qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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3,849,311 | <blockquote>
<p>What is the remainder when dividing the polynomial
<span class="math-container">$$P(x)=x^n+x^{n-1}+\cdots+x+1$$</span> with the polynomial
<span class="math-container">$$x^3-x$$</span> if <span class="math-container">$n$</span> is a natural odd number?</p>
</blockquote>
<p>So, what I know so far is:</p>
<p><span class="math-container">$$P(x)=Q(x)D(x)+R(x)$$</span></p>
<p>In this case I'll call <span class="math-container">$Q(x) = x^3-x$</span></p>
<p><span class="math-container">$$Q(x) = 0 \iff x=\pm1$$</span></p>
<p>So from here:</p>
<p><span class="math-container">$\begin{array} {l}P(1):\qquad&1^n+1^{n-1}+\cdots+1^1+1=R(1)\\P(-1):&(-1)^n+(-1)^{n-1}+\cdots+(-1)^1+1=R(-1)\end{array}$</span></p>
<p>Here is where I assumed that <span class="math-container">$R(x)=ax^2+b$</span> since <span class="math-container">$Q(x)=x^3-x$</span></p>
<p>And from the equations (assuming <span class="math-container">$n = \{2k+1 \mid k\in\mathbb{N}\}$</span>):</p>
<p><span class="math-container">$\begin{array}{l}P(1):\\&R(1)=n\\P(-1):\\&R(-1)=0\end{array}$</span></p>
<p>And here I get to:</p>
<p><span class="math-container">$$\left\{
\begin{array}{ll}
ax^2+by &=n \\
ax^2+by &=0 \\
\end{array}
\right.$$</span></p>
<p>I would like to know where did I go wrong? Is my assumption for <span class="math-container">$R(x)$</span> incorrect, my calculation of the <span class="math-container">$P(1)$</span> and <span class="math-container">$P(-1)$</span> wrong or is there something else I didn't think about?</p>
| R. J. Mathar | 805,678 | <p>The remainder is <span class="math-container">$$\left(\frac{n+1}{2}\frac{1}{x-1}-\frac{1}{x}\right)(x^3-x) = \frac{(x+1)(2+(n-1)x)}{2}$$</span></p>
|
16,831 | <p>As the title says, I'm wondering if there is a continuous function such that $f$ is nonzero on $[0, 1]$, and for which $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$. I am trying to solve a problem proving that if (on $C([0, 1])$) $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 0$, then $f$ must be identically zero. I presume then we do require the $n=0$ case to hold too, otherwise it wouldn't be part of the statement. Is there ay function which is not identically zero which satisfies $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$?</p>
<p>The statement I am attempting to prove is homework, but this is just idle curiosity (though I will tag it as homework anyway since it is related). Thank you!</p>
| Elias Costa | 19,266 | <p>There is a proof, using the Weierstrass approximation theorem, that if $f$ is continuous then $f$ is necessarily zero! </p>
<p><strong>Classical Weierstrass's Theorem :</strong> <em>If f is a continuous real valued function on $[a, b]$, then there exists a sequence of polynomials $p_n$ such that
$$ \lim_{n\rightarrow+\infty}p_n(x)=f(x)$$ uniformly on $[a, b]$</em>.</p>
<p>By the assumptions of the problem and the linearity of the integral
is easy to see that
$$
\int_{0}^1f(x)p(x)dx=0
$$
for all polynomials $p(x)$ on $C([0,1])$. Now just apply Theorem! </p>
|
2,463,421 | <p>The question is:</p>
<p>Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of $6548 for their tickets on a particular flight, how many of each type of ticket
were sold?</p>
<p>Now I set up my equation as </p>
<p>$140x+110y+78z=6548$</p>
<p>But I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for
$x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n)$</p>
<p>Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations. </p>
| Colm Bhandal | 252,983 | <p>Edit: Thanks to commenters.</p>
<p>Presumably what's being asked here is how many four digit hexadecimal numbers are there whose digits are all in increasing order. We also have the restriction that $0$ can't be chosen, because, given digits of increasing order, this would imply $0$ as the first digit, which isn't allowed for a number in the proper format. So, we must have $4$ increasing digits chosen from $1$ to $15$. Increasing implies the digits are distinct. Since any choice of $4$ digits gives rise to exactly one such number and vice versa, we just need to count the number of sets of $4$ distinct digits chosen from the set of $15$. This is simply $$15 \choose 4$$</p>
|
140,294 | <p>Generative adversarial networks (GAN) is regarded as one of "the most interesting idea in the last ten years in machine learning" by Yann LeCun. It can be used to generate photo-realistic images that are almost indistinguishable from the real ones.</p>
<p>GAN trains two competing neural networks: a generator network which generates image, and a discriminator network that distinguishes the generated image and the real training image. For example, the images shown below are generated by the network from the texts above them (taken from Han Zhang, etc., <a href="https://arxiv.org/pdf/1612.03242.pdf" rel="noreferrer">StackGAN: Text to Photo-realistic Image Synthesis
with Stacked Generative Adversarial Networks</a>).</p>
<p><a href="https://i.stack.imgur.com/r3L60.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/r3L60.jpg" alt="enter image description here"></a></p>
<p>I'm wondering whether we can implement a simplified version of that in <em>Mathematica</em>, given that the neural network framework has be enhanced greatly in version 11.1.</p>
| partida | 15,961 | <p>This is a GAN using more accurate loss function,Thanks for @Michael Curry</p>
<pre><code>mnist = ResourceData["MNIST"];
mnistDigits = First /@ mnist;
randomDim = 10;
generator = NetChain[{128, Ramp, 128, Ramp, 28*28, LogisticSigmoid,
ReshapeLayer[{1, 28, 28}]}, "Input" -> randomDim];
discriminator = NetChain[{128, Ramp, 128, Ramp, 128, Ramp,
1, LogisticSigmoid},
"Input" -> {1, 28, 28}];
</code></pre>
<p>The loss function follow this <a href="https://arxiv.org/pdf/1406.2661.pdf" rel="noreferrer">paper</a></p>
<p><a href="https://i.stack.imgur.com/iewfS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/iewfS.png" alt="enter image description here"></a></p>
<pre><code>Gloss = NetChain[{PartLayer[1], ElementwiseLayer[Log[1 - #] &]}];
Dloss = NetGraph[{PartLayer[1], ElementwiseLayer[Log[1 - #] &],
PartLayer[2], ElementwiseLayer[Log[#] &],
TotalLayer[],
ElementwiseLayer[-1 # &](*transfer Maximize to Minimize*)},
{1 -> 2, 3 -> 4, {2, 4} -> 5 -> 6}];
</code></pre>
<p><a href="https://i.stack.imgur.com/gsESz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/gsESz.png" alt="enter image description here"></a></p>
<p>In the <code>flat</code> layer,the first part estimate the probability of fake,the second part estimate the probability of real.You can add another port <code>Output2</code> to check this</p>
<pre><code>ganNet = NetInitialize[
NetGraph[<|"gen" -> generator,
"discrimop" -> NetMapOperator[discriminator],
"cat" -> CatenateLayer[],
"reshape" -> ReshapeLayer[{2, 1, 28, 28}],
"flat" -> FlattenLayer[], "G_loss" -> Gloss,
"D_loss" -> Dloss|>,
{NetPort["random"] -> "gen" -> "cat", NetPort["Input"] -> "cat", "cat" -> "reshape" -> "discrimop" -> "flat" -> {"G_loss", "D_loss"},
"G_loss" -> NetPort["Loss1"], "D_loss" -> NetPort["Loss2"]},
"Input" -> NetEncoder[{"Image", {28, 28}, ColorSpace -> "Grayscale"}]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/7d5bd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/7d5bd.png" alt="enter image description here"></a></p>
<pre><code>trainingData = <|"random" -> RandomVariate[NormalDistribution[], {Length@mnistDigits, randomDim}],
"Input" -> mnistDigits|>;
</code></pre>
<p>But when training,I want to reduce the loss of "Loss1" and "Loss2".But find nothing in docs,only got code like this <code>trained = NetTrain[net, <|"Input" -> trainingImages|>, "Loss"]</code>.But here we have two loss function.</p>
<p>In the <code>Pytorch</code>,we can use two optimizer to reduce the two loss.But in MMA,I have no idea.</p>
<p><a href="https://i.stack.imgur.com/BaY9P.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BaY9P.png" alt="enter image description here"></a></p>
<p>So this a simple way to train the net.</p>
<pre><code>ganNet = NetTrain[ganNet, trainingData, {"Loss1", "Loss2"},
Method -> {"ADAM", "Beta1" -> 0.5, "LearningRate" -> 0.00005, "WeightClipping" -> {"discrimop" -> 0.01}},
BatchSize -> 64];
</code></pre>
|
382,504 | <p>Let <span class="math-container">$n \geq 2$</span> be an integer, and let <span class="math-container">$f(x) = \prod\limits_{k = 1}^n(x - \alpha_k)$</span> be a monic irreducible polynomial in <span class="math-container">$\mathbb Z[x]$</span>, with the property that <span class="math-container">$f(-\alpha_k) \neq 0$</span> for any <span class="math-container">$k = 1, 2, \ldots, n$</span>.</p>
<p>Is there anything meaningful that we can say about <span class="math-container">$\operatorname{Res}(f(x), f(-x))$</span>, the resultant of <span class="math-container">$f(x)$</span> and <span class="math-container">$f(-x)$</span>?</p>
<p>To rephrase, what can be said about the value of the product <span class="math-container">$\prod\limits_{k = 1}^nf(-\alpha_k)$</span>? Perhaps, it can be expressed somehow through <span class="math-container">$n$</span> and the discriminant of <span class="math-container">$f(x)$</span>?</p>
<p>One thing that I can note is that <span class="math-container">$f(-\alpha_1), \ldots, f(-\alpha_k)$</span> are algebraic conjugates, which means that their product is equal to the norm of <span class="math-container">$f(-\alpha_1)$</span>. Thus, up to a sign, the product <span class="math-container">$\prod\limits_{k = 1}^nf(-\alpha_k)$</span> is equal to the constant coefficient of the minimal polynomial of <span class="math-container">$f(-\alpha_1)$</span>. But what is this constant coefficient is a mystery.</p>
| Anton | 22,733 | <p>See the discussion <a href="https://mathoverflow.net/a/382552">above</a>.</p>
<p>Also, if <span class="math-container">$f(x) = x^n + \dotsb + a_{n-1}x + a_n$</span>, then</p>
<ul>
<li><p>for <span class="math-container">$n = 2$</span>, <span class="math-container">$\operatorname{Res}(f(x), f(-x)) = 2^2a_2a_1^2$</span></p>
</li>
<li><p>for <span class="math-container">$n = 3$</span>, <span class="math-container">$\operatorname{Res}(f(x), f(-x)) = 2^3a_3(a_3 - a_1a_2)^2$</span></p>
</li>
<li><p>for <span class="math-container">$n = 4$</span>, <span class="math-container">$\operatorname{Res}(f(x), f(-x)) = 2^4a_4(a_4a_1^2 - a_1a_2a_3 + a_3^2)^2$</span>.</p>
</li>
</ul>
|
2,215,087 | <p>I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$</p>
<p>By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$</p>
<p>For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:</p>
<p>Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$</p>
<p><strong>Case 1:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$</p>
<p><strong>Case 2:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$</p>
<p><strong>Case 3:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z $ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.</p>
<p>This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $ \frac{3}{2} \leq |x-a| \leq 2$</p>
| Chan Tai Man | 876,234 | <p><strong>Division algorithm for Euclidean domain <span class="math-container">$\mathbb{Z}[\sqrt{11}]$</span></strong></p>
<p>While Oppenheim (1934) proved that there is a division algorithm for <span class="math-container">$\mathbb{Z}[\sqrt{11}]$</span>. No examples was given explicitly. Inspired by his proof, one suitable division algorithm was identified. Hence, one could compute in <span class="math-container">$\mathbb{Z}[\sqrt{11}]$</span>.</p>
<p>Let <span class="math-container">$R = \mathbb{Z}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Z} \} $</span> be a quadratic integer ring. Let <span class="math-container">$\alpha = s+t\sqrt{11} \in R$</span>. Define norm <span class="math-container">$N: R \setminus\{0\} \rightarrow \mathbb{Z}_{\geq 0}$</span> by setting <span class="math-container">$\alpha \mapsto |s^2-11t^2|$</span>, which is well defined in <span class="math-container">$R$</span>. Proof omitted. Norm <span class="math-container">$N$</span> is also well defined in <span class="math-container">$\mathbb{Q}[\sqrt{11}]$</span>. Proof omitted. There exists <span class="math-container">$\alpha/\beta \in \mathbb{Q}[\sqrt{11}]$</span> <span class="math-container">$=$</span> <span class="math-container">$\{ s+t\sqrt{11} : s,t \in \mathbb{Q} \} \supset R $</span> with a numerator and a denominator <span class="math-container">$\alpha, \beta \in R$</span>, <span class="math-container">$\beta \ne 0$</span>. Proof omitted.</p>
<p>Suppose the nearest lattice point to point <span class="math-container">$\alpha / \beta$</span> is <span class="math-container">$m+n\sqrt{11} \in R$</span> (i.e. <span class="math-container">$m,n \in \mathbb{Z}$</span>) with <span class="math-container">$\alpha / \beta = (m+n\sqrt{11}) + (a+b\sqrt{11})$</span>, <span class="math-container">$a,b \in \mathbb{Q}$</span>, <span class="math-container">$-\frac{1}{2} \leq a \leq \frac{1}{2}$</span> and <span class="math-container">$-\frac{1}{2} \leq b \leq \frac{1}{2}$</span>. We will consider lattice point <span class="math-container">$(m,n) = m+n\sqrt{11}$</span> and five other nearby lattice points. They, <span class="math-container">$(m+x,n+y)$</span> with <span class="math-container">$(x,y) \in \{ (0,0), (1,0), (-1,0), (2,0), (2,1), (-5,2) \} $</span>, are candidate quotients <span class="math-container">$q$</span> to a division algorithm. By considering only positive values of <span class="math-container">$a,b$</span> such that <span class="math-container">$0 \leq a \leq \frac{1}{2}$</span> and <span class="math-container">$0 \leq b \leq \frac{1}{2}$</span>, similar arguments could be made for the three other quadrants by symmetry. With <span class="math-container">$q=(m+x,n+y)$</span>, rewrite <span class="math-container">$\alpha = q \beta +r$</span> as:</p>
<p><span class="math-container">\begin{align}
\alpha / \beta &= q + r/\beta \\
&= \big( (m+n\sqrt{11}) + (x+y\sqrt{11}) \big) + \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)
\end{align}</span></p>
<p>Inequality <span class="math-container">$N(r) < N(\beta)$</span> is true <span class="math-container">$\iff$</span> <span class="math-container">$N(r)/N(\beta) < 1$</span> <span class="math-container">$\iff$</span> <span class="math-container">$N(r/\beta) < 1$</span> <span class="math-container">$\iff$</span> <span class="math-container">$N \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)$</span> <span class="math-container">$= |(x-a)^2 - 11(y-b)^2| < 1$</span>
. Proof omitted.</p>
<p><strong>Case 1:</strong> Consider lattice point <span class="math-container">$(m,n)$</span>, <span class="math-container">$0 \leq a^2 \leq \frac{1}{4}$</span> and <span class="math-container">$0 \leq 11b^2 < 1$</span>. The maximum possible difference occurs when <span class="math-container">$a^2 = 0$</span> and <span class="math-container">$11b^2 = 1^-$</span>. Therefore <span class="math-container">$|a^2 - 11b^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0 &\leq a \leq \frac{1}{2} \\
0 &\leq b < \sqrt{ \frac{1}{11} } \approx 0.302
\end{align}</span></p>
<p><a href="https://i.stack.imgur.com/3dUYw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3dUYw.png" alt="seven cases bar chart" /></a></p>
<p><strong>Case 2:</strong> Consider lattice point <span class="math-container">$(m+1,\ n)$</span>, <span class="math-container">$\frac{1}{4} \leq (1-a)^2 \leq 1$</span> and <span class="math-container">$1 < 11b^2 < \frac{5}{4}$</span>. The maximum possible difference occurs when <span class="math-container">$(1-a)^2 = \frac{1}{4}$</span> and <span class="math-container">$11b^2 = \frac{5}{4}^-$</span>. Therefore <span class="math-container">$|(1-a)^2 - 11b^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0 &\leq a \leq \frac{1}{2} \\
\sqrt{ \frac{1}{11} } &< b < \sqrt{ \frac{5}{44} } \approx 0.337
\end{align}</span></p>
<p><strong>Case 3:</strong> Consider lattice point <span class="math-container">$(m-1,\ n)$</span>, <span class="math-container">$1 \leq (1+a)^2 \leq \frac{9}{4}$</span> and <span class="math-container">$\frac{5}{4} < 11b^2 < 2$</span>. The maximum possible differences are <span class="math-container">$2^- - 1 < 1$</span> and <span class="math-container">$\frac{9}{4} - \frac{5}{4}^+ < 1$</span>. Therefore <span class="math-container">$|(1+a)^2 - 11b^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0 &\leq a \leq \frac{1}{2} \\
\sqrt{ \frac{5}{44} } &< b < \sqrt{ \frac{2}{11} } \approx 0.426
\end{align}</span></p>
<p><strong>Case 4:</strong> Consider lattice point <span class="math-container">$(m+2,\ n)$</span>, <span class="math-container">$\frac{9}{4} \leq (2-a)^2 < 3$</span> and <span class="math-container">$2 < 11b^2 \leq \frac{11}{4}$</span>. The maximum possible difference occurs when <span class="math-container">$(2-a)^2 = 3^-$</span> and <span class="math-container">$11b^2 = 2^+$</span>. Therefore <span class="math-container">$|(2-a)^2 - 11b^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0.270 \approx 2-\sqrt{3} &< a \leq \frac{1}{2} \\
\sqrt{ \frac{2}{11} } &< b \leq \frac{1}{2}
\end{align}</span></p>
<p><strong>Case 5:</strong> Consider lattice point <span class="math-container">$(m+2,\ n+1)$</span>, <span class="math-container">$3 < (2-a)^2 \leq 4$</span> and <span class="math-container">$3 < 11(1-b)^2 < 4$</span>. The maximum possible difference are <span class="math-container">$4^- - 3^+ < 1$</span> and <span class="math-container">$4 - 3^+ < 1$</span>. Therefore <span class="math-container">$|(2-a)^2 - 11(1-b)^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0 &\leq a < 2-\sqrt{3} \\
0.397 \approx 1-\sqrt{ \frac{4}{11} } &< b < 1-\sqrt{ \frac{3}{11} } \approx 0.478
\end{align}</span></p>
<p><strong>Case 6:</strong> Consider lattice point <span class="math-container">$(m+2,\ n+1)$</span> again with different ranges of <span class="math-container">$a,b$</span>, <span class="math-container">$\frac{9}{4} \leq (2-a)^2 < \frac{15}{4}$</span> and <span class="math-container">$\frac{11}{4} \leq 11(1-b)^2 < \frac{13}{4}$</span>. The maximum possible differences are <span class="math-container">$\frac{13}{4}^- - \frac{9}{4} < 1$</span> and <span class="math-container">$\frac{15}{4}^- - \frac{11}{4} < 1$</span>. Therefore <span class="math-container">$|(2-a)^2 - 11(1-b)^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0.064 \approx 2 - \sqrt{ \frac{15}{4} } &< a \leq \frac{1}{2} \\
0.456 \approx 1-\sqrt{\frac{13}{44}} &< b \leq \frac{1}{2}
\end{align}</span></p>
<p><strong>Case 7:</strong> Consider lattice point <span class="math-container">$(m-5,\ n+2)$</span>, <span class="math-container">$25 \leq (5+a)^2 < \frac{103}{4}$</span> and <span class="math-container">$\frac{99}{4} \leq 11(2-b)^2 < \frac{103}{4}$</span>. The maximum possible difference occurs when <span class="math-container">$(5+a)^2 = \frac{103}{4}^-$</span> and <span class="math-container">$11(2-b)^2 = \frac{99}{4}$</span>. Therefore <span class="math-container">$|(5+a)^2 - 11(2-b)^2| < 1$</span> and we have:</p>
<p><span class="math-container">\begin{align}
0 &\leq a < \sqrt{ \frac{103}{4} } - 5 \approx 0.074 \\
0.470 \approx 2 - \sqrt{\frac{103}{44}} &< b \leq \frac{1}{2}
\end{align}</span></p>
<p>All boundary cases are ignored, since <span class="math-container">$\sqrt{ \frac{1}{11} }$</span>, <span class="math-container">$\sqrt{ \frac{5}{44} }$</span>, <span class="math-container">$1-\sqrt{ \frac{4}{11} }$</span>, <span class="math-container">$\sqrt{ \frac{2}{11} }$</span>, <span class="math-container">$1-\sqrt{\frac{13}{44}}$</span>, <span class="math-container">$2 - \sqrt{\frac{103}{44}}$</span>, <span class="math-container">$1-\sqrt{ \frac{3}{11} }$</span>, <span class="math-container">$2 - \sqrt{ \frac{15}{4} }$</span>, <span class="math-container">$\sqrt{ \frac{103}{4} } - 5$</span>, <span class="math-container">$2-\sqrt{3}$</span> <span class="math-container">$\notin \mathbb{Q}$</span>.</p>
<p>Since the quadratic integer ring <span class="math-container">$\mathbb{Z}[\sqrt{11}]$</span> has (a) a well defined norm and (b) a division algorithm <span class="math-container">$\alpha = q \beta + r$</span> such that either <span class="math-container">$r=0$</span> or <span class="math-container">$N(r)<(b)$</span>, for all numerators and denominators <span class="math-container">$\alpha, \beta \in \mathbb{Z}[\sqrt{11}]$</span>, <span class="math-container">$\beta \ne 0$</span> with some quotient and remainder <span class="math-container">$q, r \in \mathbb{Z}[\sqrt{11}]$</span>, it is a Euclidean domain. <span class="math-container">$\blacksquare$</span></p>
<p><a href="https://i.stack.imgur.com/i2b2R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i2b2R.png" alt="different ranges of a,b" /></a></p>
|
2,215,087 | <p>I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$</p>
<p>By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$</p>
<p>For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:</p>
<p>Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$</p>
<p><strong>Case 1:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$</p>
<p><strong>Case 2:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$</p>
