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 |
|---|---|---|---|---|
1,002,719 | <p>If we have</p>
<p>$f: \{1, 2, 3\} \to \{1, 2, 3\}$</p>
<p>and</p>
<p>$f \circ f = id_{\{1,2,3\}}$</p>
<p>is the following then always true for every function?</p>
<p>$f = id_{\{1,2,3\}}$</p>
| amWhy | 9,003 | <p>Note that we have $f^{-1}\circ f = f\circ f^{-1}= id\{1, 2, 3\}\;\forall \text{ permutations on } \{1, 2, 3\}$.</p>
<p>By definition, all permutations on a set are bijective functions, so every permutation has an inverse.</p>
<p>You need only find a single bijective function $f$ which <em>is its own inverse,</em> (a permutation of order $2$ and you'll have your counter-example.</p>
|
201,576 | <p>Struggling with something basic. Suppose when defining f[x_] the outcome depends on Sign[x]. When calling this function, how do I tell Mathematica the sign of the argument?
My attempt:</p>
<pre><code>f[x_]=x Sign[x];
Assuming[a>0, f[a]]
</code></pre>
<p>The output I get is </p>
<pre><code>a Sign[a]
</code></pre>
<p>Obviously, I want to get simply <code>a</code> because I am trying to tell Mathematica <code>a</code> is positive</p>
<p><strong>Additional details:</strong></p>
<p>Thanks for the answers below. What I am really trying to do is as follows:</p>
<pre><code>p = x /. Solve[(a - x)/y^2 == 1, x];
q = ((3 a - 2 p)/(4 y^2));
f[x_] = x Sign[x] ;
Assuming[q > 0, Simplify[f[q]]]
</code></pre>
<p>If I delete the first line in which <code>p</code> is defined via Solve, everything works fine. If, however, the first line is present, the output is: </p>
<pre><code>{((a + 2 y^2) Sign[a + 2 y^2])/(4 y^2 Sign[y]^2)}
</code></pre>
<p>So, the Sign operator is still there.</p>
<p>What's wrong? </p>
| Bob Hanlon | 9,362 | <pre><code>$Version
(* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *)
Clear["Global`*"]
p = x /. Solve[(a - x)/y^2 == 1, x];
q = ((3 a - 2 p)/(4 y^2));
f[x_] = x Sign[x];
Assuming[q > 0, FullSimplify[f[q]]]
(* {Abs[a + 2 y^2]/(4 y^2)} *)
</code></pre>
<p>Note the presence of <a href="https://reference.wolfram.com/language/ref/List.html" rel="nofollow noreferrer"><code>List</code></a> brackets that originate in your definition of <code>p</code></p>
<pre><code>Clear["Global`*"]
</code></pre>
<p>Use <a href="https://reference.wolfram.com/language/ref/Part.html" rel="nofollow noreferrer"><code>Part</code></a> to remove the <code>List</code> brackets in the definition of <code>p</code></p>
<pre><code>p = x /. Solve[(a - x)/y^2 == 1, x][[1]];
q = ((3 a - 2 p)/(4 y^2));
f[x_] = x Sign[x];
Assuming[q > 0, Simplify[f[q]]]
(* 1/4 (2 + a/y^2) *)
</code></pre>
<p><strong>EDIT:</strong> For the more complicated expression in your comment,</p>
<pre><code>p = x /. Solve[(a - x)/(c y^2) + (b - x)/(d y^2) == 1, x][[1]];
q = ((3 a - 2 p)/(4 y^2))
(* (3 a - (2 (b c + a d - c d y^2))/(c + d))/(4 y^2) *)
f[x_] = x Sign[x];
</code></pre>
<p>If all the variables are real, </p>
<pre><code>sol = Assuming[
q > 0, (f[q] // ComplexExpand[#, TargetFunctions -> {Re, Im}] &)]
(* Sqrt[(3 a - (2 (b c + a d - c d y^2))/(c + d))^2]/(4 y^2) *)
</code></pre>
<p>This avoids <code>Abs</code> but is not the simplest form. Since the numerator is <code>Sqrt[q^2]</code> and <code>q > 0</code></p>
<pre><code>sol /. Sqrt[(z_)^2] :> z // Simplify
(* (3 a c - 2 b c + a d + 2 c d y^2)/(4 (c + d) y^2) *)
</code></pre>
|
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
<p>$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$</p>
<p>I would appreciate your input!</p>
| Daniel Franke | 86,152 | <p>When I saw your question I was immediately reminded of this:</p>
<p><a href="http://logitext.mit.edu/logitext.fcgi/tutorial" rel="nofollow">http://logitext.mit.edu/logitext.fcgi/tutorial</a></p>
<p>See also this related blog post by the author:</p>
<p><a href="http://blog.ezyang.com/2012/05/an-interactive-tutorial-of-the-sequent-calculus/" rel="nofollow">http://blog.ezyang.com/2012/05/an-interactive-tutorial-of-the-sequent-calculus/</a></p>
<p>It's essentially a rudimentary proof assistant for the LK calculus, presented with an interface that makes it feel very game-like. I think it provides strong validation for your idea of gamifying Euclidean geometry, by demonstrating that formal logic <em>in general</em> can be convincingly gamified, even absent the nice visual aids that a geometry game would allow you to provide.</p>
|
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
<p>$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$</p>
<p>I would appreciate your input!</p>
| hardmath | 3,111 | <p>Graph theory, particularly for planar graphs, seems a good area for the sort of recreational math puzzles to keep club members interested.</p>
<p>Here are two that I learned about recently:</p>
<blockquote>
<p><a href="https://en.wikipedia.org/wiki/Planarity" rel="nofollow noreferrer">Planarity</a> The goal is to move vertices
around and obtain a straight-edged embedding without crossings (computer-based)</p>
<p><a href="https://en.wikipedia.org/wiki/Sprouts_(game)" rel="nofollow noreferrer">Sprouts</a> Two players take turns
adding non-crossing edges (which may be curved) to a graph until a maximal planar
graph is reached (pencil-and-paper; normal and misère forms of play).</p>
</blockquote>
<p>Better known is <a href="https://en.wikipedia.org/wiki/Hex_(board_game)" rel="nofollow noreferrer">the game of Hex</a>,
which I learned of (many years ago) from Martin Gardner's <a href="https://en.wikipedia.org/wiki/Mathematical_Games_(column)#Mathematical_Games" rel="nofollow noreferrer">Mathematical Games
column</a>
in Scientific American, doubtless <a href="https://en.wikipedia.org/wiki/Martin_Gardner_bibliography" rel="nofollow noreferrer">another good source</a> of ideas.</p>
|
1,329,398 | <p>So, I've posted a question regarding Wikipedia's quartic page. This was from the first question.</p>
<blockquote>
<p>I'm trying to implement the general quartic solution for use in a ray tracer, but I'm having some trouble. The solvers I've found do cause some strange false negatives leaving holes in the tori I'm testing with.</p>
<p>Most implementations use the depressed quartic solutions, I don't understand the math involved and can't figure out why I'm having false non-intersections (link to layman explanation would be great). So I'm trying to implement the general solution at <a href="https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots" rel="nofollow noreferrer">this wikipedia page</a>. I got the stuff up until the special cases implemented, but at that point I have an issue.</p>
</blockquote>
<p>With lots of rays being traced most of the special cases become common. I've found a set of coefficients that <a href="http://www.wolframalpha.com/input/?i=0.999999999x%5E4-24.591812430x%5E3%2B226.372338960x%5E2-924.44347730x%2B1412.89015972" rel="nofollow noreferrer">Wolfram Alpha tells me has two real roots</a>, but my code was just returning NaN, further searching I found my S was coming up as <span class="math-container">$\sqrt{-4.9 \times 10^{-11}}$</span> Floating point precision error, means this should equate to 0, so I need the special case for S=0, it says we need to "change choice of cubic root in Q" but it does not explain how to do this. I did try changing the sign of Q when S=0, but that doesn't work. Does anyone know what this means and how I can do it?</p>
| asmeurer | 781 | <p>If you are interested in the various ways of representing the quartic solution symbolically, the <a href="https://github.com/sympy/sympy/blob/acb005ce9971ea7714b785383414258279878fd4/sympy/polys/polyroots.py#L244" rel="nofollow">SymPy code</a> goes through a few methods and tries to pick the best one for the given polynomial. There are also some useful references there for further reading. </p>
|
2,881,673 | <p>I've searched all over the internet and cannot seem to factorise this polynomial.</p>
<p>$x^4 - 2x^3 + 8x^2 - 14x + 7$</p>
<p>The result should be $(x − 1)(x^3 − x^2 + 7x − 7)$</p>
<p>What are the steps to get to that result?
I've tried grouping but doesn't seem to work...</p>
| mfl | 148,513 | <p>$x-a$ is a factor of $x^4 - 2x^3 + 8x^2 - 14x + 7$ if and only if $a^4 - 2a^3 + 8a^2 - 14a + 7=0.$ Integers that can work are the divisors of $7$ (the independent term). That is: $\pm 1,\pm 7.$ If we check with $x=1$ we get</p>
<p>$$1^4 - 2\cdot 1^3 + 8\cdot 1^2 - 14\cdot1 + 7=0.$$ So $x-1$ is a factor. That is, there exists a polynomial $p(x)$ of degree $3$ such that $$x^4 - 2x^3 + 8x^2 - 14x + 7=(x-1)p(x).$$</p>
<p>To get $p(x)$ you have to divide $(x^4 - 2x^3 + 8x^2 - 14x + 7):(x-1)$ in the way you prefer.</p>
<p>You should get </p>
<p>$$(x^4 - 2x^3 + 8x^2 - 14x + 7)=(x-1)(x^3-x^2+7x-7).$$</p>
<p>Proceeding in the same way you have</p>
<p>$$x^3-x^2+7x-7=(x-1)(x^2+7).$$ Or </p>
<p>$$(x^4 - 2x^3 + 8x^2 - 14x + 7)=(x-1)^2(x^2+7).$$</p>
|
1,415,752 | <p>I test my answer using wolfram alpha pro but it gets a different result to what I am getting. This is homework.</p>
<p>My result is z= 2(y-1)</p>
<p>partial derivative with respect to y is </p>
<pre><code> x.y^x-1
</code></pre>
<p>partial derivative with respect to x is
ln(y).y^x</p>
<p>ln(1) is zero, so the x term is zero, leaving 2^(2-1) = 2</p>
<p>so z = 2(y-1) </p>
| Yes | 155,328 | <p>In fact, the square root $\sqrt{1-x}$ is meaningless for all $x > 1$. So, in particular, the limit $\lim_{x \to 1+}\sqrt{1-x}$ does not exist at all.</p>
|
375,372 | <p>Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$</p>
<p>Working so far: </p>
<p>$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$</p>
<p>Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$</p>
<p>$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$</p>
<p>$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$</p>
<p>$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$</p>
<p>And now I'm stuck.. Any help would be great, thanks.</p>
| Santosh Linkha | 2,199 | <p>use the fact that $ \left | \frac 1 x - 0\right| < \delta $
$$\left| \frac{-3x - 17}{x^2 + x + 8}\right| < \left| \frac{-3x - 17}{x^2 }\right| \le 3\left |\frac 1 x \right | + 17\left |\frac 1 {x^2} \right | < 20 \delta = \epsilon $$</p>
|
1,918,435 | <p>Assume I have a set $S=\{1,\ldots,N\}$, where $N$ is an integer. In my case, $N$ can be zero or non-zero, depending on situation. </p>
<p>If $N=0$, does it automatically mean that $S=\emptyset$? I feel like it should, but would like to be sure. Thanks in advance for replying.</p>
| Mees de Vries | 75,429 | <p>Typically, if $S$ is informally defined by $S = \{1,\ldots,N\}$, then if $N = 1$ this means $S = \{1\}$ and if $N = 0$ this means $S = \emptyset$.</p>
|
1,918,435 | <p>Assume I have a set $S=\{1,\ldots,N\}$, where $N$ is an integer. In my case, $N$ can be zero or non-zero, depending on situation. </p>
<p>If $N=0$, does it automatically mean that $S=\emptyset$? I feel like it should, but would like to be sure. Thanks in advance for replying.</p>
| Asaf Karagila | 622 | <p>Yes. In the context of the natural numbers, if we write $S=\{1,\ldots,N\}$, then we mean really $\{x\in\Bbb N\mid 1\leq x\leq N\}$. If $N<1$, then there are no such $x$ and the set is empty.</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Per Alexandersson | 1,056 | <p>Read more papers, related to your problem. You can get inspiration, or stumble upon a proof of exactly what you are looking for. It could also be that you learn some technique in a proof.</p>
<p>I recently realized that a proof of something I needed was already proved in an earlier paper of mine, but it was not actually stated in the actual theorem (since I did not know of that particular application at that time).</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| მამუკა ჯიბლაძე | 41,291 | <p>My personal experience - it helps to listen to some talks, not necessarily very closely related to your problem, ideally not online, so you can discuss it with somebody after the talk, etc. Try to find analogies, maybe even superficial ones, the more unexpected the better. It is not only about thinking out of the box, what is more important is that somebody might be "digging from the opposite side" for the same thing you are after.</p>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| GPerez | 118,574 | <p>To be honest, I would refrain from saying that $$\sqrt{-1} = i$$
It's not even like the real numbers, where one can agree that the square root function takes on the positive values, because what is positive, in the complex plane? Even if you did establish the identity as convention, how would you extend the convention when dealing with: $$\sqrt[\Large 3]{1}$$
for example?</p>
<p>If you need to define the imaginary unit, the best choice is to say that $i$ is such that $$i^2 = -1
$$</p>
<p>There are even "better" choices, but it doesn't look like you'll be needing them.</p>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| Espen Nielsen | 45,874 | <p>This question can be approached from one of two angles. Let's do one at a time.</p>
<ol>
<li>Can $\sqrt{-1}$ be both $i$ and $-i$?</li>
</ol>
<p>The answer is in fact yes. People realized early on that some functions may take any of a number of values. This is true already in the real number system with the square root, but usually well-behaved functions like the logarithm also become "multi-valued" when extended to take as input complex numbers. The solution is to change the domain of the function, meaning that the square root takes input from "two copies of the complex numbers sewn together", something called a <a href="http://en.wikipedia.org/wiki/Riemann_surface" rel="nofollow">Riemann surface.</a></p>
<ol start="2">
<li>Are $\sqrt{-1}=i$ and $\sqrt{-1}=-i$ in some sense "the same"?</li>
</ol>
<p>Again, the answer is yes. As Meelo remarked in his answer, both $i$ and $-i$ are solutions to the polynomial equation $x^2+1=0$. However, they are equivalent solutions, in the sense that interchanging them generates a symmetry of the complex plane. This symmetry is described by flipping the plane about the real axis, also known as complex conjugation. This is the starting point of <a href="http://en.wikipedia.org/wiki/Galois_theory" rel="nofollow">Galois theory.</a></p>
|
1,869,119 | <p>Show that the Monotone Convergence Theorem may not hold for decreasing sequences of functions.</p>
<p>Suppose $\left\{f_{n}\right\}$ is a sequence of nonnegative decreasing functions converging to $f$ pointwise. I know that if $f_{1}$ is finite,we can construct the sequence say $\left\{f_{1}-f_{n}\right\}$ which is increasing to prove that MCT still hold for decreasing sequence. So the only way in which the MCT is not valid is when $f_{1}$ is not finite. I don't know how to construct such sequence.</p>
| Martín Vacas Vignolo | 297,060 | <p>Put $f_n=\infty\chi_{(0,1/n)}$</p>
|
209,856 | <p>Of course, I can use Stirling's approximation, but for me it is quite interesting, that, if we define $k = (n-1)!$, then the left function will be $(nk)!$, and the right one will be $k! k^{n!}$. I don't think that it is a coincidence. It seems, that there should be smarter solution for this, other than Stirling's approximation.</p>
| platinor | 93,324 | <p>Take $\log$ on both sides and use the $\log {n!} = \Theta(n\log n)$.
The first terms becomes $\Theta(n!\log{(n!)})$, the second one becomes $\Theta((n-1)!\log {(n-1)!}) + \Theta(n!\log{(n-1)!})$. So it's obvious that the first terms grows faster than the second one.</p>
|
424,694 | <p>Let <span class="math-container">$p$</span> be a prime, and consider <span class="math-container">$$S_p(a)=\sum_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$</span>
I have a rather complicated (15 lines) proof that <span class="math-container">$S_p(a)\equiv0\pmod{p}$</span>. This must be
extremely classical: is there a simple direct proof ?</p>
| Ira Gessel | 10,744 | <p>Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's.
We have
<span class="math-container">$$
\sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}.
$$</span>
Setting <span class="math-container">$j=(p-1)k$</span> and summing on <span class="math-container">$k$</span> gives
<span class="math-container">$$
\sum_{a=0}^\infty x^a \sum_{p-1\mid j}\binom{a}{j} = \frac{(1-x)^{p-1}}{(1-x)^p -x^{p-1}(1-x)}.
$$</span>
We subtract <span class="math-container">$\sum_{a=0}^\infty x^a \binom{a}{0} = 1/(1-x)$</span> and
<span class="math-container">$$\sum_{\substack{p-1\mid a\\ a>0}}x^a\binom{a}{a} = \frac{x^{p-1}}{1-x^{p-1}}$$</span>
to get
<span class="math-container">$$
\sum_{a=0}^\infty S_p(a) x^a = \frac{(1-x)^{p-1}}{(1-x)^p -x^{p-1}(1-x)}
-\frac{1}{1-x} -\frac{x^{p-1}}{1-x^{p-1}}.
$$</span>
Modulo <span class="math-container">$p$</span>, we may replace <span class="math-container">$(1-x)^{p-1}$</span> with <span class="math-container">$(1-x^p)/(1-x)$</span> and <span class="math-container">$(1-x)^p$</span> with <span class="math-container">$1-x^p$</span>, obtaining
<span class="math-container">\begin{align*}
\sum_{a=0}^\infty S_p(a) x^a &\equiv \frac{(1-x^p)/(1-x)}{1-x^p -x^{p-1}(1-x)}
-\frac{1}{1-x} -\frac{x^{p-1}}{1-x^{p-1}}\pmod{p}\\
&=0.
