ugly:
$$3\sum\limits_{i=1}^{\left\lfloor\frac{999}{3}\right\rfloor}i + 5\sum\limits_{j=1}^{\left\lfloor\frac{999}{5}\right\rfloor}j - 15\sum\limits_{k=1}^{\left\lfloor\frac{999}{3\cdot 5}\right\rfloor} k = 233168$$</p>
| nadia-liza | 113,971 | <p>you double added all the multiples of 15</p>
|
2,672,497 | <p>$$\lim _{n\to \infty }\sum _{k=1}^n\frac{1}{n+k+\frac{k}{n^2}}$$ I unsuccessfully tried to find two different Riemann Sums converging to the same value close to the given sum so I could use the Squeeze Theorem. Is there any other way to solve this?</p>
| Paramanand Singh | 72,031 | <p>The sum under limit is not a Riemann sum, but it differs from a Riemann sum by negligible amount.</p>
<p>To setup things let's note that by definition of Riemann integral we have <span class="math-container">$$\int_{0}^{1}f(x)\,dx=\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}f(t_k),\,\frac{k-1}{n}\leq t_k\leq \frac{k} {n} $$</span> It is easy to guess that the function <span class="math-container">$f$</span> involved here is given by <span class="math-container">$f(x) =1/(1+x)$</span> and let's choose <span class="math-container">$$t_1=0,t_k=\frac{k-1}{n}+\frac{k-1}{n^3},k=2,3\dots,n$$</span> The corresponding Riemann sum is <span class="math-container">$$S_n=\frac{1}{n}+\frac{1}{n}\sum_{k=2}^{n}\dfrac{1}{1+\dfrac{k-1}{n}+\dfrac{k-1}{n^3}}$$</span> which can be rewritten as <span class="math-container">$$S_n=\frac{1}{n}+\sum_{k=1}^{n-1}\frac{1}{n+k+(k/n^2)}$$</span> If the sum in question is denoted by <span class="math-container">$S'_n$</span> then we can see that <span class="math-container">$$S_n-S'_n=\frac{1}{n}-\frac{1}{n+n+(1/n^2)}$$</span> and clearly the above expression tends to <span class="math-container">$0$</span> as <span class="math-container">$n\to\infty $</span>. The Riemann sum <span class="math-container">$S_n$</span> tends to <span class="math-container">$\int_{0}^{1}dx/(1+x)=\log 2$</span> and hence the given sum <span class="math-container">$S'_n$</span> also tends to <span class="math-container">$\log 2$</span>.</p>
<p>The same technique has been used in <a href="https://math.stackexchange.com/a/3168198/72031">this answer</a> in a simpler manner. </p>
|
1,018,672 | <blockquote>
<p><span class="math-container">$$\int_0^{\infty} \frac{1}{x^3-1}dx$$</span></p>
</blockquote>
<p>What I did:</p>
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$</span></p>
<hr />
<blockquote>
<p>This is where my problem is, what is :</p>
<p><span class="math-container">$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$</span></p>
</blockquote>
<p>^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.</p>
| 2'5 9'2 | 11,123 | <p>This integral is not defined. You can't write $$\int_0^{\infty} \frac{1}{x^3-1}dx=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx\right)$$
(Note that although you initially write two separate limits, you combine them into one limit in a later step, so you are actually working with what I have above. Either that or you correctly started out with two limits, but then later incorrectly combined $$\lim_{\epsilon\to0^+}f(\epsilon)+ \lim_{\epsilon\to0^+}g(\epsilon)=\lim_{\epsilon\to0^+}\left(f(\epsilon)+g(\epsilon)\right)$$ without verifying the two limits each exist.)</p>
<p>This would be true if you have established the first integral exists in the first place, but it does not. Note that this setup has the limiting variables approaching the pole at $x=1$ at the same rate from either side. This is artificially creating cancellation as $x\to1^-$ in the one integral and $x\to1^+$ in the other. Something like
$$\int_0^{\infty} \frac{1}{x^3-1}dx=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon^2} \frac{1}{x^3-1}dx+\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx\right)$$ should be just as valid as the first equation, but here the result will be $-\infty$ instead of the finite answers others have found ($0$ and $-{\frac{\pi}{3\sqrt{3}}}$).</p>
<p>Instead, you can write
$$\begin{align}
\int_0^{\infty} \frac{1}{x^3-1}dx
&=\int_0^{1} \frac{1}{x^3-1}dx+\int_1^{\infty} \frac{1}{x^3-1}dx&\text{(provided both exist)}\\
&=\lim_{\epsilon_1\to0^+}\int_0^{1-\epsilon_1} \frac{1}{x^3-1}dx+\lim_{\epsilon_2\to0^+}\int_{1+\epsilon_2}^{\infty} \frac{1}{x^3-1}dx\\
&=\lim_{\epsilon_1\to0^+}\int_0^{1-\epsilon_1} \frac{1}{x^3-1}dx+\lim_{\epsilon_2\to0^+}\lim_{\epsilon_3\to\infty}\int_{1+\epsilon_2}^{\epsilon_3} \frac{1}{x^3-1}dx
\end{align}$$
Note that the limiting variables are different. Neither of these improper integrals exist, since the integrands behave like $\frac{c}{x-1}$ near $x=1$.</p>
|
251,430 | <p>Consider the measure space $(\mathbb{Z},\mathcal{P}(\mathbb{Z}),\#)$, where $\#$ is the counting measure on $\mathbb{Z}$ and $\mathcal{P}(\mathbb{Z})$ is its power set.</p>
<p>I would like to show that for any measurable function we have $\int f(n)d\#(n)=\sum_{n}f(n)$.</p>
<p>This is what I have done: Let $x\in\mathbb{Z}$ and consider the indicator function $1_{\{x\}}$. Then $$\int_\mathbb{Z} fd\#=\int_\mathbb{Z} 1_{\{x\}}d\#=\#\{x\}=1,$$ for $f=1_{\{x\}}$. Next, for a step function $f=\sum_{k=-n}^na_k1_{\{x_k\}}$ (where $x_k\in\mathbb{Z}$ and $a_k$ are real rumbers for all $k$) we have $$\int_\mathbb{Z} fd\#=\sum_{k=-n}^na_k\int_\mathbb{Z}1_{\{x\}}d\#=\sum_{k=-n}^na_k.$$</p>
<p>How do I finish this proof? I still need to prove the statement for an arbitrarily measurable function. </p>
| smiley06 | 63,954 | <p>I think the statement holds only for positive measurable functions. If you take $ f(n) = \frac{(-1)^n}{n} $ then $ \sum_n f(n) < \infty $ but $$ \int_\mathbb{Z} f^+ d\# = \int_\mathbb{Z} f^- d\# = \infty $$ So you end up with $\infty-\infty $</p>
|
1,056,045 | <blockquote>
<p>Show that </p>
<p>$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$</p>
</blockquote>
<p>I'm not really sure how to get started on this problem, but here is what I have done so far:</p>
<p>Base case $n(1)$:</p>
<p>$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$</p>
<p>But not sure where to go from here. Tips?</p>
| AJY | 192,914 | <p>$2^{3n} = 8^{n}$. Let $s_{n} = 8^{n} - 1$.</p>
<p>$$8^{n + 1} - 1 = 8(8^{n}) - 1 = 8(8^{n} - 8) + 7 = 8(8^{n} - 1) + 7 = 8s_{n} + 7$$</p>
|
1,304,971 | <p>Using a triangular facet approximation of a sphere based on <a href="http://paulbourke.net/geometry/circlesphere/" rel="nofollow">Sphere Generation by Paul Bourke</a>.</p>
<p>We take an octahedron and bisect the edges of its facets
to form 4 triangles from each triangle.</p>
<p><code>
/\ /\ /\
/ \ / \ /__\
/ \ / \ /\ /\
/ \ /______\ /__\/__\
/ \ /\ /\ /\ /\ /\
/ \ / \ / \ /__\/__\/__\
/ \ / \ / \ /\ /\ /\ /\
/______________\ /______\/______\ /__\/__\/__\/__\
0th generation 1st generation 2nd generation
</code></p>
<p>This happens for every face so 8 times for the first generation.</p>
<p>Between each generation new vertices are pushed to the surface of the sphere.</p>
<p>The number of facets will be $(4^\mathrm{generations}) \cdot 8$</p>
<p>Some facets will share vertices: in the $0^{th}$ generation there are 6 unique vertices (it's an octahedron).</p>
<p>How many unique vertices will there be in the $N^{th}$ generation?</p>
| augurar | 85,153 | <p>Observe that each vertex is shared by $6$ faces, except the six vertices of the original octahedron, which are always shared by only $4$ faces in each generation. Each face has $3$ vertices. The number of faces in generation $N$ is $8 \cdot 4^{N}$. Putting this together, we see that the number of vertices in that generation is $$\frac{3 \cdot 8 \cdot 4^{N} - 4 \cdot 6}{6} + 6 = 4^{N+1} + 2$$</p>
|
3,694,661 | <p>I was stuck on a problem from <a href="https://rads.stackoverflow.com/amzn/click/com/0821804308" rel="nofollow noreferrer" rel="nofollow noreferrer" title="Quite a Fun Recreational Math Book">Mathematical Circles: Russian Experience</a>, which reads as follows:</p>
<blockquote>
<p><em>Prove that the number <span class="math-container">$6n^3 + 3$</span> cannot be a perfect sixth power of an integer for any natural number n</em>. </p>
</blockquote>
<p>The problems previous to this dealt with proving that numbers cannot be a cube and cannot be a square. The hints offered to these problem said that a square leaves a remainder of 0 or 1 when divided by 3 or 4, and that a cube leaves a remainder of 0, 1 or 8 when divided by 9. However, for this problem, the hint states that the reader should "experiment by dividing the number by 7 and comparing it remainders of sixth powers divided by 7". </p>
<p>Where did that come from? How would the solver figure out that <span class="math-container">$6n^3 + 3$</span> should be divided by 7? Moreover, why are 3 and 4 used in proving facts about squares, and why is 9 used when proving facts about cubes? Was this mainly through trial and error over the years, or is there some obvious fact that I'm blanking out on? </p>
<p>Thanks!</p>
| Tomita | 717,427 | <p><span class="math-container">$6n^3+3=m^6\tag{1}$</span> </p>
<p><span class="math-container">$n^3\equiv {0,1,6} \pmod{7}$</span> then <span class="math-container">$6n^3+3\equiv {2,3,4} \pmod{7}$</span>.<br>
On the other hand, <span class="math-container">$m^6\equiv {0,1} \pmod{7}$</span><br>
Hence <span class="math-container">$LHS$</span> is not equal to <span class="math-container">$RHS$</span> <span class="math-container">$\pmod{7}.$</span><br>
Therefore equation <span class="math-container">$(1)$</span> has no intger solution.</p>
|
366,415 | <p>Find all the natural numbers where $ϕ(n)=110$ (Euler's totient function)</p>
<p>What the idea behind this kind of questions?</p>
| DonAntonio | 31,254 | <p>Hints: if the prime decomposition of $\,n\,$ is</p>
<p>$$n=\prod_{i=1}^np_i^{a_i}\implies \phi(n)=n\prod_{i=1}^n\left(1-\frac{1}{p_i}\right)\implies$$</p>
<p>$$2\cdot5\cdot 11=110=\phi(n)=n\prod_{i=1}^n\left(1-\frac{1}{p_i}\right)\ldots$$</p>
|
366,415 | <p>Find all the natural numbers where $ϕ(n)=110$ (Euler's totient function)</p>
<p>What the idea behind this kind of questions?</p>
| Warren Moore | 63,412 | <p><strong>Hint.</strong> Assuming you mean Euler's totient $\phi$, if you factor $n=p_1^{e_1}\cdots p_k^{e_k}$, then</p>
<p>$$
\phi(n)=\phi(p_1^{e_1})\cdots\phi(p_k^{e_k})=p_1^{e_1-1}(p_1-1)\cdots p_k^{e_k-1}(p_k-1)
$$</p>
<p>So look for primes $p$ such that $p-1\mid 110$ and go from that.</p>
|
366,415 | <p>Find all the natural numbers where $ϕ(n)=110$ (Euler's totient function)</p>
<p>What the idea behind this kind of questions?</p>
| wendy.krieger | 78,024 | <p>The totient of a prime $p$ is $p-1$. The totient of a prime power $p^n$ is $(p-1)p^{n-1}$.</p>
<p>There can't be more different odd divisors, then the power of $2$, because each new prime brings its own 'supply' of $2$, and an odd factor in a totient comes from odd primes. </p>
<p>So the solutions is a prime of the form $111$ or $11^2$. The first is composite, the second is a power of a prime. (In the case of $42$, which is $2 \cdot 3 \cdot 7$, both $43$ and $7^2$ work.)</p>
<p>Once one has addressed this issue, the next step is to consider the powers of $2$. The euler totient of $2$ itself is 1, so if $n$ is odd, then both $n$ and $2n$ both have the same totient. That applies here. </p>
|
541,926 | <p>I've been wrecking my brain with this problem and I really hope you can help me. You see I have a triangle that is either an isosceles or equilateral or right and I have to find a way to:
1)Convert it to a right one by moving one of its vertices,
2)Convert it to an isosceles one by moving one of its vertices,
3)Convert it to an equilateral one by moving two of its vertices.