<p><strong>Case 3:</strong> Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z $ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$</p>
<p>From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.</p>
<p>This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $ \frac{3}{2} \leq |x-a| \leq 2$</p>
| Chan Tai Man | 876,234 | <p>Another way to graph it for showing that <span class="math-container">$\mathbb{Z}[{\sqrt{11}}]$</span> is norm-euclidean. However, this method fails for <span class="math-container">$\mathbb{Z}[{\sqrt{19}}]$</span>. There are a few small gaps that this elementary method fails to cover.</p>
<p>Z[sqrt(11)] with max a,b ranges <br />
<a href="https://i.stack.imgur.com/H2hIl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H2hIl.png" alt="Z[sqrt(11)] with max a,b ranges" /></a></p>
<p>(cases 1,5,6) <br />
<a href="https://i.stack.imgur.com/ftE3V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ftE3V.png" alt="Z[sqrt(11)] with max a,b ranges (cases 1,5,6)" /></a></p>
<p>(cases 2,4) <br />
<a href="https://i.stack.imgur.com/yUc4W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yUc4W.png" alt="Z[sqrt(11)] with max a,b ranges (cases 2,4)" /></a></p>
<p>(case 3) <br />
<a href="https://i.stack.imgur.com/xUOdS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xUOdS.png" alt="Z[sqrt(11)] with max a,b ranges (case 3)" /></a></p>
<p>(case 7) <br />
<a href="https://i.stack.imgur.com/Bj8L1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bj8L1.png" alt="Z[sqrt(11)] with max a,b ranges (case 7)" /></a></p>
<p>(areas covered by only one algorithm) <br />
<a href="https://i.stack.imgur.com/dQxe4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQxe4.png" alt="Z[sqrt(11)] with max a,b ranges (only alg. only)" /></a></p>
|
2,444,196 | <p>The center of a group $G$ is defined as $Z(G):=\{ z\in G : gz = zg, \; \forall g \in G\}$.</p>
<p>The goal is to show that if $\vert G\vert = pq$, where $p$ and $q$ are not necessarily distinct primes then either $G$ is abelian or $Z(G) = \{ e\}$.</p>
<p>I want to suppose that $Z(G) \neq \{ e\}$ and then use the fact that $G/Z(G)$ is cyclic to imply that $G$ is abelian, which is something I have already proven. But how do I show that $G/Z(G)$ is cyclic when I am not certain what exactly $Z(G)$ looks like. I only know that it has at least one non-identity element in it, which will be of order $p$ WLOG, (the case where it is of order $pq$ is trivial).</p>
<p>Any help is appreciated. Thank you.</p>
| Community | -1 | <p>This argument admittedly doesn't use the result <span class="math-container">$G/Z(G) \space\text{cyclic}\Rightarrow G\space\text{abelian}$</span>.</p>
<p>If <span class="math-container">$q=p$</span>, it is well known that <span class="math-container">$G$</span> of order <span class="math-container">$p^2$</span> is abelian. Next, let's take <span class="math-container">$p,q$</span> distinct primes. By contradiction, let's assume <span class="math-container">$|Z(G)|=q$</span>; therefore, the <em>noncentral</em> elements <span class="math-container">$g\in G$</span> have centralizer of order <span class="math-container">$q$</span>, whence <span class="math-container">$\frac{|G|}{|C_G(g)|}=\frac{pq}{q}=p$</span> for every noncentral <span class="math-container">$g\in G$</span>, and finally (take the noncentral part of the Class Equation) <span class="math-container">$q(p-1)=kp$</span>, where <span class="math-container">$k$</span> is the number of noncentral conjugacy classes of <span class="math-container">$G$</span>: contradiction, since <span class="math-container">$p\nmid q(p-1)$</span>. Thus, <span class="math-container">$|Z(G)|\ne q$</span>. The same argument works by swapping <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, whence <span class="math-container">$|Z(G)|\ne p$</span> as well. Therefore, either <span class="math-container">$|Z(G)|=pq$</span> and <span class="math-container">$G$</span> is abelian, or <span class="math-container">$Z(G)=\{e\}$</span>.</p>
|
4,331,081 | <p>Suppose <span class="math-container">$a, b >0$</span>. I'm looking for closed expressions for the following integral:
<span class="math-container">$$\int_{-\pi}^{\pi}\sqrt{a^{2}-2ab\cos(x)+b^{2}}dx $$</span>
I tried to solve this by myself and got nowhere and even wolfram alpha couldn't get me an answer so maybe this is not easily solved, but might have some complicated closed expression. Any help is important. Thanks!</p>
| Eldar Sultanow | 993,738 | <p>Mathematica yields the result shown below when executig:</p>
<pre><code>Integrate[Sqrt[a^2 - 2 ab*Cos[x] + b^2], {x, -Pi, Pi}]
</code></pre>
<p><span class="math-container">$$4 \sqrt{a^2+2 \text{ab}+b^2} E\left(\frac{4 \text{ab}}{a^2+b^2+2 \text{ab}}\right)$$</span></p>
<p>where <span class="math-container">$E(\ldots)$</span> denote the <a href="https://reference.wolfram.com/language/ref/EllipticE.html" rel="nofollow noreferrer">complete Elliptic Integral</a>.</p>
|
2,583,454 | <p>Consider for instance the linear system:</p>
<p>$$\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$</p>
<p>This is over determined and thus has no solution. Yet, by simply multiplying both sides by $\textbf{A}^T$:</p>
<p>$$\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$</p>
<p>We find that the system now has a unique solution, which is the (x,y) that minimizes the squared error.</p>
<p>Now I understand the derivation of why multiplying by the transpose helps to find the pseudoinverse which then helps to perform OLS regression, but my question is perhaps a bit more fundamental. </p>
<p>How can multiplying both sides of an equation by a matrix change a system which previously had no solutions into one that has a unique solution? This seems to against what I assumed that the solutions to $\textbf{A}x = \textbf{B}$ were the same as the solutions to $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$.</p>
| Ian | 83,396 | <p>Consider an analogy with the 1D case: $1=0$ is false but $0(1)=0(0)$ is true. Multiplying both sides by zero turned a false equation into a true one.</p>
<p>In the case of an overdetermined system $Ax=b$, you are effectively multiplying certain <em>components</em> of the space by zero when you multiply by $A^T$. The reason is that if the rank of $A$ is less than its number of rows, then $A^T$ has a nontrivial null space. (In particular, if $A$ has more rows than columns then $A^T$ has a nontrivial null space; this is the usual situation.) Thus you are asking $Ax-b$ to be just in the null space of $A^T$, rather than actually being zero. This is a milder requirement, so new solutions can crop up.</p>
|
2,583,454 | <p>Consider for instance the linear system:</p>
<p>$$\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$</p>
<p>This is over determined and thus has no solution. Yet, by simply multiplying both sides by $\textbf{A}^T$:</p>
<p>$$\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$</p>
<p>We find that the system now has a unique solution, which is the (x,y) that minimizes the squared error.</p>
<p>Now I understand the derivation of why multiplying by the transpose helps to find the pseudoinverse which then helps to perform OLS regression, but my question is perhaps a bit more fundamental. </p>
<p>How can multiplying both sides of an equation by a matrix change a system which previously had no solutions into one that has a unique solution? This seems to against what I assumed that the solutions to $\textbf{A}x = \textbf{B}$ were the same as the solutions to $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$.</p>
| Kajelad | 354,840 | <p>One intuition for this is thinking about it geometrically. In your case, $A$ is a linearly independent $3\times 2$ matrix, so its column space is a plane in $\mathbb R^3$, and $A\vec x$ is a point in this plane. If we choose a new point $\vec b\in\mathbb R^3$, it generically won't be on the plane of $\text{col}\ A$.</p>
<p>Left multiplying by $A^T$ can be thought of as orthogonally projecting a vector in $\mathbb R^3$ into a 2-dimensional space, expressed in a new basis. In this new space, the column space of $A^TA$ is all of $\mathbb R^2$, so any choice of $\vec b\in \mathbb R^3$ will result in an $A^T\vec b\in\mathbb R^2$ than can be written as $A^TA\vec x$ for some $\vec x\in\mathbb R^2$.</p>
|
3,769,843 | <p>This is a multiple choice question from my Text Book</p>
<p>Let <span class="math-container">$A=\{1,2,3\}$</span>. The no. of relations containing <span class="math-container">$(1,2)$</span> and <span class="math-container">$(1,3)$</span> which are reflexive and Symmetric but not transitive is</p>
<p>(A) <span class="math-container">$1$</span></p>
<p>(B) <span class="math-container">$2$</span></p>
<p>(C) <span class="math-container">$3$</span></p>
<p>(D) <span class="math-container">$4$</span></p>
<hr />
<p><em><strong>My Approach:</strong></em>
<span class="math-container">$A=\{1,2,3\}$</span></p>
<p>Relation <span class="math-container">$R$</span> must contain <span class="math-container">$(1,2)$</span> and <span class="math-container">$(1,3)$</span></p>
<p>For <span class="math-container">$R$</span> to be Reflexive, it must contain <span class="math-container">$(2,2)$</span> and <span class="math-container">$(1,1)$</span></p>
<p>For <span class="math-container">$R$</span> to be symmetric, it must contain <span class="math-container">$(2,1)$</span> and <span class="math-container">$(3,1)$</span></p>
<p>For <span class="math-container">$R$</span> to not to be Transitive, it must not contain <span class="math-container">$(2,3)$</span> and <span class="math-container">$(3,2)$</span></p>
<p>\therefore, <span class="math-container">$R=\{(1,2),(1,3),(2,2),(1,1),(3,1),(2,1)\}$</span></p>
<p>Anyother addition to <span class="math-container">$R$</span> will not satisfy the stated condition.</p>
<p>Hence, option <span class="math-container">$A$</span> is correct</p>
<p>Am I right?</p>
<p>[Edit:</p>
<p>R contains <span class="math-container">$(3,3)$</span> as well]</p>
| Luke Hutchison | 365,886 | <p>The answer by "None" is the same as the answer I arrived at. I'm posting my working here:</p>
<p>To map <span class="math-container">$X_i$</span>, the local coordinates of a point <span class="math-container">$X$</span> captured in the coordinate system of camera position <span class="math-container">$L_i$</span>, to <span class="math-container">$X_{i-1}$</span>, the coordinates of the point in the coordinate system of camera position <span class="math-container">$L_{i-1}$</span>, using the known relative rotation quaternion <span class="math-container">$R_{i-1}$</span> and translation vector <span class="math-container">$T_{i-1}$</span>, we can use the following equation (this is obtained directly from the definition of how these relative rotations and translations are defined):</p>
<p><span class="math-container">$$X_{i-1} = R_{i-1} X_i R_{i-1}^{-1} + T_{i-1}\tag{1}$$</span></p>
<p>So to map <span class="math-container">$X_i$</span> (in the coordinate system of <span class="math-container">$L_i$</span>) to <span class="math-container">$X_{i-2}$</span> (in the coordinate system of <span class="math-container">$L_{i-2}$</span>), we obtain:</p>
<p><span class="math-container">$$X_{i-2} = R_{i-2} (R_{i-1} X_i R_{i-1}^{-1} + T_{i-1}) R_{i-2}^{-1} + T_{i-2}$$</span></p>
<p><span class="math-container">$$= (R_{i-2} R_{i-1} X_i R_{i-1}^{-1} R_{i-2}^{-1}) + (R_{i-2} T_{i-1} R_{i-2}^{-1} + T_{i-2})$$</span></p>
<p>where the lefthand operand of the toplevel addition is the "rotation part" and the righthand operand is the "translation part".</p>
<p>Similarly, to map <span class="math-container">$X_i$</span> (in the coordinate system of <span class="math-container">$L_i$</span>) to <span class="math-container">$X_{i-3}$</span> (in the coordinate system of <span class="math-container">$L_{i-3}$</span>), we obtain:</p>
<p><span class="math-container">$$X_{i-3} = R_{i-3} ((R_{i-2} R_{i-1} X_i R_{i-1}^{-1} R_{i-2}^{-1}) + (R_{i-2} T_{i-1} R_{i-2}^{-1} + T_{i-2})) R_{i-3}^{-1} +T_{i-3}$$</span></p>
<p><span class="math-container">$$= R_{i-3} (R_{i-2} R_{i-1} X_i R_{i-1}^{-1} R_{i-2}^{-1}) R_{i-3}^{-1} + R_{i-3} (R_{i-2} T_{i-1} R_{i-2}^{-1} + T_{i-2}) R_{i-3}^{-1} +T_{i-3}$$</span></p>
<p><span class="math-container">$$ = (R_{i-3} R_{i-2} R_{i-1} X_i R_{i-1}^{-1} R_{i-2}^{-1} R_{i-3}^{-1}) + (R_{i-3} R_{i-2} T_{i-1} R_{i-2}^{-1} R_{i-3}^{-1} + (R_{i-3} T_{i-2} R_{i-3}^{-1} +T_{i-3}))$$</span></p>
<p>We want to compute <span class="math-container">$X_0$</span>, the global coordinates of <span class="math-container">$X_i$</span>, i.e. the coordinates of <span class="math-container">$X_i$</span> in the coordinate system of camera location <span class="math-container">$L_0$</span>. Here <span class="math-container">$Q_i$</span> and <span class="math-container">$P_i$</span> represent the orientation quaternion and translation vector respectively for the camera position <span class="math-container">$L_i$</span> <em>in the coordinate system of <span class="math-container">$L_0$</span></em>:</p>
<p><span class="math-container">$$X_0 = Q_i X_i Q_i^{-1} + P_i\tag{2}$$</span></p>
<p>This yields the following orientation and position update rules:</p>
<p><span class="math-container">$$Q_i = Q_{i-1}R_{i-1} \, \, \, ; \, \, \, Q_0 = 1\tag{3}$$</span>
<span class="math-container">$$P_i = Q_{i-1} T_{i-1} Q_{i-1}^{-1} + P_{i-1} \, \, \, ; \, \, \,P_0 = [0 \,\, 0 \,\, 0]\tag{4}$$</span></p>
<p>So we started from a transformation mapping <span class="math-container">$X_i$</span> to <span class="math-container">$X_{i-1}$</span> (Eq. 1), and ended up with a pair of transformations mapping <span class="math-container">$Q_{i-1}$</span> to <span class="math-container">$Q_{i}$</span> and <span class="math-container">$P_{i-1}$</span> to <span class="math-container">$P_i$</span> (Eqs. 2, 3), i.e. <span class="math-container">$P_i$</span> and <span class="math-container">$Q_i$</span> can be computed incrementally, in a sequence starting from <span class="math-container">$\langle P_0, Q_0 \rangle$</span> and ending with <span class="math-container">$\langle P_i, Q_i \rangle$</span>. The transformations <span class="math-container">$Q_i$</span> and <span class="math-container">$P_i$</span> can then be used to map <span class="math-container">$X_i$</span> back to <span class="math-container">$X_0$</span> (Eq. 2).</p>
|
4,382,317 | <p>Show that <span class="math-container">$|z+1| = 2\cos(\frac{\theta}{2})$</span>, <span class="math-container">$z = cis(\theta)$</span> and <span class="math-container">$z \in C$</span></p>
<p>Here is what I have managed to do:</p>
<ul>
<li><span class="math-container">$r=1$</span></li>
<li><span class="math-container">$(z+1)^2 = (\cos(\theta)+i\sin(\theta)+1)^2$</span></li>
<li><span class="math-container">$(z+1)^2 = (\cos^2(\theta)+2(i\sin(\theta)+1)\cos(\theta)+(i\sin(\theta)+1)^2)$</span></li>
<li><span class="math-container">$(z+1)^2 = (\cos^2(\theta)+2i\sin(\theta)\cos(\theta)+2\cos(\theta)+(-\sin^2(\theta)+2i\sin(\theta)+1))$</span></li>
</ul>
<p>What I wanted to do was to take the square root of both sides to get the absolute value, however, I need to simplify the right-hand side first. I need help with that part</p>
| Amaan M | 860,916 | <p>Hints:</p>
<ol>
<li><p>Given a point <span class="math-container">$z = a + bi$</span> in the complex plane, the magnitude of <span class="math-container">$z$</span> is <span class="math-container">$|z| = \sqrt{a^2 + b^2}$</span>, from the Pythagorean theorem. Plot the point <span class="math-container">$z + 1 = a + bi + 1 = [a + 1] + bi$</span> on the complex plane, and determine the magnitude <span class="math-container">$|z+1|$</span> from there. Replace <span class="math-container">$a = \cos\theta$</span> and <span class="math-container">$b = \sin\theta$</span>.</p>
</li>
<li><p>Use the half-angle formula for cosine: <span class="math-container">$\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1+\cos(\theta)}{2}}$</span>.</p>
</li>
</ol>
|
3,890,382 | <blockquote>
<p>Find the locus of <span class="math-container">$z$</span> such that <span class="math-container">$\arg \frac{z-z_1}{z-z_2} = \alpha$</span>.
Use and draw <span class="math-container">$w = \frac{z-z_1}{z-z_2}$</span>.</p>
</blockquote>
<p>This exercise was discussed many times -- <a href="https://math.stackexchange.com/questions/2120597/finding-the-loci-of-arg-dfracz-az-b-theta?rq=1">1</a>, <a href="https://math.stackexchange.com/questions/1333930/reasoning-centre-and-ways-of-expressing-locus-of-arg-frac-z-az-b-c?rq=1">2</a>, <a href="https://math.stackexchange.com/questions/1573659/how-to-describe-the-locus-of-z-where-arg-left-fracz-z-1z-z-2-right-t?rq=1">3</a>, <a href="https://math.stackexchange.com/questions/1844695/locus-of-complex-number-2/1844716#1844716">4</a> -- but I was unable to find answers to my problem with <span class="math-container">$0$</span> there.</p>
<p>I believe I understand where the arcs came from, here's my work:</p>
<p><a href="https://i.stack.imgur.com/CiMR7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CiMR7.jpg" alt="picture showing my work" /></a></p>
<p>If I understand correctly, for <span class="math-container">$\alpha = \pm\pi$</span>, the locus would be the segment connecting <span class="math-container">$z_2$</span> and <span class="math-container">$z_1$</span>, not including the points themselves.</p>
<p>I can not understand what is happening when <span class="math-container">$\alpha = 0$</span>.</p>
<p><span class="math-container">$\alpha = 0 = \arg \frac{z-z_1}{z-z_2} = \arg w \Longrightarrow \frac{z-z_1}{z-z_2} = k \in \mathbb{R}, \frac{z-z_1}{z-z_2} = k\frac{z-z_2}{z-z_2}.$</span></p>
<p>Solving this for <span class="math-container">$z$</span>, <span class="math-container">$z = \frac{x_1-kx_2}{1-k} +i\frac{y_1-ky_2}{1-k}$</span>. I am having trouble understanding the locus of this <span class="math-container">$z$</span>. The textbook says it should be 'two segments with end points in <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span>, and one of this segments contains an infinitely distant point'. How to understand why is this answer right, and how to draw it? It seems the infinitely distant point matches <span class="math-container">$k=1$</span>, but why should it lie in the 'direction' of the line passing through <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span>?</p>
<p>My class notes are messy. Why is <span class="math-container">$(0, 1)$</span> special on <span class="math-container">$w$</span> plane?
<a href="https://i.stack.imgur.com/SmPAE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SmPAE.jpg" alt="messy class notes" /></a></p>
<p>Thank you.</p>
| Raffaele | 83,382 | <p>Let <span class="math-container">$z=x+iy;\;z_1=x_1+iy_1;\;z_2=x_2+iy_2$</span>
<span class="math-container">$$\frac{z-z_1}{z-z_2}= \frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}+i\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}$$</span>
<span class="math-container">$$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=\arctan\frac{{\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}}}{{\frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}}}=\\=\arctan\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}$$</span></p>
<p><span class="math-container">$$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=0\to x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)=0$$</span>
Rearrange
<span class="math-container">$$x (y_2-y_1)+y (x_1-x_2)+x_2y_1-x_1y_2=0$$</span>
which is the equation of a line.</p>
<p>In the general case, let <span class="math-container">$a=\cot\alpha$</span></p>
<p><span class="math-container">$$x^2+y^2+x (-x_1-x_2+y_1 a -y_2 a )+y (-x_1 a +x_2 a -y_1-y_2)+x_1 x_2+x_1 y_2 a -x_2 y_1 a +y_1 y_2=0$$</span>
<span class="math-container">$$x^2+y^2+px+qy+r=0$$</span>
we get a circle.</p>
<p>Hope this can be useful.</p>
|
3,984,480 | <p>Show that <span class="math-container">$-\vec{0} = \vec{0}$</span> in any vector space.</p>
<p>I know this is a seemingly obvious statement but is the following justification correct:</p>
<p>Assume <span class="math-container">$-\vec{0} \neq \vec{0}$</span>.</p>
<p><span class="math-container">$$(4): \vec{u} + \vec{0} = \vec{u}$$</span>
<span class="math-container">$$ \vec{u} + \vec{0} + (-\vec{0}) = \vec{u} + (-\vec{0})$$</span>
<span class="math-container">$$(5): \vec{u} -\vec{u} + \vec{0} + (-\vec{0}) = \vec{u} - \vec{u} + (-\vec{0})$$</span>
<span class="math-container">$$(5): \vec{0} + (-\vec{0}) = (-\vec{0}) \rightarrow -\vec{0} + \vec{0} = \vec{0}$$</span></p>
<p>Hence we clearly see that <span class="math-container">$-\vec{0} = \vec{0}$</span> QED Is this reasoning correct?</p>
| bonsoon | 48,280 | <p>Somewhat related: All subgroups of finitely generated abelian groups are finitely generated, however this is not necessarily so for subgroups of finitely generated groups.</p>
<p>Perhaps the classic example of the free group of 2 generators has subgroups that are not finitely generated, and its relation to algebraic topology: <a href="https://math.stackexchange.com/questions/983480/commutator-subgroup-of-rank-2-free-group-is-not-finitely-generated">Commutator subgroup of rank-2 free group is not finitely generated.</a></p>
|
2,797,329 | <p>The function $y_1 = x^2$ is a solution of
$x^2y'' − 3xy' + 4y = 0$.