\end{align*}</span></p>
|
1,492,027 | <p>Defining R to be the relationship on real numbers given by xRy iff x-y is rational, I've been asked to find the equivalence class of $\sqrt2$. My instincts say that the equivalence class of $\sqrt2$ would just be the empty set. But after a riveting conversation on a similar subject <a href="https://math.stackexchange.com/questions/1491971/subtraction-of-two-irrational-numbers-to-get-a-rational/1491978#1491978">subtraction of two irrational numbers to get a rational</a></p>
<p>I'm curious to if I am able to pick and choose x to suite my needs? say could I define x to be $\alpha+\sqrt2$ so the equivalence class would be $y=\alpha$?</p>
<hr>
<p>EDIT: This is just to see if I have the principle of equivalence classes down..</p>
<p>given the Relation R xRy s.t. x-y is rational, the following equivalence classes are..</p>
<p>[0] = {y: y is rational}</p>
<p>[1] = {y: y is rational}</p>
<p>[$\sqrt2$] = {y: y = $\alpha+\sqrt2$ s.t. $\alpha$ is rational}</p>
<p>To summarize to find the above equivalence classes on xRy iff x-y is rational</p>
<p>I would look at all of the ordered pair (0,y),(1,y),($\sqrt2$,y) fulfills the above stipulation?</p>
<p>Or as to ensure further confidence that I understand (I don't want to be spoon, fed but confirmation that I'm on the right track would be greatly appreciated.</p>
<p>Given the relation T given by (x,y)T(a,b) iff $x^2+y^2=a^2+b^2$</p>
<p>would the equivlanece class of (1,2) be the circle of radius $\sqrt5$?</p>
| DRF | 176,997 | <p><strong>Hint</strong> first thing you need to do is figure out what an equivalence class <strong>is</strong>. Once you have an understanding of what it is that result should be in general terms (real number, rational number, set of real numbers, set of rational numbers etc.) you can continue.</p>
|
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> "II", "value" -> 2|>
}
</code></pre>
<p>where every entry is unique in terms of <code>{#type, #subtype}</code>, </p>
<h3>I'd like to build a nested association for more handy querying, e.g. I would like to have:</h3>
<pre><code>nested["a", "II", "value"]
</code></pre>
<blockquote>
<pre><code>2
</code></pre>
</blockquote>
<h3>I can start with</h3>
<pre><code>GroupBy[dataset, {#type &, #subtype &}]
</code></pre>
<blockquote>
<pre><code><|
"a" -> <|
"I" -> {<|"type" -> "a", "subtype" -> "I", "value" -> 1|>},
"II" -> {<|"type" -> "a", "subtype" -> "II", "value" -> 2|>}|>,
"b" -> <|
"I" -> {<|"type" -> "b", "subtype" -> "I", "value" -> 3|>},
"II" -> {<|"type" -> "b", "subtype" -> "II", "value" -> 4|>}
|>|>
</code></pre>
</blockquote>
<p>But <code>nested["a", "I"]</code> points to a <strong>list with one association</strong>, what is expected but I would like to drop that list. </p>
<p>It seems that the third argument of <code>GroupBy</code> isn't generalized to handle nested grouping... </p>
<p>So basically I would like to have <code>... "I" -> <|"type" -> "a", ...</code>.</p>
<h3>What is a generic way to go?</h3>
<p>I can do:</p>
<ul>
<li><p>nested <code>GroupBy</code>: </p>
<pre><code>GroupBy[dataset, #type &, GroupBy[#, #subtype &, First] &]
</code></pre></li>
<li><p><code>Map</code> later:</p>
<pre><code>GroupBy[dataset, {#type &, #subtype &}] // Map[First, #, {-3}] &
</code></pre></li>
</ul>
<p>But the first is not handy in general while the second is ugly (and not general either). </p>
<hr>
<p>Acceptable outputs are:</p>
<pre><code><|
"a" -> <|
"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> <|"value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> 1 ,
...|>
</code></pre>
<p>but the first is the most desired one because we may have more that one ("value") key left.</p>
| WReach | 142 | <p>One approach is to employ a helper function that unwraps singleton lists:</p>
<pre><code>{delist[v_]} ^:= v
</code></pre>
<p>With this, the <code>GroupBy</code> expression is fairly succinct:</p>
<pre><code>dataset // GroupBy[{#type&, #subtype& -> delist}]
(*
<| "a" -> <| "I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>
, "II" -> <|"type" -> "a", "subtype" -> "II", "value" -> 2|>
|>
, "b" -> <| "I" -> <|"type" -> "b", "subtype" -> "I", "value" -> 1|>
, "II" -> <|"type" -> "b", "subtype" -> "II", "value" -> 2|>
|>
|>
*)
</code></pre>
<p>This generalizes to deeper nesting:</p>
<pre><code>dataset // GroupBy[{#type&, #subtype&, #type&, #subtype& -> delist}]
(*
<| "a" ->
<| "I" -> <|"a" -> <|"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>|>|>
, "II" -> <|"a" -> <|"II" -> <|"type" -> "a", "subtype" -> "II", "value" -> 2|>|>|>
|>
, "b" ->
<| "I" -> <|"b" -> <|"I" -> <|"type" -> "b", "subtype" -> "I", "value" -> 1|>|>|>
, "II" -> <|"b" -> <|"II" -> <|"type" -> "b", "subtype" -> "II", "value" -> 2|>|>|>
|>
|>
*)
</code></pre>
|
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> "II", "value" -> 2|>
}
</code></pre>
<p>where every entry is unique in terms of <code>{#type, #subtype}</code>, </p>
<h3>I'd like to build a nested association for more handy querying, e.g. I would like to have:</h3>
<pre><code>nested["a", "II", "value"]
</code></pre>
<blockquote>
<pre><code>2
</code></pre>
</blockquote>
<h3>I can start with</h3>
<pre><code>GroupBy[dataset, {#type &, #subtype &}]
</code></pre>
<blockquote>
<pre><code><|
"a" -> <|
"I" -> {<|"type" -> "a", "subtype" -> "I", "value" -> 1|>},
"II" -> {<|"type" -> "a", "subtype" -> "II", "value" -> 2|>}|>,
"b" -> <|
"I" -> {<|"type" -> "b", "subtype" -> "I", "value" -> 3|>},
"II" -> {<|"type" -> "b", "subtype" -> "II", "value" -> 4|>}
|>|>
</code></pre>
</blockquote>
<p>But <code>nested["a", "I"]</code> points to a <strong>list with one association</strong>, what is expected but I would like to drop that list. </p>
<p>It seems that the third argument of <code>GroupBy</code> isn't generalized to handle nested grouping... </p>
<p>So basically I would like to have <code>... "I" -> <|"type" -> "a", ...</code>.</p>
<h3>What is a generic way to go?</h3>
<p>I can do:</p>
<ul>
<li><p>nested <code>GroupBy</code>: </p>
<pre><code>GroupBy[dataset, #type &, GroupBy[#, #subtype &, First] &]
</code></pre></li>
<li><p><code>Map</code> later:</p>
<pre><code>GroupBy[dataset, {#type &, #subtype &}] // Map[First, #, {-3}] &
</code></pre></li>
</ul>
<p>But the first is not handy in general while the second is ugly (and not general either). </p>
<hr>
<p>Acceptable outputs are:</p>
<pre><code><|
"a" -> <|
"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> <|"value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> 1 ,
...|>
</code></pre>
<p>but the first is the most desired one because we may have more that one ("value") key left.</p>
| Ronald Monson | 2,249 | <p>If you need to retain the original keys, I'd be inclined to follow Edmund's answer; alternatively, if you are happy enough to throw away the keys (also given that "every entry is unique") one flexible approach follows a <a href="https://mathematica.stackexchange.com/a/102483/2249">nice solution of your own</a>.</p>
<pre><code> RecurAssocMerge[a : {__Association}] := Merge[a, RecurAssocMerge];
RecurAssocMerge[a_] := Last[a];
RecurAssocMerge[ini_Association, path_List, value_] :=
RecurAssocMerge[{ini, Fold[<|#2 -> #|> &, value, Reverse@path]}];
RecurAssocMerge[ini_Association, fullPath_List] := RecurAssocMerge[ini, Most@fullPath, Last@fullPath]; (* added from previous post to handle single list specification *)
</code></pre>
<p>Now build up incrementally after extracting values</p>
<pre><code>Query[All, Values]@dataset // Fold[RecurAssocMerge, <||>, #] &
(* <|"a" -> <|"I" -> 1, "II" -> 2|>, "b" -> <|"I" -> 1, "II" -> 2|>|> *)
</code></pre>
|
1,613,185 | <p>There are five red balls and five green balls in a bag. Two balls are taken out at random. What is the probability that both the balls are of the same colour</p>
| Em. | 290,196 | <p>Notice that you are interested in the event $GG$ or $RR$.
Let's consider the $GG$ case. The probability that you get first $G$
is $$\frac{5}{10}.$$ Then there are $4$ $G$ left and a total of $9$ total. Now, in the second draw, what is the probability that you get another $G$? That is
$$\frac{4}{9}.$$
Thus the probability of $GG$ is
$$\frac{5}{10}\cdot\frac{4}{9} = \frac{20}{90}.$$
Notice that this is the same probability as $RR$. Thus, the probability of same color twice is
$$2\times\frac{20}{90} = \frac{20}{45} = \frac{4}{9}.$$</p>
|
97,393 | <p>The polynomial</p>
<p>$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$</p>
<p>has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.</p>
<p>I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.</p>
| yoyo | 6,925 | <p>if $F(a)=0$, then $(x-a)|F(x)$. also, if a polynomial with real cofficients has a complex root, then the complex conjugate is also a root. so $F$ is divisible by $(x-1)$ and $x^2+1=(x-i)(x+i)$ (according to you). after dividing by these, you will have a polynomial of degree 2, which you can easily factor:
$$
F/(x^2+1)=x^3-9x^2+23x-15=G
$$
$$
G/(x-1)=x^2-8x+15=(x-5)(x-3)
$$
so
$$
F(x)=(x-1)(x-3)(x-5)(x^2+1)
$$</p>
|
97,393 | <p>The polynomial</p>
<p>$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$</p>
<p>has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.</p>
<p>I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.</p>
| guy | 231,779 | <p>As you know that x=1 is a solution, factor (x-1) out of the equation then solve the resulting quartic equation with the formula. You can see the formula at <a href="http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg" rel="nofollow">http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg</a>.</p>
|
3,172,485 | <p>Consider the familiar trigonometric identity: <span class="math-container">$\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$</span></p>
<p>Show that the identity above can be interpreted as Fourier series expansion.</p>
<p>so we know that cos is periodic between <span class="math-container">$\pi$</span> and <span class="math-container">$-\pi$</span> and <span class="math-container">$\cos$</span> is an even function, therefore, <span class="math-container">$\cos^3$</span> is even. </p>
<p>so we need to compute <span class="math-container">$a_0$</span> ( the integral of <span class="math-container">$f(x)$</span> and it will equal <span class="math-container">$0$</span>) and <span class="math-container">$a_n$</span> ( the integral from <span class="math-container">$\pi$</span> to <span class="math-container">$-\pi$</span> of <span class="math-container">$\cos^3(x) \cos(nx)$</span> )</p>
<p>how to compute <span class="math-container">$a_0$</span></p>
<p>thanks </p>
| Mostafa Ayaz | 518,023 | <p><strong>Hint</strong></p>
<p>Simply show that<span class="math-container">$$\int_{0}^{2\pi}\cos^3 x\cos nx=\int_{0}^{2\pi}\cos^3 x\sin mx=0$$</span>for <span class="math-container">$n\ne 1,3$</span>. What about <span class="math-container">$n=1$</span> or <span class="math-container">$n=3$</span>?</p>
|
2,055,559 | <blockquote>
<p>Let <span class="math-container">$a,b,c$</span> be the length of sides of a triangle then prove that:</p>
<p><span class="math-container">$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$</span></p>
</blockquote>
<p>Please help me!!!</p>
| Lelouch vi Britannia | 395,510 | <p>let</p>
<p>$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=0$ (1)</p>
<p>let</p>
<p>$x=-a+b+c; y=a-b+c; z=a+b-c$</p>
<p>$z,y,x$ are twice the length of the segments between the vertices and the touching point of the incircles. so</p>
<p>$a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$</p>
<p>substitute them to (1) and multiply the inequality by 16</p>
<p>$(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)+(x+y)^2(y+z)(x-z)\geqslant 0$</p>
<p>$x^3z+y^3x+z^2y\geqslant x^2yz+y^2zx+z^2xy$</p>
<p>and so</p>
<p>$x^3z+y^3x+z^3y-x^2yz-y^2zx-z^2xy$</p>
<p>$=zx(x-y)^2+xy(y-z)^2+yz(z-x)^2\geqslant 0$</p>
<p>as $x>0, y>o, z>0$ equality holds if and only if $x=y=z$.</p>
<p>example if $a=b=c$ then triangle is equilateral.</p>
|
2,055,559 | <blockquote>
<p>Let <span class="math-container">$a,b,c$</span> be the length of sides of a triangle then prove that:</p>
<p><span class="math-container">$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$</span></p>
</blockquote>
<p>Please help me!!!</p>
| Michael Rozenberg | 190,319 | <p>Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$ and $u\geq0$, $v\geq0$.</p>
<p>Hence, $\sum\limits_{cyc}(a^3b-a^2b^2)=(u^2-uv+v^2)x^2+(u^3+2u^2v-uv^2+v^3)x+2u^3v\geq0$.</p>
<p>Done!</p>
|
1,581,161 | <p>Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$ </p>
<p>Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle of $ABC$ on $D$ and $E$ respectively .
We suppose that $DE=BC$</p>
<p>What is the angle $\widehat{BAC}$ </p>
<p>I tried using law of sines in triangle
Also , let O be center of circle so OD=OE=r</p>
| user24142 | 208,255 | <p><a href="https://i.stack.imgur.com/r1Yal.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r1Yal.png" alt="A diagram of the setup"></a></p>
<p>The above is basically what you're given. I've labeled a lot of the angles with $\alpha, \beta, \gamma, \delta$ and $\epsilon$. Two inscribed angles are equal when they're based off the same chord, ie I'm leaning heavily on the third diagram <a href="https://www.mathsisfun.com/geometry/circle-theorems.html" rel="nofollow noreferrer">here.</a></p>
<p>Now note that $\alpha + \beta + \gamma + \delta + \epsilon = \pi$ and $\gamma + \delta = \gamma +\epsilon = \pi/2$. Furthermore, $\triangle BPC \cong \triangle DPE$ (angle side angle), so $\triangle BED \cong \triangle DCB$ (side side side) and we conclude $\gamma = \epsilon +\delta$. Therefore, $$\pi = (\gamma + \delta) + (\gamma + \epsilon) = 3\gamma$$ and so $\gamma = \pi/3$.</p>
|
356,530 | <p>I'm a really confused as to how to start this question, would really appreciate any help you guys could give me!</p>
| Paul Gustafson | 66,345 | <p>There are $2^{17}$ total strings. There are ${17 \choose {n}}$ ways of getting a string with $n$ ones in it. What's next?</p>
|
356,530 | <p>I'm a really confused as to how to start this question, would really appreciate any help you guys could give me!</p>
| Cameron Buie | 28,900 | <p>How many total bit strings of that length are there? There are $\binom{17}{k}$ bit strings of length $17$ with exactly $k$ ones, since choosing the positions of all the ones determines the string completely. It will be quicker to find the number of bit strings that <strong>don't</strong> meet your criterion, then subtract them from the total.</p>
|
273,499 | <blockquote>
<p>Show that every group $G$ of order 175 is abelian and list all isomorphism types of these groups. [HINT: Look at Sylow $p$-subgroups and use the fact that every group of order $p^2$ for a prime number $p$ is abelian.]</p>
</blockquote>
<p>What I did was this. $|G| = 175$. Splitting 175 gives us $175 = 25 \cdot 7$. Now we want to calculate the Sylow $p$-groups, i.e we want</p>
<p>$$P= n_7: n_7 \equiv 1 mod 7 \hspace{1.5cm} n_7|25$$
$$Q= n_{25}: n_{25} \equiv 1 mod 25 \hspace{1.5cm} n_{25} | 7$$</p>
<p>After listing all elements that are $\equiv 1 mod 7$ and $1 mod 25$ you see that the only (avaliable) ones are $n_7 = n_{25} = 1$. This tells us that both groups $P,Q$ are normal subgroups of $G$. I think, by definition of a normal subgroup, they are abelian and so this tells us that $G$ is abelian. To list all the isomorphism types, we want the semidirect product (SDP) such that</p>
<p>$$P \rightarrow Aut(Q) = C_7 \rightarrow C_{20}$$</p>
<p>As there are no elements of order 7 in $C_{20}$, the only SDP we have is the trivial SDP, i.e the direct product</p>
<p>$$C_7 \times C_{25} \cong C_{175}$$</p>
<p>We know that $175 = 5^2 \cdot 7$ and so multiplying the powers shows us that there are 2 non-isomorphic groups:</p>
<p>$$C_{25} \times C_7$$
$$C_5 \times C_5 \times C_7$$</p>
<p>My question for this is is my reasoning also correct for things like showing the abelian groups? I saw something which said something about $P \cap Q = I_G$ and they used this but I don't understand what it was.</p>
<p>The next question, assuming that I had to possibilites for my $p$ subgroup, i.e $n_p = 1 or x$, how would I go about answering this question? (I am doing a question like this now and am stuck as I have two Sylow $p$-subgroups).</p>
| amWhy | 9,003 | <p>The fact that $P, Q $ are both <em>normal</em> tells you that $G$ is a direct product of $P$ and $Q$. $P$ is abelian because of the hint given in the problem statement: $|P| = 25 = 5^2$, and $5$ is prime.$|Q| = 7$ with $7$ prime. All groups of prime order are cyclic, and all cyclic groups are abelian, $Q$ is therefore abelian. </p>
<p><em>Hence $G = P\times Q$, as the direct product of two abelian groups, is therefore abelian.</em></p>
<hr>
<p><em>Just to confirm/answer your first question (listing all possible abelian groups of order 175):</em></p>
<p>Yes, you correctly showed that if a group $G$ of order 175 is abelian, then it is isomorphic to <em>exactly</em> one of two (<em>non-isomorphic</em>) groups:</p>
<p>$$G \cong C_{175} \cong C_{25} \times C_7$$
$\quad\quad\quad\quad\text{**OR**}$
$$G\cong C_5 \times C_5 \times C_7$$</p>
<p>by the Fundamental Theorem of Finitely Generated Abelian Groups.</p>
<hr>
<p><strong>Note</strong> (added given comment/question below): </p>
<p>$C_{mn} \cong C_m\times C_n$ if and only if $\gcd(m, n) = 1$.</p>
<p>So we have that $C_{175} \cong C_{25}\times C_7$ since $\gcd(25, 7) = 1$.</p>
<p>On the other hand, $C_{25} \times C_7 \ncong (C_5 \times C_5) \times C_7$ since $\gcd(5, 5) \neq 1$</p>
|
2,140,192 | <p>I want to show that $C^1[0,1]$ isn't a Banach Space with the norm:</p>
<p>$$||f||=\max\limits_{y\in[0,1]}|f(y)|$$</p>
<p>Therefore, I want to show that the sequence $\left \{ |x-\frac{1}{2}|^{1+\frac{1}{n}} \right \}$ converges to $|x-\frac{1}{2}|$, but I can't find $N$ in the definition of convergence. Could anyone give me an idea?</p>
<p>Thank You. </p>
| Jonas Meyer | 1,424 | <p>To show $\left|x-\frac12\right|^{1+1/n}=\left|x-\frac12\right|\left|x-\frac12\right|^{1/n}\to \left|x-\frac12\right|$ pointwise, which is clear when $x=\frac12$, it suffices to show that when $0<t<1$, $t^{1/n}\to 1$. Note that $(t^{1/n})$ is bounded above by $1$ and monotone increasing, so it has a limit $L$, with $L>0$. The subsequence $(t^{1/(2n)})$ must converge to the same limit $L$, and the sequences of squares of this subsequence must converge to $L^2$, implying that $L^2=L$. Hence $L=1$.</p>
<p>From the above, we have monotone pointwise convergence of the sequence to $\left|x-\frac12\right|$. By Dini's Theorem, the convergence is uniform, hence convergence in the given norm.</p>
|
2,794,704 | <p>Can the following sum be further simplified? $${1\over 20}\sum_{n=1}^{\infty}\left(n^2+n\right)\left(\frac45\right)^{n-1}$$
(It's part of a probability problem)</p>
| nonuser | 463,553 | <p>$$f(x)=\sum_{n=1}^{\infty}n(n+1)x^{n-1}= (\sum_{n=1}^{\infty}(n+1)x^{n})'=(\sum_{n=1}^{\infty}x^{n+1})''$$</p>
<p>$$=\Big({x^2\over 1-x}\Big)''=\Big({2x-x^2\over (1-x)^2}\Big)'= {2\over (1-x)^3}$$ </p>
<p>So $${1\over 20}f\Big({4\over 5}\Big) = 12,5$$</p>
|
754,012 | <p>Is it possible to show that the harmonic series is divergent by showing that the sequence of partial sums is a monotone increasing sequence that is unbounded?</p>
| Steven Gubkin | 34,287 | <p>Perhaps you have seen the proof by the integral test, and you would like a more "hands on" proof, where you get a bound you can feel. Here is such a proof:</p>
<p>$$
\begin{align}1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{2^{n}-1} &> \frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right) ...+\frac{1}{2^n}\\
&=\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\\
&=\frac{n}{2}
\end{align}
$$</p>
<p>So the sum of the first $2^n-1$ terms of the harmonic series is greater than $\frac{n}{2}$. Since all the terms are positive, the sequence of partial sums is monotone increasing, and this special subsequence diverges to positive infinity, so we are done! </p>
|
548,776 | <p>I am new in Topos theory. I have actually just started learning. I am reading MacLane-Moerdijk's book, as it was suggested to me as the best introduction to the subject. Unfortunately I can not make sense of the following.</p>
<p>In section 5 of Chapter I, (page 41) they build what is needed to prove that in the presheaf category $[\mathcal{C}^{\textrm{op}},\mathbf{Set}]$, any object is the colimit of a diagram of representable objects. They define the category of elements of $P$ as follows: The objects are all pairs $(C,p)$ where $C$ is an object of $\mathcal{C}$ and $p$ is an element $p\in P(C)$. Its morphisms $(C',p')\to (C,p)$ are those morphisms $u:C'\to C$ of $\mathcal{C}$ for which $pu=p'$. I do not understand the morphisms of this category. $p$ is a set and $u$ a morphism in $\mathcal{C}$ so what does $pu$ mean? Do they mean $P(u)(p)=p'$ maybe?</p>
<p>Thanks for any help.</p>
| Slade | 33,433 | <p>Yes, that's what they mean.</p>
<p>Writing $up$ or $u(p)$ would seem to make more intuitive sense, as this feels more like $u$ is "acting" on $p$ (and since we only have one gadget, $P$, for turning $u$ into a function on sets, your definition is really the only way), but since we are using contravariant functors, these definitions would cause composition to look backwards.</p>
<p>If if $u' : C'' \to C'$ is another morphism, then writing this action as $u(p)$ would give:</p>
<p>$$u'(u(p)) = (uu')(p)$$</p>
<p>If we use the definition from the text, however, we get the much more reasonable-looking:</p>
<p>$$p(uu') = (pu)u'$$</p>
|
2,528,716 | <p>How can I prove </p>
<blockquote>
<p>$$x^2+y^2-x-y-xy+1≥0$$</p>
</blockquote>
<p>I tried $(x+y)^2-3xy-(x+y)+1≥0 \rightarrow(x+y-1)(x-y)-3xy+1≥0$ I can not continue</p>
| lab bhattacharjee | 33,337 | <p>Let $u=x^2+y^2-x-y-xy\iff x^2-x(1+y)+y^2-y-u=0$</p>
<p>As $x$ is real, the discriminant must be $\ge0$</p>
<p>$$\implies(1+y)^2-4(y^2-y-u)\ge0$$</p>
<p>$$4u\ge3y^2-6y-1=3(y-1)^2-4\ge-4$$ </p>
|
2,431,375 | <p>A continuous function $f$ on $[a,b]$, differentiable in $(a,b)$, has only 1 point where its derivative vanishes. What is true about this function?</p>
<p>A. $f$ cannot have an even number of extrema.</p>
<p>B. $f$ cannot have a maximum at one endpoint and minimum at the other.</p>
<p>C. $f$ might be monotonically increasing.</p>
<p>I think $A$ is true, since it has three extrema, but the right answer should be $C$, why?</p>
| zwim | 399,263 | <p>If $f$ is differentiable on $[a,b]$ then the example below shows only C works</p>
<p><a href="https://i.stack.imgur.com/vBaCX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vBaCX.png" alt="enter image description here"></a></p>
<p>$\bbox[5px,border:2px solid]{f(x)=-x^2\text{ on interval }[-1,0]}$ with vanishing point $f'(0)=0$</p>
<ul>
<li>A. false, $f(-1)$ minimum and $f(0)$ maximum, thus $2$ extrema</li>
<li>B. false, for the same reason</li>
<li>C. true, $f$ is $\nearrow$ </li>
</ul>
<p>Since the <em>might</em> is used for C, this is the only possible answer (there exists some $f$ that verify C).</p>
<p><br/></p>
<p><strong>But</strong>: you stated $f$ only differentiable on $]a,b[$</p>
<p>So if we are not in the previous case then we have to consider vertical tangents either in $a^+$ or $b-$ or both.</p>
<p>We shall see that even in this case, we can find functions such that only C is valid.</p>
<p><br/></p>
<p>$\bbox[5px,border:2px solid]{f(x)=(1-x)\sqrt{x}\text{ on interval }[0,\frac 13]}$ with vanishing point $f'(\frac 13)=0$</p>
<p>Is not differentiable in $0$, has a minimum in $f(0)$ and a maximum in $f(\frac 13)$ and $f\ \nearrow$.</p>
<p>So again A. false, B. false and C. true</p>
<p>I let you convince yourself that the situation is the same if we exchange the role of $a$ and $b$.</p>
<p><br/></p>
<p>Now what about the case where $f'$ is not defined nor in $a$ nor in $b$.</p>
<p><a href="https://i.stack.imgur.com/gw7LA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gw7LA.png" alt="enter image description here"></a></p>
<p>$\bbox[5px,border:2px solid]{f(x)=\sqrt{x}-\sqrt{2-x}-x\text{ on interval }[0,2]}$ with vanishing point $f'(1)=0$</p>
<p>Is not differentiable in $x=0\text{ or }2$, has minimum $f(0)$, maximum $f(2)$ and $f(1)$ is an inflexion point not an extrema, finally $f\ \nearrow$.</p>
<p>So again A. false, B. false and C. true</p>
|
1,611,506 | <blockquote>
<p>$$\int (2x^2+1)e^{x^2} \, dx$$</p>
</blockquote>
<p>It's part of my homework, and I have tried a few things but it seems to lead to more difficult integrals. I'd appreciate a hint more than an answer but all help is valued.</p>
| user304682 | 304,682 | <p>Here is a hint; you can not analytically evaluate $\int \exp(x^2) dx$.
