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 |
|---|---|---|---|---|
3,413,837 | <p>Jerry the mouse is hungry and according to some confidential information, there is a tempting piece of cheese at the end of one of the three paths after the junction he just found himself!</p>
<p>Fortunately, Tom is standing right there and Jerry hopes he can get some useful information as to which path he must get; most importantly because Spike and Tyke, the dogs, are at the end of the other two paths!</p>
<p>The only problem is that Tom gives true and false replies in alternating order. Furthermore, he has no way of knowing which will be first, the truth or the lie!</p>
<p>He is only allowed to ask Tom 2 questions that can be answered by a “yes” and a “no”.</p>
<p>What must be the two questions he must ask?</p>
<hr>
<p>No matter how hard I tried, I can't figure out anything...
I have seen several variations for the 2 doors problem but this one is different!</p>
| Caleb Stanford | 68,107 | <p>Here's another way to do it :) <strong>Ask the following question twice:</strong></p>
<blockquote>
<p>Are you telling the truth and the cheese is in the first path, OR are you lying and the cheese is in the second path?</p>
</blockquote>
<ul>
<li><p><strong>If both answers are True,</strong> then we know that in at least one of the two rounds, Tom was telling the truth. In that round, the only way this question could be true is if Tom is telling the truth and <strong>the cheese is in the first path.</strong></p></li>
<li><p><strong>If both answers are False,</strong> then we know that in at least one of the two rounds, Tom was lying. In that round, by saying False, Tom indicates that the question is true. The only way that can be is if Tom is lying and <strong>the cheese is in the second path.</strong></p></li>
<li><p><strong>If the answer switches</strong> from True to False or from False to True, then we know that the question was either true in both rounds, or false in both rounds. It can't be true in both rounds, because the cheese can only be in one path, and Tom can't tell the truth in both rounds or lie in both rounds. Therefore, the question was false in both rounds, and that means that <strong>the cheese is in the third path.</strong></p></li>
</ul>
<hr>
<p>We can also explicitly map out the 6 possibilities:</p>
<pre><code>1TF: Cheese in 1st path, telling the truth in 1st round, lying in 2nd round
2TF: Cheese in 2nd path, """
3TF: Cheese in 3rd path, """
1FT: Cheese in 1st path, lying in 1st round, telling truth in 2nd round
2FT: Cheese in 2nd path, """
3FT: Cheese in 3rd path, """
</code></pre>
<p>In each of these cases, here are the answers to our two questions:
<span class="math-container">$$
\begin{matrix}
\text{Case} & \text{Q1} & \text{Q2} \\
\hline
1TF & T & T \\
1FT & T & T \\
\hline
2TF & F & F \\
2FT & F & F \\
\hline
3TF & F & T \\
3FT & T & F
\end{matrix}
$$</span></p>
<p>As we can see, True in both rounds means first path, False in both rounds means second path, and switching means third path.</p>
|
1,736,341 | <p>As stated in the title: Find the eigenvalues and eigenfunctions for $y''+\lambda y=0$, where $y'(1)=0$ and $y'(2)=0$.</p>
<p>So I have already eliminated the cases for $\lambda=0$ and $\lambda<0$ and I'm focused on the case for $\lambda>0$ now.
The characteristic equation for this case is $r^2+\lambda=0$, so $r=\pm i \sqrt{\lambda}$. Then the possible solution is
$$y(x)=k_1\cos(\sqrt{\lambda}t)+k_2\sin(\sqrt{\lambda}t)$$
and the derivative is
$$y'(x)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}t) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda}t)$$
Substituting the boundary values into the above equation gives:
$$y'(1)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda})=0$$
and
$$y'(2)=-k_1\sqrt{\lambda}\sin(2\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(2\sqrt{\lambda})=0$$
I have no idea how to go forward from here. All the other BVP i've done had boundary values (usually one of the BVs IS zero) that led to one of the constants equaling zero or a substitution that was obvious. I'm having trouble seeing an obvious path forward with this problem. </p>
| Dylan | 135,643 | <p>The system of equations can be rewritten as</p>
<p>$$ \alpha \left( \begin{matrix} -\sin \alpha & \cos \alpha \\ -\sin 2\alpha & \cos 2\alpha \end{matrix} \right) \left( \begin{matrix} k_1 \\ k_2 \end{matrix} \right) = 0 $$</p>
<p>Since your solution vector is in the null space, a solution will exist when the determinant of the coefficient matrix is zero</p>
<p>$$ \left| \begin{matrix} -\sin \alpha & \cos \alpha \\ -\sin 2\alpha & \cos 2\alpha \end{matrix} \right| = 0 $$</p>
<p>Or </p>
<p>$$ \sin 2\alpha \cos \alpha - \sin\alpha \cos2\alpha = \sin(2\alpha -\alpha) = \sin\alpha = 0$$</p>
<p>This gives $\alpha = \sqrt\lambda = n\pi $ as your eigenvalue, and $(\cos\alpha, \sin\alpha)$ as your eigenvector. The general solution is</p>
<p>$$ y(x) = \cos(n\pi)\cos(n\pi x) + \sin(n\pi)\sin(n\pi x) = \cos(n\pi(x-1)) $$</p>
|
249,107 | <p>Im working on my thesis about semidirect products and splitting lemma. I got the following theorems to prove and Im a not sure how to start. I would appreciate any help.</p>
<p>$\\$
1. Let $f:A\to B$ be a map.</p>
<p>Show:</p>
<p>a) if $g:B\to A$ so that $gf=id_{A}$ then $f$ is injective</p>
<p>b) if $g:B\to A$ so that $fg=id_{B}$ then $f$ is surjective</p>
<p>$\\$</p>
<ol>
<li><p>$A$, $B$, $G$ groups and there is a short exact sequence</p>
<p>$1\to A\to G\to B\to 1$</p></li>
</ol>
<p>then $\alpha :A\to G$ is injective and $\beta :G\to B$ is surjective. Please show that.</p>
| Dennis Gulko | 6,948 | <p><strong>Hints:</strong><br>
For (1)a), if $f$ is not injective, then there exist $a\neq b\in A$ such that $f(a)=f(b)$. Hence $gf(a)=gf(b)$. On the other hand, $id_A(a)=a$ and $id_A(b)=b$. Is it possible that $gf=id_A$?<br>
For (1)b), if $f$ is not surjective, then there exists $b\in B$ such that for all $a\in A$, $f(a)\neq b$. Is it possible that $fg=id_B$? (What happens to $fg(b)$? could it be $b$?)<br>
For (2): By the definition of short exact sequence, you have $\ker(\alpha)=\operatorname{Im}(1)=\{1_A\}$, $\operatorname{Im}(\alpha)=\ker(\beta)$ and $\operatorname{Im}(\beta)=\ker(1)=B$.<br>
Can you prove that $\varphi:G\to H$ is injective if and only if $\ker(\varphi)=\{1_G\}$
and is surjective if and only if $\operatorname{Im}(\varphi)=H$?</p>
|
2,217,338 | <p>I am trying to define a simple function that is first concave and then convex as shown in the picture below. Since the resulting equation have to be explained/used by a non-technical audience, the function should be ideally as simple as possible, but I have been unable to find any simple form that matches the requirements below.</p>
<p><a href="https://i.stack.imgur.com/7c1II.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7c1II.jpg" alt="enter image description here"></a></p>
<p>Any help of ideas would be greatly appreciated.</p>
<p><strong>EDIT</strong>: to further clarify</p>
<ol>
<li>The dashed red line is the constant line <span class="math-container">$Y=X$</span> </li>
<li><span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> should be ideally points that I can control within the function.</li>
</ol>
| Claude Leibovici | 82,404 | <p>For sure, Marco Cantarini provided the good and formal solution of the problem.</p>
<p>I have a modest solution based on series. Starting using $$\sin^2(ax)=\frac{1-\cos(2ax)}2=\sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2 n-1} a^{2 n} }{(2 n)!}x^{2 n}$$ we then have $$x \sin^2(ax) e^{-x^2}=\sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2 n-1} a^{2 n} }{(2 n)!}x^{2 n+1}e^{-x^2}$$ $$I=\int_0^\infty x \sin^2(ax) e^{-x^2}\,dx=\sum_{n=1}^\infty\frac{(-1)^n 2^{2 n-2} a^{2 n} \Gamma (n+1)}{(2 n)!}=-\frac{\sqrt{\pi }} 4\sum_{n=1}^\infty \frac{ (-1)^n a^{2 n}}{ \Gamma \left(n+\frac{1}{2}\right)}$$ which simplifies to $$I=\frac{\sqrt{\pi }}{4} a\, e^{-a^2} \text{erfi}(a)=\frac{a F(a)}{2}$$</p>
|
691,112 | <p>Suppose $A$ and $B$ are finite sets and $f:A\rightarrow B$. Prove that if $|A|>|B|$, then $f$ is not one-to-one.</p>
<p>Scratch work:</p>
<p>Since the goal is in negation, I try to prove it by contradiction and assume that $f$ is one-to-one. Since $A$ has more elements than $B$, it can't be the case that $f$ is one-to-one because some $a\in A$ has to share images with other. But other than the false assumption $f$ is one-to-one, I have no other clue to proceed with the question. What technique should I apply? Please give hints and guidance. Thanks in advance.</p>
| Dave Clifford | 82,405 | <p>Let $f(A)$ be the set of images. Then $f(A)\subseteq B$ so $|f(A)|\le |B|$.</p>
<p>If $f$ is one-to-one then $|f(A)|=|A|$.</p>
|
466,722 | <blockquote>
<p>Find a function $f$ and a number $a$ such that:
$$
6+\int_{a}^{x}\frac{f(t)}{t^2}\:\mathrm{d}t=2\sqrt{x}
$$
For all $x>0$</p>
</blockquote>
<p>From Fundamental Theorem of Calculus section. Having some trouble with this. Any help?</p>
| what'sup | 87,411 | <p>$$ \int_a^x \frac{f(t)}{t^2} \ dt = 2\sqrt{x} - 6 $$</p>
<p>$$\mathrm{ differentiate \ using \ leibniz \ rule } $$</p>
<p>$$\frac{f(x)}{x^2} = \frac{1}{\sqrt{x} } \Rightarrow f(x) = x\sqrt{x}$$</p>
<p>$$\int_a^x \frac{\sqrt{t}}{t} \ dt = 2\sqrt{x} - 6 $$</p>
<p>$$\int_a^x t^{-\frac{1}{2}} \ dt = \left | 2\sqrt{t} \right|_a^x = 2\sqrt{x} - 2\sqrt{a} = 2\sqrt{x} - 6$$</p>
<p>$$ \Rightarrow 2\sqrt{a} = 6 \Rightarrow a = 9 $$</p>
|
1,657,664 | <p>Struggling with a homework problem here and can't understand logically which one would be correct (each has different truth tables). I need to express the following statement using quantifiers, variables, and the predicates M(s), C(s), and E(s) </p>
<blockquote>
<p>"No computer science students are engineering students" </p>
</blockquote>
<p>D = set of all students</p>
<p>C(s) = "s is a computer science major"</p>
<p>E(s) = "s is an engineering student" </p>
<p>So I'm stuck between,</p>
<p>$\forall s \in D, C(s) \implies \lnot E(s)$</p>
<p>-OR-</p>
<p>$\forall s \in D, \lnot C(s) \land E(s)$ </p>
| Q the Platypus | 264,438 | <p>The first example says "All computer students are not engineering students". People who are not computer students are free to be maths students or not be maths students.</p>
<p>The second one says "All students are not computer students and they have to be engineering students".</p>
|
2,331,191 | <p>Use either direct proof, proof by contrapositive, or proof by contradiction.</p>
<p>Using proof by contradiction method</p>
<blockquote>
<p>Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)</p>
<p>There exists integers x and y such that <span class="math-container">$n = x^2$</span> and <span class="math-container">$n+3 = y^2$</span></p>
<p>Then <span class="math-container">$x^2 + 3 = y^2$</span></p>
<p>Then <span class="math-container">$3 = y^2 - x^2$</span></p>
<p>Then <span class="math-container">$3 = (y-x)(y+x)$</span></p>
<p>Then <span class="math-container">$y+x = 3$</span> and <span class="math-container">$y-x=1$</span></p>
<p>Then <span class="math-container">$x = 1, y = 2$</span></p>
<p>Since <span class="math-container">$x = 1$</span>, that implies <span class="math-container">$n = 1$</span></p>
<p><em><strong>this is how far I got</strong></em></p>
</blockquote>
<p>Anyone know what I should do now?</p>
| David G. Stork | 210,401 | <p>Let $n=4$ (a perfect square). Note that $n+3 = 7$ is not a perfect square.</p>
<p>Is there anything more to this problem than that?</p>
<p>(Of course if $n=1$ (a perfect square) then $n+3 = 4$ happens to be a perfect square.)</p>
|
3,356,341 | <blockquote>
<p>Let <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span> ∈ <span class="math-container">$\mathbb Z$</span>, Show:</p>
<p><span class="math-container">$a_1 \mathbb Z · a_2 \mathbb Z$</span> = <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span></p>
<p>where <span class="math-container">$c\mathbb Z = \{c · n : n ∈ \mathbb Z\}$</span></p>
</blockquote>
<p>My attempt:</p>
<p>Let <span class="math-container">$x$</span> ∈ <span class="math-container">$a_1 \mathbb Z$</span> then <span class="math-container">$x$</span> can be written as <span class="math-container">$x = a_1.n$</span> for some <span class="math-container">$n$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p>Let <span class="math-container">$y$</span> ∈ <span class="math-container">$a_2 \mathbb Z$</span> then <span class="math-container">$y$</span> can be written as <span class="math-container">$y = a_2.m$</span> for some <span class="math-container">$m$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p>Let <span class="math-container">$z$</span> ∈ <span class="math-container">$(a_1.a_2) \mathbb Z$</span> then <span class="math-container">$z$</span> can be written as <span class="math-container">$z = (a_1.a_2).l$</span> for some <span class="math-container">$l$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p><span class="math-container">$x.y = a_1.n . a_2.m$</span> = <span class="math-container">$a_1. a_2. n. m$</span> since multiplication is commutative in <span class="math-container">$\mathbb Z$</span></p>
<p>But since <span class="math-container">$l$</span> ∈ <span class="math-container">$\mathbb Z$</span>, it can be written as a linear combination, say <span class="math-container">$l = n.m$</span> so <span class="math-container">$z = (a_1.a_2).l = a_1. a_2. n. m = x.y$</span>, and hence <span class="math-container">$x.y=z$</span> and thus <span class="math-container">$x.y$</span> ∈ <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span>, and therefore <span class="math-container">$a_1 \mathbb Z · a_2 \mathbb Z$</span> = <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span></p>
<p>Is my attempt correct?</p>
| Theo Bendit | 248,286 | <p>The proof is lacking a bit of clarity in its presentation. You start by defining an arbitrary element, <span class="math-container">$x \in a_1 \Bbb{Z}$</span>, then an unrelated arbitrary element <span class="math-container">$y \in a_2 \Bbb{Z}$</span>, and finally, a third arbitrary element <span class="math-container">$z \in (a_1 a_2) \Bbb{Z}$</span>, seemingly unrelated to both <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. In the final paragraph, you seem to change <span class="math-container">$z$</span> to now relate to <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, and in particular, so that it simplifies to <span class="math-container">$xy$</span>.</p>
<p>It would be better if you declared <span class="math-container">$x$</span> and <span class="math-container">$y$</span> (and deduced the existence of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> as you have), then let <span class="math-container">$z = (a_1a_2)(mn)$</span>. You can conclude that <span class="math-container">$z \in (a_1a_2)\Bbb{Z}$</span>, and that <span class="math-container">$z = xy$</span>. This means <span class="math-container">$xy \in (a_1a_2)\Bbb{Z}$</span>, as required.</p>
<p>This way, you don't have to pretend <span class="math-container">$z$</span> is arbitrary initially, and you only need to define it once it becomes useful to your proof. You could even skip defining <span class="math-container">$z$</span>, and note that <span class="math-container">$xy = (a_1a_2)(mn) \in (a_1a_2)\Bbb{Z}$</span>.</p>
<p>Also, bear in mind that all you've proven is <span class="math-container">$(a_1 \Bbb{Z})(a_2 \Bbb{Z}) \subseteq (a_1a_2) \Bbb{Z}$</span>. You also need a separate argument for the converse direction. That is, you need to be able to start with some arbitrary <span class="math-container">$z \in (a_1a_2) \Bbb{Z}$</span> and construct elements of <span class="math-container">$x \in a_1 \Bbb{Z}$</span> and <span class="math-container">$y \in a_2 \Bbb{Z}$</span>, such that <span class="math-container">$z = xy$</span>. Hint: <span class="math-container">$1$</span> is a perfectly fine integer.</p>
|
3,356,341 | <blockquote>
<p>Let <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span> ∈ <span class="math-container">$\mathbb Z$</span>, Show:</p>
<p><span class="math-container">$a_1 \mathbb Z · a_2 \mathbb Z$</span> = <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span></p>
<p>where <span class="math-container">$c\mathbb Z = \{c · n : n ∈ \mathbb Z\}$</span></p>
</blockquote>
<p>My attempt:</p>
<p>Let <span class="math-container">$x$</span> ∈ <span class="math-container">$a_1 \mathbb Z$</span> then <span class="math-container">$x$</span> can be written as <span class="math-container">$x = a_1.n$</span> for some <span class="math-container">$n$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p>Let <span class="math-container">$y$</span> ∈ <span class="math-container">$a_2 \mathbb Z$</span> then <span class="math-container">$y$</span> can be written as <span class="math-container">$y = a_2.m$</span> for some <span class="math-container">$m$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p>Let <span class="math-container">$z$</span> ∈ <span class="math-container">$(a_1.a_2) \mathbb Z$</span> then <span class="math-container">$z$</span> can be written as <span class="math-container">$z = (a_1.a_2).l$</span> for some <span class="math-container">$l$</span> ∈ <span class="math-container">$\mathbb Z$</span></p>
<p><span class="math-container">$x.y = a_1.n . a_2.m$</span> = <span class="math-container">$a_1. a_2. n. m$</span> since multiplication is commutative in <span class="math-container">$\mathbb Z$</span></p>
<p>But since <span class="math-container">$l$</span> ∈ <span class="math-container">$\mathbb Z$</span>, it can be written as a linear combination, say <span class="math-container">$l = n.m$</span> so <span class="math-container">$z = (a_1.a_2).l = a_1. a_2. n. m = x.y$</span>, and hence <span class="math-container">$x.y=z$</span> and thus <span class="math-container">$x.y$</span> ∈ <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span>, and therefore <span class="math-container">$a_1 \mathbb Z · a_2 \mathbb Z$</span> = <span class="math-container">$(a_1 · a_2)$</span> · <span class="math-container">$\mathbb Z$</span></p>
<p>Is my attempt correct?</p>
| Henno Brandsma | 4,280 | <p>Show two inclusions in a systematic way:</p>
<p>Let <span class="math-container">$z \in a \Bbb Z \cdot b \Bbb Z$</span>. This means that we can write <span class="math-container">$z=xy$</span> for some <span class="math-container">$x \in a \Bbb Z $</span> and <span class="math-container">$y \in \Bbb Z $</span>. By definition we then know there is some <span class="math-container">$n \in \Bbb Z $</span> and some <span class="math-container">$m \in \Bbb Z $</span> such that <span class="math-container">$x=an$</span> and <span class="math-container">$y=bm$</span>.
Then we can conclude that (using standard commutative ring properties of <span class="math-container">$\Bbb Z $</span>)</p>
<p><span class="math-container">$$z=xy=(an)(bm)=(ab)(nm)$$</span> and because <span class="math-container">$nm \in \Bbb Z $</span> we conclude that <span class="math-container">$z \in (ab) \Bbb Z $</span>, showing one inclusion.</p>
<p>Now let <span class="math-container">$z \in (ab) \Bbb Z $</span> be given. By definition this means that <span class="math-container">$z=(ab)n$</span> for some <span class="math-container">$n \in \Bbb Z $</span>. Again using simple ring facts:</p>
<p><span class="math-container">$$z=(ab)n= (an)(b1) \in a \Bbb Z \cdot b \Bbb Z $$</span></p>
<p>where the last step is by definitions again: <span class="math-container">$an \in a \Bbb Z $</span>, <span class="math-container">$b1 \in b \Bbb Z $</span> and so their product is in the product of the sets.</p>
<p>Both inclusions being shown, equality is shown too. The only "clever(ish)" bit is that we use <span class="math-container">$1$</span> as an integer in the reverse inclusion. The rest should write itself.</p>
|
51,757 | <p>I'm trying to find closed form for</p>
<p>$$\sum_{k=1}^{n}\sin\frac{1}{k}$$</p>
<p>I typed it in Mathematica 6.0 and WolframAlpha, but no result what i expected.</p>
<p>Any hints will be appreciated, thank you.</p>
| Henry | 6,460 | <p>I doubt you will find a closed form. </p>
<p>Your expression will give a value slightly less than than the <a href="http://en.wikipedia.org/wiki/Harmonic_number" rel="nofollow">harmonic numbers</a> $\sum_{k=1}^{n}\frac{1}{k}$ which do not have a closed form and as $k$ increases the difference will increase towards $0.191899\ldots$, and a value slightly more than $\log n$ and as $k$ increases the difference will fall towards $0.385316\ldots$. </p>
|
3,815,640 | <p>what is the most efficient way to calculate the argument of
<span class="math-container">$$
\frac{e^{i5\pi/6}-e^{-i\pi/3}}{e^{i\pi/2}-e^{-i\pi/3}}
$$</span> without calculator ?</p>
<p>i tried to use <span class="math-container">$\arg z_1-\arg z_2$</span> but the argument of <span class="math-container">$e^{i5\pi/6}-e^{-i\pi/3}$</span> take some time .