I am trying to write a program in C that will do this but unfortunately I can't figure out the math. Hope you can help. Thank you in advance.</p>
| Stefan4024 | 67,746 | <p>First of all it would be nice to post you thought on the question and what you've already done, this will give everyone a better idea of you knowledge and they can point where is your mistake.</p>
<p>Here are some hint for you questions:</p>
<p><strong>a)</strong> If you want to make a right triangle, use some properties of it. Where does circumcentre lies in right triangle? How can this help you?</p>
<p><strong>b)</strong> Where does the vertex of the isoscelec triangle opposite from the base lies? Doesn't it lie on the line that bisect the base?</p>
<p><strong>c)</strong> In equilaterial triangle all sides are equal. Can you use this property? Further hint use circles with same radii.</p>
<hr>
<p>Since you want to write code for this I'll give you additional hint.</p>
<p>First of all find the coordinates of the three given vertices. Now we can continue:</p>
<p><strong>a)</strong> Find the middle point of the non moving side, opposed to the moving vertex. Draw a circle with radius of the half of the fixed side. The equation for a circle is:</p>
<p>$$(x-x_0)^2 + (y-y_0)^2 = r^2$$</p>
<p>Where $(x_0, y_0)$ are the coordinates of the circumcentre. Now for a point $x$ such thah $(x-x_0)^2 \le r^2$ you'll find a solution for $y$ and a pair of point $(x,y)$ for the third vertex such that we'll have right triangle.</p>
<p><strong>b)</strong> Find the equation for the bisector of the fixed side. Because this line has to pass throught the middle point of the fixed side. Find the coordinates of that point. Then use the slope-formula:</p>
<p>$$y-y_0 = m(x-x_0)$$</p>
<p>where $(x_0,y_0)$ are the cooridnates of the middle point. We know that the slope of the bisector is the negative reciprocal of the slope of the line. Find the slope of the fixed side and the rest of the should be easy. Then choose some $x\neq x_0$ and you'll get one point using the equation of the bisector.</p>
<p><strong>c)</strong> Find the length of the fixed side, using the distance formula. Then find the intersection of the circles centered on the verteces of the fixed sides with radii of the length of the fixed side. To obtain the equation for the circles, use the same method I showed in <strong>a)</strong></p>
|
3,668,702 | <p>As the title says, how should I go about finding the shortest distance between all pairs of nodes (Each node has x and y co-odrinates associated with it) on a graph?</p>
<p>A brute force method is to run shortest path finding algorithms between all the pairs of the points. Is there a better way to approach this problem. (The reason I need this is I am trying to solve Traveling Salesman Problem using Ant Colony Optimziation which requires the cost matrix between each pair of nodes)</p>
| Eric Towers | 123,905 | <p>There are several algorithms, differing in the nature of the edge type, edge cost, and running time. Assume a graph <span class="math-container">$G$</span> with vertices, <span class="math-container">$V$</span>, and edge, <span class="math-container">$E$</span>. You do not say if your graph is directed or undirected. You do not describe edge weights. You do not indicate an expected number of edges in terms of vertices or number of vertices in terms of edges, so identifying a "best" algorithm is impossible.</p>
<p>Used in asymptotic estimates below, "<span class="math-container">$V$</span>" and "<span class="math-container">$E$</span>" denote the number of vertices and number of edges, respectively.</p>
<p>For undirected graphs with positive real edge weights:</p>
<ul>
<li>The <a href="https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm" rel="noreferrer">Floyd-Warshall algorithm</a> has <span class="math-container">$O(V^3)$</span> runtime.</li>
<li><a href="https://archive.org/details/proceedingsofthi2002acms/page/283" rel="noreferrer">Pettie & Ramachandran</a> has <span class="math-container">$O(EV \ln \alpha(E,V))$</span> run time, where <span class="math-container">$\alpha$</span> is Tarjan's <a href="https://en.wikipedia.org/wiki/Ackermann_function#Inverse" rel="noreferrer">inverse Ackermann function</a>.</li>
</ul>
<p>For undirected graphs with nonnegative integer edge weights:</p>
<ul>
<li><a href="https://arxiv.org/abs/1312.6680" rel="noreferrer">Williams</a> has <span class="math-container">$O\left(V^3 / 2^{\Omega(\ln V)^{1/2}} \right)$</span> run time, where <span class="math-container">$\Omega$</span> is Knuth's <a href="https://en.wikipedia.org/wiki/Big_O_notation#Big_Omega_notation" rel="noreferrer">big omega notation</a>.</li>
<li><a href="https://dl.acm.org/doi/10.1145/316542.316548" rel="noreferrer">Thorup</a>, iterated over each vertex, has <span class="math-container">$O(EV)$</span> run time.</li>
</ul>
<p>For undirected graphs having weights <span class="math-container">$1$</span> (for present edges) and <span class="math-container">$\infty$</span> (for absent edges),</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Seidel%27s_algorithm" rel="noreferrer">Seidel's algorithm</a> is <span class="math-container">$O(V^\omega \ln V)$</span>, where <span class="math-container">$\omega$</span> is the exponent in square <a href="https://en.wikipedia.org/wiki/Matrix_Multiplication#Computational_complexity" rel="noreferrer">matrix multiplication</a>. It is known <span class="math-container">$\omega < 2.373$</span>.</li>
</ul>
<p>For directed graphs with real weights and no negative cycles, </p>
<ul>
<li>The Floyd-Warshall algorithm (mentioned above) has run time <span class="math-container">$O(V^3)$</span>.</li>
<li><a href="https://en.wikipedia.org/wiki/Johnson%27s_algorithm" rel="noreferrer">Johnson's algorithm</a> followed by <a href="https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm" rel="noreferrer">Dijkstra's algorithm</a> has run time <span class="math-container">$O(EV+V^2 \ln V)$</span>.</li>
<li><a href="https://www.sciencedirect.com/science/article/pii/S030439750300402X?via%3Dihub" rel="noreferrer">Pettie</a> has run time <span class="math-container">$O(EV + V^2 \ln \ln V)$</span>.</li>
</ul>
<p>For directed graphs with nonnegative integer weights,</p>
<ul>
<li>Williams algorithm, described above can be applied.</li>
<li><a href="https://dl.acm.org/doi/10.5555/646253.686343" rel="noreferrer">Hagerup</a> has run time <span class="math-container">$O(EV + V^2 \ln ln V)$</span>.</li>
</ul>
|
131,579 | <p>I need some help solving this problem.</p>
<p>A man is about to perform a random walk. He is standing a distance of 100 units from a wall. In his pocket, he has 10 playing cards: 5 red and 5 black.</p>
<p>He shuffles the cards and draws the top card.</p>
<p>If he draws a red card, he moves 50 units (half the distance from the wall) to the right (away from the wall).</p>
<p>If he draws a black card, he moves 50 units (half the distance from the wall) to the left (towards the wall).</p>
<p>How far from the wall will he be after all 10 cards have been drawn?</p>
<p>Thank you in advance for your help!</p>
| Arturo Magidin | 742 | <p>Note that $792=8\times 9\times 11$.</p>
<p>A number is divisible by $8$ if and only if the last three digits are divisible by $8$, so we need $45z$ to be divisible by $8$. That will give you the value of $z$.</p>
<p>A number is divisible by $9$ if and only if the sum of the digits is divisible by $9$. So you need $1+3+x+y+4+5+z$ to be a multiple of $9$. This means that $4+x+y+z$ must be a multiple of $9$. You will already know the value of $z$, so this gives you information about $x+y$.</p>
<p>A number is a multiple of $11$ if and only if the sum of the odd-place digits minus the sum of the even-place digits is a multiple of $11$. So we need $1-3+x-y+4-5+z$ to be a multiple of $11$. You will already have information about $z$, so this will give you information about $x-y$.</p>
<p>From knowing stuff about $x+y$ and about $x-y$, you should be able to figure out both $x$ and $y$.</p>
|
3,480,890 | <p>I do understand pure mathematical concepts of probability space and random variables as a (measurable) functions. </p>
<p>The question is: what is the real-world meaning of probability and how can we apply the machinery of probability to the real situations?</p>
<p>Ex1: probability of heads for fair coin is 1/2. Which means that "if we will make exactly the same experiment over and over again, we will obtain heads roughly 1/2 of the time". More rigorously, there is some convergences under the hood etc.