Read more about that on <a href="https://en.wikipedia.org/wiki/Error_function" rel="nofollow">wiki</a>.
So your final answer should contain terms with $\int \exp(x^2) dx$.</p>
|
58,024 | <p><img src="https://i.stack.imgur.com/cTpA2.jpg" alt="Show that..."></p>
<p>The picture says it all. "Vis at" means "show that". My first thought was that h is 2x, which is not correct. Maybe the formulas for area size is useful? </p>
<p>EDIT: (To make the question less dependent from the <a href="https://math.meta.stackexchange.com/questions/1805/on-the-inclusion-of-pages-of-text-as-images-in-questions">picture</a>.)</p>
<p>A square with side $x$ is placed in the right triangle with legs $g$ and $h$. Show that $x=\frac{gh}{g+h}$.</p>
| Américo Tavares | 752 | <p>From the geometry of the problem (see figure and identify two <a href="http://en.wikipedia.org/wiki/Similarity_%28geometry%29#Similar_triangles" rel="nofollow noreferrer">similar triangles (Wikipedia)</a> or <a href="http://www.gcseguide.co.uk/similar_triangles.htm" rel="nofollow noreferrer">here</a>), we can get the following proportions:</p>
<p>$$\dfrac{h-x}{x}=\dfrac{x}{g-x}=\dfrac{h}{g}\tag{1}.$$ </p>
<p><img src="https://i.stack.imgur.com/czlcc.jpg" alt="enter image description here"></p>
<p>$$|AR|=h-x,|PR=x|,|PQ|=x,|BQ|=g-x,\measuredangle APR=\measuredangle PBQ,\measuredangle PAR=\measuredangle BPQ.$$</p>
<p>Relation $(1)$ comprises 3 equations. Solve one of the them: </p>
<p>$$\dfrac{h-x}{x}=\dfrac{x}{g-x},\tag{2}$$</p>
<p>$$\dfrac{h-x}{x}=\dfrac{h}{g},\tag{3}$$</p>
<p>or </p>
<p>$$\dfrac{x}{g-x}=\dfrac{h}{g}\tag{4}.$$ </p>
<ul>
<li><p>Note 1: If it is required a proof using trigonometry (<em>in the title</em>), I can reformulate my answer. <em>Added</em>. If you apply trigonometry (mentioned in the title), note that $\tan \widehat{%
ABC}=\frac{h}{g}$, $\tan \widehat{APR}=\frac{h-x}{x}$ and $\tan \widehat{APR}%
=\tan \widehat{ABC}$. Use this equality and solve for $x$.</p></li>
<li><p>Note 2: If you haven't yet learned about similar triangles, I will modify my answer.</p></li>
</ul>
|
3,444,673 | <p>How to evaluate this double integral
<span class="math-container">$$\int_{0}^1\int_{x^2}^x \frac{x}{y}e^{-\frac{x^2}{y}}dydx.$$</span></p>
<p>It seems like I am evaluating the double integral of a non-elementary function. I tried substitutions but the integral is growing. </p>
<hr>
<p>Following Fred's suggestion, the integral evaluates to <span class="math-container">$\frac{e-2}{2e}$</span></p>
| Kavi Rama Murthy | 142,385 | <p>The second one is correct. For the first one consider <span class="math-container">$a+b\int |f|d\mu =\int (a+b|f|)d\mu \leq \int \sqrt {a^{2}+b^{2}} \sqrt {1+|f|^{2}} d\mu$</span>. If <span class="math-container">$a^{2}+b^{2} \leq 1$</span> this gives <span class="math-container">$a+b\int |f|d\mu \leq \int \sqrt {1+|f|^{2}} d\mu$</span>. Maximize LHS over all <span class="math-container">$a,b$</span> with <span class="math-container">$a^{2}+b^{2} \leq 1$</span> to get <span class="math-container">$\sqrt {1+(\int|f|d\mu)^{2}} \leq \int \sqrt {1+|f|^{2}} d\mu$</span>.</p>
<p>Actually maximization occurs when <span class="math-container">$a =\frac 1 {\sqrt {1+\int (|f|d\mu)^{2}}}$</span> and <span class="math-container">$b =\frac {\int |f|d\mu} {\sqrt {1+\int (|f| d\mu)^{2}}}$</span></p>
|
19,880 | <p>I want to write down $\ln(\cos(x))$ Maclaurin polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously.</p>
<p>The known expansions of $\ln(1+x)$ and $\cos(x)$ gives:</p>
<p>$$\forall x \gt -1,\ \ln(1+x)=\sum_{n=1}^{k} (-1)^{n-1}\frac{x^n}{n} + R_{k}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}+R_{3}(x)$$
$$\cos(x)=\sum_{n=0}^{k} (-1)^{n}\frac{x^{2n}}{(2n)!} + T_{2k}(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+T_{4}(x)$$</p>
<p>Writing $\ln(1+x)$ with $t=x-1$ gives:</p>
<p>$$\forall t \gt 0,\ \ln(t)=\sum_{n=1}^{k} (-1)^{n-1}\frac{(t+1)^n}{n} + R_{k}(t)$$</p>
<p>But now I'm clueless. </p>
<ul>
<li>Do I just 'plug' $\cos(x)$ expansion in $\ln(t)$? Can I even do that?</li>
<li>Isn't it a problem that $\ln(x)$ isn't defined for $x\leq 0$ but $|\cos(x)| \leq 1$?</li>
</ul>
| Agustí Roig | 664 | <p>I think Joe Johnson's is a good idea:</p>
<p>$$
\frac{d}{dx} \ln\cos x = -\tan x
$$</p>
<p>plus the knowledge of the <a href="http://en.wikipedia.org/wiki/Taylor_series" rel="nofollow">Taylor series for the tangent</a>,</p>
<p>$$
\tan x = \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2)
$$</p>
<p>(here, $B_s$ are the <a href="http://en.wikipedia.org/wiki/Bernoulli_numbers" rel="nofollow">Bernouilli numbers</a>) gives you everything:</p>
<p>$$
\ln\cos x = C - \int \tan x dx = C - \int \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} dx = C - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n}
$$</p>
<p>Now, taking $x = 0$ on both sides, you get $C = 0$, so</p>
<p>$$
\ln\cos x = - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2) \ .
$$</p>
|
666,297 | <p>Find the value of $x$, what is the value of $x$ in this equation, step by step solution will be great.
\begin{equation}
0.4x+15=x
\end{equation}</p>
| dato datuashvili | 3,196 | <p>So we have,</p>
<p>$15=x-0.4x$</p>
<p>$15=0.6x$</p>
<p>$x=\frac{15}{0.6}=25$</p>
<p>Because, </p>
<p>$$\frac{15}{0.6}=\frac{15}{\frac{6}{10}}=\frac{(15\cdot 10)}{6}=\frac{150}{6}=25$$</p>
|
2,663,537 | <p>Suppose G is a group with x and y as elements. Show that $(xy)^2 = x^2 y^2$ if and only if x and y commute.</p>
<p>My very basic thought is that we expand such that $xxyy = xxyy$, then multiply each side by $x^{-1}$ and $y^{-1}$, such that $x^{-1} y^{-1} xxyy = xxyy x^{-1}$ , and therefore $xy=xy$.</p>
<p>I realize that this looks like a disproportionate amount of work for such a simple step, but that is what past instruction has looked like and that is perhaps why I am confused. Moreover, "if and only if" clauses have always been tricky for me since I took Foundations of Math years ago, but if I remember correctly, the goal here should be to basically do the proof from right to left and then left to right, so to speak. Anyhow, I think that I am overthinking this problem.</p>
| mfl | 148,513 | <p><strong>Hint</strong></p>
<p>$$(xy)^2=x^2y^2\iff xyxy=xxyy\iff x^{-1}xyxy=x^{-1}xxyy\iff yxy=xyy.$$</p>
<p>Can you finish?</p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature at $\mu(t)$ is given
by: $$\kappa(t) = \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$</p>
<p>But since $f$ is not a parametrization I don't know how to continue this problem. </p>
<p>Since $x=y$ and there exists a function $f$ such that $f(x)=z$ and $f(y)=z$ is it safe to assume that it is a one to one function? </p>
| FireGarden | 87,896 | <p>For a unit speed curve, the curvature can be simply $$ \kappa = || \ddot\gamma || $$
Or if it is regular, then $$ \kappa = \dfrac{||\ddot\gamma \times \dot\gamma ||}{||\dot\gamma||^3} $$</p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature at $\mu(t)$ is given
by: $$\kappa(t) = \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$</p>
<p>But since $f$ is not a parametrization I don't know how to continue this problem. </p>
<p>Since $x=y$ and there exists a function $f$ such that $f(x)=z$ and $f(y)=z$ is it safe to assume that it is a one to one function? </p>
| Christian Blatter | 1,303 | <p>Your curve lies in the plane $x=y$; therefore it is a plane curve. Denoting the length along the first main diagonal in the $(x,y)$-plane by $s$ the curve can be written as a graph in the form
$$z(s)=f\left({s\over\sqrt{2}}\right)\qquad (s_0\leq s\leq s_1)\ .$$
For such graphs the formula for the curvature reads
$$\kappa(s)={|z''(s)|\over\bigl(1+z'^2(s)\bigr)^{3/2}}={\sqrt{2}|f''(s/\sqrt{2})| \over\bigl(2+f'^2(s/\sqrt{2})\bigr)^{3/2}}\ .$$
This means that at the point $P=\bigl(x,x,f(x)\bigr)$ the curvature is given by
$$\kappa={\sqrt{2}|f''(x)|\over\bigl(2+f'^2(x)\bigr)^{3/2}}\ .$$</p>
|
87,963 | <p>Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.</p>
| Gerry Myerson | 8,269 | <p>We know that equations of degree 2 and 3 have solutions in radicals, but most equations of degree 6 (or any degree 5 or greater) don't. So let $f$ be a polynomial over $K$ of degree 6 not solvable in radicals, let $\alpha$ be a root of $f$ in some extension, let $L=K(\alpha)$. If there were an intermediate field $E$ of degree 2 or 3, then you could express $\alpha$ in radicals over $E$, and those radicals in radicals over $K$, contradicting the assumption that you couldn't solve $f$ in radicals over $K$. </p>
|
87,963 | <p>Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.</p>
| Georges Elencwajg | 3,217 | <p>You are very right when you write "I would guess this is very false": here is a precise statement. </p>
<p><strong>Proposition 1</strong><br>
For any $n\gt 1$ there exists a field extension $\mathbb Q\subset K$ of degree $[K:\mathbb Q]=n \:$ with no intermediate extension $\mathbb Q \subsetneq k\subsetneq K$.<br>
<strong>Proof</strong><br>
Let $\mathbb Q \subset L$ be a Galois extension with Galois group $S_n$, the permutation group of $n$ elements .<br>
The Galois correspondence associates to the subgroup $S_{n-1}\subset S_n$ an extension $\mathbb Q \subset K$. The intermediate fields $\mathbb Q \subsetneq k\subsetneq K$ bijectively correspond (by Galois theory again)<br>
to intermediate subgroups $S_{n-1}\subsetneq G\subsetneq S_n$. However an easy exercise in group theory shows that no such group exists and thus no intermediate field $k$ exists either. </p>
<p><strong>An analogous problem</strong><br>
It is well known that constructible numbers $c\in \mathbb C$ (those that can be constructed by straightedge and compass) have degree a power of two: $[\mathbb Q(c):\mathbb Q]=2^r$ . However the converse is false :<br>
<strong>Proposition 2</strong><br>
For every $n=2^r\geq 4$ there exists a non-constructible number $c\in \mathbb C$ which is not constructible even though it has degree $n=2^r$.<br>
<strong>Proof</strong><br>
Take an irreducible polynomial $P(X)\in \mathbb Q[X]$ whose Galois group is $S_n$. I claim that one of its roots $c $ is the required nonconstructible number of degree $n$.<br>
Else all its roots $c_1, ...,c_n$ would be constructible and all the numbers of its splitting field $K=\mathbb Q(c_1,...,c_n)$ would be constructible too (the constructible numbers form a field).<br>
But this is contradictory because a primitive element $c_0$ of $K$ (i.e. $K=\mathbb Q(c_0$)) should then also have degree some power of $2$, whereas its degree is actually $n!$, a number divisible by $3$ since $n\geq 4$. </p>
<p><strong>Bibliography</strong><br>
The existence of polynomials over $\mathbb Q$ with Galois group $S_n$ is proved for example in Dummit-Foote's <em>Abstract algebra</em>, Chapter 14, Proposition 42.</p>
|
512,037 | <p>This is a question from our reviewer for our exam for linear algebra. I just want to have some ideas how to tackle the problem.</p>
<p>If $A$ is an $n\times n$ matrix with integer coefficients, such that the sum of each row's elements is equal to $m$, show that $m$ divides the determinant.</p>
| DanielV | 97,045 | <p>The other answer is probably better, but I offer this as an alternative.</p>
<p>Let B be an upper triangular matrix of 1s and 0s. For example, 4x4:</p>
<p>$B = \begin{array} {c c c c }
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1\\
\end{array}$</p>
<p>Let $C = A \times B$. Since $det(B) = 1$ then $det(C) = det(A)$. By the definition of matrix multiplication, the last column of C is m. For example:</p>
<p>$C = \begin{array} {c c c c }
... & ... & ... & m\\
... & ... & ... & m\\
... & ... & ... & m\\
... & ... & ... & m\\
\end{array}$</p>
<p>$det(C) = m \cdot det(some sub matrix) - m \cdot det(someothersubmatrix) + \dots$</p>
<p>Since C is an integer matrix, the determinant of each submatrix is an integer, so m divides det(C), so m divides det(A).</p>
<p>Not exactly a very conceptually useful proof, but not a difficult one either.</p>
|
268,778 | <p>Consider a random process where integers are sampled uniformly with replacement from $\{1...n\}$. Let $X$ be a random variable that represents the number of samples until either a duplicate is found or both the values $1$ and $2$ have been found. So if the samples where $1,6,3,5,1$ then $X=5$ and if it was $1,6,3,2$ then $X=4$. </p>
<p>How does one find the cumulative distribution function. That is how does one find $P(X \geq x)$? </p>
| cardinal | 7,003 | <p>First, note that by the <a href="http://en.wikipedia.org/wiki/Pigeonhole_principle" rel="nofollow">pigeonhole principle</a>, $\renewcommand{\Pr}{\mathbb P}\Pr(X > n) = 0$.</p>
<p>Next let
$$
A_m := \{\text{no duplicates in first $m$ trials}\}
$$
and
$$
B_{m,i} := \{\text{The value $i$ has been seen in the first $m$ trials}\} \>.
$$ </p>
<p>Then, by decomposing into disjoint sets and employing symmetry, we have
$$
\Pr(X > m) = \Pr(A_m (B_{m,1}^c \cup B_{m,2}^c)) = \Pr(A_m B_{m,1}^c B_{m,2}^c) + 2 \Pr(A_m B_{m,1} B_{m,2}^c) \>.
$$
But the second probability on the right-hand side is (again, by symmetry)
$$
\Pr(A_m B_{m,1} B_{m,2}^c) = m \Pr(A_m \cap \{\text{$1$ is in the first position}\} \cap B_{m,2}^c) \>.
$$
These two probabilities are easy to find by simple (combinatorial) arguments.
$$
\Pr(A_m B_{m,1}^c B_{m,2}^c) = \frac{(n-2) (n-3) \cdots (n-m-1)}{n^m}
$$
and
$$
\Pr(A_m \cap \{\text{$1$ is in the first position}\} \cap B_{m,2}^c) = \frac{1 \cdot (n-2) \cdots (n-m)}{n^m} \>.
$$</p>
<p>Thus, for $1 \leq m < n$,
$$
\Pr(X > m) = \frac{(n-2)!\,(n+m-1)}{n^m (n-m-1)!} = \frac{n!}{(n-m)!\, n^m}\left(1 - \frac{m(m-1)}{n(n-1)}\right) \>.
$$</p>
|
268,778 | <p>Consider a random process where integers are sampled uniformly with replacement from $\{1...n\}$. Let $X$ be a random variable that represents the number of samples until either a duplicate is found or both the values $1$ and $2$ have been found. So if the samples where $1,6,3,5,1$ then $X=5$ and if it was $1,6,3,2$ then $X=4$. </p>
<p>How does one find the cumulative distribution function. That is how does one find $P(X \geq x)$? </p>
| Did | 6,179 | <p>Let $x\geqslant1$. There are $n^x$ samples of length $x$. Amongst these, $(n)_x=\frac{n!}{(n-x)!}$ samples have no duplicate. </p>
<p>A sample of length $x$ without duplicate where two distinct given results $i$ and $j$ appear is uniquely described by a sample of length $x-2$ without duplicate where neither result $i$ nor result $j$ appear, and the choice of a position amongst $x-1$ where one places result $i$, and the choice of a position amongst $x$ (once result $i$ is placed) where one places result $j$. There are $x(x-1)\cdot(n-2)_{x-2}$ of these samples. </p>
<p>Thus, the number $S_x$ of samples of length $x$ without duplicate where results $i$ and $j$ do not both appear is
$$
S_x=(n)_{x}-x(x-1)\cdot(n-2)_{x-2},
$$
and, for every $x\geqslant0$,
$$
\mathbb P(X\gt x)=n^{-x}S_x=\frac{n!}{(n-x)!\,n^x}\left(1-\frac{x(x-1)}{n(n-1)}\right).
$$</p>
|
721,514 | <p>Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$</p>
| Cm7F7Bb | 23,249 | <p>As mentioned in the comments, we need to show that the sequence is monotonic and bounded.</p>
<p>First, we observe that
$$
x_n-x_{n+1}=x_n-\frac12\Bigl(x_n+\frac a{x_n}\Bigr)=\frac1{2x_n}(x_n^2-a).
$$
Secondly, we obtain that
\begin{align*}
x_n^2-a
&=\frac14\Bigl(x_{n-1}+\frac a{x_{n-1}}\Bigr)^2-a\\
&=\frac{x_{n-1}^2}4-\frac a2+\frac{a^2}{4x_{n-1}^2}\\
&=\frac14\Bigl(x_{n-1}^2-2a+\frac{a^2}{x_{n-1}^2}\Bigr)\\
&=\frac{1}{4}\Bigl(x_{n-1}-\frac a{x_{n-1}}\Bigr)^2\\
&\ge0.