Find the general solution of the nonhomogeneous linear differential equation
$x^2y'' − 3xy' + 4y = x^2$</p>
<p>I know the equation $x^2y'' − 3xy' + 4y = 0$ is a Euler-Cauchy equation but I'm not sure how to proceed with this question; any help is appreciated</p>
| Rebellos | 335,894 | <p><strong>General way :</strong></p>
<p>Working over the homogeneous equation :</p>
<p>$$x^2y'' − 3xy' + 4y = 0$$</p>
<p>Since this is an Euler-Cauchy equation, assume that a solution will be proportional to $x^\lambda$ for some constant $\lambda$. Thus, this means that :</p>
<p>$$x^2 \cdot (x^\lambda)'' - 3x \cdot (x^\lambda)' + 4x^\lambda = 0 \implies (\lambda -2)^2 = 0 \Rightarrow \lambda =2$$</p>
<p>Take note that the root $\lambda = 2$ is of multiplicity $2$, thus the general solution is :</p>
<p>$$y_c(x) = c_1x^2 + c_2x^2\ln(x)$$</p>
<p>Given the fact that $y_1(x) = x^2$ is a solution to this problem from the beginning, you can simply note that $y_2(x) = x^2\ln(x)$ is also a solution and thus the basis solutions in $y_c(x)$ are $y_1$ and $y_2$.</p>
<p>To finally yield the solution for the non-homogeneous equation, calculate the <strong><em>Wronskian</em></strong> of $y_1$ and $y_2$ :</p>
<p>$$\mathcal{W}(x) = \begin{vmatrix} x^2 & x^2\ln(x) \\ 2x & x + 2x\ln(x) \end{vmatrix} = x^3 $$</p>
<p>Divide the equation by the leading term's coefficient $x^2$ :</p>
<p>$$y'' - \frac{3y'}{x} + \frac{4y}{x^2} = 1$$</p>
<p>Let $f(x)=1$ and $v_1(x), v_2(x)$ such that :</p>
<p>$$v_1(x) = - \int \frac{f(x)y_2(x)}{\mathcal{W}(x)}\mathrm{d}x, \quad v_2(x) = \int \frac{f(x)y_1(x)}{\mathcal{W}(x)}\mathrm{d}x$$</p>
<p>Then, the particular solution will be :</p>
<p>$$y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x) $$</p>
<p>and finally the general solution :</p>
<p>$$y(x) = y_c(x) + y_p(x)$$</p>
<p>I will leave the final calculations up to you.</p>
<p>You should come to the result : </p>
<p>$$\boxed{y(x) = c_1x^2 + c_2x^2\ln(x) + \frac{1}{2}x^2\ln^2(x)}$$</p>
<p>For some better feeling over the solution, I'll add some graphs that sample initial values for $y(1)$ and a general sample plot :</p>
<p><a href="https://i.stack.imgur.com/lnaSP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lnaSP.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/On3ZA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/On3ZA.jpg" alt="enter image description here"></a></p>
|
3,206,730 | <blockquote>
<p>Let <span class="math-container">$f : (-1,1)\to (-\pi/2,\pi/2)$</span> be the function defined by <span class="math-container">$f(x)= \tan^{-1}\left(\frac{2x}{1-x^2}\right)$</span> the verify that <span class="math-container">$f$</span> is bijective</p>
</blockquote>
<p>To check objectivity I assumed 2 variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to be equal and so as to prove <span class="math-container">$f(x)=f(y)$</span>. But I couldn't do so. I also wish to prove surjectivity.</p>
| egreg | 62,967 | <p>Let
<span class="math-container">$$
f(x)=\arctan\frac{2x}{1-x^2},\qquad g(x)=\frac{2x}{1-x^2}
$$</span></p>
<p>The derivative is
<span class="math-container">$$
f'(x)=\frac{1}{1+g(x)^2}g'(x)
$$</span>
Now
<span class="math-container">$$
1+g(x)^2=1+\frac{4x^2}{(1-x^2)^2}=\frac{(1+x^2)^2}{(1-x^2)^2}
$$</span>
and
<span class="math-container">$$
g'(x)=2\frac{1-x^2+2x^2}{(1-x^2)^2}
$$</span>
which means that
<span class="math-container">$$
f'(x)=\frac{(1-x^2)^2}{(1+x^2)^2}\frac{2(1+x^2)}{(1-x^2)^2}=\frac{2}{1+x^2}
$$</span>
which is positive. Since
<span class="math-container">$$
\lim_{x\to-1^+}f(x)=-\frac{\pi}{2},\qquad\lim_{x\to-1^-}f(x)=\frac{\pi}{2}
$$</span>
you're done.</p>
<p>By the way, this shows that, for <span class="math-container">$x\ne\pm1$</span>,
<span class="math-container">$$
f(x)=\begin{cases}
c_++2\arctan x & x>1 \\[4px]
c_0+2\arctan x & -1<x<1 \\[4px]
c_-+2\arctan x & x<-1
\end{cases}
$$</span>
Then <span class="math-container">$c_0=0$</span>, by evaluating at <span class="math-container">$0$</span>. Since the limits of <span class="math-container">$f$</span> at <span class="math-container">$\pm\infty$</span> are <span class="math-container">$0$</span>, we get that <span class="math-container">$c_+=-\pi$</span> and <span class="math-container">$c_-=\pi$</span>.</p>
|
2,528,306 | <p>The answer is 648 but I tried to solve this problem in reverse, so I ended up with 630. Theee are 10 ways to pick the third digit, 9 ways to pick the second digit, and 7 ways to pick the first digit. So why do these answers differ. Please do not close this question as I am trying to learn mathematics and I have stumbled upon this question.</p>
| Air Conditioner | 504,810 | <p>Let's start by picking the 1st digit. There are 9 choices, as it can be any digit but 0. The 2nd digit can be any digit but the first digit, so you have 9 choices. The 3rd digit can be any digit but digit 1 and digit 2, so you have 8 choices. So you have $9*9*8=648$. </p>
<p>If you start by picking the 3rd digit, it's less straightforward. You have 10 choices for the 3rd, and 9 choices for the 2nd. But you run into problems when picking the 1st digit, as your number of choices varies depending on whether 0 has already been picked. If it has, you have 8 choices. If it has not been picked, you have 7 choices, since it cannot be digit 2, digit 3, or 0. </p>
|
1,915,366 | <p>Suppose we have two Gaussian distributed random variable $X$~$N(0,\sigma^2)$ and $Y$~$N(0,\sigma^2)$. These variables are not independent. What will be the expected value of product of square of this random variables</p>
<p>$E[X^2Y^2]$ = ??</p>
<p>Edit 1: They are jointly Gaussian distributed with correlation coefficient $\rho$ </p>
<p>Edit 2: $X$~$N(0,\sigma^2)$, $Y$~$N(0,\sigma^2)$</p>
| Snufsan | 122,989 | <p>Try using the <a href="https://en.wikipedia.org/wiki/Law_of_total_expectation" rel="nofollow">Law of total expectation</a> - set $Z = X^2Y^2$ and use:</p>
<p>$$\mathbb{E}[X^2Y^2]=\mathbb{E}[Z]=\int_{y}\mathbb{E}[Z\mid Y=y]\cdot\Pr[Y=y]$$</p>
|
1,441,603 | <p>Solve the PDE for $u(x,y)$ $$\frac{\partial^2 u}{\partial x \, \partial y} = 0$$
I was thinking to integrated both sides in respect to $x$ first to get $$x= c(x)$$ then i will have $$c(x)-x=0$$ then i will integrate in respect to y but i think this wrong because it does not making any sense to me. </p>
| u184 | 218,127 | <p>If you integrate both sides w.r.t. x, you should get $$\frac{\partial u}{ \partial y} = c(y)$$ for some arbitrary function c. Integrating again w.r.t. y now gives $$u=a(y)+b(x)$$ where $b$ is an arbitrary function of $x$ and $a$ is the function you get when integrating $c$ which is again arbitrary since $c$ was. So the general solution is simply the sum of 2 differentiable functions, one only dependent on $x$ and the other one only dependent on $y$. For e.g $u(x,y)=x+e^y$ is one of the infinite possible solutions.</p>
|
980,941 | <p>How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?</p>
| mookid | 131,738 | <p><strong>Hint:</strong> use also that
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6
$$</p>
<p>$$
1 + (1+2) + \dots + (1 +2+\dots +n) =
\frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2
\\=\frac 12 \left[
(1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n)
\right]
\\=\frac 12 \left[
(1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n)
\right]
$$</p>
|
47,603 | <p>Is it possible to express the functions $S(x)=x+1$ and $Pd(x)=x\dot{-}1$ in terms of the functions $f_1$, $f_2$, $f_3$ and $f_4$, where $f_1(x)=0$ if $x$ is even or $1$ if $x$ is odd, $f_2(x)=\mbox{quot}(x,2)$, $f_3(x)=2x$ and $f_4(x)=2x+1$? For example, $S(x)=f_4(f_2(x))$ if x is even. Is there a similar formula if $x$ is odd?</p>
| Guillaume Brunerie | 10,217 | <p>I assume that the problem is: does there exists a sequence $(u_1,\dots,u_n)\in{}\{1,2,3,4\}$ such that for all $x\in\mathbb{N}$, $S(x) = f_{u_n}(\dots (f_{u_1}(x))\dots)$ (and similarly for $Pd$)</p>
<p>Let’s prove by induction on $n$ that every such function either is of the form $f(x)=2^kx+l$ where $k\ge 0$ and $0\le l<2^k$ or $f(2)=f(3)$ or the image of $f$ contains only two elements.</p>
<p>If there is an $f_1$ somewhere, then the image of $f$ contains only two elements, and this is preserved by composition. We suppose now that there is no $f_1$.</p>
<ul>
<li>If $n=0$, $f(x) = x$ ok.</li>
<li>If $f(2)=f(3)$ then $f_i(f(2)) = f_i(f(3))$</li>
<li>If $f(x) = 2^kx+l$, $f_3\circ f$ and $f_4\circ f$ are of the form $2^{k+1}x+l'$ with $0\le l'< 2^{k+1}$, and $f_2\circ f$ is also of the same form if $k\ge 1$, and if $k=0$ we have $f_2(f(3))=f_2(f(2))=1$.</li>
</ul>
<p>This proves that neither $S$ nor $Pd$ can be written as a composition of the functions $f_i$.</p>
|
487,084 | <p>I need to know if every group whose order is a power of a prime $p$ contains an element of order $p$? Should I proceed by picking an element $g$ of the group and proving that there is an element in $\langle g \rangle$ that has order $p$?</p>
| Mauricio Tec | 64,684 | <p>There are some results that are much stronger than that. <a href="http://en.wikipedia.org/wiki/Cauchy%27s_theorem_%28group_theory%29" rel="nofollow noreferrer">Cauchy's theorem</a> states that every finite group whose order is divisible by some prime <span class="math-container">$p$</span> has a subgroup of order <span class="math-container">$p$</span>. And from <a href="http://en.wikipedia.org/wiki/Sylow_theorems" rel="nofollow noreferrer">Sylow's theorem</a> it can be deduced (although not immediately) that if the order of the group is <span class="math-container">$p^n$</span> then there is one subgroup of order <span class="math-container">$p^k$</span> for every <span class="math-container">$k=0,1,..,n$</span>. </p>
<p>One more thing, a subgroup of order <span class="math-container">$p$</span> must be cyclic, that is, there has to be an element of order <span class="math-container">$p$</span> in it. That is because by Langrange's theorem the order of every element must divide the <span class="math-container">$p$</span> and since it is prime then the order must be <span class="math-container">$1$</span> or <span class="math-container">$p$</span>. Any element different from the identity will do the trick. </p>
<p><em>Note</em>: Lagrange's theorem alone is not enough to prove this since it only states that if the order of the group is <span class="math-container">$p^n$</span> then every subgroup is of the form <span class="math-container">$p^k$</span>. That is because what Lagrange's theorem says is that the order of every subgroup must divide the order of the group. So you actually need a little bit more I think.</p>
<p><em>Late note due to nice comments</em>: While the result doesn't follow directly from Lagrange's theorem statement. It <em>can be derived</em> from it as it is nicely shown to you in other answers. So you actually can avoid appealing to a stronger result such as Cauchy's theorem since you are in a finite <span class="math-container">$p$</span>-group (what I mean by saying that it doesn't follow directly is that Lagrange's theorem makes no reference to <span class="math-container">$p$</span>-groups, so there is math involved). </p>
|
3,342,094 | <p>I am asked to prove following proposition:</p>
<blockquote>
<p><strong>Proposition 1.</strong> If an invertible matrix <span class="math-container">$\mathbf A$</span> has a left inverse <span class="math-container">$\mathbf{B}$</span> and a right inverse <span class="math-container">$\mathbf{C}$</span>, then <span class="math-container">$\mathbf{B} = \mathbf {C}$</span></p>
</blockquote>
<p>My attempt:</p>
<p>"<span class="math-container">$\mathbf{B}$</span> is the left inverse of <span class="math-container">$\mathbf{A}$</span>" implies:
<span class="math-container">$$\mathbf{BA} = \mathbf I$$</span></p>
<p>And "<span class="math-container">$\mathbf{C}$</span> is the right inverse of <span class="math-container">$\mathbf{A}$</span>" implies:</p>
<p><span class="math-container">$$\mathbf{AC} = \mathbf I$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$\tag 1\mathbf{AC} = \mathbf{BA}$$</span></p>
<p>Premultiply <span class="math-container">$(1)$</span> by <span class="math-container">$\mathbf A^{-1}$</span>:
<span class="math-container">$$ \mathbf A^{-1}\mathbf{AC} = \mathbf A^{-1}\mathbf{BA} \implies $$</span>
<span class="math-container">$$\mathbf{C} = \mathbf A^{-1} $$</span></p>
<p>Postmultiply <span class="math-container">$(1)$</span> by <span class="math-container">$\mathbf A^{-1}$</span>:
<span class="math-container">$$\mathbf{AC}\mathbf A^{-1} = \mathbf{BA}\mathbf A^{-1} \implies$$</span>
<span class="math-container">$$\mathbf A^{-1} = \mathbf{B}$$</span></p>
<p>Hence <span class="math-container">$\mathbf B = \mathbf C.$$\Box$</span></p>
<p>Is it correct?</p>
<hr>
<p>I have one more question: </p>
<p>Assume we already proved following theorem:</p>
<blockquote>
<p>The inverse of an invertible matrix is <strong>unique</strong></p>
</blockquote>
<p>Knowing this, do we really need to prove the proposition <strong>1</strong>? My reasoning is, if we know <span class="math-container">$\mathbf{B}$</span> is the left inverse of <span class="math-container">$\mathbf{A}$</span>, and we also know theorem above, then we can infer that <span class="math-container">$\mathbf{B}$</span> will also be right inverse of <span class="math-container">$\mathbf{A}$</span>. Or am I missing something?</p>
| Sambo | 454,855 | <p><strong>EDIT</strong>: In the assumption of the theorem, we have that <span class="math-container">$A$</span> is invertible, so assuming the existence of a (left-and-right) inverse <span class="math-container">$A^{-1}$</span> is reasonable. Therefore, the original proof is correct.</p>
<p>The reason why I originally thought it was incorrect is that this assumtion is not needed, and therefore usually not used.</p>
<hr>
<p>Your proof is not quite right, but a simple modification makes it work. Let me explain.</p>
<p>When you say "multiply by <span class="math-container">$A^{-1}$</span>", you are assuming that there exists a matrix <span class="math-container">$A^{-1}$</span> which is both a left and right inverse of <span class="math-container">$A$</span>. Now, this will in fact be true, but it will follow from the theorem you are trying to prove, so you can't invoke that property.</p>
<p>Instead, when you premultiply, you can replace <span class="math-container">$A^{-1}$</span> by the left inverse, <span class="math-container">$B$</span>, which you know exists. You then end up with:
<span class="math-container">\begin{align}
AC = I && (1)\\
B(AC) = BI && (2)\\
(BA)C = B && (3)\\
IC = B && (4)\\
C = B && (5)
\end{align}</span>
Line 1 uses that <span class="math-container">$C$</span> is a right inverse; line 4 uses that <span class="math-container">$B$</span> is a left inverse. The rest is properties of identity and the associative property.</p>
<hr>
<p>To answer your second question: it all depends what you mean by "inverse". You could be saying the following:</p>
<blockquote>
<p>Let <span class="math-container">$A$</span> be a matrix, and let <span class="math-container">$B,C$</span> be matrices such that <span class="math-container">$I = AB = BA = AC = CA$</span>. Then <span class="math-container">$B=C$</span>.</p>
</blockquote>
<p>If this is the case, then it is not strictly sufficient to prove your theorem. This is because, a priori, we don't know that a left inverse is necessarily a right inverse as well, so we can't use this fact to conclude equality.</p>
|
138,698 | <p>I want to evaluate the following double summation</p>
<pre><code>Sum[(-1)^(i + j + i*j)*Exp[-Pi/2*( i^2 + j^2)], {i, -Infinity,
Infinity}, {j, -Infinity, Infinity}]
</code></pre>
<p>I am really new both in using Mathematica and in doing mathematics using computer. I don't know if there is some special technics to deal with these kind of summations (Lattice sums) in Mathematica.</p>
<p>When I evaluate the former expression, Mathematica refuses to evaluate it and just reprint it in the output.</p>
<p>Theoretically, the expected value is 0.</p>
| Szabolcs | 12 | <p><a href="https://en.wikipedia.org/wiki/Array_programming" rel="noreferrer">Vectorization</a> is one of the most effective ways to increase performance. The <code>numpy</code> code you show is fast because it uses vectorization. </p>
<p>Vectorization means working with entire arrays instead of element by element: using array arithmetic, array comparisons, etc. Array operations can be implemented very efficiently using <a href="https://en.wikipedia.org/wiki/SIMD" rel="noreferrer">SIMD processing</a>, and are also straightforwardly parallelizable. In fact Mathematica uses multiple CPU cores for vector arithmetic. This happens to be important for correct benchmarking as well because <code>Timing</code> adds up the time spent by individual CPU cores while <code>AbsoluteTiming</code> measures how much time has actually elapsed. Thus below I use only <code>AbsoluteTiming</code>. <code>Timing</code> would give <em>longer</em> and inaccurate times (try it!).</p>
<hr>
<p>The first thing we can do is replace <code>Table[Random[], n]</code> by <code>RandomReal[1, n]</code>. In the spirit of vectorization, generate the whole array in one go instead of element by element:</p>
<pre><code>RandomReal[1, 10000000]; // AbsoluteTiming
(* {0.083058, Null} *)
Table[Random[], 10000000]; // AbsoluteTiming
(* {0.312061, Null} *)
</code></pre>
<p>Then instead of <code>Select</code>, use the techniques I described here:</p>
<ul>
<li><a href="https://mathematica.stackexchange.com/q/2821/12">Does Mathematica have advanced indexing?</a></li>
</ul>
<p>Mathematica does not have built-in vector comparison, like <code>samples < 0.5</code> in <code>numpy</code>. But it is always possible to express these operations in terms of simple arithmetic, as described in the thread linked above. Unfortunately, this often results in expressions which are hard to decipher. To make it easier to use these techniques, I wrote <a href="http://szhorvat.net/mathematica/BoolEval" rel="noreferrer">a small package called BoolEval</a> that will translate expressions written in terms of relational operators (like <code><</code>, <code>></code>, <code>==</code>, etc.) into vector arithmetic.</p>
<p>This is how we can apply the package to your example:</p>
<pre><code>Select[Table[Random[], {10000000}], # < 0.5 &]; // AbsoluteTiming
(* {5.02988, Null} *)
</code></pre>
<pre><code><< BoolEval`
AbsoluteTiming[
arr = RandomReal[1, 10000000];
BoolPick[arr, arr < 0.5];
]
(* {0.310938, Null} *)
</code></pre>
<p>Behind the scenes, <code>BoolEval</code> translates this into:</p>
<pre><code>Pick[arr, 1 - UnitStep[arr - 0.5], 1]; // AbsoluteTiming
(* {0.237126, Null} *)
</code></pre>
<p>We can see this by using <code>BoolEval</code> with a symbolic expression:</p>
<pre><code>BoolEval[a < 0.5]
(* 1 - UnitStep[-0.5 + a] *)
</code></pre>
<p>The package also has other useful functions such as <code>BoolCount</code> for counting how many elements satisfy the condition:</p>
<pre><code>BoolCount[arr < 0.5] // AbsoluteTiming
(* {0.07805, 4998908} *)
</code></pre>
<p>This translates to <code>Total[1 - UnitStep[arr - 0.5]]</code> and is precisely equivalent to the <code>numpy</code> code you showed. As a comparison, your <code>numpy</code> code runs in ~0.14 seconds on my machine, which is nearly twice slower. I was using an <a href="https://en.wikipedia.org/wiki/Math_Kernel_Library" rel="noreferrer">MKL</a>-enabled numpy (from <a href="https://www.continuum.io/" rel="noreferrer">Anaconda</a>), so the comparison with Mathematica is not unfair.</p>
|
976,881 | <p>Simple and quick question. These two have to do with 90 degree angles.</p>
<p>This is the picture of the two words I have.</p>
<blockquote>
<p>Perpendicular is strictly restricted to lines.</p>
</blockquote>
<ul>
<li><p>"Line A and B are perpendicular to each other."</p></li>
<li><p>"v=(1, 1) and w=(-1, 1) -> cv and dw are perpendicular towards each other."</p></li>
</ul>
<blockquote>
<p>Orthogonal is restricted to matrices.</p>
</blockquote>
<pre><code>1 1 1 1
1 -1 1 -1
1 1 -1 -1
1 -1 -1 1
</code></pre>
<ul>
<li>The above matrix is an orthogonal matrix.</li>
</ul>
<p>So would be following statement be correct: A orthogonal matrix has vectors which are perpendicular towards each other?</p>
| rogerl | 27,542 | <p>Usually the term "orthogonal matrix" is reserved for matrices whose columns are not only mutually perpendicular, but also unit vectors. So if you were to divide each entry in your matrix above by $2$, it would be an orthogonal matrix.</p>
<p>There appears to be no standard term for a matrix whose columns are just orthogonal without any restriction on their norms.</p>
|
976,881 | <p>Simple and quick question. These two have to do with 90 degree angles.</p>
<p>This is the picture of the two words I have.</p>
<blockquote>
<p>Perpendicular is strictly restricted to lines.</p>
</blockquote>
<ul>
<li><p>"Line A and B are perpendicular to each other."</p></li>
<li><p>"v=(1, 1) and w=(-1, 1) -> cv and dw are perpendicular towards each other."</p></li>
</ul>
<blockquote>
<p>Orthogonal is restricted to matrices.</p>
</blockquote>
<pre><code>1 1 1 1
1 -1 1 -1
1 1 -1 -1
1 -1 -1 1
</code></pre>
<ul>
<li>The above matrix is an orthogonal matrix.</li>
</ul>
<p>So would be following statement be correct: A orthogonal matrix has vectors which are perpendicular towards each other?</p>
| Roland Puntaier | 390,013 | <blockquote>
<p>Perpendicular is strictly restricted to lines.</p>
</blockquote>
<p>"perp, short for perpendicular complement" can have more dimensions.
This in an article named <a href="https://en.wikipedia.org/wiki/Orthogonal_complement" rel="nofollow noreferrer">orthogonal component</a>.</p>
<p>So it is not just for lines.</p>
<ul>
<li><em>perpendicular</em> comes from Latin (hanging down)</li>
<li><em>orthogonal</em> from greek (right lines)</li>
</ul>
<p>In general, these two words seem to be used in a synonymous way.</p>
<p>Some specific contexts have settled with a specific meaning,
which uses always one of the two words,
as it is for <em>orthogonal matrix</em> meaning real matrix <span class="math-container">$Q$</span> with <span class="math-container">$QQ^T=I$</span>.
Word usage just settles over time,
possibly starting from an influential text, or more texts.</p>
<p>A different definition between the two words per text might still occur.
To avoid misunderstanding is the reason,
why every text repeats definitions.</p>
|
4,166,894 | <p>Let <span class="math-container">$N\triangleleft G$</span> and <span class="math-container">$K\leqslant G$</span>. Consider <span class="math-container">$\phi :G \mapsto G/N$</span> onto group homomorphism. Show that <span class="math-container">$\phi(K)=KN/N$</span>.</p>
<p>I thought using the equality <span class="math-container">$(G/N)^{n}=G^{n}N/N$</span> to show but I can not show. Any idea will be appreciated.</p>
| Misha Lavrov | 383,078 | <p>This is a special case of the <a href="https://en.wikipedia.org/wiki/Kruskal%E2%80%93Katona_theorem" rel="nofollow noreferrer">Kruskal–Katona theorem</a>.</p>
<p>For a set <span class="math-container">$S$</span>, let <span class="math-container">$\binom Sk$</span> denote all the <span class="math-container">$k$</span>-element subsets of <span class="math-container">$S$</span>. Lovász's simplified formulation of the Kruskal–Katona theorem says the following.</p>
<blockquote>
<p>For any <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, take any <span class="math-container">$\mathcal F \subseteq \binom{[n]}{k}$</span> with <span class="math-container">$|\mathcal F| = \binom xk$</span>. (Here, <span class="math-container">$x$</span> can be any real number, so any <span class="math-container">$|\mathcal F|$</span> can be expressed as <span class="math-container">$\binom xk$</span> for some <span class="math-container">$x$</span>.) Let the <em>shadow</em> <span class="math-container">$\Delta(\mathcal F)$</span> be defined as <span class="math-container">$\bigcup_{A \in \mathcal F} \binom{A}{k-1}$</span>. Then <span class="math-container">$|\Delta(\mathcal F)| \ge \binom{x}{k-1}$</span>.</p>
</blockquote>
<p>In this case, let <span class="math-container">$k=4$</span> and let <span class="math-container">$\mathcal F$</span> be the set of all <span class="math-container">$4$</span>-element sets that induce a copy of <span class="math-container">$K_4^3$</span> in the hypergraph. In this case, the shadow <span class="math-container">$\Delta(\mathcal F)$</span> is a subset of the edge set of the hypergraph (there may also be edges not part of any <span class="math-container">$K_4^3$</span>). The statement we want to show is just the contrapositive of what Kruskal–Katona tells us.</p>
|
1,950,809 | <p>I'm fairly certain that the probability of both dice returning an even number is $1/4$.</p>
<p>I got this by saying that since these are independent events, with each die returning an even number being $1/2$, then the probability of both being even is $1/2 \times 1/2 = 1/4$.</p>
<p>Further, there are 36 outcomes, and all possible even number combinations are $(2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)$. There are nine of them and $9/36 = 1/4$</p>
<p>What I can't seem to get over, is that there are an equal number of odd and even numbers, so, why is the answer not $1/2$?</p>
<p>I know that it's not one half, but I can't explain why. </p>
| Graham Kemp | 135,106 | <blockquote>
<p>What I can't seem to get over, is that there are an equal number of odd and even numbers, so, why is the answer not 1/2 ?</p>
</blockquote>
<p>Because they are not complementary events. There is another possibility.</p>
<p>The probability that both dice show even numbers is: $1/4$</p>
<p>The probability that both dice show odd numbers is: $1/4$</p>
<p>The probability that one die shows even and the other shows odd is: $1/2$</p>
|
270,624 | <p>For a polynomial $f(x) = \sum_{i=0}^dc_ix^i \in \mathbb Z[x]$ of degree $d$, let</p>
<p>$$
H(f):=\max\limits_{i=0,1,\ldots, d}\{|c_i|\}
$$</p>
<p>denote the naive height. Further, define</p>
<p>$$
R(M, r, d) := \#\{f(x) \colon \text{$H(f) \leq M$, $\deg f = d$ and $f(x)$ has extactly $r$ real roots}\}.
