qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,360,912 | <p>Why if we have strictly increasing, continuous and onto function its inverse must be continuous? could anyone explain this for me please?</p>
| Conrad | 298,272 | <p>Roughly speaking the sum behaves (for large <span class="math-container">$m,n$</span>) like <span class="math-container">$\sum_{m,n \ne 0}\frac{1}{m^2+n^2}$</span> and that is divergent since the sum say in m for fixed <span class="math-container">$n$</span> is about <span class="math-container">$\frac{1}{n}$</span>, so the double sum behaves like the harmonic sum. </p>
<p>(if <span class="math-container">$a \ge 1$</span>, <span class="math-container">$\int_1^{\infty}\frac{dx}{x^2+a^2}<\sum_{m \ge 1}\frac{1}{m^2+a^2}<\int_0^{\infty}\frac{dx}{x^2+a^2}$</span> so <span class="math-container">$\frac{\pi}{2a}-O(\frac{1}{a^2})< \sum_{m \ge 1}\frac{1}{m^2+a^2} < \frac{\pi}{2a}$</span></p>
|
1,244,564 | <p>For n > 1 Let $F_n = 2^{2^n} + 1$ be a fermat number and b = $2^{2^{n - 2}}$ * ($2^{2^{n - 1}}$ - 1 ).</p>
<p>Then $b^2$ $\equiv$ 2 (mod $F_n$)</p>
<p>I tried to square the original expression I got something ugly that I couldn't simplify further.</p>
<p>I got $b^2$ = $2^{2^{n - 1}}$ * ($2^{2^n}$ - $2 * 2^{2^{n - 1}}$ + 1) $\equiv$ - ($2 * 2^{2^n} * 2^{2^{n-1}}$) (mod $F_n$). I stopped here I can't simplify it anymore.</p>
| AgentS | 168,854 | <p>You're almost there! Just notice that $2^{2^n}\equiv -1\pmod{2^{2^n}+1}$ :</p>
<p>$$\begin{align}b^2&=2^{2^{n-1}}(2^{2^n}-2*2^{2^{n-1}}+1)\\&=2^{2^{n-1}}(2^{2^n}+1-2*2^{2^{n-1}})\\&\equiv2^{2^{n-1}}(0-2*2^{2^{n-1}})\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}}*2^{2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}+2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^n}\pmod{2^{2^n}+1}\\&\equiv -2*(-1)\pmod{2^{2^n}+1}\end{align}$$</p>
|
3,413,837 | <p>Jerry the mouse is hungry and according to some confidential information, there is a tempting piece of cheese at the end of one of the three paths after the junction he just found himself!</p>
<p>Fortunately, Tom is standing right there and Jerry hopes he can get some useful information as to which path he must get; most importantly because Spike and Tyke, the dogs, are at the end of the other two paths!</p>
<p>The only problem is that Tom gives true and false replies in alternating order. Furthermore, he has no way of knowing which will be first, the truth or the lie!</p>
<p>He is only allowed to ask Tom 2 questions that can be answered by a “yes” and a “no”.</p>
<p>What must be the two questions he must ask?</p>
<hr>
<p>No matter how hard I tried, I can't figure out anything...
I have seen several variations for the 2 doors problem but this one is different!</p>
| Hemant Agarwal | 645,672 | <p>Firstly go through this question, famously coined the Hardest logic puzzle : <a href="https://youtu.be/LKvjIsyYng8" rel="nofollow noreferrer">https://youtu.be/LKvjIsyYng8</a> . This YouTube link is an excellent video explaining the question and then the answer. After seeing the video, try solving your question again. If you are still not able to solve your question, then read below.</p>
<p>Notice in the YouTube video, that</p>
<ol>
<li><p>If a statement is true and you ask anybody ( truth teller
or liar),
"If I ask you if this statement is true
then would you say yes then they
will reply, "yes" "</p>
<pre><code> "If I ask you if this statement is true
then would you say "no" then reply
will be, "no".
</code></pre>
</li>
<li><p>If a statement is false and you ask anybody ( truth teller
or liar),
"If I ask you if this statement is true
then would you say yes then they
will reply, "no" "</p>
<pre><code> "If I ask you if this statement is true
then would you say "no" then reply
will be, "yes"
</code></pre>
</li>
</ol>
<p>The basic logic needed to solve the "hardest logic puzzle" is is that if it is a true statement then using the method explained in the video, both, a liar and a truthteller, will tell us that it is a true statement and if it is a false statement then both would tell us that it is a false statement. This is because, if a statement is true then "yes" gets a "yes" as a reply and "no" gets a "no" as a reply. If a statement is false then "yes" gets a "no" as a reply and "no" gets a "yes" as a reply.)</p>
<p>Now, that we know all this, it is easy to solve your question:</p>
<p>First question : if I were to ask you if you the first path leads to cheese then would you have said, "yes" ? If he says, "yes", then the cheese is at the end of the first path. If he says, "no", then ask him the second question.</p>
<p>Second question: if I were to ask you if the second path leads to cheese then would you have said, "yes" ? If he says yes then the cheese is at the end of the second path else it is at the end of the third path.</p>
|
233,075 | <p>I am trying to solve the following equation in the Natural Numbers, with the condition <span class="math-container">$a\ge1$</span>, <span class="math-container">$b\ge1$</span>, and <span class="math-container">$r\ge3$</span>:</p>
<p><span class="math-container">$$\frac{a(a + 3)(a(r - 5) + (12 - r))}{9}=\frac{b (9 + b (-14 + r) - r)}{3}\tag1$$</span></p>
<p>The method I know use is, that I solve the equation for <span class="math-container">$b$</span> and I got:</p>
<p><span class="math-container">$$b=\displaystyle\frac{1}{6} \left(\sqrt{3\cdot\frac{4 a (a+3) (r-14) (a (r-5)-r+12)+3 (r-9)^2}{(r-14)^2}}+\frac{15}{r-14}+3\right)\tag2$$</span></p>
<p>Now, I used Mathematica to check when the function under the square root is a perfect square, with the following code:</p>
<pre><code>ParallelTable[
If[IntegerQ[
FullSimplify[
Sqrt[3*((
4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)]]], {a, r}, Nothing], {a, 1,
10^5}, {r, 3, 10^5}] //. {} -> Nothing
</code></pre>
<p>And the solutions I got, I put in equation <span class="math-container">$(1)$</span> to check if I can find a solution to the original problem.</p>
<blockquote>
<p>This method takes a very very long time, but I am not knowing if there is a faster and smarter way to program this. Can you help me with this. Thanks a lot in advance.</p>
</blockquote>
| Michael E2 | 4,999 | <p>Borrowing a fast perfect-square test from <a href="https://mathematica.stackexchange.com/questions/442/fastest-square-number-test/207813#207813">Fastest square number test</a>, and shortening the length of the test case:</p>
<pre><code>(* OP's *)
Table[
If[IntegerQ[
FullSimplify[
Sqrt[3*((4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)]]], {a, r}, Nothing], {a, 1,
300}, {r, 3, 30}] // Flatten[#, 1] & // AbsoluteTiming
</code></pre>
<blockquote>
<pre><code>(* {83.9498, {{5, 19}, {117, 15}, {252, 29}}} *)
</code></pre>
</blockquote>
<pre><code>sQ[n_] := FractionalPart@Sqrt[n + 0``1] == 0;
Table[
If[IntegerQ[#] && sQ[#] &[
3*((4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)], {a, r}, Nothing], {a, 1,
300}, {r, 3, 30}] // Flatten[#, 1] & // AbsoluteTiming
</code></pre>
<blockquote>
<pre><code>(* {0.068718, {{5, 19}, {117, 15}, {252, 29}}} *)
</code></pre>
</blockquote>
<p>For the <span class="math-container">$10^5 \times 10^5$</span> search, the improved code will take on the order of 80000 seconds, but that's a lot less than the <span class="math-container">$10^8$</span> seconds that the OP's would take. (Divide by an appropriate factor if parallelized.)</p>
|
233,075 | <p>I am trying to solve the following equation in the Natural Numbers, with the condition <span class="math-container">$a\ge1$</span>, <span class="math-container">$b\ge1$</span>, and <span class="math-container">$r\ge3$</span>:</p>
<p><span class="math-container">$$\frac{a(a + 3)(a(r - 5) + (12 - r))}{9}=\frac{b (9 + b (-14 + r) - r)}{3}\tag1$$</span></p>
<p>The method I know use is, that I solve the equation for <span class="math-container">$b$</span> and I got:</p>
<p><span class="math-container">$$b=\displaystyle\frac{1}{6} \left(\sqrt{3\cdot\frac{4 a (a+3) (r-14) (a (r-5)-r+12)+3 (r-9)^2}{(r-14)^2}}+\frac{15}{r-14}+3\right)\tag2$$</span></p>
<p>Now, I used Mathematica to check when the function under the square root is a perfect square, with the following code:</p>
<pre><code>ParallelTable[
If[IntegerQ[
FullSimplify[
Sqrt[3*((
4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)]]], {a, r}, Nothing], {a, 1,
10^5}, {r, 3, 10^5}] //. {} -> Nothing
</code></pre>
<p>And the solutions I got, I put in equation <span class="math-container">$(1)$</span> to check if I can find a solution to the original problem.</p>
<blockquote>
<p>This method takes a very very long time, but I am not knowing if there is a faster and smarter way to program this. Can you help me with this. Thanks a lot in advance.</p>
</blockquote>
| Ulrich Neumann | 53,677 | <p>Try <code>NSolve</code> with restricted parameter range <code>1<= a,b,r <=50 </code></p>
<pre><code>NSolve[{1/9 a (a + 3) (a (r - 5) + 12 - r) ==1/3 b (9 + b (-14 + r) - r) , 50 >= a >= 1, 50 >= b >= 1 ,50 > r >= 1}, {a, b, r}, Integers]
(**{{a -> 3, b -> 6, r -> 24},
{a -> 5, b -> 10, r -> 31},
{a -> 5,b -> 14, r -> 19},
{a -> 9, b -> 20, r -> 46},
{a -> 12, b -> 30,r -> 45}}*)
</code></pre>
|
157,823 | <p>Check whether function series is convergent (uniformly):</p>
<p>$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n}\ln \left( \frac{x}{n} \right)$ for $x\in[1;+\infty)$</p>
<p>I don't know how to do that.</p>
| alejopelaez | 1,318 | <p>Fix an $x$ and think about this tail of the series
$$\sum_{n=3x}^{\infty}\frac{1}{n}\ln{\left(\frac{x}{n}\right)} = -\sum_{n=3x}^{\infty}\frac{1}{n}\ln{\left(\frac{n}{x}\right)} \leq -\sum_{n=3x}^{\infty}\frac{1}{n}\ln{3} \leq -\sum_{n=3x}^{\infty}\frac{1}{n} \leq 0$$
Now use what you know about $\sum_{n=1}^{\infty}\frac{1}{n}$ to conclude.</p>
|
656,458 | <p>If $A$ is an $n\times n$-matrix, $A^H$ is a Hermitian Matrix and $A^S$ is a Skew Hermitian, show $A=A^H+A^S$.</p>
<p>I am having trouble working with these so far and really cannot find many characteristics except the definitions. A Hermitian is made up of reals on the diagonal and is $A^*=A$. It is skew hermitian iff the diagonal is made up of imaginary numbers and zero and $A^H =\bar A^T$</p>
| Ben Grossmann | 81,360 | <p>I think this question can be answered in a fairly straightforward fashion using the definitions
$$
A^H = \frac 12 (A + A^*)\\
A^S = \frac 12 (A - A^*)
$$</p>
|
3,137,160 | <blockquote>
<p>Determine which of the following are subspaces of <span class="math-container">$3 \times 3$</span> matrix <span class="math-container">$M$</span>
all <span class="math-container">$3 \times 3$</span> matrices <span class="math-container">$A$</span> such that the trace of <span class="math-container">$A$</span> is <span class="math-container">$\mbox{tr}(A) = 0$</span>.</p>
</blockquote>
<p>What does trace mean?</p>
| clark | 33,325 | <p>It suffices to show that a group <span class="math-container">$G$</span> such that <span class="math-container">$|G|=p^k$</span> has an element <span class="math-container">$a$</span> of order <span class="math-container">$p$</span>.</p>
<p>We can do this by induction. For <span class="math-container">$k=1$</span> <span class="math-container">$G$</span> is cyclic hence we are done.</p>
<p>Take <span class="math-container">$k>1$</span>.
Pick <span class="math-container">$a_1\in G$</span>, such that <span class="math-container">$|a_1|=p^{k_1}$</span>, if <span class="math-container">$k_1=k$</span> then we are done since <span class="math-container">$G$</span> would be cyclic.
Otherwise <span class="math-container">$k_1<k$</span> hence by the induction the group <span class="math-container">$<a_1>$</span> has an element <span class="math-container">$b$</span> of order <span class="math-container">$p$</span>.</p>
|
1,047,263 | <p>I used to do this on my calculators and it never worked! I think it's because you can't multiply any number by itself to get a negative number. Is that even true? I think it is! I've tried it out and it never worked! Look here:$$0.5\cdot0.5=0.25$$$$0\cdot0=0$$$$-1\cdot-1=1$$$$2.1\cdot2.1=4.41$$$$-7.5\cdot-7.5=56.25$$I <em>never</em> get a negative for multiplying negative numbers because a negative times a negative is a positive and no matter what I do, I always get a nonnegative number! Tell me what you think! Is this why?</p>
| ConMan | 82,793 | <p>You are (almost) correct!</p>
<p>It is a property of the real numbers that any number squared is positive, i.e. $\forall x \in \mathbb{R}, x^2 \ge 0$. So all positive real numbers have two square roots - a positive one and a negative one (although the square root <em>function</em> is defined by convention to be the positive value) and negative numbers have none.</p>
<p>But.</p>
<p>A large amount of mathematics is about "what if". In this case, "what if we pretended that $\sqrt{-1}$ was a number and tried to treat it like one?" So we define $i$ to be one of the solutions to $i^2=-1$, and see where it takes us. As it turns out, it leads us to the <a href="http://en.wikipedia.org/wiki/Complex_numbers" rel="nofollow">Complex numbers</a>, a concept that opens up whole new possibilities in a number of fields, including ones as far reaching as number theory and electrical engineering.</p>
<p>I won't go into details here, but in moving from the real numbers to the complex numbers
you gain some things - for example, every polynomial over the real numbers has a solution in the complex numbers - and you lose others, like the good old ordering that lets you say that one number is bigger than another.</p>
<p>Then there are other extensions of the real numbers, like the quaternions, which get even weirder and more interesting because you exchange what seem to be even more fundamental properties (like the idea that it doesn't matter what order you multiply numbers) in exchange for different ones.</p>
|
225,253 | <p>Simple question, I just cannot find something that explains it right out and to the point without giving a huge confusing explanation. The question that I am struggling with is to determine a limit of a function if it exists.</p>
<blockquote>
<p>Find: <span class="math-container">$$\lim_{x\to2}{f(x)},$$</span></p>
<p>If</p>
<p><span class="math-container">$$f(x)=\begin{cases}x^{2} & \text{if } x<2 \\ 3 & \text{if } x=2 \\ 3x-2 & \text{if } x>2\end{cases}$$</span></p>
</blockquote>
<p>Now i worked out the limits from both sides and they both equal 4. But it does say that the limit at that point equals 3. Does this mean that the limit doesn't exist? Or does this just mean that the double sided limit exists at 4, but is discontinuous and equals 3 at that point?</p>
<p>Thanks in advance!</p>
| Yoni Rozenshein | 36,650 | <p>The limit from either side as $x\to2$ exists and equals $4$, so the overall limit as $x\to2$ exists and equals $4$.</p>
<p>Limits completely ignore the value of the function at the given point, as the definition of a limit only cares about what happens in a <em>punctured neighborhood</em> of that point.</p>
<p>There is a different notion, which is of <em>continuity</em>, which considers both the limit of the point and the value of the point: If the limit exists and equals the value, the function is said to be <em>continuous</em> there. However, if the limit doesn't exist or doesn't equal the value (as we have), the function is <em>incontinuous</em> at the point.</p>
|
466,722 | <blockquote>
<p>Find a function $f$ and a number $a$ such that:
$$
6+\int_{a}^{x}\frac{f(t)}{t^2}\:\mathrm{d}t=2\sqrt{x}
$$
For all $x>0$</p>
</blockquote>
<p>From Fundamental Theorem of Calculus section. Having some trouble with this. Any help?</p>
| dajoker | 72,504 | <p>Hint ::</p>
<p>Just differentiate, get an expression for $f$ and then substitute back to obtain the value of $a$.</p>
<p><em>EDIT</em> : </p>
<p>To differentiate the integral, use the property :
$$\dfrac d{dx}\large\int^{g(x)}_{f(x)}h(t)dt=h(g(x)).g'(x)-h(f(x)).f'(x)$$</p>
|
1,657,664 | <p>Struggling with a homework problem here and can't understand logically which one would be correct (each has different truth tables). I need to express the following statement using quantifiers, variables, and the predicates M(s), C(s), and E(s) </p>
<blockquote>
<p>"No computer science students are engineering students" </p>
</blockquote>
<p>D = set of all students</p>
<p>C(s) = "s is a computer science major"</p>
<p>E(s) = "s is an engineering student" </p>
<p>So I'm stuck between,</p>
<p>$\forall s \in D, C(s) \implies \lnot E(s)$</p>
<p>-OR-</p>
<p>$\forall s \in D, \lnot C(s) \land E(s)$ </p>
| ThisIsNotAnId | 24,567 | <p><strong>HINT:</strong></p>
<p>For every student, his/her being a computer science student guarantees that they are not engineering students.</p>
<p><strong>EDIT:</strong></p>
<p>Your second logical statement claims every student is an engineering major <em>and</em> not a computer science major.</p>
<p>Your first logical statement is correct. But, do say if you see how this is so or if you would like further explanation.</p>
|
3,815,640 | <p>what is the most efficient way to calculate the argument of
<span class="math-container">$$
\frac{e^{i5\pi/6}-e^{-i\pi/3}}{e^{i\pi/2}-e^{-i\pi/3}}
$$</span> without calculator ?</p>
<p>i tried to use <span class="math-container">$\arg z_1-\arg z_2$</span> but the argument of <span class="math-container">$e^{i5\pi/6}-e^{-i\pi/3}$</span> take some time .