\end{align*}
Hence, $x_n\ge x_{n+1}$ and $x_n$ is bounded from below since $x_n^2\ge a$ for each $n\ge2$.</p>
<p>Monotonic and bounded sequence converges. Denote the limit of the sequence $x=\lim_{n\to\infty}x_n$. Then we have that
$$
x=\frac12\Bigl(x+\frac ax\Bigr)\quad\iff\quad x=\sqrt a.
$$</p>
|
721,514 | <p>Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$</p>
| bobbym | 77,276 | <p>That looks a lot like the well known method for computing square roots. It is derived by using Newton's on</p>
<p>$$ x^2 - a = 0 $$</p>
<p>$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$</p>
<p>$$=x_n-\frac{x_n^2-a}{2x_n}$$</p>
<p>$$=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right) $$</p>
<p>If it converges it will converge to the $$ \sqrt{a} $$</p>
|
2,290,395 | <p>What if in Graham’s Number every “3” was replaced by “tree(3)” instead? How big is this number? Greater than Rayo’s number? Greater than every current named number?</p>
| Simply Beautiful Art | 272,831 | <p>No. If you replaced all the $3$'s in the construction of Graham's number with $\operatorname{TREE}(3)$, the resulting number would be smaller than $g_{\operatorname{TREE(3)}}$ where $g_n$ denotes the $n$th number in Graham's sequence with $g_{64}$ being Graham's number. This is much much smaller than $\operatorname{TREE}(4)$, for example.</p>
|
3,280,095 | <p>Given the function
<span class="math-container">$$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$</span>
You would need to have <span class="math-container">$u=\sqrt{x}-3$</span> and <span class="math-container">$du=\frac{1}{2 \sqrt{x}}$</span>, when I use a online calculator it suggests to rewrite the numerator as
<span class="math-container">$$\int \frac{2(u^2+6u+9)}{u}du $$</span>
and I was wondering how do you arrive to this trinomial mathematically/analytical?</p>
| cmk | 671,645 | <p>If <span class="math-container">$u=\sqrt{x}-3,$</span> then <span class="math-container">$\sqrt{x}=u+3,$</span> and <span class="math-container">$dx=2\sqrt{x} du=2(u+3)du.$</span> Making these substitutions, <span class="math-container">$$\int\frac{\sqrt{x}}{\sqrt{x}-3}\, dx=\int\frac{u+3}{u}2(u+3)\, du=\int\frac{2(u^2+6u+9)}{u}\, du $$</span> </p>
|
3,280,095 | <p>Given the function
<span class="math-container">$$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$</span>
You would need to have <span class="math-container">$u=\sqrt{x}-3$</span> and <span class="math-container">$du=\frac{1}{2 \sqrt{x}}$</span>, when I use a online calculator it suggests to rewrite the numerator as
<span class="math-container">$$\int \frac{2(u^2+6u+9)}{u}du $$</span>
and I was wondering how do you arrive to this trinomial mathematically/analytical?</p>
| Bernard | 202,857 | <p>Simple: 1) <span class="math-container">$\;\sqrt x=u+3$</span>, and 2) <span class="math-container">$\;\mathrm d u=\dfrac{\mathrm dx}{2\sqrt x}$</span>, so
<span class="math-container">$$\mathrm dx=2\sqrt x\,\mathrm d u=2(u+3)\,\mathrm d u, \quad\text{so }\quad\sqrt x\,\mathrm dx=2(u+3)^2\mathrm du.$$</span></p>
|
1,154,592 | <p>I was doing some basic Number Theory problems and came across this problem :</p>
<blockquote>
<p>Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime</p>
</blockquote>
<p><strong>My Solution : (Sloppy)</strong></p>
<blockquote>
<ul>
<li>$a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$</li>
<li>This means that $(a-1)$ | $(a^{n}-1)$ </li>
<li>But $(a^{n}-1)$ is prime</li>
<li>So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$</li>
<li>Now , let $n$ be composite</li>
<li>$n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$</li>
<li>Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$</li>
<li>This means that $2^{n} -1$ is composite </li>
<li>Hence , we have achieved a contradiction</li>
</ul>
</blockquote>
<p><strong>My Question :</strong>
Am I correct ?</p>
| Agni Chakraborty | 521,925 | <p>Since $a$ and $n$ are both positive integers so $a>0$ and $n>0$
then taking $a=1$ putting in $a^n-1$ we get $1^n-1=0$ which is not prime hence take $a=2$ where $2^n-1$ is a prime if and only if $n$ is a prime</p>
|
1,089,078 | <p>Suppose we have a deck of cards, shuffled in a random configuration. We would like to find a $k$-bit code in which we explain the current order of the cards. This would be easy to do for $k=51 \cdot 6=306$, since we could encode our deck card-by-card, using $2$ bits for the coloring and $4$ bits for the number on each card.</p>
<p>We would like to optimalise this code. There exist $52!$ ways to arrange our deck, and so $k$ will have to be at least: $\lceil\log(52!)\rceil=226$. I'm asked to find a code for a $k$-value halfway between $306$ and $226$.</p>
<p>I understand that my code cannot work card-by-card, since there exist $52$ different cards and $\lceil \log(52)\rceil=6$. Therefore any card-by-card codation will lead to $k\geq306$.</p>
<p>Therefore I concluded my strategy should encode blocks of cards, another idea I had was to encode the colouring first and the numbers second.</p>
<p>Could anyone give me a hint about where to go from here?</p>
| Arthur | 15,500 | <p>Enumerate all the cards in the deck, say by value and suit in a simple manner, e.g. let diamonds be numbered 1 to 13, ace to king, then clubs 14 to 26, then hearts 27 to 39 and lastly spades 40 to 52.</p>
<p>As you encode cards one by one, decrease the value of every card above it. So the sequence $1,2,3,4$ means the ace, three, five and seven of diamonds, while the sequence $15,15,14, 15$ means the two, three, ace and then five of clubs.</p>
<p>In the beginning, use six bits for each number. After 20 cards, use five bits per card instead (since now the highest value among the cards is 32). Sixteen cards later, you can start using four bits per card. And so on.</p>
<p>In total, you use $$20\cdot6+16\cdot5+8\cdot4+4\cdot3+2\cdot2+1=249$$ bits, which is a bit better than what you were asked for (but of course not optimal)</p>
|
4,542,985 | <p>I want to fully understand the probabilistic interpretation. As in, I know once we have a probabilistic model, we differentiate for maximum likelihood and find the weights/regressors but what i really find difficult to grasp is how exactly are we developing a probabilistic model for linear regression.
I have see that initially we we write: <span class="math-container">$y_i=\epsilon_i +w^Tx_i$</span>.---------(1)<br />
Here i want to know what is <span class="math-container">$y_i$</span>? Is it the observed value? Then how come we model it as random? Where is the randomness coming from? What is <span class="math-container">$\epsilon_i$</span>? Is it error or noise?</p>
<p>please correct if i am wrong:</p>
<p>What i understand is our measured data is noisy. i.e for the same <span class="math-container">$x_i, y_i$</span> can vary on a different draw of samples, which is due to some inherent randomness in <span class="math-container">$y_i$</span>. And this randomness is what we are quantifying using <span class="math-container">$\epsilon_i \sim N(0,\sigma^2)$</span>. Hence <span class="math-container">$y_i$</span> is a normal random variable given <span class="math-container">$x_i$</span> and it has mean <span class="math-container">$w^Tx_i$</span> so we want to maximize the likelihood, meaning maximize the probability that <span class="math-container">$y_i$</span> takes the value which we have in our current experimental data, given xi and this probability happens to be parameterized by w due to (1).</p>
| Snoop | 915,356 | <p><span class="math-container">$(\subseteq)$</span>. Note <span class="math-container">$V$</span> is real valued. Also
<span class="math-container">$$\{\omega\in \Omega:V(\omega)\leq x\}=\bigcup_{k\leq \lfloor x\rfloor }B_k\in G,\,\quad \forall x\geq 1$$</span>
<span class="math-container">$$\{\omega\in \Omega:V(\omega)\leq x\}=\emptyset \in G,\,\quad \forall x <1$$</span>
By using the generator of the Borel sets <span class="math-container">$\{(-\infty,x],x \in \mathbb{R}\}$</span> we get
<span class="math-container">$$\sigma(V)=\sigma(\{\{\omega\in \Omega:V(\omega)\leq x\},x \in \mathbb{R}\})\subseteq G$$</span>
<span class="math-container">$(\supseteq)$</span>. Note that
<span class="math-container">$$C \in G\implies C =\bigcup_{k \in I}B_k,\,I \subseteq \mathbb{N}$$</span>
for some <span class="math-container">$I$</span>, and that
<span class="math-container">$$V^{-1}(\{k\})=\{\omega \in \Omega:V(\omega)=k\}=B_k\implies C=\bigcup_{k \in I}V^{-1}(\{k\})\in \sigma(V)$$</span>
But this means that <span class="math-container">$C \in \sigma(V),\forall C \in G$</span>. So we conclude <span class="math-container">$G=\sigma(V)$</span>.</p>
|
4,050,893 | <p>Given a linear transformation <span class="math-container">$T: V \rightarrow W$</span> where <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional, then is it true that nullity(<span class="math-container">$T$</span>) = nullity(<span class="math-container">$[T]_\beta$</span>) for any basis <span class="math-container">$\beta$</span> of <span class="math-container">$V$</span>?</p>
<p>A little embarrassing that I'm 20 weeks into studying linear algebra and I've forgotten how to prove this (if I ever knew how). I think this is true, and if it is, I feel like the reasoning is extremely simple, but I can't seem to rigorously spell it out.</p>
<p>Is there a rigorous (and I'm expecting pretty simple) explanation for why this is true?</p>
<p>Thank you!</p>
| RobPratt | 683,666 | <p>You can formulate this as an integer linear programming problem, and here is a feasible solution:
<span class="math-container">\begin{align}
A&= \begin{pmatrix}
0 &0 &0 &1 \\
1 &0 &1 &0 \\
1 &1 &1 &1 \\
1 &0 &0 &1 \\
\end{pmatrix} \\
B&= \begin{pmatrix}
1 &1 &0 &0 \\
0 &1 &1 &0 \\
1 &0 &1 &1 \\
0 &0 &1 &0 \\
\end{pmatrix} \\
C&= \begin{pmatrix}
1 &1 &0 &1 \\
1 &1 &0 &0 \\
1 &0 &0 &0 \\
1 &0 &1 &0 \\
\end{pmatrix} \\
D&= \begin{pmatrix}
0 &0 &1 &1 \\
0 &0 &0 &1 \\
0 &0 &0 &0 \\
0 &0 &0 &0 \\
\end{pmatrix}
\end{align}</span>
There are 64 constraints, 64 binary decision variables <span class="math-container">$(A_{i,j},B_{i,j},C_{i,j},D_{i,j})$</span>, and 64 nonnegative integer variables (<span class="math-container">$Y_{t,i,j}$</span> for <span class="math-container">$t\in\{1,2,3,4\}$</span> and all <span class="math-container">$(i,j)$</span>). Explicitly, the constraints are
<span class="math-container">\begin{align}
\sum_{k=1}^n \left(A_{i,k} A_{k,j}
+ B_{i,k} C_{k,j}\right)
+ \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} A_{\ell,j}
+ A_{i,k} B_{k,\ell} C_{\ell,j}\right)
+ A_{i,j}
&= 2 Y_{1,i,j} + [i=j] \\
\sum_{k=1}^n A_{i,k} B_{k,j}
+ \sum_{k=1}^n \sum_{\ell=1}^n \left(B_{i,k} C_{k,\ell} B_{\ell,j}
+ A_{i,k} B_{k,\ell} D_{\ell,j}\right)
&= 2 Y_{2,i,j} \\
\sum_{k=1}^n C_{i,k} A_{k,j}
+ \sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} A_{\ell,j}
+ C_{i,k} B_{k,\ell} C_{\ell,j}\right)
&= 2 Y_{3,i,j} \\
\sum_{k=1}^n \sum_{\ell=1}^n \left(D_{i,k} C_{k,\ell} B_{\ell,j}
+ C_{i,k} B_{k,\ell} D_{\ell,j}\right)
&= 2 Y_{4,i,j} + [i=j]
\end{align}</span>
These can be linearized by replacing each product with a new binary variable, together with additional linear constraints, as shown in <a href="https://or.stackexchange.com/questions/37/how-to-linearize-the-product-of-two-binary-variables">https://or.stackexchange.com/questions/37/how-to-linearize-the-product-of-two-binary-variables</a></p>
<hr />
<p>As requested, here is the SAS code. The NOOBJ option specifies that there is no objective, and the LINEARIZE option <a href="https://go.documentation.sas.com/?cdcId=pgmsascdc&cdcVersion=v_010&docsetId=casmopt&docsetTarget=casmopt_optmodel_details57.htm&locale=en" rel="nofollow noreferrer">automatically performs the linearization</a>.</p>
<pre><code>proc optmodel;
num n = 4;
set ROWS = 1..n;
set COLS = ROWS;
set CELLS = ROWS cross COLS;
var A {CELLS} binary;
var B {CELLS} binary;
var C {CELLS} binary;
var D {CELLS} binary;
var Y {1..4, CELLS} >= 0 integer;
/* A^2 + BC + BCA + ABC + A = I_4 */
con Con1 {<i,j> in CELLS}:
sum {k in 1..n} A[i,k]*A[k,j]
+ sum {k in 1..n} B[i,k]*C[k,j]
+ sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
+ sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
+ A[i,j]
= 2*Y[1,i,j] + (i=j);
/* AB + BCB + ABD = 0 */
con Con2 {<i,j> in CELLS}:
sum {k in 1..n} A[i,k]*B[k,j]
+ sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
+ sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
= 2*Y[2,i,j];
/* CA + DCA + CBC = 0 */
con Con3 {<i,j> in CELLS}:
sum {k in 1..n} C[i,k]*A[k,j]
+ sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
+ sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
= 2*Y[3,i,j];
/* DCB + CBD = I_4 */
con Con4 {<i,j> in CELLS}:
sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
+ sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
= 2*Y[4,i,j] + (i=j);
solve noobj linearize;
print A;
print B;
print C;
print D;
quit;
</code></pre>
<p>Another approach to this feasibility problem is to introduce nonnegative surplus and slack variables and minimize their use. The original problem is feasible if and only if all surplus and slack variables are 0.</p>
<pre><code> var Surplus {1..4, CELLS} >= 0;
var Slack {1..4, CELLS} >= 0;
min Infeasibility = sum {t in 1..4, <i,j> in CELLS} (Surplus[t,i,j] + Slack[t,i,j]);
/* A^2 + BC + BCA + ABC + A = I_4 */
con Con1 {<i,j> in CELLS}:
sum {k in 1..n} A[i,k]*A[k,j]
+ sum {k in 1..n} B[i,k]*C[k,j]
+ sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*A[l,j]
+ sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*C[l,j]
+ A[i,j]
- Surplus[1,i,j] + Slack[1,i,j]
= 2*Y[1,i,j] + (i=j);
/* AB + BCB + ABD = 0 */
con Con2 {<i,j> in CELLS}:
sum {k in 1..n} A[i,k]*B[k,j]
+ sum {k in 1..n, l in 1..n} B[i,k]*C[k,l]*B[l,j]
+ sum {k in 1..n, l in 1..n} A[i,k]*B[k,l]*D[l,j]
- Surplus[2,i,j] + Slack[2,i,j]
= 2*Y[2,i,j];
/* CA + DCA + CBC = 0 */
con Con3 {<i,j> in CELLS}:
sum {k in 1..n} C[i,k]*A[k,j]
+ sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*A[l,j]
+ sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*C[l,j]
- Surplus[3,i,j] + Slack[3,i,j]
= 2*Y[3,i,j];
/* DCB + CBD = I_4 */
con Con4 {<i,j> in CELLS}:
sum {k in 1..n, l in 1..n} D[i,k]*C[k,l]*B[l,j]
+ sum {k in 1..n, l in 1..n} C[i,k]*B[k,l]*D[l,j]
- Surplus[4,i,j] + Slack[4,i,j]
= 2*Y[4,i,j] + (i=j);
solve linearize;
</code></pre>
<p>To find more than one solution, you can use the <a href="https://go.documentation.sas.com/?cdcId=pgmsascdc&cdcVersion=v_010&docsetId=casmopt&docsetTarget=casmopt_milpsolver_syntax02.htm&locale=en#casmopt.milpsolver.milpmaxpoolsols" rel="nofollow noreferrer">MAXPOOLSOLS= option</a>.</p>
<hr />
<p>Here is an example of how you can manually linearize the first product:</p>
<pre><code> /* M[i,j,k] = A[i,k]*A[k,j] */
var M {1..n, 1..n, 1..n} binary;
con Mcon1 {i in 1..n, j in 1..n, k in 1..n}:
M[i,j,k] <= A[i,k];
con Mcon2 {i in 1..n, j in 1..n, k in 1..n}:
M[i,j,k] <= A[k,j];
con Mcon3 {i in 1..n, j in 1..n, k in 1..n}:
M[i,j,k] >= A[i,k] + A[k,j] - 1;
</code></pre>
<p>You would then replace <code>A[i,k]*A[k,j]</code> with <code>M[i,j,k]</code> throughout the original nonlinear constraints.</p>
<p>You can optionally declare the new variables as <code>>= 0</code> rather than <code>binary</code>.</p>
<hr />
<p>To exclude a given set of solutions, first read them in:</p>
<pre><code>set SOLUTIONS;
num Asol {CELLS, SOLUTIONS};
num Bsol {CELLS, SOLUTIONS};
num Csol {CELLS, SOLUTIONS};
num Dsol {CELLS, SOLUTIONS};
read data out into SOLUTIONS=[_N_]
{i in 1..n, j in 1..n} <Asol[i,j,_N_]=col('A_'||i||''||j)>
{i in 1..n, j in 1..n} <Bsol[i,j,_N_]=col('B_'||i||''||j)>
{i in 1..n, j in 1..n} <Csol[i,j,_N_]=col('C_'||i||''||j)>
{i in 1..n, j in 1..n} <Dsol[i,j,_N_]=col('D_'||i||''||j)>;
</code></pre>
<p>Then impose a no-good constraint:</p>
<pre><code>con NoGood {sol in SOLUTIONS}:
sum {<i,j> in CELLS} (
(if Asol[i,j,sol] = 0 then A[i,j] else 1 - A[i,j])
+ (if Bsol[i,j,sol] = 0 then B[i,j] else 1 - B[i,j])
+ (if Csol[i,j,sol] = 0 then C[i,j] else 1 - C[i,j])
+ (if Dsol[i,j,sol] = 0 then D[i,j] else 1 - D[i,j])
) >= 1;
</code></pre>
<p>Then call the solver:</p>
<pre><code>solve noobj;
</code></pre>
<p>If you do this for the 148 solutions you found and the resulting solve comes back infeasible, then you have found all solutions.</p>
|
8 | <p>Contexts have backticks, which conflict with the normal way to enter inline code. How do I enter an inline context, since the initial approach:</p>
<pre><code>`System``
</code></pre>
<p>doesn't work ( `System`` ).</p>
| J. M.'s persistent exhaustion | 50 | <p>Backslashes before backticks seem to work too. For example,</p>
<pre><code>`?Global\`*`
</code></pre>
<p>in the comment below:</p>
|
8 | <p>Contexts have backticks, which conflict with the normal way to enter inline code. How do I enter an inline context, since the initial approach:</p>
<pre><code>`System``
</code></pre>
<p>doesn't work ( `System`` ).</p>
| Szabolcs | 12 | <p>This is so common in Mathematica that I suggest this should be specially included in the editor help or the FAQ on this site. Let me list the most common usages:</p>
<ul>
<li><p>Contexts, for example, <code>Global`</code>. </p>
<p>Markdown: <code>``Global` ``</code> (note the space before the closing <code>``</code>)</p></li>
<li><p>Precision of numbers, for example, <code>2.3`40</code>.</p>
<p>Markdown: <code>``2.3`40``</code></p></li>
<li><p>Accuracy of numbers, for example, <code>2.3``3</code>.</p>
<p>Markdown: <code>```2.3``3```</code></p></li>
<li><p><code>StringForm</code> expressions, for example, <code>StringForm["x = ``", x]</code></p>
<p>Markdown: <code>```StringForm["x = ``", x]```</code></p>
<p>And here the question comes up, how to include a double backtick (<code>``</code>) in inline code? <strong>Generally, if you want to include <code>n</code> consecutive backticks, surround the inline code span with <code>n+1</code> backticks.</strong> For this reason, if the inline code has a backtick at the very end, you need to put a space after it. Don't worry, <a href="https://meta.stackexchange.com/questions/117422/how-can-i-write-inline-code-ending-in-a-backtick">this space will be stripped when the MarkDown is rendered</a>.</p></li>
</ul>
|
3,965,455 | <p>Find <span class="math-container">$E(X^3)$</span> given <span class="math-container">$X$</span> is in <span class="math-container">$Exp(2)$</span></p>
<p>My idea is that we can use <span class="math-container">$f_X(x)=2e^{-2x}$</span> and integrate <span class="math-container">$\int_{0}^{a}x^3f_X(x)dx$</span>. But from here I don't know where to go.</p>
| Community | -1 | <p><span class="math-container">$f(x) = \dfrac{10x}{x - 10}$</span><br><br>
Let <span class="math-container">$x - 10 = k.$</span> So, <span class="math-container">$x = 10+k$</span></p>