is there a formula to calculate the argument of that kind of complex numbers ?</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>multiply the numerator and the denominator by <span class="math-container">$$e^{i\frac{\pi}{3}}$$</span></p>
<p>and use the fact that</p>
<p><span class="math-container">$$\arg(1-e^{i\theta})=\frac{\theta}{2}-\frac{\pi}{2}$$</span></p>
|
3,974,394 | <p>I am relatively new to isomorphisms and I don't understand how <span class="math-container">$\varphi$</span> is surjective in this proof. I have searched online, but I still don't understand. If anyone could straight up tell me because I feel like I'm being a bit dumb.</p>
<p><a href="https://i.stack.imgur.com/oIB4A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oIB4A.png" alt="enter image description here" /></a></p>
| PrincessEev | 597,568 | <p>You are mapping</p>
<p><span class="math-container">$$g := g_1 g_2 g_3 \cdots g_n \stackrel{\varphi}{\mapsto} (g_1,g_2,g_3,\cdots,g_n)$$</span></p>
<p>where <span class="math-container">$g_i \in G_i$</span> <span class="math-container">$\forall i \in \{1,\cdots,n\}$</span>. Every element of <span class="math-container">$G_1 \times \cdots \times G_n$</span> thus has a preimge, namely the product of the coordinates of the element of the product.</p>
|
2,340,487 | <p>I was trying to compute this limit:
$$\lim_{x \to 0}\lim_{y \to 0} (x+y)\sin{\frac{x}{y}}$$</p>
<p>And this is my solution:
$$\lim_{x \to 0}\lim_{y \to 0}|(x+y)\sin{\frac{x}{y}}|\leq\lim_{x \to 0}\lim_{y \to 0} |(x+y)|=0$$</p>
<p>So I got the limit 0.</p>
<p>The answer was different. I have no idea what is wrong with my solution?</p>
| Paramanand Singh | 72,031 | <p>You are intuitively trying to use a generalization of the following result from calculus of single variable : if $|f(x) |\leq g(x) $ as $x\to a$ and $g(x) \to 0$ as $x\to a$ then $f(x) \to 0$ as $x\to a$. This holds for limits of functions of two variables also. But you are dealing with an iterated limit and not a double limit.</p>
<p>A double limit always involves both variables moving together towards the point under consideration. An iterated limit does not work in this fashion and it is possible that double limit exists but the iterated limits fail to exist.</p>
<p>This is what happens in the current question. The double limit of the function is $(x+y) \sin(x/y) $ is $0$ precisely because of the inequality you have used. But the iterated limit $\lim_{x\to 0}\lim_{y\to 0}$ does not exist. The inequality $|(x+y) \sin(x/y) |\leq |x+y| $ is fine but when you take limits as $y\to 0$ then you can see that the RHS tends to $|x|$ which may or may not be $0$. And hence there is no guarantee that the limit of LHS exists. If you change the function to $xy\sin(x/y) $ then you don't face this problem and both the iterated limits as well as the double limit are $0$.</p>
<p>Iterated limits are nothing but single variable limits applied one after another. And their theory is much simpler. The confusion here comes because you are trying to use iterated limits but side by side also trying not to treat the variables independently. </p>
<hr>
<p><strong>Note</strong>: The definition of double limit assumes that the point under consideration (here $(0,0)$) is a point of accumulation of the domain of the function.</p>
|
4,458,863 | <p>Let <span class="math-container">$z_1,\;z_2,\;z_3\;$</span> be complex number such that <span class="math-container">$|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2\;\;$</span>. If <span class="math-container">$|z_1-z_3|=|z_1-z_2|\; \;$</span> and <span class="math-container">$z_2 \neq z_3.\; \; $</span> Then Find value of <span class="math-container">$|z_1+z_2||z_1+z_3|$</span>.</p>
<p><strong>My Thinking:</strong></p>
<p>All I could think is that <span class="math-container">$z_1, z_2, z_3 \;$</span> lies on a circle of radius <span class="math-container">$2$</span> with origin as center. Can anyone help me in how to process further.</p>
| nmasanta | 623,924 | <p><span class="math-container">$$I=\int \sqrt{\cosh(x)}~ dx=\int \sqrt{2\cosh^2\left(\frac x2\right)-1} ~dx=\int \sqrt{2\sinh^2\left(\frac x2\right)+1} ~dx$$</span>
Putting, <span class="math-container">$u=\dfrac{ix}{2}\implies du=\dfrac{i}{2}~dx$</span>,
<span class="math-container">$$I=-2i\int \sqrt{1-2\sin^2 u}~du=-2i~E(u|2)+c=-2i E\left(\frac {ix}{2}\bigg| 2\right)+c$$</span> where <span class="math-container">$~c~$</span> in integrating constant.</p>
<hr />
<p><strong>Note:</strong> Here <span class="math-container">$E(\phi|k^2)$</span> is called the <a href="https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind" rel="nofollow noreferrer">incomplete elliptic integral of the second kind</a>.</p>
<p>In trigonometric form,
<span class="math-container">$$E(\phi|k^2)=\int \sqrt{1-k^2\sin^2 \phi}~~d\phi $$</span></p>
<p>For more details about elliptic integral, you may visit the following links:</p>
<p><a href="https://en.wikipedia.org/wiki/Elliptic_integral" rel="nofollow noreferrer">Elliptic integral</a><br>
<a href="https://users.mai.liu.se/vlatk48/teaching/lect2-agm.pdf" rel="nofollow noreferrer">Elliptic functions: Introduction course</a><br>
<a href="https://www.pdfdrive.com/elliptic-functions-and-elliptic-integrals-e158364095.html" rel="nofollow noreferrer">Elliptic Functions and Elliptic Integrals</a> by Viktor Vasilʹevich Prasolov<br>
<a href="https://link.springer.com/book/10.1007/978-3-642-52803-3" rel="nofollow noreferrer">Handbook of Elliptic Integrals for Engineers and Physicists</a> by Paul F. Byrd & Morris D. Friedman<br>
<a href="https://www.gutenberg.org/files/31076/31076-pdf.pdf" rel="nofollow noreferrer">Elliptic Functions: An Elementary Text-Book for Students of Mathematics</a> by Arthur L. Baker</p>
|
649,379 | <p>I'm on the final part of my project, where I have to prove the Noether-Lasker Theorem (or copy out the following proof and "fill in the gaps"). I'm looking for an explanation of what's going on at a macro-level. I think I could follow the proof, but I don't understand how it proves what it says it proves. I've already proved that every ideal is the intersection of finitely many primary ideals, and somehow the following proves that every ideal is the reduced intersection whose radicals are unique. I feel like if I understood the "macro logic" then I'd be away. Here's the link:</p>
<p><a href="https://dl.dropboxusercontent.com/u/17606191/noether.gif" rel="nofollow">https://dl.dropboxusercontent.com/u/17606191/noether.gif</a></p>
<p>Thanks for any replies!</p>
| zcn | 115,654 | <p>If you're looking for another way to see that the radicals of the primary ideals in a primary decomposition are unique, here is a basic fact which should make the connection to associated primes clearer (this is Lemma 4.4 in Atiyah-MacDonald):
$\newcommand{q}{\mathfrak{q}}$
$\newcommand{p}{\mathfrak{p}}$</p>
<p><strong>Lemma:</strong> Let $\q$ be $\p$-primary. For any $x \in A$, we have:<br>
i) If $x \in \q$, then $(\q : x) = 1$<br>
ii) If $x \not \in \q$, then $(\q : x)$ is $\p$-primary (so $r(\q : x) = \p$)<br>
iii) If $x \not \in \p$, then $(\q : x) = \q$</p>
<p>This lemma follows directly from the definitions of radical and primaryness: if you have not done it, I would advise working it out. Once you have this though, and existence of primary decompositions, uniqueness follows. For any ideal $I$, write $I = \cap_i \q_i$ as a reduced intersection of primary ideals, and write $\p_i = r(\q_i)$ for their radicals. On the other hand, the associated primes of $I$, $\text{Ass}(R/I)$, are the prime ideals appearing in the set $\{(I : x) \mid x \in A\}$, and by definition depends only on $I$. Then for any $x$,</p>
<p>$$(I : x) = (\cap_i \q_i : x) = \cap_i (q_i : x) = \cap_{x \not \in \q_i} (q_i : x)$$</p>
<p>Thus $(I : x)$ is prime iff $(I : x) = (\q_i : x)$ is prime for some $x \not \in \q_i$ (since a prime ideal that is a finite intersection of ideals must be one of them), but then $(\q_i : x) = r(\q_i : x) = \p_i$, so every associated prime is one of the $\p_i$'s. Conversely, since the decomposition is reduced, for each $i$ there exists $x_i \not \in \q_i$, $x_i \in \cap_{j \ne i} \q_j$, so $(I : x_i) = (\q_i : x_i)$ and $r(I : x_i) = \p_i$ is prime, so each $\p_i$ is an associated prime.</p>
<p>(Note: the approach I have chosen here differs slightly from that of Atiyah-MacDonald: they define $\text{Ass}(R/I)$ as the prime ideals appearing in the set $\{r(I : x) \mid x \in A\}$. In a Noetherian ring though, it turns out a prime ideal of the form $r(I : x)$ is also of the form $(I : y)$, so the two notions coincidence in the Noetherian case).</p>
|
367,643 | <p>A Norman window has the shape of a rectangle with a semi circle on top; diameter of the semicircle exactly matches the width of the rectangle. Find the dimensions of the Norman window whose perimeter is 300 in that has maximal area.</p>
<p><a href="https://i.stack.imgur.com/WBPm7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WBPm7.png" alt="enter image description here"></a></p>
<p>The area of the semicircle would be <span class="math-container">$(\pi(w/2)^2)/2$</span>. The area of the rectangle would be <span class="math-container">$hw$</span>. I know that the perimeter is 300 in, and that the perimeter would be <span class="math-container">$2h+w+(w\pi) = 300$</span>. How would I write <span class="math-container">$h$</span> in terms of <span class="math-container">$w$</span>, and then solve for <span class="math-container">$h$</span> to specify the dimensions? The total area would be the 2 sub-areas added together. I would have to take the derivative of the combined areas to solve for the width and height. What would be the proper steps for doing this?</p>
| AnilB | 64,096 | <p>Area of the window
$$A=\frac{\pi }{8}w^2+wh$$
and the perimeter
$$P=2\, h+w+\frac12\pi\, w$$
The problem is
$$max_{w,h}\ A\qquad s.t.\, P=300$$
Using perimeter constraint you can eliminate for h
$$h=\frac{300- w\, (1+\frac12\pi)}{2}$$
The area is then
$$A=\frac{\pi }{8}w^2+150\,w-\frac{1+\frac12\pi}{2}w^2=-\frac{4+\pi}{8}w^2+150\,w$$
By taking the derivative and setting to zero
$$\frac{dA}{dw}=-\frac{4+\pi}{4}w+150=0\Rightarrow w=\frac{600}{4+\pi}\approx 84$$
and
$$h=\frac{300- \frac{600}{4+\pi}\, (1+\frac 12\pi)}{2}\approx 42$$</p>
|
184,219 | <p>I know that it is the standard functionality of <code>Merge</code> to combine the values of the same keys among associations.</p>
<p>Now I would like to deal with a situation in which, in my associations, the keys are strings (English words). And I want to define the sameness as two words having the same result from <code>WordStem</code> so that "effects" and "effect" are the same key.</p>
<p>So is it possible for <code>Merge</code> to accept such a same test function for the keys, e.g., <code>Equal@@WordStem[{##}]&</code>, to determine what keys should be considered the same and be merged?</p>
<hr>
<p>A MWE goes as below:</p>
<pre><code>mergeFunc = x \[Function] {Total[#], Union[Flatten@#2]} & @@ (x\[Transpose]);
Merge[{<|"effect" -> {5, {2, 3}}|>, <|"effects" -> {4, {1, 3, 5}}|>}, mergeFunc]
Merge[{<|"effect" -> {5, {2, 3}}|>, <|"effect" -> {4, {1, 3, 5}}|>}, mergeFunc]
</code></pre>
<blockquote>
<pre><code><|"effect" -> {5, {2, 3}}, "effects" -> {4, {1, 3, 5}}|>
<|"effect" -> {9, {1, 2, 3, 5}}|>
</code></pre>
</blockquote>
<p>The second result is desired.</p>
<hr>
<p><strong>Additional requirement</strong>: no new keys should be introduced.</p>
<p>For example, what if I now have "apple" and "apples" as the keys instead of "effect" and "effects"? </p>
<p>That means I need the word stems as the criterion but not as the keys in the merged result.</p>
| kglr | 125 | <pre><code> data = {<|"effect" -> {5, {2, 3}}|>, <|"effects" -> {4, {1, 3, 5}}|>};
Merge[KeyMap[WordStem]/@ data, mergeFunc]
</code></pre>
<blockquote>
<p><|"effect" -> {9, {1, 2, 3, 5}}|></p>
</blockquote>
<p>Also</p>
<pre><code> GroupBy[ data, First @* WordStem @* Keys -> First, mergeFunc]
</code></pre>
<blockquote>
<p> <|"effect" -> {9, {1, 2, 3, 5}}|> </p>
</blockquote>
|
802,014 | <p>For all sets $A$, $B$, $C$, if $A$ is subset of $B$, $B$ is subset of $C$, and $C$ is subset of $A$, then $A = B = C$.</p>
<p>This is a true statement and I need to provide a proof? Thus, when a statement is false I need to provide it with counterexample whereas if it is true then it has to be provided by a proof?</p>
| user152102 | 152,102 | <p>We can prove it with set size. Let $\#A$ be the size (order) of $A$, so if $\#A=n$ holds, then $A$ has $n$ elements. Then $A$ is a subset of $B$, so you get $\#A\leq\#B$. Then you get $$\#A\leq\#B\leq\#C\leq\#A. $$ So you get $\#A\leq\#A$, and it is trivial that only $\#A=\#A$ is possible. So you get $$\#A=\#B=\#C=\#A.$$ And if one set is a subset of the other, and the two sets have the same size, then they have to be the same. Otherwise there is an element in the subset which is not in the original set. But then the subset is not a subset. So $A$, $B$ and $C$ are equal. </p>
|
13,030 | <p>At work, we were discussing when is it the best time to change to winter tires for bikes and/or cars.</p>
<p>Using <code>WeatherData[]</code> and <code>DateListPlot[]</code>, it was fairly straightforward for me to create the diagram below:</p>
<p><img src="https://i.stack.imgur.com/Y5wNT.png" alt="Mean temperature per day in Stockholm"> </p>
<p><em>Fig 1 Mean temperature per day in Stockholm, red is negative and a risk without winter tires.</em></p>
<p>The code for this is</p>
<pre><code>cityTemp = WeatherData["Stockholm", "MeanTemperature", {{1977, 1, 1}, {2011, 12, 31}, "Day"}];
iceRiskDays = Select[cityTemp, Last[#] < 0 &];
yearStrip[ dataItem_] := {{0, Part[First[dataItem], 2], Last[First[dataItem]]}, Last[dataItem]}
DateListPlot[{yearStrip[#] & /@ cityTemp, yearStrip[#] & /@ iceRiskDays} ]
</code></pre>
<p>My question is: <strong>How do I calculate for each day the proportion of values for that day that are below 0 Celsius?</strong> (e.g. for a date at the end of November, the proportion is likely to be bigger than 0.5)?</p>
<p>My attempts to do this ended with trying to create separate lists for each date, but I felt that this was a less elegant way and also creates less "fit" with DateListPlot.</p>
| Murta | 2,266 | <p>Nice question. First I wrote your code in this way. Where we don't need the function <code>yearStrip</code></p>
<pre><code>{date,year,month,day,temp}={1,1,2,3,2}
cityTemp=WeatherData["Stockholm","MeanTemperature",{{1977,1,1},{2011,12,31},"Day"}];
cityTemp[[All,date,year]]=0
iceRiskDays=Select[cityTemp,#[[temp]]<0&];
DateListPlot[{cityTemp,iceRiskDays}]
</code></pre>
<p>Where we get the same result:</p>
<p><img src="https://i.stack.imgur.com/xovZz.png" alt="enter image description here"></p>
<p>Now we can group the day-month information using <code>GatherBy</code> in this way:</p>
<pre><code>dataGather=GatherBy[cityTemp,#[[date,{month,day}]]&];
getNegativeProportion[data_]:=N@Total[UnitStep[-data]]/Length[data]
negativeProportionList=Sort[{#[[1,1]],getNegativeProportion[#[[All,temp]]]}&/@dataGather];
movAvg=MovingAverage[{N@AbsoluteTime@#[[date]],#[[temp]]}&/@negativeProportionList,14];
DateListPlot[{negativeProportionList,movAvg},Joined-> True,PlotStyle-> {Blue,Red}]
</code></pre>
<p>Where we get:</p>
<p><img src="https://i.stack.imgur.com/mOqu1.png" alt="enter image description here"></p>
<p>With the red line as the moving average for 14 days!</p>
|
4,072,386 | <p>I assume this is a simple proof but i'm stuck here.</p>
<p>I need to prove that if <span class="math-container">$A^3B-B$</span> is invertible then <span class="math-container">$BA-B$</span> is invertible.</p>
<p>So <span class="math-container">$A^3B-B=(A^3-I)B$</span> and then both <span class="math-container">$(A^3-I)$</span> and <span class="math-container">$B$</span> are invertible.</p>
<p>So I need to show that <span class="math-container">$BA-B=B(A-I)$</span> are both invertible.</p>
<p>I thought maybe I could somehow show that if <span class="math-container">$(A^3-I)$</span> is invertible then <span class="math-container">$(A-I)$</span> is invertible but I guess there's a simpler way to get there. What am I missing?</p>
<p>Thanks!</p>
| Robert Lewis | 67,071 | <p>Note that</p>
<p><span class="math-container">$A^3B - B = (A^3 - I)B; \tag 1$</span></p>
<p>thus the invertibility of <span class="math-container">$A^3B - B$</span> implies both <span class="math-container">$A^3 - I$</span> and <span class="math-container">$B$</span> are invertible. Since</p>
<p><span class="math-container">$A^3 - I = (A - I)(A^2 + A + I), \tag 2$</span></p>
<p>the invertibility of <span class="math-container">$A^3 - I$</span> yields the same for <span class="math-container">$A - I$</span> and
<span class="math-container">$A^2 + A + I$</span>. Since now <span class="math-container">$A - I$</span> and <span class="math-container">$B$</span> are invertible, so is their product</p>
<p><span class="math-container">$B(A - I) = BA - B. \tag 3$</span></p>
|
126,074 | <p>I would like to know some applications of Anick's resolution in non-commutative algebras.</p>
| Ronnie Brown | 19,949 | <p>In view of Yemon's reference to group cohomology, I would like to mention Graham Ellis' work on <a href="http://hamilton.nuigalway.ie/" rel="nofollow">"Homological Algebra Programming"</a>. The key point is that he constructs free resolutions inductively together with a contracting homotopy: it is the latter that gives the computational aspect. </p>
<p>There is an explanation of some of this in Section 9.3 of the book <a href="http://pages.bangor.ac.uk/~mas010/nonab-a-t.html" rel="nofollow">Nonabelian algebraic topology</a>, in terms of constructing a "home for a contracting homotopy", as against the more traditional "killing kernels", a method which is notably non algorithmic. </p>
<p>The spirit of this derives from <a href="http://ncatlab.org/nlab/show/homological+perturbation+theory" rel="nofollow">Homological Perturbation Theory</a>, in which also the homotopies are crucial. </p>
|
1,390,382 | <p>I have a problem that comes from absorbing random walks on a connected undirected graph $G$ with two types of nodes, absorbing nodes and free nodes. We randomly pick a node to start, once the random walk reaches an absorbing node, it will never leave the node again. But if we are at a free node, we will pick an outgoing edge with probability proportional to the edge weight and go to one of the neighbouring nodes. With proper labelling of the nodes, the transition matrix $T(G)$ can be written as $$T = \begin{pmatrix} T_{aa} & T_{af} \\ T_{fa} & T_{ff} \end{pmatrix}, $$ where the matrix $T_{aa}$ corresponds to the block of probabilities corresponding to transitions from an absorbing node to another absorbing node, and so on. For the computation of the overall limiting distribution, I need to show that $\lim_{n \rightarrow \infty}{T_{ff}^{n} = 0}$ so that I can have a nice looking result $$\lim_{n \rightarrow \infty}{T^n} = \begin{pmatrix} I & 0 \\ (I-T_{ff})^{-1}T_{fa} & 0 \end{pmatrix}$$ </p>
<p>I now believe that the limit is indeed zero and I think the easiest way is probably proving that all eigenvalues have absolute value strictly smaller than 1. And I can already show that the absolute values of all eigenvalues are no larger than 1. Can somebody help me to prove that there cannot be an eigenvalue whose absolute value is 1? Also, once we know no eigenvalue of $T_{ff}$ is 1, we would immediately have the fact that $I-T_{ff}$ is invertible. Other approaches are also welcome. Thanks.</p>
| Aiden | 96,554 | <p>Just come up with an argument from another direction.</p>
<p>Let the number of columns in $T_{ff}$ be $k$.
Suppose, to derive a contradiction, that for any given $i$, $0\le i\le k$, the sum of entries of $i$-th row in $T_{ff}^{n}$ is positive when $n$ goes to infinity. And let us agree for the time being that the limit exists since the absolute values of all eigenvalues are no larger than 1. In other words, we are trying to derive a contradiction starting from the assumption that for any given vertex $v_i$, there is a random walk starting $v_i$, such that with a positive probability, it will never terminate. Since the graph is connected, there are some node(s) that have a absorbing neighbour. We denote by $B$ the set of nodes that are connected to at least one absorbing nodes. Let $B = \{ b_1, b_2, ..., b_k \}.$ At any $b_i$, let $p_i$ be the probability of the random walk staying within the subgraph $H$ that is induced by the free nodes. Since $p_i<1$, for a random walk of arbitrary length $n$, if it arrives at some node in $B$ for infinite number of times as $n$ goes to infinity, the probability it stays within $H$ will go to $0$. Therefore, all vertices in $B$ are visited only a finite number of times in order for the ith row in $T_{ff}^{n}$ to sum up to a positive value. When $n$ goes beyond a certain threshold, the nodes in $B$ will no longer be visited. Inductively, we can treat the nodes in $B$ as absorbing and apply the argument above. Eventually, we will be confined at a free node and have no where to go, which contradicts to the assumption. Thus the limit of $T_{ff}^{n}$ goes to the zero matrix.</p>
|
501,678 | <p>Let $f(x)=x-\cos(x)$. Find all points on the graph of $y=f(x)$ where the tangent line has slope 1. (In each answer $n$ varies among all integers).</p>
<p>So far I've used the Sum derivative rule for which I have $1+\sin(x)$. So do I put in 1 in for $x$ for sin$(x)$.</p>
<p>Please Help!!</p>
| Community | -1 | <p>Let $y = \dfrac1x$. We then have
$$L = \lim_{x \to 0^+} \left(\sqrt{\dfrac1x+2} - \sqrt{\dfrac1x}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) \times \dfrac{\sqrt{y+2} + \sqrt{y}}{\sqrt{y+2} + \sqrt{y}}$$
This gives us
$$L = \lim_{y \to \infty} \dfrac{y+2 - y}{\sqrt{y+2} + \sqrt{y}} = \lim_{y \to \infty} \dfrac2{\sqrt{y+2} + \sqrt{y}}$$
Can you now finish it off?</p>
|
688,782 | <p>$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$</p>
<p>I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit. </p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations. It will be useful to do this explicitly for say the product of the first $5$ terms. </p>
|
688,782 | <p>$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$</p>
<p>I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit. </p>
| Satish Ramanathan | 99,745 | <p>Answer:</p>
<p>$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$</p>
<p>After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$</p>
<p>$$\frac{n+1}{2}\cdot\frac{1}{n}$$</p>
<p>Limit of this function tending to infinity $= 1/2$.</p>
|
3,360,879 | <p>Given the localised ring <span class="math-container">$\mathbb{Z}_{(2)}=\{\frac{a}{b}:a,b \in \mathbb{Z}, 2 \nmid b \}$</span>, I want to show that this is an integral domain.</p>
<p>We choose some fraction <span class="math-container">$ \frac{a}{b}\in \mathbb{Z}_{(2)}$</span>,where <span class="math-container">$a \in \mathbb{Z}$</span> and <span class="math-container">$b \in \mathbb{Z}$</span>, such that <span class="math-container">$2 \nmid b$</span> and pick up another fraction <span class="math-container">$ \frac{a'}{b'}\neq 0\in \mathbb{Z}_{(2)}$</span> with <span class="math-container">$a' \in \mathbb{Z}$</span> and <span class="math-container">$b' \in \mathbb{Z}$</span> such that <span class="math-container">$2 \nmid b'$</span>. We look at the term <span class="math-container">$ \frac{a}{b}*\frac{a'}{b'}=0$</span>. Can we just conclude that <span class="math-container">$a$</span> and <span class="math-container">$b$</span> have to be zero because the only zero divisors in <span class="math-container">$\mathbb{Z}$</span> are the zeroes? How could I argue alternatively with the prime ideal <span class="math-container">$(2)$</span> ?</p>
| Noah Riggenbach | 482,732 | <p>I think that the slickest proof of this is that <span class="math-container">$\mathbb{Z}_{(2)}$</span> embeds into <span class="math-container">$\mathbb{Q}$</span>. Then since <span class="math-container">$\mathbb{Q}$</span> is a field, all subrings must be domains.</p>
|
3,333,924 | <p>I am wondering about solutions to the following differential equation:
<span class="math-container">$f(x)=C_1 \cdot f'(x+C_2) \; \forall x \in \mathbb{R} \; \exists \; C_1, C_2 \in \mathbb{R}$</span>. With <span class="math-container">$C_1, C_2$</span> being constant. Are the solutions uniquely in the family of sin/cos functions? It bugs me that I was not able to come up with a counterexample except for the trivial solution <span class="math-container">$f(x)=0$</span>. </p>
| Community | -1 | <p>A polynomial is an expression obtained by combining constants and variables by means of a <em>finite number</em> of additions and multiplications.</p>
<p>E.g. <span class="math-container">$3xy^3+2x-1$</span>.</p>
<p>An <em>algebraic function</em> of a single variable <span class="math-container">$x$</span> is such that the relation to the dependent variable <span class="math-container">$y(x)$</span> can be expressed by a bivariate polynomial equation with integer coefficients.</p>
<p>E.g. <span class="math-container">$3x(y(x))^3+2x-1=0$</span>, which can also be written <span class="math-container">$y(x)=\sqrt[3]{\dfrac{1-2x}{3x}}$</span>.</p>
<p>In particular, the ratio of two polynomials in <span class="math-container">$x$</span> is an algebraic function, as is any expression involving only the four operations and radicals.</p>
<p>A <em>transcendental</em> function is one that is not algebraic.</p>
<p>E.g. <span class="math-container">$\sin(x)$</span> is transcendental because there is no polynomial <span class="math-container">$P$</span> such that <span class="math-container">$P(x,\sin(x))=0$</span>.</p>
|
1,841,882 | <p>By Cayley's theorem, we know that for any finite group $G$, there exists $N \in \mathbb{N}$ such that $G$ is isomorphic to a subgroup of $S_N$, the symmetric group on $N$ letters. Can we prove that for every finite group $G$ there is some symmetric group $S_N$ such that $G$ is isomorphic to a $normal$ subgroup of $S_N$?</p>
| Stefan4024 | 67,746 | <p><strong>HINT:</strong> Try to prove that for $n \ge 5$, $A_n$, the alternating group of $n$ elements is the only proper and nontrivial normal subgroup of $S_n$.</p>
<p><strong>UPDATE:</strong> This has to do something with the fact that $A_n$ is simple for $n \ge 5$. After proving this and checking the cases $n \le 4$ we can conclude that a normal subgroup of symmetric group has order $1, 4, \frac{n!}{2}$ or $n!$. Hence...</p>
|
1,641,076 | <p>Let's say I have the following decomposition: </p>
<p>$$\{100,10011,00110\}^*$$</p>
<p>How would I determine if the decomposition is ambiguous or unambiguous?</p>
| Brian Tung | 224,454 | <p>One way is to recognize that the intersection must satisfy the equation for both planes, and must therefore satisfy their sum:</p>
<p>$$
(3x-y+z)+(y+z) = 4+2
$$</p>
<p>$$
3x+2z = 6
$$</p>
<p>You can then let $x = t$, and then $3t+2z = 6$, whence we get $z = 3-\frac{3}{2}t$. You can then rewrite your first equation as</p>
<p>$$
y = 3x+z-4
$$</p>
<p>to obtain an expression for $y$ in terms of $t$. Well, that is more or less what the solution did, except that they chose to set $x = 2t$, in order to avoid some unsightly fractions. Otherwise, the parametrization is the same.</p>
|
2,521,710 | <p>I am trying to do a proof for convergence. But I am stuck in my proof not getting any further... What is missing to finish that proof?</p>
<p>$$a_n = \frac{1}{(n+1)^2}$$
Show that: $$\lim_{n \to \infty}a_n=0$$</p>
<p>Let $e > 0$ and $\forall n \ge n_0 = \lceil \frac{1}{\sqrt{\epsilon}}\rceil+1 \in \mathbb Z^+:$</p>
<p>$\begin{align}
|a_n-0| &\equiv |a_n| \\
&\equiv | \frac{1}{(n+1)^2}|\\
&\equiv|\frac{1}{(n+1)} \cdot \frac{1}{(n+1)}| \\
&\equiv |\frac{1}{n+1}| \cdot |\frac{1}{n+1}| \\
&< |\frac{1}{n}| \cdot |\frac{1}{n}| \\
&(\text{because } n \in \mathbb N) \\
&= \frac{1}{n^2} < \epsilon \text{ (by definition of convergence) }
\end{align}$</p>
<p>$\begin{align}
n \ge n_0 &= \lceil \frac{1}{\sqrt{\epsilon}}\rceil +1 \\
&> \lceil \frac{1}{\sqrt{\epsilon}} \rceil \\
&\ge \frac{1}{\sqrt{\epsilon}}
\end{align}$</p>
<p>thus</p>
<p>$\begin{align}
n &> \frac{1}{\sqrt{\epsilon}} \\
&\equiv \frac{1}{n} > \sqrt{\epsilon} \\
&\equiv \frac{1}{n^2} > \epsilon
\end{align}$</p>
<p>But the line that should follow: </p>
<p>$\begin{align}
\frac{1}{n^2} &< \epsilon
\equiv \frac{1}{n^2} < \frac{1}{n^2}
\end{align}$</p>
<p>which is wrong.. </p>
| NotAMathematician | 485,701 | <p>Do you know about squeezing methods? Since $$0<a_n$$
You can show that the limit of the sequence is "trapped" between $0$, which only means it's equal to zero. Now, since
$$n+1>n$$ then
$$\frac{1}{n+1}<
\frac{1}{n}$$
$$\frac{1}{(n+1)^2}<\frac{1}{n^2}$$
Take in count here we're using only natural numbers, so these inequalities hold (otherwise, they don't necessarily do).