is there a formula to calculate the argument of that kind of complex numbers ?</p>
| Bernard | 202,857 | <p>First factor out <span class="math-container">$\mathrm e^{-\tfrac{i\pi}3}$</span> both in the numerator and denominator:
<span class="math-container">$$\frac{\mathrm e^{\tfrac{5i\pi}6}-\mathrm e^{-\tfrac{i\pi}3}}{\mathrm e^{\tfrac{i\pi}2}-\mathrm e^{-\tfrac{i\pi}3}} =
\frac{\mathrm e^{\tfrac{5i\pi}6+\tfrac{i\pi}3}-1}{\mathrm e^{\tfrac{i\pi}2+\tfrac{i\pi}3}-1}=\frac{\mathrm e^{\tfrac{7i\pi}6}-1}{\mathrm e^{\tfrac{5i\pi}6}-1}.$$</span>
Then factor the exponentials of half the arguments
<span class="math-container">$$\frac{\mathrm e^{\tfrac{7i\pi}6}-1}{\mathrm e^{\tfrac{5i\pi}6}-1}= \frac{\mathrm e^{\tfrac{7i\pi}{12}}}{\mathrm e^{\tfrac{5i\pi}{12}}}\cdot\frac{\mathrm e^{\tfrac{7i\pi}{12}}-\mathrm e^{\tfrac{-7i\pi}{12}}}{\mathrm e^{\tfrac{5i\pi}{12}}-\mathrm e^{-\tfrac{5i\pi}{12}}}. $$</span>
Can you end the calculations?</p>
|
2,232,390 | <p>$X$ is a subset of $\mathbb{R}$. It contains $\sqrt{3}$ and $\sqrt{2}$ and is closed under addition, subtraction and multiplication.</p>
<ol>
<li>Prove that $X$ contains $\sqrt{8}$.</li>
<li>Prove that $X$ contains $1$.</li>
<li>Prove that $X$ contains $\frac{1}{\sqrt{2}+1}$.</li>
<li>Is it true that $X$ is necessarily a field?</li>
</ol>
<p>So far I have only been successful with 1 and 2. I've just used the elements that are already in the group and performed operations on the them to try and get $\sqrt{8}$ and $1$. </p>
| Brenton | 226,184 | <p>Why not just compute the cdf and take a derivative:</p>
<p>$P(\frac{X_1}{X_2} < a) = P(X_2 > \frac{1}{a}X_1) = \int_0^\infty P(X_2 > \frac{1}{a}x)f_{X_1}(x)dx = \int_0^\infty e^{-3x/a} 3e^{-3x} dx = \frac{a}{a+1}$</p>
<p>Take a derivative with respect to a and you have the pdf of $f_{\frac{X_1}{X_2}}(a)$</p>
|
3,202,955 | <blockquote>
<p><strong>Note:</strong> Please do not give a solution; I would prefer guidance to help me complete the question myself. Thank you.</p>
</blockquote>
<hr>
<p>I am having trouble understanding and finding the continuous and residual spectrum. I am working through the following problem:</p>
<p>Let <span class="math-container">$\alpha = (\alpha_{i})\in\ell^{\infty}$</span> and let <span class="math-container">$T_{\alpha}:\ell^{2}\rightarrow\ell^{2}$</span> with <span class="math-container">$T_{\alpha}x=(\alpha_{1}x_{1},\alpha_{2}x_{2},\ldots)$</span>.</p>
<p>(i) Compute the spectrum, <span class="math-container">$\sigma(T_{\alpha})$</span>, of <span class="math-container">$T_{\alpha}$</span>.</p>
<p>(ii) Identify the point spectrum, <span class="math-container">$\sigma_{p}(T_{\alpha})$</span>, the continuous spectrum, <span class="math-container">$\sigma_{c}(T_{\alpha})$</span>, and the residual spectrum, <span class="math-container">$\sigma_{r}(T_{\alpha})$</span>, of <span class="math-container">$T_{\alpha}$</span>.</p>
<p><strong>My Solution</strong></p>
<p>(i) Suppose <span class="math-container">$\lambda\in\rho(T_{\alpha})$</span> (where <span class="math-container">$\rho(T_{\alpha})$</span> is the resolvent set of <span class="math-container">$T_{\alpha}$</span>) then <span class="math-container">$\lambda I-T_{\alpha}$</span> is bijective. Hence, for every <span class="math-container">$y\in\ell^{2}$</span> <span class="math-container">$\exists ! x\in\ell^{2}$</span> such that,
<span class="math-container">\begin{align}
(\lambda I-T_{\alpha})x = y\implies x = (\lambda I-T_{\alpha})^{-1}y,
\end{align}</span>
and
<span class="math-container">\begin{align}
x_{n} = \frac{y_{n}}{\lambda - \alpha_{n}},
\end{align}</span>
for each element, <span class="math-container">$n\in\mathbb{N}$</span>. Since <span class="math-container">$x\in\ell^{2}$</span>,
<span class="math-container">\begin{align}
|x|_{2}^{2} = \sum_{n=1}^{\infty}|x_{n}|^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}<\infty.
\end{align}</span>
Therefore, if <span class="math-container">$\lambda\in\overline{\{\alpha_{n}\}}$</span> then <span class="math-container">$|x|_{2} = \infty$</span>. Hence <span class="math-container">$\sigma(T_{\alpha}) = \{\lambda\in\mathbb{C}|\lambda\in\overline{\{\alpha_{n}\}},n\in\mathbb{N}\}$</span>.</p>
<p>(ii) Using the definition of point spectrum,
<span class="math-container">\begin{align}
\sigma_{p}(T_{\alpha}):=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha} \text{is not injective}\},
\end{align}</span>
then there exists <span class="math-container">$x_{1},x_{2}\in\ell^{2}$</span>, with <span class="math-container">$x_{1}\neq x_{2}$</span>, such that <span class="math-container">$(\lambda I-T_{\alpha})x_{1} = (\lambda I-T_{\alpha})x_{2}$</span>. This occurs when <span class="math-container">$|x|_{2}=\infty$</span>, that is, <span class="math-container">$\lambda = \alpha_{n}$</span> for some <span class="math-container">$n\in\mathbb{N}$</span>. Hence, <span class="math-container">$\sigma_{p}(T_{\alpha})=\{\lambda\in\mathbb{C}|\lambda = \alpha_{n}\text{ for some }n\in\mathbb{N}\}$</span>.</p>
<p>At this stage I do not know how to continue. Particularly, I don't understand how to use the definitions of continuous and residual spectrum:
<span class="math-container">\begin{align}
\sigma_{c}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2},(\lambda I-T_{\alpha})^{-1}\text{ is unbounded}\},\\
\sigma_{r}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}\}
\end{align}</span></p>
<p><strong>Finding continuous and residual spectrum</strong></p>
<p>Assume <span class="math-container">$\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}$</span> then <span class="math-container">$\text{im}(\lambda I-T_{\alpha})^{\perp}\neq\{0\}$</span>. Hence there is <span class="math-container">$y\in\text{im}(\lambda I-T_{\alpha})^{\perp}$</span>, <span class="math-container">$y\neq 0$</span>, such that,
<span class="math-container">\begin{align}
\sum_{n=1}^{\infty}(\lambda x_{n}-\alpha_{n}x_{n})y_{n} = 0,
\end{align}</span>
for all <span class="math-container">$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$</span>. This implies <span class="math-container">$\lambda=\alpha_{n}$</span> for each <span class="math-container">$n^{\text{th}}$</span> element of <span class="math-container">$x$</span>, which is impossible. Otherwise, <span class="math-container">$x=0$</span>, however this contradicts our assumption that holds for all <span class="math-container">$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$</span>. Hence by contradiction, <span class="math-container">$\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2}$</span> and <span class="math-container">$\text{im}(\lambda I-T_{\alpha})^{\perp}=\{0\}$</span>. Therefore, <span class="math-container">$\sigma_{r}(T_{\alpha})=\emptyset$</span>.</p>
<p>By above <span class="math-container">$(\lambda I-T_{\alpha})$</span> has trivial kernel and hence injective. Given <span class="math-container">$\lambda\in\sigma_{c}(T_{\alpha})$</span> we need <span class="math-container">$(\lambda I-T_{\alpha})^{-1}$</span> unbounded. From above this implies,
<span class="math-container">\begin{align}
|x|_{2}^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}\rightarrow\infty.
\end{align}</span>
Hence, for all <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$\exists N\in\mathbb{N}$</span> such that for <span class="math-container">$n>N$</span>,
<span class="math-container">\begin{align}
|\lambda-\alpha_{n}|<\epsilon.
\end{align}</span>
Now <span class="math-container">$(\alpha_{n})\in\ell^{\infty}$</span> so they are uniformly bounded. Taking <span class="math-container">$\lambda=\sup_{n\in\mathbb{N}}|\alpha_{n}|$</span> we satisfy the criteria and hence <span class="math-container">$(\lambda I-T_{\alpha})^{-1}$</span> is unbounded.</p>
<p>Therefore,
<span class="math-container">\begin{align}
\sigma_{p}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda=\alpha_{n}\text{ for some }n\in\mathbb{N}\},\\
\sigma_{c}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda = \sup_{n\in\mathbb{N}}|\alpha_{n}|\},\\
\sigma_{r}(T_{\alpha}) &= \emptyset.
\end{align}</span></p>
| Kavi Rama Murthy | 142,385 | <p><span class="math-container">$\lambda \in \sigma_r(T_{\alpha})$</span> iff <span class="math-container">$\lambda I-T_{\alpha}$</span> is injective and there is a non zero element <span class="math-container">$y$</span> orthogonal to <span class="math-container">$im(\lambda I-T_{\alpha})$</span> which means <span class="math-container">$\sum (\lambda x_i-\alpha_i x_i) y_i=0$</span> for all <span class="math-container">$x \in \ell^{2}$</span>. It should be easy to determine when this happens. For the continuous spectrum see what it means to have <span class="math-container">$\|\lambda I-T_{\alpha}(x)\|^{2}\geq C \|x\|^{2}$</span> for some <span class="math-container">$C>0$</span>. </p>
|
13,030 | <p>At work, we were discussing when is it the best time to change to winter tires for bikes and/or cars.</p>
<p>Using <code>WeatherData[]</code> and <code>DateListPlot[]</code>, it was fairly straightforward for me to create the diagram below:</p>
<p><img src="https://i.stack.imgur.com/Y5wNT.png" alt="Mean temperature per day in Stockholm"> </p>
<p><em>Fig 1 Mean temperature per day in Stockholm, red is negative and a risk without winter tires.</em></p>
<p>The code for this is</p>
<pre><code>cityTemp = WeatherData["Stockholm", "MeanTemperature", {{1977, 1, 1}, {2011, 12, 31}, "Day"}];
iceRiskDays = Select[cityTemp, Last[#] < 0 &];
yearStrip[ dataItem_] := {{0, Part[First[dataItem], 2], Last[First[dataItem]]}, Last[dataItem]}
DateListPlot[{yearStrip[#] & /@ cityTemp, yearStrip[#] & /@ iceRiskDays} ]
</code></pre>
<p>My question is: <strong>How do I calculate for each day the proportion of values for that day that are below 0 Celsius?</strong> (e.g. for a date at the end of November, the proportion is likely to be bigger than 0.5)?</p>
<p>My attempts to do this ended with trying to create separate lists for each date, but I felt that this was a less elegant way and also creates less "fit" with DateListPlot.</p>
| image_doctor | 776 | <p>Here is a slightly different approach using <code>GatherBy</code>.</p>
<p>Reformat dates to be just {month, day}:</p>
<pre><code>shortDates = {DateString[First@#, { "Month","Day",}], Last@#} & /@ cityTemp;
</code></pre>
<p>Group the values by day and month, count the days on which the temperature is not positive:</p>
<pre><code>probs = Count[#[[All, 2]], t_?NonPositive]/Length@# & /@
GatherBy[Sort@shortDates, #[[1]] &];
</code></pre>
<p>Plot the likelihoods:</p>
<pre><code> ListLinePlot[probs, AxesLabel -> { "Day Number","Likelihood"}]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/soYxj.png" alt="Mathematica graphics"></p>
</blockquote>
<p>You can calculate different likelihoods by adjusting the short date conversion string, for a month value just switch the date reformatting to months:</p>
<pre><code>shortDates = {DateString[First@#, {"Month"}], Last@#} & /@ cityTemp;
</code></pre>
|
1,449,776 | <p>I have always known that $a^n=a*a*a*.....$(n times)</p>
<p>Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?</p>
<p>I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$. </p>
<p>But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?</p>
<p>Any help is highly appreciated.</p>
| Enrico M. | 266,764 | <p>You can see that in this way:</p>
<p>$$a^0 = a^{m - m}$$</p>
<p>for every value of $m$. Using the properties of powers we have:</p>
<p>$$a^{m-m} = \frac{a^m}{a^m} = 1$$</p>
<p>Because the two terms are identical so they are canceled.
So</p>
<p>$$a^0 = 1$$</p>
|
707,193 | <p>The inequality to solve:
$$\left[\frac{-K^2+13K+44}{14-K}\right] > 0$$</p>
<p>How do I solve this?
I tried this:
$$
-K^2+13K+44 > 0 \quad \text{(multiply both sides by $14-K$)}\\
K^2-13K < 44\\
K(K-13) < 44
$$
Is this correct? Any way to get a more precise $K$ value? Thanks. </p>
| gt6989b | 16,192 | <p><strong>Hints</strong>
You have 3 cases: $14-K < 0, 14-K >0$, and the easy one $14-K = 0$.
In the first two, you end up with a different sign after multiplication. Take the one you used ($14-K>0$).</p>
<p>Then indeed $-K^2+13K+44>0$ but if you factor the left-hand side to get $(K-a)(K-b)<0$ for some $a,b$ you can find. When is a product negative? Can you take it from here?</p>
|
374,881 | <p>I'd like to know how I can recursively (iteratively) compute variance, so that I may calculate the standard deviation of a very large dataset in javascript. The input is a sorted array of positive integers.</p>
| Rose Perrone | 32,062 | <p>I ended up using this incremental approach:</p>
<pre><code>function mean(array) {
var i = -1, j = 0, n = array.length, m = 0;
while (++i < n) if (a = array[i]) m += (a - m) / ++j;
return j ? m : undefined;
}
function variance(array, mean_value) {
if (!mean_value) return undefined;
var i = -1, j = 0, n = array.length, v = 0;
while (++i < n) {
a = Math.pow((array[i] - mean_value), 2)
v += (a - v) / ++j;
}
return v * (n/(n-1));
}
</code></pre>
|
688,782 | <p>$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$</p>
<p>I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit. </p>
| Community | -1 | <p>We write</p>
<p>$$a_n=\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{(k-1)(k+1)}{k^2}=\prod_{k=2}^n\frac{v_k}{v_{k+1}}$$
where
$$v_k=\frac{k-1}{k}$$
so by change of index
$$a_n=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=2}^nv_{k+1}}=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=3}^{n+1}v_{k}}=\frac{v_2}{v_{n+1}}=v_2\times\frac{n+1}{n}\to v_2=\frac12$$</p>
|
1,179,843 | <p>Proving $\sum_{n=1}^\infty \frac{\xi ^n}{n}$ is not uniformly convergent for $\xi \in (0,1)$.</p>
<p>I am trying to do the above. I have attempted to show it is not a cauchy sequence by considering $||\frac{\xi ^n}{n} ||_{\sup}$ but no avail. Any help please</p>
| RRL | 148,510 | <p>The series fails to converge uniformly if you can show there is some $\epsilon_0 > 0$ such that for every $N \in \mathbb{N}$,</p>
<p>$$\sup_{\xi \in (0,1)} \sum_{n=N}^\infty\frac{\xi^n}{n}\geqslant \epsilon_0.$$</p>
<p>Notice that</p>
<p>$$\sup_{\xi \in (0,1)} \sum_{n=N}^\infty\frac{\xi^n}{n}\geqslant \sup_{\xi \in (0,1)} \sum_{n=N+1}^{2N}\frac{\xi^n}{n}\geqslant \sup_{\xi \in (0,1)}N\frac{\xi^{N+1}}{2N} = \frac{1}{2}.$$</p>
<p>Hence, using $\epsilon_0 = 1/2$ confirms the negation and the series is not uniformly convergent on $(0,1)$.</p>
|
4,589,921 | <p>Let <span class="math-container">$X$</span> be a set and <span class="math-container">$d : X \times X \to \mathbb R$</span> be a function satisfying the following conditions:</p>
<p><span class="math-container">$d(x, x) = 0$</span> whatever <span class="math-container">$x \in X$</span>;</p>
<p><span class="math-container">$d(x, y) = d(y, x)$</span> whatever <span class="math-container">$x, y \in X$</span>;</p>
<p><span class="math-container">$d(x, z) \le d(x, y) + d(y, z)$</span> whatever <span class="math-container">$x, y, z \in X$</span>.</p>
<p>Show that <span class="math-container">$d(x, y) \ge 0$</span> whatever <span class="math-container">$x, y \in X$</span>.</p>
<hr />
<p>My attempt</p>
<p><span class="math-container">$d(x, y) \le d(x, y) + d(x, y)$</span></p>
<p><span class="math-container">$d(x, y) \le d(x, y) + d(x, y)$</span></p>
<p><span class="math-container">$d(x, y) - d(x,y) \le d(x, y)$</span></p>
<p><span class="math-container">$0 \le d(x, y)$</span></p>
<p>I know it shouldn't be too difficult but I'm not sure about my answer, is that correct?</p>
| Thomas | 89,516 | <p><span class="math-container">$0=d(x,x)\le d(x,y)+d(y,x)=d(x,y)+d(x,y)=2d(x,y)$</span></p>
<p>where all properties 1,2 and 3 have been used.</p>
|
4,589,921 | <p>Let <span class="math-container">$X$</span> be a set and <span class="math-container">$d : X \times X \to \mathbb R$</span> be a function satisfying the following conditions:</p>
<p><span class="math-container">$d(x, x) = 0$</span> whatever <span class="math-container">$x \in X$</span>;</p>
<p><span class="math-container">$d(x, y) = d(y, x)$</span> whatever <span class="math-container">$x, y \in X$</span>;</p>
<p><span class="math-container">$d(x, z) \le d(x, y) + d(y, z)$</span> whatever <span class="math-container">$x, y, z \in X$</span>.</p>
<p>Show that <span class="math-container">$d(x, y) \ge 0$</span> whatever <span class="math-container">$x, y \in X$</span>.</p>
<hr />
<p>My attempt</p>
<p><span class="math-container">$d(x, y) \le d(x, y) + d(x, y)$</span></p>
<p><span class="math-container">$d(x, y) \le d(x, y) + d(x, y)$</span></p>
<p><span class="math-container">$d(x, y) - d(x,y) \le d(x, y)$</span></p>
<p><span class="math-container">$0 \le d(x, y)$</span></p>
<p>I know it shouldn't be too difficult but I'm not sure about my answer, is that correct?</p>
| latrys | 1,126,389 | <p>You don't say how you establish
<span class="math-container">$$
d(x,y)≤d(x,y)+d(x,y)
$$</span>
but once you <em>do</em> have this, the following steps are valid. In general, not all quantities <span class="math-container">$r \in \mathbb{R}$</span> satisfy <span class="math-container">$r \le r + r$</span>; for example <span class="math-container">$r = -1$</span>. So we need to do a bit more work to get an inequality like that, from the information given.</p>
<p>From the axioms, the only one with a "<span class="math-container">$\le$</span>" in it is the third. We should look to see if there is a choice of <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> that will let us get something of the form "<span class="math-container">$0 \le \dots$</span>", which is what we need. There is a <span class="math-container">$0$</span> in the first axiom, which we can reach by letting <span class="math-container">$x = z$</span> in axiom (3), like this:
<span class="math-container">$$
d(x, x) \le d(x, y) + d(y, x)
$$</span>
Now, <span class="math-container">$d(x, x) = 0$</span>, and <span class="math-container">$d(x, y) = d(y, x)$</span>, so you should be able to take it from here.</p>
|
3,929,703 | <p>so, we know how to solve if the question was only <span class="math-container">$5$</span> different rings in <span class="math-container">$4$</span> different fingers, which is <span class="math-container">$4^5$</span>. but what if internal order of rings within the finger matters, is this counted in this answer or we have a different answer?</p>
| saulspatz | 235,128 | <p>There are <span class="math-container">$5!=120$</span> ways to arrange the rings in order. Divide them into <span class="math-container">$4$</span> groups and place the first group, in order, on the first finger, the second group on the second finger, and so on.</p>
<p>We have five rings to place on <span class="math-container">$4$</span> fingers.