But this is nonsense: if I will make exactly the same initial conditions, velocity, humidity, wind etc, then I will always obtain the same result. Moreover, I cannot guarantee the same circumstances: we have some planets flying around, and molecules are not in the same places etc. So it is not really possible to make the same experiment twice. Like, you know, one cannot step twice in the same river.</p>
<p>Ex2: probability of rain is 0.9 for the next day. The "frequency" idea is really absurd in this case. </p>
<p>Ex3: I'm throwing a coin and see the side. You don't see it. I ask you, what is the probability that it comes Heads. This is really somewhat vague...</p>
| kccu | 255,727 | <p>This is really more of a philosophy of math question than a math question. You might find this Wikipedia page interesting as a starting point: <a href="https://en.wikipedia.org/wiki/Probability_interpretations" rel="noreferrer">https://en.wikipedia.org/wiki/Probability_interpretations</a>. I don't think there is one accepted "correct" interpretation of probability when it comes to the "real world." </p>
|
3,480,890 | <p>I do understand pure mathematical concepts of probability space and random variables as a (measurable) functions. </p>
<p>The question is: what is the real-world meaning of probability and how can we apply the machinery of probability to the real situations?</p>
<p>Ex1: probability of heads for fair coin is 1/2. Which means that "if we will make exactly the same experiment over and over again, we will obtain heads roughly 1/2 of the time". More rigorously, there is some convergences under the hood etc.
But this is nonsense: if I will make exactly the same initial conditions, velocity, humidity, wind etc, then I will always obtain the same result. Moreover, I cannot guarantee the same circumstances: we have some planets flying around, and molecules are not in the same places etc. So it is not really possible to make the same experiment twice. Like, you know, one cannot step twice in the same river.</p>
<p>Ex2: probability of rain is 0.9 for the next day. The "frequency" idea is really absurd in this case. </p>
<p>Ex3: I'm throwing a coin and see the side. You don't see it. I ask you, what is the probability that it comes Heads. This is really somewhat vague...</p>
| Ripi2 | 688,039 | <p>We don't know well how to do the weather forecase, just because we don't know the exact independat variables nor their full relation between them. Our mathematical models are not accurate enough.</p>
<p>But we have zillions of daily measures: wind, temperature, moist, etc.
<br>And we do this asertion:
"For the N times we have observed the same measures as now, we've been right P·N times at saying 'tomorrow rains'". So, tomorrow rains with a probability of P.</p>
<p>This is just a small example of how stadistics were born as a useful tool.</p>
|
1,326,816 | <p>How to find the Maclaurin series of the function $$f(x)=\frac{1}{(9-x^2)^2}$$
I guess we are gonna use derivatives but i have no idea how the final answer should be formed.</p>
| user84413 | 84,413 | <p>This answer uses ideas similar to the previous answers, but in a different order:</p>
<p>$\displaystyle\frac{1}{9-x^2}=\frac{\frac{1}{9}}{1-\frac{x^2}{9}}=\frac{1}{9}\sum_{n=0}^{\infty}\left(\frac{x^2}{9}\right)^{n}=\sum_{n=0}^{\infty}\frac{1}{9^{n+1}}x^{2n},\;\;$ so differentiating gives</p>
<p>$\displaystyle\frac{2x}{(9-x^2)^2}=\sum_{n=1}^{\infty}\frac{2n}{9^{n+1}}x^{2n-1}$ and therefore $\displaystyle\frac{1}{(9-x^2)^2}=\sum_{n=1}^{\infty}\frac{n}{9^{n+1}}x^{2n-2}$.</p>
|
2,485,447 | <p>My attempt:</p>
<p>3x≡1 mod 7 (1)</p>
<p>4x≡1 mod 9 (2)</p>
<p>Multiply (1) by 5</p>
<p>Multiply (2) by 7</p>
<p>x≡5 mod 7</p>
<p>x≡7 mod 9</p>
<p>So x≡9k+7</p>
<p>9k+7=5(mod7)</p>
<p>k=5(mod7)</p>
<p>k=7j+5</p>
<p>x=9(7j+5)+7</p>
<p>=63j+52</p>
<p>x≡52(mod63)</p>
| lab bhattacharjee | 33,337 | <p>Notice that we need $x+2$ divisible by $7$ and by $9$, hence by LCM$(7,9)$</p>
|
2,239,058 | <blockquote>
<p>Find the spherically symmetric solution to $$\nabla^2u=1$$ in the
region $r=|\mathbf{r}|\le a$ for $a>0$ that satisfies the following
boundary condition at $r=a$:</p>
<p>$\frac{\partial u}{\partial n}=0$</p>
</blockquote>
<p>The solution I have looked at states to begin with $\frac{\partial u}{\partial r}=0$ and I can go from here, my question is not actually the above question (but rather I'm using that to illustrate my actualy question) which is:</p>
<p>Why does $\frac{\partial u}{\partial n}=0 \implies \frac{\partial u}{\partial r}=0$? I thought maybe the chain rule, i.e. $\frac{\partial r}{\partial n}\frac{\partial u}{\partial r}=\frac{\partial u}{\partial n}=0$ but I have no idea what $\frac{\partial r}{\partial n}$ actually is and so I don't know that it's not equal to $0$, in fact I don't have a great grasp on what $\frac{\partial u}{\partial n}$ actually $is$, so i would also greatly appreciate if someone could just explain to me what these actually represent- at the moment they're very much just notation for me.</p>
<p>Sorry one more thing: we also arrive at $\frac{\partial u}{\partial r}=\frac13a$ at $r=a$ which I'm fine with, but than it says $>0$ so contradiction. Where has this come from?</p>
<p>Thanks in advance and sorry there are so many parts to my question, they just all are related to the above question and hence why I did not ask each separately.</p>
| Exodd | 161,426 | <p>You can't prove it because it is false.</p>
<p>$$A = \begin{pmatrix}1 & 0\\ 0 &1\end{pmatrix}$$</p>
<p>$$B = \begin{pmatrix}-1 & 0\\ 0 &-1\end{pmatrix}$$</p>
<p>aren't congruent, since $PAP^T = PP^T$ is definite positive, so can't be $B$. </p>
<p>Another example? let $A$ be the identity matrix, and $B$ the zero matrix. $A\ne PBP^T=B$.</p>
|
1,956,855 | <p>I'm doing some math work involving proofs, and one of the definitions is:</p>
<p>|a| = -a when a < 0</p>
<p>Isn't the absolute value of a, positive a no matter what a is in the beginning? Am I looking at this wrong? Could use an explanation.</p>
| Edward Evans | 312,721 | <p>If $a < 0$ then $-a > 0$, so $-a$ is positive.</p>
<p>The point is that if $a = -2$ for instance, then $\lvert -2 \vert = -(-2) = 2.$</p>
|
1,956,855 | <p>I'm doing some math work involving proofs, and one of the definitions is:</p>
<p>|a| = -a when a < 0</p>
<p>Isn't the absolute value of a, positive a no matter what a is in the beginning? Am I looking at this wrong? Could use an explanation.</p>
| fleablood | 280,126 | <p>"Isn't the absolute value of $a$, positive $a$ no matter what $a$ is in the beginning?"</p>
<p>Yes. $-a $ is a positive number. </p>
<p>"Am I looking at this wrong?" Yes. $-a$ is a positive number.</p>
<p>"Could use an explanation?" </p>
<p>$a < 0$. So $a$ is negative. Which means $-a > 0$ and $-a$ is positive. </p>
<p>Also $a = -|a|$ (an equation with 2 negative values). So if you negate both sides you get $-a = -(-|a|) = |a|$. </p>
<p>Or a specific example. Let a = -25. Then $|a| = 25$. $-a = -(-25) = 25$. SO $|a| = 25 = -a$.</p>
<p>Moral: when we say a "negative times a negative is a positive" we <em>mean</em> it. $-a$ is <em>positive</em> if $a$ is negative.</p>
|
16,797 | <p>Is there any good way to approximate following integral?<br>
$$\int_0^{0.5}\frac{x^2}{\sqrt{2\pi}\sigma}\cdot \exp\left(-\frac{(x^2-\mu)^2}{2\sigma^2}\right)\mathrm dx$$<br>
$\mu$ is between $0$ and $0.25$, the problem is in $\sigma$ which is always positive, but it can be arbitrarily small.<br>
I was trying to expand it using Taylor series, but terms looks more or less this $\pm a_n\cdot\frac{x^{2n+3}}{\sigma^{2n}}$ and that can be arbitrarily large, so the error is significant.</p>
| Mose Wintner | 5,523 | <p>How about some good old-fashioned trapezoid rule?</p>
|
2,796,618 | <p>I am trying to</p>
<p>i) determine the infimum</p>
<p>ii) show that there's a function for which $\int_{0}^{1} {f'(x)}^2 dx$ is the infimum</p>
<p>iii) show if such function is unique.</p>
<p>I tried out several functions that suit the given condition, but couldn't see how $\int_{0}^{1} {f'(x)}^2 dx$ changes as $f(x)$ changes. How could we solve this problem?</p>
| Yiorgos S. Smyrlis | 57,021 | <p><strong>Claim.</strong> <em>Let $\,\mathscr X=\{\,g\in C^1[0,1]: g(0)=0\,\,\& \,\,g(1)=1.\}$. Then then functional $\,\varPhi(\,g)=\int_0^1 \big(\,g'(x)\big)^2\,dx$, attains a global minimum at $\,f(x)=x$, i.e.
$$
\min_{g\in\mathscr X}\varPhi(g)=\varPhi(\,f)=1.
$$</em></p>
<p><em>Proof.</em> If $g\in\mathscr X$, then $g$ can be expressed as $g(x)=x+h(x)$, where $g\in C^1[0,1]$ and $g(0)=g(1)=0.$ We have that
$$
\varPhi(g)=\int_0^1 \big(g'(x)\big)^2\,dx=\int_0^1 \big(1+h'(x)\big)^2\,dx=1+2\int_0^1\,h'(x)\,dx+\int_0^1 \big(h'(x)\big)^2\,dx \\=1+\int_0^1 \big(h'(x)\big)^2\,dx\ge 1=\varPhi(\,f),
$$
since $\int_0^1 h'(x)\,dx=h(1)-h(0)=0$. Hence $\,\varPhi(g)\ge\varPhi(\,f)$, for all $g\in\mathscr X$. </p>
<p><strong>Uniqueness.</strong> In particular, if $g(x)\not\equiv x$, then $h'\ne 0$, where $h(x)=g(x)-x$, and hence $\int_0^1\big(h'(x)\big)^2\,dx>0$, which implies that $\varPhi(g)>1=\varPhi(\,f)$. Thus $\varPhi$ possesses a unique minimiser.</p>
<p><strong>Note.</strong> It can be shown with Calculus of Variation as well, but its quite complicated, if one wants a rigorous proof.</p>
|
2,687,932 | <p>Let $x\in\mathbb{R}$. Prove that $x=-1$ if and only if $x^3+x^2+x+1=0$.