$$</p>
<p>I wonder if anything is known about the quantity $R(M,r,d)$. More precisely, I am interested how it changes as $r$ and $d$ remain fixed and $M$ varies. Has this question been explored at all? I would be thankful for any references.</p>
| Stanley Yao Xiao | 10,898 | <p>The quadratic case can be dealt with as follows. A quadratic polynomial $f(x) = ax^2 + bx + c \in \mathbb{Z}[x]$ has two distinct real roots if and only if $\Delta(f) = b^2 - 4ac > 0$, and a pair of complex conjugate roots if and only if $\Delta(f) < 0$. </p>
<p>We now let $a,b,c$ vary in the box $[-X,X]^3$. We first pick a pair $(a,c) \in \mathbb{Z}^2 \cap [-X,X]^2$. If $ac < 0$, that is, if the pair $(a,c)$ lies in two of four quadrants, then $\Delta(f) > 0$; hence 100% of quadratic polynomials with $a,c$ coming from those two quadrants have two real roots. The remaining two quadrants are symmetric to each other, so we might as well consider only the positive quadrant. We can exploit symmetry once more to assume that $a \leq c$, and from density considerations we can carve out the 0-density sets corresponding to $a = 0$ and $a =c$; whence we assume $0 < a < c$. </p>
<p>The count of triples $(a,b,c)$ satisfying $0 < a < c \leq X$ and $\Delta(f) < 0$ can be estimated by the triple integral</p>
<p>$$\displaystyle \int_1^X \int_1^c \int_{-2\sqrt{ac}}^{2 \sqrt{ac}} db da dc = \frac{8}{9} X^3 + O(X^2).$$</p>
<p>Multiplying by 4 to account for the assumption that $a \leq c$ and $a,c > 0$ (and using standard geometry of numbers arguments), we see that the total number of negative discriminant quadratic polynomials of height at most $X$ is </p>
<p>$$\displaystyle N^+(X) = \frac{32}{9} X^3 + O(X^2).$$</p>
<p>The number of positive discriminant forms is then </p>
<p>$$\displaystyle N^{-}(X) = 8X^3 - \frac{32}{9} X^3 + O(X^2) = \frac{40}{9} X^3 + O(X^2).$$</p>
<p>One can do a similar (but much more difficult) argument for cubic polynomials (binary forms), by exploiting the fact that for a cubic binary form $g(x,y) \in \mathbb{Z}[x,y]$, its Hessian covariant $q_g(x,y)$ (which is a quadratic form) has discriminant $-3\Delta(g)$; and hence the problem of counting cubic binary forms with three or one real linear factors is reduced to dealing with the Hessians. However, the inequalities involved are no longer linear in general, and hence the application of geometry of numbers methods will be more complicated. Cremona also worked out the exact conditions for quartic polynomials to have 0, 2, or 4 real roots in <a href="https://homepages.warwick.ac.uk/~masgaj/papers/r34jcm.pdf" rel="nofollow noreferrer">https://homepages.warwick.ac.uk/~masgaj/papers/r34jcm.pdf</a></p>
<p>I suspect the methods I used above become intractable very quickly with respect to the degree, so perhaps a different formulation is necessary to make progress.</p>
<p>Addendum: I should add that the answer to the question is known for degrees 3 and 4 if instead one counts $\operatorname{GL}_2(\mathbb{Z})$-classes of binary forms (of degrees 3 and 4 respectively) with respect to an appropriate $\operatorname{GL}_2(\mathbb{Z})$-invariant height. In particular, when $d = 3$ and we put the height as the discriminant, we have</p>
<p>\begin{align*} N_3(X)& = \# \{F = a_3 x^3 + a_2 x^2 y + a_1 xy^2 + a_0 y^3 \in \mathbb{Z}[x,y]: |\Delta(F)| \leq X\} \\
& = \frac{\pi^2}{18} X + O(X^{5/6}),
\end{align*}
and $N_3^{\pm}(X)$ (which counts the number of forms of bounded positive/negative discriminant, respectively) is given by</p>
<p>$$\displaystyle N_3^+(X) = \frac{\pi^2}{72} X + O(X^{5/6}), N_3^-(X) = \frac{\pi^2}{24} X + O(X^{5/6}).$$</p>
<p>The main term was first obtained by Davenport, and the error term as given was obtained by Shintani. Taniguchi and Thorne and Bhargava-Shankar-Tsimerman independently obtained a secondary main term of order $X^{5/6}$. If one includes this secondary term, then the error term is $O(X^{3/4+\varepsilon})$. </p>
<p>For the degree 4 case, if we put the height as $H(F) = \max\{|I(F)|^3, J(F)^2/4\}$, as in Bhargava-Shankar (<a href="http://annals.math.princeton.edu/2015/181-1/p03" rel="nofollow noreferrer">http://annals.math.princeton.edu/2015/181-1/p03</a>), and put $N_4^{(0)}(X), N_4^{(1)}(X), N_4^{(2)}(X)$ for the number of $\operatorname{GL}_2(\mathbb{Z})$-classes of integral quartic forms of height at most $X$ with 0 pairs of complex conjugate linear factors, 1 pair of complex conjugate linear factors, and 2 pairs of complex conjugate linear factors respectively. They showed that</p>
<p>$$\displaystyle N_4^{(0)}(X) = \frac{4 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}),$$
$$\displaystyle N_4^{(1)}(X) = \frac{32 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}),$$
and
$$\displaystyle N_4^{(2)}(X) = \frac{8 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}).$$</p>
<p>Sorting by a $\operatorname{GL}_2(\mathbb{Z})$-invariant height is likely a more natural question and possibly easier than the original question. </p>
|
2,724,744 | <p>I have a basic math question.</p>
<p>If I have the following inequality:
$$-a-b > -1$$
and I want to flip (or reverse) the sign. What is the correct way of the following? And why?</p>
<p>i) $a+b \le 1$<br>
ii) $a+b < 1$</p>
<p>Many thanks! (:</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Multiplying the given inequality by $-1$ we get $$a+b<1$$</p>
|
33,005 | <p>Let n and p be any positive integer, make $p$ the subject of the equation: $(3n + p)\bmod4 = 0$. How is it done?</p>
<p>I've worked out that the only values for p are 1, 2, 3 and 0.</p>
<p>This formula is for calculating the amount of padding required in a bitmap's pixel array:</p>
<blockquote>
<p>Padding bytes (not necessarily 0) must be appended to the end of the rows in order to bring up the length of the rows to a multiple of four bytes. When the Pixel Array is loaded into memory, each row must begin at a memory address that is a multiple of 4. This address/offset restriction is mandatory only for Pixel Array's loaded in memory. For file storage purposes, only the size of each row must be a multiple of 4 bytes while the file offset can be arbitrary.[1]
A 24-bit bitmap with Width=1, would have 3 bytes of data per row (blue, green, red) and 1 byte of padding, while Width=2 would have 2 bytes of padding, Width=3 would have 3 bytes of padding, and Width=4 would not have any padding at all.</p>
</blockquote>
| Bill Dubuque | 242 | <p>$\rm\: mod\ 4\::\ \ 3\equiv -1\ $ so $\rm\ 0\ \equiv\ p + 3\:n\ \equiv\ p - n\ \iff\ p\ \equiv\ n\ \ $ by basic <a href="http://en.wikipedia.org/wiki/Modular_arithmetic" rel="nofollow">modular arithmetic.</a> </p>
<p>Alternatively $\rm\ 4\ |\ p + 3\:n\ \iff\ 4\ |\ p - n + 4\:n\ \iff\ 4\ |\ p - n\ \ $ where $\rm\ a\ |\ b\ $ means $\rm\:a\:$ divides $\rm\:b\:.$</p>
|
1,757,092 | <p>I want to find an explicit formula for $\sum_{n=0}^\infty n^3x^n$ for $|x|\le1$.Is the idea that first to show that this series is convergent and then we can find the number that it converges to? I tried to use ratio test, but it didn't work. Any suggestion? Thanks!</p>
| Nicholas Stull | 28,997 | <p>The Ratio Test will tell you for what $x$ the series converges:</p>
<p>$$L = \lim_{n\to\infty} \left|\frac{(n+1)^3x^{n+1}}{n^3x^n}\right| = |x|\lim_{n\to\infty} \frac{(n+1)^3}{n^3} = |x|$$</p>
<p>And the ratio test tells us that the series converges absolutely if $|x|<1$ (you should check that it diverges at the endpoints).</p>
<p>To compute its value, first notice
$$n^3 = n(n-1)(n-2) + 3n(n-1) + n$$</p>
<p>Next, we start with the geometric series: $$S_0=\sum_{n=0}^\infty x^n = \frac{1}{1-x},\quad |x|<1$$
Differentiate termwise to get:
$$\sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2},\quad |x|<1$$
Multiply by $x$ to get
$$S_1 = \sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2},\quad |x|<1$$
Repeat two more times to get:
$$S_2=\sum_{n=2}^\infty n(n-1) x^{n} = \frac{2x^2}{(1-x)^3},\quad |x|<1$$</p>
<p>$$S_3=\sum_{n=3}^\infty n(n-1)(n-2) x^{n} = \frac{6x^3}{(1-x)^4},\quad |x|<1$$</p>
<p>See if you can use the above information to find a combination of these that gives $\sum_{n=0}^\infty n^3x^n$. </p>
<hr>
<p>By the way, even though the bottom indeces on $S_1$, $S_2$, $S_3$ are not $0$, we can simply change them all to $0$, because we're adding multiple copies of $0$. (We change indeces at every step, since the constant terms vanish rather than introducing negative powers of $x$. But in this case, there's nothing to worry about.)</p>
|
6,562 | <p>I want to make some button shaped graphics that would essentially be a rectangular shape with curved edges. In the example below I have used <code>Polygon</code> rather than <code>Rectangle</code> so as to take advantage of <code>VertexColors</code> and have a gradient fill. The code below illustrates the sort of thing I want in so far as the <code>Frame</code> with <code>RoundingRadius</code> shows where I want the boundaries of the <code>Graphic</code> to be cut off (for example).</p>
<pre><code>Framed[Graphics[{
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}},
VertexColors -> {Red, Red, Blue, Blue}]
},
AspectRatio -> 0.2,
ImagePadding -> 0,
ImageMargins -> 0,
ImageSize -> 200,
PlotRangePadding -> 0],
ContentPadding -> True,
FrameMargins -> 0,
ImageMargins -> 0,
RoundingRadius -> 20]
</code></pre>
<p>I'm thinking there is probably a very straight forward way of accomplishing this. Is there some way to exclude parts of the <code>Graphic</code> that fall outside the <code>Frame</code> from displaying? Any alternative methods would be welcome.</p>
<p><strong>Edit</strong></p>
<p>I had been expecting that this was going to be possible with existing options rather than having to write functions. @Mr.Wizard provided a concise solution from existing built in functionality but I ultimately didn't want a raster solution. @Heike used <code>RegionPlot</code> like the others, but in a way in which the user, i.e. me, could implement it by simply changing a rounding radius parameter, so that makes it a more straight forward solution IMO.</p>
| Alec Titterton | 65,882 | <p>I've evolved the vector-based approach.
Rather than converting the object being rounded to an image, I needed to interact with maps and 3D graphics so have used an Overlay with a transparent interior. This means the object in question can exist as its original head and behaves the same, it just has rounded edges overlaid.</p>
<pre><code>Options[roundedOverlay] = {Background -> White};
roundedOverlay[g_, r_, opts : OptionsPattern[]] :=
Module[
{w, h, bgColor = OptionValue[Background],
commonOptions =
Sequence[PlotRange -> {{0, 1}, {0, 1}}, AspectRatio -> Full],
passepartout},
{w, h} = ImageDimensions[g];
passepartout =
FilledCurve[{{BezierCurve[{{0, #1/h}, {0, 1 - #2/h}, {0,
1 - #2/h}, {0, 1}, {#2/w, 1}, {1 - #3/w, 1}, {1 - #3/w,
1}, {1, 1}, {1, 1 - #3/h}, {1, #4/h}, {1, #4/h}, {1,
0}, {1 - #4/w, 0}, {#1/w, 0}, {#1/w, 0}, {0,
0}, {0, #1/h}}, SplineDegree -> 2] & @@
Apply[PadRight[#, 4, Last[#]] &, {Flatten[{r}]}]}, {Line[{{0,
0}, {0, 1}, {1, 1}, {1, 0}}]}}];
Overlay[{g,
Graphics[{FaceForm[bgColor], EdgeForm[bgColor], passepartout },
Background -> Transparent, ImageSize -> ImageDimensions[g],
commonOptions]}, All, 1]
]
</code></pre>
<p>Here's some examples:</p>
<pre><code>im = GeoGraphics[
Tooltip[Polygon[Entity["Country", "UnitedStates"]], "USA"]];
roundedOverlay[im, 20]
</code></pre>
<p><a href="https://i.stack.imgur.com/u1HBv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u1HBv.png" alt="Tooltipped map"></a></p>
<pre><code>test = Graphics3D[{Yellow, Cuboid[{0, 0, 0}, {1, 3, 1}], Blue,
Cuboid[{2, 1, 1}, {4, 2, 3}]}, SphericalRegion -> True,
Background -> LightBlue];
roundedOverlay[test, 50]
</code></pre>
<p>(Note the SphericalRegion->True is needed as the overlay does not resize)
<a href="https://i.stack.imgur.com/S2Q6h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S2Q6h.png" alt="enter image description here"></a></p>
|
388,225 | <p>If we have a random graph $G \in g(n,\frac{1}{2})$ how do we show that the expected number of edges is $\frac{1}{2} {{n}\choose{2}}$</p>
<p>Thanks in advance</p>
| Community | -1 | <p>Let $\mathbb{I}_k$ be the indicator function for the edge $k$, i.e.,
$$\mathbb{I}_k = \begin{cases}1 & \text{ if $k^{th}$ edge is present}\\ 0 & \text{ if $k^{th}$ edge is not present} \end{cases}$$
The quantity you are interested in is
\begin{align}
\mathbb{E}\left(\sum_{k=1}^{\binom{n}2} \mathbb{I}_k\right) & = \sum_{k=1}^{\binom{n}2} \mathbb{E}\left(\mathbb{I}_k\right) \left(\because \text{By linearity of expectation} \right)\\
& = \sum_{k=1}^{\binom{n}2} \mathbb{P}\left(\mathbb{I}_k=1\right) = \sum_{k=1}^{\binom{n}2} \mathbb{P}\left(\text{$k^{th}$ edge is present}\right)\\
& = \sum_{k=1}^{\binom{n}2} \dfrac12 \left(\text{Assuming that it is equally likely for an edge to be present or absent} \right)\\
& = \dfrac12 \dbinom{n}2
\end{align}</p>
|
2,694,525 | <p>I came across this exercise</p>
<p>$f(x,y)= \lim_{y\to\infty}{{1-y\sin{\pi x\over y}}\over \arctan x}$</p>
<p>The result I get is ${1-\pi x \over \arctan x}$, which depends on the value of $x$.</p>
<p>However, the question I have is that whatever $x$ is, since it's in the $\sin()$, which is a bounded function, shouldn't lay any effect on the limit. For example: $\lim_{x\to\infty}{1 \over x}\sin ax = 0, a\in(-\infty,+\infty)$, whose limit doesn't depend on $a$. </p>
<p>Is there any intuitive understanding for this ?</p>
| Community | -1 | <p>You appear to be under the delusion that the correct way to apply the change of variable $y=\frac1x$ to $\lim_{y\to \infty} y\sin\frac{a}{y}$ is $$\lim_{x\to\infty}\frac1x\sin(ax)$$ while it should rather be $$\lim_{x\to 0}\frac1x\sin(ax)$$ (if we wanted to be pedantic, it should actually be $x\to 0^+$, but whatever).</p>
|
1,641,922 | <p>I've came accros this excersize:<br>
Suppose that $D=\{z:|z| \le 1\}\subset \mathbb C$ and $$f:D\rightarrow\mathbb C$$
suppose that for every $z\in D$ such that $|z|<1$ $$|f(z)-\bar z|<0.9$$ where $\bar z$ is the complex conjugate of $z$. Prove that $f$ cannot be analytic in $D$.<br>
I started with assuming that $f$ is indeed analytic in order to get a contradiction. My first attempt was to try and get some similarities between $f$ and $g(z)=\bar z$ since they are relatively close to each other, and $g$ is not analytic. This idea quickly failed. Also I tried to integrate $f$ around $D$, or see if it possible that $f$ satisfies the Cauchy–Riemann equations, which also did not get me any further.</p>
| DeepSea | 101,504 | <p><strong>hint</strong>: Let $C,J$ be Kathy's and Jason's ages as of now. Thus: $C = 2J , C-6 = 5(J-6)$. Can you finish it off?</p>
|
1,266,210 | <p>Hello everybody my query is regarding the number of positive integral solution.</p>
<blockquote>
<p>In the sport of cricket, find the number of ways in which a batsman can score $14$ runs in $6$ balls not scoring more than $4$ runs in any ball.</p>
</blockquote>
| Brian M. Scott | 12,042 | <p>If they’re the $6$ balls of a single over, and we’re talking about one batsman, either he scores nothing but $0,2$, and $4$, or he scores $3$ twice and $4$ twice, with the other batsman taking strike twice and scoring an odd number of runs each time. I suspect, though, that you’re intended to assume that the same batsman receives all $6$ balls, and that that $6$ balls are not necessarily consecutive.</p>
<p>You can solve it as a <a href="http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29" rel="noreferrer">stars and bars</a> problem, though you’ll have to use an <a href="http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="noreferrer">inclusion-exclusion argument</a> to take into account the limitation that he never scores better than a $4$. Specifically, if $x_k$ is his score on the $k$-th ball, you want the number of solutions of</p>
<p>$$x_1+x_2+x_3+x_4+x_5+x_6=14\tag{1}$$</p>
<p>in non-negative integers, subject to the condition that each $x_k\le 4$. Ignoring that last restriction, the stars and bars computation (which is explained reasonably well at the link in case you’ve not seen it before) gives a total of </p>
<p>$$\binom{14+6-1}{6-1}=\binom{19}5\tag{2}$$</p>
<p>solutions in non-negative integers. However, some of these solutions aren’t wanted, because one or more of the scores is over $4$. How many have $x_1\ge 5$? If we replace $x_1$ by $y_1=x_1-5$, every solution in non-negative integers to</p>
<p>$$y_1+x_2+x_3+x_4+x_5+x_6=9\tag{3}$$</p>
<p>corresponds to a solution of $(1)$ in non-negative integers with $x_1\ge 5$. The stars and bars calculation yields a total of </p>
<p>$$\binom{9+6-1}{6-1}=\binom{14}5$$</p>
<p>solutions in non-negative integers to $(3)$, so we should subtract them from $(1)$. Moreover, the same thing can happen with each of the other five scores, so we should reduce $(2)$ to</p>
<p>$$\binom{19}5-6\binom{14}5\;.\tag{4}$$</p>
<p>Unfortunately, any solution to $(1)$ that had <em>two</em> scores over $4$ has been removed twice in $(4)$ and should be added back in once.</p>
<p>Suppose that $x_1$ and $x_2$ are both over $4$; then we can replace $x_1$ by $y_1=x_1-5$ and $x_2$ by $y_2=x_2-5$ and count the non-negative solutions to</p>
<p>$$y_1+y_2+x_3+x_4+x_5+x_6=4\;.$$</p>
<p>The same calculation that we’ve already made twice tells us that there are</p>
<p>$$\binom{4+6-1}{6-1}=\binom95$$</p>
<p>of them. There are $\binom62$ pairs of scores, so we have to add $\binom95$ back in $\binom62$ times, getting</p>
<p>$$\binom{19}5-6\binom{14}5+\binom62\binom95\;.\tag{5}$$</p>
<p>And this is as far as we need to go, since it’s impossible to have more than two scores over $4$ and a total score of $14$: $(5)$ is the desired number.</p>
|
332,603 | <p>I've passed by this article:
<a href="http://gauravtiwari.org/2011/12/11/claim-for-a-prime-number-formula/" rel="noreferrer">http://gauravtiwari.org/2011/12/11/claim-for-a-prime-number-formula/</a></p>
<p>and this paper:
<a href="http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/kaddouraAMS73-76-2012.pdf" rel="noreferrer">http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/kaddouraAMS73-76-2012.pdf</a></p>
<p>They say that there is a formula such that when you give it (n) then it returns the n-th prime number. Where other articles states that no formula discovered so far that does such thing.</p>
<p>If the formula exists indeed, then why from time to time they discover a new largest prime number known ever. It would be very simple using the formula to find a larger one.</p>
<p>I just want to ensure whether such formula exists or not.</p>
| Inceptio | 63,477 | <p>If formulae for computing primes existed, there wouldn't be a a thing called 'Largest known primes' . And moreover, there are few primes like Mersenne primes and Fermat Prime. But eventually, their converse isn't true <strong>ALWAYS</strong>.</p>
<p>For eg: Mersenne Prime.</p>
<p>$q$ is $prime$ which is equal to $2^p-1$, this shows $p$ is a prime. But the converse, take a prime $a$, it is always not true that $2^a-1$ is a prime. </p>
<p>Similarily for the Fermat primes. So, there are no thing called 'Formula' to find primes.</p>
|
332,603 | <p>I've passed by this article:
<a href="http://gauravtiwari.org/2011/12/11/claim-for-a-prime-number-formula/" rel="noreferrer">http://gauravtiwari.org/2011/12/11/claim-for-a-prime-number-formula/</a></p>
<p>and this paper:
<a href="http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/kaddouraAMS73-76-2012.pdf" rel="noreferrer">http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/kaddouraAMS73-76-2012.pdf</a></p>
<p>They say that there is a formula such that when you give it (n) then it returns the n-th prime number. Where other articles states that no formula discovered so far that does such thing.</p>
<p>If the formula exists indeed, then why from time to time they discover a new largest prime number known ever. It would be very simple using the formula to find a larger one.</p>
<p>I just want to ensure whether such formula exists or not.</p>
| Barry K | 212,888 | <p>Well there have been a # of functions developed all of which-I believe-have started at 1 & gone up. The problem is: they are sloooooow! If it takes l000 steps to obtain all primes up to #7 ... well! The thing about them though is they show weird relationships to other things: One uses pentagonal #s; one only pi & e; & I found a relationship between primes & the Fibonacci #s.So they are fascinating to mathematicians.</p>
<p>Now, I just took a look at Kaddoura's formula in the website you gave. I'm impressed that he appears to be giving exactly what you are asking for. You can now see what the key Q is: how fast is it, assuming it works. But he gives us a way to check: the Mathematica info to run it. Most people do not have it; so connect up with a local University & befriend a grad student!</p>
<p>We might at least imagine K's formula to be faster than E's sieve, but we can ck it. Possibly it's faster for a lower n, but not for higher; but who knows?</p>
<p>Another big Q for you is what range of #s you are interested in. Oh, say you are around 10^6=1,000,000 & your function-with a certain computer-takes 3 secs to solve f(n). Then we estimate it will take 1,000,000 times as long to find f(n) for 10^12. Modern problems related to cryptography were in the range of 10^150 ten yrs ago(for the product of 2 primes) & have steadily increased. To give you an idea, there are 10^80th protons in the universe. And 10^150 is 10^70 times larger! </p>
<p>Note that the commentator, on the other site mentioned, claimed it a threat wrt cryptography. Totally not, unless it can deal with very, very large #s. </p>
<p>OK! lets' just suppose K's formula gives f(n) quickly for any #, no matter how large. In that case it could bust certain old ones because of their structure. RSA, of course, would immediately change such in defense. It would probably also necessitate an increase in n. And that would increase times for coding & decoding. Offhand I would not expect a serious increase.</p>
<p>Hamzeh, I hope this has helped.</p>
|
333,360 | <p>I know the series for $\cos(x)$ it is $\sum \limits_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$ </p>
<p>which will result in $\sum \limits_{n=0}^\infty \dfrac{\left(-1\right)^ n x^{2n+1}}{\left(2n\right)!}$ </p>
<p>Which is great when you already know the series; however, my question is how does one find Maclaurin series when you don't already know the series? </p>
| Henrik Finsberg | 67,299 | <p>If you know that your'e function satisfies a differential equation you can use Picard iteration to find the McLaurin series. If your functions satisfy $y' = f(t,y)$, you can use the formula $ y(x) = y(0) + \int_0^x f(t,y(t)) dt$ to calculate the series.</p>
|
1,803,205 | <p>I am trying to find $\int_0^{\infty} \frac{dx}{1 + x^n}$ using contour integration. I did the computation by taking the contour $[0,R] \cup \gamma_R \cup [R e^{2i\pi/n}, 0]$, with $\gamma_R$ the arc joining $R$ to $Re^{2i\pi/n}$, and found that (for $n \ge 2$):</p>
<p>$$\int_0^{\infty} \frac{dx}{1 + x^n} = \frac{2\pi i}{(1 - e^{2i\pi/n})\prod_{k=0, k \neq 1}^{n-1} (e^{i\pi/n} - e^{i(2k-1)\pi/n})}$$</p>
<p>Apparently, this should be equal to:</p>
<p>$$\frac{\pi/n}{\sin(\pi/n)}$$</p>
<p>How to get that?</p>
| reuns | 276,986 | <p>$$\int_{[0,Re^{2i \pi /n}]} \frac{1}{1+z^n} dz = \int_0^R \frac{1}{1+(e^{2i \pi /n}x)^n} d(e^{2i \pi /n}x) = e^{2i \pi /n} \int_0^R \frac{1}{1+x^n} dx$$ the poles of $\frac{1}{1+z^n}$ are at $e^{i\pi (2k+1)/ n}, k = 0 \ldots n-1$. the only one inside the whole contour $\Gamma_R : [0,R] \cup [R, R e^{2 i \pi /n}] \cup [R e^{2 i \pi / n}, 0]$ is the one at $e^{i \pi /n }$. </p>
<p>and $\displaystyle\frac{1}{1+z^n} = \frac{h(z)}{z-e^{i \pi / n}}$ with $ \displaystyle h(z) = \frac{-1}{\prod_{k=1}^{n-1} (z-e^{i \pi (2 k +1)/n})}$ so that $$Res(\frac{1}{1+z^n} , e^{i \pi /n}) = h(e^{i \pi / n}) = \frac{-1}{\prod_{k=1}^{n-1} (e^{i\pi /n}-e^{i \pi (2 k +1)/n})}$$ </p>
<p>hence $$\lim_{R \to \infty}\int_{\Gamma_R} \frac{1}{1+z^n}dz = 2 i \pi \frac{-1}{\prod_{k=1}^{n-1} (e^{i\pi /n}-e^{i \pi (2 k +1)/n})} = (1-e^{2i \pi /n}) \int_0^\infty \frac{1}{1+x^n} dx$$ and $$\int_0^\infty \frac{1}{1+x^n} dx = \frac{-2 i \pi }{(1-e^{2i \pi /n})\prod_{k=1}^{n-1} (e^{i\pi /n}-e^{i \pi (2 k +1)/n})}$$</p>
<p>note that another way to compute the residue $Res(\frac{1}{g(z)},a)$ when $g(z)$ has a simple zero at $a$ is $$Res(\frac{1}{g(z)},a) = \frac{1}{g'(a)}$$</p>
<p>hence with $g(z) = 1+z^n, g'(z) = n z^{n-1}$ $$Res(\frac{1}{1+z^n} , e^{i \pi /n}) = \frac{1}{g'(e^{i \pi /n})} = \frac{1}{n(e^{i \pi (n-1)/n})} = \frac{-e^{i \pi /n }}{n}$$
so that </p>
<p>$$\frac{-2 i \pi }{(1-e^{2i \pi /n})\prod_{k=1}^{n-1} (e^{i\pi /n}-e^{i \pi (2 k +1)/n})} = \frac{-2 i \pi e^{i \pi /n }}{n(1-e^{2i \pi /n})}$$</p>
<p>and your assertion follows (up to errors of sign, probably on my side)</p>
|
3,366,158 | <blockquote>
<p>find a linear transformation such that <span class="math-container">$M^2 (1- M) =0$</span> but <span class="math-container">$M$</span> is not idempotent ?</p>
</blockquote>
<p>My attempt : i take take the vector space generated by the base <span class="math-container">$\{e_1, e_2\}$</span> and define <span class="math-container">$T(e_1)=e_2$</span>,and <span class="math-container">$T(e_2)=e_1 $</span></p>
| pancini | 252,495 | <p>We know <span class="math-container">$M$</span> is idempotent if and only if <span class="math-container">$M^2=M$</span>, and the assumption is that <span class="math-container">$M^3=M^2$</span>. So is it possible that <span class="math-container">$M^2\neq M$</span> but <span class="math-container">$M^3=M^2$</span>? Clearly not if <span class="math-container">$M$</span> is invertible, because <span class="math-container">$M^3=M^2$</span> would imply <span class="math-container">$M^2=M$</span>.</p>
<p>Now we know that <span class="math-container">$x^2$</span> divides the minimal polynomial of <span class="math-container">$M$</span>, and thus we necessarily have a nontrivial Jordan block with eigenvalue <span class="math-container">$0$</span>. For example, consider</p>
<p><span class="math-container">$$M=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$</span></p>
|
1,257,900 | <p>This is a question about cyclotomic polynomials and I have already shown that $x^n-1 =\Pi\Phi_d(x)$, taking the product over all divisors d of n. </p>
| quid | 85,306 | <p>Each of $\zeta^j$ for $j=0, \dots, n-1$ is a root of $X^n -1$. Thus $(X-\zeta^j)$ must divide it. Since all the $\zeta^j$ are distinct, the linear factors are co-prime and thus the product divides $X^n - 1$ too. </p>
<p>As the degree of both is $n$ and they are both normed, equality follows.</p>
|
1,257,900 | <p>This is a question about cyclotomic polynomials and I have already shown that $x^n-1 =\Pi\Phi_d(x)$, taking the product over all divisors d of n. </p>
| Timbuc | 118,527 | <p>Hints:</p>
<p>1) Show that for $\;1\le j,\,k\le n\;,\;\;\zeta^j=\zeta^k\iff j=k\;$ </p>
<p>2) Show that $\;\left(\zeta^k\right)^n=1\;,\;\;\forall\,k=1,2,...,n\;$</p>
<p>3) Show that all the roots of $\;x^n-1\;$ are different</p>
<p>There you go...</p>
|
779,042 | <p>Any pointers on how should I start?</p>
<p>$$I:=\int_ {0}^{\infty} \frac {\cos(ax)} {(x^2 + b^2)^n} \ \mathrm{d}x$$</p>
| Bennett Gardiner | 78,722 | <p>I will leave some details to the reader. W.L.o.G. assume $a,b>0$. Consider
$$
J(a,b) = \int_0^{\infty} \frac{\cos(ax)}{x^2+b^2} \ \mathrm{d}x,
$$
then
$$
\frac{\partial^{n-1} J}{\partial b^{n-1}} = (-2b)^{n-1}(n-1)!\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x.