<p>Therefore, <span class="math-container">$\dfrac{10x}{x-10} = \dfrac{10(k+10)}{k} = 10 + \dfrac{100}{k}$</span><br><br></p>
<p>So <span class="math-container">$k$</span> divides <span class="math-container">$100$</span> and therefore <span class="math-container">$k = 1,2,4,5,10,...$</span><br>
You can find respective values of <span class="math-container">$x$</span> since <span class="math-container">$x = 10 + k$</span></p>
<p>EDIT: Solutions for <span class="math-container">$f(x) = \dfrac{10x}{3x-10}$</span><br></p>
<p>Let <span class="math-container">$k = 3x-10$</span></p>
<p><span class="math-container">$\dfrac{10x}{3x-10} = \dfrac{100+10k}{3k}=t$</span> (Lets say it is equal to t)<br><br>
<span class="math-container">$100 = k(3t -10)$</span></p>
<p>Now basically you can write 100 as a product of 2 factors like</p>
<p>100 = 1 x 100...</p>
<p>Now check for which values you get t as an integer and you will get your answer.</p>
|
43,611 | <p>I posted this on Stack Exchange and got a lot of interest, but no answer.</p>
<p>A recent <a href="http://people.missouristate.edu/lesreid/POW12_0910.html" rel="nofollow">Missouri State problem</a> stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. <strong>Question:</strong> So can the plane be decomposed into unit open intervals? closed intervals?</p>
| Jeremy West | 10,204 | <p>Start with the collection of half-open intervals of the form $[a,a+1) \times 0$ where $a \geq 0$ is an integer. This decomposes the positive $x$-axis into half-open intervals. Now, for every value of $0 < \theta < 2\pi$, decompose the ray whose angle with the positive $x$-axis is $\theta$ into half-open intervals with the open end of the interval placed at the endpoint nearest the origin. </p>
|
4,031,476 | <p>I recently completed a variation of a problem I found from a mathematical olympiad which is as follows:</p>
<p>Prove that, for all <span class="math-container">$n \in \mathbb{Z}^+$</span>, <span class="math-container">$n \geq 1$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span></p>
<p>I would like to hear comments on my method of proving the inequality above as well as ways of improving it or making it more rigorous. Thank you!</p>
<p>Proof: We begin by noticing that the series above is a Riemann sum (or approximation by areas of rectangles) for the integral <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx$$</span> where the right endpoints are taken, hence <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < \int_1^n \frac{x}{2^x} dx$$</span> for all <span class="math-container">$n$</span>.</p>
<p>By evaluating the original integral, it can be shown that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx = \frac{n \ln 2 + 1}{2^n (\ln 2)^2} + \frac{\ln 2 + 1}{2 (\ln 2)^2}$$</span></p>
<p>As <span class="math-container">$n$</span> approaches infinity, the integral approaches the value <span class="math-container">$\frac{\ln 2 + 1}{2 (\ln 2)^2}$</span>, which is less than <span class="math-container">$2$</span>. Hence, it can be deduced that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> for a large enough <span class="math-container">$n$</span>.</p>
<p>By inspection and calculation, it is obvious that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> when <span class="math-container">$n > 5, n \in \mathbb{Z}^+$</span>. Hence, when <span class="math-container">$1 \leq n \leq 5$</span>, the integral has a value that is greater than <span class="math-container">$2$</span>.</p>
<p>However, it can be verified by calculation that for <span class="math-container">$1 \leq n \leq 5$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span> Since the statement holds true for <span class="math-container">$1 \leq n \leq 5$</span> as well as for <span class="math-container">$n > 5$</span>, where <span class="math-container">$n \in Z^+$</span>, the statement must be true for all <span class="math-container">$n \geq 1$</span>. Hence, the proof is complete.</p>
<p>Hope to hear your thoughts on this!</p>
| Brian M. Scott | 12,042 | <p>Note that no calculus is needed:</p>
<p><span class="math-container">$$\begin{align*}
\sum_{k=1}^n\frac{k}{2^k}&=\sum_{k=1}^n\sum_{\ell=1}^k\frac1{2^k}\\
&=\sum_{\ell=1}^n\sum_{k=\ell}^n\frac1{2^k}\\
&=\sum_{\ell=1}^n\frac{\frac1{2^{\ell}}-\frac1{2^{n+1}}}{1-\frac12}\\
&=\sum_{\ell=1}^n\left(\frac1{2^{\ell-1}}-\frac1{2^n}\right)\\
&=\sum_{\ell=0}^{n-1}\frac1{2^\ell}-\frac{n}{2^n}\\
&=\frac{1-\frac1{2^n}}{1-\frac12}-\frac{n}{2^n}\\
&=2\left(1-\frac1{2^n}\right)-\frac{n}{2^n}\\
&=2-\frac{n+2}{2^n}\\
&<2\,.
\end{align*}$$</span></p>
<p>For the visually oriented this has a nice schematic representation:</p>
<p><span class="math-container">$$\begin{array}{cccccc|c}
\frac12&\frac14&\frac18&\frac1{16}&\cdots&\frac1{2^n}&2\left(\frac12-\frac1{2^{n+1}}\right)=1-\frac1{2^n}\\
&\frac14&\frac18&\frac1{16}&\cdots&\frac1{2^n}&2\left(\frac14-\frac1{2^{n+1}}\right)=\frac12-\frac1{2^n}\\
&&\frac18&\frac1{16}&\cdots&\frac1{2^n}&2\left(\frac18-\frac1{2^{n+1}}\right)=\frac14-\frac1{2^n}\\
&&&\frac1{16}&\cdots&\frac1{2^n}&2\left(\frac1{16}-\frac1{2^{n+1}}\right)=\frac18-\frac1{2^n}\\
&&&&\ddots&\vdots&\vdots\\
&&&&&\frac1{2^n}&2\left(\frac1{2^n}-\frac1{2^{n+1}}\right)=\frac1{2^{n-1}}-\frac1{2^n}\\\hline
\frac12&\frac2{2^2}&\frac3{2^3}&\frac4{2^4}&\cdots&\frac{n}{2^n}&2\left(1-\frac1{2^n}\right)-\frac{n}{2^n}=2-\frac{n+2}{2^n}
\end{array}$$</span></p>
<p>Instead of summing the columns to the left of the vertical line to get the terms of the summation and then adding, I’ve summed the rows of the triangular matrix, each of which is a geometric progression, to get the totals in the righthand column above the horizontal line, and then summed those row sums to get the desired total.</p>
|
2,279,281 | <blockquote>
<p>$\displaystyle F(x)=\int_0^x (t-2)f(t)\; dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$?</p>
</blockquote>
<p>The title explains a lot. Given
$$
\displaystyle F(x)=\int_0^x (t-2)f(t)\,dt
$$
with $f(0)=1$, $f(1)=0$ and $f:\mathbb{R}\to\mathbb{R}$ is a strictly decreasing differentiable function. Then:</p>
<p>1) $F$ is strictly increasing in $[0,3]$<br>
2) $F$ has a unique maximum but no minimum in $(0,3)$<br>
3) $F$ has a unique minimum but no maximum in $(0,3)$<br>
4) $F$ has both maximum and minimum in $(0,3)$</p>
<p>I have done $F''(x)=(x-2)f'(x)+f(x)$ but no idea how to verify the options. Any help is appreciated. </p>
| Jay Zha | 379,853 | <p>$$F'(x)=(x-2)f(x)$$</p>
<p>$f$ is strictly decreasing and differentiable (thus continuous). So $f(x) > 0$ on $[0,1)$, $f(1)=0$, and $f(x) <0$ on $(1,3]$.</p>
<p>$F'(x)<0$ on $[0,1)$, $F'(1) =0$, $F'(x) >0$ on $(1,2)$, $F'(2)=0$, and $F'(x)<0$ on $(2,3]$</p>
<p>So $1$ is the minimum point, $2$ is the maximum point. $F(x)$ first decreases, and then reaches local minimum, then increases, then reaches local maximum, and then decreases again. $F''(1) >0$ and $F''(2) < 0$ also confirms this.</p>
|
2,279,281 | <blockquote>
<p>$\displaystyle F(x)=\int_0^x (t-2)f(t)\; dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$?</p>
</blockquote>
<p>The title explains a lot. Given
$$
\displaystyle F(x)=\int_0^x (t-2)f(t)\,dt
$$
with $f(0)=1$, $f(1)=0$ and $f:\mathbb{R}\to\mathbb{R}$ is a strictly decreasing differentiable function. Then:</p>
<p>1) $F$ is strictly increasing in $[0,3]$<br>
2) $F$ has a unique maximum but no minimum in $(0,3)$<br>
3) $F$ has a unique minimum but no maximum in $(0,3)$<br>
4) $F$ has both maximum and minimum in $(0,3)$</p>
<p>I have done $F''(x)=(x-2)f'(x)+f(x)$ but no idea how to verify the options. Any help is appreciated. </p>
| Kevin Limanta | 152,696 | <p>Since $F'(x) = (x-2)f(x)$, then if $F'(x) = 0$, it implies that either $x = 2$ or $f(x) = 0$. We also know that $f$ is strictly decreasing, so $f$ has at most one root. Given that $f(1) = 0$, then the only solution for $f(x) = 0$ is $x = 1$. Hence $F'(x) = 0$ implies that either $x = 1$ or $x = 2$.</p>
<p>Now, on $(0,3)$, we divide into three cases:</p>
<ol>
<li>If $0 < x < 1$, then $x-2<0$ and $f(x) > 0$, so $F'(x) < 0$. [Note: $f(x) > 0$ since $f$ is strictly decreasing and $f(0) = 1$ and $f(1) = 0$]</li>
<li>If $1 < x < 2$, then $x-2<0$ and $f(x) < 0$, so $F'(x) > 0$.</li>
<li>If $2 < x < 3$, then $x-2>0$ and $f(x) < 0$, so $F'(x) < 0$.</li>
</ol>
<p>Hence, $F$ cannot be possibly strictly increasing on $(0,3)$. $F$ has a minimum when $x = 1$ and maximum when $x = 2$. The only true statement is (4).</p>
|
3,358,449 | <blockquote>
<p>I have 8 variables; <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-container">$F$</span>, <span class="math-container">$G$</span>, and <span class="math-container">$H$</span>.</p>
<p>What are they all equal to?</p>
<p>Rules:</p>
<ol>
<li><p>All the variables are equal to integer values between one and eight.</p>
</li>
<li><p>None of the variables are equal to each other.</p>
</li>
<li><p><span class="math-container">$D + E = F$</span></p>
</li>
<li><p><span class="math-container">$B + C = D$</span></p>
</li>
<li><p><span class="math-container">$B + H = A$</span></p>
</li>
<li><p><span class="math-container">$C + G = F$</span></p>
</li>
<li><p><span class="math-container">$E + G = H$</span></p>
</li>
<li><p><span class="math-container">$B + E = G$</span></p>
</li>
</ol>
</blockquote>
<p>I know that <span class="math-container">$1$</span> and <span class="math-container">$2$</span> both must be either <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, or <span class="math-container">$E$</span> since there is no <span class="math-container">$0$</span> so <span class="math-container">$1$</span> must be a letter that does not have an equation answer. For <span class="math-container">$2$</span> I know that there is no equation that is the a variable plus itself.</p>
| gandalf61 | 424,513 | <p>Some hints:</p>
<p>Since <span class="math-container">$8$</span> cannot occur on the left hand side of an equation, we know that either <span class="math-container">$A=8$</span> or <span class="math-container">$F=8$</span>.</p>
<p>From equations <span class="math-container">$1$</span> and <span class="math-container">$2$</span> we know that <span class="math-container">$B+C+E=F$</span>. So <span class="math-container">$F=6,7$</span> or <span class="math-container">$8$</span>.</p>
<p>From equations <span class="math-container">$5,7,8$</span> we know that <span class="math-container">$A=B+H = B+E+G=2G$</span>. Since <span class="math-container">$G \ge 3$</span> we have either <span class="math-container">$G=3, A=6$</span> or <span class="math-container">$G=4, A=8$</span>.</p>
<p>Suppose <span class="math-container">$G=3$</span>. Then <span class="math-container">$A=6$</span> and so <span class="math-container">$F=8$</span>. From equation <span class="math-container">$8$</span> we know that <span class="math-container">$B$</span> and <span class="math-container">$E$</span> are <span class="math-container">$1$</span> and <span class="math-container">$2$</span> (either way round), so <span class="math-container">$C=5$</span> (because <span class="math-container">$B+C+E=F=8$</span>). Since <span class="math-container">$E+G=H$</span> we know that <span class="math-container">$E$</span> cannot be <span class="math-container">$2$</span> (otherwise <span class="math-container">$H=C=5$</span>) so <span class="math-container">$E=1$</span> and <span class="math-container">$B=2$</span>.</p>
<p>This should be enough for you to find one solution.</p>
<p>You should also consider the case "Suppose <span class="math-container">$G=4$</span> ..." to see if there are any other solutions.</p>
|
25,488 | <p>I have noticed a common pattern followed by many students in crisis:</p>
<ul>
<li>They experience a crisis or setback (injury, illness, tragedy, etc)</li>
<li>This causes them to miss a lot of class.</li>
<li>They may stay away from class longer than they "need to" because of shame: they feel that since they have been absent, coming back to class will cause their teacher to be critical of them.</li>
<li>Once they do come back, they are overwhelmed. They might need to do twice the amount of work in all of their courses just to catch up. In classes where early material becomes a needed prerequisite for later material, it can become impossible to follow the course content being presented when they do return.</li>
<li>This experience of failure can create long term patterns of giving up under pressure.</li>
</ul>
<p>I have known many <strong>very</strong> capable students who fall into this trap and end up flunking out of University. I could have fallen into this trap myself. As a high school student and college student I got close on several occasions. I have a lot of empathy for this kind of situation.</p>
<p>Question: What kinds of policies and practices can institutions and individual faculty members adopt which help students out of this trap?</p>
| guest troll on a phone | 20,297 | <p>Policies differ a lot. My dad died December 1 of my second year at USNA. I got 3 days off and zero allowances for turning in final papers or doing final exams. My sister got sent home from a civilian school, total forgiveness on finals and assignments, and credit to move on. Both semester system. Neither one seemed to affect us much academically and next semester it was onward. My littlest sister was in high school and got ongoing attention from teachers throughout the year. And it seemed to derail her.</p>
<p>I guess to boil it down, I think best to cut any allowances short at the end of the term. After that life goes on. Death of parents is something that all of us will encounter unless we die young. I don't have a good suggestion for dealing with death in the family earlier in the term.</p>
<p>This is not to say you can't be someone to talk to. But even here, I would use your judgment and let the student come to you. At the end, he or she is just going to have to deal with it. You may or more likely are not be someone for them to connect with. Use your spidey sense.</p>
<p>I read Stef's answer and violently agree. None of us, students or schools can prevent death if a parent or have a time machine. The important thing is to make a plan and go forward. You can't take away the suck factor of losing a parent. Just be pragmatic about what you do control, school.</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Sam Hopkins | 25,028 | <p>I am one of the moderators of the <a href="https://realopacblog.wordpress.com/" rel="noreferrer">Open Problems in Algebraic Combinatorics blog</a>.</p>
<p>First of all, we welcome submissions from anyone who has a good open problem in algebraic combinatorics that they want to advertise (see the top post for info about how to submit).</p>
<p>Secondly, hopefully this blog shows you how it is easy to set up an open problem repository like this for your own subfield.</p>
|
143,070 | <p>Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case. </p>
<p><img src="https://i.stack.imgur.com/yhrcd.jpg" alt="square"></p>
<p>Now we seek some addition condition on $X\to Y$ that forces the right square is a pullback too. </p>
<p>My question: is epic a sufficient condition? (If the category is Sets, then yes.)</p>
<p><strong>Added</strong>: Let $P$ be the pullback of the right square, then there exists $B\to P$, and the square $A\to P \to Y$ // $A\to X \to Y$ is a pullback, so we have the following diagram in which the bottom and the whole squares are pullback, so is the upper square. If the category is Sets, $X\to Y$ is surjective then $A\to P $ is also surjective. Since the pullback of $B\to P$ along a surjective map is an bijection, $B\to P$ must be a bijection. This shows the right square of the original diagram is a pullback. We can also see why we consider some nice condition on $X\to Y$.</p>
<p><img src="https://i.stack.imgur.com/Uepcb.jpg" alt="2nd square"></p>
| Peter LeFanu Lumsdaine | 2,273 | <p><strong>No, epic is not sufficient;</strong> there is a counterexample in <strong>Pos</strong>, the category of posets. Take <em>A</em>, <em>B</em>, and <em>X</em> each to be the discrete two-element poset $\{0,1\}$; take <em>C</em>, <em>Y</em>, and <em>Z</em> to be the same elements with their natural ordering. $\{ 0 < 1 \}$; and take all maps to be identities on underlying sets.</p>
<p>On the other hand, <strong>split epic is sufficient in any category;</strong> this is a reasonably straightforward diagram chase.</p>
<p>Trying to weaken this, I would suspect (though I haven’t checked it) that in <a href="http://ncatlab.org/nlab/show/regular+category">regular categories</a>, regular epic may well be a sufficient condition; certainly regular categories (and strengthenings like <a href="http://ncatlab.org/nlab/show/exact+category">exact categories</a>) give a natural setting for these sorts of interactions between epis and pullbacks.</p>
|
754,583 | <p>Write <span class="math-container">$$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$</span> so that <span class="math-container">$\phi_n=n+\frac{n}{\phi_n},$</span> which gives <span class="math-container">$\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$</span> We know <span class="math-container">$\phi_1=\phi$</span>, the <a href="http://en.wikipedia.org/wiki/Golden_ratio" rel="nofollow noreferrer">Golden Ratio</a>, so let's take <span class="math-container">$\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$</span>. (Is that justified?)</p>
<p><a href="http://m.wolframalpha.com/input/?i=%28n%2B%E2%88%9A%28n%28n%2B4%29%29%29%2F2&x=0&y=0" rel="nofollow noreferrer">Wolfram Alpha</a> states that, with <span class="math-container">$(2)$</span>, <span class="math-container">$$\lim\limits_{n\to -\infty}\phi_n=-1.$$</span> Why? Can I infer that this is true for <span class="math-container">$(1)$</span> and, if so, <em>why</em>?</p>
<p><strong>I wonder what happens in <span class="math-container">$(1)$</span> for <span class="math-container">$n\in\mathbb{C}\backslash\mathbb{Z}$</span> too</strong>. I got something horrendous looking in <span class="math-container">$(2)$</span> for <span class="math-container">$n=i$</span>.</p>
<hr>
<p><em>Clarification:</em> I'm trying to <strong>find <span class="math-container">$\phi_n$</span> in terms of <span class="math-container">$n$</span></strong>. See the comments below.</p>
| mercio | 17,445 | <p>If $f_n(x) = n+n/x$, I'm not sure about your notation, but I think $(1)$ defines $\phi_n$ as the limit (provided it exists) of $f_n \circ f_n \circ \ldots \circ f_n (n)$.</p>
<p>Such a limit would be a fixpoint of $f_n$, which means one of the two complex numbers $y_n,z_n = \frac {n \pm \sqrt {n(n+4)}}2$.</p>
<p>To guess which one it is, we should wonder which one is the attractive fixpoint ,and which one is repulsive.