This is a bi-conditional statement, thus to prove it we need to prove:
<a href="https://i.stack.imgur.com/PRPSC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PRPSC.png" alt="enter image description here"></a></p>
| StackTD | 159,845 | <p>You did the hard(er) part.</p>
<p>To prove the other implication, you simply need to show that $x=-1$ satisfies the cubic equation.</p>
<p>Substitute and simplify!</p>
|
60,081 | <p>I have a stochastic matrix $A \in R^{n \times n}$ whose sum of the entries in each row is $1$. When I found out the eigenvalues and eigenvectors for this stochastic matrix, it always happens that one of the eigenvalues is $1$. </p>
<p>Is it true that for any square <a href="https://en.wikipedia.org/wiki/Stochastic_matrix" rel="nofollow noreferrer">row- or right-stochastic matrix</a> (i.e. each row sums up to $1$) one of the eigenvalues is $1$?
If so, how do we prove it?</p>
| leonbloy | 312 | <p>That is a <a href="http://en.wikipedia.org/wiki/Stochastic_matrix#Definition_and_properties" rel="nofollow noreferrer">basic</a> and important property of stochastic matrices. It's also non-obvious, unless you are aware of the <a href="http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem" rel="nofollow noreferrer">Perron-Frobenius theorem</a>.</p>
|
1,760,148 | <p>If I have a connected metric space $X$, is any ball around a point $x\in X$ also connected?</p>
| Brian M. Scott | 12,042 | <p>No. The <a href="https://en.wikipedia.org/wiki/Knaster%E2%80%93Kuratowski_fan">Knaster-Kuratowski fan</a> is a connected subspace of the plane that becomes totally disconnected when a certain point is removed, so open balls centred at the other points cannot be connected if they are small enough to exclude the explosion point.</p>
|
2,987,071 | <blockquote>
<p>I have to show that the set <span class="math-container">$$\{1, 1 + X, (1 + X)^2 , . . . , (1 + X)^n \}$$</span> is a basis
for <span class="math-container">$\Bbb{R}_n [X]$</span>, where <span class="math-container">$\Bbb{R}_n [X]$</span> denotes the vectorspace of all polynomials of degree less than or equal to <span class="math-container">$n$</span> over <span class="math-container">$\Bbb{R}$</span>. </p>
</blockquote>
<p>I guess I need to show that its spanning set and linear independent</p>
| lhf | 589 | <p>The map <span class="math-container">$p(X) \mapsto p(X+1)$</span> is a bijective linear transformation <span class="math-container">$\Bbb{R}_n [X] \to \Bbb{R}_n [X]$</span>.</p>
<p>The set in question is the image of the canonical basis of <span class="math-container">$\Bbb{R}_n [X]$</span> and so is a basis of <span class="math-container">$\Bbb{R}_n [X]$</span>.</p>
|
483,131 | <p>How can you prove that if $t$ is less than or equal to $1$, that the probability of the sum of a sequence of uniform random variables being less than or equal to $t$ equals $t^k/k!$ ?</p>
<p>In other words:</p>
<p>Prove if $t \leq 1$, $$P(U_1+U_2 + \dots +U_k \leq t)=\frac{t^k}{k!}$$</p>
<p>My thought is that integration is needed <em>CDF</em>, but I can't figure out how to analytically solve this. </p>
<p>Thanks!</p>
| user2566092 | 87,313 | <p>If you are familiar with the Chinese remainder theorem, it gives a one line proof of this, because it implies that the unique solution to $(a-b) = 0 \hbox { mod } p_1 p_2 \cdots p_n$ for $(a-b)$ is the solution to the system $(a-b) = 0 \hbox { mod } p_i$ for each $i$, since the $p_i$ are distinct.</p>
|
1,918,674 | <p>For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$</p>
<p>I was thinking about multiplying the numerator by it's conjugate, but that makes the denominator very messy, so I don't rlly know what to do.</p>
| Claude Leibovici | 82,404 | <p>Beside the simple solutions already given,you could change variable $x=y+2$ which makes
$$\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}=\frac{\sqrt{2 y+9}-\sqrt{y+9}}{y}$$ Now, using the generalized binomial theorem or Taylor series, you should find $$\frac{\sqrt{2 y+9}-\sqrt{y+9}}{y}=\frac{1}{6}-\frac{y}{72}+O\left(y^2\right)$$ with $y \to 0$.</p>
|
1,319,761 | <p>I'm stuck trying to figure out how to solve the following integral:</p>
<p>$\int_{C(0,1)^+}\sin(z)dz$</p>
<p>I've tried parameterizing z(t) but then I get</p>
<p>$\int_0^{2\pi}\sin(e^{it})ie^{it}$ which I don't know how to integrate.</p>
<p>So then I'm looking to use Cauchy's Integral formula but I'm not sure if I can use it because it doesnt fit the structure.</p>
<p>I'm not sure where to go from here so any help would be greatly appreciated!</p>
| Timbuc | 118,527 | <p><strong>First way:</strong></p>
<p>Parametrize the unit circle as $\;\gamma(t):=e^{it}\;,\;\;0\le t<2\pi\;\implies \gamma'(t)=ie^{it}dt $</p>
<p>so that your integral becomes</p>
<p>$$i\int_0^{2\pi}e^{it}\sin e^{it}dt=\left.-i\cos e^{it}\right|_0^{2\pi}=0$$</p>
<p><strong>Second way:</strong></p>
<p>The function $\;\cos z\;$ is a primitive function of $\;\sin z\;$ in the whole complex plane (this is what sometimes's called "a potential function" of a two-variable function in real analysis in the plane, and thus we get a conservative field...), so that directly</p>
<p>$$\oint_{S^1}\sin z\;dz=\left.-\cos z\right|_{z_1=0,z_2=0}=-(\cos 0-\cos 0)=0$$</p>
<p>or, of course, directly noting that when we've a primitive function (potential) and integrate on a closed, simple rectifiable path, the integral is zero.</p>
<p><strong>Third way:</strong></p>
<p>A third way is mentioned in the other answer: using the power series of sine, which has infinite convergence radius and is thus possible to integrate it elementwise in the whole complex plane, and then observing that with the usual parametrization of the unit circle (which you use)</p>
<p>$$\oint_{S^1} z^n dz=\int_0^{2\pi} ie^{it}e^{nit}\,dt=i\int_0^{2\pi} e^{(n+1)it}\,dt=\left.\frac{-i}{n+1}e^{(n+1)it}\right|_0^{2\pi} =0\;\;,\;\;\;n=0,1,2,...$$</p>
<p><strong>The last way:</strong></p>
<p>I only mention it for completeness as you obviously haven't yet studied it, but it is Cauchy's Theorem: the integral over a closed, simple path of an <em>analytic function</em> (analytic on the path and inside the region it encloses) is zero.</p>
|
692,376 | <p>Can a vector space over an infinite field be a finite union of proper subspaces ?</p>
| George Turcas | 119,544 | <p>Denote by $K$ the field over which $V$ is a vector space.
Suppose the answer is true and consider write $V=\bigcup\limits_{i=1}^{n} V_i$, where $V_i$ are proper subspaces of $V$ and $n \in \mathbb{N}$ is minimal.</p>
<p>Because $n$ is minimal, there exists an element $v_n \in V_n \setminus \bigcup\limits_{i=1}^{n-1}V_i $. There exists also an element $v \in \bigcup\limits_{i=1}^{n-1}V_i \setminus V_n$. </p>
<p>Now if we look at the infinite set $M=\{ v_n+k \cdot v : k \in K \}$, using Pigeonhole Principle we can see that there exists $i \in \{ 1, \dots, n\}$ such that $|M \cap V_i|\geq 2$ (we can actually say that the cardinal is infinite, but we don't need that for the rest of the proof).</p>
<p>If $V_n \cap M$ contains some element other than $v_n$, then there exists $k \in K \setminus \{ 0 \}$ such that $k \cdot v \in V_n$, which contradicts the choice of $v$.</p>
<p>If $V_i \cap M$ has more than two elements for some $i \in \{1,\dots, n-1\}$ then, looking at their difference, we obtain that $v_n \in V_i$ which contradicts the choice of $v_n$.</p>
<p>Therefore, $V$ is never a finite union of proper subspaces.</p>
<p>Edit: Repetition was not intended, I was posting while BlackAdder posted.</p>
|
441,448 | <p><strong>Contextual Problem</strong></p>
<p>A PhD student in Applied Mathematics is defending his dissertation and needs to make 10 gallon keg consisting of vodka and beer to placate his thesis committee. Suppose that all committee members, being stubborn people, refuse to sign his dissertation paperwork until the next day. Since all committee members will be driving home immediately after his defense, he wants to make sure that they all drive home safely. To do so, he must ensure that his mixture doesn't contain too much alcohol in it! </p>
<p>Therefore, his goal is to make a 10 gallon mixture of vodka and beer such that the total alcohol content of the mixture is only $12$ percent. Suppose that beer has $8\%$ alcohol while vodka has $40\%$. If $x$ is the volume of beer and $y$ is the volume of vodka needed, then clearly the system of equations is </p>
<p>\begin{equation}
x+y=10 \\ 0.08 x +0.4 y = 0.12\times 10
\end{equation}</p>
<p><strong>My Question</strong></p>
<p>The eigenvalues and eigenvectors of the corresponding matrix</p>
<p>\begin{equation}
\left[
\begin{array}{cc}
1 & 1\\
0.08 & 0.4
\end{array}
\right]
\end{equation} </p>
<p>are</p>
<p>\begin{align}
\lambda_1\approx 1.1123 \\
\lambda_2\approx 0.2877 \\
v_1\approx\left[\begin{array}{c}
0.9938 \\ 0.1116
\end{array}
\right] \\
v_2\approx\left[\begin{array}{c}
-0.8145 \\ 0.5802
\end{array}
\right]
\end{align}</p>
<p>How do I interpret their physical meaning in the context of this particular problem?</p>
| awkward | 76,172 | <p>You don't really need eigenvectors to solve this problem. Just treat it as a linear programming problem and let the Simplex Algorithm (or your algorithm of choice) come up with a feasible solution. Since you need an objective function, you can choose one arbitrarily; for example, you might choose to minimize the total cost. If you don't know the costs of beer and vodka, just make some up.</p>
|
2,882,678 | <p>Why are the morphisms of the category of sets functions?
Shouldn't the morphisms take an object in a category and turn it into another object of the category, i.e. map Set to Set. I don't understand how e.g.