$$
The integral $J$ can be calculated via the <a href="http://en.wikipedia.org/wiki/Residue_theorem#Example" rel="nofollow">Residue Theorem</a> as
$$
J = \frac{\pi}{2b}\mathrm{e}^{-ab},
$$
so we have
$$
\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x = \frac{(-2b)^{1-n}}{(n-1)!}\frac{\partial^{n-1} }{\partial b^{n-1}}\left(\frac{\pi}{2b}\mathrm{e}^{-ab}\right),
$$
where
$$
\frac{\partial^{n-1} }{\partial b^{n-1}}\left(\frac{\mathrm{e}^{-ab}}{b}\right) = (-1)^{n-1}\frac{\mathrm{e}^{-ab}}{b^{n}}\displaystyle \sum_{i=0}^{n-1}\frac{(n-1)!}{(n-i-1)!} (ab)^{n-i-1},
$$
so the final result is, with some cancellation -
$$
\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x = \frac{\pi\mathrm{e}^{-ab}}{2^{n}b^{2n-1}}\displaystyle \sum_{k=1}^{n}\frac{(ab)^{n-k}}{(n-k)!} , \qquad n \ge 1.
$$</p>
|
399,948 | <p>How do you in general find the trigonometric function values? I know how to find them for 30 45, and 60 using the 60-60-60 and 45-45-90 triangle but don't know for, say $\sin(15)$ or $\tan(75)$ or $\csc(50)$, etc.. I tried looking for how to do it but neither my textbook or any other place has a tutorial for it. I want to know how to find the exact values for all the trigonometric functions like $\sin x$, $\csc x$, ... opposed to looking it up or using calculator. According to my textbook, $\sin(15)=0.26$, $\tan(75)=3.73$, and $\csc(50)=1.31$ but doesn't show where those numbers came from, as if it was dropped from the Math heaven!</p>
| Peter Košinár | 77,812 | <p>For some angles (such as 15 or 75 degrees), you can apply the formulas for <a href="http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae" rel="nofollow">doubling and halving the angles</a> and those for <a href="http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities" rel="nofollow">sums and differences of angles</a>. For example, if you know that $\sin x=0.4$, these formulas allow you to calculate $\sin 2x$ or $\sin 17x$ or $\sin \frac{11}{32}x$, ...</p>
<p>Unfortunately, such formulas are not powerful enough to calculate the values for arbitrary angle (e.g. $\sin 43$). For those, one needs to employ the real-analysis definition of $\sin x$ using the Taylor series which might be beyond the scope of what you have learned so far.</p>
|
2,718,495 | <p>$$\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n]),$$</p>
<p>where $\xi_{\big| \Bbb{N}}$ is the counting measure on $\Bbb{N}$.</p>
<p>I am looking for $A \subset \Bbb{N}$ for which $\lim_{n \to \infty}\frac{1}{n} \xi_{\big| \Bbb{N}} (A \cap[1,n])$ is not defined. So I need to find $A$ so that the limit above does not converge but I don't come to any solution.</p>
| GNUSupporter 8964民主女神 地下教會 | 290,189 | <p>Let <a href="https://en.wikipedia.org/wiki/Natural_density#Properties_and_examples" rel="nofollow noreferrer">$A=\bigcup\limits_{k=0}^\infty \{2^{2k},\ldots,2^{2k+1}-1\}$</a> (be a set of integers whose binary expansion contains an odd number of digits).</p>
<p>\begin{align}
n &= 2^{2m+1} - 1,
\frac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n])
= \frac{1+2^2+\cdots +2^{2m}}{2^{2m+1}-1}
= \frac{2^{2m+2}-1}{3(2^{2m+1}-1)}
\xrightarrow[m \to \infty]{} \frac 23; \\
n &= 2^{2m+2} - 1,
\frac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n])
= \frac{1+2^2+\cdots +2^{2m}}{2^{2m+2}-1}
= \frac{2^{2m+2}-1}{3(2^{2m+2}-1)}
\xrightarrow[m \to \infty]{} \frac 13.
\end{align}</p>
<p>Therefore, the limit of $\dfrac{1}{n} \xi_{\big| \mathbb{N}} (A \cap[1,n])$ when $m \to \infty$ doesn't exist.</p>
|
2,543,123 | <p>Let $\gcd(a, 11) = 1$. If $3a^7 \equiv 5 \pmod{11}$, show that $a \equiv 3 \pmod{11}$.</p>
<p>My first approach was to use Euler's theorem:</p>
<p>$a^{10} \equiv 1 \pmod{11}$</p>
<p>$3a^7 \equiv 5 \pmod{11}$ implies that $a^{-3} \equiv 9 \pmod{11}$ </p>
<p>I feel i'm not on the right track, hints are appreciated.</p>
| JayTuma | 506,755 | <p>If you sum the number you get then the answer is no, since</p>
<ol>
<li>there is no way you can achieve $1$ (the least possible value is $2$)</li>
<li>$P(X+Y = 2) = \frac{1}{36}$ while $P(X+Y = 7) = \frac{1}{6}$ as you can easily check, where $X$ is a random variable representing the result of one dice roll</li>
</ol>
|
1,368,455 | <p>"I take a journey and, due to heavy traffic, crawl along the first half of the complete distance of my journey at an average speed of $10$ mph. How fast would I have to travel over the second half of the journey to bring my average speed to $20$ mph?"</p>
<p>At work, this has been a topic of a long debate.<br>
Proposed answers included:
{infinity, $0, 15, 30, 45$, speed of light~}.</p>
<p>What do you think? </p>
| Ángel Mario Gallegos | 67,622 | <p>Let $t$ hours the time spent in the first half of the distance, namely $d$ miles, it is known that $$d/2=10t$$
The time needed in order to travel the remaining half of distance, with a speed $v$, is $t_2=(d/2)/v$ and must satisfy
\begin{align*}
\frac{d}{\frac{d}{20}+\frac{d}{2v}}&=20\\
\frac{20vd}{vd+10d}&=20\\
\frac{v}{v+10}&=1
\end{align*}</p>
|
1,220,800 | <blockquote>
<p>Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $</p>
</blockquote>
<p>$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get</p>
<p>$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.</p>
<p>$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$</p>
<p>But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)</p>
<p>So we get no solution.</p>
<p>My Question is : Can we solve above question by using comparision of expressions?</p>
<p>Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$ </p>
<p>If that way possible, please help me solve it. Thanks.</p>
| Hagen von Eitzen | 39,174 | <p>For $x\ge1$ we have $$\sqrt{4x-1}\ge \sqrt {3x} $$
and $$\sqrt{x+1}\le \sqrt {2x}$$
hence
$$\sqrt{x+1}-\sqrt{x-1}\le \sqrt{2x}<\sqrt{3x}\le\sqrt{4x-1} $$</p>
|
3,170,871 | <p>Could anyone please give me a hint on how to compute the following integral?</p>
<p><span class="math-container">$$\int \sqrt{\frac{x-2}{x^7}} \, \mathrm d x$$</span></p>
<p>I'm not required to use hyperbolic/ inverse trigonometric functions.</p>
| logo | 587,007 | <p>Try the substitution
<span class="math-container">$$
u=\frac{x-2}{x}
$$</span> </p>
<p>or equivalently</p>
<p><span class="math-container">$$
x=\frac{2}{1-u}
$$</span></p>
|
1,507,190 | <p>Prove that if the vector $v$ belongs both to the nullspace and the row space of a matrix $A$, then $v = 0$.</p>
<p>I did this solution but I don't know if it's right.</p>
<p>$V=A^Tb$ and $Av=0\Rightarrow AV=AA^Tv\Rightarrow A^2b=0\Rightarrow b=0 \Rightarrow v=0$</p>
| Michael Biro | 29,356 | <p>Hint: If $A\mathbf{v} = \mathbf{0}$ and $\mathbf{x}^TA = \mathbf{v}^T$, then what is $\mathbf{v} \cdot \mathbf{v}$?</p>
|
1,507,190 | <p>Prove that if the vector $v$ belongs both to the nullspace and the row space of a matrix $A$, then $v = 0$.</p>
<p>I did this solution but I don't know if it's right.</p>
<p>$V=A^Tb$ and $Av=0\Rightarrow AV=AA^Tv\Rightarrow A^2b=0\Rightarrow b=0 \Rightarrow v=0$</p>
| Ned | 67,710 | <p>Since $v$ is in Nul A, $<v,r> = 0$ for each row $r$ of A. So for any linear combination $w$ of the rows, $<w,v>=0$. But $v$, being in Row A, is itself such a linear combination $w$. So $<v,v>=0$.</p>
|
71,952 | <p><strong>Background</strong></p>
<p>Take 2 convex sets in $\mathbb{R}^2$, or 3 convex sets in $\mathbb{R}^3$, or generally, $n$ convex sets in $\mathbb{R}^n$. "Mixed volume" assigns to such a family $A_1, \ldots, A_n$ a real number $V(A_1, \ldots, A_n)$, measured in $\mathrm{metres}^n$. </p>
<p>As I understand it, mixed volume is a kind of cousin of the determinant. I'll give the definition in a moment, but first here are some examples. </p>
<ol>
<li><p>$V(A, \ldots, A) = \mathrm{Vol}(A)$, for any convex set $A$.</p></li>
<li><p>More generally, suppose that $A_1, \ldots, A_n$ are all scalings of a single convex set (so that $A = r_i B$ for some convex $B$ and $r_i \geq 0$). Then $V(A_1, \ldots, A_n)$ is the geometric mean of $\mathrm{Vol}(A_1), \ldots, \mathrm{Vol}(A_n)$.</p></li>
<li><p>The previous examples don't show how mixed volume typically depends on the interplay between the sets. So, taking $n = 2$, let $A_1$ be an $a \times b$ rectangle and $A_2$ a $c \times d$ rectangle in $\mathbb{R}^2$, with their edges parallel to the coordinate axes. Then
$$
V(A_1, A_2) = \frac{1}{2}(ad + bc).
$$
(Compare and contrast the determinant formula $ad - bc$.)</p></li>
<li><p>More generally, take axis-parallel parallelepipeds $A_1, \ldots, A_n$ in $\mathbb{R}^n$. Write $m_{i1}, \ldots, m_{in}$ for the edge-lengths of $A_i$. Then
$$
V(A_1, \ldots, A_n) = \frac{1}{n!} \sum_{\sigma \in S_n} m_{1, \sigma(1)} \cdots m_{n, \sigma(n)}.
$$
(Again, compare and contrast the determinant formula.)</p></li>
</ol>
<p>The definition of mixed volume depends on a theorem of Minkowski: for any compact convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$, the function
$$
(\lambda_1, \ldots, \lambda_m) \mapsto \mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m)
$$
(where $\lambda_i \geq 0$ and $+$ means Minkowski sum) is a polynomial, homogeneous of degree $n$. For $m = n$, the mixed volume $V(A_1, \ldots, A_n)$ is defined as the coefficient of $\lambda_1 \lambda_2 \cdots \lambda_n$ in this polynomial, divided by $n!$.</p>
<p>Why pick out this particular coefficient? Because it turns out to tell you everything, in the following sense: for any convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$,
$$
\mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) = \sum_{i_1, \ldots, i_n = 1}^m V(A_{i_1}, \ldots, A_{i_n}) \lambda_{i_1} \cdots \lambda_{i_n}.
$$</p>
<p><strong>Properties of mixed volume</strong></p>
<p>Formally, let $\mathscr{K}_n$ be the set of nonempty compact convex subsets of $\mathbb{R}^n$. Then mixed volume is a function
$$
V: (\mathscr{K}_n)^n \to [0, \infty),
$$
and has the following properties:</p>
<ol>
<li><p><em>Volume:</em> $V(A, \ldots, A) = \mathrm{Vol}(A)$. (Here and below, the letters $A$, $A_i$ etc. will be understood to range over $\mathscr{K}_n$, and $\lambda$, $\lambda_i$ etc. will be nonnegative reals.)</p></li>
<li><p><em>Symmetry:</em> $V$ is symmetric in its arguments.</p></li>
<li><p><em>Multilinearity:</em>
$$
V(\lambda A_1 + \lambda' A'_1, A_2, \ldots, A_n) = \lambda V(A_1, A_2, \ldots, A_n) + \lambda' V(A'_1, A_2, \ldots, A_n).
$$
(These first three properties closely resemble a standard characterization of determinants.) </p></li>
<li><p><em>Continuity:</em> $V$ is continuous with respect to the Hausdorff metric on $\mathscr{K}_n$.</p></li>
<li><p><em>Invariance:</em> $V(gA_1, \ldots, gA_n) = V(A_1, \ldots, A_n)$ for any isometry $g$ of Euclidean space $\mathbb{R}^n$ onto itself.</p></li>
<li><p><em>Multivaluation:</em>
$$
V(A_1 \cup A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots) + V(A'_1, A_2, \ldots) - V(A_1 \cap A'_1, A_2, \ldots)
$$
whenever $A_1, A'_1, A_1 \cup A'_1 \in \mathscr{K}_n$.</p></li>
<li><p><em>Monotonicity:</em> $V(A_1, A_2, \ldots, A_n) \leq V(A'_1, A_2, \ldots, A_n)$ whenever $A_1 \subseteq A'_1$.</p></li>
</ol>
<p>There are other basic properties, but I'll stop there.</p>
<p><strong>Questions</strong></p>
<p>Is $V$ the unique function $(\mathscr{K}_n)^n \to [0, \infty)$ satisfying properties 1--7?</p>
<p>If so, does some subset of these properties suffice? In particular, do properties 1--3 suffice? </p>
<p>If not, is there a similar characterization involving different properties?</p>
<p>(Partway through writing this question, I found a recent paper of Vitali Milman and Rolf Schneider: <a href="http://home.mathematik.uni-freiburg.de/rschnei/CharMixVol.rev.pdf">Characterizing the mixed volume</a>. I don't think it answers my question, though it does give me the impression that the answer might be unknown.)</p>
| Tom Leinster | 586 | <p>Sorry to answer my own question, but asking this in public seems to have spurred me into thought.</p>
<p>As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying</p>
<ol>
<li><p>$V(A, \ldots, A) = Vol(A)$</p></li>
<li><p>$V$ is symmetric</p></li>
<li><p>$V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$.</p></li>
</ol>
<p>In other words, we don't need multilinearity, just multiadditivity. </p>
<p>The proof is along the lines suggested by auniket. </p>
<p>Fix $n$ and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write $\mathbf{n} = \{1, \ldots, n\}$, and for sets $R$ and $S$, write $\mathrm{Surj}(R, S)$ for the set of surjections $R \to S$.</p>
<p>I claim that for all subsets $S$ of $\mathbf{n}$,
$$
\sum_{f \in \mathrm{Surj}(\mathbf{n}, S)} V(A_{f(1)}, \ldots, A_{f(n)})
$$
is uniquely determined by the properties above. The proof will be by induction on the cardinality of $S$. When $S = \mathbf{n}$, this sum is
$$
n! V(A_1, \ldots, A_n),
$$
so this claim will imply the characterization theorem.</p>
<p>To prove the claim, take $S \subseteq \mathbf{n}$. Then
$$
Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)})
$$
by the three properties. This in turn is equal to
$$
\sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R)} V(A_{f(1)}, \ldots, A_{f(n)}).
$$
By the inductive assumption, all but one of the summands in the first summation - namely, $R = S$ - is uniquely determined. Hence the $S$-summand is uniquely determined too. This completes the induction, and so completes the proof.</p>
<p>The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the $A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book <em>Convex Bodies: The Brunn-Minkowski Theory</em>). </p>
<p>This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1. </p>
|
3,377,031 | <p>I am stuck on a homework question and I don't know how to approach this. Please help! The question is as follows:</p>
<p>"Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be two events. </p>
<p><span class="math-container">$P(A)= 0.3$</span><br>
<span class="math-container">$P(B)= 0.4$</span><br>
<span class="math-container">$P(A\cap B)= 0.2$</span> </p>
<p>Denote the complement of <span class="math-container">$A$</span> by <span class="math-container">$A'$</span>. Then <span class="math-container">$P(A'\cap(A\cup B))$</span> is closest to?</p>
<p>a.) 0.1 b.)0.2 c.) 0.3 d.) 0.4</p>
<p>(The correct answer is 0.2, but I don't get how they got it)</p>
| Community | -1 | <p>Hint: <span class="math-container">$P(A'\cap(A\cup B))$</span> represents the probability that <span class="math-container">$A$</span> doesn't happen and also that either <span class="math-container">$A$</span> or <span class="math-container">$B$</span> happens. How can you simplify that statement?</p>
|
3,377,031 | <p>I am stuck on a homework question and I don't know how to approach this. Please help! The question is as follows:</p>
<p>"Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be two events. </p>
<p><span class="math-container">$P(A)= 0.3$</span><br>
<span class="math-container">$P(B)= 0.4$</span><br>
<span class="math-container">$P(A\cap B)= 0.2$</span> </p>
<p>Denote the complement of <span class="math-container">$A$</span> by <span class="math-container">$A'$</span>. Then <span class="math-container">$P(A'\cap(A\cup B))$</span> is closest to?</p>
<p>a.) 0.1 b.)0.2 c.) 0.3 d.) 0.4</p>
<p>(The correct answer is 0.2, but I don't get how they got it)</p>
| Bram28 | 256,001 | <p>Just as in logic you have that <span class="math-container">$A'(A+B)=A'B$</span>, in set theory you have that <span class="math-container">$A'\cap(A\cup B)=A'\cap B$</span></p>
<p>And I assume you can find <span class="math-container">$P(A'\cap B)$</span> yourself</p>
|
1,083,841 | <p>I have extracted the below passage from the wikipedia webpage - Point (geometry): </p>
<blockquote>
<p>In particular, the geometric points do not have any length, area, volume, or any other dimensional attribute. </p>
</blockquote>
<p>I think the above passage imply\ies that the point is zero dimensional. If it is zero dimensional, how can it form a one dimensional line? </p>
<p>Physics texts sometimes talk of lines' being made up of points, planes' being made up of lines and so forth. Clearly a line segment, thought of as a connected interval of the real numbers, cannot be built as a countable union of points. What axiom systems define the building up of a line from points, or, how do we rigorously define the building of a line from points? </p>
<hr>
<p><em>Links:</em></p>
<ol>
<li>The section one (<a href="https://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch01.htm" rel="noreferrer">Physical meaning of geometrical propositions</a>) of part one of the book "Relativity: The Special and General Theory" seems to be giving Einsteins view on this matter. </li>
<li><a href="https://hsm.stackexchange.com/questions/8400/what-was-the-intended-utility-of-euclids-definitions-of-lines-and-points">What was the intended utility of Euclid's definitions of lines and points?</a></li>
</ol>
<p>Related: History of Euclidean and Non-Euclidean Geometry</p>
| James Herndon | 811,806 | <p>The simplest answer I believe is that a line is not "made up" of points, in the sense that a chain is made of links. Rather, points are best thought of as positions, and positions have no size. It is possible to define a set of possible positions in such a way that one can derive from this set all the properties that a line would have, and thus the two are logically equivalent: we might as well say that this set of points is a line. But the construction of this set is a logical procedure and not a physical process; there is no starting with one point and then adding "the next one", because of course there is no next point: between any two points on line lie an infinity of other points.