$f_n'(x) = -n/x^2 = y_nz_n/x^2$, and so $|f'_n(y_n)| = |z_n/y_n| = 1/|f'_n(z_n)|$. To be attractive, the derivative at that point has to be small ($< 1$ in modulus), hence the solution with the bigger modulus will be attractive, and that's the one the sequence will converge to (unless the sequence accidentally ends up exactly on the smaller fixpoint).</p>
<p>One can check that the two fixpoints have the same modulus when $n \in [-4;0]$, so we can't really define what's $\phi_n$ for those values of $n$. Just look at the sequence for such an $n$, you will see it has chaotic behaviour. On the rest of the complex plane, by picking the bigger fixpoint you obtain a continuous function $n \mapsto \phi_n$ defined on $\Bbb C \setminus [-4;0]$.</p>
<p>So $(2)$ is only correct for $n>0$, and you have to switch the sign of the square root for $n < -4$</p>
|
36,774 | <p>Do asymmetric random walks also return to the origin infinitely?</p>
| Yuval Filmus | 1,277 | <p>Proof sketch: let $P(x,y)$ be the generating function of all walks which end up at the origin for the first time, with $x$ meaning left and $y$ meaning right. You can write a recurrence relation for the walks and deduce an expression for $P$ by solving a quadratic. Now substitute $pt$ for $x$ and $1-p$ for $y$.</p>
|
36,774 | <p>Do asymmetric random walks also return to the origin infinitely?</p>
| Charles | 1,778 | <p>No. Heuristic: If the walk goes right with probability $1/2+\alpha/2>1/2$ then the expected position after $n$ steps is $\alpha n,$ while the expected variation is only $O(\sqrt n).$ Thus the walk crosses the origin only finitely often.</p>
|
680,364 | <p>I want some verification for my proof to a homework problem. (Is it correct? Is there a simpler way to do this?)</p>
<p>Let $G$ be a finite group of odd order and suppose there is an element $g$ that is conjugate to its own inverse. In other words, there is $h \neq e$ such that $h^{-1}gh = g^{-1}$. We will show $g=e$ by supposing it's not and finding a contradiction.</p>
<p>We see that $gh = hg^{-1}$ and $hg = g^{-1}h$. This means given any word composed of $g$, $h$, $g^{-1}$, and $h^{-1}$, we can always "push" the $g^r$'s to one side and the $h^s$'s to the other, giving us a canonical spelling $g^rh^s$. That is to say, $\langle g, h\rangle \cong \langle g \rangle \oplus \langle h \rangle$. </p>
<p>By Lagrange, finding any non-trivial even subgroup will prove that $|G|$ was even after all, giving us the desired contradiction. Since both $\langle g \rangle$ and $\langle h \rangle$ are nontrivial, neither may be even. But if that is the case, then they are both odd, and $\langle g \rangle \oplus \langle h \rangle \cong \langle g, h \rangle$ is now non-trivial even.</p>
<p><strong>EDIT</strong> Critical error at the end. The order of the direct sum is the product, not the sum.</p>
| zyy | 1,019,350 | <p>Assume there exists <span class="math-container">$x \neq 1$</span> in <span class="math-container">$G$</span> with <span class="math-container">$g^{-1}xg=x^{-1}$</span> for some <span class="math-container">$g \in G$</span>. Then <span class="math-container">$g^{-1}x^{m}g = x^{-m}$</span> for any <span class="math-container">$m$</span>, so we can let <span class="math-container">$G$</span> act on <span class="math-container">$\langle x \rangle$</span> by conjugation.</p>
<p>Consider the conjugacy class <span class="math-container">$x^G$</span>, we know if <span class="math-container">$y \in x^G$</span> then also <span class="math-container">$y^{-1} \in x^G$</span>. In addition, <span class="math-container">$y \neq y^{-1}$</span>, otherwise <span class="math-container">$y^2=1$</span>, contradicting <span class="math-container">$|G|$</span> is odd.
Thus any element in <span class="math-container">$x^G$</span> can be paired with its inverse, hence <span class="math-container">$x^G$</span> must contain even elements. Therefore, <span class="math-container">$|G|$</span> is also even since it is a multiple of <span class="math-container">$|x^G|$</span>, a contradiction.</p>
<p><strong>Remark:</strong> <span class="math-container">$\,$</span> Actually we can let <span class="math-container">$G$</span> act on itself by conjugation, and deduce that <span class="math-container">$x^G$</span> has even elements, because if <span class="math-container">$x$</span> is conjugate to <span class="math-container">$x^{-1}$</span> then any other element <span class="math-container">$g^{-1}xg$</span> conjugating to <span class="math-container">$x$</span> must also be conjugate to <span class="math-container">$x^{-1}$</span>, and henceforth to its inverse <span class="math-container">$g^{-1}x^{-1}g$</span>. But I did not realize that at first, and I think in order to force all elements in <span class="math-container">$x^G$</span> to conjugate to its inverse we must put some constraint on <span class="math-container">$x^G$</span> so that any other element will be in some relation with <span class="math-container">$x$</span>.</p>
|
3,062,597 | <p>For the statement of Rouché's theorem, I've always seen that both <span class="math-container">$f$</span> and <span class="math-container">$g$</span> have to be holomorphic on and inside a simple closed curve <span class="math-container">$ C $</span>. However, I am solving a problem which seems to suggest that I should use Rouché's theorem even though I only know that <span class="math-container">$ f $</span> is holomorphic in the unit disk <span class="math-container">$ D $</span> and continuous in <span class="math-container">$ \bar{D} $</span>. I also check Wikipedia's page on Rouché's theorem which says that <span class="math-container">$ f $</span> and <span class="math-container">$ g $</span> only need to be holomorphic inside the region, not on the boundary. Is this sufficient?</p>
| mathcounterexamples.net | 187,663 | <p>Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset <span class="math-container">$U\subseteq \mathbb C$</span> and a compact <span class="math-container">$K \subset U$</span> whose boundary is a closed simple curve positively oriented.</p>
|
134,407 | <p>Some shapes, such as the disk or the <a href="http://en.wikipedia.org/wiki/Reuleaux_triangle" rel="nofollow noreferrer">Releaux triangle</a> can be used as manholes,
that is, it is a curve of constant width.
(The width between two parallel tangents to the curve are independent of the orientation of the curve.)</p>
<p><strong>(1) Is it possible to tile the plane with such shapes?</strong></p>
<p>The shapes should be simply connected, and all must have an area greater than some fixed uniform $\epsilon.$
Otherwise, we'd just tile the plane with disks of various diameters where some are arbitrarily small, similar to the Apollonian circle.
By scaling the tiling, we can take $\epsilon=1$.</p>
<p>As a note,
<a href="https://mathoverflow.net/questions/31810/">an earlier question here on MO gave a positive answer on the existence of non-convex simply connected manholes.</a></p>
<p><em>Some clarifications:</em> By tiling, we mean that all manholes used are closed sets,
and there is no open ball that is simultaneously in the interior of two different manholes. Thus, boundaries of manholes may intersect.
Note, we may use several different manholes in a tiling (otherwise, we are essentially asking for a solution for the open einstein problem).</p>
<p>If the answer is negative, a more general question is the following.
Define the roundness of an object as the minimum width divided by the maximum width of the shape (width is the distance between two non-equal parallel tangents).
The roundness of a circle or a Releaux triangle is 1,
and the square has roundness $1/\sqrt{2}.$ Define the roundness of a tiling as the minimum of the roundness of all shapes in the tiling. </p>
<p>For non-convex shapes, roundness can be defined as follows: It is the factor I need to re-size the hole with, so that the original shape cannot fall through that hole.
For example, a square with sides $1$ cannot fall through a square hole with sides $1/\sqrt{2}.$</p>
<p><strong>(2) What is the best possible roundness $R$ a tiling of the plane can have?</strong></p>
<p>Trivially, $\sqrt{3}/2 \leq R \leq 1$ since we can tile the plane with equilateral triangles, and a positive answer to question 1 gives $R=1.$</p>
<p>The applications are evident: This would be a nice way to tile a ceiling,
instead of using regular square tiles, that sometimes falls down.</p>
| Włodzimierz Holsztyński | 8,385 | <p>(<strong>EDIT (remarks after the actual answer below)</strong> Technically word <em>cover</em> allows for arbitrary overlaps. However, despite the so familiar definition of notion <em>cover</em>, I was still under impression that the Question meant <em>holes</em> which had pairwise disjoint interiors. Thus my (modest) theorem makes this assumption. <strong>Actually</strong>, @Per himself (in his <em>Question</em>) had this clarification: <em>there is no open ball in the interior of two different manholes</em>.</p>
<hr>
<p>I'll answer <strong>Question 1</strong> for more general shapes, and in $\mathbb R^n\ (n>1)$:</p>
<p><strong>(1) Is it possible to cover the plane with such shapes?</strong></p>
<p>By sloppy pronoun <strong><em>such</em></strong> in the question above we should understand arbitrary bounded convex bodies (closed sets) of constant width, where each body can have its own width.</p>
<p>Instead I'll prove more, by answering the question for arbitrary $n$-dimensional strictly convex proper subsets $B\subseteq \mathbb R^n$ (i.e. different from $\mathbb R^n$ itself)--they do not have to be bounded either:</p>
<p><strong>THEOREM</strong> Let $\ F\ $ be an arbitrary family of <strong>proper</strong> strictly convex closed sets $\ B\subseteq \mathbb R^n\ (n>1)\ $ such that the intersection of the interiors of every different two of these sets is empty. Then there exist arbitrarily small closed Euclidean balls $\ K\ $ such that $\ K\ $ is not contained in $\ \bigcup F$.</p>
<p><strong>PROOF</strong> If family $\ F\ $ is empty then the theorem holds. Thus let $\ B_0\in F\ $ and let $\ a\ $ belong to the boundary of $\ B_0.\ $ Then there exists an arbitrarily small real $\ r > 0\ $ (i.e. for every $\ \epsilon > 0\ $ there exists positive real $\ r < \epsilon\ $) such that</p>
<p>$$\forall_{B\ C\in F,\ B\ne C}\quad S\cap B\cap C\ =\ \emptyset$$</p>
<p>where $\ S\ $ is the boundary sphere of $\ K.\ $ <strong>Indeed</strong>, (1) F is countable due to the empty interior intersections; (2) the set of all intersections of the pairs of sets from $\ F\ $ is countable because each pair has at the most one intersection point, and (3) there are continuum of positive $\ r<\epsilon\ $ as needed. Now let's recall a classic Sierpiński's theorem (hm, all Sierpiński theorems are classic though):</p>
<blockquote>
<p>If a connected (metric) compact space $\ X\ $ is covered by a countable family $\ F\ $ of pairwise disjoint closed subsets of $\ X\ $ then $\ F\ $ has at the most one member.</p>
</blockquote>
<p>Since $\ S\ $ is compact and connected, and $\ S\ $ is not contained in $\ B_0,\ $ it follows from the Sierpiński's theorem that</p>
<p>$$\ \bigcup_{B\in F}\ (S\cap B)\ \ \ne\ \ S$$</p>
<p>Consequently, $\ F\ $ does not cover $\ K$.
<strong>END of PROOF</strong></p>
<hr>
<p><strong>REMARK 1</strong> Wacław Sierpiński formulated and proved his theorem for the metric compact spaces (also in general, Siepiński was not shy to use metric methods). But the theorem holds for arbitrary compact spaces. I am sure that Sierpiński proved his theorem before general compact spaces were introduced by Alexandrov and Urysohn (as <em>bicompact spaces</em>--perhaps even today general compact spaces are called bicompacts most of the time?).</p>
<p><strong>REMARK 2</strong> The assumption from the (full text of the) original question:</p>
<blockquote>
<p> <em>all must have an area greater than some fixed uniform</em> ϵ</p>
</blockquote>
<p>was not necessary (as seen from my proof).</p>
|
150,809 | <p>An Iwasawa manifold is a compact quotient of a 3-dimensional complex Heisenberg group by a cocompact, discrete subgroup.
We can also refer to Griffiths and Harris's Principles of Algebraic Geometry p. 444 for simpler description.</p>
<p>I want to compute the automorphism of Iwasawa manifold,i.e.the group of biholomorphisms of X onto itself.
But I don't know how to set out to do this.Is there any already known result about this queation?If not,how should I start to study this problem?
Thanks in advance!</p>
| abx | 40,297 | <p>Here are some easy remarks to start with. The group $\mathrm{Aut}(X)$ is a complex Lie group, its Lie algebra is $H^0(X,T_X)$. Since $X=G/\Gamma $, the tangent bundle $T_X$ is trivial, so $H^0(X,T_X)$ has dimension 3, and is naturally identified with the Lie algebra of $G$; this implies that the neutral component of $\mathrm{Aut}(X)$ is $G$. It remains to compute the finite group $\pi _0(\mathrm{Aut}(X))$; this is probably more subtle.</p>
|
4,620,319 | <p>Let's assume that for <span class="math-container">$0<\beta<\alpha<\frac{\pi}{2}$</span>, <span class="math-container">$\sin(\alpha+\beta) = \frac{4}{5}$</span>, and <span class="math-container">$\sin(\alpha-\beta) = \frac{3}{5}$</span>. Then, how could we find <span class="math-container">$\cot(\beta)$</span>?</p>
<p><span class="math-container">$$\sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin(\alpha)\cos(\beta) = \frac{7}{5}$$</span></p>
<p><span class="math-container">$$\sin(\alpha+\beta)-\sin(\alpha-\beta) = 2\sin(\beta)\cos(\alpha) = \frac{1}{5}$$</span></p>
<p><span class="math-container">$$\tan(\alpha)\cot(\beta) = 7$$</span></p>
<p>But I am not sure where this would lead us.</p>
| mrtechtroid | 960,957 | <p><span class="math-container">$\displaystyle sin( a+b) =4/5\ \ \ sin( a-b) =3/5.$</span><br />
<span class="math-container">$\displaystyle So\ cos( a+b) \ =\ 3/5$</span> but that means <span class="math-container">$\displaystyle sin\left(\frac{\pi }{2} -( a+b)\right) =sin( a-b) =3/5$</span><br />
So <span class="math-container">$ $$\displaystyle \frac{\pi }{2} =a+b+a-b$</span><br />
Or <span class="math-container">$\displaystyle a\ =\ \frac{\pi }{4}$</span>
So <span class="math-container">$cot(a) = 1 $</span>
And since you have already figured out <span class="math-container">$tan(a)cot(b)=7$</span> so <span class="math-container">$cot(b)=7$</span></p>
|
2,977,645 | <p>I'm trying to prove that <span class="math-container">$\sqrt[n]{\frac{s}{t}}$</span> is irrational unless both s and t are perfect nth powers. I have found plenty of proofs for nth root of an integer but cannot find anything for rationals. Also trying to work up from the proofs I have found is rather difficult.</p>
<p>My lecturer wants me to prove this using uniqueness of prime factorisation but I also have no idea where to start with that. </p>
<p>Any help is greatly appreciated.</p>
| Calum Gilhooley | 213,690 | <p>This is more of a comment than an answer, because it doesn't use the uniqueness of prime factorisation. Instead, it uses the more basic result that if a positive integer <span class="math-container">$c$</span> divides the product of two positive integers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, and <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are relatively prime, i.e. they have no common factors other than <span class="math-container">$\pm1$</span>, then <span class="math-container">$c$</span> divides <span class="math-container">$a$</span>.</p>
<p>Suppose that the fraction <span class="math-container">$\tfrac{s}{t}$</span> is in its lowest terms, i.e. <span class="math-container">$s$</span> and <span class="math-container">$t$</span> are relatively prime; and suppose that <span class="math-container">$\left(\tfrac{u}{v}\right)^n = \tfrac{s}{t}$</span>, where <span class="math-container">$\tfrac{u}{v}$</span> also is in its lowest terms, i.e. <span class="math-container">$u$</span> and <span class="math-container">$v$</span> are relatively prime. Then:
<span class="math-container">\begin{equation}
\tag{$*$}\label{eq:simple}
tu^n = sv^n.
\end{equation}</span></p>
<p>The hypothesis that <span class="math-container">$u$</span> and <span class="math-container">$v$</span> are relatively prime implies that <span class="math-container">$u^n$</span> and <span class="math-container">$v^n$</span> are also relatively prime:</p>
<p><em>Proof.</em> Suppose that an integer <span class="math-container">$r > 1$</span> divides both <span class="math-container">$u^n$</span> and <span class="math-container">$v^n$</span>. Every integer <span class="math-container">$> 1$</span> has a prime factor. (This fact is admittedly an ingredient of the proof of uniqueness of prime factorisation, but still, it is a simpler theorem.) Let <span class="math-container">$p$</span> be a prime factor of <span class="math-container">$r$</span>. By the hypothesis that <span class="math-container">$u$</span> and <span class="math-container">$v$</span> are relatively prime, <span class="math-container">$p$</span> cannot divide both <span class="math-container">$u$</span> and <span class="math-container">$v$</span>. But if <span class="math-container">$p$</span> does not divide <span class="math-container">$u$</span>, then by repeated application of the theorem quoted above, <span class="math-container">$p$</span> does not divide <span class="math-container">$u^n$</span>. Similarly, if <span class="math-container">$p$</span> does not divide <span class="math-container">$v$</span>, then <span class="math-container">$p$</span> does not divide <span class="math-container">$v^n$</span>. Therefore, <span class="math-container">$p$</span> does not divide both <span class="math-container">$u^n$</span> and <span class="math-container">$v^n$</span>. This contradicts the hypothesis that <span class="math-container">$p$</span> divides <span class="math-container">$r$</span>. Therefore, no such <span class="math-container">$p$</span> and no such <span class="math-container">$r$</span> can exist. <span class="math-container">$\square$</span></p>
<p>Now, four applications of our favourite theorem to \eqref{eq:simple} show that <span class="math-container">$s$</span> divides <span class="math-container">$u^n$</span>, <span class="math-container">$u^n$</span> divides <span class="math-container">$s$</span>, <span class="math-container">$t$</span> divides <span class="math-container">$v^n$</span>, and <span class="math-container">$v^n$</span> divides <span class="math-container">$t$</span>. Therefore, <span class="math-container">$s = u^n$</span> and <span class="math-container">$t = v^n$</span>. <span class="math-container">$\square$</span></p>
<p>(Perhaps in an effort to be simple, this has ended up being too laborious? Oh, well.)</p>
|
234,866 | <p>Problem: when I draw a rectangle and put a coloured edge around it, the displayed edge is centred along the nominal edge and if it follows the same course as one of the axes then it does not show up. For example:</p>
<pre><code>Graphics[{EdgeForm[Red], FaceForm[], Rectangle[{0, 0}, {4, 3}]}, Axes -> True]
</code></pre>
<p>yields this:</p>
<p><a href="https://i.stack.imgur.com/eBZNY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eBZNY.jpg" alt="Enter image description here" /></a></p>
<p>The problem also arises when for example I want to show a grid of 1x1 squares with some edged in red and others in blue: along any given line-segment, only one colour shows.</p>
<p>How can I make the coloured "edge" of a rectangle sit inside the shape, touching the shape's boundary, but not running outside it?</p>
| kglr | 125 | <p>We can mimic thick edges overlaying a white scaled rectangle over a colored rectangle:</p>
<pre><code>t = .1;
Graphics[Table[
{EdgeForm[], If[EvenQ[i + j], Red, Blue], Rectangle[{i, j}, 1 + {i, j}],
FaceForm[White], Scale[Rectangle[{i, j}, 1 + {i, j}], 1 - 2 t]},
{i, 7}, {j, 5}],
ImageSize -> Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/RmO9o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RmO9o.png" alt="enter image description here" /></a></p>
<p>Use <code>t = .2</code> to get</p>
<p><a href="https://i.stack.imgur.com/VAZou.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VAZou.png" alt="enter image description here" /></a></p>
|
2,423,569 | <p>I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers. </p>
<p>So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see if I get some information about $a,b,c,d$ considering that $T(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ but this doesn't really work. Also I've considered $SL(2,\mathbb{C}) / \{ \pm I\}$ which is isomorphic to the group of Mobius transformations but this doesn't really help either.</p>
| Angina Seng | 436,618 | <p>Prove first that
$$T(z)=\frac{a'z+b'}{c'z+d'}$$
for real $a'$, $b'$, $c'$ and $d'$ and that then $a=\pm a'/\sqrt{a'd'-b'c'}$ etc.</p>
|
234,851 | <p>Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.</p>
<p>My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.</p>
<p>Next, I used this formula, </p>
<p>$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$</p>
<p>this attempt leaves me with such a messy long equation that eventually took up 2 pages, and still left me unsolved. I am convinced there must be an easier way.</p>
<p>Any hints please? thanks in advance.</p>
| Community | -1 | <p>You are asked to find the length of the curve from $y=1$ to $y=2$ and $x$ is a function of $y$, so you need to use this equation (page 585 of 5th edition of Stewart's calculus):</p>
<p>$$L=\int_1^2\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y$$</p>
|
234,851 | <p>Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.</p>
<p>My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.</p>
<p>Next, I used this formula, </p>
<p>$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$</p>
<p>this attempt leaves me with such a messy long equation that eventually took up 2 pages, and still left me unsolved. I am convinced there must be an easier way.</p>
<p>Any hints please? thanks in advance.</p>
| Dilawar | 1,674 | <p>Make the following substitution,
$ y = \sec w $ (since y is between 1 and 2, it should be fine) and change the limits of integration accordingly.</p>
<p>Use the trigonometric identities to simplify intermediate expressions. </p>
|
811,856 | <p>Let $X,Y,H$ be the standard base for the Lie algebra $\mathrm{sl}_2({\mathbb{C}})$, i.e. $H=\begin{pmatrix} 1 & 0\\ 0 &-1\end{pmatrix}$, $X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Let $V$ be some finite dimensional irreducible representation of $\mathrm{sl}_2$. I'm sorry if my question is for children but why the action of $H$ is diagonalizable?</p>
| Rene Schipperus | 149,912 | <p>There are two ways to see this one is the fact that the Jordan decomposition is preserved under representations of semisimple Lie algebras. The second is Weyl's unitary trick which is based on the fact that all representations of compact groups are completely reducible.