$f(x)=x^2$
is a map from a set to a set. </p>
<p>If the morphism had been the image of a set under a function it would make more sense to me.</p>
| BallBoy | 512,865 | <blockquote>
<p>I don't understand how e.g. $f(x)=x^2$ is a map from a set to a set.</p>
</blockquote>
<p>In fact it is a map from a set to a set: you haven't given a precise definition of the function, so there are multiple options what the sets are, but $f$ may, e.g., map from the domain $\mathbb R$ (the set of all real numbers) to the codomain $\mathbb R_{\geq 0}$ (the set of all nonnegative real numbers). Then, in the category of sets, $f$ would constitute a morphism between the objects $\mathbb R$ and $\mathbb R_{\geq 0}$.</p>
<p>In this context, you probably want to be thinking of functions as rules taking an input value from one set and returning an output value from another set.</p>
|
1,021,631 | <p>Does anybody know how to solve the equation</p>
<p>$\mathbf{a} + \mathbf{b} \times \hat{\mathbf{v}} = c \hat{\mathbf{v}},$</p>
<p>where $\mathbf{a}$ and $\mathbf{b}$ are given real vectors, for the unit vector $\hat{\mathbf{v}}$ and the real number $c$?</p>
| Janak | 184,121 | <p>Here is an illustration of how to use Cauchy Scwarz inequality in this case:</p>
<p>You want to show: $\sum(X_i-\bar X)(Y_i-\bar Y) \leq\sqrt{\sum(X_i-\bar X)^2\sum(Y_i -\bar Y)^2}$ </p>
<p>CS inequality gives us:
$$\sum a_ib_i \leq\sqrt{\sum a_i^2\sum b_i^2},$$ here equality occurs if $a_i=c_1+c_2b_i$ $\forall i$, where $c_1,c_2\in R$.
Take $a_i=(X_i-\bar X)$ and $b_i=(Y_i-\bar Y)$, then you are done. </p>
|
1,045,941 | <p>Usually this is just given as a straight up definition in a calculus course. I am wondering how you prove it?
I tried using the limit definition, $$\lim\limits_{h\rightarrow 0} \dfrac{\log(x+h)-\log(x)}{h}$$
but this led to no developments.</p>
| Steven Gubkin | 34,287 | <p>This depends on how you define $\log(x)$. As you say, this could be taken to be the definition of $\log(x)$: define $y=\log(x)$ as the unique solution to the first order diff EQ $y'=\frac{1}{x}$ satisfying $y(1)=0$.</p>
<p>You might define $e^x$ as a solution to a diff EQ, and $\log$ as its inverse. In this case, you can prove the theorem by the standard method of finding derivatives of inverse functions.</p>
<p>What definition of $\log$ do you have in mind? Only then can we provide you with a computation of its derivative.</p>
|
2,025,007 | <p>One can show that if $n \geq 3$ is a positive integer, $d=n^2-4$, and $\varepsilon = 1$ if $n$ is odd and $\varepsilon = 0$ if $n$ is even, then the continued fraction expansion of $\frac{\sqrt{d}+\varepsilon}{2}$ has period of even length of the form $(1,n-2)$. One can show that such a continued fraction has period of odd length if and only if the negative Pell equation $t^2 - du^2 = -4$ has a solution. Thus it has no solution when $d=n^2-4$ except when $d=5$, since then the period degenerates to $(1)$. It takes some work to prove all of the details here.</p>
<p>However, in Barbeau's book "Pell's equation", Exercise 3.11, he says to take an odd positive integer $n$ and set $d = n^2-4$. Part (b) of the exercise says to show that $d$ has a prime factor congruent to 3 modulo 4 and hence show that the negative Pell equation $t^2 - du^2 = -4$ has no solution. But the claim is false: when $n=3$ or $n=15$, for instance, $d$ has no prime factor congruent to 3 modulo 4. </p>
<p>It gets me wondering, though, if he had something else in mind. Is there a short, simple proof that $t^2 - du^2 = -4$ has no solution for $d = n^2-4$, excepting the case $d=5$, perhaps just using some tricky algebra, congruence conditions, quadratic reciprocity, etc.?</p>
| Will Jagy | 10,400 | <p>We need consider only $d = n^2 - 4$ when $n = 12 w + 3.$ That is, if $n \neq 0 \pmod 3,$ then one of $(n+2)(n-2)$ is divisible by $3.$ Next, if $n \equiv 1 \pmod 4,$ then both of $(n+2),(n-2) \equiv 3 \pmod 4.$</p>
<p>The cycle of Gauss reduced forms equivalent to $x^2 - d y^2$ is of length 6, as you have found. The discriminant is $4 d.$</p>
<p>By the theorem of Lagrange, the numbers primitively represented, with absolute value up to $\sqrt d \approx n,$ are
$$ 1, 4, 2 - n. $$ Also primitively represented is
$$ 5 - 2n $$
Sometimes numbers represented in the cycle have absolute value larger than half the square root of the discriminant. A more careful search would be needed to confirm these in-between size numbers are the only one represented of that size.</p>
<p>I remember mentioning this to you and asking whether Zagier has something similar. I have his book, I believe he does, although I'm not sure he comes out and states the simplest version.</p>
|
1,329,214 | <p>I'm having difficulty solving a linear algebra problem:<br>
Let $A,B,C,D$ be real $n \times n$ matrices. Show that there is a non-zero $n \times n$ matrix $X$ such that $AXB$ and $CXD$ are both symmetric. </p>
<p>There is an accompanying hint:<br>
Show that the set of all matrices $X$ for which $AXB$ is symmetric is a vector space, and compute its dimension. </p>
<p>I introduced some notation:
let $S(A,B)$ denote the set of all $n \times n$ real matrices such that $AXB$ is symmetric. It's easy to prove that for any $A,B$ real $n \times n$ matrices $S(A,B)$ is a subspace of $\cal {M}_{n \times n} (\mathbb{R})$-it's closed under addition, scalar multiplication, and contains the zero matrix. Now I'm not entirely sure where to go. I'd like to be able to show that the space $S(A,B) \cap S (C,D)$ has dimension greater than $0$, but I haven't made any progress thus far. </p>
<p>Any help would be greatly appreciated!</p>
| Meni Rosenfeld | 153,429 | <p>Let $T$ map every matrix $X$ to $AXB$.</p>
<ol>
<li><p>Show that $T$ is a linear transformation.</p></li>
<li><p>What is the preimage of the space of symmetric matrices under $T$?</p></li>
<li><p>What is the dimension of the space of symmetric matrices?</p></li>
<li><p>What can you deduce from 1 and 2 about the dimension of $S(A,B)$?</p></li>
<li><p>Use 4 to prove the intersection is nontrivial.</p></li>
</ol>
|
989,740 | <p>How do I prove the following statement?</p>
<blockquote>
<p>If $x^2$ is irrational, then $x$ is irrational. The number $y = π^2$ is irrational. Therefore, the number $x = π$ is irrational</p>
</blockquote>
| André Nicolas | 6,312 | <p>To prove the first assertion, we can use a proof by contradiction.</p>
<p>Suppose to the contrary that $x$ is rational:
$$\exists (a, b) : b\ne 0\land x=\frac{a}{b}$$
$$\implies x^2=\frac{a^2}{b^2}$$
$$\therefore x^2 \text{ is rational. }$$
This contradicts the given fact that $x^2$ is irrational.</p>
<p><strong>Added:</strong> If you are only asking for how one goes from the first (universal) assertion to the particular case about $\pi$, it is by rule often called <a href="http://en.wikipedia.org/wiki/Universal_instantiation" rel="nofollow noreferrer">Universal Instantiation.</a></p>
|
4,204,282 | <p>How do I calculate <span class="math-container">$$\int_{-\infty}^{\infty} \frac{dw}{1+iw^3}$$</span> using complex path integrals?</p>
<p>I just need a hint on how to start, not the actual computation, because I need to understand how to deal with similar questions.</p>
<p><strong>Edit:</strong> Following @Tavish's comments, I used the residue theorem:</p>
<p>The function has poles at <span class="math-container">$w=\pm 0.866 +0.5i$</span> and <span class="math-container">$w=i$</span>. Now the integral is equal to <span class="math-container">$$2\pi i [Res(f,0.866 +0.5i)+Res(f,-0.866 +0.5i)]=\frac{2\pi}3.$$</span></p>
<p>However, I'm more interested in understanding the steps here than finding the answer. For example, how does one know which contour to take? Moreover, is there an alternate way to find this integral?</p>
| Svyatoslav | 869,237 | <p>I would recommend to implement some transformations of the integral first.</p>
<p>For example,
<span class="math-container">$$I=\int_{-\infty}^\infty\frac{dw}{1+iw^3}=-i\int_{-\infty}^\infty\frac{dw}{w^3-i}=-i\int_{0}^\infty\frac{dw}{1+iw^3}-i\int_{-\infty}^0\frac{dw}{1+iw^3}$$</span>
Making change in the second integral <span class="math-container">$t=-w$</span>
<span class="math-container">$$I=-i\Big(\int_{0}^\infty\frac{dw}{w^3-i}-\int_{0}^\infty\frac{dw}{w^3+i}\Big)=2\int_{0}^\infty\frac{dw}{w^6+1}$$</span>
Now, we can go different ways.</p>
<p>For example, this one:
<span class="math-container">$$I=2\int_{0}^\infty\frac{w^5dw}{w^5(w^6+1)}=\frac{2}{6}\int_{0}^\infty\frac{d(w^6)}{w^5(w^6+1)}=\frac{1}{3}\int_{0}^\infty\frac{t^{-5/6}}{(t+1)}dt$$</span>
Taking a keyhole contour in the complex plane with the cut from zero along the positive part of axis <span class="math-container">$X$</span> (we need it to make the function single-valued).</p>
<p><a href="https://i.stack.imgur.com/30LyV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/30LyV.png" alt="enter image description here" /></a></p>
<p>Integral along a big circle <span class="math-container">$\to0$</span> as <span class="math-container">$R\to \infty$</span> (function decreases rapidly enough). Also, integral along a small circle (around zero) <span class="math-container">$\to0$</span> as <span class="math-container">$r\to0$</span>.</p>
<p>Integral along the upper bank of the cut is the desired integral <span class="math-container">$I= \frac{1}{3}\int_0^\infty\frac{z^{-5/6}}{z+1}dx$</span></p>
<p>And integral along the lower bank of the cut is integral <span class="math-container">$-Ie^{2\pi i(-5/6)}$</span> (minus is due to integration in the negative direction, and the factor <span class="math-container">$e^{2\pi i(-5/6)}$</span> - due to the full turn counter-clockwise around the branch point <span class="math-container">$z=0$</span>)</p>