By stars and bars, there are <span class="math-container">$\binom83=56$</span> ways to do this, so in all we have <span class="math-container">$$120\cdot 56 = 6720$$</span> ways.</p>
|
181,855 | <p>In the latest <a href="http://what-if.xkcd.com/113/" rel="noreferrer">what-if</a> Randall Munroe ask for the smallest number of geodesics that intersect all regions of a map. The following shows that five paths of satellites suffice to cover the 50 states of the USA:
<img src="https://i.stack.imgur.com/gyfYt.png" alt="from what-if.xkcd.com"></p>
<p>A similar configuration where the lines are actually great circles is claimed by the author:</p>
<blockquote>
<p>They're all slightly curved, since the Earth is turning under the satellites, but it turns out that this arrangement of lines also works for the much simpler version of the question that ignores orbital motion: "How many straight (great-circle) lines does it take to intersect every state?" For both versions of the question, my best answer is a version of the arrangement above.</p>
</blockquote>
<p>There has been quite some work on similar sounding problems. For stabbing (or finding transversals of) line segments see as an example <a href="http://link.springer.com/article/10.1007%2FBF01934440" rel="noreferrer">Stabbing line segments</a> by H. Edelsbrunner, H. A. Maurer,
F. P. Preparata, A. L. Rosenberg, E. Welzl and D. Wood (and papers which reference it.)
or L.M. Schlipf's <a href="http://www.diss.fu-berlin.de/diss/receive/FUDISS_thesis_000000096077" rel="noreferrer">dissertation</a> with examples of different kinds. </p>
<p><strong>Is there an algorithmic approach known to tackle this problem (or for the
simpler problem when all regions of the map are convex)?</strong></p>
<p>In the case of the 50 states of the USA, it is of course easy to see that one great circle does not suffice: take two states (e.g. New York and Louisiana) such that all great circles that intersect those do not pass through a third state (e.g. Alaska). Similarly one can show that we need at least 3 great circles. </p>
<p>Maybe it would be helpful to consider all triples of regions that do not lie on a great circle and use this hypergraph information to deduce lower bounds.</p>
<p><strong>What are good methods to find lowers bounds?</strong></p>
<p>Randall Munroe's conjectures that 5 is optimal:</p>
<blockquote>
<p>I don't know for sure that 5 is the absolute minimum; it's possible
there's a way to do it with four, but my guess is that there isn't.
[...] If
anyone finds a way (or proof that it's impossible) I'd love to see it!</p>
</blockquote>
| David Eppstein | 440 | <p>It is known to be NP-hard to cover regions (or even just points) with a minimum number of lines. For the Euclidean plane, see Megiddo, Nimrod and Tamir, Arie (1982), "On the complexity of locating linear facilities in the plane", <em>Oper. Res. Lett.</em> 1 (5): 194–197, <a href="https://doi.org/10.1016/0167-6377(82)90039-6" rel="nofollow noreferrer">doi:10.1016/0167-6377(82)90039-6</a>. Their construction is flexible enough that, at least for the region version, it should extend to the approximations to Euclidean geometry that one gets in small patches of spherical geometry.</p>
|
3,106,550 | <p>Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', '<span class="math-container">$a = $</span>', or '<span class="math-container">$a \neq$</span>', then specify a value or comma-separated list of values.
<span class="math-container">$$
\begin{align}
2x−6y−4z &= 16\\
ax−y−4z &= 6\\
2x−3y−4z &= 10
\end{align}
$$</span>
This is a problem for my Linear Algebra class, and I can't seem to figure out how to get it into RREF and evaluate.</p>
| Bernard | 202,857 | <p>The criterion to have solutions is is that the matrix of coefficients of the linear system and the augmented matrix have the same rank. This common rank is then the <code>codimension</code> of the affine space of solutions. </p>
<p>So let's perform row reduction on the augmented matrix:
<span class="math-container">\begin{align}
&\left[\begin{array}{ccc|c}
2&-6&-4&16\\
a&-1&-4&6\\2&-3&-4&10
\end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c}
1&-3&-2&8\\ 2&-3&-4&10 \\
a&-1&-4&6
\end{array}\right]\rightsquigarrow\left[\begin{array}{ccc|c}
1&-3&-2&8\\ 0&3&0&-6 \\
0&3a-1&2a-4&6-8a
\end{array}\right]\rightsquigarrow\\
\rightsquigarrow&\left[\begin{array}{ccc|c}
1&-3&-2&8\\ 0&1&0&-2 \\
0&3a-1&2a-4&6-8a
\end{array}\right]\rightsquigarrow
\left[\begin{array}{ccc|c}
1&-3&-2&8\\ 0&1&0&-2 \\
0&0&2a-4&4-2a
\end{array}\right]\rightsquigarrow
\left[\begin{array}{ccc|c}
1&0&-2&2\\ 0&1&0&-2 \\
0&0&a-2&2-a
\end{array}\right]
\end{align}</span></p>
<ul>
<li>if <span class="math-container">$a\ne2$</span>, we can proceed as follows:
<span class="math-container">$$\left[\begin{array}{ccc|c}
1&0&-2&2\\ 0&1&0&-2 \\
0&0&1&-1
\end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c}
1&0&0&0\\ 0&1&0&-2 \\
0&0&1&-1
\end{array}\right] $$</span>
There is a unique solution (the last column) since the matrix has full rank.</li>
<li>If <span class="math-container">$a=2$</span>, we have a row reduced echelon form:
<span class="math-container">$$\left[\begin{array}{ccc|c}
1&0&-2&2\\ 0&1&0&-2 \\
0&0&0&0
\end{array}\right]$$</span>
which shows, as the matrix and the augmented matrix have rank <span class="math-container">$2$</span> has; as a set of solutions, an affine space of dimension <span class="math-container">$3-2=1$</span>.</li>
</ul>
|
1,580,270 | <p>Consider the groups $G = \{0,1,2\} = \mathbb Z_3$ and $H = \{a,b,c\}$
given by the following multiplication tables:</p>
<p><a href="https://i.stack.imgur.com/hXgBb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hXgBb.jpg" alt="enter image description here"></a></p>
<p>The first one isn't really multiplication but in my notes it said it doesn't really matter.</p>
<p>So how do I show an isomorphism? The groups have the same size so they can be bijective right? But it just seems so abstract to show if there's an isomorphism... Exactly what do we have to check.</p>
| Lubin | 17,760 | <p>First, find the identity in each. In your first example, it’s “$0$”, while in the second it’s “$b$”. How to see this? It’s the element whose row and column matches the labeling rows. Now see whether you can match some nonidentity element in the first example to a nonidentity element in the second so that you force a homomorphism.</p>
|
659,254 | <p>Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.</p>
<p>What is the product distribution of two of such random variables, e.g.,
$Z_2 = X_1 \cdot X_2$?</p>
<p>What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?</p>
<p>What if there are $n$ of such uniform variables?
$Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?</p>
| heropup | 118,193 | <p>We can at least work out the distribution of two IID ${\rm Uniform}(0,1)$ variables $X_1, X_2$: Let $Z_2 = X_1 X_2$. Then the CDF is $$\begin{align*} F_{Z_2}(z) &= \Pr[Z_2 \le z] = \int_{x=0}^1 \Pr[X_2 \le z/x] f_{X_1}(x) \, dx \\ &= \int_{x=0}^z \, dx + \int_{x=z}^1 \frac{z}{x} \, dx \\ &= z - z \log z. \end{align*}$$ Thus the density of $Z_2$ is $$f_{Z_2}(z) = -\log z, \quad 0 < z \le 1.$$ For a third variable, we would write $$\begin{align*} F_{Z_3}(z) &= \Pr[Z_3 \le z] = \int_{x=0}^1 \Pr[X_3 \le z/x] f_{Z_2}(x) \, dx \\ &= -\int_{x=0}^z \log x dx - \int_{x=z}^1 \frac{z}{x} \log x \, dx. \end{align*}$$ Then taking the derivative gives $$f_{Z_3}(z) = \frac{1}{2} \left( \log z \right)^2, \quad 0 < z \le 1.$$ In general, we can conjecture that $$f_{Z_n}(z) = \begin{cases} \frac{(- \log z)^{n-1}}{(n-1)!}, & 0 < z \le 1 \\ 0, & {\rm otherwise},\end{cases}$$ which we can prove via induction on $n$. I leave this as an exercise.</p>
|
3,276,984 | <p>Why we divide the small difference 'd√x' by the difference in area 'dx' Where we normally divide the difference in area by the small difference ?<a href="https://i.stack.imgur.com/LDKB7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LDKB7.jpg" alt="enter image description here"></a></p>
<p>My huge confusion is, when we try to find the derivative of x^2 we <strong>divide the difference in area 'dx^2' by the small difference dx</strong> -> <strong>dx^2 / dx</strong> . But while i try to find the derivative of √x, i get a correct answer only when i <strong>divide the small difference by the difference in area</strong> -> <strong>d√x / dx</strong> ??</p>
<p>Thanks in advance! :)</p>
| Fareed Abi Farraj | 584,389 | <p>When talking about derivatives here, we will be studying the variation of a certain function(<span class="math-container">$x^2$</span>, <span class="math-container">$\sqrt{x}$</span>, ...) with respect to the variable <span class="math-container">$x$</span>. In other words, we will be studying how the function changes, according to a very small variation <span class="math-container">$dx$</span> on <span class="math-container">$x$</span>.</p>
<p>That is <span class="math-container">$\frac{df(x)}{dx}$</span> or in your examples <span class="math-container">$\frac{dx^2}{dx}$</span> and <span class="math-container">$\frac{d\sqrt{x}}{dx}$</span>.</p>
<p>It is not a rule to divide what you call the "difference in area" by the "small difference" to get the derivative. If you put <span class="math-container">$\frac{dx}{d\sqrt{x}}$</span> you'll be studying the variation of <span class="math-container">$x$</span> according to a small variation of <span class="math-container">$\sqrt{x}$</span> and thats not the definition of the derivative you're talking about.</p>
|
2,698,357 | <p><strong>Draw a graph on $10$ vertices with no more than $20$ edges that contains no independent set of size $3$.</strong></p>
<p>So I was trying to draw the above graph but I was kind of stuck. What I basically did was draw a bipartite graph with $5$ vertices on each side and then $20$ edges randomly connecting between them. I suppose it was similar to $K5, 5$, though I don't believe it is quite the same. I was also a little confused on how to identify an independent set in a graph so it was hard for me to verify my graph.</p>
<p>If anyone knows how to approach this problem, some assistance will be highly appreciated!</p>
| Robert Z | 299,698 | <p>Take $G=K_5+ K_5$ (the disjoint union of two 5-cliques). Then $G$ has $2\cdot 5=10$ vertices and $2\cdot \binom{5}{2}=20$ edges and any independent set of $G$ has cardinality less than $3$ because it has no more than one vertex from each $K_5$. </p>
<p>P.S. I don't think that by selecting $20$ edges in $K_{5,5}$ we can obtain a graph $G$ that contains no independent set of size $3$. You should be able to get a contradiction by noting that the average degree in $G$ is $2\cdot 20/10=4$.</p>
|
1,023,830 | <p><img src="https://i.stack.imgur.com/X5aCr.png" alt="AB,AC,BC and h are known"></p>
<p><strong>AB,AC,BC</strong> and <strong>h</strong> are known and its a isosceles triangle
<strong>how to find angle BAC?</strong></p>
| Mark Fantini | 88,052 | <p>One of the properties of the Laplace transform is that</p>
<p>$$\mathcal{L} \left( \frac{f(t)}{t} \right) = \int_s^{+ \infty} F(s) \, ds.$$</p>
<p>This means that</p>
<p>$$\mathcal{L} \left( \frac{\cosh(at)}{at} \right) = \frac{1}{a} \int_s^{+ \infty} \mathcal{L}(\cosh(at)) \, ds.$$</p>
<p>Take from here.</p>
|
1,023,830 | <p><img src="https://i.stack.imgur.com/X5aCr.png" alt="AB,AC,BC and h are known"></p>
<p><strong>AB,AC,BC</strong> and <strong>h</strong> are known and its a isosceles triangle
<strong>how to find angle BAC?</strong></p>
| graydad | 166,967 | <p>By definition,$$\mathcal{L}\left\{\frac{\cosh(at)}{at}\right\}=\int_0^\infty e^{-st}\frac{\cosh(at)}{at}dt \\ = \frac{1}{2a}\int_0^\infty e^{-st}\frac{e^{at}+e^{-at}}{t}dt \\ = \frac{1}{2a}\int_0^\infty \frac{e^{t(a-s)}+e^{-t(a+s)}}{t}dt \\=\frac{1}{2a} \int_0^\infty \frac{e^{t(a-s)}}{t}dt+ \frac{1}{2a} \int_0^\infty \frac{e^{-t(a+s)}}{t}dt$$ Also, the Laplace transform is not defined for some $a \in \Bbb{R}$; you should find those values before integrating.</p>
|
28,955 | <p>I need to crack a stream cipher with a repeating key.</p>
<p>The length of the key is definitely 16. Each key can be any of the characters numbered 32-126 in ASCII.</p>
<p>The algorithm goes like this:</p>
<p>Let's say you have a plain text:</p>
<p>"Welcome to Q&A for people studying math at any level and professionals in related fields."</p>
<p>Let's say that the password is:</p>
<p>"0123456789abcdef"</p>
<p>Then, to encrypt the plaintext, just XOR them together. If the key isn't long enough, just repeat it. e.g., </p>
<p>Welcome to Q&A for people studying math at any level and professionals in related fields.</p>
<pre><code> XOR
</code></pre>
<p>0123456789abcdef0123456789abcdef0123456789abcdef0123456789abcdef0123456789ab</p>
<p>I have 2 English messages encrypted with the above algorithm and with the same key.
I know about the communicative property of xor and that it can be exploited for the example above.