You can find more on both in Fulton and Harris: Representation Theory A First Course.</p>
|
811,856 | <p>Let $X,Y,H$ be the standard base for the Lie algebra $\mathrm{sl}_2({\mathbb{C}})$, i.e. $H=\begin{pmatrix} 1 & 0\\ 0 &-1\end{pmatrix}$, $X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Let $V$ be some finite dimensional irreducible representation of $\mathrm{sl}_2$. I'm sorry if my question is for children but why the action of $H$ is diagonalizable?</p>
| Olivier Bégassat | 11,258 | <p>An elementary proof relies on the folllowing observation:</p>
<blockquote>
<p>Let $\lambda\in\Bbb C$ be an eigenvalue of $H$ acting on $V$ (<em>any</em> finite dimensional representation of $\mathfrak{sl}_2(\Bbb C)$). Then the following sum of eigenspaces of $H$
$$W=\bigoplus_{n\in\Bbb Z}E_{\lambda+2n}(H)$$
is a subrepresentation of $V$.</p>
</blockquote>
<p>Here, for every complex number $\alpha$, $E_{\alpha}(H)=\lbrace v\in V\mid H\cdot{}v=\alpha v\rbrace$ is either $0$ or an eigenspace of $H$ when $\alpha$ is an eigenvalue.</p>
<p>The fact that $\lambda$ is eigenvalue is actually irrelevant to the proof of the highlighted claim. Of course, if it isn't an eigenvalue of $H$, then $W$ might very well be $0$.</p>
<hr>
<p>Let $\lambda\in\Bbb C$ be a complex number and $W$ defined as above. Let us show that $W$ is a subrepresentation of $V$, that is, let us show that it is stable under the action of $\mathfrak{sl}_2(\Bbb C)$. By construction it is stable under $H$. Let $n\in\Bbb Z$ be an integer, and $v\in E_{\lambda+2n}(H)$. Then</p>
<p>$$
\begin{array}{rvl}
H\cdot\big(X\cdot v\big)&=&[H,X]\cdot v+X\cdot\big(H\cdot v\big)\\
&=& 2X\cdot v+X\cdot(\lambda+2n)v\\
&=&(\lambda+2(n+1))X\cdot v
\end{array}$$
that is, $X\cdot v\in E_{\lambda+2(n+1)}(H)\subset W$. Similarly, $Y\cdot v\in E_{\lambda+2(n-1)}(H)\subset W$, and the claim follows. This shows that $W$ is stable under $X, H$ and $Y$, so that it is indeed a subrepresentation of $V$.</p>
<hr>
<p>Now suppose $V$ is irreducible and $\lambda$ is an eigenvalue of the action of $H$ on $V$ (because $V$ is finite dimensional over $\Bbb C$, there exists at least one such eigenvalue). From what precedes we know that $W$ is a (nonzero) subrepresentation of of $V$ so that it is all of $V$. By definition, $H$ is diagonalizable on $W$, hence $H$ acts diagonally on $V$.</p>
<p>If $V$ is irreducible, then $W$ is all of $V$ and we further have that
$$\mathrm{Spec}(H)\subset\lbrace\lambda+2n\mid n\in\Bbb Z\rbrace$$
Actually if $0\leq m,n$ are maximal such that $\lbrace\lambda+2k\mid -m\leq k\leq n\rbrace\subset\mathrm{Spec}(H)$ then
$$\mathrm{Spec}(H)=\lbrace\lambda+2k\mid -m\leq k\leq n\rbrace)$$</p>
<p>The slightly stronger claim follows at once and is very easily established. It requires more work to show that all those subspaces $E_{\lambda+2k}$, for $-m\leq k\leq m$ are one dimensional.</p>
<p>I warmly recommend <strong>Problem 1.55</strong> from this free and wonderful <a href="http://math.mit.edu/~etingof/replect.pdf" rel="nofollow">Introduction to representation theory</a>.</p>
|
2,416,071 | <p>I have this integral: $\displaystyle \int^{\infty}_0 kx e^{-kx} dx$.</p>
<p>I tried integrating it by parts:</p>
<p>$\dfrac{1}{k}\displaystyle \int^{\infty}_0 kx e^{-kx} dx = ... $. But I'm stuck </p>
<p>now. Can you help me please?</p>
| DeepSea | 101,504 | <p><strong>hint:</strong> $\displaystyle \int kxe^{-kx}dx = -\displaystyle \int xd(e^{-kx})$</p>
|
572,276 | <p>I am trying to write the definition of greatest common divisor using symbolic notation. Here is my current attempt: </p>
<p>$d = gcd(m,n) \Leftrightarrow d \in Z \wedge max(d | m \wedge d | n)$</p>
<p>Any help or hints are greatly appreciated! Thanks!</p>
| amWhy | 9,003 | <p>Let's try to say what you are saying on the right-hand side: $$d = \gcd(m, n) \iff \Big((d\in \mathbb Z \land d\mid m \land d\mid n) \land \forall d' \in \mathbb Z\left((d'\mid m \land d'\mid n) \rightarrow d' \leq d\right)\Big)$$</p>
|
572,276 | <p>I am trying to write the definition of greatest common divisor using symbolic notation. Here is my current attempt: </p>
<p>$d = gcd(m,n) \Leftrightarrow d \in Z \wedge max(d | m \wedge d | n)$</p>
<p>Any help or hints are greatly appreciated! Thanks!</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is an alternative definition: $\;\gcd(m,n)\;$ is the (unique) non-negative number that satisfies
$$
\langle \forall d \in \mathbb Z :: d \mid \gcd(m,n) \;\equiv\; d \mid m \:\land\: d \mid n \rangle
$$
See <a href="https://math.stackexchange.com/search?q=is%3Aanswer+user%3A11994+gcd">some answers of mine which use this definition</a>.</p>
|
4,247,268 | <p><strong>Q:</strong></p>
<blockquote>
<p>If <span class="math-container">$f\left(x\right)=-\frac{x\left|x\right|}{1+x^{2}}$</span> then find <span class="math-container">$f^{-1}\left(x\right)$</span></p>
</blockquote>
<p>My approach:</p>
<ol>
<li>Dividing the cases when <span class="math-container">$x\ge0$</span> and when <span class="math-container">$x\le0$</span> to break free of modulus.</li>
<li>Re-arranging the terms to get the expression of x in terms of y.</li>
<li>Here's what I got:</li>
</ol>
<blockquote>
<p>When <span class="math-container">$x\ge0$</span>:
<span class="math-container">$$x=\sqrt{\frac{-y}{1+y}}$$</span> <span class="math-container">$$\to\ y\ ∈\ \left(-1,0\right] Now, y\to x$$</span>
so, <span class="math-container">$f^{-1}\left(x\right)=\sqrt{-\frac{x}{1+x}}$</span> when <span class="math-container">$x\le0$</span></p>
</blockquote>
<blockquote>
<p>When <span class="math-container">$x\le0$</span>:
<span class="math-container">$$x=-\sqrt{\frac{y}{1-y}}$$</span>
when <span class="math-container">$y\ ∈\ \left[0.1\right)$</span> Now replacing <span class="math-container">$y\to x$</span>
We get, <span class="math-container">$f^{-1}\left(x\right)=-\sqrt{\frac{x}{1-x}}\ ;\ x\ge0$</span></p>
</blockquote>
<p>But I have to show that the inverse function <span class="math-container">$f^{-1}\left(x\right)$</span>=<span class="math-container">$\operatorname{sgn}\left(-x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$</span></p>
<p>This is where I'm getting stuck. I am unable to convert my answer into this form, mainly because I'm not able to convert the cases into this expression. Is there any step-by-step systematic way in which I can do the same? Any help or guide will be greatly appreciated.</p>
<p><strong>Edit:</strong></p>
<p>Since we got <span class="math-container">$f^{-1}\left(x\right)$</span> and the cases,:</p>
<p><span class="math-container">$f^{-1}\left(x\right)=-\sqrt{\frac{x}{1-x}}\ ;\ x\ge0$</span> and
<span class="math-container">$f^{-1}\left(x\right)=\sqrt{-\frac{x}{1+x}}$</span> when <span class="math-container">$x\le0$</span>,</p>
<p>to write it in given form we need something that will give - sign when <span class="math-container">$x>0$</span> so we will use sgn(-x), and rest is just use of modulus so that we can make the general answer.</p>
| Alessio K | 702,692 | <p>The sign function is given by</p>
<p><span class="math-container">$$\operatorname{sgn}(x)=\begin{cases}-1, \space\text{if}\space x<0\\ 0, \space\space\text{if}\space x=0\\1, \space\text{if}\space x>0\end{cases}$$</span></p>
<p>and the modulus of <span class="math-container">$x$</span> is given by</p>
<p><span class="math-container">$$|x|=\begin{cases}x, \space\text{if}\space x\geq0\\ -x, \space\text{if}\space x<0\\\end{cases}$$</span></p>
<p>Thus the inverse can be written as</p>
<p><span class="math-container">$$f^{-1}\left(x\right)=\operatorname{sgn}\left(-x\right)\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}=\begin{cases}-\sqrt{\frac{x}{1-x}}, \space\text{if} \space x\geq0 \\ \sqrt{\frac{-x}{1+x}}, \space\text{if} \space x<0 \end{cases}$$</span></p>
|
5,528 | <p>Let H be a subgroup of G. (We can assume G finite if it helps.) A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1. Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.</p>
<p>Contrary to my initial naive expectation, it is neither necessary nor sufficient that one of H and K be normal. I ran across both of the following counterexamples in Dummit and Foote:</p>
<ul>
<li><p>It is not necessary that H or K be normal. An example is S<sub>4</sub> which can be written as the product of H=⟨(1234), (12)(34)⟩≅D<sub>8</sub> and K=⟨(123)⟩≅ℤ<sub>3</sub>, neither of which is normal in S<sub>4</sub>.</p></li>
<li><p>It is not sufficient that one of H or K be normal. An example is Q<sub>8</sub> which has a normal subgroup isomorphic to Z<sub>4</sub> (generated by i, say), but which cannot be written as the product of that subgroup and a subgroup of order 2.</p></li>
</ul>
<p>Are there any general statements about when a subgroup has a complement? The <A href="http://en.wikipedia.org/wiki/Complement_%28group_theory%29">Wikipedia page</A> doesn't have much to say. In practice, there are many situations where one wants to work with a transversal of a subgroup, and it's nice when one can find a transversal that is also a group. Failing that, one can ask for the smallest subgroup of G containing a transversal of H.</p>
| William DeMeo | 9,124 | <p>(This is a follow up question rather than an answer.) </p>
<p>Wouldn't it make more sense if the complement of a subgroup $H \leq G$ were defined to be a subgroup $K\leq G$ such that $H \cap K = 1$ and $\langle H, K \rangle = G$? That is, shouldn't we allow for the possibility that the set $HK$, consisting of all products $h k$ ($h\in H, k\in K$), is not a group? (Yet we may still have $HK \subsetneq \langle H, K \rangle = G$ in this situation.)</p>
<p>..or does everyone take $HK$ to mean $\langle H, K \rangle$ in this context?</p>
|
2,118,931 | <p>If $A$ is a $4\times2$ matrix and $B$ is a $2\times 3$ matrix, what are the possible values of $\operatorname*{rank}(AB)$?</p>
<p>Construct examples of $A$ and $B$ exhibiting each possible value of $\operatorname*{rank}(AB)$ and explain your reasoning.</p>
| Fernando Revilla | 401,424 | <p><strong>Hint.</strong> Use that in general, $$\text{rank}(AB)\leq \min\{\text{rank }A, \text{rank }B\},$$ and in our case, $$0\le \text{rank }A \le 2,\quad 0\le \text{rank }B \le 2.$$</p>
|
3,886,523 | <p>Each of <span class="math-container">$3$</span> urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?</p>
<p>I think, this is <span class="math-container">$\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$</span> by Bayes'theorem and Law of total probability, but can't be sure.</p>
<p>Thanks for any help.</p>
| Théophile | 26,091 | <p>Any polygon can be decomposed into triangles, giving <span class="math-container">$A = A_1 + \cdots + A_n$</span>.</p>
<p>Thus,
<span class="math-container">$$\begin{align}
V &= V_1 + \cdots + V_n\\
&=\frac13A_1h + \cdots + \frac13A_nh\\
&=\frac13(A_1 + \cdots + A_n)h\\
&=\frac13Ah.
\end{align}$$</span></p>
<hr />
<p>For the base case, you can decompose a prism like so. To see that the three pyramids have the same volume, note that the left and middle pyramids share a blue base and have the same height; similarly, the middle and right pyramids share a red base and have the same height.</p>
<p><a href="https://i.stack.imgur.com/tmmQX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tmmQX.png" alt="enter image description here" /></a></p>
|
1,482,644 | <p>I am trying to find the shortest equivalent expression of the following:</p>
<p>((C → D) $\wedge$ (D → C)) $↔$ (C $\wedge$ D ∨ ¬C $\wedge$ ¬D)</p>
<p>I have "simplified" the expression into the following:</p>
<p>(($\neg$C $\vee$ D) $\wedge$ ($\neg$D $\vee$ C)) $↔$ ((C $\wedge$ D) ∨ (¬C $\wedge$ ¬D))</p>
<p>I am able to figure our that the expression on the left hand side of the $↔$ operator requires either D or C to be true to evaluate to true. However, the expression on the right hand side requires both to be either true or both to be false.</p>
<p>However I am not sure on how to find an even shorter equivalent expression than I already have.
Also, would the resulting expression be a tautology or contradiction?</p>
| Jeevan Devaranjan | 220,567 | <p>Your simplification is correct. One key quality of the material conditional(if) in classical logic is that $C \rightarrow D$ is equivalent to $\lnot C \lor D$. You can read more about the material conditional <a href="https://en.wikipedia.org/wiki/Material_conditional" rel="nofollow">here</a>
Since your simplication is correct it can be shown that the statement is a tautology.
Use De Morgan's law to simplify the second portion and you will see
\begin{equation}
(C \land D) \lor (\lnot C \land \lnot D)
\end{equation}
which is equivalent to
\begin{equation}
(C \land D) \lor \lnot(C \lor D)
\end{equation}
You can see that this is only true if C and D are both true or both false. This is quivalent to the not exclusive or (XNOR) which is also equivalent to the if and only if condition. Therefore the statement becomes
\begin{equation}
(\lnot C \lor D) \land (C \lor \lnot D) \leftrightarrow (C \leftrightarrow D)
\end{equation}
But you can also see that on the left side both C and D need to both be true or both be false. This is just the XNOR which is just the if and only if. The final expression of your statment is
\begin{equation}
(C \leftrightarrow D) \leftrightarrow (C \leftrightarrow D)
\end{equation}
Which is always true and therefore a tautology.</p>
|
2,155,180 | <p>Let $f,g$ be analytic on some domain $\Omega \subset \mathbb{C}$. By Cauchy's formula, we have
$$
\frac{1}{2\pi i} \oint_{\partial\Omega}
\frac{f(z) \, g(z)}{z - z_0}
\, dz
=
f(z_0) \, g(z_0)
=
-\frac{1}{4\pi^2}
\oint_{\partial\Omega}
\frac{f(u)}{u - z_0}
\, du
\,
\oint_{\partial\Omega}
\frac{g(v)}{v - z_0}
\, dv
.