Therefore:
$$0<\frac{1}{(n+1)^2}<\frac{1}{n^2}$$
So
$$\lim_{n\to\infty} 0\le \lim_{n\to\infty} a_n\le\lim_{n\to\infty} \frac{1}{n^2}$$
$$0\le \lim_{n\to\infty} a_n\le0 $$
Which necessarily implies that
$$\lim_{n\to\infty} a_n=0$$
If you're looking for a proof involing the limit definition, I think you've already got a useful bunch. </p>
|
118,701 | <p>I have two vectors of 134 elements each ($mu$, and $gt$). $mu$ contains Integers, and $gt$ contains machine precision Reals. I execute the following simple expression multiple times without changing either mu or gt:
$$
(mu/2*gt).gt
$$
I will get one of two different results: $88474.52216839303$ or $88474.52216839301$ (they differ in the last digit). What's up with that? I have never seen Mathematica behave non-deterministically on simple mathematical expressions before (I'm using version 9.0.1). Could it be performing the dot product or the multiplication in parallel or something? I really need deterministic results (at least on the same computer). I'm using Windows 7 on a Dell laptop.</p>
<p><strong>Update 2.</strong>
More fun (but apparently only when n = 134).</p>
<pre><code>n = 134;
</code></pre>
<p>Manually evaluate the following expression over and over again in a notebook. I get a different output for Hash[a.a] about every 5 or 6 evaluations (I realize that I'm using RandomReal, but I'm resetting the seed every time so $a$ always receives the same value).</p>
<pre><code>SeedRandom[1]
a = RandomReal[{0, 1}, n];
Hash[a]
Hash[a.a]
</code></pre>
<p><strong>Update 1.</strong>
Here's the output of two supposedly identical computations:</p>
<pre><code>InputForm[(mu/2 gt).gt]
88474.52216839301
InputForm@Sum[(mu[[i]]/2 gt[[i]]) gt[[i]], {i, Length@mu}]
88474.52216839303
</code></pre>
<p>Sometimes the first expression comes out identical to the second expression.</p>
<p>The two vectors are:</p>
<pre><code>InputForm[mu];
{1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10};
InputForm[gt]
{-0.004653655618380714, -0.009080939192239956, -0.031034220835509885,
0.06650237015337201, -0.009447749078904444, 0.03336916426793746,
-0.4631539984577655, -6.265611574451846, 0.5050454470606143,
-0.19087962150762205, -4.3828201756709735, 0.49876272299955815,
0.3861679596859371, -6.1635962937354645, -0.15452549964472695,
0.3473530604725559, -4.199256945736951, -0.040039918046105516,
0.004474673856913913, -0.08203350223995054, 0.00755080417433085,
0.19742429370731282, -0.058492983677093546, -0.07179532877025352,
-0.06654467292071248, 0.053278032391796154, 0.021824097049037344,
-0.15494663566691225, 0.0490691048534343, 0.06064592126063756,
-0.07745582064698114, -0.00746967588173042, 0.03559995124065775,
-0.013295758759910827, -0.0021771809232924225, 0.014751625280327196,
-0.012490576622492892, -0.004059719351428424, 0.02260878725186556,
0.00906537308368245, 0.00005640969962929232, 0.013322793809346312,
0.01939688322252025, 0.0012921413237499124, 0.005353410478874665,
0.013744515181619552, -0.0008260952031691714, 0.0019523689330869926,
0.0036484573285596087, 0.0002894048077433703, 0.0009607045972313589,
-0.0005226991236028443, -0.002322046639359741, -0.024682514037441098,
0.027220500404569047, 0.0545967335062756, 0.05642841641401902,
0.004367474285221214, -0.0005337849541497026, -0.0002226867162167549,
0.00019518810603202322, 4.710483244170917*^-6,
-0.00016670488036184576, -0.00011284285936618362,
-0.00008617428801798452, -0.00009997965667755927,
-0.00011813749998897688, -0.00011200191449290103,
-0.00003504307476243085, -0.00007170970464401294,
8.314045093271716*^-6, -0.000012108637423846602,
-0.00008318583029878757, 6.872184070694232*^-6,
0.000039598867001222615, -0.00030723457677017674,
0.000021126291712914025, 0.00017867733864129764, -0.00132238312878738,
0.00008352828516677846, 0.00016343481838493323,
-0.0009495187710002947, 0.000014501494743246468,
0.00010295794136475078, -0.0007041704456466069,
-0.000038074551413580515, 0.000024071562009758685,
-0.000368866036143653, -0.000035157347339220824,
-7.738001399044264*^-6, -0.00012744527523419168,
-0.00001767740166705445, -0.00006515075020033605,
-0.0005012315885292151, -0.00009246376382040004,
0.000035397133843528654, 0.00010514304758398063, 0.000164070573737746,
0.00033034132631488825, 0.0002563940375876836, 0.0001581254676494522,
0.00010340418143852661, 0.0001271214003914325, 0.000144619937452181,
-6.445829617805765*^-8, -7.918633438308738*^-8,
-5.331850464748234*^-10, -5.395049641407357*^-11,
-9.692675239369703*^-10, -2.3443641234029144*^-10,
-2.741707881994002*^-9, -1.8602797897764294*^-9,
1.852265547443208*^-10, -8.268466237349803*^-9,
-2.873207919632894*^-10, -8.965460213556384*^-9,
-2.7681125512775745*^-9, -6.542104162408522*^-9,
-4.0174693373664825*^-9, -3.665420456657473*^-9,
-2.843221227293432*^-9, -1.5467319689177169*^-9, 0.12984908695319053,
0.3495235744679681, 0.31365238179137667, 0.5566921439809294,
0.09068947055843457, 78.52618761194245, 1.462920938774925,
0.13819238950337862, 0.24562317270333922, 0.28816819236661184,
0.02950510971255533, 0.19955099011468305}
</code></pre>
| rcollyer | 52 | <p>They are not identical computations. With the first form,</p>
<pre><code>(mu/2 gt).gt
</code></pre>
<p>Mathematica can take advantage of vector arithmetic, usually going through specialized routines like LAPACK. The second form, </p>
<pre><code>Sum[(mu[[i]]/2 gt[[i]]) gt[[i]], {i, Length@mu}]
</code></pre>
<p>however, will usually be calculated term by term because there is a possibility that the input can change from term to term. The fact that they differ only in the last decimal place is surprising as various <a href="https://en.wikipedia.org/wiki/Kahan_summation_algorithm" rel="nofollow">numerical errors</a> can creep into summations if they are not carefully accounted for. </p>
|
2,146,508 | <blockquote>
<p>Let $K$ be the algebraic closure of a finite field $k$. Prove that $Gal(K/k) \cong \hat{\mathbb{Z}}$.</p>
</blockquote>
<p>From the definition in the book, here is how $\hat{\mathbb{Z}}$ is defined:
Let $D = Cr(\mathbb{Z}_{p} | \; p \; prime)$, let $\delta: \mathbb{Z} \rightarrow D$ be the map taking $x \in \mathbb{Z}$ to the vector with all coordinates equal to $x$. Then the group $D$ together with the map $\delta$ is the profinite completion of $\mathbb{Z}$, denoted $\hat{\mathbb{Z}}$.</p>
<p>There seem to be many sources online that cite this result as true, but I'm having trouble finding anywhere that shows a proof. This question is from Profinite Groups (Wilson), so I doubt that the solution is all that straight-forward. Could anyone offer me a solution or perhaps some insight on how to tackle this problem?</p>
| nguyen quang do | 300,700 | <p>For any field $k$ and a fixed algebraic closure $K$ of $k$, practically by definition, $K$ is the inductive (=direct) limit of its subextensions $L/k$ of finite degree. By (infinite) Galois theory, $Gal(K/k)$ is then the projective (=inverse) limit of its subgroups of finite index. If $k$ is a finite field, for any integer $n$, we know that $K$ admits a unique subextension $L/k$ of degree $n$, and this extension is cyclic. Hence $Gal(K/k)$ is the projective limit of the $(Z/nZ, +)$, i.e. $(\hat Z, +)$ .</p>
|
181,855 | <p>In the latest <a href="http://what-if.xkcd.com/113/" rel="noreferrer">what-if</a> Randall Munroe ask for the smallest number of geodesics that intersect all regions of a map. The following shows that five paths of satellites suffice to cover the 50 states of the USA:
<img src="https://i.stack.imgur.com/gyfYt.png" alt="from what-if.xkcd.com"></p>
<p>A similar configuration where the lines are actually great circles is claimed by the author:</p>
<blockquote>
<p>They're all slightly curved, since the Earth is turning under the satellites, but it turns out that this arrangement of lines also works for the much simpler version of the question that ignores orbital motion: "How many straight (great-circle) lines does it take to intersect every state?" For both versions of the question, my best answer is a version of the arrangement above.</p>
</blockquote>
<p>There has been quite some work on similar sounding problems. For stabbing (or finding transversals of) line segments see as an example <a href="http://link.springer.com/article/10.1007%2FBF01934440" rel="noreferrer">Stabbing line segments</a> by H. Edelsbrunner, H. A. Maurer,
F. P. Preparata, A. L. Rosenberg, E. Welzl and D. Wood (and papers which reference it.)
or L.M. Schlipf's <a href="http://www.diss.fu-berlin.de/diss/receive/FUDISS_thesis_000000096077" rel="noreferrer">dissertation</a> with examples of different kinds. </p>
<p><strong>Is there an algorithmic approach known to tackle this problem (or for the
simpler problem when all regions of the map are convex)?</strong></p>
<p>In the case of the 50 states of the USA, it is of course easy to see that one great circle does not suffice: take two states (e.g. New York and Louisiana) such that all great circles that intersect those do not pass through a third state (e.g. Alaska). Similarly one can show that we need at least 3 great circles. </p>
<p>Maybe it would be helpful to consider all triples of regions that do not lie on a great circle and use this hypergraph information to deduce lower bounds.</p>
<p><strong>What are good methods to find lowers bounds?</strong></p>
<p>Randall Munroe's conjectures that 5 is optimal:</p>
<blockquote>
<p>I don't know for sure that 5 is the absolute minimum; it's possible
there's a way to do it with four, but my guess is that there isn't.
[...] If
anyone finds a way (or proof that it's impossible) I'd love to see it!</p>
</blockquote>
| Moritz Firsching | 39,495 | <p>Looking at this old question again, I'm now fairly convinced that the easiest route of solving this problem is using similar ideas to the one suggested by <a href="https://mathoverflow.net/users/297/david-e-speyer">David E Speyer</a> in a <a href="https://mathoverflow.net/questions/181855/what-if-xkcd-com-stabbing-simply-connected-regions-on-the-2-sphere-with-few-g#comment457260_181855">comment</a>, namely basically setting up a integer program after some combinatorial information (like inclusion maximal sets of states that lie on a geodesic). For example if <code>longest_geodesics</code> is a such a set of maximal geodesics, then the following <a href="https://www.sagemath.org/" rel="noreferrer">sage</a> code will give the answer <span class="math-container">$5$</span>.</p>
<pre><code>p = MixedIntegerLinearProgram(maximization=False)
m = p.new_variable(binary=True)
for states in all_states:
p.add_constraint(p.sum(m[line]*(1 if (states in line) else 0)
for line in longest_geodesics)>=1)
p.set_objective(p.sum(m[line] for line in longest_geodesics))
p.solve()
</code></pre>
<p>If we assume that the set of <code>longest_geodesics</code> actually contained all geodesics and that the solver actually found a correct solution, then this should be a proof that it is impossible to do with <span class="math-container">$4$</span> geodesics. Obtaining a complete set of geodesics (or a superset thereof) could be done by finding a complete set of "forbidden triples", i.e. triples of states that do not lie on a geodesic and then build a lattice out of all sets that do not contain any forbidden triple.
I took the map data <a href="https://www2.census.gov/geo/tiger/GENZ2018/shp/cb_2018_us_state_20m.zip" rel="noreferrer">cb_2018_us_state_20m.zip</a> from this page <a href="https://www.census.gov/geographies/mapping-files/time-series/geo/carto-boundary-file.html" rel="noreferrer">https://www.census.gov/geographies/mapping-files/time-series/geo/carto-boundary-file.html</a> and found a (hopefully) complete set of <code>longest_geodesics</code>. And in fact the integer program didn't find a solution with only <span class="math-container">$4$</span> great circles. Here's one with <span class="math-container">$5$</span>:</p>
<img src="https://mo271.github.io/mo/181855/five_great_circles_US_gnomonic.svg" alt="map covering all 50 states with 5 great circles">
<p>This is using a gnomonic projections (suggested by <a href="https://mathoverflow.net/users/1227/martin-m-w">Martin M. W.</a> in a <a href="https://mathoverflow.net/questions/181855/what-if-xkcd-com-stabbing-simply-connected-regions-on-the-2-sphere-with-few-g/374512#comment456280_181855">comment</a>) to make all the great circles appear as line. A more conventional map would look like this:</p>
<img src="https://mo271.github.io/mo/181855/five_great_circles_US.svg" alt="map covering all 50 states with 5 great circles">
<p>However some people might prefer a more <em>human checkable proof</em>, so I provide an argument using techniques and ideas also mentioned in other answers/comments. Throughout the argument the only assumption that is made is that certain triples of states cannot lie on a great circle.</p>
<p>I want to show that it is impossible to cover all <span class="math-container">$50$</span> states with <span class="math-container">$4$</span> great circles. First consider the following set of states: <code>{'Alaska', 'Delaware', 'Hawaii', 'Kentucky', 'Vermont', 'Washington'}</code>. I claim that any three of them cannot lie on a great circle. Therefore if those <span class="math-container">$6$</span> states are covered by <span class="math-container">$4$</span> lines, either they are split <span class="math-container">$2, 2, 2$</span> or <span class="math-container">$1, 1, 2, 2$</span>, since this are the only partitions of <span class="math-container">$6$</span> into at most <span class="math-container">$4$</span> parts with size of at most <span class="math-container">$3$</span>.</p>
<p>First let's consider splitting <code>{'Alaska', 'Delaware', 'Hawaii', 'Kentucky', 'Vermont', 'Washington'}</code> into <span class="math-container">$3$</span> pairs, i.e. the split <span class="math-container">$2, 2, 2$</span>.
Hence we assume that <span class="math-container">$3$</span> great circles each cover two of those those 6 states (and potentially many more states). We proof that one more great circle is not enough to cover all states by giving a triple of states which itself cannot lie on one great circle and each state in the triple cannot lie on any of the <span class="math-container">$3$</span> great circles covering the pairs in the split. For each partition we first write the split and then the triple. For example</p>
<p><code>{{'Alaska', 'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont', 'Washington'}}: {'Florida', 'South Carolina', 'Wyoming'}</code></p>
<p>means that there is no great circle through <code>{'Florida', 'South Carolina', 'Wyoming'}</code> and each of these three states does not lie on a great circle containing any of the three pairs. So for example <code>{'Florida', 'Alaska', 'Delaware'}</code> cannot lie on a great circle and the same for <code>{'Florida', 'Hawaii', 'Kentucky'}</code> and <code>{'Florida', 'Vermont', 'Washington'}</code>. Analogously for <code>'South Carolina'</code> and <code>'Wyoming'</code>.</p>
<pre><code>{{'Alaska', 'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont', 'Washington'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Delaware'}, {'Hawaii', 'Vermont'}, {'Kentucky', 'Washington'}}: {'Rhode Island', 'Arkansas', 'Colorado'}
{{'Alaska', 'Delaware'}, {'Hawaii', 'Washington'}, {'Kentucky', 'Vermont'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Hawaii'}, {'Delaware', 'Kentucky'}, {'Vermont', 'Washington'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Hawaii'}, {'Delaware', 'Vermont'}, {'Kentucky', 'Washington'}}: {'Louisiana', 'Rhode Island', 'New Mexico'}
{{'Alaska', 'Hawaii'}, {'Delaware', 'Washington'}, {'Kentucky', 'Vermont'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Kentucky'}, {'Delaware', 'Hawaii'}, {'Vermont', 'Washington'}}: {'Connecticut', 'New Mexico', 'Louisiana'}
{{'Alaska', 'Vermont'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Washington'}}: {'Louisiana', 'Michigan', 'New Mexico'}
{{'Alaska', 'Washington'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Vermont'}}: {'Connecticut', 'South Carolina', 'Minnesota'}
{{'Alaska', 'Kentucky'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Washington'}}: {'Rhode Island', 'Arkansas', 'Colorado'}
{{'Alaska', 'Kentucky'}, {'Delaware', 'Washington'}, {'Hawaii', 'Vermont'}}: {'Rhode Island', 'Arkansas', 'Colorado'}
{{'Alaska', 'Vermont'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Washington'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Washington'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Vermont'}}: see argument below
{{'Alaska', 'Vermont'}, {'Delaware', 'Washington'}, {'Hawaii', 'Kentucky'}}: {'Florida', 'South Carolina', 'Wyoming'}
{{'Alaska', 'Washington'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Kentucky'}}: {'Rhode Island', 'Iowa', 'North Dakota'}
</code></pre>
<p>For the split <code>{{'Alaska', 'Washington'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Vermont'}}</code>, we first notice that <code>'Connecticut'</code> and <code>'South Carolina'</code> can't lie on any of the three great circles going through the three pairs in the split, therefore the <span class="math-container">$4$</span>th great circle has to contain both of those. Then we look at <code>'Indiana'</code> and <code>'Tennessee'</code>: the following triples cannot lie on a great circle:</p>
<pre><code>{'Indiana', 'Washington', 'Alaska'}
{'Indiana', 'Hawaii', 'Vermont'}
{'Indiana', 'Connecticut', 'South Carolina'}
{'Tennessee', 'Washington', 'Alaska'}
{'Tennessee', 'Hawaii', 'Vermont'}
{'Tennessee', 'Connecticut', 'South Carolina'}
</code></pre>
<p>Therefore both <code>'Indiana'</code> and <code>'Tennessee'</code> must lie on the great circle containing <code>'Delaware'</code> and <code>'Kentucky'</code>, and a contradiction is reached because the triple <code>{'Tennessee', 'Indiana', 'Delaware'}</code> does not lie on a great circle.</p>
<p>Second let's consider splitting <code>{'Alaska', 'Delaware', 'Hawaii', 'Kentucky', 'Vermont', 'Washington'}</code> into <span class="math-container">$4$</span> non-empty parts, i.e. splits of the type <span class="math-container">$1, 1, 2, 2$</span>.
Here we assume that <span class="math-container">$4$</span> great circles cover those <span class="math-container">$6$</span> states (and potentially many more states). We now consider all possible such partition (there are 45).
For some of these partitions we prove that they are impossible to complete by giving a triple of states that have the following properties:</p>
<ol>
<li>Each of the three states in the triple does not lie on a great circle going through any of the two parts with two elements.</li>
<li>The three states in the triple cannot lie on one great circle.</li>
<li>Each combination of two states from the triple cannot lie on the great circle through each of the two parts with one element.</li>
</ol>
<p>From (1) it follows that the three states have to be assigned to the two parts with one element. Not all them can go to only one of them because of (2). But splitting "<span class="math-container">$3$</span>" into two nonempty parts leaves one part with <span class="math-container">$2$</span> elements and this lead to a contradiction by (3).
Take for example</p>
<pre><code>{{'Alaska', 'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont'}, {'Washington'}}: {'Rhode Island', 'South Carolina', 'Arkansas'}
</code></pre>
<p>We have (1) because <code>'Rhode Island'</code> cannot lie on the great circle through <code>{'Alaska', 'Delaware'}</code> nor <code>{'Hawaii', 'Kentucky'}</code> and analogously for <code>'South Carolina'</code> and <code>'Arkansas'</code>.
We have (2) because <code>{'Rhode Island', 'South Carolina', 'Arkansas'}</code> cannot lie on a great circle.
We have (3) because now matter what combination of two states you take from <code>{'Rhode Island', 'South Carolina', 'Arkansas'}</code>, this pair cannot lie on a great circle with <code>'Vermont'</code> nor <code>'Washington'</code>.
Here's the list of all the partitions we can exlcude with that argument:</p>
<pre><code>{{'Alaska', 'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont'}, {'Washington'}}: {'Rhode Island', 'South Carolina', 'Arkansas'}
{{'Alaska', 'Delaware'}, {'Hawaii', 'Vermont'}, {'Kentucky'}, {'Washington'}}: {'Rhode Island', 'Arizona', 'Florida'}
{{'Alaska', 'Delaware'}, {'Hawaii'}, {'Kentucky', 'Vermont'}, {'Washington'}}: {'Utah', 'South Carolina', 'Florida'}
{{'Alaska', 'Delaware'}, {'Hawaii', 'Washington'}, {'Kentucky'}, {'Vermont'}}: {'Arizona', 'Louisiana', 'Wyoming'}
{{'Alaska', 'Delaware'}, {'Hawaii'}, {'Kentucky', 'Washington'}, {'Vermont'}}: {'Louisiana', 'Oklahoma', 'Connecticut'}
{{'Alaska', 'Hawaii'}, {'Delaware', 'Vermont'}, {'Kentucky'}, {'Washington'}}: {'Arizona', 'Rhode Island', 'Wyoming'}
{{'Alaska', 'Hawaii'}, {'Delaware'}, {'Kentucky', 'Vermont'}, {'Washington'}}: {'Arizona', 'Rhode Island', 'Wyoming'}
{{'Alaska', 'Hawaii'}, {'Delaware', 'Washington'}, {'Kentucky'}, {'Vermont'}}: {'Arizona', 'Louisiana', 'Wyoming'}
{{'Alaska', 'Hawaii'}, {'Delaware'}, {'Kentucky', 'Washington'}, {'Vermont'}}: {'Arkansas', 'Michigan', 'Connecticut'}
{{'Alaska', 'Hawaii'}, {'Delaware'}, {'Kentucky'}, {'Vermont', 'Washington'}}: {'Arizona', 'Louisiana', 'Wyoming'}
{{'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Vermont'}, {'Washington'}}: {'New Mexico', 'Michigan', 'Florida'}
{{'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Washington'}, {'Vermont'}}: {'Oklahoma', 'Wisconsin', 'Connecticut'}
{{'Alaska', 'Kentucky'}, {'Delaware', 'Washington'}, {'Hawaii'}, {'Vermont'}}: {'Louisiana', 'Oklahoma', 'Rhode Island'}
{{'Alaska', 'Kentucky'}, {'Delaware'}, {'Hawaii', 'Washington'}, {'Vermont'}}: {'Arizona', 'Louisiana', 'Wyoming'}
{{'Alaska', 'Vermont'}, {'Delaware'}, {'Hawaii', 'Kentucky'}, {'Washington'}}: {'South Carolina', 'Michigan', 'Arkansas'}
{{'Alaska', 'Washington'}, {'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont'}}: {'Iowa', 'Connecticut', 'North Dakota'}
{{'Alaska'}, {'Delaware', 'Washington'}, {'Hawaii', 'Kentucky'}, {'Vermont'}}: {'Rhode Island', 'South Carolina', 'Arkansas'}
{{'Alaska', 'Vermont'}, {'Delaware'}, {'Hawaii', 'Washington'}, {'Kentucky'}}: {'Arizona', 'Louisiana', 'Wyoming'}
{{'Alaska', 'Vermont'}, {'Delaware'}, {'Hawaii'}, {'Kentucky', 'Washington'}}: {'Louisiana', 'Oklahoma', 'Wisconsin'}
{{'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Washington'}, {'Kentucky'}}: {'Rhode Island', 'Arizona', 'South Dakota'}
{{'Alaska'}, {'Delaware', 'Washington'}, {'Hawaii', 'Vermont'}, {'Kentucky'}}: {'Rhode Island', 'Arizona', 'Florida'}
{{'Alaska'}, {'Delaware', 'Washington'}, {'Hawaii'}, {'Kentucky', 'Vermont'}}: {'Utah', 'South Carolina', 'Florida'}
{{'Alaska'}, {'Delaware'}, {'Hawaii', 'Washington'}, {'Kentucky', 'Vermont'}}: {'Rhode Island', 'Arizona', 'South Dakota'}
</code></pre>
<p>For the rest of these partitions, we consider a pair such that</p>
<ol>
<li>Each state the pair does not lie on a great circle going through any of the two parts with two elements.</li>
<li>The two states in the pair cannot lie on a great circle together with any of the two part with one element.</li>
</ol>
<p>Therefore the two states in the pair have to be distributed on the two parts with one element one way or the other. We list both possible ways and the resulting partitions of the <span class="math-container">$8$</span> states into <span class="math-container">$4$</span> great circles with at least <span class="math-container">$2$</span> elements. Then we proof that this partition cannot be completed in either of the following two ways:</p>
<ul>
<li>(a) Giving a single state which cannot lie in a great circle with any of its parts</li>
<li>(b) Giving three states which cannot lie in a great circle, but have to lie in one of the parts. (Which part can be seen because one of the three states will already be a state in the corresponding part).</li>
</ul>
<p>Let's take as an example <code>{{'Alaska', 'Kentucky'}, {'Delaware', 'Hawaii'}, {'Vermont'}, {'Washington'}}</code>. Here a pair would be <code>('Arkansas', 'Rhode Island')</code>. We have (1) because each triple <code>{'Arkansas', 'Alaska', 'Kentucky'}, {'Arkansas', 'Delaware', 'Hawaii'}, {'Rhode Island', 'Alaska', 'Kentucky'}</code> and <code>{'Rhode Island', 'Delaware', 'Hawaii'}</code> cannot lie on a great circle and we have (2) because
<code>{'Arkansas', 'Rhode Island', 'Vermont'}</code> and <code>{'Arkansas', 'Rhode Island', 'Washington'}</code> cannot lie on a great circle. Hence we consider the two partitions where <code>'Arkansas'</code> and <code>'Rhode Island'</code> are paired with <code>'Vermont'</code> and <code>'Washington'</code>.
For the candidate</p>
<pre><code>[{'Kentucky', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Arkansas', 'Vermont'}, {'Washington', 'Rhode Island'}]
</code></pre>
<p>we are in case (b). Each of <code>'Florida'</code> and <code>'South Carolina'</code> cannot lie in any of the great circles given by <code>{'Delaware', 'Hawaii'}</code>, <code>{'Arkansas', 'Vermont'}</code> or <code>{'Washington', 'Rhode Island'}</code> Therefore they must both lie on the great circle going through <code>{'Kentucky', 'Alaska'}</code>, but this is a contradiction since <code>{'Florida', 'South Carolina', 'Alaska'}</code> cannot lie on a common great circle.
For the candidate</p>
<pre><code>[{'Kentucky', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Rhode Island', 'Vermont'}, {'Washington', 'Arkansas'}]
</code></pre>
<p>we are in case (a): the state <code>'New Mexico'</code> cannot lie in a great circles going any of the pairs in the candidate.