By the way, the idea of a unique physical position is somewhat problematic in physics. The uncertainty principle and the problem of "singularities" come into play here - another topic.</p>
|
96,110 | <p><span class="math-container">$A = \begin{pmatrix}
0 & 1 &1 \\
1 & 0 &1 \\
1& 1 &0
\end{pmatrix} $</span></p>
<p>The matrix <span class="math-container">$(A+I)$</span> has rank <span class="math-container">$1$</span> , so <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of at least <span class="math-container">$2$</span> .</p>
<p>I was reviewing my notes and I don't understand how the first statement implies the second one. </p>
<p>Can anyone please explain how rank 1 of <span class="math-container">$(A + I)$</span> implies <span class="math-container">$-1$</span> is an eigenvalue with an algebraic multiplicity of <span class="math-container">$2$</span>?</p>
<p>Thank you in advance.</p>
| Student | 19,118 | <p>An other way to see that is the following (maybe less clear, but I try anyway) : $A+I$ is a $3\times 3$ matrix with rank one, and thus has for eigenvalues $0$ with multiplicity two and an other one which is given by ${\rm Tr}(A+I)$ [is that clear for you ?]. </p>
<p>Then, since $A+I$ and $I$ trivially commutes, they have common basis for diagonalization, so that the eigenvalues of $A$ are the one of $A+I$ minus $I$, namely $-1$ with multiplicity two, and ${\rm Tr}A$. </p>
|
3,277,171 | <p>I am a ninth grader and I would like to learn mathematics on my own. I have already learned algebra, geometry, trigonometry, some precalculus, number theory and tried to understand some calculus. Apart from those I learnt a bit from other areas of mathematics but not enough to be worth mentioning. </p>
<p>I have learned several things from books but those couldn't answer all questions so I had to turn to the internet (sometimes I can't be sure if something is correct or not). Besides I like to have proof for everything which is not always given in books. (I want understand it properly not just use a procedure).</p>
<p>I have tried using online lectures but the ones I have found according to my level were either going too slowly or didn't have complicated problems. Besides most didn't have the proofs. I use Khan Academy sometimes although it too lacks the difficult problems.</p>
<p>Could you please tell me what I can do to learn further mathematics(eg. More number theory, proof writing, calculus and... maybe analysis altough I suppose I am not prepared well enough for that)? Do you know any books I could read (normal high school /college /university books included just give me the name please) or lectures I could watch? Any other things I could do? I would like a good understanding of the subject as I would like to become a mathematician. </p>
| runway44 | 681,431 | <p>Your classification is missing spirals: images of one-parameter homomorphisms <span class="math-container">$\exp(zt)$</span> when the complex number <span class="math-container">$z$</span> is not purely real or purely imaginary. Or discrete spirals, such as the image of the integers under such a homomorphism, or "dusty" spirals as the image of subgroups of <span class="math-container">$\Bbb Q$</span> under the homomorphism.</p>
<p>The subgroups of <span class="math-container">$\Bbb Q$</span> can be classified by "supernatural" rational numbers <span class="math-container">$n=\prod p^v$</span>, which are formal infinite products of prime powers with exponents <span class="math-container">$v$</span> in <span class="math-container">$\Bbb Z\cup\{+\infty\}$</span>. The subgroup corresponding to <span class="math-container">$n$</span> would be the set of all rationals with denominator dividing <span class="math-container">$n$</span> (with divisibility defined prime by prime, and every <span class="math-container">$p^v$</span> dividing <span class="math-container">$p^{\infty}$</span>).</p>
<p>Note that <span class="math-container">$S^1\cong\Bbb R/\Bbb Z\cong(\Bbb Q^{\frak \oplus c})/\Bbb Z\cong(\Bbb Q/\Bbb Z)\times\Bbb Q^{\oplus\frak c}$</span>, but picking a <span class="math-container">$\Bbb Q$</span>-basis for <span class="math-container">$\Bbb R$</span> seems pretty wild - the axiom of choice guarantees the existence of a basis, but I don't know if anybody's ever constructed an explicit one. (Although apparently the "complexity" of such a basis is <a href="https://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers">not as extreme</a> as it could be made out to be, whatever that means.) And <span class="math-container">$\Bbb R_{>0}\cong\Bbb R\cong\Bbb Q^{\oplus\frak c}$</span> too, so the group <span class="math-container">$\Bbb C^{\times}$</span> is abstractly isomorphic to <span class="math-container">$(\Bbb Q/\Bbb Z)\times \Bbb Q^{\oplus\frak c}$</span>.</p>
<p>Now any subgroup of this direct product will pull back (wrt the projection <span class="math-container">$\Bbb Q^{\oplus\frak c}\to(\Bbb Q/\Bbb Z)\times\Bbb Q^{\oplus\frak c}$</span>) to a <span class="math-container">$\Bbb Z$</span>-submodules (i.e. subgroups) of <span class="math-container">$\Bbb Q^{\oplus\frak c}$</span> containing <span class="math-container">$1$</span> in the "first" factor.</p>
<p>Probably we can go further with this by incorporating a version of Goursat's lemma, which would predict the spiral subgroups and likely classify all subgroups as products of spirals (where you count the circle and rays, as well as their discrete or "dust" versions, as generalized spirals).</p>
|
3,277,171 | <p>I am a ninth grader and I would like to learn mathematics on my own. I have already learned algebra, geometry, trigonometry, some precalculus, number theory and tried to understand some calculus. Apart from those I learnt a bit from other areas of mathematics but not enough to be worth mentioning. </p>
<p>I have learned several things from books but those couldn't answer all questions so I had to turn to the internet (sometimes I can't be sure if something is correct or not). Besides I like to have proof for everything which is not always given in books. (I want understand it properly not just use a procedure).</p>
<p>I have tried using online lectures but the ones I have found according to my level were either going too slowly or didn't have complicated problems. Besides most didn't have the proofs. I use Khan Academy sometimes although it too lacks the difficult problems.</p>
<p>Could you please tell me what I can do to learn further mathematics(eg. More number theory, proof writing, calculus and... maybe analysis altough I suppose I am not prepared well enough for that)? Do you know any books I could read (normal high school /college /university books included just give me the name please) or lectures I could watch? Any other things I could do? I would like a good understanding of the subject as I would like to become a mathematician. </p>
| P Vanchinathan | 28,915 | <p>More groups can be identified by taking a subring of <span class="math-container">$A\subset\mathbf{C}$</span> and then the group <span class="math-container">$A^*$</span> of invertible elements of that ring. For example <span class="math-container">$A$</span> could be the ring of integers of some algebraic number field, and Dirichlet's theorem gives their structure. Possibly <span class="math-container">$A$</span> can be a non-Noetherian ring.</p>
|
2,894,376 | <blockquote>
<p>$2$ different History books, $3$ different Geography books and $2$ different Science books are placed on a book shelf. How many different ways can they be arranged? How many ways can they be arranged if books of the same subject must be placed together?</p>
</blockquote>
<p>For the first part of the question I think the answer is </p>
<p>$$(2+3+2)! = 5040 \text{ different ways}$$</p>
<p>For the second part of the question I think that I will need to multiply the different factorials of each subject. There are $2!$ arrangements for science, $3!$ for geography and $2!$ for history. Am I correct in saying that the number of different ways to place the books on the shelf together by subject would be
$$2! \times 3! \times 2! = 24 \text{ different ways}$$ </p>
| Key Flex | 568,718 | <p>All the books can be arranged in $(2+3+2)!=7!$ ways</p>
<p>There are $3$ branches, three units of books: $\{$History$\}$,$\{$Geography$\}$,$\{$Science$\}$- Arranging branches $=3!$ ways.</p>
<p>Arranging the books within the branches:</p>
<p>History: $2!$</p>
<p>Geography: $3!$</p>
<p>Science:$2!$</p>
<p>Total $=3!(2!\times3!\times2!)=144$ ways</p>
|
2,297,421 | <p>Probably a really simple question, but I am trynig to fit an air bed in a tent.</p>
<p>Circular tent with a diameter of $3$m and a central vertical pole in the middle.</p>
<p>The air bed measures $1.41$ m $\times$ $1.9$ m.</p>
<p>Will the air bed fit fully inside the tent without being obstructed by the central pole?</p>
| Ben G. | 334,171 | <p>I believe the simplest way to solve this problem is to simply count the number of distinct polynomials of that form that factor, and see that it is less than $n$. Thus some polynomials of that form do not factor.</p>
<p>For every odd n, 2 is a unit modulo $n$. Thus it is sufficient to show that some $x^2+x+i$ is irreducible modulo $n$ for $i$ ranging from $0$ to $n-1$.</p>
<p>Suppose that all $n$ such (distinct) polynomials factored into $(x-a_i)(x-b_i)$ where $i$ ranges from $0$ to $n-1$. Then by expanding, we get $x^2+x+i = x^2 + (-a_i-b_i)x + a_ib_i$. In particular, $-a_i-b_i = 1$. Since each $b_i$ is fixed with respect to $a_i$, there can only be $n$ such pairs, corresponding to $(x-a_i)(x+a_i+1)$, where $a_i$ ranges from 1 to $n$. However, the polynomials given by $a_i = 0$ and $a_i = -1$ are both the same polynomial: $(x-0)(x+1)$. In fact, we can group each $m, -m+1$ pair in a like manner to find that the number of distinct polynomials of the form $x^2+x+i$ that factor is $(n+1)/2$. It is then easy to see that the number of distinct polynomials that do not factor is $(n-1)/2$. Since this is less than $n$, there are some polynomials that do not factor.</p>
|
2,297,421 | <p>Probably a really simple question, but I am trynig to fit an air bed in a tent.</p>
<p>Circular tent with a diameter of $3$m and a central vertical pole in the middle.</p>
<p>The air bed measures $1.41$ m $\times$ $1.9$ m.</p>
<p>Will the air bed fit fully inside the tent without being obstructed by the central pole?</p>
| Arpan1729 | 444,208 | <p>Say, no such $k$, exists.</p>
<p>So there exists numbers $x_1,x_2\dots ,x_{n} \in \{0,1,2,3...n\}$, such that</p>
<p>$x_1^2+x_1=0$ <strong>(mod n)</strong></p>
<p>$x_2^2+x_2=2$ <strong>(mod n)</strong></p>
<p>$x_3^2+x_3=4$ <strong>(mod n)</strong>
.</p>
<p>.</p>
<p>.</p>
<p>.
$x_{n}^2+x_{n}=2(n-1)$ <strong>(mod n)</strong> </p>
<p>Claim, $x_1,x_2 \dots ,x_{n}$, are all distinct, else if say $x_i=x_j$, then subtract the corresponding two equation, we will have $n|2(i-j)$, which is not possible as $n$ is odd and $|i-j|<n$, hence $\{x_1,x_2 \dots ,x_{n}\}$ is an enumeration of $\{0,1,2,3,\dots ,n-1\}$</p>
<p>So there exists $p,q \in \{1,2,3,\dots ,n\}$ such that $x_p=0$ and $x_q=n-1$, but $0^2+0=(n-1)^2+(n-1)$ <strong>(mod n)</strong> , hence a condradiction.</p>
|
35,230 | <p>This happens frequently, both on the main site and on meta: An old question pops back up on the front page, I open it, the text under the title says "Modified today", but when I check the timeline the last event is years ago, even if I show vote summaries.</p>
<p>Here's the <a href="https://math.stackexchange.com/questions/924266/definition-of-bilinear-maps">latest one</a> for me. This is happening as of 12:00 noon, EDT.</p>
<p>My guesses are</p>
<ul>
<li>I don't have high enough rep to see the latest event. (I have ~5k.)</li>
<li>The event showing up in the timeline query is lagging behind the update of the "Last Modified" field, or some other database jiggery-pokery.</li>
<li>???</li>
</ul>
<p>Do others see this happening too? Does anyone know why it happens?</p>
<hr />
<p>Adding the key points from MartinSleziak's thorough answer below, for future readers:</p>
<ul>
<li>Questions get bumped by events on answers, which have their own, separate timelines.</li>
<li>Use the "Modified" value, under the title, to take you to the latest activity. It is a link, despite not looking at all like a link.</li>
</ul>
| Martin Sleziak | 8,297 | <p>As already mentioned in the other answer, a question can be bumped for various reasons - in particular, it might be caused by activity on an answer (rather than a question). See the <a href="https://math.meta.stackexchange.com/tags/bumping/info">corresponding tag-info</a> and the FAQ post: <a href="https://meta.stackexchange.com/q/48578">What can cause a question to be bumped?</a></p>
<p>If you click on the timestamp in "modified today", you'll go directly to the <a href="https://meta.stackexchange.com/tags/last-activity/info">last activity</a>. (With some exceptions - for example, if the last activity was on a deleted answer, the users below 10k won't be able to see it.) This link <a href="https://meta.stackexchange.com/q/187026">used to be in the sidebar</a>, but it was later moved <a href="https://meta.stackexchange.com/q/331349">below the question title</a>.</p>
<p>You can find the last activity also if you click on a timestamp when the question is displayed in the list of questions showing the recent activity. This was a result of this feature request: <a href="https://meta.stackexchange.com/q/4391">Add link to most recent activity of a question</a>.</p>
<p>In many places on Stack Exchange sites, the timestamps are clickable. (And quite often the exact time and date is shown when hovering over the timestamp.)</p>
<hr />
<p>Tangentially, since I sometimes want to see recent events for some posts, I made some <a href="https://math.meta.stackexchange.com/tags/data-explorer/info">SEDE</a> queries that combines these event both for the question and all answers. However, SEDE is only updated once a week - so it is more useful in situations where some time passed since the most recent activity. For the posts you've mentioned you get this:</p>
<ul>
<li><a href="https://data.stackexchange.com/math/query/1253323/combined-posthistory-question-and-all-the-answers?qid=924266" rel="nofollow noreferrer">Combined PostHistory (question and all the answers)</a></li>
<li><a href="https://data.stackexchange.com/math/query/1253327/combined-posthistory-question-and-all-the-answers-only-edits?qid=924266" rel="nofollow noreferrer">Combined PostHistory (question and all the answers) - only edits</a></li>
</ul>
|
136,453 | <p>For every $k\in\mathbb{N}$, let
$$
x_k=\sum_{n=1}^{\infty}\frac{1}{n^2}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}.
$$
Calculate the limit $\displaystyle\lim_{k\rightarrow\infty}x_k$.</p>
| lhf | 589 | <p>The additive structure of the integers is trivial: it's generated by 1. This is in essence the Peano axioms.</p>
<p>The multiplicative structure of the integers is not trivial: it's generated by prime numbers. In other words, prime numbers are the multiplicative building blocks of the integers in the sense that every nonzero number is either a prime or a product of primes (the empty product gives 1).</p>
<p>This gives a rich ring structure to the integers. Number theory is about that structure. Add to this the amazing regular irregularity of the primes and you get the richness of analytic methods.</p>
|
4,011,581 | <p>I know that a function is continuous at a point if the limit from left and right side exists and are equal and for a function to be continuous, the function should be continuous at all points. My question is that if I want to check continuity of a function, I cannot practically check continuity at each and every point in <span class="math-container">$\mathbb{R}$</span>. So, how to do it?</p>
| GBA | 791,969 | <p>One way is to take an arbitrary point <span class="math-container">$x_0$</span> with no prior assumptions and show that for every <span class="math-container">$\varepsilon>0$</span> there exists <span class="math-container">$\delta>0$</span> (which may depend on <span class="math-container">$x_0$</span>) s.t for every <span class="math-container">$x\in (x_0-\delta,x_0+\delta)$</span>, <span class="math-container">$|f(x)-f(x_0)|<\varepsilon$</span>. This is the case in <span class="math-container">$\Bbb{R}$</span>, but this is the basic definition.</p>
<p>For example if we wanted to show that the function <span class="math-container">$f(x)=x$</span> is continuous in <span class="math-container">$\Bbb{R}$</span>, we may say the following: let <span class="math-container">$x_0\in\Bbb{R}$</span>, <span class="math-container">$\varepsilon>0$</span>. So we take <span class="math-container">$\delta=\varepsilon$</span>, so for every <span class="math-container">$x$</span> s.t <span class="math-container">$|x-x_0|<\delta$</span> we get <span class="math-container">$|f(x)-f(x_0)|=|x-x_0|<\delta=\varepsilon$</span>, so <span class="math-container">$x$</span> is continuous in <span class="math-container">$x_0$</span> and therefore because we made no prior assumptions, it's continuous in <span class="math-container">$\Bbb{R}$</span>.</p>
|
178,666 | <p>For every natural number n, let:</p>
<ul>
<li><p>Gn be the number of distinct group structures with at most n elements;</p></li>
<li><p>An be the number of distinct abelian group structures wit at most n elements;</p></li>
<li><p>Sn be the number of distinct solvable group structures with at most n elements.</p></li>
</ul>
<p>Question 1: Is there a known limit for the quotient An/Gn ?</p>
<p>Question 2: Is there a known limit for the quotient Sn/Gn ?</p>
| Emil Jeřábek | 12,705 | <p>The number of abelian groups of order at most $n$ is $O(n)$, whereas if $n=2^k$, the number of class $2$ nilpotent groups of order $n$ is $2^{(2/27)k^3+O(k^{8/3})}=n^{\Omega(\log^2n)}$ by a result of Sims, hence the answer to question 1 is $0$. It is conjectured that the global asymptotic density of $2$-groups of nilpotent class $2$, and a fortiori of solvable groups, is $1$, but as far as I know, this has not been proved.</p>
|
52,480 | <p>The question comes from a statement in Concrete Mathematics by Graham, Knuth, and Patashnik on page 465.</p>
<p>$$\sum_{k \geq n} \frac{(\log k)^2}{k^2} = O \left(\frac{(\log n)^2}{n} \right).$$</p>
<p>How is this calculated?</p>
| davidlowryduda | 9,754 | <p>I think the easiest way here is to simply find $\displaystyle \int_k ^{\infty} \frac {(\log x)^2}{x^2}dx$. After some work, it turns out to be $\dfrac{\log k(\log k + 2) + 2}{k}$. Oh - but I'm also somewhat confident that Qiaochu's suggestion to Sum by Parts would work as well (and give almost the same answer).</p>
|
4,251,161 | <p><strong>Objective</strong><br />
I need to find roots of <span class="math-container">$$f(x)=c$$</span> in interval <span class="math-container">$[a,b]$</span>, where</p>
<ul>
<li><span class="math-container">$f(a)=0$</span> and <span class="math-container">$c<f(b)<1$</span></li>
<li><span class="math-container">$f(x)$</span> is unknown outside of the interval</li>
<li><span class="math-container">$f(x)$</span> is monotonuous and has decreasing derivative (i.e., <span class="math-container">$f''(x)<0$</span>)</li>
</ul>
<p>(Overall it resembles flipped hockey stick.)</p>
<p><strong>What I tried</strong><br />
Right now I am using bisection method, which is somewhat slow for my purposes, as evaluating the function is expensive.</p>
<p>I tried other bracketing methods:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Regula_falsi" rel="nofollow noreferrer"><em>false position (regula falsi)</em></a> is slower due to the stagnant bound. <a href="https://en.wikipedia.org/wiki/Regula_falsi#The_Illinois_algorithm" rel="nofollow noreferrer"><em>Illinois</em></a> improves performance a bit, but still slower than bisection.</li>
<li><a href="https://en.wikipedia.org/wiki/Ridders%27_method" rel="nofollow noreferrer"><em>Ridder's method</em></a> converges in less steps, but overall slower, as it requires more function evaluations (perhaps also due to the stagnant bound).</li>
</ul>
<p><strong>Question</strong><br />
Could I use my knowledge of the function properties outline in the beginning, to achieve faster convergence (less function evaluations)?</p>
<hr />
<p><strong>Possibilities</strong>.