I've read that this is a pretty weak cipher and it has been cracked. However, I have no idea how to do it. So, where can I find a cryptanalysis tool to do it for me?</p>
| Seth | 229,250 | <p>The other answers can work but they make this a much harder problem than it is.</p>
<p>The first step it to determine the key length with IC or the Chi Test, but you seem to have done and determined the key to be 16 bytes.</p>
<p>After that, take the first byte and one every 16 after that. If the message is standard English the most common byte in that group should be a space (0x20) so xor that common byte with 0x20 to find the first byte of the key. Repeat that starting with the next byte, etc. until you have the entire key.</p>
|
1,134,510 | <p>Regarding My Background I have covered stuff like </p>
<p>1.Single Variable Calculus</p>
<p>2.Multivariable Calculus (Multiple Integration,Vector Calculus etc) (Thomas Finney)</p>
<p>3.Basic Linear Algebra Course (Containing Vector spaces,Linear Transformation)</p>
<p>4.Ordinary Differential Equation</p>
<p>5.Real Analysis (Sequences And series)</p>
<p>I am interested In Number theory and i am big fan of Ramanujan .I have not been through rigorous proofs before . But i want to dig deep into number theory especially area where Ramanujan was working .Anyone researcher,Professor can advice me about which preparations are needed for going into number theory and which books should i need .I will be highly obliged .I am asking this for self study or you can say pursuing research at home .</p>
<p>Note - All stuff i have covered is with help of youtube videos and self study .Thanks </p>
| Tamim Addari | 83,311 | <p>Well , the answer is </p>
<p>$$(26*25)*(\frac{7!}{3!*4!})$$</p>
<p>Now , why would your answer be wrong ? cause you are not considering that you can you can choose only 2 at a time , then the rearrangement depends on only what number you choosed for the first time . after you choose 2 letters , it is the problem how 4 and 3 similar things can be arranged. </p>
<p>if you consider $26^7$ , here is what happening,</p>
<p>The first place can be filled by 26 letters , second place also 26 , third place also , so you get a combination like <code>ABCDEFG</code> which is not correct. $26^7$ , is all the permutation you can get from 26 letters , when you multiply it by 7! I don't know what you get.</p>
|
1,830,989 | <p>so while playing around with circles and triangles I found 2-3 limits to calculate the value of $ \pi $ using the <em>sin, cos and tan</em> functions, I am not posting the formula for obvious reasons.<br>
My question is that is there another infinite series or another way to define the trig functions when the value of the angle is in <strong>degrees</strong> without converting it to radians, I know of the <em>taylor series</em> but it takes the value of x in radians and to convert the angle to radians you obviously need $ \pi $, So is there another way to convert the or maybe find the angles in radians without using $ \pi $ or maybe a series for the trig functions which uses degrees? Also I know as a rule of thumb you always use radians in calculus can anyone explain to me why??<br>
Sorry if i asked a really dumb question.<br>
Regards,<br>
Kinshuk </p>
| Andrei | 331,661 | <p>For $x$ in radians
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...=\sum_{i=0}^\infty\frac{(-1)^i x^{2i+1}}{(2i+1)!}$$
To transform the angle $x$ into degrees, use $y=x\cdot180/\pi$. So for the angle in degrees:
$$\sin(y)=\sum_{i=0}^\infty\frac{(-1)^i (180 y/\pi)^{2i+1}}{(2i+1)!}$$</p>
|
130,914 | <p>I dont know how to proceed with solving $$\sum_{i=1}^{n}i^{k}(n+1-i).$$ Please give advise.</p>
| andreimf | 28,948 | <p>$H_n^{(k)}=\sum_{i=1}^n i^{-k}$ is by definition the $n$-th harmonic number of order $k$. Thus,
$$
\sum_{i=1}^n i^{k} (n+1-i) = (n+1) H_n^{(-k)}- H_n^{(-k-1)}.
$$
I don't think it can be simplified further, at least considerably and for any $n$ and $k$. Is this what you meant by "solving"?</p>
|
130,914 | <p>I dont know how to proceed with solving $$\sum_{i=1}^{n}i^{k}(n+1-i).$$ Please give advise.</p>
| pravin | 29,006 | <p>This is a prefix sum of natural numbers. The solution to this is</p>
<p>$$\binom{n+k+1}{k+2}$$</p>
|
2,952,028 | <p>The question asks: Find the values of k for which the line</p>
<p><span class="math-container">$y=2x-k$</span> is tangent to the circle with equation <span class="math-container">$x^2+y^2=5$</span></p>
<p>So I started by substituting,</p>
<p><span class="math-container">$x^2+(2x-k)^2=5$</span></p>
<p><span class="math-container">$x^2+4x^2-4xk+k^2=5$</span></p>
<p><span class="math-container">$5x^2+k^2-4xk-5=0$</span></p>
<p>But after this I couldn't see a way to factor it that would make able to find the discriminant and set that equal to <span class="math-container">$0$</span>.</p>
<p>Help would be appreciated.</p>
| Michael Rozenberg | 190,319 | <p>The distance from the point <span class="math-container">$(0,0)$</span> to the line <span class="math-container">$2x-y-k=0$</span> should be equal to the radius of the circle.</p>
<p>Thus, <span class="math-container">$$\frac{|2\cdot0-0-k|}{\sqrt{2^2+1^2}}=\sqrt5$$</span> or
<span class="math-container">$$|k|=5.$$</span></p>
|
3,787,111 | <p>Sets can have minimal and least elements, and they are two different things, for example:
given the set <span class="math-container">$A=\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\},\{1,2,3\}\}$</span> and the subset relation, this set has <span class="math-container">$\{1\}$</span> as a minimal element but not as a least element. Now, my question is, what is <span class="math-container">$\min(A)$</span>? I would say that it doesn't exists since there are not least elements, but does <span class="math-container">$\min(A)$</span> really mean "least element of the set <span class="math-container">$A$</span>"? I'm asking because the notation "<span class="math-container">$\min$</span>" looks like it's referring to the minimal element and not the least one. And is this notation really referring to the least element of a set or instead it's a completely different things and we're talking about a unary operation used for the natural numbers, integers and real numbers?</p>
| Charlie Vanaret | 741,916 | <p>You can use a very simple "interval analysis" evaluation approach:</p>
<p>For <span class="math-container">$X = [-2, 3]$</span>, we have <span class="math-container">$f(X) = [-2, 3]^2 + 3 = [0, 9] + 3$</span> (since <span class="math-container">$x \mapsto x^2$</span> maps to nonnegative numbers).</p>
|
3,787,111 | <p>Sets can have minimal and least elements, and they are two different things, for example:
given the set <span class="math-container">$A=\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\},\{1,2,3\}\}$</span> and the subset relation, this set has <span class="math-container">$\{1\}$</span> as a minimal element but not as a least element. Now, my question is, what is <span class="math-container">$\min(A)$</span>? I would say that it doesn't exists since there are not least elements, but does <span class="math-container">$\min(A)$</span> really mean "least element of the set <span class="math-container">$A$</span>"? I'm asking because the notation "<span class="math-container">$\min$</span>" looks like it's referring to the minimal element and not the least one. And is this notation really referring to the least element of a set or instead it's a completely different things and we're talking about a unary operation used for the natural numbers, integers and real numbers?</p>
| Arthur | 15,500 | <p>The answer is A because the end points of the domain don't necessarily correspond to the end points of the range. Consider <span class="math-container">$f(0)$</span> compared to <span class="math-container">$f(-2)$</span> or <span class="math-container">$f(3)$</span>.</p>
|
152,582 | <p>I have tried to use <code>OpenRead</code> for my application as it appears to be less of a burden on the memory.</p>
<p>I <code>OpenRead</code> a .csv file to extract coordinates in the form {x,y} in the following way</p>
<pre><code>ClearAll[f];
f = OpenRead["mathfile.csv"];
g = ReadList[f, String, RecordLists -> True];
Close[f];
</code></pre>
<p>I can access each record seperately by just calling <code>g[[1]]</code>. However the format is not as numbers and so a function like <code>ListPlot[g[[1]]]</code> doesn't work.</p>
<p>If I use <code>"Number"</code> or <code>"Real"</code> as an option within <code>ReadList</code> I get an error:</p>
<p><strong>"Read:readn: Invalid real number found when reading from 'f'."</strong></p>
<p>Using <code>"String"</code> or <code>"Record"</code> within <code>ReadList</code> doesn't cause a problem. However I would have to convert the <code>"String"</code> or <code>"Record"</code> into a number format.</p>
<p>What am I doing wrong ? I would like to call <code>g[[1]]</code> and have the co-ordinate in a form that can be read by <code>ListPlot</code></p>
<p>Thanks</p>
| BRS | 51,362 | <p>I got some good performance in the following manner. My data was originally an <code>XLSX</code>. I saved it as a MSDOS txt file. Unfortunately, the {x,y} data had long runs punctuated by a text saying 'Segment 1:' e.t.c.. This was stopping the <code>ReadList</code> from working..So after removing the text and whitespace within Excel, it was saved as .txt.I didn't use <code>Import</code>, as it just takes too long and eats up memory (I have a total 300 000 rows of {x,y} data that I want to build a dynamic animation with) with a <code>CSV</code>....</p>
<p>So I basically open the txt for reading and then use <code>ReadList</code> to extract a number of entries in blocks... this is quite quick as my data has a high resolution and so not every data point is necessary for a good looking plot.</p>
<pre><code>ClearAll[str];
str = OpenRead["mathfile1.txt"];
g = Table[ReadList[str, {Real, Number}, 1000], {100}];
Close[str];
</code></pre>
<p>Entries can be accessed as <code>g[[1,x]]</code></p>
<p>I found that if <code>ReadList</code> reads over 15000 entries, my computer starts to seriously hang. I have 8 GB of RAM</p>
|
367,669 | <p><img src="https://i.stack.imgur.com/zQFyC.jpg" alt="enter image description here"></p>
<p>This is probably a very simple questions but I am not clear on Möbius transformations and how to solve this problem. I'd appreciate if somebody can point me towards a method to do these sort of questions or a webpage that explains what I need to solve this problem.</p>
| Davide Giraudo | 9,849 | <p>Consider the sequence
$$0,0,1,\color{green}{0,1,2},\color{red}{0,1,2,3},0,1,2,3,4, 0,1,2,3,4,5,\dots$$</p>
|
2,316,362 | <p>If $G$ is a Lie Group, a representation of $G$ is a pair $(\rho,V)$ where $V$ is a vector space and $\rho : G\to GL(V)$ is a group homomorphism.</p>
<p>Similarly, if $\mathfrak{g}$ is a Lie Algebra, a representation of $\mathfrak{g}$ is a Lie Algebra homomorphism $\rho : \mathfrak{g}\to \mathfrak{gl}(V)$ to the Lie Algebra of endomorphisms.</p>
<p>Now, many Physics books treating Quantum Field Theory, immediately relate the representations of Lie Groups and Lie Algebras without citing the result being used nor explaining how is it used really.</p>
<p>This is quite common in order to find the representations of the Lorentz group $SO(1,3)$ in terms of elements of its Lie Algebra.</p>
<p>Now since physicists don't clear this in the books, I'm asking here. What is actually the relation between representations of Lie Groups and Lie Algebras that allows one to find the representations of the Lie Group in terms of the representations of the Lie Algebra?</p>
| José Carlos Santos | 446,262 | <p>Each finite-dimensional linear representation of a Lie group <span class="math-container">$G$</span> on a vector space <span class="math-container">$V$</span> induces a representation of its Lie algebra <span class="math-container">$\mathfrak g$</span> on <span class="math-container">$V$</span>. If <span class="math-container">$X\in\mathfrak g$</span> and if <span class="math-container">$v\in V$</span>, define<span class="math-container">$$X.v=\left.\frac{d}{dt}\exp(tX).v\right|_{t=0}$$</span>Note that not all representations of <span class="math-container">$\mathfrak g$</span> can be obtained by this process, although this is actually true if <span class="math-container">$G$</span> turns out to be simply connected.</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| GEdgar | 442 | <p>idm's solution written differently... </p>
<p>For $x \ge 0$,
$$
\arctan x = \int_0^x \frac{dt}{1+t^2}
$$
(in some texts, this may even be the definition)</p>
<p>For $t>0$,
$$
\frac{1}{1+t^2} < 1
$$
so we conclude
$$
\arctan x \le \int_0^x 1\,dt = x
$$
with strict inequality except for $x=0$. </p>
<p>Similar arguments when $x<0$. I assume you do not intend to prove something for complex $x$... For example, $|\arctan x| > |x|$ for $x$ purely imaginary (and nonzero).</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| copper.hat | 27,978 | <p>Note that $\tan 0 = 0$, $\tan' 0 = 1$ and $\tan'x = {1 \over \cos^2 x} > 1$ for $0<|x|< {\pi \over 2}$.</p>
<p>Then $\arctan 0 = 0$; the inverse function theorem gives
$\arctan' 0 = 1$ and $0<\arctan' x < 1$ for $x \neq 0$.</p>
<p>Since $\arctan x = \arctan 0 + \arctan'( \xi ) x$ for some $ \xi \in (0,x)$ (or $(x,0)$, depending on sign of $x$),
we see that $\arctan x < x$ for $x >0$ and $\arctan x > x$ for $x <0$.</p>
|
2,929,025 | <p>Bill gave exams for the entrance at some specific gymnasium. <span class="math-container">$602$</span> students took part, which were classified, after the exams, in an ascending order, and the first <span class="math-container">$108$</span> students will be taken, which will accept to enter. Every student that has the possibility to enter will not enter with a small possibility <span class="math-container">$p=0.02$</span>, same for all, and independent from the rest. Bill is at the position <span class="math-container">$113$</span>, so he will be accepted if at least <span class="math-container">$5$</span> students from the first <span class="math-container">$112$</span> will not enter at the gymnasium. I want to give an exact expression for the probability <span class="math-container">$q$</span> that Bill gets accepted. I also want to give an approximate expression for the probability <span class="math-container">$q$</span> .</p>
<p>Is the probability that Bill get accepted equal to</p>
<p><span class="math-container">$$5 \cdot 0.02?$$</span></p>
<p>Or do we have to take also something else into consideration?</p>
| Phil H | 554,494 | <p>You have to take into consideration the number of ways you can choose <span class="math-container">$5$</span> from <span class="math-container">$112$</span> and the probability of remaining chosen <span class="math-container">$(.98)$</span>. Also its the probability of at least <span class="math-container">$5$</span> not being chosen. This makes it more complicated as it then becomes one minus the probability of from <span class="math-container">$0$</span> to <span class="math-container">$4$</span> not chosen. The probability of Bill being chosen is therefore:</p>
<p><span class="math-container">$P_B =1 - (^{112}C_0\cdot .98^{112} + ^{112}C_1\cdot .02\cdot .98^{111} + ^{112}C_2\cdot .02^2\cdot .98^{110} + ^{112}C_3\cdot .02^3\cdot .98^{109} + ^{112}C_4\cdot .02^4\cdot .98^{108})$</span></p>
<p><span class="math-container">$P_B = 1 - 0.9251 = 0.0749$</span></p>
|
1,023,463 | <p>I know $x \sin(1/x)$ is not Lipschitz on $[0,1]$, but some experimentation makes me conjecture that it is $1/2$-Holder. What is a good way to prove this?</p>
| DuFong | 193,997 | <p>Check this paper and the first reference paper. <a href="https://arxiv.org/pdf/1407.6871.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/1407.6871.pdf</a></p>
|
3,348,178 | <p>I have three vectors in 3d that originate at a point. If I look at them along a line perpendicular to a plane that intersects two of them, how do I find the angles between those two vectors and the third one?</p>
<p>Clarification because this is frickin difficult to explain:</p>
<p><a href="https://i.stack.imgur.com/HNyOB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HNyOB.png" alt="enter image description here"></a></p>
<p>I'll also accept a way to find how much of the longest line is on either side of the shorter line, because that's ultimately what I need.</p>
| FFjet | 597,771 | <p><strong>You are wrong at the first step:</strong>
<span class="math-container">$$
x=y \nLeftrightarrow \sqrt{x^2} =\sqrt{y^2}
$$</span></p>
|
300,944 | <p>Show that there are no intergers $x$ and $y$ such that</p>
<p>$P(x,y)=x^2-5y^2=2$</p>
<p>Hint from professor:</p>
<p>Consider the equation in a convenient $\mod (n)$ so that you end up with a polynomial in a single variable. Then proceed as solving number of congruence.</p>
<hr>
<p>Im not sure how to approach this question</p>
<p>Since $P(x,y)=x^2-5y^2=2$</p>
<p>then $x^2-5y^2=0$ $\to$ $x^2=5y^2$</p>
<p>we have $5y^2\equiv0\mod(x)$</p>
<p>then how do I continue..?</p>
<p>Thank you!!</p>
| pritam | 33,736 | <p>Note that, every perfect square is either $0$ or $1$ modulo $4$. Now check all the cases to see that $x^2-5y^2$ is never equal to $2$ modulo $4$.</p>
|
1,599,890 | <blockquote>
<p>Let $a_n$ be the number of those permutation $\sigma $ on $\{1,2,...,n\}$ such that $\sigma $ is a product of exactly two disjoint cycles. Then find $a_4$ and $a_5$.</p>
</blockquote>
<p>Calculating $a_4$: Possible cases which can happen are $(12)(34),(13)(24),(14)(23)$, any cycle of the form $(123)$ or $(12)$ i.e. two-cycles and three cycles thus we have in total $3+\frac{1}{3}4P_3+\frac{1}{2}4P_2=3+8+6=17$ but the correct answer is given to be either $11$ or $14$.</p>
<p>Where am I wrong? Please help.</p>
| joriki | 6,622 | <p>From the possible answers, it appears that the question intended to ask how many permutations have exactly two cycles in their cycle decomposition, including the $1$-cycles.</p>
<p>If the smaller cycle has $k$ elements, the greater has $n-k$, and $j$ particular elements can form $(j-1)!$ different $j$-cycles. Thus the desired count is</p>
<p>$$
\frac12\sum_{k=1}^{n-1}\binom nk(k-1)!(n-k-1)!=\frac{n!}2\sum_{k=1}^{n-1}\frac1{k(n-k)}\;,
$$</p>
<p>where the factor $\frac12$ compensates for the double-counting of all products, including in the case of even $n$ the products of two cycles of length $\frac n2$. So we have</p>
<p>$$
a_4=4!\left(\frac1{1(4-1)}+\frac12\frac1{2(4-2)}\right)=24\left(\frac13+\frac18\right)=11
$$</p>
<p>and</p>
<p>$$
a_5=5!\left(\frac1{1(5-1)}+\frac1{2(5-2)}\right)=120\left(\frac14+\frac16\right)=50\;.