Here's list of all but one of the remaining partitions of the <span class="math-container">$6$</span> states into <span class="math-container">$4$</span> with proofs as described above:</p>
<pre><code>{{'Alaska', 'Delaware'}, {'Hawaii'}, {'Kentucky'}, {'Vermont', 'Washington'}}: ('Wyoming', 'Arizona')
[{'Delaware', 'Alaska'}, {'Hawaii', 'Wyoming'}, {'Kentucky', 'Arizona'}, {'Washington', 'Vermont'}]: Mississippi
[{'Delaware', 'Alaska'}, {'Hawaii', 'Arizona'}, {'Kentucky', 'Wyoming'}, {'Washington', 'Vermont'}]: Connecticut
{{'Alaska', 'Hawaii'}, {'Delaware', 'Kentucky'}, {'Vermont'}, {'Washington'}}: ('Michigan', 'South Carolina')
[{'Hawaii', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Michigan', 'Vermont'}, {'Washington', 'South Carolina'}]: Florida
[{'Hawaii', 'Alaska'}, {'Delaware', 'Kentucky'}, {'South Carolina', 'Vermont'}, {'Washington', 'Michigan'}]: Wyoming
{{'Alaska', 'Kentucky'}, {'Delaware', 'Hawaii'}, {'Vermont'}, {'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Kentucky', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Arkansas', 'Vermont'}, {'Washington', 'Rhode Island'}]: ['Florida', 'South Carolina', 'Alaska']
[{'Kentucky', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Rhode Island', 'Vermont'}, {'Washington', 'Arkansas'}]: New Mexico
{{'Alaska', 'Vermont'}, {'Delaware', 'Hawaii'}, {'Kentucky'}, {'Washington'}}: ('New Mexico', 'South Dakota')
[{'Alaska', 'Vermont'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'New Mexico'}, {'Washington', 'South Dakota'}]: Louisiana
[{'Alaska', 'Vermont'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'South Dakota'}, {'Washington', 'New Mexico'}]: Michigan
{{'Alaska', 'Washington'}, {'Delaware', 'Hawaii'}, {'Kentucky'}, {'Vermont'}}: ('North Dakota', 'Connecticut')
[{'Washington', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'North Dakota'}, {'Connecticut', 'Vermont'}]: ['Louisiana', 'New Mexico', 'Alaska']
[{'Washington', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Connecticut'}, {'Vermont', 'North Dakota'}]: South Carolina
{{'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky'}, {'Vermont', 'Washington'}}: ('North Carolina', 'Connecticut')
[{'North Carolina', 'Alaska'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'Connecticut'}, {'Washington', 'Vermont'}]: ['Delaware', 'Kansas', 'Wyoming']
[{'Alaska', 'Connecticut'}, {'Delaware', 'Hawaii'}, {'Kentucky', 'North Carolina'}, {'Washington', 'Vermont'}]: ['Oklahoma', 'North Carolina', 'South Dakota']
{{'Alaska', 'Kentucky'}, {'Delaware', 'Vermont'}, {'Hawaii'}, {'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Kentucky', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Arkansas'}, {'Washington', 'Rhode Island'}]: Wyoming
[{'Kentucky', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Rhode Island'}, {'Washington', 'Arkansas'}]: New Mexico
{{'Alaska', 'Kentucky'}, {'Delaware'}, {'Hawaii', 'Vermont'}, {'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Kentucky', 'Alaska'}, {'Delaware', 'Arkansas'}, {'Hawaii', 'Vermont'}, {'Washington', 'Rhode Island'}]: Kansas
[{'Kentucky', 'Alaska'}, {'Delaware', 'Rhode Island'}, {'Hawaii', 'Vermont'}, {'Washington', 'Arkansas'}]: New Mexico
{{'Alaska', 'Kentucky'}, {'Delaware'}, {'Hawaii'}, {'Vermont', 'Washington'}}: ('Oklahoma', 'Connecticut')
[{'Kentucky', 'Alaska'}, {'Delaware', 'Oklahoma'}, {'Hawaii', 'Connecticut'}, {'Washington', 'Vermont'}]: Louisiana
[{'Kentucky', 'Alaska'}, {'Delaware', 'Connecticut'}, {'Oklahoma', 'Hawaii'}, {'Washington', 'Vermont'}]: Wyoming
{{'Alaska', 'Vermont'}, {'Delaware', 'Kentucky'}, {'Hawaii'}, {'Washington'}}: ('Michigan', 'South Carolina')
[{'Alaska', 'Vermont'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Michigan'}, {'Washington', 'South Carolina'}]: Florida
[{'Alaska', 'Vermont'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'South Carolina'}, {'Washington', 'Michigan'}]: Wyoming
{{'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Vermont'}, {'Washington'}}: ('North Carolina', 'Connecticut')
[{'North Carolina', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Vermont'}, {'Washington', 'Connecticut'}]: ['Delaware', 'Oklahoma', 'Louisiana']
[{'Alaska', 'Connecticut'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Vermont'}, {'Washington', 'North Carolina'}]: Florida
{{'Alaska', 'Washington'}, {'Delaware', 'Kentucky'}, {'Hawaii'}, {'Vermont'}}: ('Michigan', 'South Carolina')
[{'Washington', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Michigan'}, {'South Carolina', 'Vermont'}]: ['Iowa', 'North Dakota', 'Hawaii']
[{'Washington', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'South Carolina'}, {'Michigan', 'Vermont'}]: Connecticut
{{'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Washington'}, {'Vermont'}}: ('Iowa', 'Connecticut')
[{'Iowa', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Washington', 'Hawaii'}, {'Connecticut', 'Vermont'}]: Michigan
[{'Alaska', 'Connecticut'}, {'Delaware', 'Kentucky'}, {'Washington', 'Hawaii'}, {'Iowa', 'Vermont'}]: South Carolina
{{'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii'}, {'Vermont', 'Washington'}}: ('North Carolina', 'Connecticut')
[{'North Carolina', 'Alaska'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'Connecticut'}, {'Washington', 'Vermont'}]: ['Delaware', 'Oklahoma', 'Louisiana']
[{'Alaska', 'Connecticut'}, {'Delaware', 'Kentucky'}, {'Hawaii', 'North Carolina'}, {'Washington', 'Vermont'}]: Wyoming
{{'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Kentucky'}, {'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Arkansas', 'Alaska'}, {'Delaware', 'Vermont'}, {'Kentucky', 'Hawaii'}, {'Washington', 'Rhode Island'}]: ['Hawaii', 'New Mexico', 'Ohio']
[{'Rhode Island', 'Alaska'}, {'Delaware', 'Vermont'}, {'Kentucky', 'Hawaii'}, {'Washington', 'Arkansas'}]: Michigan
{{'Alaska'}, {'Delaware'}, {'Hawaii', 'Kentucky'}, {'Vermont', 'Washington'}}: ('Connecticut', 'Arkansas')
[{'Alaska', 'Connecticut'}, {'Delaware', 'Arkansas'}, {'Kentucky', 'Hawaii'}, {'Washington', 'Vermont'}]: South Carolina
[{'Arkansas', 'Alaska'}, {'Delaware', 'Connecticut'}, {'Kentucky', 'Hawaii'}, {'Washington', 'Vermont'}]: ['Hawaii', 'New Mexico', 'Ohio']
{{'Alaska', 'Vermont'}, {'Delaware', 'Washington'}, {'Hawaii'}, {'Kentucky'}}: ('Wyoming', 'Arizona')
[{'Alaska', 'Vermont'}, {'Washington', 'Delaware'}, {'Hawaii', 'Wyoming'}, {'Kentucky', 'Arizona'}]: Mississippi
[{'Alaska', 'Vermont'}, {'Washington', 'Delaware'}, {'Hawaii', 'Arizona'}, {'Kentucky', 'Wyoming'}]: ['Maine', 'Connecticut', 'Alaska']
{{'Alaska', 'Washington'}, {'Delaware', 'Vermont'}, {'Hawaii'}, {'Kentucky'}}: ('North Dakota', 'Rhode Island')
[{'Washington', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'North Dakota'}, {'Kentucky', 'Rhode Island'}]: Iowa
[{'Washington', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Rhode Island'}, {'Kentucky', 'North Dakota'}]: ['Oklahoma', 'Washington', 'Arizona']
{{'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii'}, {'Kentucky', 'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Arkansas', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Rhode Island'}, {'Washington', 'Kentucky'}]: New Mexico
[{'Rhode Island', 'Alaska'}, {'Delaware', 'Vermont'}, {'Hawaii', 'Arkansas'}, {'Washington', 'Kentucky'}]: Michigan
{{'Alaska', 'Washington'}, {'Delaware'}, {'Hawaii', 'Vermont'}, {'Kentucky'}}: see argument below
{{'Alaska'}, {'Delaware'}, {'Hawaii', 'Vermont'}, {'Kentucky', 'Washington'}}: ('Arkansas', 'Rhode Island')
[{'Arkansas', 'Alaska'}, {'Delaware', 'Rhode Island'}, {'Hawaii', 'Vermont'}, {'Washington', 'Kentucky'}]: New Mexico
[{'Rhode Island', 'Alaska'}, {'Delaware', 'Arkansas'}, {'Hawaii', 'Vermont'}, {'Washington', 'Kentucky'}]: Colorado
{{'Alaska', 'Washington'}, {'Delaware'}, {'Hawaii'}, {'Kentucky', 'Vermont'}}: ('Michigan', 'South Carolina')
[{'Washington', 'Alaska'}, {'Delaware', 'Michigan'}, {'Hawaii', 'South Carolina'}, {'Kentucky', 'Vermont'}]: Rhode Island
[{'Washington', 'Alaska'}, {'Delaware', 'South Carolina'}, {'Hawaii', 'Michigan'}, {'Kentucky', 'Vermont'}]: ['Washington', 'New Mexico', 'Kansas']
</code></pre>
<p>For the only remaining partition, namely <code>{{'Alaska', 'Washington'}, {'Delaware'}, {'Hawaii', 'Vermont'}, {'Kentucky'}}</code> we provide the following ad-hoc argument:
We use the fact that following states cannot lie on the great circles through any of the the part <code>{'Alaska', 'Washington'}</code> nor <code>{'Hawaii', 'Vermont'}</code>:</p>
<pre><code>['South Carolina', 'West Virginia', 'Tennessee', 'Illinois', 'Rhode Island'].
</code></pre>
<p>Therefore <code>'South Carolina'</code> must either lie on a great circle with <code>'Delaware'</code> or <code>'Kentucky'</code>. Assume <code>'South Carolina'</code> and <code>'Delaware'</code> lie on a great circle with <code>'Delaware'</code>. On this great circle we cannot also have any of <code>'West Virginia', 'Tennessee'</code> and <code>'Illinois'</code>, which therefore must lie on the great circle through <code>'Kentucky'</code>, but this is a contradiction, since <code>'West Virginia', 'Tennessee'</code> and <code>'Illinois'</code> cannot lie on a great circle.
Therefore <code>'South Carolina'</code> has to lie on the same great circle as Kentucky. Since <code>'Rhode Island'</code> does not lie on a great circle with <code>'Kentucky'</code> and <code>'South Carolina'</code>, we have to assume that it lies on the same great circle as <code>'Delaware'</code>, which leaves us with the following partition:</p>
<pre><code>[{'Alaska', 'Washington'}, {'Delaware', 'Rhode Island'}, {'Hawaii', 'Vermont'}, {'Kentucky', 'South Carolina'}],
</code></pre>
<p>which can be seen to be absurd with the method described above by considering <code>['Arkansas', 'New Mexico', 'Washington']</code>, i.e. both <code>'Arkansas'</code> and <code>'New Mexico'</code> can only lie on the part with <code>'Alaska'</code>, but at the same time cannot lie on a great circle with it.</p>
<p>This now got a little longer that I expected; I do like the proof using integer programming better.</p>
|
2,955,780 | <p>The midpoint of a chord of length <span class="math-container">$2a$</span> is at a distance <span class="math-container">$d$</span> from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the circle is <span class="math-container">$\frac{a^2+d^2}{d}$</span> .</p>
<p>I know I have to find similar triangles, I cannot see them...</p>
| Bernard | 202,857 | <p>Let <span class="math-container">$A$</span>, <span class="math-container">$B$</span> be the end points of the chord (and of the chords cut out on the circle), <span class="math-container">$C$</span> the mid point of the minor arc, <span class="math-container">$D$</span> the midpoint of the major arc, so that CD is a diameter of the circle, and <span class="math-container">$H$</span> the intersection point of <span class="math-container">$CD$</span> with the chord <span class="math-container">$AB$</span>.</p>
<p>In the right triangle <span class="math-container">$CAD$</span>, the altitude <span class="math-container">$AH$</span> cuts out two segments <span class="math-container">$CH$</span> and <span class="math-container">$HD$</span> on the hypotenuse <span class="math-container">$CD$</span> and we know the altitude is the geometric mean of these two segments, in other words
<span class="math-container">$$AH^2=a^2=CH\cdot HC=d(CD-d),\enspace\text{so }\;d\, CD=a^2+d^2,$$</span>
whence the formula.</p>
|
3,184,802 | <p>Are there any <span class="math-container">$C^\infty$</span> real functions except the exponential family and gamma function family which has all the derivatives of same sign on an interval [a,<span class="math-container">$\infty$</span>) with a<span class="math-container">$\gt$</span>0 ?
I speculate the function is always uses exponential as building blocks and it is unique defining property of exponential functions.</p>
<p>Please provide some instances otherwise.</p>
<p>I have not been able to find any so far.</p>
| Kavi Rama Murthy | 142,385 | <p>Your function is necessarily a 'mixture' of exponentials. All you have to do is apply Bernstein's Theorem (<a href="https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions</a> ) to <span class="math-container">$g(x)=f(-x)$</span>. Thus <span class="math-container">$f(x)=\int e^{ax} dg(x)$</span> for some <span class="math-container">$g$</span>.</p>
|
43,743 | <p>An alternative title is: When can I homotope a continuous map to a smooth immersion?</p>
<p>I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated.</p>
<p>The set-up is the following:
Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$. In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion. </p>
<p>I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion.</p>
<p>Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.</p>
<p>References would be greatly appreciated.</p>
<p>Thanks!</p>
<p>Edit:</p>
<p>As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves. This is where my intuition says that there should be such a smooth immersion.</p>
| Tom Goodwillie | 6,666 | <p>Reading between the lines, I suppose you have oriented manifolds $\Sigma_i$, so that each of them determines an element of the integral homology group $H_k(M)$. In fact the assumption that these two classes are equal does not even imply that there is a compact oriented $(k+1)$-manifold $\Gamma$ with boundary such that the boundary is $\Sigma_1\coprod \Sigma_2$. For example, $\Sigma_1$ could be $\mathbb CP^2$, $\Sigma_2$ could be $S^4$, and they could both be embedded in $S^n$ for $n>>0$. The boundary of a compact oriented $5$-manifold always has signature zero.</p>
<p>Maybe modify the question, so that insteading of assuming homologous you assume cobordant, meaning that you assume there is a compact oriented $\Gamma_0$ mapped into $M$ in such a way that $\partial \Gamma_0\to M$ is an immersion, and you ask whether there exists another thing $\Gamma\to M$ with $\partial\Gamma\to M$ isomorphic to $\partial \Gamma_0\to M$ but this time with $\Gamma\to M$ an immersion.</p>
|
304 | <p>Per <a href="http://blog.stackoverflow.com/2010/07/moderator-pro-tempore/">this post on the SO/SE blog</a> (which, curiously, does not include math.SE in its graphic list), it looks like the admins will choose moderators pro tempore at about 7 days into the public beta. In the roughly 24 hours that we've been in public beta, I've wondered several times: should we push to have the admins choose moderators pro tempore sooner (e.g. <em>now</em>)?</p>
<p><em>edit</em>: to try to get a bit more clarity in response to this question, I've created CW answers "YES" and "NO" below--please up/down vote those as you see fit.</p>
<p><em>edit2</em>: see also: <a href="https://math.meta.stackexchange.com/questions/150/elect-our-provisional-moderators">elect our (Provisional) Moderators</a></p>
| Isaac | 72 | <p><strong>YES</strong> we do need moderators pro tempore now</p>
|
1,830,989 | <p>so while playing around with circles and triangles I found 2-3 limits to calculate the value of $ \pi $ using the <em>sin, cos and tan</em> functions, I am not posting the formula for obvious reasons.<br>
My question is that is there another infinite series or another way to define the trig functions when the value of the angle is in <strong>degrees</strong> without converting it to radians, I know of the <em>taylor series</em> but it takes the value of x in radians and to convert the angle to radians you obviously need $ \pi $, So is there another way to convert the or maybe find the angles in radians without using $ \pi $ or maybe a series for the trig functions which uses degrees? Also I know as a rule of thumb you always use radians in calculus can anyone explain to me why??<br>
Sorry if i asked a really dumb question.<br>
Regards,<br>
Kinshuk </p>
| Noble Mushtak | 307,483 | <p>The summation for $\sin$ with degrees requires $\pi$:
$$\sin(x^\circ)=\sum_{i=1}^\infty (-1)^n\frac{\pi^{2n+1}x^{2n+1}}{180^{2n+1}(2n+1)!}$$
Notice how all I did was substitute $\frac{\pi x}{180}$ into the power series for sin with radians since that's the conversion from degrees to radians. There's just no way to avoid $\pi$ in the series if you're going with a Taylor series and degrees.</p>
<p>The reason we use radians in calculus all the time because it's easier. For example, these are the derivatives for $\sin$ and $\cos$ in radians:
$$\frac{d}{dx}\sin(x)=\cos(x)$$
$$\frac{d}{dx}\cos(x)=-\sin(x)$$
Now, here it is for degrees:
$$\frac{d}{dx}\sin(x^\circ)=\frac{\pi}{180}\cos(x^\circ)$$
$$\frac{d}{dx}\cos(x^\circ)=-\frac{\pi}{180}\sin(x^\circ)$$
That's why we use radians instead of degrees or gradians: It gets rid of the factor in front of the derivative and makes everything a lot simpler, especially in more complicated calculus.</p>
|
4,567,410 | <p>I know that when calculating the pre-image of a function <span class="math-container">$f:X \rightarrow Y$</span> in a given subset <span class="math-container">$B$</span>, that is, <span class="math-container">$f^{-1}(B)=\{x \in X\mid f(x) \in B\}$</span>,the <span class="math-container">$f^{-1}$</span> simbol is just notation because it does not imply the existance of the inverse of the function.</p>
<p>But, taking into account this, can we conclude that <span class="math-container">$f(A)=B \implies A=f^{-1}(B)$</span> is not necessarily true if <span class="math-container">$f$</span> is not invertible and the reciprocal is also not true? Because I have been trying to find a counterexample but I haven't been able to. Thanks.</p>
| Átila Correia | 953,679 | <p>The proposed relation does not hold in general.</p>
<p>In order to conclude so, consider the function <span class="math-container">$f := \{(0,1),(1,1)\}$</span> as well as the sets <span class="math-container">$A = \{0\}$</span> and <span class="math-container">$B = \{1\}$</span>. On the one hand, we have that <span class="math-container">$f(A) = B$</span>. On the other hand, <span class="math-container">$f^{-1}(B) = \{0,1\}$</span>, which is different from <span class="math-container">$A$</span>. Therefore the proposed claim is false.</p>
<p>Hopefully this helps!</p>
|
4,194,611 | <p>Let <span class="math-container">$A \in \mathbb{R}^{m \times n}$</span>.</p>
<p><span class="math-container">$\forall i \in \mathbb{N} \; [ x_i \in \mathbb{R}^n \; \mbox{and} \; x_i \geq \textbf{0}] $</span></p>
<p>Assume that the sequence <span class="math-container">$Ax_1, Ax_2, ...$</span> converges to <span class="math-container">$p$</span>. Show that <span class="math-container">$ \exists x \in \mathbb{R}^n, x \geq \textbf{0}$</span>, such that <span class="math-container">$p=Ax$</span>.</p>
<p>I am trying to prove this statement to prove that a cone formed by finitely many vectors in <span class="math-container">$\mathbb{R}^n$</span> is closed.</p>
<p>Edit (additional context): I got the question while reading the proof of Farkas's Lemma here: <a href="https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf" rel="nofollow noreferrer">https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf</a>. <br />
The problem statement is assumed in this proof.</p>
| Yuval Filmus | 1,277 | <p>Let <span class="math-container">$\ell$</span> be some linear dependence of the columns of <span class="math-container">$A$</span>, and consider some <span class="math-container">$x_t$</span>. Thus <span class="math-container">$A(x_t + \gamma \ell) = Ax_t$</span> for all <span class="math-container">$\gamma$</span>. Choosing an appropriate <span class="math-container">$\gamma$</span>, we can ensure that <span class="math-container">$y_t = x_t + \gamma \ell$</span> satisfies the following two properties: <span class="math-container">$y_t \geq 0$</span>, and <span class="math-container">$(y_t)_i = 0$</span> for some <span class="math-container">$i$</span> in the support of <span class="math-container">$\ell$</span> (i.e., <span class="math-container">$\ell_i \neq 0$</span>). Let <span class="math-container">$A'$</span> be <span class="math-container">$A$</span> with the <span class="math-container">$i$</span>'th column removed, and let <span class="math-container">$x'_t$</span> be <span class="math-container">$y_t$</span> with the <span class="math-container">$i$</span>'th entry removed. Then <span class="math-container">$A x_t = A' x'_t$</span>. Continuing in this way, we find a matrix <span class="math-container">$B_t$</span>, obtained from <span class="math-container">$A$</span> by removing a subset of the columns, and a vector <span class="math-container">$y_t \geq 0$</span>, such that the columns of <span class="math-container">$B_t$</span> are independent, and <span class="math-container">$A x_t = B_t y_t$</span>.</p>
<p>There are finitely many options for <span class="math-container">$B_t$</span>, and so we can find an infinite subsequence <span class="math-container">$(t_s)_{s \in \mathbb{N}}$</span> and a matrix <span class="math-container">$B$</span> obtained from <span class="math-container">$A$</span> by removing a subset of the columns such that the columns of <span class="math-container">$B$</span> are independent and <span class="math-container">$B y_{t_s} \to p$</span>.</p>
<p>Suppose that <span class="math-container">$B$</span> has <span class="math-container">$m$</span> rows and <span class="math-container">$n'$</span> columns. Since all columns of <span class="math-container">$B$</span> are independent, necessarily <span class="math-container">$m \geq n'$</span>. Let <span class="math-container">$C$</span> be an arbitrary <span class="math-container">$n' \times n'$</span> submatrix of <span class="math-container">$B$</span> obtained by removing <span class="math-container">$m-n'$</span> rows, so that <span class="math-container">$C$</span> is invertible. Let <span class="math-container">$q$</span> be the corresponding restriction of <span class="math-container">$p$</span>, so that <span class="math-container">$C y_{t_s} \to q$</span>. Then <span class="math-container">$y_{t_s} \to C^{-1} q$</span>, and so <span class="math-container">$C^{-1} q \geq 0$</span> and <span class="math-container">$B C^{-1} q = p$</span>. Adding back zeroes to <span class="math-container">$C^{-1} q$</span>, we obtain the vector you are looking for.</p>
|
152,582 | <p>I have tried to use <code>OpenRead</code> for my application as it appears to be less of a burden on the memory.</p>
<p>I <code>OpenRead</code> a .csv file to extract coordinates in the form {x,y} in the following way</p>
<pre><code>ClearAll[f];
f = OpenRead["mathfile.csv"];
g = ReadList[f, String, RecordLists -> True];
Close[f];
</code></pre>
<p>I can access each record seperately by just calling <code>g[[1]]</code>. However the format is not as numbers and so a function like <code>ListPlot[g[[1]]]</code> doesn't work.</p>
<p>If I use <code>"Number"</code> or <code>"Real"</code> as an option within <code>ReadList</code> I get an error:</p>
<p><strong>"Read:readn: Invalid real number found when reading from 'f'."</strong></p>
<p>Using <code>"String"</code> or <code>"Record"</code> within <code>ReadList</code> doesn't cause a problem. However I would have to convert the <code>"String"</code> or <code>"Record"</code> into a number format.</p>
<p>What am I doing wrong ? I would like to call <code>g[[1]]</code> and have the co-ordinate in a form that can be read by <code>ListPlot</code></p>
<p>Thanks</p>
| terrygarcia | 49,372 | <p>If you use <code>OpenRead</code> as described above you'll have to iterate over your records a second time in order to convert strings to pairs of numbers. This is not memory-efficient. Also, if your input stream will no longer be needed after you load the records, it's better to invoke <code>With[{f=OpenRead["mathfile.csv"]},...]</code> so as to localize the variable as a constant.</p>
<p>If you have a CSV with records of the form:</p>
<pre><code>x1, y1
x2, y2
...