<p>You will get <span class="math-container">$I(1-e^{2\pi i(-5/6)})=2\pi i \,Res _{(z=e^{\pi i})}\frac{1}{3}\frac{z^{-5/6}}{z+1}=\frac{2\pi i}{3}e^{-5\pi i/6}$</span> - residual in one simple pole inside the contour.</p>
<p>Finally,
<span class="math-container">$$I=\frac{1}{3}\frac{2\pi i}{e^{5\pi i/6}-e^{-5\pi i/6}}=\frac{\pi}{3\sin(5\pi/6)}=\frac{2\pi}{3}$$</span></p>
<p>In another way we can evaluate this integral via beta-function. Making change <span class="math-container">$x=\frac{1}{1+t}$</span>
<span class="math-container">$$I=\frac{1}{3}\int_{0}^\infty\frac{t^{-5/6}}{(t+1)}dt=\frac{1}{3}\int_0^1\frac{(1-x)^{-5/6}x}{x^{-5/6}x^2}dx=\frac{1}{3}\int_0^1(1-x)^{-5/6}x^{-1/6}dx$$</span>
<span class="math-container">$$=B\big(1/6;5/6\big)=\frac{\Gamma(1/6)\Gamma(5/6)}{\Gamma(1/6+5/6)}$$</span>
Using Euler reflection formula <span class="math-container">$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$</span>
<span class="math-container">$$I=\frac{1}{3}\frac{\pi}{\sin(5/6)}=\frac{2\pi}{3}$$</span></p>
|
1,517,086 | <p>I spent a long time trying to find a natural deduction derivation for the formula $\exists x(\exists y A(y) \rightarrow A(x))$, but I always got stuck at some point with free variables in the leaves. Could someone please help me or give me some hints to find a proof. </p>
<p>Thanks.</p>
| BrianO | 277,043 | <p>You can derive it this way:</p>
<ol>
<li>$\exists y\,A(y) \qquad\qquad\textsf{assumption}$</li>
<li>$A(a) \qquad\quad\qquad\textsf{$\exists$ new parameter introduction}\text{ ($a$)}$</li>
<li>$\exists y\,A(y) \to A(a) \quad\quad\textsf{discharge 1.}$</li>
<li>$\exists x\,(\exists y\,A(y) \to A(x)) \quad\textsf{$\exists$ introduction}$</li>
</ol>
|
1,260,945 | <p>$\textbf{My understanding of divergence:}$ Consider any vector field $\textbf{u}$, then $\operatorname{div}(u) = \nabla \cdot u$. More conceptually, if I place an arbitrarily small sphere around any point of the vector field $\textbf{u}$, divergence measures the amount of "particles" exiting the sphere, i.e. positive divergence represent a vector field which is "moving faster" as we move to the right. However, how do I interpret $$ \int_U \operatorname{div}(u) \, dx$$ where $U$ is any bounded open subset of $\mathbb{R}^n$. </p>
| Elaqqad | 204,937 | <p>Let:
$$A=\{x^2|x\in \Bbb Z_p\},\ \ \ \ \ \ \ B=\{-(1+y^2)|y\in \Bbb Z_p\}$$</p>
<p>it is known that $$|A|=|B|=\frac{p+1}{2}$$ (maybe you can try to prove this ), if $A\cap B=\varnothing$ then $|A\cup B|=|A|+|B|=p+1>|\Bbb Z_p |$ which is impossible. as a conclusion $A\cap B$ is not empty and you're done.</p>
|
2,400,900 | <p>Let $f(x) : \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex and differentiable function, and let $P$ be a point in $\mathbb{R}^n$. </p>
<p>Define a function $g(m): R \rightarrow R$ to be the distance between point $P$ and the sub-level set $ K_m = \{ x \in \mathbb{R}^n \mid f(x) \le m\}$, i.e., $g(m) = d(P, K_m)$. Is the function $g$ a continuous function? If not, is there any restriction on $f$ that will make $g$ continuous? Thanks!</p>
| orangeskid | 168,051 | <p>HINT: the function $g$ is convex. </p>
<p>Indeed, we have </p>
<p>$$\lambda_1 K_{m_1} + \lambda_2 K_{m_2} \subset K_{\lambda_1 m_1 + \lambda_2 m_2}$$ since $f$ is convex, and</p>
<p>$$d(P, \lambda_1 K_{m_1} + \lambda_2 K_{m_2}) \le \lambda_1 d( P, K_{m_1}) + \lambda_2 d(P, K_{m_2} ) $$ since $Q\mapsto d(P, Q)$ is convex.</p>
|
2,400,900 | <p>Let $f(x) : \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex and differentiable function, and let $P$ be a point in $\mathbb{R}^n$. </p>
<p>Define a function $g(m): R \rightarrow R$ to be the distance between point $P$ and the sub-level set $ K_m = \{ x \in \mathbb{R}^n \mid f(x) \le m\}$, i.e., $g(m) = d(P, K_m)$. Is the function $g$ a continuous function? If not, is there any restriction on $f$ that will make $g$ continuous? Thanks!</p>
| haydn_c | 472,778 | <p>Another way to look at this is note that $ h(x, \lambda, m) = d(x, P) + \lambda (f(x) - m)$ is an affine function of $m$ and a convex function for $x$, for each given $\lambda$. </p>
<p>Hence $h_1(x, \lambda) = max_{\lambda \ge 0} h(x, \lambda, m)$ is a convex function of $m$ and $x$ since point-wise supremum preserves convexity, and $g(m) = \min_x h_1(x,m)$ is convex in $m$ since minimization preserves convexity. </p>
|
4,417,325 | <p>When I say "divisibility trick" I mean "a recursive algorithm designed to show that, after multiple iterations, if the final output is a multiple of the desired number, then the original was also a multiple of the same number." Here's an example for a divisibility trick for 17.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|q-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Here's another.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$100q+r$</span>, with <span class="math-container">$r<100$</span>. Then, evaluate <span class="math-container">$|r-2q|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Just to show these both work (or, at least, work for one particular number, let's try both on <span class="math-container">$31382$</span>.</p>
<blockquote>
<p>METHOD ONE:
<span class="math-container">$31382\rightarrow3128\rightarrow272\rightarrow17$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<blockquote>
<p>METHOD TWO:
<span class="math-container">$31382\rightarrow544\rightarrow34$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<p>These divisibility tricks rely on breaking the number down into groups of digits and applying some linear operation to them. However, when we try to use a <em>non-linear</em> function, things seem to break down. For example, much as how method one here draws off the fact that <span class="math-container">$17\ |\ 51$</span> and the second relies off of <span class="math-container">$17\ |\ 119$</span>, let's try to do something with <span class="math-container">$17\ |\ 34$</span>. Namely:</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|6q^2-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>We can try this with <span class="math-container">$34$</span> and see, yes, <span class="math-container">$6(9)-5(4)=34$</span>, so <span class="math-container">$17\ |\ 34$</span>. But this fails for most numbers. For <span class="math-container">$51$</span>, we have <span class="math-container">$51\rightarrow145$</span>, and it diverges from there (also note that <span class="math-container">$17\not|\ 145$</span>). Even with <span class="math-container">$17$</span>, which is obviously a multiple of <span class="math-container">$17$</span>, we have <span class="math-container">$17\rightarrow29$</span>.</p>
<p>What separates the wheat from the chaff here, so to speak? Why is it that if we break down the digits of the multiple of some prime and make a linear relation around it, it seems to be true for all other multiples of the prime, but the same doesn't work for, say, a quadratic relation?</p>
| mjqxxxx | 5,546 | <p>I wouldn't say that it's generally true that these linear reductions preserve divisibility; specific choices are being made for that to work out. For the first relation, where <span class="math-container">$n=10q+r$</span>:
<span class="math-container">$$
q-5r = q-5(n-10q) = 51q-5n\equiv-5n\;\text{(mod 17)}.
$$</span>
So <span class="math-container">$q-5r$</span> is divisible by <span class="math-container">$17$</span> iff <span class="math-container">$n$</span> is divisible by <span class="math-container">$17$</span>. The key is the vanishing of the extra term (in this case, <span class="math-container">$51q$</span>) when working modulo <span class="math-container">$17$</span>. This can't happen with an expression that is quadratic in <span class="math-container">$q$</span> and linear in <span class="math-container">$r$</span>... <span class="math-container">$r$</span> is linear in <span class="math-container">$q$</span> and <span class="math-container">$n$</span>, so there is no new <span class="math-container">$q^2$</span> term to cancel out the one you're starting with. That's leaving aside the fact that this kind of method is only <em>helpful</em> if the values are getting smaller... going from <span class="math-container">$n=10q+r$</span> to <span class="math-container">$|6q^2-5r|$</span>, for instance, doesn't have that guarantee.</p>
|
4,417,325 | <p>When I say "divisibility trick" I mean "a recursive algorithm designed to show that, after multiple iterations, if the final output is a multiple of the desired number, then the original was also a multiple of the same number." Here's an example for a divisibility trick for 17.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|q-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Here's another.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$100q+r$</span>, with <span class="math-container">$r<100$</span>. Then, evaluate <span class="math-container">$|r-2q|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Just to show these both work (or, at least, work for one particular number, let's try both on <span class="math-container">$31382$</span>.</p>
<blockquote>
<p>METHOD ONE:
<span class="math-container">$31382\rightarrow3128\rightarrow272\rightarrow17$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<blockquote>
<p>METHOD TWO:
<span class="math-container">$31382\rightarrow544\rightarrow34$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<p>These divisibility tricks rely on breaking the number down into groups of digits and applying some linear operation to them. However, when we try to use a <em>non-linear</em> function, things seem to break down. For example, much as how method one here draws off the fact that <span class="math-container">$17\ |\ 51$</span> and the second relies off of <span class="math-container">$17\ |\ 119$</span>, let's try to do something with <span class="math-container">$17\ |\ 34$</span>. Namely:</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|6q^2-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>We can try this with <span class="math-container">$34$</span> and see, yes, <span class="math-container">$6(9)-5(4)=34$</span>, so <span class="math-container">$17\ |\ 34$</span>. But this fails for most numbers. For <span class="math-container">$51$</span>, we have <span class="math-container">$51\rightarrow145$</span>, and it diverges from there (also note that <span class="math-container">$17\not|\ 145$</span>). Even with <span class="math-container">$17$</span>, which is obviously a multiple of <span class="math-container">$17$</span>, we have <span class="math-container">$17\rightarrow29$</span>.</p>
<p>What separates the wheat from the chaff here, so to speak? Why is it that if we break down the digits of the multiple of some prime and make a linear relation around it, it seems to be true for all other multiples of the prime, but the same doesn't work for, say, a quadratic relation?</p>
| Bill Dubuque | 242 | <p>No, divisibility tests are <em>not</em> restricted to <em>linear</em> forms. As explained <a href="https://math.stackexchange.com/a/16015/242">here</a> & <a href="https://math.stackexchange.com/a/2989299/242">here</a> the rule for casting out nines: <span class="math-container">$\,9\mid 10a+b\!\iff\! 9\mid a+b\,$</span> extends to higher degree as <span class="math-container">$\,9\mid p(10)\!\iff\! 9\mid p(1)\,$</span> [by <span class="math-container">$\!\bmod 9\!:\ p(10)\equiv p(1)\,],\:\!$</span> for any polynomial <span class="math-container">$p(x)$</span> with integer coef's (by <span class="math-container">$\rm\color{#0a0}{PCR}$</span> below). When <span class="math-container">$\,n = p(10)\,$</span> then <span class="math-container">$p(1)$</span> is the sum of the decimal digits of <span class="math-container">$n$</span>. Similarly <span class="math-container">$\,11\mid 10a\!+\!b\!\iff\! 11\mid a\!-\!b\,$</span> extends to <span class="math-container">$\,11\mid p(10)\!\iff\! 11\mid p(-1) =$</span> alternating digit sum.</p>
<p>The common tests you refer to correspond to <em>reversed</em> forms of the above divisibility tests, e.g. <span class="math-container">$\bmod 17\!:\ 10a+b\equiv 0 \!\iff\!$</span> <span class="math-container">$ 10(a+b\color{#c00}{/10})\equiv 0\!\iff\!$</span> <span class="math-container">$ a\color{#c00}{-5}b\equiv 0\,$</span> by <span class="math-container">$\,\color{#c00}{1/10\equiv -5},\,$</span> i.e. it arises via scaling by <span class="math-container">$\,\color{#c00}{10^{-1}\equiv -5}.\,$</span> Similarly, if <span class="math-container">$\,\deg p = k\,$</span> then scaling by <span class="math-container">$(-5)^k\equiv 10^{-k}$</span> changes all powers of <span class="math-container">$10$</span> in <span class="math-container">$\,p(10)\,$</span> into powers of <span class="math-container">$-5$</span>, effectively <em>reversing</em> the coef's, e.g. for a quadratic</p>
<p><span class="math-container">$\ \ \ \bmod \color{#c00}{17}\!:\,\ \ 0\equiv \overbrace{a\:\!10^2+b\:\!10+c}^{\large p(\color{#c00}{10})} \overset{\times\ (\color{c00}{-5})^{\large 2}\!\!}\iff\ \overbrace{0\equiv c(-5)^2+b(-5)+a}^{\large \tilde p(\color{#c00}{-5})}$</span></p>
<p>thus <span class="math-container">$\,\color{#c00}{17}\mid p(\color{#c00}{10})\!\iff\! 17\mid\tilde p(\color{#c00}{-5}) =\,$</span> reversed poly in radix <span class="math-container">$-5,\,$</span> by <span class="math-container">$\,\color{#c00}{10(-5)\equiv_{17} 1},\,$</span> e.g.</p>
<p><span class="math-container">$$ \color{#c00}{17}\mid 901\,\ \ {\rm by}\ \ 17\mid 109_{\color{#c00}{-5}} = 1(\color{#c00}{-5})^2+0(\color{#c00}{-5})+9 = 34 \quad$$</span></p>
<p>Such radix reciprocity divisibility by <span class="math-container">$\,d\,$</span> tests exist for any radices <span class="math-container">$r_1,\, r_2$</span> being reciprocal <span class="math-container">$\!\bmod d,\,$</span> i.e. when <span class="math-container">$\,r_1 r_2\equiv 1,\,$</span> e.g. for binary <span class="math-container">$\,r_2\!:\ \color{#0a0}{10(2)\equiv_{19} 1}$</span> and <span class="math-container">$\,\color{#c00}{10(-2)\equiv_{21} 1},\,$</span> so</p>
<p><span class="math-container">$$\begin{align} &\color{#0a0}{19}\mid 912\,\ \ {\rm by}\ \ 19\mid219_{\,\color{#0a0}2} \, =\ 2\,(\color{#0a0}2)^2\ +\ 1\,(\color{#0a0}2)\ +\ 9 = 19\\[.4em]
&\color{#c00}{21}\mid 924\,\ \ {\rm by} \ \ 21\mid 429_{\color{#c00}{-2}}= 4(\color{#c00}{-2})^2+2(\color{#c00}{-2})+9 = 21\end{align}\quad $$</span></p>
<p>This is but one of numerous examples of higher-degree divisibility inferences that are ubiquitous in number theory and algebra. Such inferences become obvious once one masters congruences and modular arithmetic (see esp. <span class="math-container">$\rm\color{#0a0}{PCR}$</span> = <a href="https://math.stackexchange.com/a/879262/242">Polynomial Congruence Rule</a>, i.e. <span class="math-container">$\,a\equiv b\Rightarrow p(a)\equiv p(b)).\,$</span></p>
<p>See <a href="https://math.stackexchange.com/a/3830244/242">here</a> for more on reverse (reciprocal) polynomials, and <a href="https://math.stackexchange.com/a/4373787/242">here</a> for a similar application of such.</p>
<p><strong>Note</strong> <span class="math-container">$ $</span> The reason that these divisibility tests can be expressed as iterations of <em>linear</em> operations is because that is how polynomials can be generated (nested Horner form), e.g.</p>
<p><span class="math-container">$$ a_0 + a_1 x + a_2 x^2 + a_3 x^3 =\, a_0 + x(a_1 + x (a_2 + x(a_3)))\qquad$$</span></p>
<p>i.e. polynomials can be generated by iterating linear operations <span class="math-container">$\,f_{n+1} = c_{n+1}+ x f_n,\,$</span> so any polynomial operation (e.g. evaluation <span class="math-container">$\!\bmod d$</span>) can be performed by recursively piggy-backing on this inductive generation process (see <a href="https://math.stackexchange.com/a/3575086/242">structural induction</a>).</p>
<p>In this way, recursive evaluation <span class="math-container">$\!\bmod d\,$</span> of a polynomial (representation of an integer in radix notation) leads to a <a href="https://math.stackexchange.com/a/52815/242"><em>universal</em> test for divisibility by <span class="math-container">$\,d,\,$</span></a> that works by repeatedly modding out leading chunks of digits <span class="math-container">$\!\bmod d\,$</span> (like longhand division but ignoring quotients). Your tests can be viewed as a reversed form of such a test. The forward form has the advantage over the reversed form that it yields the exact remainder so it can be used for much more than just divisibility testing.</p>
<p>Let's use the forward universal test to compute <span class="math-container">$\, 43211\bmod 7.\,$</span> The algorithm consists of repeatedly replacing the first two leading digits <span class="math-container">$\rm\ \color{#0a0}{d_n\ d_{n-1}}\ $</span> by <span class="math-container">$\rm\, \color{#0a0}{(\color{#000}3\, d_n + d_{n-1})}\bmod 7,\,$</span> since <span class="math-container">$\,10d_n+d_{n-1}\equiv 3d_n+d_{n-1}\pmod{\!7}$</span></p>
<p><span class="math-container">$$\begin{array}{rrl}\bmod 7\!:\ &\color{#0A0}{4\ 3}\ 2\ 1\ 1^{\phantom{|^{|}}}\!\!\!&\\
\equiv\!\!\!\! &\color{#c00}{1\ 2}\ 1\ 1 &\!{\rm by}\ \ \:\! \smash[t]{\overbrace{3\cdot \color{#0a0}4 +
\color{#0a0}3}^{\rm\textstyle\color{#0a0}{\,\color{#000} 3\,\ d_n\! + d_{n-1}}\!\!\!\!\!\!\!}} \equiv\ \color{#c00}1\\
\equiv\!\!\!\! &\color{#0af}{5\ 1}\ 1&\!{\rm by}\ \ \ 3\cdot \color{#c00}1 +
\color{#c00}2\ \equiv\ \color{#0af}5\\
\equiv\!\!\!\! & \color{#f60}{2\ 1}&\!{\rm by}\ \ \ 3\cdot \color{#0af}5 +
\color{#0af}1\ \equiv\ \color{#f60}2\\
\equiv\!\!\!\! &\color{#8d0}0&\!{\rm by}\ \ \ 3\cdot \color{#f60}2 +
\color{#f60}1\ \equiv\ \color{#8d0}0
\end{array}\qquad\qquad\quad\ \, $$</span></p>
<p>Hence <span class="math-container">$\rm\ 43211\equiv 0\pmod{\!7},\,$</span> indeed <span class="math-container">$\rm\ 43211 = 7\cdot 6173.\:$</span> Generally the modular arithmetic is simpler if we use <em>least magnitude</em> residues, e.g. <span class="math-container">$\rm\, \pm\{0,1,2,3\}\ \:(mod\ 7),\,$</span> by allowing <em>negative</em> digits, e.g. <a href="https://math.stackexchange.com/a/2119845/242">here</a>. Note that for modulus <span class="math-container">$11$</span> or <span class="math-container">$9\:$</span> the above method reduces to the well-known divisibility tests by <span class="math-container">$11$</span> or <span class="math-container">$9\:$</span> (a.k.a. "casting out nines" for modulus <span class="math-container">$9$</span>).</p>
|
2,375,023 | <p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
| Χpẘ | 309,642 | <p>HINT: Determine how many real roots there are. If there is only one root, how big is the interval?</p>
|
2,375,023 | <p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
| David K | 139,123 | <p>Note that
$$ f(x) = x^5+x^4+x^3+x^2+1 = (x + 1)(x^4 + x^2) + 1. $$</p>
<p>Now try answering some question about $f(x)$ for numbers that are relatively easy to work with:</p>
<p>Can $f(x)$ be zero if $x > 0$?
Can $f(x)$ be zero if $x < 0$?</p>
<p>Can $f(x)$ be zero if $x > -1$?
Can $f(x)$ be zero if $x < -1$?</p>
<p>Can $f(x)$ be zero if $x > -2$?