As an idea, one could try using transformation:
<span class="math-container">$$F(x) = \log (1-f(x)), C=\log (1-c),$$</span>
In a hope that <em>false position</em> converges faster for equation
<span class="math-container">$$F(x)=C.$$</span></p>
<p><strong>Supporting information:</strong> more details about the curve can be found <a href="https://math.stackexchange.com/q/4252758/765359">here</a>. Mine is not exactly teh same, but very similar. Calculating it directly is expensive, and one resorts to resampling and averaging.</p>
| Simply Beautiful Art | 272,831 | <p>The largest bottleneck is the function evaluations, but luckily for us the root can be bracketed with something like bisection, ensuring we can avoid interpolation iterations if we know they will likely be unhelpful.</p>
<p>This is exactly where <a href="http://dl.acm.org/citation.cfm?id=248563" rel="nofollow noreferrer">Chandrupatla's method</a> shines. You can check <a href="https://tio.run/##lVVNb@M2EL3nV0zdQ0yt7Vpq04MRBygWRbFAT/04bT9Ay1RMVCa1FLWRsOhvT9@Q@lx4DzUQhORwZt57M0NVnb9Y8@1r4eyVfFdp80z6Wlnn6a0sS3kq1YbeeeWkt@7u7qwKqvWzWbcHKkorvaDtE2njD3eEn1O@cYZS0gW19ER7UmWtaNsfPA4H@1cOlF@kObumkr6U6@IwJnz/PoT@c0PD/zbt02GZjUvlnHX9jo6Uqm36XcAz4O3jHO4CuK/prS1LlXvg1V7LkioL5PUuWLsUIYp1m4q4zeI2E4PzL@pDo50iearXXSZAJqxSQWARReHNV8d@nYkYWNa1gpq3LgQ7lBkCPQ7Bo5r8AyDmvAE@/CHtEUcZK8JbPh4B/qQM01a36WlVnuE132SD5@8oib8ouippqLAubArtakgVpNTW7Og3@Y@iXEICWyC2VyZksR@Vg84vMU8LgG1Kb2i/e6CEsdIWB4JpxsbB@thLAHX7fugvs@NM8p@trahBmpJya5DnWZlcxUQvF13GYsBry17ot9ARHIkDBZuI5UE5ofu@74QY/YeqKjuSVF/RdySvFpmYmrclOCMRVqRM3YCx@gi2jbyBI5J@g@C7jEnchJBEvj1UAUsvjBBjkC42nJhDHEtqVOtJ1V5fsZsS94Wcu/z6IqsZcKTRNeayKmWuznTqqJ3ch5p0U0nQT1Pz/a8GnFopNEwdQPSIqdRGSad9xwIzy5l4f4f7IW5fyG9oLOpEdbrWwdTFa10a13MI7wruX1QW9OumKHSuUTyUOoLYUFPzjKCM@P@hkWcOnPMjplxly9jtc42G1ElCGWZ0APxInHwdEMQjEa4s9ZNA3NGElcTCfGJztqAyN@eTd3bD@8zmdDKn4rPacaFiyTbcpUCTIGVCs07MsT1PoyqWIcI72H1JXm2@KPBJ13hqF1LyqC/FaTl6eyvh4lYcr4cZ7uGFGGdY5dJMExIesYt@vmwrp3JdAwbeM4P5Xc5QnB@WPqFAH9mT8FzNChK@VcXyixdWi28emjdhQbeEZ7Vy6KZ1sfp0/@OACDzvD4/Z/l/6w9On@2I9WLg87V@cPIsXVqL3Xx1XiPmwB1vm00Luz76ZG9pvaPhcjEnbw1O236XfFzEXj9t4shKv/wE" rel="nofollow noreferrer">here for an example implementation</a>.</p>
<p>Output of the above example:</p>
<p><span class="math-container">\begin{array}{r|r}
x & f(x)=x^2-2 \\ \hline
0.0000000000000000 & -2.0000000000000000 \\
2.0000000000000000 & 2.0000000000000000 \\
1.0000000000000000 & -1.0000000000000000 \\
1.5000000000000000 & 0.2500000000000000 \\
1.4095238095238036 & -0.0132426303855042 \\
1.4142641817017454 & 0.0001431756445074 \\
1.4142135574926071 & -0.0000000138041043 \\
1.4142135623731014 & 0.0000000000000178 \\
1.4142135623730892 & -0.0000000000000167 \\
1.4142135623730951 & 0.0000000000000004
\end{array}</span></p>
<p>It can be seen that Chandrupatla's method intelligently chooses to use bisection during the initial iterations to ensure convergence while it uses interpolation for later iterations where it is confident it should be accurate.</p>
<p>The downside is that in some cases it does not alternate sides of the root each iteration. This leads to a slower order of convergence. When alternating, the order of convergence is <span class="math-container">$\approx1.839$</span>. When not alternating, the order of convergence is <span class="math-container">$\approx1.618$</span>. This means the number of digits accurate increases by roughly <span class="math-container">$83.9\%$</span> and <span class="math-container">$61.8\%$</span> respectively every iteration.</p>
<hr />
<p>If your function is stochastic i.e. you are estimating <span class="math-container">$f(x)$</span> as an average of <span class="math-container">$f_i(x)$</span> for many values of <span class="math-container">$i$</span>, then you may want to resort to stochastic root-finding algorithms, at least for the initial iterations.</p>
<p>You can check <a href="https://tio.run/##dVNNj5swEL3zK0b0EJwAhVTbAyp7qaqqUm9tT9s28oKdWDU2MmYLWvW3p2MMZLWbjRQFZt689@Yj7WhPWr07c6MbEJYZq7XsQDStNhYq3SsbTDk7tkIdl8RHKiW9lyyGL1hDrTbnmnHorK5OtLOiOhitbcSLFXl3x6Wm9lcMy@9Q@McYRHcQqjKMdihRwD1agBK@mx75a1bRcUZiME@zGFr9l5lLLEtvCCS3q5VZqQgAP2EYfltdgXOVcKFq1wvXBh0bUVk5wsUARBhH2fmVAO9VZYVWXYpkwcT6Bt3R6g/YEwP6gLJHBgwVGmpZOiGGA304TuaypeSr1i3q4KCEwlHL0SOdDYFxP@0oJ4Uv8EU/2ho5rwpNlUO6gr3kroRogMS/EXgL4indZ6bckDxhq4Wyl/pRMFn7uisOXlOtB@ySR4OTinLYYS9bvzUC263f1QoW/Nmy18zUACQl8q0xJjv2HLFziPN8hoaqWjfB9eMVnRQVCwJ3mOhvvpfpUqYnz2yY7Y1yI9sC9pDAHlvwvGmvBDbbREmW5jFuMidB0BqcWcTDx82nZSAlDJviwz77Bz8tPG54tGSIS/3eO1IPCMlcH5Yh6t1kSDiN063f@41e/IdiyAl@s4x4x6uDobjdo6n33As/aWFNhOT8Hw" rel="nofollow noreferrer">here for an example implementation</a> of the Robbins-Monro algorithm with Polyak-Ruppert averaging.</p>
<p>Output of the above example:</p>
<p><span class="math-container">\begin{array}{r|r}
x & f(x)=x^2-2~~(\pm0.1) \\ \hline
1.0000000000000000 & -1.0000000000000000 \\
1.3208425751018684 & -0.2553748917982650 \\
1.2931735764314634 & -0.3277021012194583 \\
1.3337128053831124 & -0.2212101527571080 \\
1.3501779010773112 & -0.1770196354424665 \\
1.3656025721082210 & -0.1351296150514110 \\
1.3758861134900708 & -0.1069374027051879 \\
1.3847629867055029 & -0.0824314706504552 \\
1.3880601300384776 & -0.0732890753975646 \\
1.3900420572765071 & -0.0677830790024958 \\
1.3914381380169945 & -0.0638999080717995 \\
1.3909224968549345 & -0.0653346077428347 \\
1.3916674836639233 & -0.0632616149125236 \\
1.3918200228103006 & -0.0628370241043343 \\
1.3941212662850380 & -0.0564258948918022 \\
1.3964265279275152 & -0.0499929521003046 \\
1.3987379168844518 & -0.0435322398697444 \\
1.4008372481138009 & -0.0376550042969532 \\
1.4009627374051223 & -0.0373034084023462 \\
1.4012682932051819 & -0.0364471704578364 \\
1.4010721993805688 & -0.0369966921228957 \\
1.4015011630538421 & -0.0357944899587279
\end{array}</span></p>
<p>Once you feel like you have converged well enough for the stochastic approximation, you can use that estimate to kick-start your bracketing method to reach high-precision results in only a few additional iterations (at the cost of needing to spend more computations on those iterations).</p>
|
2,992,127 | <p>I know that <span class="math-container">$ad\neq bc $</span> is sufficient for <span class="math-container">$z$</span> irrational because if <span class="math-container">$ad = bc$</span> then <span class="math-container">$\frac{ax+b}{cx+d} = \frac{ax+b}{cx+d} \frac{cb}{ad} = \frac{cax+cb}{cax+da}\frac{b}{d}$</span> because <span class="math-container">$cb=da$</span> nominator and denominator are the same. Hence <span class="math-container">$\frac{cax+cb}{cax+da}\frac{b}{d} = \frac{b}{d}$</span> (Contradiction).</p>
<p>But I don't know to prove that <span class="math-container">$ad \neq bc$</span> is also necessary for <span class="math-container">$z$</span> being irrational. </p>
| Community | -1 | <p>Assume by contradiction that</p>
<p><span class="math-container">$$\frac{ax+b}{cx+d}=\frac pq,$$</span> or</p>
<p><span class="math-container">$$(qa-pc)x+qb-pd=0,$$</span> and <span class="math-container">$x$</span> is rational !</p>
<p>Note that this doesn't work when <span class="math-container">$qa-pc=qb-pd=0$</span>, which is equivalent to <span class="math-container">$ad=
bc$</span>.</p>
|
3,471,684 | <p>Is always correct statement that if natural numbers <span class="math-container">$a,b \in \Bbb N$</span> for which LCM<span class="math-container">$(a,b)=16\cdot(a,b)$</span>, then <span class="math-container">$a|b$</span> or <span class="math-container">$b|a$</span>?</p>
<p>I used formula that LCM<span class="math-container">$(a,b)=\frac{a\cdot b}{(a,b)}$</span></p>
<p><span class="math-container">$\frac{a\cdot b}{(a,b)}=16\cdot(a,b) \implies a\cdot b= (4\cdot(a,b))^2$</span></p>
<p>Is it somehow useful? What should I do next?</p>
| nonuser | 463,553 | <p>Let <span class="math-container">$d= (a,b)$</span>. We know that <span class="math-container">$LCM (a,b) ={ab\over d}$</span>. Then as you noted we have <span class="math-container">$$ab =16d^2$$</span> We can also write <span class="math-container">$a=dx$</span> and <span class="math-container">$b=dy$</span> where <span class="math-container">$x,y$</span> are coprime. So <span class="math-container">$$xy=16$$</span> Now it is not difficult to finish. </p>
|
3,990,195 | <p>I need some assistance solving what seems to be a very <a href="https://i.stack.imgur.com/arMiv.png" rel="nofollow noreferrer">intuitive problem</a>, but becomes tough when only using strict natural deduction and not assuming De Morgan laws.</p>
<p>Laws allowed: Implication, And, Or, MT, PBC, Copy Rule, Negation, Double Negation, Contradictions, law of excluded middle</p>
<p>I'm thinking it uses the law of excluded middle but I can't quite figure it out.</p>
<p><span class="math-container">$$ \lnot(P \land \lnot Q), \; (\lnot P \to S) \land \lnot Q \;\;\; \text{premises} \tag{1} $$</span></p>
<p><span class="math-container">$$ T \lor S \;\;\; \text{conclusion} \tag{2} $$</span></p>
| algebruh | 875,254 | <p><a href="https://i.stack.imgur.com/jPvEk.png" rel="nofollow noreferrer">My proof here</a>. Let me know if you see anything wrong with it, or if there's a more efficient way</p>
|
1,195,175 | <p>In <a href="http://en.wikipedia.org/wiki/Solid_angle" rel="nofollow">wikipedia</a> the solid angle is defined as follows:</p>
<blockquote>
<p>In geometry, a solid angle (symbol: Ω) is the two-dimensional angle in three-dimensional space that an object subtends at a point. </p>
</blockquote>
<p>Why solid angle is a two dimensional angle?</p>
| Mark Bennet | 2,906 | <p>You should look at it as follows:</p>
<p>$$\sqrt {x+\delta x}-\sqrt x =\left(\sqrt {x+\delta x}-\sqrt x\right) \cdot \frac {\sqrt {x+\delta x}+\sqrt x }{\sqrt {x+\delta x}+\sqrt x}=\frac {\delta x}{\sqrt {x+\delta x}+\sqrt x }$$</p>
<p>Now the denominator is "large" compared with $\delta x$ and you don't need to get rid of the square root.</p>
<p>It looks as though you have done the equivalent of $(a-b)^2=a^2-b^2$, which is not true, and what you need instead is $(a-b)(a+b)=a^2-b^2$, as I have used above.</p>
|
105,535 | <p>In a thread in <a href="https://math.stackexchange.com/questions/186292/derivatives-of-the-riemann-zeta-function-at-s-0">MSE</a> I proposed an older routine of mine for the efficient computation of coefficients; I use a very similar routine for the quick&dirty computation of the Stieltjes-constants. </p>
<p>This motivated me to try to improve my earlier toy-computations to calculate now the first 512 Stieltjes to 1000 dec digits precision. I'm unable to estimate the number of correct digits by analytical arguments; at least wolframalpha allowed me to display StieltjesGamma[511] to 400 digits, which met my own computations. </p>
<p>The only freely available table around seems to be that of S. Plouffe (linked via <a href="http://en.wikipedia.org/wiki/Stieltjes_constants" rel="nofollow noreferrer">wikipedia</a>) but they display only the first 78 numbers to 256 digits precision. </p>
<p><strong>Update2:</strong> This is the effective formula to which the Pari/GP code reduces: </p>
<p>Let $ \qquad h_c = {1\over c!} \sum_{k=0}^\infty (-1)^k {\ln(1+k)^c\over1+k}$ This is done using the <em>sumalt</em>-procedure. </p>
<p>Next let $ \qquad r_c = - {\ln(2)^{c-1}\over c!} b_c$ where $b_c$ are the bernoulli numbers </p>
<p>Then $ \qquad \gamma_c = c! \sum_{d=0}^{c+1} h_d \cdot r_{c+1-d} $ </p>
<p><em>So my question:</em></p>
<blockquote>
<p>how could I possibly get an educated guess for the number of correct digits based on my Pari/GP-routine?* </p>
</blockquote>
<p>Alternatively: </p>
<blockquote>
<p>is there some table with comparable precision around such that I can at least check the match for the first m digits (where m should optimally go to 1000)?</p>
</blockquote>
<p><em>(here is the table with my current computations of <a href="http://go.helms-net.de/math/tables/stieltjes_512x1000.zip" rel="nofollow noreferrer">512 coeffs by 1000 digits</a>)</em><br>
<hr>
<strong>Update1:</strong><br>
Heuristically I find, that beginning with some precision, say $300$ dec digits at the first $\gamma_0$ , I simply lose one digit precision per step in the index, so in $\gamma_k$ are roughly $300-k$ digits correct, maybe a handful less.<br>
For this I used differences when computed with precision $200,300,400,500,600,700$ from that with precision $800$, $\gamma_0$ had just nearly all leading digits constant, when precision was increased, so that was always correct to the full precision.<br>
That would mean, that if I want $1000$ correct digits for $\gamma_{511}$ I need dec precision of (at least) $1550$ . Simple, if that is true...</p>
<hr>
<p>Here is my routine. I reduced the precision-parameter so that this can just be copied & pasted to a Pari/GP-environment. For precision of 1000 dec digits and 512 coefficients this must be optimized due to exorbitant increase of stack and computation-time otherwise</p>
<p>Prepare computations with parameters for precision of computation</p>
<pre><code>termsforseries = 32
digitstocompute = 200; digitstoshow = 12;
default(realprecision,digitstocompute)
default(format,Str("g0.",digitstoshow))
default(seriesprecision,termsforseries)
</code></pre>
<p>Compute the coefficients of the Laurent-expansion of the zeta by conversion from the same series-type of the eta-function (the alternating zeta) </p>
<pre><code>\\ ========= Zeta Laurent-expansion providing Stieltjes-coefficients ====
ps_eta = sumalt(k=0,taylor((-1)^k/(1+k)^(1-x),x))
tmp = Vec(1-2*2^(-(1-x)));
tmp[1]=0; \\ make the first zero exact. this step is needed for
\\ allowing the reciprocal of the powerseries
ps_etatozeta=1/Ser(tmp)
ps_zeta = ps_eta * ps_etatozeta \\ contains now the Stieltjes-coefficients
tmp=Vec(ps_zeta);tmp=vector(#tmp-1,c,tmp[1+c]) \\ remove the first coefficient (at 1/x)
sti = vector(#tmp,r,tmp[r]*(r-1)!) \\ extract Stieltjes-constants by mult with factorials
</code></pre>
| Fredrik Johansson | 4,854 | <p>You could compare with the output from mpmath:</p>
<pre><code>sage: import mpmath
sage: mpmath.mp.dps = 1000
sage: %time mpmath.stieltjes(511)
CPU times: user 123.17 s, sys: 0.02 s, total: 123.19 s
Wall time: 123.40 s
mpf('673581492593841075447052270498937988033439947306384442967711559788996269245614412865378751092398327114199475672304543519558074203937367354475627304841991475249868411079091195038704370379319922304314968920977080.4218186954910530966341150821211999689800345913062006500416130863993252444286525401536530609127800808358611180051913954061786113778487954768827917318185861285728540852470806490244553130800206629709991267757983837666355484638397085316115099902138453930569718675294835237821298508690226519561229169443578986238598614523990440226172962706436119188515904391443174279895106345752233034115379099381680958168062786627389335290431416199037643058641914376639305675292168558263653044141610653456719446309980037732502545489019580865593535176949757824659484296855986638635532332512794555243036229273521585906314889067495562018805980518215400448131311489588531760771126389926309367463577942595344292677230759234541824332220012416082001221662802813469321335808232095303910714771240349667445255785796410716571')
</code></pre>
<p>This only seems to agree with your result up to about 715 digits. Mathematica 8 agrees with mpmath, so presumably you will need to increase the precision in your algorithm.</p>
|
1,731,382 | <p>Notice that the parabola, defined by certain properties, is also the trajectory of a cannon ball. Does the same sort of thing hold for the catenary? That is, is the catenary, defined by certain properties, also the trajectory of something?</p>
| J. M. ain't a mathematician | 498 | <p>As I've shown in <a href="https://math.stackexchange.com/a/63075">a previous answer</a>, the focus of a parabola rolling on a straight line traces a catenary. Similarly, the directrix of the same rolling parabola will envelope another catenary, a reflection of the one being traced by the focus.</p>
<p>Here is a modern (as in done with the current version of <em>Mathematica</em>) version of the cartoon I did for that previous answer:</p>
<p><img src="https://i.stack.imgur.com/8rcYK.gif" alt="thus a parabola rolls"></p>
|
878,517 | <p>Is there any more solutions to this functional equation $f(f(x))=x$?</p>
<p>I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.</p>
| doraemonpaul | 30,938 | <p>In fact this belongs to a functional equation of the form <a href="http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf" rel="nofollow">http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf</a>.</p>
<p>Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,</p>
<p>Then $u(t+2)=u(t)$</p>
<p>$u(t)=\theta(t)$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$</p>
<p>$\therefore\begin{cases}x=\theta(t)\\f=\theta(t+1)\end{cases}$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$</p>
|
4,643,832 | <p><strong>Question</strong></p>
<p><a href="https://i.stack.imgur.com/XSeNR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XSeNR.png" alt="enter image description here" /></a></p>
<p>I am trying to prove this using balls (that is what we use in my school). The definition is that a subset <span class="math-container">$A$</span> is open if <span class="math-container">$\forall a \in A$</span> <span class="math-container">$ \exists$</span> r such that <span class="math-container">$B_r(a) \subseteq A $</span>.</p>
<p><strong>Textbook Solution</strong></p>
<p><a href="https://i.stack.imgur.com/qNnal.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qNnal.png" alt="enter image description here" /></a></p>
<p><strong>Confusion</strong></p>
<p>This is the answer my book gave but the problem is that I don't understand why we chose <span class="math-container">$\delta=1-||(x-1)^2-(y+2)^2||$</span> . Looking at the condition for <span class="math-container">$(x,y)$</span> to be in <span class="math-container">$A$</span> I realized that for <span class="math-container">$a=(-1,2)$</span> then we could have <span class="math-container">$r=1$</span> and <span class="math-container">$B_1((-1,2))={(x,y)\in R :||(x,y)-(-1,2)||<1 }$</span> so <span class="math-container">$a=(-1,2)$</span> is open. but I don't see why we chose that <span class="math-container">$\delta$</span>.</p>
<p>Note: I am not saying I don't agree with it. I just don't understand the trick to know the <span class="math-container">$\delta$</span> to choose.</p>
| Mastrem | 253,433 | <p>With Bernoulli's inequality:
<span class="math-container">$$
\left(1+\frac1y\right)^{x-y}\ge 1+(x-y)\cdot\frac 1y=1+\frac xy-1=\frac xy.
$$</span></p>
|
1,085,702 | <p>It's said that a computer program "prints" a set <span class="math-container">$A$</span> (<span class="math-container">$A \subseteq \mathbb N$</span>, positive integers.) if it prints every element of <span class="math-container">$A$</span> in ascending order (even if <span class="math-container">$A$</span> is infinite.). For example, the program can "print":</p>
<ol>
<li>All the prime numbers.</li>
<li>All the even numbers from <span class="math-container">$5$</span> to <span class="math-container">$100$</span>.</li>
<li>Numbers including "<span class="math-container">$7$</span>" in them.</li>
</ol>
<p>Prove there is a set that no computer program can print.</p>
<p>I guess it has something to do with an algorithm meant to manipulate or confuse the program, or to create a paradox, but I can't find an example to prove this. Any help?</p>
<p>Guys, this was given to me by my Set Theory professor, meaning, this question does not regard computer but regards an algorithm that cannot exhaust <span class="math-container">$\mathcal{P}(\mathbb{N})$</span>. Everything you say about computer or number of programs do not really help me with this... The proof has to contain Set Theory claims, and I probably have to find a set with terms that will make it impossible for the program to print. I am not saying your proofs including computing are not good, on the contrary, I suppose they are wonderful, but I don't really understand them nor do I need to use them, for it's about sets.</p>
| WGroleau | 203,965 | <p>If you can define a function to determine whether a given integer is in the set, then I can write a program that can print integers in the set until it runs out of storage.</p>
<p>Your "question" is actually a statement, but assuming you ask whether the statement is true, the answer is "yes." For example, I know of a way to generate truly random numbers. No program can predict which numbers will not come from this device.</p>
|
370,599 | <p>If A is an invertible $nxn$ matrix prove that:$ adj(adjA)=(A)(detA)^{n-2}$
I have done this but it somewhere went wrong:
$ adj(adjA)=adj(A^{-1} detA)=(A^{-1}detA)^{-1} det(A^{-1}detA)=AdetA det(A^{-1}detA)= Adet(AA^{-1}detA)=A (detA)^n $ </p>
| Ross Millikan | 1,827 | <p>How about $x=12,y=15,n=3?$</p>
<p>For the update, $x=27,y=189,n=6$</p>
|
1,341,385 | <p>I want to be a mathematician or computer scientist. I'm going to be a junior in high school, and I skipped precalc/trig to go straight to AP Calc since I've studied a lot of analysis and stuff on my own. My dad wants me to memorize about 30 trig identities (though some of them are very similar) since I'm missing trig. I've gone through and proved all of them, but memorizing them seems like a waste of effort. My dad is a physicist, so he is good at math, but I think he may be wrong here. Can't one just use deMoivre's theorem to get around memorizing the identities?</p>
| Plutoro | 108,709 | <p>Usually, yes, though I prefer Euler's identity. Pretty much every trig identity can be derived from
$$e^{ix}=\cos(x)+i\sin(x).$$
However, it is useful to memorize some of the common ones because they will help you a lot in calculus and beyond to quickly identify when an expression can be simplified. I would start with memorizing the angle addition formulas. From there you can quickly derive the double and half angle formulas as well as some others.</p>
|
2,674,799 | <blockquote>
<p>Let $X_{2n}$ be the group whose presentation is$\langle x,y\,|\,x^n=y^2=1, xy=yx^2\rangle$. From $x=xy^2$, it is seen that $x^3=1$, hence $X_{2n}$ has at most $6$ elements. I have to show that if $n=3k$, then $X_{2n}$ has exactly $6$ elements. </p>
</blockquote>
<p>I can't see where I am having problem if I assume $x=1$.</p>
| Derek Holt | 2,820 | <p>The elements $x=(1,2,3)$ and $y=(2,3)$ of $S_3$ satisfy the three relations when $n=3k$, and $S_3$ has order $6$ and is generated by $x$ and $y$. So $|X_{2n}| \ge 6$ when $n=3k$.</p>
<p>Also, the relations can be used to write any element of $X_{2n}$ in the form $x^ay^b$, the fact that $x^3=y^2=1$ in $X_{2n}$ proves that $|X_{2n}| \le 6$ for all $n$.</p>
<p>So $|X_{2n}|=6$ when $n=3k$.</p>
|
65,002 | <p>I am a programmer, so to me $[x] \neq x$—a scalar in some sort of container is not equal to the scalar. However, I just read in a math book that for $1 \times 1$ matrices, the brackets are often dropped. This strikes me as very sloppy notation if $1 \times 1$ matrices are not at least <em>functionally equivalent</em> to scalars. As I began to think about the matrix operations I am familiar with, I could not think of any (tho I am weak on matrices) in which a $1 \times 1$ matrix would not act the same way as a scalar would when the corresponding scalar operations were applied to it. So, is $[x]$ functionally equivalent to $x$? And can we then say $[x] = x$? (And are those two different questions, or are entities in mathematics "duck typed" as we would say in the coding world?)</p>
| Xander Henderson | 468,350 | <h2>Some Background</h2>
<p>There are three basic kinds of mathematical spaces that are being asked about in the original question: scalars, vectors, and matrices. In order to answer the question, it is perhaps useful to develop a better understanding of what each of these objects is, or how they should be interpreted.</p>
<ol>
<li><p>In this context, a <strong>scalar</strong> is a an element of a field. Without getting into too much detail, a field is an object made from three ingredients $(k,+,\cdot)$, where $k$ is a set, and $+$ and $\cdot$ are two binary operations (addition and multiplication). These operations are associative and commutative, there are distinguished identity elements (0 and 1) for each operation, every element has an additive inverse, every nonzero element has a multiplicative inverse, and multiplication distributes over addition.</p>
<p>Examples of fields include (among others) the real numbers, as well as the rationals, the complex numbers, and (if you are feeling a little more esoteric) the $p$-adic numbers.</p></li>
<li><p>A <strong>vector</strong> is an element of a vector space. A vector space is an object made from four ingredients $(V,k,+,\cdot)$, where $V$ is a set, $k$ is a field (the <em>base field</em>), $+$ is a binary operation which acts on two vectors, and $\cdot$ is a binary operation (called <em>scalar multiplication</em>) which acts on a scalar from $k$ and a vector from $V$. The addition is associative and commutative, there is a distinguished 0 vector (the additive identity), and every vector has an additive inverse. Scalar multiplication allows scalars to "act on" vectors; a vector can be multiplied by a scalar, and this multiplication "plays nice" with the addition (e.g. $a(b\mathbf{v}) = (ab)\mathbf{v}$ and $a(\mathbf{u}+\mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ for scalars $a$ and $b$, and vectors $\mathbf{u}$ and $\mathbf{v}$).</p>
<p>Examples of vector spaces include $\mathbb{R}^n$ (the base field is $k = \mathbb{R}$), $\mathbb{C}^n$ (the base field is $\mathbb{C}$), and the space of continuous functions from $[0,1]$ to $\mathbb{C}$ (i.e. the space $C([0,1])$–the base field here is $\mathbb{C}$ again).</p></li>
<li><p>Finally, a <strong>matrix</strong> is an element of a specific kind of algebra over a field. An algebra has two ingredients: $(V,\times)$, were $V$ is a vector space and $\times$ is a binary operation called a <em>bilinear product</em>. This product is both left- and right-distributive over the addition in $V$, and plays nice with scalar multiplication. In detail, if $a$ and $b$ are scalars and $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are vectors, then</p>
<ul>
<li>$\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = (\mathbf{u}\times \mathbf{v}) + (\mathbf{u}\times \mathbf{w})$, </li>
<li>$(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = (\mathbf{u}\times \mathbf{w}) + (\mathbf{v}\times \mathbf{w})$, and</li>
<li>$(a\mathbf{u})\times (b\mathbf{v}) = (ab)(\mathbf{u}\times \mathbf{v}).$</li>
</ul>
<p>Note that the underlying vector space (with its underlying field) is still running around, so there is a <em>ton</em> of structure here to play around with. </p>
<p>Examples of algebras include $\mathbb{R}^3$ with the usual cross product, and matrix algebras such as the space of $n\times n$ matrices over $\mathbb{R}$ with the usual matrix multiplication. The structure of an algebra is actually more general than this, and there are much more interesting examples (such as $C^{\ast}$-algebras), but these are not really germane to this question.</p></li>
</ol>
<hr>
<h2>The Question Itself</h2>
<p>Now to attempt to answer the question:</p>
<p>If $k$ is any field, then we can regard $k$ as a (one-dimensional) vector space over itself by taking the scalar multiplication to coincide with the multiplication in the field. That is, if we let $a$ denote a scalar and $\langle a \rangle$ denote a vector, then
$$ a \cdot \langle b \rangle := \langle a\cdot b \rangle. $$
In a mathematically meaningful sense, this is a way of identifying the field $(k,+,\cdot)$ with the vector space $(V,k,+,\cdot)$ (though something is lost in the translation–it doesn't make sense to multiply two vectors, though it does make sense to multiply a vector by a scalar). This is a bit outside my area of expertise, but I think that the right language (from category theory) is that there is an <em>faithful functor</em> from the category of fields to the category of vector spaces. In more plain language, this says that fields and (certain) vector spaces are equivalent to each other. Specifically, we can sometimes regard a one-dimensional vector as a scalar–while they are not quite the same objects, they can, in the right context, be <em>treated</em> as though they are the same. In this sense, a scalar is "functionally equivalent" to a one-dimensional vector.</p>
<p>By a similar argument, we can regard a field $k$ as an algebra over itself, with the underlying vector space being that obtained by viewing $k$ as a vector space, and the bilinear product being the product obtained by making the identification
$$ [a]\times [b] = [a\cdot b]. $$
In this case, the algebra obtained is the algebra of $1\times 1$ matrices over $k$. Hence there is a faithful function identifying any field with an algebra of $1\times 1$ matrices. Again, this gives us a way of identifying scalars with matrices, so (again) we may meaningfully assert that scalars can be identified with matrices. In contrast to the previous example, we really don't lose anything in the translation. Even more surprising, the bilinear product ends up being a commutative operation, which is a neat property for an algebra to have.</p>
<p>It is worth observing that, in this setting, $[a]$ is not "the scalar $a$ in some kind of container." The notation $[a]$ denotes a $1\times 1$ matrix with entry $a$. The brackets denote a great deal of structure–more than is implied by the simple statement "a scalar in a container."</p>
<blockquote>
<p><strong>Long Answer Short:</strong> A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. It might be a little sloppy to do so, but a little bit of sloppiness is forgivable if it introduces no confusion or ambiguity, and if it aids brevity or clarity of exposition.</p>
</blockquote>
|
3,567,662 | <p>I have just learned how to convert a plane in R3 from Cartesian to parametric form, by setting 2 variables to 0 and solving for the 3rd one in order to obtain 3 points on the plane, and solve from there. However, this does not work when 1 or 2 of the variables are 0, as it is not possible to find 3 points on the plane in the same way (for example in the picture). How can this be solved for the particular question, and for other cases where there are variables that are 0?