</code></pre>
<p>you can simply use the <code>Import</code> function's CSV support, like so:</p>
<pre><code>g = Import["mathfile.csv","CSV"]
</code></pre>
<p>This will return a two-dimensional list of records of the form <code>{{x1,y1},{x2,y2},...}</code>. Then you get</p>
<pre><code>g[[1]] == {x1,y1}
</code></pre>
<p>Use the <code>Import</code> option <code>HeaderLines->n</code> to skip <code>n</code> lines if you have named columns etc.</p>
<p>In my opinion, the capabilities of <code>Import</code> are quite robust and efficient-- even for large files-- and should be considered before falling back to low-level streams, particularly for simple record formats like 2-tuples.</p>
|
2,316,362 | <p>If $G$ is a Lie Group, a representation of $G$ is a pair $(\rho,V)$ where $V$ is a vector space and $\rho : G\to GL(V)$ is a group homomorphism.</p>
<p>Similarly, if $\mathfrak{g}$ is a Lie Algebra, a representation of $\mathfrak{g}$ is a Lie Algebra homomorphism $\rho : \mathfrak{g}\to \mathfrak{gl}(V)$ to the Lie Algebra of endomorphisms.</p>
<p>Now, many Physics books treating Quantum Field Theory, immediately relate the representations of Lie Groups and Lie Algebras without citing the result being used nor explaining how is it used really.</p>
<p>This is quite common in order to find the representations of the Lorentz group $SO(1,3)$ in terms of elements of its Lie Algebra.</p>
<p>Now since physicists don't clear this in the books, I'm asking here. What is actually the relation between representations of Lie Groups and Lie Algebras that allows one to find the representations of the Lie Group in terms of the representations of the Lie Algebra?</p>
| Community | -1 | <p>I think the answer on your question is given probably from a geometric point of view. There is a beautiful theorem from <strong>Lie</strong> himself and is usually referred as <strong>Lie's 3rd Theorem</strong>, and states something which nowadays is rephrased as follows (over the complex numbers)</p>
<p><strong>Theorem:</strong>
There is an equivalence between the category of complex simply connected Lie groups and category of complex Lie algebras.</p>
<p>The above theorem isn't difficult to get proved and what it says in fact is that any complex Lie Algebra can be thought as the Lie Algebra of some Lie group. Think of a complex Lie Algebra $\mathfrak{g}$, and get the group generated by $\{ exp(A) \thinspace | \thinspace A \in \mathfrak{g} \}$, inside $GL_n(\mathbf{C})$; with some topological reasonable assumptions such that admits a cover, take its universal covering space (hence simply connected). Then this new <em>manifold</em> can be endowed with a group structure, hence it can be proven that this new Lie group has $\mathfrak{g}$ as its associated algebra. </p>
<p>Moreover, because of the equivalence you get a correspondence
$$Hom_{\mathbb{C}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C})) = Hom_{\mathsf{Lie}}(G, GL_n(\mathbb{C})),$$
so the representations of one structure correspond to the representations of the other.</p>
<p>Hope the above helps! Cheers!</p>
|
84,249 | <p>I have a data set with evenly spaced data points. The plot is frequency vs. intensity. The overall shape of the plot is an upwards curve into a plateau, this cannot be seen in the data as this is an unimportant feature. There is also an oscillation in this curve. This can be seen in the plot.</p>
<p><img src="https://i.stack.imgur.com/1d2a7.png" alt="enter image description here"></p>
<p>There are a few things I would like to do here…</p>
<ol>
<li><p>Find out the frequency of this oscillation via a Fourier transform. I have done this already but I have encountered some problems. There are large peaks at the very end of the Fourier transformed data, I believe this can be sorted by zero filling using PadLeft/Right. I have also had quite a few attempts at this but have so far been unsuccessful.</p></li>
<li><p>I would like to filter this oscillation (frequency) from the data and then re-plot the data with this frequency removed. Again, I have had a few attempts at this but to no avail. I also believe that the data can be back-converted using the inverse Fourier transform.</p></li>
</ol>
<p>The data that I am using has been attached below:</p>
<pre><code>{{2.96536, 0.104234}, {2.98246, 0.0969915}, {2.99966,
0.102057}, {3.01696, 0.0921243}, {3.03436, 0.119644}, {3.05186,
0.111209}, {3.06945, 0.114199}, {3.08716, 0.109548}, {3.10496,
0.131311}, {3.12286, 0.11789}, {3.14087, 0.136387}, {3.15898,
0.156646}, {3.1772, 0.14701}, {3.19552, 0.170584}, {3.21395,
0.135949}, {3.23248, 0.155617}, {3.25112, 0.169365}, {3.26987,
0.177859}, {3.28873, 0.166621}, {3.30769, 0.16418}, {3.32676,
0.176456}, {3.34595, 0.194153}, {3.36524, 0.191821}, {3.38465,
0.16664}, {3.40417, 0.19331}, {3.4238, 0.197461}, {3.44354,
0.190734}, {3.4634, 0.218241}, {3.48337, 0.22238}, {3.50346,
0.218784}, {3.52366, 0.215578}, {3.54398, 0.23948}, {3.56442,
0.245256}, {3.58497, 0.226847}, {3.60564, 0.219071}, {3.62643,
0.223373}, {3.64735, 0.225021}, {3.66838, 0.225226}, {3.68953,
0.233922}, {3.71081, 0.232157}, {3.73221, 0.228506}, {3.75373,
0.203525}, {3.77538, 0.244966}, {3.79715, 0.241911}, {3.81904,
0.21011}, {3.84107, 0.208428}, {3.86322, 0.227731}, {3.88549,
0.222106}, {3.9079, 0.237606}, {3.93043, 0.221255}, {3.9531,
0.19241}, {3.9759, 0.221645}, {3.99882, 0.243768}, {4.02188,
0.217034}, {4.04507, 0.203556}, {4.0684, 0.205594}, {4.09186,
0.224882}, {4.11546, 0.213087}, {4.13919, 0.205046}, {4.16306,
0.216099}, {4.18706, 0.225207}, {4.21121, 0.222689}, {4.23549,
0.214728}, {4.25992, 0.23614}, {4.28448, 0.240632}, {4.30919,
0.224024}, {4.33404, 0.239854}, {4.35903, 0.242658}, {4.38417,
0.27057}, {4.40945, 0.258658}, {4.43488, 0.265637}, {4.46045,
0.259903}, {4.48617, 0.269462}, {4.51204, 0.283}, {4.53806,
0.289011}, {4.56423, 0.30783}, {4.59055, 0.297366}, {4.61702,
0.299034}, {4.64365, 0.311518}, {4.67042, 0.305525}, {4.69736,
0.313848}, {4.72444, 0.330391}, {4.75169, 0.329848}, {4.77909,
0.322798}, {4.80665, 0.351296}, {4.83437, 0.347589}, {4.86224,
0.370153}, {4.89028, 0.35712}, {4.91848, 0.357503}, {4.94685,
0.369257}, {4.97537, 0.342726}, {5.00406, 0.361666}, {5.03292,
0.353577}, {5.06194, 0.361541}, {5.09113, 0.355642}, {5.12049,
0.356704}, {5.15002, 0.333525}, {5.17971, 0.373981}, {5.20958,
0.365135}, {5.23963, 0.354161}, {5.26984, 0.338223}, {5.30023,
0.332814}, {5.33079, 0.34522}, {5.36153, 0.339017}, {5.39245,
0.335138}, {5.42355, 0.304693}, {5.45482, 0.3446}, {5.48628,
0.302124}, {5.51791, 0.319763}, {5.54973, 0.322771}, {5.58174,
0.330913}, {5.61392, 0.32405}, {5.6463, 0.356941}, {5.67886,
0.334621}, {5.7116, 0.342564}, {5.74454, 0.371111}, {5.77767,
0.34261}, {5.81098, 0.388414}, {5.84449, 0.384681}, {5.8782,
0.399269}, {5.91209, 0.387332}, {5.94619, 0.384739}, {5.98048,
0.379754}, {6.01496, 0.400486}, {6.04965, 0.45052}, {6.08453,
0.421576}, {6.11962, 0.426608}, {6.15491, 0.426233}, {6.1904,
0.436149}, {6.2261, 0.455608}, {6.262, 0.45478}, {6.29811,
0.458522}, {6.33443, 0.471644}, {6.37096, 0.469424}, {6.4077,
0.450665}, {6.44465, 0.441694}, {6.48181, 0.4626}, {6.51919,
0.477626}, {6.55678, 0.435283}, {6.59459, 0.429262}, {6.63262,
0.44671}, {6.67087, 0.403679}, {6.70934, 0.444082}, {6.74803,
0.450073}, {6.78694, 0.442818}, {6.82608, 0.431519}, {6.86544,
0.429014}, {6.90503, 0.446844}, {6.94485, 0.439155}, {6.9849,
0.441625}, {7.02517, 0.448944}, {7.06569, 0.469756}, {7.10643,
0.467907}, {7.14741, 0.48434}, {7.18863, 0.505442}, {7.23008,
0.493814}, {7.27177, 0.480819}, {7.31371, 0.509904}, {7.35588,
0.510856}, {7.3983, 0.520773}, {7.44096, 0.539145}, {7.48387,
0.541484}, {7.52703, 0.552256}, {7.57043, 0.566155}, {7.61409,
0.579046}, {7.65799, 0.55074}, {7.70215, 0.560432}, {7.74657,
0.546139}, {7.79124, 0.568914}, {7.83617, 0.549678}, {7.88136,
0.520845}, {7.9268, 0.531402}, {7.97251, 0.529267}, {8.01849,
0.545642}, {8.06473, 0.540622}, {8.11123, 0.534639}, {8.15801,
0.527247}, {8.20505, 0.517973}, {8.25237, 0.5166}, {8.29995,
0.51883}, {8.34782, 0.534269}, {8.39595, 0.535702}, {8.44437,
0.537064}, {8.49307, 0.550537}, {8.54204, 0.552024}, {8.5913,
0.53197}, {8.64084, 0.576196}, {8.69067, 0.590832}, {8.74078,
0.602607}, {8.79119, 0.585699}, {8.84188, 0.578139}, {8.89287,
0.57396}, {8.94415, 0.588638}, {8.99573, 0.55418}, {9.0476,
0.578538}, {9.09978, 0.588798}, {9.15225, 0.618243}, {9.20503,
0.61227}, {9.25811, 0.623181}, {9.3115, 0.614365}, {9.36519,
0.582252}, {9.4192, 0.591002}, {9.47351, 0.582036}, {9.52814,
0.57551}, {9.58309, 0.579221}, {9.63835, 0.601598}, {9.69393,
0.583821}, {9.74983, 0.601753}, {9.80605, 0.616571}, {9.8626,
0.623343}, {9.91948, 0.625228}, {9.97668, 0.646208}, {10.0342,
0.640938}, {10.0921, 0.652611}, {10.1503, 0.662041}, {10.2088,
0.667227}, {10.2677, 0.676957}, {10.3269, 0.668459}, {10.3864,
0.676449}, {10.4463, 0.663062}, {10.5066, 0.671728}, {10.5671,
0.658277}, {10.6281, 0.645957}, {10.6894, 0.672471}, {10.751,
0.628736}, {10.813, 0.632566}, {10.8754, 0.636309}, {10.9381,
0.65611}, {11.0012, 0.615158}, {11.0646, 0.649256}, {11.1284,
0.632535}, {11.1926, 0.640524}, {11.2571, 0.649521}, {11.322,
0.678226}, {11.3873, 0.692686}, {11.453, 0.69979}, {11.519,
0.707412}, {11.5854, 0.702047}, {11.6523, 0.691649}, {11.7195,
0.70039}, {11.787, 0.708576}, {11.855, 0.688944}, {11.9234,
0.701633}, {11.9921, 0.679641}, {12.0613, 0.699239}, {12.1308,
0.673699}, {12.2008, 0.677758}, {12.2711, 0.671892}, {12.3419,
0.682907}, {12.4131, 0.689255}, {12.4847, 0.695831}, {12.5566,
0.708726}, {12.6291, 0.706843}, {12.7019, 0.700961}, {12.7751,
0.708156}, {12.8488, 0.735001}, {12.9229, 0.721247}, {12.9974,
0.727045}, {13.0724, 0.711983}, {13.1477, 0.747442}, {13.2236,
0.743774}, {13.2998, 0.703271}, {13.3765, 0.74117}, {13.4536,
0.739813}, {13.5312, 0.6996}, {13.6093, 0.698404}, {13.6877,
0.723868}, {13.7667, 0.709428}, {13.8461, 0.752924}, {13.9259,
0.747375}, {14.0062, 0.723421}, {14.087, 0.768379}, {14.1682,
0.756985}, {14.2499, 0.7739}, {14.3321, 0.780965}, {14.4147,
0.751474}, {14.4978, 0.797507}, {14.5814, 0.724096}, {14.6655,
0.744758}, {14.7501, 0.712119}, {14.8352, 0.740484}, {14.9207,
0.729569}, {15.0067, 0.696926}, {15.0933, 0.715435}, {15.1803,
0.716336}, {15.2679, 0.755648}, {15.3559, 0.785479}, {15.4445,
0.780429}, {15.5335, 0.810293}, {15.6231, 0.776618}, {15.7132,
0.806514}, {15.8038, 0.794361}, {15.8949, 0.768466}, {15.9866,
0.758793}, {16.0788, 0.799427}, {16.1715, 0.713711}, {16.2647,
0.779966}, {16.3585, 0.724277}, {16.4529, 0.743149}, {16.5477,
0.759417}, {16.6432, 0.783702}, {16.7391, 0.771348}, {16.8357,
0.833143}, {16.9328, 0.865346}, {17.0304, 0.838565}, {17.1286,
0.849294}, {17.2274, 0.772466}, {17.3267, 0.791529}, {17.4266,
0.768608}, {17.5271, 0.763986}, {17.6282, 0.741564}, {17.7299,
0.7283}, {17.8321, 0.748744}, {17.9349, 0.768283}, {18.0383,
0.803771}, {18.1424, 0.793137}, {18.247, 0.851768}, {18.3522,
0.887429}, {18.458, 0.864324}, {18.5645, 0.862401}, {18.6715,
0.789341}, {18.7792, 0.773965}, {18.8875, 0.827819}, {18.9964,
0.815678}, {19.106, 0.773111}, {19.2161, 0.849946}, {19.3269,
0.850296}, {19.4384, 0.834852}, {19.5505, 0.883059}, {19.6632,
0.881796}, {19.7766, 0.888633}, {19.8907, 0.862076}, {20.0054,
0.859558}, {20.1207, 0.852859}, {20.2367, 0.844693}, {20.3534,
0.792461}, {20.4708, 0.836719}, {20.5889, 0.834452}, {20.7076,
0.844025}, {20.827, 0.855403}, {20.9471, 0.864835}, {21.0679,
0.852573}, {21.1894, 0.88278}, {21.3116, 0.854075}, {21.4345,
0.823129}, {21.5581, 0.863766}, {21.6824, 0.787823}, {21.8074,
0.837579}, {21.9332, 0.832601}, {22.0596, 0.843718}, {22.1869,
0.8772}, {22.3148, 0.872614}, {22.4435, 0.884425}, {22.5729,
0.860168}, {22.7031, 0.843206}, {22.834, 0.861999}, {22.9657,
0.827936}, {23.0981, 0.842105}, {23.2313, 0.83025}, {23.3653,
0.843434}, {23.5, 0.866308}, {23.6355, 0.887162}, {23.7718,
0.879083}, {23.9089, 0.885303}, {24.0468, 0.872957}, {24.1854,
0.875103}, {24.3249, 0.851151}, {24.4652, 0.872132}, {24.6062,
0.893851}, {24.7481, 0.876361}, {24.8908, 0.891544}, {25.0344,
0.881993}, {25.1787, 0.867608}, {25.3239, 0.902699}, {25.47,
0.880534}, {25.6168, 0.883281}, {25.7646, 0.894037}, {25.9131,
0.865991}, {26.0626, 0.894638}, {26.2129, 0.884609}, {26.364,
0.902747}, {26.516, 0.894505}, {26.669, 0.891673}, {26.8227,
0.894257}, {26.9774, 0.868452}, {27.133, 0.894673}, {27.2894,
0.880052}, {27.4468, 0.89386}, {27.6051, 0.926104}, {27.7643,
0.89179}, {27.9244, 0.910141}, {28.0854, 0.889865}, {28.2474,
0.891172}, {28.4103, 0.873922}, {28.5741, 0.878855}, {28.7389,
0.890755}, {28.9046, 0.912344}, {29.0713, 0.904055}, {29.2389,
0.903534}, {29.4075, 0.892855}, {29.5771, 0.881583}}
</code></pre>
<p>I apologise for importing my data like this, if there is a more convenient way of doing so please let me know and I can import it that way.</p>
<p>Thank you very much for your help.
Stuart.</p>
<p>Addition:</p>
<p><img src="https://i.stack.imgur.com/JRUWd.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/2oU67.png" alt="enter image description here"></p>
| Mausy5043 | 29,714 | <p>The data looks like it follows an exponential curve y=Ae^(x+t).</p>
<ol>
<li>find a least-squares fit y=f(x) of the data on such a curve. So, find the function f(x) for the curve.</li>
<li>for each point on the graph do y0 = y-f(x). This flattens out the graph.</li>
<li>then do a FFT analysis on y0 to find the frequency.</li>
<li>filter out the frequency to find y0 -(FFT)-> y' and transform back using y'+f(x)</li>
</ol>
<p>something like that?</p>
|
1,765,946 | <p>$\newcommand{\Sig}{\Sigma}$
Let $\Sig$ be a diagonal matrix with strictly positive entries on the diagonal. Define $V=\{B \in M_n\mid B\Sig +\Sig B^T=\Sig B +B^T \Sig \}$ (where $M_n$ is the vector space of $n \times n$ real matrices).</p>
<p><strong>What is the dimension of $V$? Is there a "nice" basis for it?</strong></p>
<p>Clearly $V$ contains the subspace of the symmetric matrices.
So $\dim V \ge \frac{n(1+n)}{2}$.</p>
<p>Of course the anser depends on $\Sig$. If $\Sig =Id$ then $V=M_n$. </p>
<p>It seems that the more identical entries, the larger is $V$.</p>
<p><strong>Is $\dim V$ only dependent on the number of different entries of $\Sig$?</strong> If so, what is the functional dependence?</p>
| joriki | 6,622 | <p>Since $V$ contains the subspace of symmetric matrices, we can write $B$ as the sum of a symmetric matrix and an antisymmetric matrix that's also in $V$. Antisymmetric matrices $A$ in $V$ fulfil $A\Sigma-\Sigma A=\Sigma A-A\Sigma$ and thus $A\Sigma=\Sigma A$, that is, they commute with $\Sigma$. That means an entry $A_{ij}$ can be non-zero iff $\Sigma_{ii}=\Sigma_{jj}$. Thus, if $\Sigma$ has multiplicities $l_i$ (that is, $l_i$ identical elements $\sigma_i$ on the diagonal), the dimension of $V$ is</p>
<p>$$
\frac{n(n+1)}2+\sum_i\frac{l_i(l_i-1)}2\;.
$$</p>
<p>A basis is given by the canonical basis of the symmetric matrices together with those elements of the canonical basis of the antisymmetric matrices that correspond to rows and columns with equal diagonal elements in $\Sigma$.</p>
|
3,716,619 | <p>Evaluating
<span class="math-container">$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$</span>
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.</p>
| md2perpe | 168,433 | <p>Using Maclaurin expansion we get:
<span class="math-container">$$
\frac{\sin\pi x}{\pi x}
= \frac{\pi x - \frac16 (\pi x)^3 + O(x^5)}{\pi x}
= 1 - \frac{\pi^2 x^2}{6} + O(x^4)
\\
\frac{\pi x}{\sin\pi x} = 1 + \frac{\pi^2 x^2}{6} + O(x^4)
\\
\left(\frac{\pi x}{\sin\pi x}\right)^2 = 1 + \frac{\pi^2 x^2}{3} + O(x^4)
\\
\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}
= \frac{1}{x^2}\left[\left(\frac{\pi x}{\sin\pi x}\right)^2-1\right]
= \frac{\pi^2}{3} + O(x^2)
\to \frac{\pi^2}{3}
$$</span></p>
|
3,716,619 | <p>Evaluating
<span class="math-container">$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$</span>
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.</p>
| J.G. | 56,861 | <p>Yet another approach: as it's an even function, assume <span class="math-container">$x>0$</span>. Cut a sector in a radius-<span class="math-container">$\sqrt{2}$</span> angle subtending <span class="math-container">$\pi x$</span> radians at the centre so <span class="math-container">$\pi x-\sin\pi x$</span> is the area in the sector outside the triangle with the same vertices. We'll approximate the arc as a parabola, in Cartesian coordinates with the line segment of the same endpoints a part of the <span class="math-container">$X$</span>-axis, with extrema at <span class="math-container">$X=\pm\sqrt{2}\sin\frac{\pi x}{2}\sim\pm\frac{\pi x}{\sqrt{2}}$</span>. The peak is at<span class="math-container">$$X=0,\,Y=\sqrt{2}(1-\cos\frac{\pi x}{2})=2\sqrt{2}\sin^2\frac{\pi x}{4}\sim\frac{\pi^2x^2\sqrt{2}}{8}.$$</span>At leading order, the parabola is <span class="math-container">$Y=\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)$</span>, so the area below it is<span class="math-container">$$\int_{-\pi x/\sqrt{2}}^{\pi x/\sqrt{2}}\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)dX=\frac{\pi^3x^3}{6}.$$</span>So for small <span class="math-container">$x$</span>,<span class="math-container">$$\pi x-\sin\pi x\sim\frac{\pi^3x^3}{6}\implies\frac{1}{\sin\pi x}-\frac{1}{\pi x}\sim\frac{\pi^2x^2}{6\sin\pi x}\sim\frac{\pi x}{6}\implies\frac{1}{\sin^2\pi x}-\frac{1}{\pi^2x^2}\sim\frac{\pi x}{2}\cdot\frac{2}{\pi x}=\frac13.$$</span></p>
|
4,621,390 | <p>I'm studying Linear Algebra and have come to think of a column vector as an ordered bunch of objects, where each object is the product of a scalar and a basis vector, vis:</p>
<p><span class="math-container">$\begin{bmatrix}
a \\
b \\
\vdots \\
\end{bmatrix} = a \hat{i} + b \hat{j} + \dots$</span></p>
<p>Where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are scalars and <span class="math-container">$\hat{i}$</span> and <span class="math-container">$\hat{j}$</span> are <em>unique</em> basis vectors. By that I mean <span class="math-container">$\hat{i}$</span> must not equal <span class="math-container">$\hat{j}$</span>.</p>
<p>Is that correct?</p>
| helixer | 1,073,720 | <p>First of all, basis vectors are linearly independent, so they better be unique, or else they're not basis vectors.</p>
<p>To clarify Seeker's answer:</p>
<blockquote>
<p>The entries in a column vector come from a linear combination of the basis vectors.</p>
</blockquote>
<p>This is referring to all entries at once, i.e. the whole resultant column vector. A linear combination of (basis) vectors will obviously produce a vector. For example using the unit basis vectors:</p>
<p><span class="math-container">$$
a\hat{i} + b\hat{j} = a\begin{bmatrix}1\\0\end{bmatrix} + b\begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}a\\b\end{bmatrix}
$$</span></p>
<p>Each <em>element</em> in a vector is a linear combination of certain <em>elements</em> in the basis vectors. For example <span class="math-container">$a=a(1)+b(0)$</span>.</p>
<p>What you said is true if <span class="math-container">$\hat{i}, \hat{j}, ...$</span> are simple unit basis vectors e.g. <span class="math-container">$(1,0,...), (0,1,...), ...$</span>. But it's not necessarily true for other basis vectors (as Seeker said).</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| rar | 196,345 | <p>We want to show that $\arctan(x) \leq x$ for all positive x (or vice-versa for negative x). Notice that at $x=0$, we can evaluate $\arctan(x) = 0$, so the functions are equal. Now, the derivative of $\arctan$ is $1/(1+x^2) < x' = 1$, and paired with our former observation, by a well-known theorem from calculus, this means that $\arctan(x) \leq x$ for positive x. (By the same theorem, this is in fact a strict inequality).</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| marwalix | 441 | <p>For $0\lt x \lt \frac{\pi}{2}$ we have $x\leq\tan{x}$ This leads to $\frac{\arctan{x}}{x}\leq 1$
Similarly for $0\gt x \gt -\frac{\pi}{2}$ we have $x\geq\tan{x}$ and this leads to the same inequality. </p>
|
2,929,025 | <p>Bill gave exams for the entrance at some specific gymnasium. <span class="math-container">$602$</span> students took part, which were classified, after the exams, in an ascending order, and the first <span class="math-container">$108$</span> students will be taken, which will accept to enter. Every student that has the possibility to enter will not enter with a small possibility <span class="math-container">$p=0.02$</span>, same for all, and independent from the rest. Bill is at the position <span class="math-container">$113$</span>, so he will be accepted if at least <span class="math-container">$5$</span> students from the first <span class="math-container">$112$</span> will not enter at the gymnasium. I want to give an exact expression for the probability <span class="math-container">$q$</span> that Bill gets accepted. I also want to give an approximate expression for the probability <span class="math-container">$q$</span> .</p>
<p>Is the probability that Bill get accepted equal to</p>
<p><span class="math-container">$$5 \cdot 0.02?$$</span></p>
<p>Or do we have to take also something else into consideration?</p>
| karakfa | 14,900 | <p>easier to compute by hand (or calculator) if you re-write it as nested terms</p>
<p><span class="math-container">$$
1-p^n(1+rn(1+r\frac{(n-1)}2(1+r\frac{(n-2)}3(1+r\frac{(n-3)}4))))
$$</span></p>
<p>where
<span class="math-container">$$
p=0.98 \\
n=112 \\
r=(1-p)/p \\
$$</span></p>
|
3,348,178 | <p>I have three vectors in 3d that originate at a point. If I look at them along a line perpendicular to a plane that intersects two of them, how do I find the angles between those two vectors and the third one?</p>
<p>Clarification because this is frickin difficult to explain:</p>
<p><a href="https://i.stack.imgur.com/HNyOB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HNyOB.png" alt="enter image description here"></a></p>
<p>I'll also accept a way to find how much of the longest line is on either side of the shorter line, because that's ultimately what I need.</p>
| mathcounterexamples.net | 187,663 | <p><span class="math-container">$\sqrt{x^2}= \sqrt{y^2}$</span> doesn’t imply <span class="math-container">$x=y$</span>!!!</p>
<p>Take <span class="math-container">$x=2$</span> and <span class="math-container">$y=-2$</span>.</p>
|
3,348,178 | <p>I have three vectors in 3d that originate at a point. If I look at them along a line perpendicular to a plane that intersects two of them, how do I find the angles between those two vectors and the third one?</p>
<p>Clarification because this is frickin difficult to explain:</p>
<p><a href="https://i.stack.imgur.com/HNyOB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HNyOB.png" alt="enter image description here"></a></p>
<p>I'll also accept a way to find how much of the longest line is on either side of the shorter line, because that's ultimately what I need.</p>
| cansomeonehelpmeout | 413,677 | <blockquote>
<p><span class="math-container">$$x=y\iff -x=y\lor x=-y\tag{1}$$</span></p>
</blockquote>
<p>Doesn't makes sense since <span class="math-container">$-x=y\iff x=-y$</span>, that is, they're equivalent. This reduces <span class="math-container">$(1)$</span> to <span class="math-container">$$x=y\iff -x=y\tag{2}$$</span> Which is only true when <span class="math-container">$x=y=0$</span> (since it implies that <span class="math-container">$-x=x$</span>).
In your approach, the first step:</p>
<blockquote>
<p><span class="math-container">$$x=y\iff \sqrt{x^2}=\sqrt{y^2}\tag{3}$$</span></p>
</blockquote>
<p>is false. This is because <span class="math-container">$\sqrt{x^2}=|x|$</span>, so <span class="math-container">$(3)$</span> becomes <span class="math-container">$$x=y\iff |x|=|y|$$</span></p>
<p>This is only true when both <span class="math-container">$x,y>0$</span> in which case <span class="math-container">$|x|=x$</span>. For instance <span class="math-container">$|2|=|-2|$</span>, but <span class="math-container">$2\neq -2$</span>.</p>
<hr>
<p>Do you mean</p>
<blockquote>
<p><span class="math-container">$$x^2=y^2\Rightarrow x=y\lor x=-y\tag{4}$$</span></p>
</blockquote>
<p>?</p>
<p>In which case, notice that you can write <span class="math-container">$$x^2=y^2\iff x^2-y^2=0\iff (x-y)(x+y)=0$$</span></p>
|
3,348,178 | <p>I have three vectors in 3d that originate at a point. If I look at them along a line perpendicular to a plane that intersects two of them, how do I find the angles between those two vectors and the third one?</p>
<p>Clarification because this is frickin difficult to explain:</p>
<p><a href="https://i.stack.imgur.com/HNyOB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HNyOB.png" alt="enter image description here"></a></p>
<p>I'll also accept a way to find how much of the longest line is on either side of the shorter line, because that's ultimately what I need.</p>
| Allawonder | 145,126 | <p>You made your very first mistake in the first line. The statements <span class="math-container">$x=y$</span> and <span class="math-container">$\sqrt{x^2}=\sqrt{y^2}$</span> are not equivalent as claimed, since the latter means <span class="math-container">$|x|=|y|.$</span> Thus, if you take <span class="math-container">$x$</span> to be negative, the equivalence is seen to be false. In particular, although <span class="math-container">$x=y\implies |x|=|y|,$</span> is true, it's not in general true that <span class="math-container">$|x|=|y|\implies x=y$</span> is also true.</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Martin Brandenburg | 1,650 | <p>The blog entry <a href="http://rs.io/why-category-theory-matters/">"Why Category Theory Matters"</a> by Robert Seaton ends with a quite impressive reference list of applications of category theory to the sciences:</p>
<ul>
<li>Category theory has been used to study <a href="http://reyes-reyes.com/1999/06/01/count-nouns-mass-nouns-and-their-transformations-a-category-theoretic-unified-semantics/">grammar and human language</a>.</li>
<li>In <a href="http://arxiv.org/ftp/arxiv/papers/0803/0803.2027.pdf">building a spreadsheet application</a>.</li>
<li>As a <a href="http://arxiv.org/PS_cache/math/pdf/0306/0306223v1.pdf">descriptive tool in neuroscience</a>.</li>
<li>In the <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.32.2635">analysis and design of cognitive neural network architectures</a>.</li>
<li>In programming languages, especially <a href="https://en.wikibooks.org/wiki/Haskell/Category_theory">Haskell and most famously monads</a>, but also, for instance, a <a href="http://www.cs.cornell.edu/~ross/publications/italx/">typed assembly language</a> and work on the <a href="http://www.cs.nott.ac.uk/~txa/publ/lics01.pdf">typed lambda calculus</a>.</li>
<li><a href="http://www.cs.cornell.edu/~ross/publications/proofgen/proofgen_tate_popl10.pdf">Generating program optimizations</a>.</li>
<li>To <a href="http://www.amazon.co.uk/Space-Motion-Communicating-Agents/dp/0521738334/ref=sr_1_1?ie=UTF8&s=books&qid=1283624498&sr=8-1">model systems of interacting agents</a>.</li>
<li>To <a href="https://www.cs.ox.ac.uk/people/ralf.hinze/publications/WGP12.pdf">generalize sorting algorithms</a>.</li>
<li>To <a href="http://arxiv.org/abs/1311.3903">understand collaborative text editing</a>. See also <a href="https://bosker.wordpress.com/2014/06/19/revisiting-on-editing-text/">this blog</a>.</li>
<li>To <a href="http://www.cs.bham.ac.uk/~mhe/papers/selection-escardo-oliva.pdf">understand optimal play in sequential games</a> like chess.</li>
<li>To <a href="http://arxiv.org/pdf/math/0602053v3.pdf">formalize the notion of algorithm</a>.</li>
<li>In <a href="http://link.springer.com/article/10.1023/A:1018963029743">the study of analogy</a>.</li>
<li>As <a href="http://math.mit.edu/~dspivak/CT4S.pdf">“a language for experimental design patterns” and “a new vocabulary in which to think and communicate.”</a></li>
<li>In definitions of <a href="http://www.nbi.dk/~emmeche/coPubl/97d.NABCE/ExplEmer.html">emergence</a> and discussions of <a href="https://golem.ph.utexas.edu/category/2007/11/category_theory_and_biology.html">biology</a>.</li>
</ul>
|
1,452,121 | <p>I very well know that every open ball is an open set. and that every open set need not be an open ball. But illustrate me some counter example.</p>
| layman | 131,740 | <p>Here is a counterexample.</p>
<p>Consider the set $\Bbb R$ of real numbers. Let $B(0, 2)$ represent the ball of radius $2$ around $0$, i.e, the interval $(0-2, 0+2) = (-2,2)$.</p>
<p>Now let $B(5,2)$ represent the ball of radius $2$ around $5$, i.e., the interval $(5-2, 5+2) = (3,7)$.</p>
<p>The union of these two balls, $B(0,2) \cup B(5,2)$, is open, but it is not itself a ball because every open ball in $\Bbb R$ can be expressed as an open interval, but how do you express that union as a single open interval? If it helps, feel free to draw a picture of the real line $\Bbb R$ and the intervals I mentioned.</p>
|
4,942 | <p>I'd like to control more aspects of a <code>DateListPlot</code>, for example: shading for weekend days, and/or indicators for daytime/nighttime areas. </p>
<p>By way of illustration, here's a simple example of a set of time data points (recent questions on mathematica.stackexchange):</p>
<pre><code>questions =
First[Rest[
Import["http://api.stackexchange.com/2.0/questions?page=1&\
pagesize=100&order=desc&sort=creation&site=mathematica", "JSON"]]];
questionTimes =
Cases[questions, HoldPattern["creation_date" -> value_] :> value,
Infinity];
hours = (Mod[#, 86400] /86400) 24 & /@ questionTimes ;
DateListPlot[Transpose[{questionTimes, hours}], Filling -> Bottom,
GridLines -> False, Frame -> {True, True, False, False}]
</code></pre>
<p>and the plot is like this:</p>
<p><img src="https://i.stack.imgur.com/xwQlR.png" alt="stack exchange"></p>
<p>I can't see how to show the necessary information along the x-axis, nor how to shade different areas of the graph to show day/night.</p>
<p><em>Edit</em>: I now realise that the date/times returned by the SE API are in Unix Epoch (1970), and I hadn't noticed because I wasn't able to see the day numbers or years on my first attempt at a plot...</p>
| Markus Roellig | 135 | <p>Somehow DateListPlot is resistant to many styling options. Without fiddling with the internals, here is a starting point.</p>
<pre><code>WeekendQ[date_] :=
With[{d = DateString[date, "DayName"]},
MatchQ[d, "Saturday" | "Sunday"]]
weekStyle = {Blue};
weekendStyle = {PointSize -> .015, Directive[Red]};
styleList =
Map[If[WeekendQ[#], weekendStyle, weekStyle] &, questionTimes];
DateListPlot[Partition[Transpose[{questionTimes, hours}], 1],
PlotStyle -> styleList, Filling -> Bottom, GridLines -> False,
FillingStyle -> ColorData[1][1],
Frame -> {True, True, False, False},
PlotRange -> {0, 24},
Prolog -> {LightGray, Rectangle[Scaled[{0, 0}], Scaled[{1, 6/24}]],
Rectangle[Scaled[{0, 20/24}], Scaled[{1, 1}]]}]
</code></pre>
<p>Looks like:</p>
<p><img src="https://i.stack.imgur.com/QJevU.jpg" alt="enter image description here"></p>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| Deane Yang | 613 | <p>Another thing that is surprisingly rare is a long term collaboration within a single math department. Erwin Lutwak, Gaoyong Zhang, and I, all at NYU-Poly, have a long term relationship that has yielded about jointly authored 21 papers so far, some with an additional author.</p>
|
3,760,450 | <p>I have a class in numerical mathematics, and I received several tasks I should answer. I am not a mathematician, and this is a bit out of my mind range, and I would be grateful for answers.