$$</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Godot | 38,875 | <p>Category theory is a good and powerful language capable of expressing various concepts of purely algebraical nature.</p>
<p>But it is a terrible tool for actually solving problems.</p>
<p>To convince yourself that the last statement is true try to think about a proof of a theorem from another branch of mathematics that depends on a category theory in a crucial, non-linguistic way. I doubt you will find any such proof. Thus it is hardly a surprising fact that you will not find any serious application of category theory in engineering problems. </p>
<p>On the other hand, you will find plenty of applications of category theory inside category theory. Category theory fights with problems originating in category theory, with problems of no practical relevance for mathematicians, not to mention engineers. </p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Community | -1 | <p>The state-of-the-art real world dimensionality reduction technique UMAP is based among other things on category theory:</p>
<p><a href="https://arxiv.org/abs/1802.03426" rel="nofollow noreferrer">https://arxiv.org/abs/1802.03426</a></p>
<p><a href="https://github.com/lmcinnes/umap" rel="nofollow noreferrer">https://github.com/lmcinnes/umap</a></p>
|
41,836 | <p>Nakayama's lemma is as follows:</p>
<blockquote>
<p>Let <span class="math-container">$A$</span> be a ring, and <span class="math-container">$\frak{a}$</span> an ideal such that <span class="math-container">$\frak{a}$</span> is contained in every maximal ideal. Let <span class="math-container">$M$</span> be a finitely generated <span class="math-container">$A$</span>-module. Then if <span class="math-container">$\frak{a}$$M=M$</span>, we have that <span class="math-container">$M = 0$</span>.</p>
</blockquote>
<p>Most proofs of this result that I've seen in books use some non-trivial linear algebra results (like Cramer's rule), and I had come to believe that these were certainly necessary. However, in Lang's Algebraic Number Theory book, I came across a quick proof using only the definitions and induction. I felt initially like something must be wrong--I thought perhaps the proof is simpler because Lang is assuming throughout that all rings are integral domains, but he doesn't use this in the proof he gives, as far as I can see.</p>
<p>Here is the proof, verbatim: We do induction on the number of generators of <span class="math-container">$M$</span>. Say M is generated by <span class="math-container">$w_1, \cdots, w_m$</span>. There exists an expression <span class="math-container">$$w_1 = a_1w_1 + \cdots + a_mw_m$$</span> with <span class="math-container">$a_i \in \frak{a}$</span>. Hence <span class="math-container">$$(1-a_1)w_1 = a_2w_2 + \cdots +a_mw_m$$</span> If <span class="math-container">$(1-a_1)$</span> is not a unit in A, then it is contained in a maximal ideal <span class="math-container">$\frak{p}$</span>. Since <span class="math-container">$a_1 \in \frak{p}$</span> by hypothesis, we have a contradiction. Hence <span class="math-container">$1-a_1$</span> is a unit, and dividing by it shows that <span class="math-container">$M$</span> can be generated by <span class="math-container">$m-1$</span> elements, thereby concluding the proof.</p>
<p>Is the fact that <span class="math-container">$A$</span> is assumed to be a domain being smuggled in here in some way that I missed? Or is this really an elementary proof of Nakayama's lemma, in full generality?</p>
| Arrow | 69,037 | <p>I think the following proof is valid and avoids both determinants and maximal ideals. The cost is induction over all <span class="math-container">$A$</span>-modules generated by <span class="math-container">$n$</span> elements.</p>
<p><strong>Nakayama.</strong> Let <span class="math-container">$J$</span> be the Jacobson radical. Let <span class="math-container">$M$</span> be a finitely generated <span class="math-container">$A$</span>-module satisfying <span class="math-container">$M=JM$</span>. Then <span class="math-container">$M=0$</span>.</p>
<p><em>Proof.</em> Induction on size of generating set. If <span class="math-container">$M$</span> is generated by zero elements then it is zero. Assume the assertion holds for modules generated by <span class="math-container">$n-1$</span> elements and let <span class="math-container">$M$</span> be generated by <span class="math-container">$n$</span> elements <span class="math-container">$m_1,\dots ,m_n$</span>. Then <span class="math-container">$m_n=\sum _{i=1}^n \varepsilon _im_i$</span> with <span class="math-container">$\varepsilon _i\in J$</span> so <span class="math-container">$(1-\varepsilon _n)m_n=\sum_{i=1}^{n-1}m_i$</span>. Dividing by the unit <span class="math-container">$1-\varepsilon _n$</span> we obtain <span class="math-container">$m_n$</span> as a linear combination of <span class="math-container">$m_1,\dots ,m_{n-1}$</span>. Proceeding in this fashion we find <span class="math-container">$M=0$</span>.</p>
|
148,809 | <p>Let $M$ be a complete Riemannian manifold, does there exists a positive non-constant harmonic function $f \in L^1(M)$? Who can answer me or give me a counter example? Thank you very much!</p>
| Neal | 20,569 | <p>No. </p>
<p>The second result on Google gives <a href="http://intlpress.com/JDG/archive/1984/20-2-447.pdf" rel="nofollow">http://intlpress.com/JDG/archive/1984/20-2-447.pdf</a>, "Uniqueness of $L^1$ solutions for the Laplace equation and heat equation on Riemannian manifolds" by Peter Li, J Diff Geo 20 (1984) 447-457. It has the following result:</p>
<blockquote>
<p><strong>Theorem 1</strong>: If $M$ is a complete noncompact Riemannian manifold without boundary, and if the Ricci curvature of $M$ has a negative quadratic lower bound, then any $L^1$ nonnegative subharmonic function on $M$ is identically constant. In particular, any $L^1$ nonnegative harmonic function on $M$ is identically constant.</p>
</blockquote>
|
4,613,449 | <p>I first calculated it using the substitution method and got the result <span class="math-container">$\frac 1 2\ln^2x+C$</span>, <span class="math-container">$C\in \Bbb R$</span>, but I am getting a wrong result attempting to use per parts on the same problem. Is there something I am missing? Is there a reason per parts cannot be applied here?</p>
| Henry | 6,460 | <p>Ignoring the indefinite integration constant temporarily,</p>
<p>if you are attempting to apply <span class="math-container">$\int u\, dv = uv-\int v \, du$</span> here,</p>
<p>then with <span class="math-container">$u=\log_e(x)$</span> and <span class="math-container">$dv=\frac1x\, dx$</span>, and thus <span class="math-container">$v=\log_e(x)$</span> and <span class="math-container">$du = \frac1x\, dx$</span>,</p>
<p>you would get <span class="math-container">$\int \log_e(x) \frac1x\, dx= \left(\log_e(x)\right)^2 -\int \log_e(x) \frac1x\, dx$</span> and rearranging <span class="math-container">$$\int \frac{\log_e(x)}x\, dx = \tfrac12 \left(\log_e(x)\right)^2$$</span> as expected.</p>
<p>The indefinite integration constant comes into the <span class="math-container">$uv$</span> term so you can make this <span class="math-container">$\int \frac{\log_e(x)}x\, dx = \tfrac12 \left(\log_e(x)\right)^2+C.$</span></p>
|
4,613,449 | <p>I first calculated it using the substitution method and got the result <span class="math-container">$\frac 1 2\ln^2x+C$</span>, <span class="math-container">$C\in \Bbb R$</span>, but I am getting a wrong result attempting to use per parts on the same problem. Is there something I am missing? Is there a reason per parts cannot be applied here?</p>
| Abezhiko | 1,133,926 | <p>Integration by parts brings :
<span class="math-container">$$
I := \int\frac{1}{x}\,\ln(x)\,\mathrm{d}x = \ln^2(x) - \int\ln(x)\,\frac{1}{x}\,\mathrm{d}x
$$</span>
hence <span class="math-container">$2I = \ln^2(x)$</span> and the desired result.</p>
|
179,659 | <p>I want to assign each cell in my current notebook a tag so that I can rerun cells with specific tags later in the notebook <a href="https://mathematica.stackexchange.com/a/71333/44420">(Using a method similar to this answer)</a>.</p>
<p>My questions are</p>
<ul>
<li>How do I assign cells in my current notebook tags?</li>
<li>How do I view the tags that I have assigned?</li>
</ul>
| Carl Woll | 45,431 | <p>You can use the menu item <code>Cell | CellTags</code> to do this. If you want to inspect and modify <a href="http://reference.wolfram.com/language/ref/CellTags" rel="noreferrer"><code>CellTags</code></a> programmatically, you can use <a href="http://reference.wolfram.com/language/ref/CurrentValue" rel="noreferrer"><code>CurrentValue</code></a> and <a href="http://reference.wolfram.com/language/ref/SetOptions" rel="noreferrer"><code>SetOptions</code></a>. For example:</p>
<pre><code>cell = PreviousCell[];
CurrentValue[cell, CellTags]
</code></pre>
<blockquote>
<p>{}</p>
</blockquote>
<p>Change the tag:</p>
<pre><code>SetOptions[cell, CellTags->"FOO"]
</code></pre>
<p>Check:</p>
<pre><code>CurrentValue[cell, CellTags]
</code></pre>
<blockquote>
<p>"FOO"</p>
</blockquote>
<p>Another useful function is <a href="http://reference.wolfram.com/language/ref/Cells" rel="noreferrer"><code>Cells</code></a>, which will return a list of <a href="http://reference.wolfram.com/language/ref/Cell" rel="noreferrer"><code>Cell</code></a> objects satisfying various criteria.</p>
|
234,409 | <p>I'm trying to obtain the coordinates of the border of the continents. I need this information to be ordered such that when I do, for example,</p>
<pre><code>ListLinePlot[data]
</code></pre>
<p>It does not yield a messed up image, as happens for disordered points. Initially I was trying by highlighting points on images of maps, and the detecting the points. This approach was really problematic. However, I find that Mathematica has a built in functionality which gives me just what I need, but for countries. That is for example,</p>
<pre><code>CountryData["Iran", "FullCoordinates"][[1]]
</code></pre>
<p>I'm exploring if this can be done for continents as entities, and I have found that the information for a continent as a polygon can be obtained as,</p>
<pre><code>africa = Entity["GeographicRegion", "Africa"]["Polygon"]
</code></pre>
<p>Is there any way by which I can obtain a list of points corresponding to the coordinates of this continent?</p>
| creidhne | 41,569 | <p>Although <code>ListLinePlot</code> can plot coordinates for geographic data, I recommend that you avoid this method and use <code>GeoGraphics</code> instead. Some of the advantages are:</p>
<ul>
<li>selecting map features, e.g, islands</li>
<li>using map projections (with the <code>GeoProjection</code> option)</li>
<li>locating points on the map (using <code>GeoMarker</code>)</li>
</ul>
<p>But first, to answer your question about coordinates, you can get them for the outline of a continent, Africa for example, from its polygon:</p>
<pre><code>continentPoly = Entity["GeographicRegion", "Africa"]["Polygon"];
data = continentPoly[[1, 1, 1]];(*a long list of coordinate pairs*)
</code></pre>
<p>There's other useful information available from <code>GeoBounds</code> and <code>GeoBoundingBox</code>.</p>
<pre><code>GeoBounds[Entity["GeographicRegion", "Africa"]]
(* {{-34.8341, 37.5423}, {-25.36, 57.8092}} *)
GeoBoundingBox[Entity["GeographicRegion","Africa"]]
(* {GeoPosition[{-34.8341, -25.36}], GeoPosition[{37.5423, 57.8092}]} *)
</code></pre>
<p><strong>Using GeoGraphics</strong></p>
<p>Let's look at some of the advantages of using <code>GeoGraphics</code>. If you use <code>ListLinePlot</code>, you lose ease-of-use capabilities. For example, here's a map of the continent of Africa including the island of Madagascar.</p>
<pre><code>GeoGraphics[{EdgeForm[Black], FaceForm[None],
Polygon[GeoPosition[continentPoly[[1, 1, {1,2}]]]]},
GeoBackground -> None, GeoProjection -> "Mercator"]
</code></pre>
<p><a href="https://i.stack.imgur.com/6IK4S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6IK4S.png" alt="Africa map" /></a></p>
<p>It's simple to choose the orthographic projection instead of Mercator:</p>
<pre><code>GeoGraphics[{EdgeForm[Black], FaceForm[None],
Polygon[GeoPosition[continentPoly[[1, 1, {1, 2}]]]]},
GeoBackground -> None, GeoProjection -> "Orthographic"]
</code></pre>
<p><a href="https://i.stack.imgur.com/Vspe0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vspe0.png" alt="orthographic projection" /></a></p>
<p>When you want to highlight a point on the map, <code>GeoMarker</code> does the work for you.</p>
<pre><code>GeoGraphics[{EdgeForm[Black], FaceForm[Red],
Polygon[GeoPosition[continentPoly[[1, 1, {1, 2}]]]],
GeoMarker[Entity["City", {"Pretoria", "Gauteng", "SouthAfrica"}]]},
GeoBackground -> None, GeoProjection -> "Mercator"]
</code></pre>
<p><a href="https://i.stack.imgur.com/q73fa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q73fa.png" alt="map with a location" /></a></p>
<p>Here's a map of several locations with the orthographic projection.</p>
<pre><code>GeoGraphics[{EdgeForm[Black], FaceForm[Orange],
Polygon[GeoPosition[continentPoly[[1,1,{1,2}]]]],
GeoMarker[Entity["City", {"Pretoria", "Gauteng", "SouthAfrica"}]],
GeoMarker[Entity["GeographicRegion", "Africa"]["HighestFeature"], "Color" -> Green],
GeoMarker[Entity["GeographicRegion", "Africa"]["CenterCoordinates"], "Color" -> Blue]},
GeoBackground -> None, GeoProjection -> "Orthographic"]
</code></pre>
<p><a href="https://i.stack.imgur.com/NfBKy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NfBKy.png" alt="several map locations" /></a></p>
<p>All of these features would be tricky or difficult to implement with <code>ListLinePlot</code>.</p>
<p><strong>An additional feature unique to GeoGraphics</strong></p>
<p>Another advantage with <code>GeoGraphics</code> is that you can use <code>Get Coordinates</code> to read positions directly from the graphic in a notebook (see the left-hand image). If you click on a point (the small red point on the right-hand image), you can copy and paste the point's geo-position on the map.</p>
<p><a href="https://i.stack.imgur.com/1Wsqe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Wsqe.png" alt="get coordinates from notebook graphic" /></a></p>
<pre><code>{GeoPosition[{18.293290236969238`, -1.4117877048275167`}, "ITRF00"]}
</code></pre>
|
4,516,356 | <p>In Ch. 20 of Spivak's <em>Calculus</em>, he shows that the remainder terms for <span class="math-container">$\arctan$</span> and <span class="math-container">$\log{(1+x)}$</span> become large with the order of the Taylor polynomial used to approximate these functions. Thus these approximations</p>
<blockquote>
<p>are of no use whatsoever in computing <span class="math-container">$\arctan{x}$</span> and <span class="math-container">$\log{(1+x)}$</span>.