<a href="https://i.stack.imgur.com/C7fEp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C7fEp.png" alt="enter image description here"></a></p>
| infinity | 719,182 | <p>Denote by <span class="math-container">$f:\Bbb N \to X$</span> , <span class="math-container">$g:\Bbb N \to Y$</span> the bijections from <span class="math-container">$\Bbb N$</span> to <span class="math-container">$X,Y$</span>.</p>
<p>Then <span class="math-container">$h : \Bbb N \times \Bbb N \to X \times Y$</span> which is defined by <span class="math-container">$h(n,m)= (f(n),g(m))$</span> is the bijection you are looking for.</p>
|
2,895,382 | <p>Let $J_n=\{1,\dots,n\}$. How do I show that the set of all functions $J_n\to \mathbb N$ is countable? Any function is given by specifying the images of $1,\dots,n$. There are $|\mathbb N|$ options for the image of each $i=1,
\dots, n$. So intuitively, the set of such functions is the union of $n$ copies of $\mathbb N$, hence countable. But how to formalize it?</p>
| Robert Israel | 8,508 | <p>It's not the union, it's the Cartesian product.</p>
|
2,660,316 | <p>$$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac 2{\sqrt{2\pi}}\int_0^{\sqrt y} \exp\left(-{\frac{x^2}{2}}\right) \,\mathrm{d}x\right).$$</p>
<p>I try to integrate first and then do the differentiation but it's not easy. I want to know other way to do it. Thank you.</p>
| Lubin | 17,760 | <p>I think the most useful way of looking at this situation is that the series
$$
H(x)=1+\sum_1^\infty\binom{1/2}nx^n
$$
is the Binomial Series for $(1+x)^{1/2}$, as @Hurkyl has already noted. Please note that the only denominators in the coefficients are powers of $2$: they are $p$-integral for all $p\ne2$. This means that the series is $p$-adically convergent for all values of $x$ with $\vert x\vert_p<1$.</p>
<p>I’ll omit the argument that the series is a <em>formal</em> square root of the series $(1+x)$. That is, that if we call the ring $R=\Bbb Z[1/2]$, we have a formal identity $H(x)^2=1+x$, in the formal power-series ring $R[[x]]$.</p>
<p>But granting that, you have to notice that when you evaluate $x$ to an element of $z\in p\Bbb Z_p$, you certainly get a result that is $\equiv1\pmod p$. In particular, in either $\Bbb Z_3$ or $\Bbb Z_5$, you get the result that $H(z)$ not only is a square root of $1+z$, but <strong>it’s also congruent to 1 modulo p</strong>.</p>
<p>That’s what does it: the square root of $16$ that’s $\equiv1\pmod3$ is $+4$, while the square root of $16$ that’s $\equiv1\pmod5$ is $-4$.</p>
|
1,761,857 | <p>I need to show that, for $f:X\to \mathbb{R}$ bounded, we have:</p>
<p>$$\sup\{|f(x)-f(y)|, x,y\in X\}= \sup f - \inf f$$</p>
<p>Well, I know that </p>
<p>$$\sup\{|f(x)-f(y)|, x,y\in X\}\ge |f(x)-f(y)|$$ but in what this helps? I really have no idea in how to prove this one</p>
| RRL | 148,510 | <p>We have</p>
<p>$$\inf f \leqslant f(x) \leqslant \sup f, \\ -\inf f \geqslant -f(y) \geqslant -\sup f.$$</p>
<p>Hence, </p>
<p>$$f(x) - f(y) \leqslant \sup f - \inf f.$$</p>
<p>Now interchange $x$ and $y$.</p>
<p>$$f(x) - f(y) \geqslant -(\sup f - \inf f).$$</p>
|
2,135,228 | <p>Find</p>
<p>(a) $P\{A \cup B\}$</p>
<p>(b) $P\{A^c\}$</p>
<p>(c) $P\{A^c \cap B\}$</p>
<p>This is what I have right now:</p>
<p>(a) $P\{A \cup B\}=0.4+0.5=0.90$</p>
<p>(b) $P\{A^c\}= 1-0.4=0.60$</p>
<p>(c) $P\{A^c \cap B\}= (0.6)\cdot(0.5)=0.30$</p>
<p>Am I doing it correctly?</p>
| Community | -1 | <p>When I have issues like this, I often like to completely change the variables I'm using. $u$ and $v$ are the dummy variables you use in the pointwise definition of $\oplus$? Then use dummy variables $a,b,c$ in the formulation of associativity.</p>
<p>When you're trying to compute $(a \oplus b) \oplus c$, the fact you use a completely different set of variables removes one source of difficulty in matching up the arguments of the operation to the dummy variables in the pointwise formula.</p>
|
108,200 | <p>If you are to calculate the hypotenuse of a triangle, the formula is:</p>
<p>$h = \sqrt{x^2 + y^2}$</p>
<p>If you don't have any units for the numbers, replacing x and y is pretty straightforward:
$h = \sqrt{4^2 + 6^2}$</p>
<p>But what if the numbers are in meters?<br />
$h = \sqrt{4^2m + 6^2m}$ <em>(wrong, would become $\sqrt{52m}$)</em><br />
$h = \sqrt{4m^2 + 6m^2}$ <em>(wrong, would become $\sqrt{10m^2}$)</em><br />
$h = \sqrt{(4m)^2 + (6m)^2}$ <em>(correct, would become $\sqrt{52m^2}$)</em><br /></p>
<p>Or should I just ignore the unit of measurement in these cases?</p>
| Jose Garcia | 17,341 | <p>if you know the Hamiltonian of the system then</p>
<p>$ Tr(exp(-t \Delta)\sim \int dq \int dp exp(-t\sum_{ab}p^{2}_{a}p^{2}_{b}-tV(q))$</p>
<p>the integeral is taken over ALL the p's and the q's momenta and position of the particle.</p>
<p>if the operator is a Laplacian operator then </p>
<p>$ Tr(exp(-t \Delta)\sim \int dq \int dp exp(-t\sum_{ab}p^{2}_{a}p^{2}_{b}g^{ab}(q))$</p>
<p>here the metric $ ds^{2}= \sum_{ab}g_{ab}dx_{a}^{2}dx_{b}^{2} $</p>
|
2,965,193 | <p>Basically the question is asking us to prove that given any integers <span class="math-container">$$x_1,x_2,x_3,x_4,x_5$$</span> Prove that 3 of the integers from the set above, suppose <span class="math-container">$$x_a,x_b,x_c$$</span> satisfy this equation: <span class="math-container">$$x_a^2 + x_b^2 + x_c^2 = 3k$$</span> So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?</p>
| Ricky Tensor | 583,074 | <p>Any square integer must be congruent to either 0 or 1 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: <span class="math-container">$0+0+0\equiv 0$</span> or <span class="math-container">$1+1+1\equiv0$</span> mod 3.</p>
|
2,390,670 | <p>For an array with range $n$ filled with random numbers ranging from 0 (inclusive) to $n$ (exclusive), what percent of the array contains unique numbers?</p>
<p>I was able to make a program that tries to calculate this with repeated trials and ended up with ~63.212%.</p>
<p>My Question is what equation could calculate this instead of me just repeating trials.</p>
| Henno Brandsma | 4,280 | <p>If my interpretation is correct there are $n^n$ equiprobable array fillings, of which $n!$ have all numbers occurring at most once.</p>
<p>So you get as the answer: $$\frac{n!}{n^n}$$ and <a href="http://mathworld.wolfram.com/StirlingsApproximation.html" rel="nofollow noreferrer">Stirling's approximation formula</a> says that:</p>
<p>$$n! \sim n^{n+\frac{1}{2}}\sqrt{2\pi} e^{-n}$$ This should allow one to compute the limit of this fraction as $n$ gets larger.</p>
|
4,193,578 | <p>This is maybe a very easy one, but I can't find a solution...</p>
<p>I'm looking for a sequence <span class="math-container">$a_1,...,a_n$</span> such that <span class="math-container">$0\leq a_1<\cdots<a_n<1$</span> and <span class="math-container">$\sum_{k=1}^na_k=1$</span>. Of course, this should work for any choice of <span class="math-container">$n\in\mathbb{N}$</span>.</p>
<p>My first option was a finite geometric series, but couldn't come up with the right parameters.</p>
<p>Another option is to use some discrete probability distribution like Binomial, but this does not guarantee the increasing condition.</p>
<p>Any help will be appreciated.</p>
| Alan | 175,602 | <p>Assuming you are okay with a piecewise rule, this is simply done with a modification of the geometric sequence with base <span class="math-container">$\frac 1 2$</span>. First note that this will be impossible for <span class="math-container">$n=1$</span> no matter what sequence you choose. Since you want the sequence ascending, we need to reverse the normal direction of <span class="math-container">$(\frac 1 2)^n$</span> which is decreasing.
So, for <span class="math-container">$1\leq i <n$</span> let <span class="math-container">$a_i=(\frac 1 2)^{n-i}$</span> Finally, let <span class="math-container">$a_n=a_{n-1}+a_{1}$</span>. That gets you that final jump over the Zeno's paradox problem.</p>
<p>Note that this doesn't work for <span class="math-container">$n=2$</span> since you won't have inequality, but for <span class="math-container">$n=2$</span> you can just use whatever sequence you want like <span class="math-container">$\frac 1 4, \frac 3 4$</span></p>
|
182,316 | <p>I am trying to find an example of a separable Hausdorff space which has a non-separable subspace. This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or non-separable?</p>
<p>A space is separable if it contains a countable dense subset. A subset $A$ of a space $X$ is dense in $X$ if $\bar{A}=X$.</p>
<p>It's easy to come up with a dense set and a countable set in $\mathbb{R}\setminus\mathbb{Q}$, since (trivially) $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}$ in the subspace topology, and as a countable set we can take something like $\{k\pi \mid k \in \mathbb{Z}\}$. But is there a subset that is both dense AND countable? And of course, how do we prove the result?</p>
| Asaf Karagila | 622 | <p>Every subspace of a separable metric space is separable.</p>
<p>For the irrationals, take the irrational algebraic numbers, those are dense in $\mathbb R$ and therefore in the irrationals too. As Jacob remarks below, if $\alpha$ is irrational, then $\{\alpha+q\mid q\in\mathbb Q\}$ is also dense.</p>
<p>Generally speaking, the irrationals are homeomorphic to the Baire space, the set of sequences of natural numbers, equipped with the product topology $\mathbb N^\mathbb N$. </p>
<p>This has a countable dense subset: eventually constant sequences.</p>
|
3,911,221 | <p>I am working on a probability exercice and I am trying to calculate E(Y) which comes down to this expression :</p>
<p><span class="math-container">$$ E(Y) = \int_{-∞}^{+∞} y\frac{e^{-y}}{(1+e^{-y})^{2}} \, \mathrm{d}y $$</span></p>
<p>I tried to use integrals by part but it diverges and I can't find a good change of variables.</p>
<p><strong>Any idea ?</strong></p>
| DatBoi | 734,160 | <p>Note that the integrand is an odd function<span class="math-container">$\big(f(y)=-f(-y)\big)$</span> <span class="math-container">$\to$</span>which means that integral is <span class="math-container">$0$</span></p>
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3,911,221 | <p>I am working on a probability exercice and I am trying to calculate E(Y) which comes down to this expression :</p>
<p><span class="math-container">$$ E(Y) = \int_{-∞}^{+∞} y\frac{e^{-y}}{(1+e^{-y})^{2}} \, \mathrm{d}y $$</span></p>
<p>I tried to use integrals by part but it diverges and I can't find a good change of variables.</p>
<p><strong>Any idea ?</strong></p>
| Z Ahmed | 671,540 | <p><span class="math-container">$$E=\int_{-\infty}^{\infty} y\frac{e^{-y}}{(1+e^{y})^2}dy=\int_{-\infty}^{\infty} \frac{y}{4} \text{sech}^2(y/2) dy=0.$$</span>
As the integrand is an odd function.</p>
<p>However, <span class="math-container">$$\int_{0}^{\infty} \frac{y}{4} \text{sech}^2(y/2) dy$$</span>
is convergent and finite.</p>
|
1,782,558 | <p><strong>Problem</strong></p>
<p>I have two differential equations</p>
<p>$ \frac{dx}{dt} + \frac{dy}{dt} + x + y = 0$</p>
<p>$ 2 \frac{dx}{dt} + \frac{dy}{dt} + x = 0 $</p>
<p>initial conditions: $y(0) = 1$ and $x(0) = 0$</p>
<p><strong>Attempt</strong></p>
<p>I've solved the system via the Matrix method of setting determinant to $0$, and I got $w = -1$ --> a repeated root.</p>
<p>I'm not sure why I have to use the extra $t$ factor</p>
<p>$x = (At + B) e^{-t}$ </p>
<p>$y = (Ct + D) e^{-t} $ </p>
<p>to solve these equations. </p>
<p>In addition when I tried using the extra factor, using the initial conditions, I just got $0 = A\cdot 0+ B$ and $1= C\cdot 0 + D$, which I can't use to find $C$ and $A$. I found $B$ and $D$ to be $0$ and $1$ respectively.</p>
| JJacquelin | 108,514 | <p>In a so simple case, the method with matrix appears a bit overmuch.
$$\begin{cases}
x'+y'+x+y=0 \\
2x'+y'+x=0
\end{cases} \quad\to\quad (2x'+y'+x)-(x'+y'+x+y)=0 \quad\to\quad y=x'$$
$2x'+y'+x=2x'+x''+x=0\quad\to\quad x=Ae^{-t}+Bte^{-t}\quad\to\quad y=(-A+B)e^{-t}-Bte^{-t}$</p>
<p>Conditions : $\begin{cases}
x(0)=A=0 \\
y(0)=-A+B=1\quad\to\quad B=1
\end{cases} \quad\to\quad \begin{cases}
x=te^{-t} \\
y=(1-t)e^{-t}
\end{cases}$</p>
|
664,152 | <p>I have a homework problem that I'm very stuck on. The problem statement is as follows:</p>
<p>"Suppose that $X$ is a metric space, and that for any sets $E,F \subseteq X$, if dist$(E,F) > 0$ then $\mu^*(E \cup F) = \mu^*(E) + \mu^*(F)$. Prove that every open set is a splitting set. (Recall that the distance between subsets $E$ and $F$ of a metric space is defined to be dist$(E,F) = \inf \{ d(x,y) : x \in E, y \in F \}$.)"</p>
<p>Our professor defines a "splitting set" as follows: Let $\mu^*$ be an outer measure on a nonempty set $X$. A set $A \subseteq X$ is called a splitting set if, for all $E \subseteq X$, $\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c)$, where $A^c = X \backslash A$. (This is what Folland's Real Analysis calls a $\mu^*$-measurable set.)</p>
<p>Here are a couple of my (failed) attempts:</p>
<p>My first try was to let $U$ be an arbitrary open set in $X$, let $G \subseteq X$ be arbitrary, and define $E = G \cap U$, $F = G \cap U^c$. If I could somehow show that dist$(E,F) > 0$ in this case, then the result would follow, but in general, this is not true (take $\mathbb{R}$ with the standard metric, let $G = [0,1]$, $U = (-1/2,1/2)$).</p>
<p>My next attempt was by contradiction: suppose there is an open set $U$ such that $U$ is not a splitting set. Then there is some $E \subseteq X$ such that $\mu^*(E) \neq \mu^*(E \cap U) + \mu^*(E \cap U^c)$... and by monotonicity this means that $\mu^*(E) < \mu^*(E \cap U) + \mu^*(E \cap U^c)$. But then I only have one set to work with, and with the assumptions, I need two sets $E,F$ to work with in order to get anywhere.</p>
<p>I also tried exploring what I could do with closed sets, since if dist$(E,F) > 0$, then the closures of $E$ and $F$ respectively are disjoint. But I'm still stuck. Any hints would be appreciated!!! Thanks in advance.</p>
| Quinn Culver | 11,030 | <p>Here's a strategy for making your second attempt work: since $E = (E\cap U) \cup (E \cap U^c)$, you can conclude that $d(E\cap U, E \cap U^c)=0$. Now you should be able to shrink $U$ just a little, yielding $U'$ so that $d(E \cap U', E \cap U^c)>0$ but still $\mu^{*}(E)<\mu^{*}(E\cap U')+ \mu^{*}(E\cap U^c)$.</p>
|
1,801,946 | <p>I need to find the equation of tangent line passing $(2,3)$ and perpendicular to $3x+4y=8$. Need help in this and also show me how you got the answer. I will be very thankful.</p>
| StackTD | 159,845 | <p>Your approach my depend on what you have seen about slopes and/or normal vectors and their relations with respect to being perpendicular.</p>
<p><strong>Slope approach</strong></p>
<p>A line through $(x_1,y_1)$ with slope $m$ has the following equation:
$$y = m(x-x_1)+y_1$$
If $m_1$ and $m_2$ are slopes corresponding to perpendicular lines, then $m_1m_2 = -1$. The slope of the given line is is $-3/4$, so a perpendicular line has slope $4/3$; filling in:
$$y = \tfrac{4}{3} (x-2)+3$$</p>
<p><strong>Normal vector approach</strong></p>
<p>The normal vector of a line $ax+by+c=0$ is $(a,b)$ so any line with normal vector $(b,-a)$, or any non-zero multiple of this vector, is perpendicular to it (since their dot product is zero); so you're looking for a line of the form:
$$4x-3y=C$$
Substitution of $(2,3)$ gives you $C$. Or you may know the standard form to go directly to:
$$a(x-x_1)+b(y-y_1) = 0 \longrightarrow 4(x-2)-3(y-3)=0$$</p>
|
788,245 | <p>$$\sum_{n=1}^{\infty}\frac{(n+2)!}{(3n-1)}$$ I know this series does not converge. Can someone show me how to prove that? Should i use criteria of Dalamber or any other criteria?</p>
| Zook | 135,276 | <p>The Ratio Test is usually the best way to approach factorials in series, however with this one you can just use the Test for Divergence by showing that the terms of the series do not approach $0$. In fact, the terms should all be greater than $1$, because the numerator is greater than the denominator. That's pretty straightforward with induction.</p>
<p><strong>In general</strong>, your method should be:</p>
<p>Verify that the terms approach $0$ and that the absolute value of the terms are decreasing, either by showing that $a_{n+1}<a_n$ or by showing that the derivative is negative. If not, then use the Test for Divergence.</p>
<p>If the series is alternating, use the Alternating Series Test. If the terms approach $0$ and their absolute value is decreasing, it is at least conditionally convergent.</p>
<p>Check if the series is a geometric series, a telescoping series (usually by partial fraction decomposition), or a $p$-series (of the form $\frac1{n^p}$).</p>
<p>Geometric series converge when the ratio $|r|<1$. $p$-series converge when $p>1$.</p>
<p>If the series looks like a rational function, use the Comparison Tests against a $p$-series. Pull the leading terms out and cancel them down to find the appropriate exponent to use.</p>
<p>If the Comparison Test fails in the last case, because it's greater when it should be less than or vice versa, then use the Limit Comparison Test (or just jump straight here and skip the last one).</p>
<p>If the series has $\sin(n)$ or $\cos(n)$ in it, use the Comparison Test, replacing the trig function with $\pm1$.</p>
<p>If the series has factorials in it, use the Ratio Test (also apparently known as the criteria of Dalamber).</p>
<p>If the series has an exponent around a nasty term, use the Root Test.</p>
<p>If all else fails, or if you think the integral is easy enough, use the Integral Test. That's how most of these tests are proved anyway.</p>
|
3,773,933 | <p>Question goes like this:</p>
<p>In a box containing <span class="math-container">$36$</span> strawberries, <span class="math-container">$2$</span> of them are rotten. Kyle randomly picked <span class="math-container">$5$</span> of these strawberries.<br>
a. What is the probability of having at least 1 rotten strawberry among the 5?<br>
b. How many strawberries should be picked so that the probability of having exactly <span class="math-container">$2$</span> rotten strawberries among them equals <span class="math-container">$2/35$</span>?</p>
<p>My Work:
a) <span class="math-container">$C(36,3) = 7140$</span> and <span class="math-container">$C(36,5) = 37692$</span>.
<span class="math-container">$7140 / 37692 = .0189$</span>, which is the probability of having at least <span class="math-container">$1$</span> rotten strawberry. Is this correct? If not, what did I do wrong?<br>
b) I have no clue where to start, any help would be appreciated.</p>
<p>Thank You</p>
| Robert Shore | 640,080 | <p>It's often easier to approach a problem like this backwards. Let's compute the probability that you picked no rotten strawberries.</p>
<p>There are <span class="math-container">$\binom{34}{5}$</span> ways to choose <span class="math-container">$5$</span> good strawberries and <span class="math-container">$\binom{36}{5}$</span> ways to choose any <span class="math-container">$5$</span> strawberries, so the probability that you choose <span class="math-container">$5$</span> good strawberries (after some cancellation) is <span class="math-container">$\frac{31 \cdot 30}{36 \cdot 35}= \frac{31}{42},$</span> which means the probability that you choose at least one rotten strawberry is <span class="math-container">$\frac{11}{42}$</span>.</p>
<p>Picking <span class="math-container">$n$</span> strawberries that include <span class="math-container">$2$</span> rotten strawberries means that you're picking <span class="math-container">$n-2$</span> good ones from among <span class="math-container">$34$</span> choices, so the probability that this happens is:</p>
<p><span class="math-container">$$\frac{\binom{34}{n-2}}{\binom{36}{n}}.$$</span></p>
<p>There will be cancellations that make this fraction easier to handle.</p>
|
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