Question is as follows:</p>
<p>Let <span class="math-container">$x, y \in \Bbb{C}^n$</span> be arbitrary vectors. Show what can be the rank of matrix <span class="math-container">$xy^* \in \Bbb{C}^{n×n}$</span>. If rank of matrix <span class="math-container">$xy^*$</span> is equal to zero, describe (in most general terms) what should vectors <span class="math-container">$x$</span> and <span class="math-container">$y$</span> look like.</p>
<p>Edit:</p>
<p>From what I know, multiplying vector <span class="math-container">$x$</span> with conjugate transpose of vector <span class="math-container">$y$</span>, both of which are dimension <span class="math-container">$n$</span>, will result in square matrix of dimension <span class="math-container">$n\times n$</span>. Maximum number of diagonal elements that can be independent are <span class="math-container">$n$</span>, so rank of that matrix can vary from zero to n. (Please correct me if I am completely wrong in this)
And only way the rank can be zero is if all diagonal elements are zero, so <span class="math-container">$x$</span> and <span class="math-container">$y$</span> should be null vectors. These are my thoughts, but I am not sufficiently familiar with linear algebra to be sure. Please help</p>
| heropup | 118,193 | <p>Your answer is correct if:</p>
<ol>
<li>The area is also bounded by the line <span class="math-container">$x = 3$</span></li>
<li>We are interested in the unsigned (absolute) area enclosed.</li>
</ol>
<p>I think the most challenging part of the question is determining <span class="math-container">$g$</span>. For <span class="math-container">$-2 \le x < 0$</span>, we must show that <span class="math-container">$$\max_{-2 \le t < x} t^2 - |t| = 2,$$</span> which we observe by noting that when <span class="math-container">$t < x < 0$</span>, <span class="math-container">$f(t) = t^2 + t$</span>, hence the global extrema occur at <span class="math-container">$f'(t) = 2t + 1 = 0$</span> or at an endpoint of the interval <span class="math-container">$[-2, x]$</span>. Moreover, if <span class="math-container">$x < -1/2$</span> the critical point <span class="math-container">$t = -1/2$</span> is not allowed and the global extremum is at <span class="math-container">$t = -2$</span>. If <span class="math-container">$-1/2 \le x < 0$</span>, we have candidates <span class="math-container">$$t \in \{-2, -1/2, x\},$$</span> at which the function takes on values <span class="math-container">$$(t, f(t)) \in \{(-2, 2), (-1/2, -1/4), (x, x^2 + x)\}.$$</span> Again, the global maximum occurs at <span class="math-container">$t = -2$</span>, so for all <span class="math-container">$-2 \le x < 0$</span>, <span class="math-container">$g(x) = 2$</span>.</p>
<p>Similarly, we consider the case where <span class="math-container">$1/2 \le x \le 3$</span> to find that the minimum occurs at <span class="math-container">$x = 1/2$</span>, hence <span class="math-container">$g(x) = -1/4$</span> on this interval. The trickiest part is when <span class="math-container">$0 \le x < 1/2$</span>, in which case there are no critical points to <span class="math-container">$f(x) = x^2 - x$</span> on this interval, and the global minimum for <span class="math-container">$0 \le t \le x$</span> is attained at the variable right endpoint <span class="math-container">$t = x$</span>. All of this, put together, results in your piecewise function <span class="math-container">$$g(x) = \begin{cases} 2, & -2 \le x < 0 \\ x^2 - x, & 0 \le x \le 1/2 \\ -1/4, & 1/2 < x \le 3. \end{cases}$$</span></p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| Micah | 30,836 | <p>Here's a way of finding, at the very least, a large class of rational solutions. It seems plausible to me that these are all the rational solutions, but I don't actually have a proof yet...</p>
<p>Say we want to solve $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$ for some fixed $n$. The left-hand side looks an awful lot like the root of a depressed cubic (as it would be given by <a href="http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method">Cardano's formula</a>). So let's try to build some specific depressed cubic having $n$ as a root, where the cubic formula realizes $n$ as $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$.</p>
<p>The depressed cubics having $n$ as a root all take the following form:
$$(x-n)(x^2+nx+b) = x^3 + (b-n^2)x-nb$$
where $b$ is arbitrary. If we want to apply the cubic formula to such a polynomial and come up with the root $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$, we must have:
\begin{eqnarray}
p&=&\frac{nb}{2}\\
3q^2&=& \frac{(nb)^2}{4}+\frac{(-n^2+b)^3}{27}\\
&=&\frac{b^3}{27}+\frac{5b^2n^2}{36}+\frac{bn^4}{9}-\frac{n^6}{27}\\
&=&\frac{1}{108}(4b-n^2)(b+2n^2)^2
\end{eqnarray}
(where I cheated and used Wolfram Alpha to do the last factorization :)).</p>
<p>So the $p$ that arises here will be rational iff $b$ is; the $q$ that arises will be rational iff $4b-n^2$ is a perfect rational square (since $3 * 108=324$ is a perfect square). That is, we can choose rational $n$ and $m$ and set $m^2=4b-n^2$, and then we will be able to find rational $p,q$ via the above formulae, where
$(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is a root of the cubic
$$
(x-n)\left(x^2+nx+\frac{m^2+n^2}{4}\right)=(x-n)\left(\left(x+\frac{n}{2}\right)^2+\left(\frac{m}{2}\right)^2\right) \, .
$$</p>
<p>The quadratic factor of this cubic manifestly does not have real roots unless $m=0$; since $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is real, it must therefore be equal to $n$ whenever $m \neq 0$.</p>
<p>To summarize, we have found a two-parameter family of rational solutions to the general equation $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$. One of those parameters is $n$ itself; if the other is $m$, we can substitute $b=\frac{m^2+n^2}{4}$ into the above relations to get
\begin{eqnarray}
p&=&n\left(\frac{m^2+n^2}{8}\right)\\
q&=&m\left(\frac{m^2+9n^2}{72}\right) \, .
\end{eqnarray}</p>
<p>To make sure I didn't make any algebra errors, I randomly picked $n=5$, $m=27$ to try out. These give $(p,q)=\left(\frac{1885}{4},\frac{1431}{4}\right)$, and indeed Wolfram Alpha <a href="http://www.wolframalpha.com/input/?i=%281885/4%2b1431/4%20%2a%20sqrt%283%29%29%5E%281/3%29%2b%281885/4-1431/4%20%2a%20sqrt%283%29%29%5E%281/3%29&a=%5E_Real">confirms</a> that
$$
\left(\frac{1885}{4}+\frac{1431}{4} \sqrt{3}\right)^{1/3}+\left(\frac{1885}{4}-\frac{1431}{4} \sqrt{3}\right)^{1/3}=5 \, .
$$</p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| mercio | 17,445 | <p>This kind of simplification occurs if and only if $p \pm q\sqrt d$ has a cube root of the form $x \pm y\sqrt d$ with rational $x,y$. So, to get all instances of this, start by choosing $x+y\sqrt d$, and cube it to get the values for $p$ and $q$.</p>
<p>Setting up the system $(x + y\sqrt d)^3 = p + q\sqrt d$, we get $x^3+3dxy^2 = p$ and $3x^2y+dy^3 = q$. This system has $9$ solutions in $\Bbb C^2$. </p>
<p>To get a particular solution, pick a cube root $a$ of $p + q\sqrt d$, a cube root $b$ of $p - \sqrt d$, and build $x = (a+b)/2, y = (a-b)/2\sqrt d$ :
Then, $8(x^3+3dxy^2) = (a+b)^3+3(a+b)(a-b)^2 = 4a^3 + 4b^3 = 8p$, and $8\sqrt d(3x^2y+dy^3) = 3(a+b)^2(a-b)+(a-b)^3 = 4a^3-4b^3 = 8q\sqrt d$, which proves that those $9$ couples are solution.</p>
<p>To show that it has only $9$ solution, we can show $x$ is a root of a degree $9$ polynomial :
squaring the second equation we get $q^2 = 9x^4y^2 + 6dx^2y^4 + d^2y^6$. Multiply by $27dx^3$ and use the first equation to get $27dx^3q^2 = 81x^6(p-x^3) + 18x^3(p-x^3)^2 + (p-x^3)^3$, which reduces to $$64x^9-48px^6+(27dq^2-15p^2)x^3-p^3 = 0$$</p>
<p>Finally, given a value for $x$, we can solve for $y$ : The first equation gives $y^2 = \frac{p-x^3}{3dx}$, and plugging this into the second we get $y = \frac{3qx}{8x^3+p}$. Hence the $9$ solutions I gave earlier are the only solutions to the system.</p>
<p>More importantly, if we can find a pair of cube roots such that $x$ is rational, then the corresponding $y$ is also rational, and what we really have found is that $p+q\sqrt d$ has a cube root in $\Bbb Q(\sqrt d)$.</p>
<p>Now, this makes all problems of the form "show that $\sqrt[3]{p+q\sqrt d} + \sqrt[3]{p-q\sqrt d} = 2x$" solvable immediately by computing $y= \frac{3qx}{8x^3+p}$ and then checking that $(x+y\sqrt d)^3 = p + q\sqrt d$. If this doesn't work, then the problem was wrong to begin with.</p>
<hr>
<p>I should also point out that the degree $9$ polynomial in $x$ factors over $\Bbb Q(\sqrt{-3},\sqrt[3]{p^2-dq^2})$ into a product of $3$ cubics :
$$(4x^3 - 3\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j^2\sqrt[3]{p^2-dq^2}x - p) = 0$$ (where $j$ is a primitive cube root of $1$). </p>
<p>Attempting to use Cardan's formula on those will give you back the original expression $2x = \sqrt[3]{p+q\sqrt d}+\sqrt[3]{p-q\sqrt d}$.<br>
Attempting to use Cardan's formula on the cubic for $x^3$ and then taking a cube root seems to be an even worse idea, and the moral of the story is : unlike square roots, you can't algebraically find the cube roots of $p + q\sqrt d$.</p>
|
1,744,698 | <p>How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.</p>
<p>$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$</p>
<p>$B=\begin{bmatrix}
-1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$</p>
<p>Where $A$ and $B$ are symmetric matrices of order $n+1$.</p>
| Solumilkyu | 297,490 | <p>Use induction on $n$, we only consider the matrix $A$ because solving $B$ is similar. Before starting, I think that the characteristic polynomial of $A$ and $B$ should be of the form $\color{red}{(-1)^{n+1}}\lambda^{n-1}(\lambda^2-\lambda-n)$ and
$\color{red}{(-1)^{n+1}}\lambda^{n-1}(\lambda^2+\lambda-n)$, respectively. </p>
<p>For $n=1$, then
$$\det(A-tI_2)=\left\vert\begin{matrix}1-t&1\\1&-t\end{matrix}\right\vert=t^2-t-1,$$
which is done.
Now, suppose that the result holds for some $n\in\mathbb{N}$, then for $n+1$,
we have
\begin{align*}
\det(A-tI_{n+2})&=
\left\vert
\begin{matrix}
1-t&1&1&\cdots&1\\
1&-t&0&\cdots&0\\
1&0&-t&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\color{blue} 1&0&0&\cdots&\color{blue}{-t}
\end{matrix}
\right\vert_{n+2}\\
&=(-1)^{n+3}
\left\vert
\begin{matrix}
1&1&1&\cdots&1&1\\
-t&0&0&\cdots&0&0\\
0&-t&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&0\\
0&0&0&\cdots&-t&0
\end{matrix}
\right\vert_{n+1}
-t
\left\vert
\begin{matrix}
1-t&1&1&\cdots&1\\
1&-t&0&\cdots&0\\
1&0&-t&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&\cdots&-t
\end{matrix}
\right\vert_{n+1}\\
&=(-1)^{n+3}
\left\vert
\begin{matrix}
1&1&1&\cdots&1&\color{blue}1\\
-t&0&0&\cdots&0&0\\
0&-t&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&0\\
0&0&0&\cdots&-t&0
\end{matrix}
\right\vert_{n+1}
-t(-1)^{n+1}t^{n-1}(t^2-t-n)\\
&=
(-1)^{n+3}(-1)^{n+2}
\left\vert
\begin{matrix}
-t&0&\cdots&0\\
0&-t&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&-t
\end{matrix}
\right\vert_{n}
+(-1)^{n+2}t^n(t^2-t-n)\\
&=-(-1)^nt^n+(-1)^{n+2}t^n(t^2-t-n)\\
&=(-1)^{n+2}t^n[(t^2-t-n)-1]\\
&=(-1)^{(n+1)+1}t^{(n+1)+1}[t^2-t-(n+1)].
\end{align*}
This completes the proof for the matrix $A$.</p>
|
226,449 | <p>Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:</p>
<blockquote>
<p>If $A$ is an associative and commutative ring, then we can define an
unary operation on the set of all the finite subsets of our ring,
denoted by $+ \left(A\right) $ and $\times \left(A\right)$. While it
is intuitive to define $+ \left( \emptyset \right) =0$, why should
the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?</p>
</blockquote>
| kcrisman | 24,113 | <p>As pointed out in one of the answers to <a href="https://math.stackexchange.com/questions/20969/prove-0-1-from-first-principles">this math.SX question</a>, you can get the Gamma function as an extension of factorials, and then this falls out from it (though this isn't a very combinatorial answer).</p>
|
300,253 | <p>I'm interested in invertible matrices that are built out of invertible sub-blocks. For example, four sub-blocks from $GL_n(F)$ (i.e. the group of $n \times n$ invertible matrices over a field $F$) can be assembled into a $2n \times 2n$ matrix, which may or may not be invertible.</p>
<p>Suppose that a $kn \times kn$ matrix, $M$, composed of $k^2$ invertible sub-blocks is invertible. Is it true that each sub-block of $M^{-1}$ is invertible?</p>
<p>I think that it is true (although I am happy to be shown otherwise), but I am having difficulty constructing a general proof.</p>
<p>In the case $k=1$, there is nothing to prove.</p>
<p>For the case $k=2$, we can use block-wise row reduction, as described <a href="http://en.wikipedia.org/wiki/Matrix_inversion_lemma#Blockwise_inversion" rel="nofollow">here</a>.</p>
<p>For $k>2$, I'm stuck. I've tried examining the maps $X \mapsto MX$ and $Y \mapsto M^{-1}Y$ where I suppose that some sub-block $B_{ij}$ of $M^{-1}$ is not invertible. My thought was that this might tell me something about block $B_{ji}$ of $M$, but I can't seem to make any conclusions.</p>
<p>Any suggestions?</p>
<p>Note that I am mainly interested in the case where $F = \mathrm{GF}(2)$.</p>
| Did | 6,179 | <p>Dunno, maybe that $\mathrm{var}(X_t)=1$ for every $t$.</p>
|
542,951 | <p>I am trying show that the function $f:[0,1]\to \mathbb{R}$ defined by $f(x)=\sin \dfrac{1}{x}$ if $x\neq 0$ and $f(0)=0$ possesses IVP. Though it looks easy, but I am not getting any clue how to start with. Any help would be appreciated.</p>
| Michael Hoppe | 93,935 | <p>You want to show -- despite $f$ is not continuos -- that every value between $f(0)=0$ and $f(1)=\sin(1)$ is attached by $f$. That's simple: consider the restriction of $f$ on $[1/\pi,1]$. Can you work from here applying IVT on that restriction?</p>
|
1,356,932 | <blockquote>
<p><strong>Problem.</strong> Let $h\in C(\mathbb{R})$ be a continuous function, and let
$\Phi:\Omega:=[0,1]^{2}\rightarrow\mathbb{R}^{2}$ be the map defined
by \begin{align*}
\Phi(x_{1},x_{2}):=\left(x_{1}+h(x_{1}+x_{2}),x_{2}-h(x_{1}+x_{2})\right) \tag{1}
\end{align*}
What is the (Lebesgue) measure (denoted by $\left|\cdot\right|$) of the set $\Phi(\Omega)$?</p>
</blockquote>
<p>Implicit in the problem statement is that $\Phi(\Omega)$ is Lebesgue measurable, but this is obvious as $\Phi$ is the composition of continuous functions, so $\Phi(\Omega)$ is a compact subset of $\mathbb{R}^{2}$.</p>
<p>If we knew that $h$ was $C^{1}$, then the Jacobian of $\Phi$ is
\begin{align*}
J\Phi(x_{1},x_{2})=\begin{bmatrix} {1+h'(x_{1}+x_{2})} & {h'(x_{1}+x_{2})} \\ {-h'(x_{1}+x_{2})} & {1-h'(x_{1}+x_{2})} \end{bmatrix}
\end{align*}
which has determinant $1$ everywhere. Whence,
\begin{align*}
\left|\Phi(\Omega)\right|=\int_{\mathbb{R}^{2}}\chi_{\Phi(\Omega)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}=\int_{\mathbb{R}^{2}}\chi_{\Omega}(x_{1},x_{2})\left|J\Phi(x_{1},x_{2})\right|\mathrm{d}x_{1}\mathrm{d}x_{2}=1
\end{align*}</p>
<p>My thought was to approximate $h$ uniformly in a neighborhood of $[-2,2]$ by a smooth function $g$, so that $\left|h(x_{1}+x_{2)}-g(x_{1}+x_{2})\right|<\epsilon$ for all $(x_{1},x_{2})\in\Omega$, given $\epsilon>0$. Denote the analogue of $\Phi$ with $g$ instead of $h$ by $\Phi_{\epsilon}$. One can verify that</p>
<p>$$\forall x=(x_{1},x_{2})\in\Omega,\quad \left|\Phi(x)-\Phi_{\epsilon}(x)\right|=\sqrt{2}\left|h(x_{1}+x_{2})-g(x_{1}+x_{2})\right|<\sqrt{2}\epsilon$$</p>
<p>So $\Phi(\Omega)\subset U_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ and $\Phi_{\epsilon}(\Omega)\subset V_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi(\Omega))<\sqrt{2}\epsilon\right\}$.</p>
<p>We obtain that
\begin{align*}
1=\left|\Phi_{\epsilon}(\Omega)\right|\leq\int_{\mathbb{R}^{2}}\chi_{V_{\epsilon}}(x)\mathrm{d}x
\end{align*}
Since $\Phi(\Omega)$ is compact, in particular closed, $\chi_{V_{\epsilon}}\rightarrow\chi_{\Phi(\Omega)}$ a.e. as $\epsilon\downarrow 0$. From monotone convergence, we obtain that
\begin{align*}
1\leq\left|\Phi(\Omega)\right| \tag{2}
\end{align*}</p>
<p>My issue is with establishing the reverse inequality. I know that
\begin{align*}
\left|\Phi(\Omega)\right|\leq\left|U_{\epsilon}\right|=1+\left|\left\{x\in\mathbb{R}^{2}:0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}\right|
\end{align*}
But I do not know how to control the measure of set $\left\{x\in\mathbb{R}^{2}: 0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ as $\epsilon\downarrow 0$, as $\Phi(\Omega)\setminus\Phi_{\epsilon}(\Omega)$ is contained in this set for all $\epsilon>0$.</p>
| Theo Bendit | 248,286 | <p>The domain of "all functions" seems a little too broad here. I think we should restrict ourselves to functions in $L^2[0, 1]$, so at least we may work within a Hilbert space.</p>
<p>Notice that $\phi : f \mapsto \int_0^1 x^2 f(x) \mathrm{d}x = \langle f(x), x^2 \rangle$ is a bounded linear functional, mapping $L^2[0, 1] \rightarrow \mathbb{R}$. We want to maximise $\phi$ on the unit ball, restricted to the orthogonal complement of $\lbrace 1, x \rbrace$.</p>
<p>Now, $V = \lbrace 1, x \rbrace^\perp$ is a Hilbert Space itself, and $\phi\rvert_V$ is still a bounded linear functional. By the Riesz Representation Theorem, there ought to be some $f \in V$ such that $\phi(g) = \langle g, f \rangle = \int_0^1 f(x)g(x) \mathrm{d}x$ for all $g \in V$. We need to find this $f$. Note that $f(x) = x^2$ will not do, since $x^2 \notin V$.</p>
<p>To find $f$, all we need to do is apply Gram-Schmidt to $1, x, x^2$. I can't be bothered doing all the integrals, but your third polynomial should be a quadratic, that is, in the form $ax^2 + bx + c$ for some (potentially horrible) $a, b, c$. Take this quadratic, and scale it by $1/a$ (equivalently, while performing Gram-Schmidt, do not scale the final polynomial, so that it remains monic). This resulting monic quadratic is the $f$ you're looking for.</p>
<p>Why? Certainly $f$ is perpendicular to $1$ and $x$, by the Gram-Schmidt construction. Moreover, it has the form $f(x) = x^2 + bx + c$, so for any $g$ that is perpendicular to $1$ and $x$, we have,
$$\langle g, f \rangle = \int_0^1 g(x)(x^2 + bx + c) \mathrm{d}x = \int_0^1 x^2g(x) \mathrm{d}x + b \int_0^1 x g(x) \mathrm{d}x + c \int_0^1 g(x) \mathrm{d}x,$$
but the final two integrals are $0$, since $g \in V$. So, for $g \in V$, we have,
$$\phi(g) = \int_0^1 x^2g(x) \mathrm{d}x = \langle g, f \rangle,$$
as needed.</p>
<p>Now, finally, for the optimisation. The maximum value on the unit ball of $\phi(\cdot) = \langle \cdot, f \rangle$ is achieved at the unit vector $f/\|f\|$, with a maximum value of $\|f\|$. This should be easy to calculate, given $f$.</p>
|
2,150,832 | <p>I don't understand this equation $\int_0^t ds \int_0^{t'} ds' \delta(s-s')= \min(t,t')$.
I tried to work with the property of the dirac delta function that $\int_a^b \delta(x-c)dx = 1$ if $c \in [a,b]$, but I can't see how I can obtain the minimum. Can someone help me? </p>
<p>Thank you in advance!</p>
| Qmechanic | 11,127 | <p>Using the notation for <a href="https://en.wikipedia.org/wiki/Positive_and_negative_parts" rel="nofollow noreferrer">positive and negative parts</a>, we calculate</p>
<p>$$I(t,t^{\prime})~:=~ \int_0^t \!\mathrm{d}s \int_0^{t^{\prime}} \! \mathrm{d}s^{\prime} ~\delta(s\!-\!s^{\prime})
~=~{\rm sgn}(t)~{\rm sgn}(t^{\prime}) ~J(t,t^{\prime}), \tag{1}$$
where
$$J(t,t^{\prime})~:=~ \iint_{\mathbb{R^2}}\!\mathrm{d}s~\mathrm{d}s^{\prime} ~1_{[-t^{-},t^{+}]}(s)~1_{[-t^{\prime -},t^{\prime +}]}(s^{\prime})~\delta(s\!-\!s^{\prime})~=~ \int_{\mathbb{R}}\!\mathrm{d}s ~1_{[-t^{-},t^{+}]}(s)~1_{[-t^{\prime -},t^{\prime +}]}(s)$$
$$
~=~ \int_{\mathbb{R}}\!\mathrm{d}s ~1_{[-\min(t^{-},t^{\prime -}),\min(t^{+},t^{\prime +})]}(s) ~=~\min(t^{-},t^{\prime -})+\min(t^{+},t^{\prime +}).\tag{2}$$
Combining eqs. (1) & (2), we find that OP's double integral reads
$$I(t,t^{\prime})~=~\theta(tt^{\prime})\min(|t|,|t^{\prime}|),\tag{3}$$
where $\theta$ denotes the <a href="http://en.wikipedia.org/wiki/Heaviside_step_function" rel="nofollow noreferrer">Heaviside step function</a>.</p>
|
2,942,879 | <p>Suppose that <span class="math-container">$lim_{n\rightarrow \infty} a_n = L$</span> and <span class="math-container">$L \neq 0$</span>. Prove there is some <span class="math-container">$N$</span> such that <span class="math-container">$a_n \neq 0$</span> for all <span class="math-container">$n \geq N$</span>.</p>
<p>We know by the definition of convergence of a sequence, <span class="math-container">$\forall \epsilon > 0, \exists\ N \in \mathbb{N}$</span> such that <span class="math-container">$\forall n \geq N$</span>, <span class="math-container">$|x_n - L| < \epsilon$</span>.</p>
<p>So take such an <span class="math-container">$N$</span> which we know exists since we're given the limit <span class="math-container">$L$</span>.