This is no tragedy, <strong>because the values of these functions can be
found for any <span class="math-container">$x$</span> once they are known for all <span class="math-container">$x$</span> with <span class="math-container">$|x|<1$</span></strong>.</p>
</blockquote>
<p>How do we find the values of these functions for any <span class="math-container">$x$</span> if we know the functions for <span class="math-container">$|x|<1$</span>?</p>
| Community | -1 | <p>It is easy to show, using differentiation the formula</p>
<p><span class="math-container">$arctanx+arctan(\dfrac{1}{x})=\dfrac{\pi}{2}$</span>. Hence, knowing the value</p>
<p>for <span class="math-container">$0<x<1$</span> we obtain the value of <span class="math-container">$arctanx$</span> for <span class="math-container">$x\geq1$</span>. For negative <span class="math-container">$x$</span></p>
<p>we have <span class="math-container">$arctanx+arctan(\dfrac{1}{x})=-\dfrac{\pi}{2}$</span> and we follow the same method.</p>
<p>Now, for <span class="math-container">$log(1+x)$</span>. We assume that <span class="math-container">$x\in\,[1,2)$</span> then <span class="math-container">$x=1+y$</span> where <span class="math-container">$0\leq\,y\,<1$</span>.</p>
<p>So <span class="math-container">$log(1+x)=log(2+y)=log2(1+\dfrac{y}{2})$</span>=<span class="math-container">$log2+log(1+\dfrac{y}{2})$</span> where <span class="math-container">$\dfrac{y}{2}<1$</span>.</p>
<p>If <span class="math-container">$x\in[2,3)$</span> then <span class="math-container">$x=1+z$</span> where <span class="math-container">$1\leq z<2$</span>, hence</p>
<p><span class="math-container">$log(1+x)=log(2+z)=log2(1+\dfrac{z}{2})=log2+log(1+\dfrac{z}{2})$</span> where</p>
<p><span class="math-container">$\dfrac{z}{2}<1$</span>.</p>
<p>Continuing this process on <span class="math-container">$[3,4)$</span> e.t.c. we get the value of <span class="math-container">$log(1+x)$</span> for all <span class="math-container">$x>1$</span>.</p>
|
216,803 | <p>Let $X$ be a nonempty set. Find the topology $\tau$ on $X$ satisfying one of the following conditions:</p>
<ul>
<li><p>$(X, \tau)$ has the largest number of compact subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of compact subsets.</p></li>
<li><p>$(X, \tau)$ has the largest number of connected subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of connected subsets.</p></li>
<li><p>$(X, \tau)$ has the largest number of dense subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of dense subsets.</p></li>
</ul>
<p>I would like to thank all for their helping and comments.</p>
| user123123 | 45,121 | <p>Try thinking about when $X$ is given the discrete topology, and see if any of them become trivial. For example, if $X$ is discrete, then every subset is dense. Then you should be able to use arguments involving the discrete and trivial topologies to answer the other questions.</p>
|
216,803 | <p>Let $X$ be a nonempty set. Find the topology $\tau$ on $X$ satisfying one of the following conditions:</p>
<ul>
<li><p>$(X, \tau)$ has the largest number of compact subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of compact subsets.</p></li>
<li><p>$(X, \tau)$ has the largest number of connected subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of connected subsets.</p></li>
<li><p>$(X, \tau)$ has the largest number of dense subsets.</p></li>
<li><p>$(X, \tau)$ has the least number of dense subsets.</p></li>
</ul>
<p>I would like to thank all for their helping and comments.</p>
| joriki | 6,622 | <p>In each case, the property in question has a definite behaviour with respect to <a href="http://en.wikipedia.org/wiki/Comparison_of_topologies" rel="nofollow">comparison of topologies</a>, which you can determine by going through its definition and checking whether its conditions are easier or harder to fulfil when there are additional open sets. The set of topologies on a set is a complete lattice with least element the trivial topology and greatest element the discrete topology, so each of the questions is answered by one of these two.</p>
|
1,851,084 | <p>I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.</p>
| boaz | 83,796 | <p>Note that once you have one matrix that satisfies $A^2+A=I$, you have infinity many, since for every invertible matrix $P$ we get
$$
(P^{-1}AP)^2+(P^{-1}AP)=I\qquad\text{and}\qquad \det(P^{-1}AP)=1
$$
Now, if $4\mid N$, then choose the matrix
$$
A=diag(\phi,-1-\phi,\ldots,\phi,-1-\phi)
$$
where $\phi=\frac{\sqrt{5}-1}{2}$. Since $\phi^2+\phi=1$, we have that $A^2+A=I$. In addition, $\det A=(\phi(-1-\phi))^{N/2}=(-1)^{N/2}=1$ because $4\mid N$.</p>
<p>Suppose now that If $4\nmid N$. Since $f(A)=O$ for $f(x)=x^2+x-1$ and since $f(x)$ is simple above $\mathbb{R}$ it follows that $A$ is similar to a diagonal matrix $D$ that has $\phi$ and $-1-\phi$ along its main diagonal. But unless $4\mid N$, we get that $\det A=\det D\neq 1$. </p>
|
1,356,932 | <blockquote>
<p><strong>Problem.</strong> Let $h\in C(\mathbb{R})$ be a continuous function, and let
$\Phi:\Omega:=[0,1]^{2}\rightarrow\mathbb{R}^{2}$ be the map defined
by \begin{align*}
\Phi(x_{1},x_{2}):=\left(x_{1}+h(x_{1}+x_{2}),x_{2}-h(x_{1}+x_{2})\right) \tag{1}
\end{align*}
What is the (Lebesgue) measure (denoted by $\left|\cdot\right|$) of the set $\Phi(\Omega)$?</p>
</blockquote>
<p>Implicit in the problem statement is that $\Phi(\Omega)$ is Lebesgue measurable, but this is obvious as $\Phi$ is the composition of continuous functions, so $\Phi(\Omega)$ is a compact subset of $\mathbb{R}^{2}$.</p>
<p>If we knew that $h$ was $C^{1}$, then the Jacobian of $\Phi$ is
\begin{align*}
J\Phi(x_{1},x_{2})=\begin{bmatrix} {1+h'(x_{1}+x_{2})} & {h'(x_{1}+x_{2})} \\ {-h'(x_{1}+x_{2})} & {1-h'(x_{1}+x_{2})} \end{bmatrix}
\end{align*}
which has determinant $1$ everywhere. Whence,
\begin{align*}
\left|\Phi(\Omega)\right|=\int_{\mathbb{R}^{2}}\chi_{\Phi(\Omega)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}=\int_{\mathbb{R}^{2}}\chi_{\Omega}(x_{1},x_{2})\left|J\Phi(x_{1},x_{2})\right|\mathrm{d}x_{1}\mathrm{d}x_{2}=1
\end{align*}</p>
<p>My thought was to approximate $h$ uniformly in a neighborhood of $[-2,2]$ by a smooth function $g$, so that $\left|h(x_{1}+x_{2)}-g(x_{1}+x_{2})\right|<\epsilon$ for all $(x_{1},x_{2})\in\Omega$, given $\epsilon>0$. Denote the analogue of $\Phi$ with $g$ instead of $h$ by $\Phi_{\epsilon}$. One can verify that</p>
<p>$$\forall x=(x_{1},x_{2})\in\Omega,\quad \left|\Phi(x)-\Phi_{\epsilon}(x)\right|=\sqrt{2}\left|h(x_{1}+x_{2})-g(x_{1}+x_{2})\right|<\sqrt{2}\epsilon$$</p>
<p>So $\Phi(\Omega)\subset U_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ and $\Phi_{\epsilon}(\Omega)\subset V_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi(\Omega))<\sqrt{2}\epsilon\right\}$.</p>
<p>We obtain that
\begin{align*}
1=\left|\Phi_{\epsilon}(\Omega)\right|\leq\int_{\mathbb{R}^{2}}\chi_{V_{\epsilon}}(x)\mathrm{d}x
\end{align*}
Since $\Phi(\Omega)$ is compact, in particular closed, $\chi_{V_{\epsilon}}\rightarrow\chi_{\Phi(\Omega)}$ a.e. as $\epsilon\downarrow 0$. From monotone convergence, we obtain that
\begin{align*}
1\leq\left|\Phi(\Omega)\right| \tag{2}
\end{align*}</p>
<p>My issue is with establishing the reverse inequality. I know that
\begin{align*}
\left|\Phi(\Omega)\right|\leq\left|U_{\epsilon}\right|=1+\left|\left\{x\in\mathbb{R}^{2}:0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}\right|
\end{align*}
But I do not know how to control the measure of set $\left\{x\in\mathbb{R}^{2}: 0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ as $\epsilon\downarrow 0$, as $\Phi(\Omega)\setminus\Phi_{\epsilon}(\Omega)$ is contained in this set for all $\epsilon>0$.</p>
| Batominovski | 72,152 | <p>Since $\int_0^1\,g(x)\,\text{d}x=0$ and $\int_0^1\,x\,g(x)\,\text{d}x=0$, we have
$$\int_0^1\,x^2\,g(x)\,\text{d}x=\int_0^1\,\left(x^2-x+\frac{1}{6}\right)\,g(x)\,\text{d}x\leq\int_0^1\,\left|x^2-x+\frac{1}{6}\right|\,\big|g(x)\big|\,\text{d}x\,.$$
By the Cauchy-Schwarz Inequality,
$$\int_0^1\,x^2\,g(x)\,\text{d}x\leq\int_0^1\,\left|x^2-x+\frac{1}{6}\right|\,\big|g(x)\big|\,\text{d}x \leq \sqrt{\int_0^1\,\left|x^2-x+\frac{1}{6}\right|^2\,\text{d}x}\sqrt{\int_0^1\,\big|g(x)\big|^2\,\text{d}x}\,.$$
From $\int_0^1\,\big|g(x)\big|^2\,\text{d}x=1$, we conclude that
$$\int_0^1\,x^2\,g(x)\,\text{d}x\leq \sqrt{\int_0^1\,\left(x^2-x+\frac{1}{6}\right)^2\,\text{d}x}=+\frac{1}{6\sqrt{5}}\,.$$
The equality holds if and only if $g(x)=+6\sqrt{5}\left(x^2-x+\frac{1}{6}\right)=+\sqrt{5}\left(6x^2-6x+1\right)$ for almost every $x\in[0,1]$ (which coincidentally satisfies all three required conditions).</p>
<p>Similarly, $\int_0^1\,x^2\,g(x)\,\text{d}x\geq-\frac{1}{6\sqrt{5}}$. The inequality becomes an equality if and only if $g(x)=-6\left(x^2-x+\frac{1}{6}\right)=-\sqrt{5}\left(6x^2-6x+1\right)$ for almost every $x\in[0,1]$.</p>
<p><strong>How did I find $f(x)=x^2-x+\frac{1}{6}$?</strong></p>
<p>Well, I wanted a quadratic function $f(x)=x^2+ax+b$ for $x\in[0,1]$ such that $\int_0^1\,f(x)\,\text{d}x=0$ and $\int_0^1\,x\,f(x)\,\text{d}x=0$. Hence, I had to solve the simultaneous system of equations $\frac{1}{3}+\frac{1}{2}a+b=0$ and $\frac{1}{4}+\frac{1}{3}a+\frac{1}{2}b=0$, which would give $a=-1$ and $b=\frac{1}{6}$.</p>
|
548,902 | <p>I need to find the limit:
$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$ </p>
<p>any ideas here? I've tried to use "squeeze theorem" but with no luck.. </p>
| Luiz Cordeiro | 58,818 | <p>Maybe you're not familiar with the definition of differentiability in higher dimensions: see <a href="http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions" rel="nofollow">Wikipedia</a>, for example. The idea is this: in one-variable calculus, the derivative of a function at a point gives the tangent line to the graph of a function at that point. With more variables, the derivative (the linear function) gives the tangent line/plane/space to the graph of the function at the point.</p>
<p>Now, to the problem. Youhave to show that the differential of $f$ at $(0,0)$ is the zero-linear function, that, is,
$$\lim_{(x,y)\rightarrow (0,0)}\dfrac{f((0,0)+(x,y))-f(0,0)-\mathbf{0}(x,y)}{\Vert (x,y)\Vert}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{\sqrt{9-x^2-y^2}-3}{\Vert (x,y)\Vert}=0$$</p>
<p>You can calculate this limit directly, and attain the result you want (you can use the fact that $\lim_{h\rightarrow 0}\dfrac{\sqrt{9-h^2}-3}{h}=0$, in one-variable calculus!).</p>
|
3,104,051 | <p>I have the work of the proof done, but at the end after showing
<span class="math-container">$3^{2(n+1)} - 1=9(3^{2n} - 1)+8$</span>
I make the statement that since <span class="math-container">$9(3^{2n} - 1)$</span> is a multiple of 8 and 8 is a multiple of 8 then <span class="math-container">$3^{2n} - 1$</span> is a multiple of 8. I need to quote the facts about divisibility I use in this statement and I'm not sure exactly what to put for that, I have already included the definition of divisibility in the proof</p>
| nonuser | 463,553 | <p>Induction hypothesis: <span class="math-container">$$8\mid 9^n-1$$</span>so <span class="math-container">$$ 8\mid 9(9^n-1) = 9^{n+1}-9$$</span></p>
<p>but then <span class="math-container">$$8\mid (9^{n+1}-9) +8 = 9^{n+1}-1$$</span></p>
|
3,390,979 | <blockquote>
<p>Let <span class="math-container">$\sum_i^na_i=n$</span>, <span class="math-container">$a_i>0$</span>. Then prove that <span class="math-container">$$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$</span></p>
</blockquote>
<p>I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I struggle to see what can be done.</p>
<p>My attempts didn’t lead me to any result ... Simply cauchy , where <span class="math-container">$a_i=x,$</span> <span class="math-container">$b_i=1$</span> to find an inequality involving <span class="math-container">$\sum x^2$</span> . I also tried finding an inequality involving <span class="math-container">$\sum x^3$</span> using <span class="math-container">$a_i=\frac{x^3}{2}$</span> and <span class="math-container">$b_i=x^{\frac{1}{2}}$</span></p>
| Michael Rozenberg | 190,319 | <p>You need to use another queue: </p>
<p>By Rearrangement, AM-GM and C-S we obtain: <span class="math-container">$$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$</span></p>
<p>I used the following.
<span class="math-container">$$\frac{x^3+1}{x^2+1}\geq\frac{x+1}{2}$$</span> it's
<span class="math-container">$$2(x^3+1)\geq(x^2+1)(x+1)$$</span> or
<span class="math-container">$$x^3+1\geq x^2+x$$</span> or
<span class="math-container">$$x^2\cdot x+1\cdot1\geq x^2\cdot1+x\cdot1,$$</span> which is true by Rearrangement because <span class="math-container">$(x^2,1)$</span> and <span class="math-container">$(x,1)$</span> have the same ordering.</p>
<p>Also, by AM-GM <span class="math-container">$$\left(\frac{x+1}{2}\right)^4\geq\left(\sqrt{x}\right)^4=x^2.$$</span></p>
|
4,552,955 | <p>I'm solving a probability problem, and I've ended up with this sum:</p>
<p><span class="math-container">$$\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$$</span></p>
<p>WolframAlpha says I should get the answer <span class="math-container">$\frac{n!}{a\binom{a+b}{a}}$</span>, but I don't see how to get there. I tried to get to something containing <span class="math-container">$\binom{a+k-1}{k}$</span> so that I could use the Hockey Stick Theorem, but I wasn't successful.</p>
<p>So any hints would be very welcome, thanks for any help</p>
| epi163sqrt | 132,007 | <blockquote>
<p>We obtain
<span class="math-container">\begin{align*}
\color{blue}{\sum_{k=0}^{n-a-b}}&\color{blue}{\binom{n-a-b}{k}(a+k-1)!(n-a-k)!}\\
&=(n-1)!\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\binom{n-1}{n-a-k}^{-1}\tag{1}\\
&=n!\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\int_{0}^1z^{n-a-k}(1-z)^{a+k-1}\,dz\tag{2}\\
&=n!\int_{0}^1z^{n-a}(1-z)^{a-1}\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\left(\frac{1-z}{z}\right)^k\,dz\\
&=n!\int_{0}^1z^{n-a}(1-z)^{a-1}\left(1+\frac{1-z}{z}\right)^{n-a-b}\,dz\\
&=n!\int_{0}^1z^b(1-z)^{a-1}\,dz\\
&=\frac{n!}{a+b}\binom{a+b-1}{b}^{-1}\tag{2}\\
&\,\,\color{blue}{=\frac{n!}{a}\binom{a+b}{a}^{-1}}\tag{3}
\end{align*}</span>
in accordance with the claim and other answers.</p>
</blockquote>
<p><em>Comment:</em></p>
<ul>
<li><p>In (1) we use a representation using binomial coefficients.</p>
</li>
<li><p>In (2) we write the reciprocal of a binomial coefficient using the <em><a href="https://en.wikipedia.org/wiki/Beta_function" rel="nofollow noreferrer">beta function</a></em>
<span class="math-container">\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}</span></p>
</li>
<li><p>In (3) we use the binomial identity
<span class="math-container">\begin{align*}
(a+b)\binom{a+b-1}{b}=(a+b)\binom{a+b-1}{a-1}=a\binom{a+b}{a}
\end{align*}</span></p>
</li>
</ul>
|
2,150,832 | <p>I don't understand this equation $\int_0^t ds \int_0^{t'} ds' \delta(s-s')= \min(t,t')$.
I tried to work with the property of the dirac delta function that $\int_a^b \delta(x-c)dx = 1$ if $c \in [a,b]$, but I can't see how I can obtain the minimum. Can someone help me? </p>
<p>Thank you in advance!</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>Note that
$\ds{\int_{b}^{c}\delta\pars{x - a}\dd x} =
\ds{\bracks{b < a < c} - \bracks{c < a < b}}$ where $\ds{\bracks{\cdots}}$ is an <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow noreferrer">Iverson Bracket</a>.</p>
</blockquote>
<p>\begin{align}
&\int_{0}^{t}\dd s\int_{0}^{t'}\dd s'\,\delta\pars{s - s'} =
\int_{0}^{t}\int_{0}^{t'}\delta\pars{s' - s}\dd s'\,\dd s
\\[5mm] = &\
\int_{0}^{t}\bracks{0 < s < t'}\dd s - \int_{0}^{t}\bracks{t' < s < 0}\dd s =
\int_{0}^{t}\bracks{0 < s < t'}\dd s + \int_{t}^{0}\bracks{t' < s < 0}\dd s
\\[1cm] = &\
\bracks{t > 0}\bracks{t' > 0}\braces{\vphantom{\Large A}%
\bracks{t' < t}t' + \bracks{t' > t}t}
\\[5mm] + &\
\bracks{t < 0}\bracks{t' < 0}\braces{\vphantom{\Large A}%
\bracks{t' < t}\pars{-t} + \bracks{t' > t}\pars{-t'}}
\\[1cm] = &\
\bracks{t > 0}\bracks{t' > 0}\min\braces{t,t'} -
\bracks{t < 0}\bracks{t' < 0}\max\braces{t,t'}
\\[5mm] = &\
\bracks{t > 0}\bracks{t' > 0}\min\braces{t,t'} +
\bracks{t < 0}\bracks{t' < 0}\min\braces{-t,-t'}
\\[5mm] = &\
\braces{\vphantom{\Large A}%
\bracks{t > 0}\bracks{t' > 0} + \bracks{t < 0}\bracks{t' < 0}}
\min\braces{\verts{t},\verts{t'}}
\\[5mm] = &\
\bbox[#ffe,20px,border:2px dotted navy]{%
\ds{\bracks{tt' > 0}\min\braces{\verts{t},\verts{t'}}}}
\end{align}</p>
|
2,964,897 | <blockquote>
<p>Using Leibniz on <span class="math-container">$\sum_{n=1}^\infty \sin(\pi \sqrt{n^2+1})$</span></p>
</blockquote>
<p>So the question actually is how to rewrite <span class="math-container">$\sin(\pi\sqrt{n^2+1})$</span> in the form of <span class="math-container">$(-1)^n\times a_n$</span> so that I can apply Leibniz and decide the convergence or divergence?</p>
<p>I'm sorry but I'm pretty new in studying <em>series</em>.</p>
| Community | -1 | <p><span class="math-container">$$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(\pi n\sqrt{1+\frac1{n^2}}\right)\sim\sin\left(\pi n\left(1+\frac1{2n^2}\right)\right)=(-1)^n\sin\left(\frac\pi{2n}\right)\sim(-1)^n\frac{\pi}{2n}.$$</span></p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| Math Gems | 75,092 | <p>Put <span class="math-container">$\,\rm (a,b)=1\,$</span> below.</p>
<p><strong>Theorem</strong> <span class="math-container">$\rm\,\ \ (a\!+\!b,\ a^2\!+\!b^2)\, =\, (\color{#c00}{2a^2,\ \ 2ab,\ \ 2b^2},\ a\!+\!b)\, \overset{\rm\color{#c00}E}=\, (\color{#c50}{2(a,b)^2}\!,\ a\!+\!b)$</span></p>
<p><strong>Proof</strong> <span class="math-container">$\rm\ mod\ a\!+\!b\!:\ a^2\!+\!b^2 \equiv \color{#c00}{2a^2\! \equiv -2ab \equiv 2b^2}\ $</span> by <span class="math-container">$\rm\,a\!+\!b\,$</span> divides <span class="math-container">$\rm\color{#0A0}{green}$</span> terms below</p>
<p><span class="math-container">$$\quad\ \ \ \rm a^2\!+\!b^2 = (\color{#0A0}{b^2\!-\!a^2})+\color{#c00}{2a^2} = (\color{#0A0}{a\!+\!b})^2\!\color{#c00}{-2ab} = (\color{#0A0}{a^2\!-\!b^2})+\color{#c00}{2b^2} $$</span></p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| N. S. | 9,176 | <p><span class="math-container">$$ \gcd(a+b,a^2+b^2) \mid \gcd((a+b)(a-b), a^2+b^2) = \gcd(a^2-b^2, a^2+b^2) \mid \gcd [ ( a^2+b^2)+ (a^2-b^2) , ( a^2+b^2)+ (a^2-b^2) ]=2 \gcd(a^2,b^2)=2$$</span></p>
<p>Now it is easy to check that both 1 and 2 are possible...</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| Jens | 245 | <p>The methods suggested by David and István already do the job perfectly, but one could add something to the collection:</p>
<p>As described in the documentation on <a href="http://reference.wolfram.com/mathematica/tutorial/OperatorsWithoutBuiltInMeanings.html">Operators without Built-in Meanings</a>, there are some two-dimensional forms such as <code>Subscript</code> and <code>UnderBar</code> which can be re-defined to your liking. This is particularly useful if you need the 'N' with additional decorations anyway, such as $\text{N}_1$, $\text{N}_\text{total}$ or $\underline{\text{N}}$.</p>
<p>For these kinds of symbols you don't re-define <code>N</code> directly, but instead add a definition that is associated with the two-dimensional form in which the symbol appears. For example, I can set the following (note that the form <code>Subscript["N", 0]</code> can be entered more conveniently using keyboard shortcuts, but that wouldn't display so nicely here):</p>
<pre><code>ClearAll[Subscript];
Subscript["N", 0] = 6
(* ==> 6 *)
7 Subscript["N", 0]
(* ==> 42 *)
DownValues[Subscript]
(* ==> {HoldPattern[Subscript["N", 0]] :> 6} *)
</code></pre>
<p>The last line shows how the definition is stored as a definition associated with <code>DownValues[Subscript]</code>. The fact that we get <code>42</code> from the multiplication shows that the definition of the subscript works as expected. </p>
<p>To erase the definition, I can either use <code>ClearAll[Subscript]</code> or more specifically </p>
<pre><code>Subscript["N", 0] =.