Can $f(x)$ be zero if $x < -2$?</p>
<p>If you haven't found an interval by this time, you can try other numbers.</p>
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1,109,918 | <p>Is it always possible to add terms into limits, like in the following example? (Or must certain conditions be fulfilled first, such as for example the numerator by itself must converge etc)</p>
<p>$\lim_{h \to 0} {f(x)} = \lim_{h \to 0} \frac{e^xf(x)}{e^x}$</p>
| abel | 9,252 | <p>i will use the fundamental theorem of calculus and integration by parts to derive
$\lim_{h \to 0} \frac{f(x + h) + f(x - h) - 2f(x)}{h^2}= f''(x)$</p>
<p>the left hand side limit may exist even for a function that is twice differentiable. take for example $f(x) = |x|, x = 0.$ the left hand side is $2$ and right hand side does not exist.</p>
<p>we will assume that $f$ is three times differentiable.</p>
<p>$\begin{align}
f(x+h) - 2f(x) + f(x-h) &=
\int_x^{x+h}f^\prime(t) dt - \int_{x-h}^{x}f^\prime(t) dt\\
&= (t - x - h)f^\prime(t) \rvert_x^{x+h} - \int_x^{x+h} (t - x - h) f^{\prime \prime}(t) dt \\
& \ \ - (t - x + h)f^\prime(t)\rvert_{x-h}^x + \int_x^{x+h} (t - x + h) f^{\prime \prime}(t) dt \\
& =- \int_x^{x+h} (t - x - h) f^{\prime \prime}(t) dt + \int_x^{x+h} (t - x + h) f^{\prime \prime}(t) dt \\
&= -\frac{1}{2}(t-x-h)^2 f^{\prime \prime}(t) \rvert_x^{x+h} + \frac{1}{2}(t-x-h)^2 f^{\prime \prime}(t) \rvert_{x-h}^{x} \\
& \ \ + \int_x^{x+h}\frac{1}{2}(t-x-h)^2 f^{\prime \prime \prime}(t) dt -
\int_x^{x+h}\frac{1}{2}(t-x-h)^2 f^{\prime \prime \prime}(t) dt\\
&=h^2f^{\prime \prime}(x)+Kh^3/6 \text{ where $K = f^{\prime \prime \prime}(c)$ for some $x-h < c < x + h$}
\end{align}$</p>
<p>moving things around and taking the limit gives you the desired result.</p>
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1,929,977 | <p>Let $f\colon\mathbb R\to\mathbb R$ satisfy the Lipschitz condition: there exists $K\geq 0$ such that for all $x,y\in\mathbb R$, we have $|f(x)-f(y)|\leq K\cdot |x-y|$. Is it true that $f$ has one-sided derivatives everywhere? I.e., that the limits
$$\lim_{h\nearrow 0}\frac{f(x+h)-f(x)}{h}\quad\text{and}\quad\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h}$$
exist for every $x$?</p>
<p>I know that both exist a.e. since the Lipschitz condition implies that $f$ is differentiable a.e., but I would like to know if it is not only true a.e., but <em>everywhere</em>. After working with this for some time, I got the feeling that it might be.</p>
| Community | -1 | <p>No. Counterexample: </p>
<p>Let $g : [1,2] \to \mathbb R$ be the graph which linearly connects the points
$$(1, 1), (1.5, -1.5), (2, 2).$$</p>
<p>Define $f [0,1] \to \mathbb R$ be </p>
<p>$$f(x) = \begin{cases} 0 & \text{if } x=0,\\ \frac{1}{2^{n+1}}\ g(2^{n+1} x) &\text{if }x\in [2^{-(n+1)}, 2^{-n}), n =0, 1, 2\cdots.\\
\end{cases}$$</p>
<p>Then $f$ is Lipschitz since it has bounded derivatives (except at some countable points where the derivatives does not exists). However, </p>
<p>$$\lim_{h\to 0^+} \frac{f(h) - f(0)}{h}$$</p>
<p>does not exist since </p>
<p>$$ \frac{f(2^{-(n+1)}) - f(0)}{2^{-(n+1)}} =1, \ \ \ \frac{f(2^{-(n+1)} 1.5) - f(0)}{2^{-(n+1)} 1.5} = -1$$</p>
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1,015,264 | <p>This is a worked out example in my book, but I am having a little trouble understanding it:</p>
<p>Consider the system of equations:</p>
<p>$$x'=y+x(1-x^2-y^2)$$
$$y'=-x+y(1-x^2-y^2)$$</p>
<p>The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?)</p>
<p>The polar coordinate satisfies $r^2=x^2+y^2$ so by differentiating with respect to t and using the differential equations we get:</p>
<p>$r\cdot r'=x\cdot x'+y\cdot y'$ (I am unclear about how the book even got this first equation from $r^2=x^2+y^2$) </p>
<p>$=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)$ Substitute in $x'$ and $y'$ and then multiple out and replace with $r$, I get this step</p>
<p>$=r^2(1-r^2)$ cancel terms, I get this step too</p>
<p>$r'=r(1-r)$</p>
<p>similarly, the angle variable $\theta$ satisfies $\tan\theta=\frac yx$, so the derivative with respect to $t$ yields $\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}$ so $\theta=1$</p>
<p>Thus the solution goes clockwise around the origin at unit angular speed.</p>
<p>I don't understand the $\theta$ step at all or how they reached the conclusion of clockwise around the origin with unit angular speed..</p>
<p>But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion..</p>
| Mark Fischler | 150,362 | <p>When you see an $x^2 + y^2$ in a problem that is probably going to be tractable, one good first thing to try is to see how the problem looks transformed to polar coordinates. (Even in a problem that comes about in real life, where you don't know the solution will be possible to obtain, this is a good first shot.)</p>
<p>When $r^2 = x^2 + y^2$ it is valid to take the derivative of each side with respect to $t$.
For example,
$${d \over dt} (r^2) = 2r {dr \over dt} = 2rr'
$$
The book got the equation you present by doing this on both sides of the equation, and dividing by 2.</p>
<p>I see another answer and that you are unclear on how $\tan \theta$ comes in. That is easy: In polar coordinates, $y/x = \tan \theta$. Draw yourself a right triangle with one point at the origin, the right angle on the X axis and the third point at $(x,y)$ and ask yourself what $\tan \theta$ would be in that picture.</p>
<p>By the way, the solution that approaches $r=1$ is in fact a stable orbit, since if $r = 1+ \epsilon$, $r' = -\epsilon r$ so the solution will approach $r=1$ in an exponential fachion.</p>
<p>Then you have seen how to get from that to
$$
r' = r(1-r) $$
The first thing to notice about that equation is that considering $r$ alone (without $\theta$) it will have fixed points wherever $r' = 0$, that is, at $r=0$ and at $r=1$. But then $r=0$ is a fixed point of the whole system since at $r=0$, $\theta$ is irrelevant.</p>
<p>However, that fixed point is an unstable fixed point, since near the origin with $r(t=0) = \epsilon$, the solution behaves like $r = \epsilon e^t$, exploding away from the origin.</p>
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2,077,275 | <p>Let $U$ be the set $U$ of quaternions of unit length. I know that $U\times S^1$ is compact, connected and is a $2n$ manifold in a $2n+1$ dimensional vector space $V$.</p>
<blockquote>
<p>How can I construct a differentiable tangent vector field on $U\times S^1$ that has no zeroes?</p>
</blockquote>
<p>What I know:<br>
A tangent vector field is a $C^k$ map $F:U\times S^1\rightarrow \mathbb{R}^7$ s.t. $F(x)\in T_x(U\times S^1)$ for all $x\in U\times S^1$. But how can I construct one? </p>
| Jean Marie | 305,862 | <p>We assume that the initial system has a non-zero determinant (see remark below for a discussion).</p>
<p>Let us consider the following system:</p>
<p>$$\tag{1}\left\{
\begin{aligned}
x e^{ia} + y e^{2ia} + ze^{3ia}&=e^{4ia}\\
x e^{ib} + y e^{2ib} + ze^{3ib}&=e^{4ib}\\
x e^{ic} + y e^{2ic} + ze^{3ic}&=e^{4ic}
\end{aligned}
\right.
\qquad\iff\qquad
\begin{cases}
\begin{aligned}
x + y e^{ia} + z e^{2ia} &= e^{3ia}\\
x + y e^{ib} + z e^{2ib} &= e^{3ib}\\
x + y e^{ic} + z e^{2ic} &= e^{3ic}
\end{aligned}\end{cases}
$$</p>
<p>Let us define polynomial $P(T):=x+yT+zT^2-T^3$.</p>
<p>With this notation, system $(1)$ is equivalent to </p>
<p>$$P(e^{ia})=P(e^{ib})=P(e^{ic})=0.$$</p>
<p>The three roots of $P$ being $e^{ia}$, $e^{ib}$, $e^{ic}$, Vieta's formulas give:</p>
<p>$$\tag{2}
\begin{cases}
\begin{aligned}
x&= \phantom{-}e^{ia}e^{ib}e^{ic}=e^{i(a+b+c)}\\
y&= -(e^{i(a+b)}+e^{i(a+c)}+e^{i(b+c)})\\
z&= \phantom{-}e^{ia}+e^{ib}+e^{ic}
\end{aligned}
\end{cases}
$$</p>
<p>It suffices now to take the imaginary part of system (1) whose solutions will be the imaginary parts of solutions above, i.e.,</p>
<blockquote>
<p>$$\tag{3}
\begin{cases}
\begin{aligned}
x&= \phantom{-}\sin(a+b+c)\\
y&=-\bigl(\sin(a+b)+\sin(a+c)+\sin(b+c)\bigr)\\
z&= \phantom{-}\sin(a)+\sin(b)+\sin(c)
\end{aligned}
\end{cases}
$$</p>
</blockquote>
<p>Remark: the determinant of system $(1)$, under its second form, is a Vandermonde determinant; it is non-zero unless, two values among $e^{ia}$, $e^{ib}$, $e^{ic}$ are identical, which means two angles $a,b,c$ are equal modulo $2\pi$. This is the only case of indeterminacy.</p>
<hr>
<p><strong>Edit</strong>: a different proof, not using complex numbers, although using the same 3rd degree polynomial trick is as follows.</p>
<p>Consider the following formulas:</p>
<p>$$\begin{cases}\sin 2a&=&\sin a \times (2 \alpha)\\ \sin 3a&=&\sin a \times (4 \alpha^2-1)\\ \sin 4a&=&\sin a \times (8 \alpha^3-4\alpha)\end{cases} \ \ \ \text{with} \ \ \ \alpha:=\cos a$$</p>
<p>which all are particular cases of the fact that $\frac{\sin(na)}{\sin(a)}$ can be expressed as a degree $n-1$ polynomial in $\cos(a)$ (which is named 2nd order Chebyshev polynomial).</p>
<p>Plugging these values of $\sin 2a$, $\sin 3a$,$\sin 4a$ in the first equation of the initial system, and simplifying by $\sin a$ (assumed non-zero) we obtain </p>
<p>$$x+y2\alpha+z(4 \alpha^2-1)=8 \alpha^3-4\alpha.$$</p>
<p>In other words, </p>
<p>$$8 \alpha^3-4z\alpha^2-(2y+4)\alpha+(z-x)=0$$</p>
<p>We can do the same for the second and third equations, with $\beta:=\cos b$ and $\gamma:=\cos c$. Gathering these 3 results, we have:</p>
<p>$$\tag{4}\left\{
\begin{aligned}
8 \alpha^3-4z\alpha^2-(2y+4)\alpha+(z-x)=0\\
8 \beta^3-4z\beta^2-(2y+4)\beta+(z-x)=0\\
8 \gamma^3-4z\gamma^2-(2y+4)\gamma+(z-x)=0
\end{aligned}
\right.$$</p>
<p>This means that $\alpha,\beta,\gamma$ are roots of polynomial:</p>
<p>$$P(T):=8T^3-4zT^2-(2y+4)T+(z-x)$$</p>
<p>Viete's formulas give:</p>
<p>$$\tag{5}
\begin{cases}
\begin{aligned}
z/2&=&\alpha+\beta+\gamma\\
(2y+4)/8&=&\alpha\beta+\beta\gamma+\gamma\alpha\\
(z-x)/8&=&\alpha\beta\gamma
\end{aligned}
\end{cases} \ \ \ \implies$$</p>
<blockquote>
<p>$$\begin{cases}
\begin{aligned}
x&=& 2(\cos(a)+\cos(b)+\cos(c))+8\cos(a)\cos(b)\cos(c)\\
y&=&-2-4(\cos(a)\cos(b)+\cos(b)\cos(c)+\cos(c)\cos(a))\\
z&=& 2(\cos(a)+\cos(b)+\cos(c))\\
\end{aligned}
\end{cases}
$$</p>
</blockquote>
<p>Remark: it must be understood that this is the same set of solutions as presented in the yellow cartridge above, expressed in terms of cosines instead of sines of angles $a,b,c$.</p>
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