So for the condition <span class="math-container">$|a_n - L| < \epsilon$</span> to hold <span class="math-container">$\forall \epsilon > 0$</span>, <span class="math-container">$a_n \neq 0$</span>, as otherwise if <span class="math-container">$a_n = 0$</span>, since <span class="math-container">$L \neq 0$</span>, we could find an <span class="math-container">$\epsilon$</span> such that <span class="math-container">$\epsilon < |-L| = L$</span>.</p>
<p>Proof seems rather short and maybe not rigorous enough. Wanted thoughts or if there's a better way to do this.</p>
| James | 549,970 | <p>WLOG assume <span class="math-container">$L > 0$</span>, so in the definition of limit, we may take <span class="math-container">$\epsilon = L$</span> and then an <span class="math-container">$N$</span> can be found so that for all <span class="math-container">$n \geq N$</span>,</p>
<p><span class="math-container">$$ 0 < a_n < 2 L $$</span></p>
<p>In particular, <span class="math-container">$a_n \neq 0$</span></p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| Andreas Caranti | 58,401 | <p>We have $a^2 + b^2 - a(a+b) = b^2 - ab = -b (a-b)$ and $a^2 + b^2 - b(a+b) = a^2 - ba = a (a-b)$. </p>
<p>So if $d$ divides both $a+b$ and $a^2+b^2$, then $d$ divides $$\gcd(a (a-b), b (a-b)) = \gcd(a, b) (a-b) = a - b.$$</p>
<p>So $d$ divides $a+b + a - b = 2a$ and $a+b - (a - b) = 2b$.</p>
<p>So $d$ divides $2\gcd(a,b)=2$. </p>
<p>So the possibilities for the $\gcd$ appear to be $1$ and $2$, and both clearly occur.</p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| robjohn | 13,854 | <p>Since <span class="math-container">$\gcd(a,b)=1$</span>, <a href="https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" rel="nofollow noreferrer">Bezout's Identity</a> says we have an <span class="math-container">$x$</span> and <span class="math-container">$y$</span> so that
<span class="math-container">$$
ax+by=1\tag{1}
$$</span>
Note that
<span class="math-container">$$
\begin{align}
2a^2&=(a^2+b^2)+(a+b)(a-b)\\
2ab&=(a+b)^2-(a^2+b^2)\\
2b^2&=(a^2+b^2)-(a+b)(a-b)\\
\end{align}\tag{2}
$$</span>
Therefore, incorporating <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>,
<span class="math-container">$$
\begin{align}
2
&=2(ax+by)^2\\
&=2a^2x^2+4abxy+2b^2y^2\\
&=\Big((a^2+b^2)+(a+b)(a-b)\Big)x^2\\
&+2\Big((a+b)^2-(a^2+b^2)\Big)xy\\
&+\Big((a^2+b^2)-(a+b)(a-b)\Big)y^2\\
&=\color{#00A000}{(x-y)^2}\color{#C00000}{(a^2+b^2)}
+\color{#00A000}{((x^2-y^2)(a-b)+2xy(a+b))}\color{#C00000}{(a+b)}\tag{3}
\end{align}
$$</span>
Equation <span class="math-container">$(3)$</span> says that
<span class="math-container">$$
\gcd(a+b,a^2+b^2)\,|\,2\tag{4}
$$</span>
Note that <span class="math-container">$\gcd(1+2,1^2+2^2)=1$</span> and <span class="math-container">$\gcd(1+3,1^2+3^2)=2$</span>, so both <span class="math-container">$1$</span> and <span class="math-container">$2$</span> are possible.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| AlvinL | 229,673 | <p>It helps to discretise the problem. E.g red and blue marbles. Suppose there are <span class="math-container">$M$</span> red marbles in container <span class="math-container">$1$</span> and <span class="math-container">$N$</span> blue marbles in container <span class="math-container">$2$</span>. We can assume <span class="math-container">$M>>N$</span> but it is unnecessary. So we start with
<span class="math-container">$$ \begin{array}{c|c|c}
& \text{Container }1 & \text{Container }2 \\ \hline
\text{Red} &M &0 \\
\text{Blue} &0 &N
\end{array} $$</span></p>
<p>Pick <span class="math-container">$a\leqslant M$</span> red marbles and toss them in with the blue marbles. Thus,
<span class="math-container">$$ \begin{array}{c|c|c}
& \text{Container }1 & \text{Container }2 \\ \hline
\text{Red} &M-a &a \\
\text{Blue} &0 &N
\end{array} $$</span>
Finally, pick <span class="math-container">$a$</span> marbles from container <span class="math-container">$2$</span> and toss them in container <span class="math-container">$1$</span>. Suppose <span class="math-container">$x\leqslant a$</span> of them are blue so
<span class="math-container">$$ \begin{array}{c|c|c}
& \text{Container }1 & \text{Container }2 \\ \hline
\text{Red} &M-a+x &a-(a-x)=x \\
\text{Blue} &x &N-x
\end{array} $$</span>
Even discretisation is unnecessary, but there are some philosophical debates bound to take place due to some intuition about liquids.</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| Peter Gorry | 8,273 | <p>There is actually an easy way to achieve this (almost) perfectly. Just start the variable name with a none printing character. I use this technique to have variables for atomic orbitals that start with numbers e.g. 3d or 4f - also not allowed by Mathematica. I just write the name as <code>\[Null]3d</code>. Mathematica is now fine with this as a variable. It works for <code>\[Null]N</code> or any other "protected" name. The output looks fine because the Null character doesn't take up space.</p>
<p>A variation is <code>\[LetterSpace]</code> - an almost-invisible space treated as a letter, e.g. in a symbol name - but it has a very faint bar showing </p>
|
3,952,702 | <p>The problem is to make the following integral stationary:
<span class="math-container">$$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$</span>
to simplify the Euler equation, I tried to change the independent variable:
<span class="math-container">$$ \int_{y_1}^{y_2} \frac{\sqrt{1+x'^2}}{y^2}dy, \: \: \: \: \: \: \: \: y=y\left( x \right),\: y'=\frac{dy}{dx} $$</span>
with the correspondent Euler equation:
<span class="math-container">$$ \frac{d}{dy}\frac{\partial F}{\partial x'}-\frac{\partial F}{\partial x}=0 $$</span>
thus
<span class="math-container">$$
\begin{aligned}
\frac{\partial F}{\partial x}=0 \Rightarrow \frac{\partial F}{\partial x'} &= k\\
\frac{x'}{y^2 \sqrt{1+x^2}} &= k\\
\frac{dx}{dy} = x' &= \frac{ky^2}{\sqrt{1-k^2y^4}}
\end{aligned}
$$</span>
and I get:
<span class="math-container">$$ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $$</span>
Now, can I change <span class="math-container">$ ky^2 $</span> to a new single arbitrary variabel to simplify the integrand? Or are there a more effective method?</p>
| Parcly Taxel | 357,390 | <p>Do it, substitute <span class="math-container">$u=ky^2,\frac{du}{dy}=2\sqrt{ku}$</span> (assuming <span class="math-container">$k>0$</span>):
<span class="math-container">$$x=\frac1{2\sqrt k}\int\sqrt{\frac u{1-u^2}}\,du$$</span>
This is an elliptic integral. Say we're integrating from <span class="math-container">$0$</span> to <span class="math-container">$u$</span>, then Byrd and Friedman 235.06 gives
<span class="math-container">$$x=\sqrt{\frac2k}\left(E(\varphi,m)-F(\varphi,m)/2-\sqrt{\frac{u(1-u)}{2(1+u)}}\right)+C$$</span>
where <span class="math-container">$\sin^2\varphi=\frac{2u}{1+u}$</span> and <span class="math-container">$m=1/2$</span>. You will have to use numerical methods to get <span class="math-container">$y$</span> in terms of <span class="math-container">$x$</span>.</p>
|
35,393 | <p>I have some data that represents small-scale surface topography. The data is planar but not flat because it's impractical to level this microscope to a tolerance of microns. It looks like this:</p>
<p><img src="https://i.stack.imgur.com/nQcLB.jpg" alt="An example of 3D profilometric data with a nonzero planar slope">
I'd like to postprocess that slope out, such that the major axis is colinear with the long side of the blue rectangle and the vertical axis can then be rescaled to show meaningful topographic variation instead of the (really large) planar slope.</p>
<p>One approach here might be to do an orthogonal transformation via (basically) principal component analysis, eg. find the three eigenvalue/eigenvector pairs, and then treat the eigenvectors corresponding to the largest two eigenvalues as the new X and Y axes. I'd be very interested to see how this (or some better way) might be done efficiently in Mathematica (this data is generally in the millions of rows).</p>
<p>An example of my data is <a href="https://dl.dropboxusercontent.com/u/28347262/6000s.txt" rel="nofollow noreferrer">here.</a></p>
<p>and this is the code I've written so far (which depends on version 9+ for <code>Downsample[]</code>):</p>
<pre><code> pathToDatafile = "path\\to\\file\\6000s.xyz";
data = Import[pathToDatafile,"Table","HeaderLines"->15];
data = DeleteCases[data,{__,_String,_String}]; (*remove the "no data" lines that the instrument creates*)
data=Delete[data,-1]; (*throw away the instrument's pound sign at the end of the file*)
data = Downsample[data,{100,1}];(*downsample by a lot because we have way too many datapoints*)
ListPlot3D[data,
Mesh->None,
BoxRatios->Automatic,
PlotStyle->LightBlue,
Boxed->False,
Axes->{True,True,True},
AxesLabel->{"\[Mu]m","","\[Mu]m"},
PlotRange->{Automatic,Automatic,{0,250}},
ImageSize->Large
]
</code></pre>
| ssch | 1,517 | <p>You can use <a href="http://reference.wolfram.com/mathematica/ref/LinearModelFit.html" rel="nofollow noreferrer"><code>LinearModelFit</code></a> to create a linear-fit of the model, and use that to find an appropriate rotation.</p>
<pre><code>model = LinearModelFit[
data,
{x, y},
{x, y}];
(* We have: z = c + a x + b y *)
{c, a, b} = model["BestFitParameters"];
rt = Composition[
RotationTransform[{
{1, 0, a},
{1, 0, 0}},
{0, 0, c}],
RotationTransform[{
{0, 1, b},
{0, 1, 0}},
{0, 0, c}]];
corrected = rt@data;
ListPlot3D@corrected
</code></pre>
<p><img src="https://i.stack.imgur.com/L5W0P.png" alt="model"></p>
<p>(Note: As I don't have Mathematica v9 I downsampled by <code>data = data[[;; ;;100, All]]</code> which I assume is more-or-less the same in this case)</p>
<p>Edit: Answer used to be just <code>corrected = data;
corrected[[All, 3]] -= model["Function"] @@@ corrected[[All, ;; 2]];</code> but as @bobthechemist points out that will squish together the plane.</p>
|
2,247,900 | <p>Prove the following inequality</p>
<p>$$\ln \frac{\pi + 2}{2} \cdot \frac{2}{\pi} < \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \ln \frac{\pi + 2}{2}$$</p>
<p>I can prove that $\frac{\sin\ x}{x^2 + x} < \frac{1}{x + 1} \ \forall x \in (0, +\infty) \Rightarrow \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \int \limits_0^{\pi/2} \frac{1}{x + 1} = \ln \frac{\pi + 2}{2}$.</p>
<p>Unfortunately, I don't know what happens when $x = 0$.</p>
<p>I can prove $\leqslant$ on LHS using Mean Value Theorem, but I have no idea how to prove that the sign is strict.</p>
| Jack D'Aurizio | 44,121 | <p>The given inequality just depends on the convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad \frac{2}{\pi}x\leq \sin(x)\leq x.$$
We may use another convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad x-\frac{1}{6}x^3\leq \sin(x)\leq x-\left(\frac{4}{\pi^2}-\frac{8}{\pi^3}\right)x^3$$
and derive a much better inequality:
$$\small \frac{1}{48} \left(4 \pi -\pi ^2+40 \log\left(\frac{2+\pi }{2}\right)\right)\leq I\\ I \leq \frac{1}{2\pi^3}\left(\pi(\pi-2)(4-\pi)+2 \left(8-4 \pi +\pi ^3\right) \log\left(\frac{2+\pi }{2}\right)\right).$$</p>
|
2,306,709 | <p>Say I have a set of buildings $B$ with a single building $b \in B$ and a set of people $P$ with single persons $p \in P$.</p>
<p>What would be the proper notation for a subset of people in $P$, that live in a building $p$ which comprehensibly associates this subset to the respective building? </p>
| Georgios | 311,495 | <p>You could define a function $f \colon P \to B$ that maps every person to the building he or she lives in. Then, $f^{-1} (b)$ would be the set of people that live in building $b$.</p>
|
2,306,709 | <p>Say I have a set of buildings $B$ with a single building $b \in B$ and a set of people $P$ with single persons $p \in P$.</p>
<p>What would be the proper notation for a subset of people in $P$, that live in a building $p$ which comprehensibly associates this subset to the respective building? </p>
| Mark S. | 26,369 | <p>You could write many different notations, depending on your emphasis/context. I would consider something like "For each $b\in B $, define $P_b $ to be the set of people who live in building $b $.".</p>
|
160,169 | <p>Consider the following implementation of the complex square root:</p>
<pre><code>f[z_]:=Sqrt[(z - I)/(z + I)]*(z + I);
</code></pre>
<p>This implementation has branch points at $\lambda=\pm i$ and a (vertical) branch cut connecting them.</p>
<p>Then</p>
<pre><code>g[z_]:=Sinc[f[z]];
</code></pre>
<p>(recalling $\mathrm{sinc}(x)=\sin(x)/x$ ) has no branch cut and it is analytic on the entire complex plane, and admits power series expansions at $\lambda=\pm i$. </p>
<p>Indeed, using Mathematica 11.0.0 (Mac OS 10.10.5) gives:</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>$1-\frac{1}{3} i (z-i)-\frac{1}{5} (z-i)^2+\frac{11}{315} i (z-i)^3+\frac{61
(z-i)^4}{5670}+O\left((z-i)^5\right)$</p>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>gives $\frac{61}{5670}$.</p>
<p>Now, using Mathematica 11.1.1 (both on Mac OS 10.12 Sierra and Linux Ubuntu 16 LTS)</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
</blockquote>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}].
</code></pre>
</blockquote>
<p>So neither of these stock functions work in properly in Mathematica 11.1.1. Does anyone know what is going on? Will this be fixed? They worked properly even in Mathematica 9 and also in Mathematica 11.0.0</p>
<p>Besides any information, I'd also appreciate if anyone has a workaround for this.</p>
| Daniel Lichtblau | 51 | <p>It is a problematic result to be sure. In the version under development there is modest improvement to the <code>Series</code>, and no change of note for <code>SeriesCoefficient</code>. The underlying issue I think is an inability to deduce a needed simplification.</p>
<pre><code>f[z_] := Sqrt[(z - I)/(z + I)]*(z + I);
v1 = Sinc[f[z]];
InputForm[v2 = Normal[Series[v1, {z, I, 4}]]]
(* Out[40]//InputForm=
Sinc[I*Sqrt[2]*Sqrt[-1 - I*z] + (Sqrt[-1 - I*z]*(-I + z))/(2*Sqrt[2]) +
((I/16)*Sqrt[-1 - I*z]*(-I + z)^2)/Sqrt[2] - (Sqrt[-1 - I*z]*(-I + z)^3)/
(64*Sqrt[2]) - (((5*I)/1024)*Sqrt[-1 - I*z]*(-I + z)^4)/Sqrt[2]] *)
</code></pre>
<p>The above is not exactly wrong but also not the desired bona fide <code>SeriesData</code> result. Numerically the evaluations work out though.</p>
<pre><code>Transpose[Chop[v1/v2 /. z->I+.01*Exp[2*I*Pi*Range[0,7]/8]]
(* Out[41]= {1., 1., 1., 1., 1., 1., 1., 1.} *)
</code></pre>
<p>If instead of using <code>Sinc</code> we have <code>Sin</code> then it becomes a different matter. Version 11.0 still delivers the "nice" form and the current version does not...</p>
<pre><code>w1 = Sin[f[z]];
InputForm[w2 = Normal[Series[w1, {z, I, 4}]]]
(* Out[49]//InputForm=
I*Sinh[Sqrt[2]*Sqrt[-1 - I*z] - ((I/2)*Sqrt[-1 - I*z]*(-I + z))/Sqrt[2] +
(Sqrt[-1 - I*z]*(-I + z)^2)/(16*Sqrt[2]) +
((I/64)*Sqrt[-1 - I*z]*(-I + z)^3)/Sqrt[2] - (5*Sqrt[-1 - I*z]*(-I + z)^4)/
(1024*Sqrt[2])] *)
Chop[w1/w2 /. z->I+.01*Exp[2*I*Pi*Range[0,7]/8]]
(* Out[50]= {1., 1., 1., 1., 1., 1., 1., 1.} *)
</code></pre>
<p>...but versions 11.0 and prior simply got this wrong, due to branch point issues. That numerical check has some results being the negative of what they should be (result below is from 11.0).</p>
<pre><code>(* Out[16]= {1., 1., 1., 1., 1., -1., -1., 1.} *)
</code></pre>
<p>I'll give some thought into how to improve in the case where an improvement exists. I will say this does not look promising though.</p>
|
1,009,082 | <p>Given is a linear map f from V to W, whereby V has dimension n and W has dimension m.</p>
<p>Now given n > m, can the map be injective,surjective or invertible?
And what about the same questions, given that m > n?</p>
<p>My thoughts so far: </p>
<ul>
<li><p>Invertibility should be possible in the second case, if we think of the map e.g. as a quadratic matrix which maps from the V to a subspace of W, but not in the first case (the matrix would not be quadratic.</p></li>
<li><p>Injectivity could be possible in the second case - else basically the a vector in W must have two pre-images</p></li>
<li><p>Surjectivity should be possible in the first case - the image of a linear map can not have a bigger dimensionaly than the space V</p></li>
</ul>
<p>Is this correct?</p>
| Seth | 31,659 | <p>The <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Stone.E2.80.93Weierstrass_theorem.2C_real_version" rel="nofollow">Stone-Weierstrass theorem</a> (a natural generalization of the Weierstrass approximation theorem) says that polynomials with all terms of even degree are uniformly dense in continuous functions on $[0,1]$, since they form a subalgebra which separates points and contains the constants. Using this you may proceed as in the proof of the first part. Note that the Stone-Weierstrass theorem does not apply to polynomials with all terms of odd degree as these don't even form an algebra. </p>
|
172,139 | <p>I have a data set, I am trying to join all the data by a line. But I am afraid the plot is not doing it properly. </p>
<p>This is the example (or almost) of my problem:</p>
<pre><code>data = {{0, π}, {π/2, π/2}, {π/2,
3 π/2}, {π, 0}, {π,
2 π}, {3 π/2, π/2}, {3 π/2,
3 π/2}, {2 π, π}};
</code></pre>
<p>Now, trying to interpolate the data.
Result is</p>
<p><a href="https://i.stack.imgur.com/VcL7A.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/VcL7A.jpg" alt="enter image description here"></a> </p>
<p>instead of</p>
<p><a href="https://i.stack.imgur.com/57Orh.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/57Orh.jpg" alt="enter image description here"></a></p>
<p>Sorry, I just downloaded and directly uploaded (as I couldn't get this result). $x,y$ are from $[0,2\pi]$.</p>
| corey979 | 22,013 | <pre><code>ListLinePlot[data[[FindShortestTour[data][[2]]]],
Frame -> True, AspectRatio -> 1, Epilog -> {Red, PointSize[Large], Point[data]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/un934.png" rel="noreferrer"><img src="https://i.stack.imgur.com/un934.png" alt="enter image description here"></a></p>
|
3,077,312 | <p>The proof given in my book (and I came up with as well) is:</p>
<p><a href="https://i.stack.imgur.com/H6eqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6eqf.png" alt="Proof"></a></p>
<p>However, the part that throws me off is line #3 where they do <span class="math-container">$\Sigma A_{jk} B_{ki} = \Sigma B_{ki} A_{jk}$</span></p>
<p>I understand that multiplication of real numbers is commutative and you can do the last step. However, you could've just as easily left the expression without swapping the two and arrived at the expression: <span class="math-container">$(AB)^t = A^tB^t$</span>, right?</p>
<p>What am I missing here?</p>
| b00n heT | 119,285 | <p>You are not missing much, just the fact that by convention we usually write
<span class="math-container">$$(AB)_{ij}=\sum_k A_{ik}B_{kj}$$</span>
and not (although equivalent)
<span class="math-container">$$(AB)_{ij}=\sum_k B_{kj}A_{ik}$$</span>
because in the first case “ikkj becoms ij”</p>
|
3,077,312 | <p>The proof given in my book (and I came up with as well) is:</p>
<p><a href="https://i.stack.imgur.com/H6eqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6eqf.png" alt="Proof"></a></p>
<p>However, the part that throws me off is line #3 where they do <span class="math-container">$\Sigma A_{jk} B_{ki} = \Sigma B_{ki} A_{jk}$</span></p>
<p>I understand that multiplication of real numbers is commutative and you can do the last step. However, you could've just as easily left the expression without swapping the two and arrived at the expression: <span class="math-container">$(AB)^t = A^tB^t$</span>, right?</p>
<p>What am I missing here?</p>
| user408858 | 408,858 | <p>What they want to make clear in line 3 is, that when you want to calculate the entry <span class="math-container">$ij$</span> of the transposed matrix, you can sum over the <span class="math-container">$i$</span>-th column of <span class="math-container">$B$</span> and <span class="math-container">$j$</span>-th row of <span class="math-container">$A$</span>. So this is the same when you sum over the <span class="math-container">$i$</span>-th row of <span class="math-container">$B^T$</span> and the <span class="math-container">$j$</span>-th column of <span class="math-container">$A^T$</span>. And that's what they do in the next line.</p>
<p>Again, this is the calculation:</p>
<p><span class="math-container">$$(AB)^T_{ij}=(AB)_{ji}=\sum_{k=1}^n A_{jk} B_{ki}=\sum_{k=1}^n B_{ki}A_{jk}=\sum_{k=1}^n B^T_{ik}A^T_{kj}=(B^TA^T)_{ij}$$</span></p>
|
59,932 | <p>So I need to make a plot of some ~2000 data points I have from a spreadsheet. I'm able to import the data just fine, it stores it like so</p>
<pre><code>sundat
</code></pre>
<blockquote>
<p>{{280.,0.082},{280.5,0.099},......{3995.,0.0087},{4000.,0.00868}}</p>
</blockquote>
<p>And I can call individual data sets or points like so</p>
<pre><code>sundat[[1]]
</code></pre>
<blockquote>
<p>{280.,0.082}</p>
</blockquote>
<pre><code>sundat[[1,1]]
</code></pre>
<blockquote>
<p>280.</p>
</blockquote>
<p>There are 2002 sets of two points of data. Now I can plot it fine using ListPlot, but the raw data I have is in messy units that I want to convert to mks units. I have the conversion factor for each column, but I have no idea how to do this. I presume I need to make a new list to apply this to, but trying something like</p>
<pre><code>data2 = {{sundat[[All,1]]*(conversion1)},{sundat[[All,2]]*(conversion2)}}
</code></pre>
<p>just totally messes up the table.</p>
<p>Any help would be appreciated.</p>
| Simon Woods | 862 | <p>You can also use <a href="http://reference.wolfram.com/language/ref/Dot.html" rel="noreferrer"><code>Dot</code></a>:</p>
<pre><code>sundat.{{2., 0}, {0, .5}}
(* {{560., 0.041}, {561., 0.0495}, {7990., 0.00435}, {8000., 0.00434}} *)
</code></pre>
|
76,420 | <p>The following code segment shows what I'd like to do. I'm a procedural programmer trying to learn the Mathematica functional style. Any help on this would be appreciated.</p>
<pre><code>B = {{1, 3}, {1, 5}, {4, 2}, {5, 2}, {5, 5}}
u = SparseArray[{{1, 2} -> 5, {1, 3} -> 9, {1, 4} -> 6,
{1, 5} -> Infinity, {3, 2} -> 2, {3, 4} -> 4,
{4, 2} -> 9, {5, 2} -> Infinity, {5, 3} -> Infinity,
{5, 4} -> Infinity, {5, 5} -> Infinity}];
r = {3, 0, -2, 10, 19}
(* Want to convert this python-looking code to functional Mathematica code
for {i, j} in B :
r[i] = r[i] - u[[i, j]]
r[j] = r[j] + u[[i, j]]
*)
</code></pre>
| C. E. | 731 | <p>This is just a way to rewrite the data into a - for our purposes - better format:</p>
<pre><code>u = {{1, 2} -> 5, {1, 3} -> 9, {1, 4} -> 6, {1, 5} -> Infinity, {3, 2} -> 2, {3, 4} -> 4, {4, 2} -> 9, {5, 2} -> Infinity, {5, 3} -> Infinity, {5, 4} -> Infinity, {5, 5} -> Infinity};
rules = B //. {a___, {i_Integer, j_Integer}, b___} :> {a, {i -> -({i, j} /. u), j -> ({i, j} /. u)}, b}
(* Out: {{1 -> -9, 3 -> 9}, {1 -> -∞, 5 -> ∞}, {4 -> -9, 2 -> 9}, {5 -> -∞, 2 -> ∞}, {5 -> -∞, 5 -> ∞}} *)
</code></pre>
<p>Using this set of rules:</p>
<pre><code>replace[r_, rules_] := r + SparseArray[rules, Length@r]
Fold[replace, r, rules]
(* Out: {-∞, ∞, 7, 1, Indeterminate} *)
</code></pre>
<p>In one way this is more functional than what belisarius and Bill are proposing because it doesn't update <code>r</code> in place. This is typical for functional programming.</p>
|
76,420 | <p>The following code segment shows what I'd like to do. I'm a procedural programmer trying to learn the Mathematica functional style. Any help on this would be appreciated.</p>
<pre><code>B = {{1, 3}, {1, 5}, {4, 2}, {5, 2}, {5, 5}}
u = SparseArray[{{1, 2} -> 5, {1, 3} -> 9, {1, 4} -> 6,
{1, 5} -> Infinity, {3, 2} -> 2, {3, 4} -> 4,
{4, 2} -> 9, {5, 2} -> Infinity, {5, 3} -> Infinity,
{5, 4} -> Infinity, {5, 5} -> Infinity}];
r = {3, 0, -2, 10, 19}
(* Want to convert this python-looking code to functional Mathematica code
for {i, j} in B :
r[i] = r[i] - u[[i, j]]
r[j] = r[j] + u[[i, j]]
*)
</code></pre>
| djp | 25,325 | <p>I don't think this is a good example for learning functional style. Of course it can be done as other answers show, but they are cryptic for two reasons:</p>
<p>(1) Mathematica doesn't accommodate "for {i, j} in B" (though Simon Woods' answer is pretty close)</p>
<p>(2) your code is actually depending on side effects (it is changing <code>r</code> each time the loop iterates)</p>
<p>The following is clear and fairly canonical:</p>
<pre><code>Do[
With[{i = x[[1]], j = x[[2]]},
r[[i]] = r[[i]] - u[[i, j]];
r[[j]] = r[[j]] + u[[i, j]];
],
{x, B}]
</code></pre>
<p>It translates into a functional style like this:</p>
<pre><code>f[{i_, j_}] :=
(
r[[i]] = r[[i]] - u[[i, j]];
r[[j]] = r[[j]] + u[[i, j]];
)
</code></pre>
<p>(to be clear, <code>f</code> has no output, and only side effects). We then apply <code>f</code> like this</p>
<pre><code>f /@ B
</code></pre>
<p>or to make it clearer that we are depending on side effects:</p>
<pre><code>Scan[f, B]
</code></pre>
|
119,696 | <p>Let us suppose i have two graphs for sequences A and B as follows</p>
<pre><code>a1 := {{0.9, 0.086133}, {0.086133, 0.0082432}, {0.0082432,
0.0007889}, {0.0007889, 0.0000755}, {0.0000755,
7.2256*10^-6}, {7.2256*10^-6, 6.9151*10^-7}, {6.9151*10^-7,
6.618*10^-8}, {6.618*10^-8, 6.3336*10^-9}, {6.3336*10^-9,
6.0615*10^-10}, {6.0615*10^-10, 5.801*10^-11}};
plot1 = ListPlot[a1, Joined -> True,
PlotStyle -> {Red, Thick, Dashing[{0.01}]}]
</code></pre>
<p>and </p>
<pre><code>a2 := {{0.9, 0.21797}, {0.21797, 0.052789}, {0.052789,
0.012785}, {0.012785, 0.0030963}, {0.0030963,
0.0007499}, {0.0007499, 0.00018162}, {0.00018162,
0.000043985}, {0.000043985, 0.000010653}, {0.000010653,
2.5799*10^-6}, {2.5799*10^-6, 6.24831*10^-7}};
plot2 = ListPlot[a2, Joined -> True, PlotStyle -> {Yellow, Thick}]
</code></pre>
<p>I combine both the graphs by </p>
<pre><code>Show[plot1, plot2, PlotRange -> All]
</code></pre>
<p>But cant do any things with usings Legends.