</code></pre>
<p>The shortcut to input the subscript is to type <code>"N"</code> followed by <kbd>Ctrl</kbd><code>-</code> and then <code>0</code>.</p>
<p>You can also use strings in the subscript, instead of numbers. That allows you to get a "variable" that displays as $N$ without decoration by entering </p>
<pre><code>Subscript["N", ""]
</code></pre>
<p>Again you could assign things to this subscripted form as well. Another use for a string subscript would be (here I'll copy the two-dimensional form in box notation, it should look nicer when copied into the notebook):</p>
<pre><code>\!\(\*SubscriptBox[\("\<N\>"\), \("\<total\>"\)]\) = 100
(* ==> 100 *)
Sum[i, {i,
\!\(\*SubscriptBox[\("\<N\>"\), \("\<total\>"\)]\)}]
(* ==> 5050 *)
</code></pre>
<p>Of course one can use these re-defined operators as a substitute for <code>Format</code> as well. For example, here is some symbolic input and symbolic output:</p>
<pre><code>nn = Subscript["N", ""]
</code></pre>
<blockquote>
<p>$\text{N}$</p>
</blockquote>
<pre><code>Sum[i, {i, nn}]
</code></pre>
<blockquote>
<p>$\frac{1}{2} \text{N}_{\text{}}\left(\text{N}_{\text{}}+1\right)$</p>
</blockquote>
<p>You can then keep working with <code>nn</code> as the symbolic variable for computations but get the capital $\text{N}$ in the displayed output. </p>
<p>But you can also do numerical substitutions in the usual way, as here:</p>
<pre><code>% /. Subscript["N", ""] -> 100
(* ==> 5050 *)
</code></pre>
<p>(again, all the expressions looking like <code>Subscript</code> above can be made to appear more naturally by using the two-dimensional keyboard input methods).</p>
<p>Finally, everything I said here about <code>Subscript</code> also applies to things like <code>UnderBar["N"]</code> which displays as $\underline{\text{N}}$.</p>
<p>The <code>Subscript</code> combined with strings can also be useful when defining a whole set of similarly named variables automatically using something like this:</p>
<pre><code>normalVector =
Array[Subscript["N", {"x", "y", "z"}[[#]]] &, 3]
</code></pre>
<blockquote>
<p>$\left\{\text{N}_{\text{x}},\text{N}_{\text{y}},\text{N}_{\text{z}}\right\}$</p>
</blockquote>
<pre><code>Norm[normalVector] /. {
\!\(\*SubscriptBox[\("\<N\>"\), \("\<x\>"\)]\) -> 1,
\!\(\*SubscriptBox[\("\<N\>"\), \("\<y\>"\)]\) -> 0,
\!\(\*SubscriptBox[\("\<N\>"\), \("\<z\>"\)]\) -> 0}
(* ==> 1 *)
</code></pre>
<p>For even better formatting, you could replace <code>normalVector</code> by <code>"N"</code> <kbd>Ctrl</kbd>'7' <kbd>esc</kbd><code>vec</code><kbd>esc</kbd> which displays as $\vec{\text{N}}$.</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| Sjoerd C. de Vries | 57 | <p>To continue with Ajasja's land mine theme, it's not so problematic to use <code>N</code> as a <em>symbol</em> in equations like this:</p>
<p><img src="https://i.stack.imgur.com/snqa9.png" alt="Mathematica graphics"></p>
<p>as long as you keep the following in mind:</p>
<ul>
<li>Don't try to use it in an assignment (because you can't)</li>
<li>Don't use <code>[...]</code> in its neighborhood, nor <code>#&@</code></li>
<li>Keep it far away from <code>Map</code>, <code>Apply</code>, <code>Fold</code>, <code>Nest</code> etc.</li>
</ul>
<p>But frankly, this probably isn't the best idea to go for. Land mines may be safer.</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| Mr.Wizard | 121 | <p>The question of reassigning build-in Symbols has come up a number of times. There is an approach not yet posted here, however <em>it is not recommended</em>. Nevertheless for the sake of completeness I shall illustrate it.</p>
<p>Because <a href="http://reference.wolfram.com/language/ref/N.html" rel="nofollow noreferrer"><code>N</code></a> is not <a href="http://reference.wolfram.com/language/ref/Locked.html" rel="nofollow noreferrer"><code>Locked</code></a> it is not the more difficult example, as one can (perilously) <a href="http://reference.wolfram.com/language/ref/Unprotect.html" rel="nofollow noreferrer"><code>Unprotect</code></a> and redefine it. Instead I shall use <a href="http://reference.wolfram.com/language/ref/I.html" rel="nofollow noreferrer"><code>I</code></a> as it is Locked, and redefinition comes up from time to time.<sup><a href="https://mathematica.stackexchange.com/q/17617/121">(1)</a><a href="https://mathematica.stackexchange.com/q/124644/121">(2)</a></sup></p>
<p>It is possible to change the Context of a System Symbol, even a Locked one. Therefore:</p>
<pre><code>Context[I] = "foo`";
System`I = 3;
E^(I*Pi) + 1 // N
</code></pre>
<blockquote>
<pre><code>12392.6
</code></pre>
</blockquote>
<p>The original Symbol is still usable with a qualified context:</p>
<pre><code>E^(foo`I*Pi) + 1
</code></pre>
<blockquote>
<pre><code>0
</code></pre>
</blockquote>
<p>Functions that return <a href="http://reference.wolfram.com/language/ref/Complex.html" rel="nofollow noreferrer"><code>Complex</code></a> values still work:</p>
<pre><code>out = FactorInteger[Prime[100], GaussianIntegers -> True]
</code></pre>
<blockquote>
<pre><code>{{-I, 1}, {10 + 21 I, 1}, {21 + 10 I, 1}}
</code></pre>
</blockquote>
<p>Note the <a href="http://reference.wolfram.com/language/ref/InputForm.html" rel="nofollow noreferrer"><code>InputForm</code></a> and <a href="http://reference.wolfram.com/language/ref/FullForm.html" rel="nofollow noreferrer"><code>FullForm</code></a> of that output:</p>
<pre><code>out // InputForm
</code></pre>
<blockquote>
<pre><code>{{-foo`I, 1}, {10 + 21*foo`I, 1}, {21 + 10*foo`I, 1}}
</code></pre>
</blockquote>
<pre><code>out // FullForm
</code></pre>
<blockquote>
<pre><code>List[List[Complex[0,-1],1],List[Complex[10,21],1],List[Complex[21,10],1]]
</code></pre>
</blockquote>
<p>I think messing with System Context Symbols in this manner is a <strong>Bad Idea</strong> apart from a few specific cases<sup><a href="https://mathematica.stackexchange.com/a/103553/121">(3)</a></sup> but I also think users should be aware of the full set of tools at their disposal and make individual decisions.</p>
|
4,181,442 | <p>I know this is a dumb question but...</p>
<p>Is <span class="math-container">$x ≠ 1,3,5$</span> the same as <span class="math-container">$x$</span> does not belong to {<span class="math-container">$1,3,5$</span>}, for example ?</p>
<hr />
<p>Sorry for the formatting.</p>
<p>Btw, anyone has a link with mathjax's commands I guess?</p>
<hr />
<p>Thanks in advance.</p>
| herb steinberg | 501,262 | <p><span class="math-container">$x\notin (1,3,5)$</span> means that <span class="math-container">$x$</span> is not a member of a set. The set is defined as elements, which could be letters, intervals or anything else. <span class="math-container">$x\ne 1$</span>, etc. means that equality is defined which is no necessarily true, unless equality is defined by identity.</p>
<p>Simple example <span class="math-container">$1+2=3$</span>, but for set membership <span class="math-container">$\{1+2\}$</span> is not the same as <span class="math-container">$\{3\}$</span>.</p>
|
160,169 | <p>Consider the following implementation of the complex square root:</p>
<pre><code>f[z_]:=Sqrt[(z - I)/(z + I)]*(z + I);
</code></pre>
<p>This implementation has branch points at $\lambda=\pm i$ and a (vertical) branch cut connecting them.</p>
<p>Then</p>
<pre><code>g[z_]:=Sinc[f[z]];
</code></pre>
<p>(recalling $\mathrm{sinc}(x)=\sin(x)/x$ ) has no branch cut and it is analytic on the entire complex plane, and admits power series expansions at $\lambda=\pm i$. </p>
<p>Indeed, using Mathematica 11.0.0 (Mac OS 10.10.5) gives:</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>$1-\frac{1}{3} i (z-i)-\frac{1}{5} (z-i)^2+\frac{11}{315} i (z-i)^3+\frac{61
(z-i)^4}{5670}+O\left((z-i)^5\right)$</p>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>gives $\frac{61}{5670}$.</p>
<p>Now, using Mathematica 11.1.1 (both on Mac OS 10.12 Sierra and Linux Ubuntu 16 LTS)</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
</blockquote>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}].
</code></pre>
</blockquote>
<p>So neither of these stock functions work in properly in Mathematica 11.1.1. Does anyone know what is going on? Will this be fixed? They worked properly even in Mathematica 9 and also in Mathematica 11.0.0</p>
<p>Besides any information, I'd also appreciate if anyone has a workaround for this.</p>
| José Antonio Díaz Navas | 1,309 | <p>You can try with:</p>
<pre><code>Series[PowerExpand@g[z], {z, I, 4}]
</code></pre>
<p>$1-\frac{1}{3} i (z-i)-\frac{1}{5} (z-i)^2+\frac{11}{315} i (z-i)^3+\frac{61 (z-i)^4}{5670}+O\left((z-i)^5\right)$</p>
<p>and</p>
<pre><code>SeriesCoefficients[PowerExpand@g[z], {z, I, 4}]
(* 61/5670 *)
</code></pre>
|
631,053 | <p>I have a container of 100 yellow items.</p>
<p>I choose 2 at random and paint each of them blue.</p>
<p>I return the items to the container.</p>
<p>If I repeat this process, on average how many cycles will I make before all 100 items are painted?</p>
<p>It is obviously 50 (100/2) if there is no replacement. But in this case, the items are returned to the container, so the same item could be chosen often.</p>
<p>What if we choose 3?</p>
| Henry | 6,460 | <p>You could calculate the exact figure for the expected number using recursion and find the mean is about $258.32$ with a standard deviation of about $62.59$.</p>
<p>Alternatively, if you took the balls one at a time, this would be the <a href="http://en.wikipedia.org/wiki/Coupon_collector%27s_problem" rel="nofollow">Coupon Collector's Problem</a>, with an expected value of $100H_{100} \approx 518.74$: you then need to halve this as you are taking items two at a time, so giving about $259.37$. The actual answer differs from this for two reasons: your pair of items are distinct, reducing the number needed, but this is slightly offset by the possibility that the first item of the final pair is the last that needs to be painted. </p>
|
1,424,198 | <p>My mathematical logic textbook defines $\{x \ | \ \text {_} x \text {_} \ \}$, but I'm not sure what the $\text {_} x \text {_}$ means. </p>
<p>Do the _ just mean 'for any expression involving $x$', or is there something I'm missing?</p>
| Community | -1 | <p>The symbols in question just mean "any expression involving $x$."</p>
|
172,139 | <p>I have a data set, I am trying to join all the data by a line. But I am afraid the plot is not doing it properly. </p>
<p>This is the example (or almost) of my problem:</p>
<pre><code>data = {{0, π}, {π/2, π/2}, {π/2,
3 π/2}, {π, 0}, {π,
2 π}, {3 π/2, π/2}, {3 π/2,
3 π/2}, {2 π, π}};
</code></pre>
<p>Now, trying to interpolate the data.