$$
Is there a way how I can get from the first expression to the last without the intermediate step? </p>
| Mark Fischler | 150,362 | <p>For this specific case, of course it is easiest to notice that $20\equiv -3$ and $49 \equiv +3$, as in the comments.</p>
<p>In the general case, it is useful to first apply Euclid's algorithm on the multiplier of $x$ (here, $23$ and the modulus. So let's say we needed to solve $23x=2\pmod {79}$; we would have
$$
79 = 3(23)+10\\23 = 2(10)+3\\10=3(3)+1
$$
and then
$$
1 = 10 - 3(3) = 10-3(23-2(10))\\
= 7(10)-3(23) = 7(79-3(23))-3(23) \\= 7(79) - 24(23) \equiv -24(23)
$$
So $$24(23) \equiv -1 \pmod{79}$$
and $x = 24(-2) \equiv 31 \pmod{79}$.</p>
|
499,840 | <p>I have three points $A=(2,3), B=(6,4)$ and $C=(6,6).$
Given $\vec{AB}=\vec v$ and $\vec{BC}={0 \choose 2}$. I have also that for every $t\in [0,1]$ there is a point $D$ given as $\vec{AD}=t\vec{v}.$ </p>
<p>My question is determine $t$ such that the area of triangle $ADC$ equals area of the triangle $DBC$.</p>
<p>My suggestions is Can I say that $D$ is on line $AB$ dividing the area of $ABC$ in to two equal parts, namely $ADC$ and $DBC$? If this is true then why is it true?</p>
<p>Thanks a lot.</p>
| njguliyev | 90,209 | <p>Yes, $t=\frac12$. Hint: The triangles $ADC$ and $DBC$ have the same altitude.</p>
|
343,281 | <p>Consider the following note written by Gerhard Gentzen in early 1932, on the onset of his work on a consistency proof for arithmetic:</p>
<blockquote>
<p>The axioms of arithmetic are obviously correct, and the principles of proof obviously preserve correctness. Why cannot one simply conclude consistency, i.e., what is the meaning of the second incompleteness theorem, the one by which consistency of arithmetic cannot be proved by arithmetic means? Where is the Godel-point hiding?</p>
</blockquote>
<p>The first question one might ask when reading this statement (plus three questions) is, how is it that Gentzen concludes that, "The axioms of arithmetic [read 'arithmetic' as meaning, <span class="math-container">$PA$</span>--my comment] are obviously correct."? Well, one might infer that Gentzen infers that "The axioms of arithmetic are obviously correct" by virtue of the fact that the axioms of <span class="math-container">$PA$</span> satisfy the following structure:</p>
<p><span class="math-container">$$\langle \mathfrak N, +, \times, = \rangle$$</span></p>
<p>where <span class="math-container">$\mathfrak N = \{ |, ||, |||,\ldots\},$</span> '<span class="math-container">$+$</span>' as meaning concatenation, '<span class="math-container">$\times$</span>' as meaning the Hilbert-Bernays definition of multiplication (e.g., || <span class="math-container">$\times$</span> ||| means replacing each | in || by |||, i.e., ||||||), and '=' as simply meaning equality as defined by the axioms of equality, i.e. for the axiom of equality 'a=a' one has, for the elements of <span class="math-container">$\mathfrak N$</span>, the following equalities:</p>
<p>{ |=|, ||=||, |||=|||,...} [given this, and the closure of <span class="math-container">$\mathfrak N$</span> under <span class="math-container">$+$</span> and <span class="math-container">$\times$</span>, how is it possible that <span class="math-container">$PA$</span>, satisfying this structure, could ever derive, say, '||=|||'?]</p>
<p>In his answer to Noah Schweber's mathoverflow question, "What are some proofs of Godel's Theorem which are <em>essentially different</em> from the original proof?", Ron Maimon mentions the "Jech/Woodin Set theory model proof". In regards to Gentzen's point of view (at least in early 1932), it might behoove one to take a close look at Prof. Jech's three-page paper (<em>Proceedings of the American Mathematical Society</em>, Volume 121, Number 1, May 1994, pp. 311-313).</p>
<p>Why? Because of "Remark 2" on pg. 312 which states:</p>
<blockquote>
<p>Even though our proof of Godel's Theorem [Second Incompleteness Theorem--my comment] uses the Completeness Theorem, it can be modified to apply to weaker theories such as Peano Arithmetic (<span class="math-container">$PA$</span>). To prove that <span class="math-container">$PA$</span> does not prove its own consistency, (unless it is inconsistent), we argue as follows:</p>
<p>Assume that <span class="math-container">$PA$</span> is consistent and that "<span class="math-container">$PA$</span> is consistent" is provable in <span class="math-container">$PA$</span>. There is a conservative extension <span class="math-container">$\Gamma$</span> [let it be <span class="math-container">$ACA_0$</span> as in Noah Schweber's answer--my comment] of <span class="math-container">$PA$</span> in which the Completeness Theorem is provable [Theorem 5.5, p. 443, of Takeuti's <em>Proof Theory</em>, 2nd ed.--my expansion of his comment by his reference], and moreover, <span class="math-container">$PA$</span> <span class="math-container">$\vdash$</span> (<span class="math-container">$\Gamma$</span> is a conservative extension of <span class="math-container">$PA$</span>). Therefore, <span class="math-container">$\Gamma$</span> <span class="math-container">$\vdash$</span> (<span class="math-container">$\Gamma$</span> is a conservative extension of a consistent theory) and thus proves its own consistency. Consequently, <span class="math-container">$\Gamma$</span> proves that <span class="math-container">$\Gamma$</span> has a model.</p>
<p>Now let <span class="math-container">$\Sigma$</span> be a sufficiently strong finite subset of of <span class="math-container">$\Gamma$</span> that proves that <span class="math-container">$\Sigma$</span> has a model; the proof above leads to a contradiction.</p>
</blockquote>
<p>Is this where the Godel-point is hiding with regards to Gentzen's statement and first question?</p>
<blockquote>
<p>The axioms of arithmetic are obviously correct, and the principles of proof obviously prove correctness. Why cannot one simply conclude consistency....?</p>
</blockquote>
<p>Would the 'Godel-point' in question be, following Prof. Jech's Main Theorem,</p>
<blockquote>
<p>It is unprovable in <span class="math-container">$ACA_0$</span> (unless <span class="math-container">$ACA_0$</span> is inconsistent) that there exists a model of <span class="math-container">$PA$</span>. ?</p>
</blockquote>
<p>Now as regards Noah Schweber's very nice answer, I have two questions regarding the following passage</p>
<blockquote>
<p>...However, we are <em>not</em> guaranteed that our model <span class="math-container">$\mathfrak M$</span> [of <span class="math-container">$ACA_0$</span>-- my comment] thinks that its first-order part actually satisfies <span class="math-container">$PA$</span>. That is, the "obvious truth" of the <span class="math-container">$PA$</span> axioms is not actually that obvious.</p>
<p>This is an example of a failure on an <span class="math-container">$\omega$</span>-rule: while for each axiom <span class="math-container">$\varphi$</span> of <span class="math-container">$PA$</span> we do in fact have "<span class="math-container">$Num$</span>(<span class="math-container">$\mathfrak M$</span>) <span class="math-container">$\vDash$</span> <span class="math-container">$\varphi$</span>" (appropriately phrased) is true in <span class="math-container">$\mathfrak M$</span>, we do <em>not</em> get from this that "<span class="math-container">$Num$</span>(<span class="math-container">$\mathfrak M$</span>) <span class="math-container">$\vDash$</span> each <span class="math-container">$PA$</span> axiom" is true in <span class="math-container">$\mathfrak M$</span>. And this is just like how being able to check each individual derivation in <span class="math-container">$PA$</span> doesn't give us a way to check all derivations at once, so it really shouldn't be suprising.</p>
</blockquote>
<ol>
<li>How does the above passage relate to Gentzen's note, especially the phrase</li>
</ol>
<blockquote>
<p>That is, the "obvious truth" of the <span class="math-container">$PA$</span> axioms is not actually that obvious.</p>
</blockquote>
<ol start="2">
<li>What perspective is Gentzen taking in his note (external or internal) and why does it matter what <span class="math-container">$\mathfrak M$</span> 'thinks' (so to speak) as regards Gentzen's note?</li>
</ol>
<p>Now two questions for Panu Raatikainen: as regards your statement</p>
<blockquote>
<p>In general, we just cannot see that they [the theories "which include elementary arithmetic and happen to be consistent"--my paraphrase of your earlier comment] are consistent.</p>
</blockquote>
<ol>
<li><p>Why not?</p></li>
<li><p>What was Gentzen 'seeing' when he made his statement ("The axioms of arithmetic are obviously correct, and the principles of proof obviously preserve correctness"), and why was his 'seeing' incorrect (i.e., leading to inconsistency)?</p></li>
</ol>
| Noah Schweber | 8,133 | <p>Your question seems to boil down to <em>(after fixing an error)</em> the following:</p>
<blockquote>
<p>Any model <span class="math-container">$\mathfrak{M}$</span> of ACA<span class="math-container">$_0$</span> has a first-order part <span class="math-container">$Num(\mathfrak{M})$</span>, which satisfies PA; why doesn't this mean that ACA<span class="math-container">$_0$</span> proves "PA has a model" <em>(indeed, the a priori stronger "PA is sound")</em> and hence its own consistency?</p>
</blockquote>
<p>Here ACA<span class="math-container">$_0$</span> is a conservative extension of the type Jech mentions (what he calls "<span class="math-container">$\Gamma$</span>"): it can talk about models and prove basic model-theoretic results <em>(e.g. the soundness and completeness theorems)</em>, and is PA-provably a conservative extension of PA.</p>
<p>The issue is the following. Fix a model <span class="math-container">$\mathfrak{M}$</span> of ACA<span class="math-container">$_0$</span>. We can indeed talk about <span class="math-container">$Num(\mathfrak{M})$</span> as a structure in <span class="math-container">$\mathfrak{M}$</span>, and have the following:</p>
<blockquote>
<p><span class="math-container">$(*)\quad$</span> Any set of sentences true in <span class="math-container">$Num(\mathfrak{M})$</span> is consistent.</p>
</blockquote>
<p>However, we are <em>not</em> guaranteed that our model <span class="math-container">$\mathfrak{M}$</span> thinks that its first-order part actually satisfies PA. That is, the "obvious truth" of the PA axioms is not actually that obvious. From the existence of such models we conclude </p>
<blockquote>
<p>PA <span class="math-container">$\not\vdash Sound($</span>PA),</p>
</blockquote>
<p>and so in particular the obvious argument for PA<span class="math-container">$\vdash$</span> "PA has a model" breaks down.</p>
<p>This is an example of a failure of the <span class="math-container">$\omega$</span>-rule: while for each axiom <span class="math-container">$\varphi$</span> of PA we do in fact have that "<span class="math-container">$Num(\mathfrak{M})\models \varphi$</span>" (appropriately formalized) is true in <span class="math-container">$\mathfrak{M}$</span>, we do <em>not</em> get from this that "<span class="math-container">$Num(\mathfrak{M})\models$</span> each PA axiom" is true in <span class="math-container">$\mathfrak{M}$</span>. That is:</p>
<blockquote>
<p>ACA<span class="math-container">$_0$</span> doesn't prove that PA is <strong>sound</strong>; indeed, no conservative extension of PA can <em>(since then it would prove that PA is consistent, which is in the language of PA and not PA-provable)</em>.</p>
</blockquote>
<p>This is just like how being able to check each individual derivation in PA doesn't give us a way to check all derivations at once, so it really shouldn't be surprising.</p>
<hr>
<p>Re: your edits, the point is that knowing that ACA<span class="math-container">$_0$</span> doesn't prove the soundness of PA indicates that we can be "smart enough" to know each specific axiom of PA, yet still not know that PA as a whole is true. So when Gentzen says that that PA is "obviously correct," that's a slightly <em>weaker</em> level of obviousness than any of the individual PA axioms.</p>
<p>This saves us from the circle Gentzen is gesturing at. While focusing on PA, Gentzen is more generally describing a hypothetical situation where - in a sufficiently rich language (e.g. that of second-order arithmetic) - we have some notion of "obviousness" with the following properties:</p>
<ul>
<li><p>The set of obvious sentences is consistent,</p></li>
<li><p>Every axiom of PA is obvious,</p></li>
<li><p>The set of obvious sentences is c.e., and</p></li>
<li><p>It's obvious that every obvious arithmetic sentence is true.</p></li>
</ul>
<p>Godel's theorem implies that no such property exists; the fact that ACA<span class="math-container">$_0\not\vdash Sound($</span>PA) is a particular example of this phenomenon, showing that PA doesn't correspond to the first-order part of such a notion of obviousness and hopefully making it more intuitively clear why the above situation can't happen despite its face-value-plausibility.</p>
|
1,119,634 | <p>Find the point on the curve $y=x^2+2$ where the tangent is parallel to the line $2x+y-1=0$</p>
<p>I understand the answer is $(-1,3)$ but I can't find a way to get there... Thanks </p>
| abel | 9,252 | <p>pick a point $(a, a^2 + 2)$ on the graph of $y = x^2 + 2.$ the tangent at this point has slope $2a$ so the tangent has the equation $y = a^2 + 2 + 2a(x-a) = 2ax + 2 - a^2$ if this line is to match the line $2x + y - 1 = 0$ then $a = -1$ and the point at which this line is tangent is $(-1, 3).$</p>
|
3,285,036 | <p>Obviously this cannot happen in a right rectangle, but otherwise - as Sin(0) or 180 or 360 equals 0, I guess there is no way to find out what the original angle was?</p>
| fleablood | 280,126 | <p>In general if <span class="math-container">$\sin x = M$</span> and <span class="math-container">$-180 < x \le 180$</span> degrees there are two possible answers to what <span class="math-container">$x$</span> is. </p>
<p>There is one <span class="math-container">$k$</span> where <span class="math-container">$0 \le k \le 90$</span> or <span class="math-container">$-90 \le k < 0$</span>. That one is called <span class="math-container">$k=\arcsin M$</span>.</p>
<p>Then there is another one <span class="math-container">$m$</span> where <span class="math-container">$x \ge 90$</span> or <span class="math-container">$k \le -90$</span>. These are related in that <span class="math-container">$m = 180 - \arcsin M> 90$</span> if <span class="math-container">$0\le \arcsin M < 90$</span>; or <span class="math-container">$m = -180 - \arcsin M$</span> if <span class="math-container">$-90\le \arcsin M < 0$</span>.</p>
<p>(The only time these equal each other and there is only one answer is when <span class="math-container">$\pm 1$</span> and <span class="math-container">$\pm 180 -\arcsin M = \arcsin M = \pm 90$</span>.)</p>
<p>However, it's easier to consider that as angles revolve every <span class="math-container">$360$</span> degrees and trig functions repeat with a period of <span class="math-container">$360$</span> degrees we think of angles <span class="math-container">$x$</span> and <span class="math-container">$x + 360$</span> and <span class="math-container">$x - 360$</span> and <span class="math-container">$x +360k$</span> for any integer <span class="math-container">$k$</span> are considered equivalent. (So <span class="math-container">$0^\circ$</span> and <span class="math-container">$360^\circ$</span> are considered to be the "same" thing. As are <span class="math-container">$... \equiv - 450^{\circ} \equiv-90^\circ \equiv 270^{\circ}\equiv 630^{\circ} \equiv ....$</span>)</p>
<p>And if we have <span class="math-container">$\sin x = M$</span> we say there are two solutions <span class="math-container">$\arcsin M$</span> and <span class="math-container">$180 - \arcsin M$</span> where <span class="math-container">$\arcsin M$</span> is the value in the first and fourth quadrant and <span class="math-container">$180 - \arcsin M$</span> is the value in the second or third. And whether we refer to angles in the third and fourth quadrant as greater than <span class="math-container">$180$</span> or as negative is just a matter of convenience.</p>
<p>Also, once you start do any serious trig you should just stop doing angles and start doing radians.</p>
<p><span class="math-container">$\pi \approx 180^\circ$</span>; <span class="math-container">$\frac \pi 2 \approx 90^\circ$</span>; <span class="math-container">$2\pi \approx 360^\circ$</span> etc.</p>
<p>And <span class="math-container">$x \equiv x + 2k\pi; k \in \mathbb Z$</span> and <span class="math-container">$-\frac \pi 2 \equiv \frac {3\pi} 2$</span>.</p>
<p>So If <span class="math-container">$\sin x = 0$</span> then either <span class="math-container">$x = \arcsin 0 = 0$</span> or <span class="math-container">$x = \pi - \arcsin 0 = \pi - 0 =\pi$</span>. So <span class="math-container">$x = 0$</span> or <span class="math-container">$ \pi$</span>.</p>
|
2,083,460 | <p>While trying to answer <a href="https://stackoverflow.com/questions/41464753/generate-random-numbers-from-lognormal-distribution-in-python/41465013#41465013">this SO question</a> I got stuck on a messy bit of algebra: given</p>
<p>$$
\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)
$$</p>
<p>I need to solve for $m$. I no longer remember enough logarithmic identities to attempt to do this by hand. Maxima can’t do it at all, and Wolfram Alpha <a href="http://www.wolframalpha.com/input/?i=solve+[log+m+%3D+log+n+%2B+%283%2F2%29*log%281+%2B+v%2F%28m%5E2%29%29]+for+m" rel="nofollow noreferrer">coughs up a hairball</a> that appears to be the zeroes of a quartic, with no obvious relationship to the original equation.</p>
<p>Is there a short, tidy solution? Failing that, an explanation of how WA managed to turn this into a quartic, and the quartic itself, would be ok.</p>
| Arnaldo | 391,612 | <p>$$\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)=\log n+\log \biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}=\log n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}$$</p>
<p>Then</p>
<p>$$m=n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2} \rightarrow m^{2}=\frac{n^2(m^2+v)^3}{m^6}\rightarrow m^8=n^2(m^2+v)^3$$</p>
<p>I just don't know if the above equation is solvable for $m$.</p>
|
233,238 | <p>I am just practicing making some new designs with Mathematica and I thought of this recently. I want to make a tear drop shape (doesn't matter the orientation) constructed of mini cubes. I am familiar with the preliminary material, I am just having some difficulty getting it to work.</p>
| NonDairyNeutrino | 46,490 | <p>You were <em>really</em> close. You just need to add another underscore to <code>y_</code> as</p>
<pre><code>list = {
{Position, {Code}},
{1, {0000, 0001}},
{2, {0100, 0011}},
{3, {0110, 0111}},
{4, {1000, 1001}},
{5, {1100,1011}},
{6, {1110, 1111}}
};
list /. {x_, {y__}} :> {x, y}
</code></pre>
<blockquote>
<p>{{Position, Code},
{1, 0000, 0001},
{2, 0100, 0011},
{3, 0110, 0111},
{4, 1000, 1001},
{5, 1100, 1011},
{6, 1110, 1111}}</p>
</blockquote>
|
257,567 | <p>Assuming <a href="http://www.springer.com/gp/book/9783642649059" rel="nofollow noreferrer">Bishop's</a> constructive mathematics, is it true that any real-valued square matrix with <strong>distinct</strong> roots of the characteristic polynomial can be diagonalized? By distinct, I mean <strong>apart</strong>: $x \neq y \triangleq \exists q \in \mathbb{Q}.|x - y| > q$. (There may be similar defnitions)</p>
<p>For the case $n=2$, it seems true since we can deduce $a_{11} - \lambda_j \neq 0 \lor a_{22} - \lambda_j \neq 0$ for any $j = 1, 2$ where $a_{ii}, i=1,2$ are the diagonal elements and $\lambda_j$s are the roots of the characteristic polynomial. It then allows multiplying by a nonzero number and, using the property that $(a_{11} - \lambda_j)(a_{22} - \lambda_j) = a_{12}a_{21}$, solving the respective system of linear equations and consequently finding an eigenvector: $A v = \lambda_j v$. </p>
<p>Already for the case $n=3$, the argument seems not to work and one cannot proceed without some case distinction on reals. <a href="http://ac.els-cdn.com/0168007286900060/1-s2.0-0168007286900060-main.pdf?_tid=f7d5faaa-c5ca-11e6-9674-00000aab0f27&acdnat=1482138801_34d65d1543ed9d625fe261e2606b9cd1" rel="nofollow noreferrer">This work</a> addresses the problem in Lemma 1.5, but they seem to assume $xy = 0 \implies x =0 \lor y=0$ which is not valid constructively.</p>
<p><a href="http://www.cse.chalmers.se/~coquand/FISCHBACHAU/AlgebraLogicCoqLom.pdf" rel="nofollow noreferrer">Coquand and Lombardi</a> in Theorem 2.3 constructed an effective procedure of finding eigenvectors of a projection matrix. I suspected that something like this could be done for general matrices where it is known beforehand that the roots of the characteristic polynomial are distinct. </p>
<p><a href="https://arxiv.org/pdf/1605.04832.pdf" rel="nofollow noreferrer">Lombardi and Quitte</a> addressed the problem on page 100 (top). Also in Proposition 5.3, they claim that </p>
<p>$$\prod_{i=1}^n ( (A - \lambda_i I)_{1 \dots n-1, 1 \dots n-1} ) \neq 0$$</p>
<p>by "exhibiting" the companion matrix of $\lambda^n - 1$. Their language is a bit obscure. However, it seems they imply the following lemma:</p>
<p><em>Let $A = M(\mathbb{R},n)$, and suppose that the roots $\lambda_1, \dots, \lambda_n$ of the charteristic polynomial for $A$ are mutually apart. Then, for every root $\lambda_j$ of the characteristic polynomial, there is a principal submatrix of $A - \lambda_j I$ of size $(n-1) \times (n-1)$ which is invertible.</em></p>
| Suvrit | 8,430 | <p>As suspected, the desired inequality actually holds for all hyperbolic polynomials; the inequality in the OP follows as corollary (Corollary 1) to Theorem 2 (which seems to be new). </p>
<p>We will need the following remarkable theorem to obtain our result.</p>
<blockquote>
<blockquote>
<p><strong>Theorem 1 (<a href="https://legacy.orie.cornell.edu/aslewis/publications/01-hyperbolic.pdf" rel="nofollow noreferrer">Bauschke, Güler, Lewis, Sendov, 2001</a>)</strong> <em>Let $p$ be a homogenous hyperbolic polynomial; let $v$ be a vector in the strict interior of the hyperbolicity cone $\Lambda_{++}$ of $p$. Then,
\begin{equation*}
g(x) := \frac{p(x)}{Dp(x)[v]}
\end{equation*}
is concave on $\Lambda_{++}$.</em></p>
</blockquote>
</blockquote>
<p>This theorem helps prove the more general inequality (also conjectured by <strong>Denis Serre</strong> above).</p>
<blockquote>
<blockquote>
<p><strong>Theorem 2.</strong> <em>Let $p$ be a homogenous hyperbolic polynomial with hyperbolicity cone $\Lambda_{++}$. Let $a, b, c \in \Lambda_{++}$. Then, $p$ satisfies the</em> (conic log-submodularity) <em>inequality:</em>
\begin{equation*}
\tag{0}
p(a)p(a+b+c) \le p(a+b)p(a+c).
\end{equation*}</p>
</blockquote>
</blockquote>
<p><em>Proof.</em>
Let $c \in \Lambda_{++}$ be arbitrary. Consider the function $f(a) := \frac{p(a+c)}{p(a)}$. Inequality (0) amounts to showing that $f(a)$ is monotonically decreasing on the cone $\Lambda_{++}$. Equivalently, we consider $\log f$ and show that its derivative is negative in the direction $v$. That is, for an arbitrary direction vector $v\in \Lambda_{++}$, we show that
\begin{equation}
\tag{1}
\frac{Dp(a+c)[v]}{p(a+c)} - \frac{Dp(a)[v]}{p(a)} \le 0\quad\Longleftrightarrow\quad
\frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]}.
\end{equation}
But from Theorem 1, we know that $\frac{p(x)}{Dp(x)[v]}$ is concave. Moreover, since $p$ is homogenous, from its concavity we obtain its <em>superadditivity</em>
\begin{equation*}
\frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]} + \frac{p(c)}{Dp(c)[v]},
\end{equation*}
which is stronger than the desired monotonicity inequality (1) (since all terms are nonnegative).</p>
<blockquote>
<blockquote>
<p><strong>Corollary 1.</strong> <em>Let $E_k(A) = e_k \circ \lambda(A)$ denote the $k$-th elementary symmetric polynomial of a positive definite matrix $A$. Then for any positive definite $A, B, C$ we have
\begin{equation*}
E_k(A)E_k(A+B+C) \le E_k(A+B)E_k(A+C).
\end{equation*}
This log-submodularity, immediately implies the log-submodularity of $f_k(S) := E_k(A+\sum\nolimits_{i\in S}v_iv_i^T)$.</em></p>
</blockquote>
</blockquote>
|
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