I want to use Legends to show that "Red colour box"=A and "yellow colour box"=B</p>
<p>Thanks in Advance</p>
| e.doroskevic | 18,696 | <h2>Example</h2>
<p><strong>Code</strong></p>
<pre><code>ListPlot[{a, b}, Joined -> True, PlotStyle -> {Red, Yellow}, PlotLegends -> Automatic]
</code></pre>
<p><strong>Note:</strong> <code>a</code> and <code>b</code> are your original data lists <code>a1</code> and <code>a2</code> respectively.</p>
<p><strong>Output</strong></p>
<p><a href="https://i.stack.imgur.com/o4QDN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o4QDN.png" alt="example"></a></p>
<p>With <code>SwatchLegend</code></p>
<p><strong>Code</strong></p>
<pre><code>ListPlot[
{a, b},
Joined -> True,
PlotStyle -> {Red, Yellow},
PlotLegends -> SwatchLegend[{"a", "b"}]
]
</code></pre>
<p><strong>Output</strong></p>
<p><a href="https://i.stack.imgur.com/Jp7M7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jp7M7.png" alt="example"></a></p>
<p>With <code>Show</code></p>
<p><strong>Code</strong></p>
<pre><code>Show[{
ListPlot[a, Joined -> True, PlotStyle -> Green],
ListPlot[b, Joined -> True, PlotStyle -> Red]
},
Epilog -> Inset[Panel @ SwatchLegend[{Green, Red}, {"a", "b"}], {0.2, .01}]
]
</code></pre>
<p><strong>Output</strong></p>
<p><a href="https://i.stack.imgur.com/5NWlO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5NWlO.png" alt="example output"></a></p>
|
3,607,430 | <blockquote>
<p>Given that <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> are the angles of a right-angled triangle, prove that:
<span class="math-container">$$\begin{align}
\sin a\sin b\sin(a-b) &+\sin b\sin c\sin(b-c)+\sin c\sin a\sin(c-a) \\ &+\sin(a-b)\sin(b-c)\sin(c-a)=0
\end{align}$$</span></p>
</blockquote>
<p>I know I'm supposed to use the properties of polynomials for this, as this question was found on the chapter on polynomials. I've tried having these values be roots of some function but don't know how to carry it out.</p>
<p>I also know that I can consider one of the angles to be <span class="math-container">$90^\circ$</span>, so the sine of that would be <span class="math-container">$1$</span>, but that dosen't really simplify it too much.</p>
| John Omielan | 602,049 | <p>You have the right idea. Since the lines involved in the question must have a <span class="math-container">$y$</span>-intercept, they can't be vertical and so they have a specific slope. Thus, they can be written in the form</p>
<p><span class="math-container">$$y = mx + b \tag{1}\label{eq1A}$$</span></p>
<p>The <span class="math-container">$y$</span>-intercept is where <span class="math-container">$x = 0$</span>, so it's</p>
<p><span class="math-container">$$y_i = b \tag{2}\label{eq2A}$$</span></p>
<p>and the <span class="math-container">$x$</span>-intercept is where <span class="math-container">$y = 0$</span>, i.e.,</p>
<p><span class="math-container">$$0 = mx_i + b \implies x_i = -\frac{b}{m} \tag{3}\label{eq3A}$$</span></p>
<p>As you want their product to be some number <span class="math-container">$c$</span>, you thus get</p>
<p><span class="math-container">$$x_i y_i = c \implies \frac{-b^2}{m} = c \implies b^2 = -mc \tag{4}\label{eq4A}$$</span></p>
<p>Next, as you also require the lines to pass through some point, let's say in general <span class="math-container">$(x_0, y_0)$</span>, you thus get from \eqref{eq1A} that</p>
<p><span class="math-container">$$y_0 = mx_0 + b \implies b = - mx_0 + y_0 \tag{5}\label{eq5A}$$</span></p>
<p>Substituting this into \eqref{eq4A} gives</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
(- mx_0 + y_0)^2 & = -mc \\
(x_0)^2m^2 - 2(x_0)(y_0)m + y_0^2 & = -mc \\
(x_0)^2m^2 + (c - 2(x_0)(y_0))m + y_0^2 & = 0
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$</span></p>
<p>If <span class="math-container">$c = 0$</span>, then you have <span class="math-container">$x_0 = 0$</span> and/or <span class="math-container">$y_0 = 0$</span>. I'll leave it to you to handle that case from \eqref{eq6A}. Otherwise, assuming <span class="math-container">$c \neq 0$</span>, then <span class="math-container">$x_0 \neq 0$</span>, so the <a href="https://en.wikipedia.org/wiki/Quadratic_formula" rel="nofollow noreferrer">quadratic formula</a> gives</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
m & = \frac{2(x_0)(y_0) - c \pm \sqrt{(c - 2(x_0)(y_0))^2 - 4(x_0^2)(y_0^2)}}{2x_0^2} \\
& = \frac{2(x_0)(y_0) - c \pm \sqrt{c^2 - 4c(x_0)(y_0) + 4(x_0^2)(y_0^2) - 4(x_0^2)(y_0^2)}}{2x_0^2} \\
& = \frac{2(x_0)(y_0) - c \pm \sqrt{c^2 - 4c(x_0)(y_0)}}{2x_0^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$</span></p>
<p>You can substitute these <span class="math-container">$m$</span> values into \eqref{eq5A} to get <span class="math-container">$b$</span> and, thus, can then substitute these <span class="math-container">$2$</span> values into \eqref{eq1A} to get the line equations. I'll leave it to you to plug into any specific values you want to use.</p>
|
3,607,430 | <blockquote>
<p>Given that <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> are the angles of a right-angled triangle, prove that:
<span class="math-container">$$\begin{align}
\sin a\sin b\sin(a-b) &+\sin b\sin c\sin(b-c)+\sin c\sin a\sin(c-a) \\ &+\sin(a-b)\sin(b-c)\sin(c-a)=0
\end{align}$$</span></p>
</blockquote>
<p>I know I'm supposed to use the properties of polynomials for this, as this question was found on the chapter on polynomials. I've tried having these values be roots of some function but don't know how to carry it out.</p>
<p>I also know that I can consider one of the angles to be <span class="math-container">$90^\circ$</span>, so the sine of that would be <span class="math-container">$1$</span>, but that dosen't really simplify it too much.</p>
| Peter Szilas | 408,605 | <p>Intercept equation of a line:</p>
<p><span class="math-container">$x/a+y/b=1$</span>;</p>
<p>This line passes through <span class="math-container">$(x_0,y_0):$</span></p>
<p><span class="math-container">$x_0/a+y_0/b=1$</span>;</p>
<p>Given: <span class="math-container">$c=ab$</span>;</p>
<p>Then </p>
<p><span class="math-container">$x_0/a+(ay_0/c)=1$</span>;</p>
<p>Solve for <span class="math-container">$a$</span>:</p>
<p><span class="math-container">$y_0 a^2-ac+cx_0=0$</span>;</p>
<p>Quadratic in <span class="math-container">$a$</span>: <span class="math-container">$a_{1,2}$</span>:</p>
<p>Lines: </p>
<p><span class="math-container">$x/a_{1,2} +y/(c/a_{1,2})=1.$</span></p>
|
300,531 | <p>Prove that : $$ \gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.$$</p>
<p>where $\gamma$ is Euler's constant ($\gamma \approx 0.57721$).</p>
<hr>
<p>This integral was mentioned in <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant">Wikipedia</a> as in <a href="http://mathworld.wolfram.com/Euler-MascheroniConstant.html">Mathworld</a> , but the solutions I've got uses corollaries from <a href="http://en.wikipedia.org/wiki/Harmonic_number">this theorem</a>. Can you give me a simple solution (not using much advanced theorems) or at least some hints.</p>
| robjohn | 13,854 | <p>In <a href="https://math.stackexchange.com/a/300574">this answer</a>, it is shown that since $\Gamma$ is log-convex,
$$
\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)\tag{1}
$$
Setting $x=1$ yields
$$
\Gamma'(1)=-\gamma\tag{2}
$$
The integral definition of $\Gamma$ says
$$
\begin{align}
\Gamma(x)&=\int_0^\infty t^{x-1}\,e^{-t}\,\mathrm{d}t\\
\Gamma'(x)&=\int_0^\infty\log(t)\,t^{x-1}\,e^{-t}\,\mathrm{d}t\\
\Gamma'(1)&=\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t\tag{3}
\end{align}
$$
Putting together $(2)$ and $(3)$ gives
$$
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag{4}
$$
Substituting $t\mapsto\log(1/t)$ transforms $(4)$ to
$$
\int_0^1\log(\log(1/t))\,\mathrm{d}t=-\gamma\tag{5}
$$</p>
|
103,970 | <p>Here's a very bizarre inconsistency I've just struggled with and I'm wondering why it exists or if I'm missing something.</p>
<p>I have some noisy data and I wish to make a framed plot of the data but allow the data to extend outside the vertical limits of the frame (for stylistic reasons). Like so:</p>
<pre><code> xs = Range[0, 0.5, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
opts = Sequence[PlotRange -> {{0, 0.5}, {0, 1}}, Frame -> True,
PlotRangeClipping -> False, ImagePadding -> 20];
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> All]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/vjuqv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vjuqv.png" alt="enter image description here"></a></p>
<p>However, my dataset contains points outside my desired x limits, thus my data is more accurately given by:</p>
<pre><code> xs = Range[-0.1, 0.6, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
</code></pre>
<p>Which when plotted with exactly the same code gives:</p>
<p><a href="https://i.stack.imgur.com/XstNd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XstNd.png" alt="enter image description here"></a></p>
<p>which obviously extends in both the x and y directions when I only want the extension in the y. </p>
<p>My solution is to change the value of <code>PlotRange -> All</code> in the 'Prolog' 'ListLinePlot'. However, this only works in the y-direction, observe:</p>
<pre><code> Grid[{{
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {All, {0, 1}}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, All}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, {0, 1}}]
]
}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/YuJ95.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YuJ95.png" alt="enter image description here"></a></p>
<p>As you can see the data never obeys the PlotRange in the x direction! Looking into the content of the <code>First@ListLinePlot[data, ...]</code> we can see that the graphics items describing the data do get clipped in the y-direction:</p>
<p><a href="https://i.stack.imgur.com/Gn4wW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gn4wW.png" alt="enter image description here"></a></p>
<p>You can see there are several instances near the beginning where the <code>Line</code>'s y-coordinate has been clipped to <code>0.</code> and many at the end where it is <code>1.</code>. </p>
<p>However if we try to restrict the graphics in the x-direction:</p>
<p><a href="https://i.stack.imgur.com/F0Zdy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F0Zdy.png" alt="enter image description here"></a></p>
<p>We see no such clipping occurs, leading to the problems described above.</p>
<p>Why is this happening? Is there an elegant workaround? My current method to circumvent this problem is to create and <code>Interpolation</code> of may data and then use <code>Plot</code> as opposed to <code>ListLinePlot</code> in the <code>Prolog</code> but this seems like overkill for what should be a simple fix?</p>
<p>I note that merely taking a subset of my data won't work for my real data as the x-values do not lie nicely on my coordinates, ie I might not have a value at 0 so I would be left with a gap either side. </p>
<p>Many thanks. </p>
| LLlAMnYP | 26,956 | <p>Here's some fake data:</p>
<pre><code>pts = RandomReal[{-.2, .2}, 37] + (Cos[.1 #] & /@ Range[-1, 35]) // Thread[{Range[-1, 35], #}] &
</code></pre>
<p>Make a plot without the frame that clips the data in the x direction but not in the y direction (by having an extended y range):</p>
<pre><code>plot = ListPlot[pts, InterpolationOrder -> 3, Joined -> True,
Frame -> False, Axes -> False, PlotRangePadding -> None,
ImagePadding -> None, PlotRange -> {{2, 32}, {-1.2, 1.2}},
AspectRatio -> Full]
</code></pre>
<p><a href="https://i.stack.imgur.com/DDvxo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DDvxo.png" alt="enter image description here"></a></p>
<p>Apparently <code>AspectRatio -> Full</code> is very important here, but I realized that with trial and error and am too lazy to investigate further. <code>ListPlot</code> with <code>Joined -> True</code> seems to behave nicer with <code>Inset</code> and gives the same result.</p>
<p>Make an empty framed plot with the correct frame range (defined by <code>PlotRange</code>) and add the line as an inset:</p>
<pre><code>Plot[{}, {x, 2, 32}, Frame -> True, PlotRange -> {{2, 32}, {-1, 1}},
PlotRangeClipping -> False,
Epilog -> Inset[plot, {2, 0}, {2, 0}, {30, 2.4}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/czvxv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/czvxv.png" alt="enter image description here"></a></p>
<p>Here's a functional form:</p>
<pre><code>xClipListPlot[pts_List, {{x1_, x2_}, {y1_, y2_}},
yRange_List: Automatic] :=
Module[{x,
yR = If[yRange ==
Automatic, {(y1 + y2)/2 - (y1 - y2)/0.4, (y1 + y2)/2 + (
y1 - y2)/0.4}, yRange], plot},
plot = ListPlot[pts, InterpolationOrder -> 3, Joined -> True,
Frame -> False, Axes -> False, PlotRangePadding -> None,
ImagePadding -> None, PlotRange -> {{x1, x2}, yR},
AspectRatio -> Full];
Plot[{}, {x, x1, x2}, Frame -> True,
PlotRange -> {{x1, x2}, {y1, y2}}, PlotRangeClipping -> False,
Epilog ->
Inset[plot, {0, 0}, {0, 0}, {x2 - x1, First@Differences[yR]}]]]
</code></pre>
|
2,333,857 | <p>I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!</p>
<p>Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid. </p>
<p>$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$</p>
<p><a href="https://i.stack.imgur.com/FtgoC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FtgoC.jpg" alt="Surface area problem worked out"></a></p>
| MPW | 113,214 | <p>Your second alternative is the correct one.</p>
<p>The value of $x$ is not a <em>set</em> of numbers, rather it is <em>in a set</em> of numbers.</p>
|
347,315 | <p>How can I find </p>
<ol>
<li>The image of the upper half plane $\mathrm{Im}(z)>0$ under the linear fractional transformation $w=\dfrac{3z + i}{-iz + 1}$.</li>
<li>The image of the set {${z∈C-0:\{\mathrm{Im}(z)} = \mathrm{Re}(z)\}$} under $w=z + \dfrac{1}{z}$.</li>
</ol>
<p>For 1.,
I consider $y>0, x \in R$ and then find the inverse $z=\dfrac{w-i}{3+iw}$. This gave me $x=\dfrac{2u}{{(3-v)}^2+u^2}$ and $y=\dfrac{-v^2-u^2+4v-3}{(3-v)^2+u^2}$. Then since $y>0$,
${-v^2-u^2+4v-3}>0$ which gave me ${(v-2)}^2+u^2<1$.
I therefore knew that the image is the interior of the unit disk centered at $(0,2)$ and drew it. So I need confirmation.
But for the second one, I am confused as how the image should be because if $\text{Im}(z)=\text{Re}(z)$ and $w=z+\dfrac{1}{z}$, then $z$ should not be $0$, but I thought the the line $\text{Im}(z)=\text{Re}(z)$ must pass through the origin. I am waiting for help!</p>
| sebigu | 32,185 | <p>I can answer the question like this:</p>
<p>Looking at the Fratini subgroups using GAP shows that the Fratini subgroup of $SL_2(3)$ is of order $2$, where the one of $PGL_2(3)$ is trivial.</p>
<p>Assuming the algorithms are correct (which I guess, since they have been testet much), the groups are not isomorphic. You might want to prove this by hand.</p>
<p>So you get a counterexample for your claim.</p>
|
69,961 | <p>I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s.</p>
<p>$$S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}$$</p>
<p>I want to check whether that would be:
0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on.</p>
<p>Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set?</p>
| Brian M. Scott | 12,042 | <p>You can write $$S = \{0,3,5\} \cup \{n\in\mathbb{N}:n\ge 8\}.$$ You can also write $$S = (\mathbb{N}\cup\{0\})\setminus \{1,2,4,7\}.$$</p>
<p>Once you know that $8,9,10\in S$, you can prove that every natural number $n\ge 8$ belongs to $S$ by showing that if $n\ge 8$, there is a non-negative integer $k$ such that $n$ is $3k+8$, $3k+9$, or $3k+10$: then you can start with $3$’s and $5$’s adding up to $8$, $9$, or $10$ and add $k$ more $3$’s.</p>
<p>HINT: Look at the quotient and remainder when you divide $n$ by $3$.</p>
|
3,547,989 | <p><strong>An ordinary deck of cards is dealt randomly to four players so that each player receives 13 cards.
Find the probability that each player is dealt exactly one ace.</strong></p>
<p>I was wondering if I could compare this scenario with the following:
Given 4 indistinguishable balls and 4 distinguishable boxes, if we distribute the balls randomly, what is the probability that each box has exactly one ball?
The answer to that is <span class="math-container">$\frac{1}{4+4-1\choose 4}=\frac{1}{7\choose 4}=\frac{1}{35}$</span>. So is that also the answer to the question above? If not then why?</p>
| Robert Lewis | 67,071 | <p>We have</p>
<p><span class="math-container">$(X(X'X)^{-1}X')^2 = (X(X'X)^{-1}X')(X(X'X)^{-1}X') = X(X'X)^{-1}(X'X)(X'X)^{-1}X'=X(X'X)^{-1}X', \tag 1$</span></p>
<p>which shows that <span class="math-container">$X(X'X)^{-1}X'$</span> is idempotent. Now note that for any idempotent <span class="math-container">$P$</span>,</p>
<p><span class="math-container">$P^2 = P, \tag 2$</span></p>
<p>we have</p>
<p><span class="math-container">$(I - P)^2 = I - 2P + P^2 = I - 2P + P = I - P \tag 3$</span></p>
<p>is also idempotent; applying this fact to <span class="math-container">$X(X'X)^{-1}X'$</span> shows that <span class="math-container">$I_n - X(X'X)^{-1}X'$</span> is idempotent as well.</p>
<p>We compute the trace of <span class="math-container">$X(X'X)^{-1}X'$</span> as follows: in <a href="https://en.wikipedia.org/wiki/Trace_(linear_algebra)" rel="nofollow noreferrer">the wikipedia article on traces</a>, it is shown that
for <span class="math-container">$A$</span> and <span class="math-container">$n \times m$</span> matrix and <span class="math-container">$B$</span> and <span class="math-container">$m \times n$</span> matrix, </p>
<p><span class="math-container">$\text{tr}(AB) = \text{tr}(BA); \tag 4$</span></p>
<p>applying this to <span class="math-container">$X(X'X)^{-1}$</span> and <span class="math-container">$X'$</span> yields</p>
<p><span class="math-container">$\text{tr}((X(X'X)^{-1})X') = \text{tr}(X'(X(X'X)^{-1}))$</span>
<span class="math-container">$= \text{tr}((X'X)(X'X)^{-1}) = \text{tr}(I_p) = p, \tag 5$</span></p>
<p>where <span class="math-container">$I_p$</span> is the <span class="math-container">$p \times p$</span> identity matrix. Then</p>
<p><span class="math-container">$\text{tr}(I_n - X(X'X)^{-1})X') = \text{tr}(I_n) - \text{tr}(I_p) = n - p. \tag 6$</span></p>
|
3,335,892 | <p>If I have two injective functions <span class="math-container">$f : A \to B$</span> and <span class="math-container">$g : B \to A$</span>, as Schröder-Bernstein (SB) says, then there is a function <span class="math-container">$h : A \to B$</span> which is bijective.</p>
<p>As for a proof, my reasoning goes something like this:</p>
<p>The injectivity of <span class="math-container">$f \implies |A| \leq |B|$</span>.
Similarly, the injectivity of <span class="math-container">$g \implies |B| \leq |A|$</span>. At this point I would say that it is perhaps obvious that <span class="math-container">$|B| = |A|$</span> in order for the prior statements to remain true.</p>
<p>With that being said, the final question is whether or not <span class="math-container">$|A| = |B| \implies $</span> that there exists a function <span class="math-container">$h : A \to B$</span> which is bijective? I am reading (perhaps somewhat naively) on wikipedia that if X and Y are finite sets then a bijection exists <span class="math-container">$ \leftrightarrow$</span> <span class="math-container">$|A| = |B|$</span>.</p>
<p>Taking the if and only if symbol as a statement of equivalence means that, at least in the finite case, considering the cardinalities of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> proves the existence of <span class="math-container">$h$</span>?</p>
| Asaf Karagila | 622 | <p>This is not obvious at all that <span class="math-container">$|A|\leq|B|$</span> and <span class="math-container">$|B|\leq|A|$</span> imply that <span class="math-container">$|A|=|B|$</span>.</p>
<p>Consider a different notion of equivalence defined, say, on linearly ordered sets, with <span class="math-container">$|A|\leq|B|$</span> meaning now that there is an order embedding of <span class="math-container">$A$</span> into <span class="math-container">$B$</span>, and <span class="math-container">$|A|=|B|$</span> means that there is an order isomorphism. (Of course, there is an implicit order, <span class="math-container">$<_A$</span> and <span class="math-container">$<_B$</span> given on these sets.)</p>
<p>Now consider <span class="math-container">$A=[0,1]$</span> and <span class="math-container">$B=(0,1)$</span> as ordered by the standard ordering of the real numbers. Then <span class="math-container">$A$</span> injects into <span class="math-container">$B$</span> by <span class="math-container">$f(x)=\frac14+\frac x2=\frac{2+x}4$</span>, and <span class="math-container">$B$</span> injects into <span class="math-container">$A$</span> by the identity function. But <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are not isomorphic.</p>
<p>Well, you might argue, they are kinda almost isomorphic. Then we can instead take <span class="math-container">$A=[0,1]$</span> and <span class="math-container">$B=[0,1]\cup[2,3]$</span> and now <span class="math-container">$A$</span> maps into <span class="math-container">$B$</span> using the identity function, and <span class="math-container">$B$</span> maps into <span class="math-container">$A$</span> by <span class="math-container">$x\mapsto\frac x3$</span>. Again, these two are not isomorphic.</p>
<hr>
<p>The reason you say that this is trivial, perhaps, is that you're thinking about the finite cardinals as a model of how sets should behave. There, if there are two injections then the equality is indeed an easy consequence of the pigeonhole principle.</p>
<p>This might have been Cantor's intuition, who also considered this somewhat trivial, by using an implicit assumption: <span class="math-container">$A$</span> and <span class="math-container">$B$</span> can be well-ordered, and then we can take the shortest order type of each and prove that they must be order isomorphic, similar to the pigeonhole principle argument.</p>
<p>But this appeals to the well-ordering principle, which many people have claimed to be "counter-intuitive",<sup>1</sup> and at the very least is not trivially obvious. And without the axiom of choice the Cantor–Bernstein theorem is still provable, so this cannot be the reason de jure for its proof to work. And infinite sets, in general, are <em>very much</em> not like finite sets.</p>
<p>Just to satisfy some of your curiosity, we can replace injections by surjections, and ask what happens if there are surjections from <span class="math-container">$A$</span> onto <span class="math-container">$B$</span> and from <span class="math-container">$B$</span> onto <span class="math-container">$A$</span>. Does that entail the existence of a bijection? For finite sets, obviously yes, for the same reasons. But in general, without assuming the axiom of choice, we cannot prove this. We know that it is consistent that the axiom of choice fails, and there are two sets which surject onto one another, but there is no bijection between them.</p>
<p>Lest you worry too much, these are the kind of mistakes that many a great mathematician once made. You're in good company.</p>
<p>At the end of the day, the claim that <span class="math-container">$|A|\leq|B|$</span> and <span class="math-container">$|B|\leq|A|$</span> implies <span class="math-container">$|A|=|B|$</span> <em>is</em> restating the Cantor–Bernstein theorem. </p>
<hr>
<p>Footnotes:</p>
<ol>
<li>I disagree with that claim of course, and attribute it to historically bad teaching around this subject that treats this as somehow witchcraft is involved in producing this well-ordering. While it's true that it is not an explicit proof, having a well-ordering of the real numbers is no more counter-intuitive than proving that the rational numbers can be enumerated.</li>
</ol>
|
1,978,532 | <p>A sequence with a limit that is a real number is called a convergent sequence. </p>
<p>A sequence which does not converge is said to be divergent. </p>
<p>Find two divergent sequences like $\{x_k\},\{y_k\}$ such that $\lim_{k \rightarrow \infty}|y_k-x_k|=0$ . </p>
<p><strong>Notice that we know $\forall k \in \mathbb N \space 0 \lt |y_k-x_k|$</strong> </p>
<p>Note ( For those who ask about my try ) : There is nothing to try! If my try was successful, I wouldn't be asking this question.</p>
| ziggurism | 16,490 | <p>The rationals in the interval are one class. $\pi$ plus rationals which lie in the interval are another ($\pi-3$, $\pi-22/7$, etc). $\sqrt{2}-1$ represents a class, so does $e-2$. But I can't construct a list of representatives of all equivalence classes without appealing to the axiom of choice. </p>
<p>The easiest way to get the list using the axiom of choice is to view the reals as a vector space over the rationals, and remember that the axiom of choice is equivalent to the assertion that every vector space has a basis. A basis for the reals over the rationals gives representatives for the equivalence classes under your equivalence relation. You'll want to check that you can arrange for them to be in the prescribed interval of course. </p>
<p>You say $x-y$ is rational only if both $x$ and $y$ are rational. This is not true. Consider $\pi$ and $\pi+1$. Neither are rational but their difference is rational.</p>
|
1,978,532 | <p>A sequence with a limit that is a real number is called a convergent sequence. </p>
<p>A sequence which does not converge is said to be divergent. </p>
<p>Find two divergent sequences like $\{x_k\},\{y_k\}$ such that $\lim_{k \rightarrow \infty}|y_k-x_k|=0$ . </p>
<p><strong>Notice that we know $\forall k \in \mathbb N \space 0 \lt |y_k-x_k|$</strong> </p>
<p>Note ( For those who ask about my try ) : There is nothing to try! If my try was successful, I wouldn't be asking this question.</p>
| Piquito | 219,998 | <p>For all irrational $I\in [0,1)$ you have by definition
$$\widetilde I=\text { class of } I=\{I+r;\space r\in \Bbb Q\}$$</p>
|
1,657,115 | <p>$$\int \frac{2x + 10}{(x^2 + 5x + 8) ^ 2}dx$$
we can rewrite as:
$\int \frac{2x + 5}{(x^2 + 5x + 8) ^ 2}dx$ + $\int \frac{5}{(x^2 + 5x + 8) ^ 2}dx$</p>
<p>first one is easy. What can we do with the second one?</p>
| Narasimham | 95,860 | <p>Slope of PD =-1. Slope of AB is +1 . Your last equation is quite correct. After finding x, find the y.</p>
|
4,302,855 | <p>I have tried setting up multiple systems of equations using many known volumes but I always seem to come up short. My last attempt was a hollow cylinder but that leaves you with three unknowns in only two sim. equations (for V and S.A). Can anyone help?</p>
| PC1 | 960,197 | <p>We can use <a href="https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem" rel="nofollow noreferrer">Pappus's centroid theorems</a> for this purpose.</p>
<p>I will take a torus for example, with a centroid at a distance <span class="math-container">$R$</span> from the origin axis. Let's say that my torus radius is <span class="math-container">$r$</span>. So using the 2 theorems, I find that the area <span class="math-container">$A$</span> and the volume <span class="math-container">$V$</span> are:</p>
<p><span class="math-container">$$A = (2\pi R)(2\pi r)$$</span>
<span class="math-container">$$V = (2\pi R)(\pi r^2)$$</span></p>
<p>From your hypothesis, <span class="math-container">$\frac V A = 2$</span> so <span class="math-container">$r=4$</span>, and we can find from there that <span class="math-container">$R=\frac 9{4\pi}$</span>. There might be other ways to find another solution.</p>
<p>EDIT:</p>
<p>Actually <span class="math-container">$r>R$</span> so I don't thinks the solution works. It might be that there is no such solid in the way I tried to solve it...</p>
|
3,051,207 | <p>In the very first chapter in my class "mathematical analysis 1" I've seen something called <strong>the triangle inequality</strong>, which is <span class="math-container">$||a| - |b|| \leq |a \pm b| \leq |a| + |b|$</span>. Now the thing is that I do understand why this is true, but i fail to see what this actually has to do with triangles. It was never explained nor proven. I'd like to know how this inequality relates to triangles to get a better understanding of it, because it's also used a lot further on. Would someone be able to explain this?</p>
| Bernard | 202,857 | <p>Suppose <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are complex numbers, so that they correspond in the Argand-Cauchy plane, to two sides of a triangle, the third side corresponding to <span class="math-container">$a-b$</span>. Then these inequalities correspond to well-known from middle school inequalities about triangles – namely that the length of a side of a triangle is between the difference and the sum of the lengths of the two other sides.</p>
|
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