Result is</p>
<p><a href="https://i.stack.imgur.com/VcL7A.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/VcL7A.jpg" alt="enter image description here"></a> </p>
<p>instead of</p>
<p><a href="https://i.stack.imgur.com/57Orh.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/57Orh.jpg" alt="enter image description here"></a></p>
<p>Sorry, I just downloaded and directly uploaded (as I couldn't get this result). $x,y$ are from $[0,2\pi]$.</p>
| Hugh | 12,558 | <p>Your data do lie on the boundary of a square. However their order is not going around the square. Here I plot them and number them. </p>
<pre><code>Graphics[{Point[data],
Table[Text[ToString[n], data[[n]], {1, 1}], {n, Length@data}]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/Tq3C4.png" alt="Mathematica graphics"></p>
<p>By inspection I can see a better order and then I reorder your data and plot.</p>
<pre><code>ord = {1, 2, 4, 6, 8, 7, 5, 3, 1};
data2 = data[[ord]];
Graphics[{Line[data2]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/99nnd.png" alt="Mathematica graphics"></p>
<p>Is this what you need or are you looking to do this automatically? That would be much more advanced.</p>
|
4,292,427 | <p><span class="math-container">$$ \frac{d^{2}y}{dt^2}+ 2t \frac{dy}{dt}+ t y=0 ~~ \tag{1} $$</span></p>
<p>At least I know that in this case of ODE can be solved by finding out 2 particular solutions.</p>
<p>As those 2 particular solutions are known, the general solution for this ODE can be written as below form.</p>
<p><span class="math-container">$$ y\left(x\right)=C_{1}y_{1}\left(x\right)+ C_{2} y_{2}\left(x\right) $$</span></p>
<p>I've been completely struggling to find it.</p>
<p>Is there some cheat sheet(s) which can be used to find out particular solutions so that I don't have to ask about it in MSE?</p>
<p>Or can you give me some hint(s) so that I can find the particulr solutions?</p>
<p>I've found of it <a href="https://slideplayer.com/slide/13806843/" rel="nofollow noreferrer">one(at 13th page)</a> (however not applicable to this ODE though).</p>
<p><strong>I assume computer cannot be used and this problem is solved in a offline paper test.</strong></p>
<p>The wolfram showed the below result but what are Ai and Bi ?</p>
<p><a href="https://i.stack.imgur.com/zdkyN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zdkyN.png" alt="enter image description here" /></a></p>
| Aleksas Domarkas | 562,074 | <p>Maple:
<span class="math-container">$$y \left( t \right) ={\it \_C1}\,{{\rm e}^{-t/2}}{{ \rm KummerM}\left(1/16,\,1
/2,\,-1/4\, \left( 2\,t-1 \right) ^{2}\right)}+{\it \_C2}\,{{\rm e}^{-
t/2}}{{\rm KummerU}\left(1/16,\,1/2,\,-1/4\, \left( 2\,t-1 \right) ^{2}
\right)}
$$</span></p>
|
3,518,719 | <p>Evaluate <span class="math-container">$$\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$$</span></p>
<p>We know that as <span class="math-container">$n\to \infty$</span> we have <span class="math-container">$2^n<<2^{2n^2}$</span> and therefore the limit is <span class="math-container">$4$</span></p>
<p>In a more formal way I started with:</p>
<p><span class="math-container">$$\log(L)=\lim_{n \to \infty} \log(2^n+4^{n^2})^{\frac{1}{n^2}}=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n+2^{2n^2})$$</span></p>
<p>Continuing to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\log\left[2^n(1+2^{2n})\right]$$</span></p>
<p>Did not help much</p>
<p>As I arrived to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\left[\log(2^n)+\log(1+2^{2n})\right]=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n)+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})=0+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})$$</span></p>
| Narasimham | 95,860 | <p>We need to look at how Frenet-Serret relations are derived. Imagine a circular band or belt in a horizontal plane . All vectors <span class="math-container">$n$</span> normal to the band remain in a plane. and derivative <span class="math-container">$\dot b $</span> of bi-normal that is vertical up rotating in the normal plane of <span class="math-container">$(n,b)$</span> is always directed perpendicular to <span class="math-container">$ n$</span>. Dot product of unit vectors gives a zero torsion.</p>
<p>If the band twists, angle between<span class="math-container">$(\dot b, n) $</span> becomes less or more than <span class="math-container">$90^{\circ}$</span> in the normal plane of <span class="math-container">$(n,b)$</span> by an out-of-plane rotation of <span class="math-container">$b$</span> vector so that their dot product is non-zer0...by vector definition torsion is non-zero.</p>
|
3,348,780 | <p>I worked through <span class="math-container">$\int \frac{e^x}{(1-e^x)^2}dx$</span> using u-substitution, but my answer, <span class="math-container">$(1-e^x)^{-1}+C$</span> is incorrect. It should be <span class="math-container">$- \ln|1-e^x|+C$</span></p>
<blockquote>
<p><span class="math-container">$$\int \frac{e^x}{(1-e^x)^2}dx$$</span></p>
</blockquote>
<p>Let <span class="math-container">$u = 1 - e^x$</span></p>
<p>Then <span class="math-container">$du = -e^x$</span> and <span class="math-container">$dx = \frac{-du}{e^x}$</span></p>
<p>So <span class="math-container">$\int \frac{e^x}{u^2} \frac{-1}{e^x}du = -\int \frac{1}{u^2}du = -\int u^{-2}du$</span></p>
<p><span class="math-container">$$-\int u^{-2} du$$</span></p>
<p><span class="math-container">$$=-(\frac{u^{-1}}{-1})+C$$</span></p>
<p><span class="math-container">$$=u^{-1}+C$$</span></p>
<p><span class="math-container">$$=(1-e^x)^{-1}+C$$</span></p>
<p>What did I do wrong?</p>
| Allawonder | 145,126 | <p>Your answer is the correct one, as you can confirm by differentiating both. In general you can always confirm whether you got the right primitive by differentiation.</p>
|
3,077,312 | <p>The proof given in my book (and I came up with as well) is:</p>
<p><a href="https://i.stack.imgur.com/H6eqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6eqf.png" alt="Proof"></a></p>
<p>However, the part that throws me off is line #3 where they do <span class="math-container">$\Sigma A_{jk} B_{ki} = \Sigma B_{ki} A_{jk}$</span></p>
<p>I understand that multiplication of real numbers is commutative and you can do the last step. However, you could've just as easily left the expression without swapping the two and arrived at the expression: <span class="math-container">$(AB)^t = A^tB^t$</span>, right?</p>
<p>What am I missing here?</p>
| Community | -1 | <p>What you have is that the <span class="math-container">$(i,j)$</span>-th element of <span class="math-container">$C^t$</span> is</p>
<p><span class="math-container">$$ \sum_k A_{jk}B_{ki} $$</span></p>
<p>where <span class="math-container">$A_{jk}$</span> is the <span class="math-container">$(j,k)$</span>-th element of <span class="math-container">$A$</span> and <span class="math-container">$B_{ki}$</span> is the <span class="math-container">$(k,i)$</span>-th element of <span class="math-container">$B$</span></p>
<p>Then because of the commutativity of the multiplication of the underlying field, you get that <span class="math-container">$A_{jk}B_{ki} = B_{ki}A_{jk}$</span>, since both <span class="math-container">$A_{jk}$</span> and <span class="math-container">$B_{ki}$</span> are just elements of your field. Then you get in total:</p>
<p><span class="math-container">$$ \begin{split}
\sum_k A_{jk}B_{ki} &= A_{j1}B_{1i} + A_{j2}B_{2i} + \dots \\
&= B_{1i} A_{j1} + B_{2i} A_{j2} + \dots \\
&= \sum_k B_{ki} A_{jk}
\end{split}$$</span></p>
<p><strong>EDIT</strong>:</p>
<blockquote>
<p>you could've just as easily left the expression without swapping the
two and arrived at the expression: <span class="math-container">$(AB)^t = A^tB^t$</span>, right?</p>
</blockquote>
<p>No, the textbooks wants to show that <span class="math-container">$(AB)^t = B^t A^t$</span>. Because generally speaking <span class="math-container">$(AB)^t \neq A^t B^t$</span>. Example:
<span class="math-container">$$\left(
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
\right)^t
=
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
^t
=
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
$$</span></p>
<p>But</p>
<p><span class="math-container">$$
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}^t
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}^t
=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
$$</span></p>
<p>And as José Carlos Santos already pointed out, <span class="math-container">$\sum_k A_{jk}B_{ki}$</span> is not the <span class="math-container">$(i,j)$</span>-th entry of <span class="math-container">$A^tB^t$</span>, but as your textbooks shows, it is the <span class="math-container">$(i,j)$</span>-th entry of <span class="math-container">$B^tA^t$</span></p>
|
59,932 | <p>So I need to make a plot of some ~2000 data points I have from a spreadsheet. I'm able to import the data just fine, it stores it like so</p>
<pre><code>sundat
</code></pre>
<blockquote>
<p>{{280.,0.082},{280.5,0.099},......{3995.,0.0087},{4000.,0.00868}}</p>
</blockquote>
<p>And I can call individual data sets or points like so</p>
<pre><code>sundat[[1]]
</code></pre>
<blockquote>
<p>{280.,0.082}</p>
</blockquote>
<pre><code>sundat[[1,1]]
</code></pre>
<blockquote>
<p>280.</p>
</blockquote>
<p>There are 2002 sets of two points of data. Now I can plot it fine using ListPlot, but the raw data I have is in messy units that I want to convert to mks units. I have the conversion factor for each column, but I have no idea how to do this. I presume I need to make a new list to apply this to, but trying something like</p>
<pre><code>data2 = {{sundat[[All,1]]*(conversion1)},{sundat[[All,2]]*(conversion2)}}
</code></pre>
<p>just totally messes up the table.</p>
<p>Any help would be appreciated.</p>
| kglr | 125 | <pre><code>sundat = {{280., 0.082}, {280.5, 0.099}, {3995., 0.0087}, {4000., 0.00868}};
conversionfactors = {2., .5};
data2 = Transpose[conversionfactors Transpose[sundat]]
</code></pre>
<p>or</p>
<pre><code>data2 = conversionfactors # & /@ sundat
</code></pre>
<p>both give</p>
<pre><code>(* {{560.,0.041},{561.,0.0495},{7990.,0.00435},{8000.,0.00434}} *)
</code></pre>
|
59,932 | <p>So I need to make a plot of some ~2000 data points I have from a spreadsheet. I'm able to import the data just fine, it stores it like so</p>
<pre><code>sundat
</code></pre>
<blockquote>
<p>{{280.,0.082},{280.5,0.099},......{3995.,0.0087},{4000.,0.00868}}</p>
</blockquote>
<p>And I can call individual data sets or points like so</p>
<pre><code>sundat[[1]]
</code></pre>
<blockquote>
<p>{280.,0.082}</p>
</blockquote>
<pre><code>sundat[[1,1]]
</code></pre>
<blockquote>
<p>280.</p>
</blockquote>
<p>There are 2002 sets of two points of data. Now I can plot it fine using ListPlot, but the raw data I have is in messy units that I want to convert to mks units. I have the conversion factor for each column, but I have no idea how to do this. I presume I need to make a new list to apply this to, but trying something like</p>
<pre><code>data2 = {{sundat[[All,1]]*(conversion1)},{sundat[[All,2]]*(conversion2)}}
</code></pre>
<p>just totally messes up the table.</p>
<p>Any help would be appreciated.</p>
| Mr.Wizard | 121 | <p>Although I prefer kguler's double <code>Transpose</code> you could also use <a href="http://reference.wolfram.com/mathematica/ref/Inner.html" rel="nofollow noreferrer"><code>Inner</code></a>:</p>
<pre><code>sundat = {{280., 0.082}, {280.5, 0.099}, {3995., 0.0087}, {4000., 0.00868}};
Inner[Times, sundat, {2., .5}, List]
</code></pre>
<blockquote>
<pre><code>{{560., 0.041}, {561., 0.0495}, {7990., 0.00435}, {8000., 0.00434}}
</code></pre>
</blockquote>
<p>Or leveraging Simon's solution from: <a href="https://mathematica.stackexchange.com/q/23395/121">How can I make threading more flexible?</a></p>
<pre><code>smartThread[sundat {2., .5}]
</code></pre>
<blockquote>
<pre><code>{{560., 0.041}, {561., 0.0495}, {7990., 0.00435}, {8000., 0.00434}}
</code></pre>
</blockquote>
<hr>
<p>A related question: <a href="https://mathematica.stackexchange.com/q/58971/121">Mapping a function over the parts of a deeply nested Array</a></p>
<p>See also: <a href="https://mathematica.stackexchange.com/q/3069/121">Elegant operations on matrix rows and columns</a></p>
|
2,466,527 | <p>Let $A$ be the matrix of $T:P_2\to P_2$ with respect to basis $B=\{v_1,v_2,v_3\}$. Find $T(v_1)$</p>
<p>$$A=\begin{bmatrix}
1 & 3 & -1 \\
2 & 0 & 5 \\
6 & -2 & 4
\end{bmatrix}$$</p>
<p>$v_1=3x+3x^2$</p>
<p>$v_2=-1+3x+2x^2$</p>
<p>$v_3=3+7x+2x^2$</p>
<hr>
<p>First part of question asks to find $T(v_1)_B$.</p>
<p>$$T(v_1)_B=[T]_{P_2}^{B}\cdot (v_1)_{P_2}=\begin{bmatrix}
1 & 3 & -1 \\
2 & 0 & 5 \\
6 & -2 & 4
\end{bmatrix}\cdot\begin{bmatrix}
0 \\
3 \\
3
\end{bmatrix}=\begin{bmatrix}
6 \\
15 \\
6
\end{bmatrix}$$</p>
<p>Where $v_1$ is our vector, $B$ is the basis, $P_2=\{1,x,x^2\}$, $A=[T]_{P_2}^{B}$</p>
<p>Assuming I did this part of the question correct, now the next part asks be to find $T(v_1)$, with respect to <strong>no basis</strong>.</p>
<hr>
<p>I think this makes sense: since we already found $T(v_1)_B$, that means that the vector $[6,15,6]$ can be written as a linear combinations of the vectors in the basis $B$. Hence we have:</p>
<p>$$6+15x+6x^2=r_1(3x+3x^2)+r_2(-1+3x+2x^2)+r_3(3+7x+2x^2)$$</p>
<p>Cleaning up a bit we get:</p>
<p>$$6+15x+6x^2=(-r_2+3r_3)1+(3r_1+3r_2+7r_3)x+(3r_1+2r_2+2r_3)x^2$$</p>
<p>Giving us the solution set:</p>
<p>$$\color{red}{-r_2+3r_3=5}, \color{green}{3r_1+3r_2+7r_3=15}, \color{blue}{3r_1+2r_2+2r_3=6}$$</p>
<p>Solving for $r_1,r_2,r_3$, we should have $T(v_1)$. Is this correct?</p>
| Patrick Stevens | 259,262 | <p>Let $c=ab+a$ to make the numbers nicer; then $x_1 = 2$ and $x_n = c x_{n-1} - 1$.</p>
<p>Then $x_2 = 2c-1, x_3 = 2c^2-c-1, x_4 = 2c^3-c^2-c-1$.</p>
<p>I formulate an inductive hypothesis! $x_i = 2c^{i-1} - c^{i-2} - \dots - c - 1$.</p>
<p>This is easy to prove inductively. You can also simplify it a bit if you like, noting that the $- c^{i-2} - \dots - c - 1$ terms are the [negation of a] sum of finitely many terms of a geometric series.</p>
|
817,934 | <p>How to prove</p>
<p>$$\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$</p>
<p>where $\sin^{-1}x$ and $\tan^{-1}x$ are inverse of trig functions. I don't know how to find the integral because of inverse of trig functions. I missed calc class twice. Please help me. Thanks.</p>
| Pranav Arora | 117,767 | <p>$$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\sin^{-1}x\int \frac{12x}{(3x^2+1)^2}\,dx -\int \left(\frac{1}{\sqrt{1-x^2}}\int \frac{12x}{(3x^2+1)^2}\,dx\right)\,dx$$
To evaluate $\displaystyle \int \frac{12x}{(3x^2+1)^2}$, use the substitution $3x^2+1=u \Rightarrow 6x\,dx=du$ to get: $\frac{-2}{3x^2+1}$, i.e
$$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\frac{-2\sin^{-1}x}{3x^2+1}+2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx$$
To evaluate the last integral, use the substitution $x=\sin\theta \Rightarrow dx=\cos\theta d\,\theta$ to get:
$$2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx=2\int \frac{1}{3\sin^2\theta+1}\,d\theta=2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta$$
$$\Rightarrow 2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta=\tan^{-1}(2\tan\theta)+C$$
Since $\sin\theta=x$, it is easy to see that $\tan\theta=x/\sqrt{1-x^2}$, hence
$$\tan^{-1}(2\tan\theta)=\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
Putting everything together, the final answer is:
$$\frac{-2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$</p>
|
119,696 | <p>Let us suppose i have two graphs for sequences A and B as follows</p>
<pre><code>a1 := {{0.9, 0.086133}, {0.086133, 0.0082432}, {0.0082432,
0.0007889}, {0.0007889, 0.0000755}, {0.0000755,
7.2256*10^-6}, {7.2256*10^-6, 6.9151*10^-7}, {6.9151*10^-7,
6.618*10^-8}, {6.618*10^-8, 6.3336*10^-9}, {6.3336*10^-9,
6.0615*10^-10}, {6.0615*10^-10, 5.801*10^-11}};
plot1 = ListPlot[a1, Joined -> True,
PlotStyle -> {Red, Thick, Dashing[{0.01}]}]
</code></pre>
<p>and </p>
<pre><code>a2 := {{0.9, 0.21797}, {0.21797, 0.052789}, {0.052789,
0.012785}, {0.012785, 0.0030963}, {0.0030963,
0.0007499}, {0.0007499, 0.00018162}, {0.00018162,
0.000043985}, {0.000043985, 0.000010653}, {0.000010653,
2.5799*10^-6}, {2.5799*10^-6, 6.24831*10^-7}};
plot2 = ListPlot[a2, Joined -> True, PlotStyle -> {Yellow, Thick}]
</code></pre>
<p>I combine both the graphs by </p>
<pre><code>Show[plot1, plot2, PlotRange -> All]
</code></pre>
<p>But cant do any things with usings Legends.
I want to use Legends to show that "Red colour box"=A and "yellow colour box"=B</p>
<p>Thanks in Advance</p>
| Sumit | 8,070 | <p>I would suggest creating the legend separately and combine with <code>Grid</code>.</p>
<pre><code>Grid[{{Show[plot1, plot2, PlotRange -> All],
SwatchLegend[{Red, Yellow}, {"plot1", "plot2"}]}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/f0uH1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f0uH1.png" alt="enter image description here"></a></p>
|
3,607,430 | <blockquote>
<p>Given that <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> are the angles of a right-angled triangle, prove that:
<span class="math-container">$$\begin{align}
\sin a\sin b\sin(a-b) &+\sin b\sin c\sin(b-c)+\sin c\sin a\sin(c-a) \\ &+\sin(a-b)\sin(b-c)\sin(c-a)=0
\end{align}$$</span></p>
</blockquote>
<p>I know I'm supposed to use the properties of polynomials for this, as this question was found on the chapter on polynomials. I've tried having these values be roots of some function but don't know how to carry it out.</p>
<p>I also know that I can consider one of the angles to be <span class="math-container">$90^\circ$</span>, so the sine of that would be <span class="math-container">$1$</span>, but that dosen't really simplify it too much.</p>
| Rezha Adrian Tanuharja | 751,970 | <p>Points <span class="math-container">$P,Q,R$</span> are collinear if and only if they satisfy the expression <span class="math-container">$m\vec{P}+(1-m)\vec{Q}=\vec{R}$</span>. Substitute <span class="math-container">$\vec{P}=\{a,0\}, \vec{Q}=\{0,\frac{3}{a}\}, \vec{R}=\{6,-1\}$</span> To obtain the following expressions</p>
<p><span class="math-container">$$
\begin{aligned}
am&=6\\
a&=\frac{6}{m}\\
\\
(1-m)\frac{3}{a}&=-1\\
(1-m)\frac{m}{2}&=-1\\
m^{2}-m-2&=0\\
m&\in\{-1,2\}
\end{aligned}
$$</span>
Substitute <span class="math-container">$m$</span> to obtain <span class="math-container">$a$</span>, we got: <span class="math-container">$\{-6,0\},\{0,-\frac{1}{2}\}$</span> or <span class="math-container">$\{3,0\},\{0,1\}$</span></p>
<p>For other point <span class="math-container">$\{x_{i},y_{i}\}$</span> and <span class="math-container">$c$</span> the intercepts are <span class="math-container">$\{a,0\}, \{0,\frac{c}{a}\}$</span>, <span class="math-container">$a=\frac{x_{i}}{m}$</span>, <span class="math-container">$m=\frac{1\pm\sqrt{1-4\frac{x_{i}y_{i}}{c}}}{2}$</span></p>
|
95,965 | <p>Joyal's <a href="http://en.wikipedia.org/wiki/Combinatorial_species" rel="nofollow">combinatorial species</a>, endofunctors in the category of finite sets with bijections $\mathbf B$ have found numerous applications. One generalisation is given by so-called "tensor species" (also "tensorial species", or, "linear species" - not to be confused with the species on totally ordered sets in the book by Bergeron, Labelle and Leroux) which are defined as functors from $\mathbf B$ into the category of finite dimensional vector spaces (say, over the complex numbers) with linear transformations $\mathbf{Vect}$.</p>
<p>I wonder whether there have been any "practical" applications of tensor species? I know of a very short list of articles dealing with them (eg. by <a href="http://www.pnas.org/content/88/21/9892.full.pdf" rel="nofollow">Méndez</a>) but hardly any spelled out examples. I wonder whether I overlooked something.</p>
<p>Note that for any combinatorial species $F$ we cann regard $F[\{1,2,\dots,n\}]$ as a finite set with an action of the symmetric group. Similarly, if $F$ is a tensor species, we can regard $F[{1,2,\dots,n}]$ as a linear representation of the symmetric group. Thus, I am mostly interested in examples that use the combinatorial operations for greater clarity of a construction.</p>
| Jan Weidner | 2,837 | <p>In the theory of algebraic operads, the language of "tensor species" is often used,
see Chapter 5 of "Algebraic Operads, Jean-Louis Loday & Bruno Vallette, Grundlehren der mathematischen Wissenschaften, Volume 346, Springer-Verlag (2012).</p>
<p>For example one can define an operad very concise as a monoid in species under a certain monoidal structure. Without this language, it takes quite a while to write down all the compatibilities with the various $S_n$ actions (Although I find it illuminating to write down an "elementary" definition never the less).</p>
<p>There are in fact many definitions in the theory of operads, which are a bit cumbersome to write down without talking about "tensor species".</p>
<p>And of course things like generating series for operads are of interest and operadic phenomena/constructions like Koszul duality give constraints/relations for their generating series.</p>
|
23,192 | <p>The question if there is an upper bound known for Brun's constant was discussed briefly here: <a href="http://gowers.wordpress.com/2009/05/22/what-is-wolfram-alpha-good-for/" rel="nofollow">http://gowers.wordpress.com/2009/05/22/what-is-wolfram-alpha-good-for/</a> but no sure answer was given. </p>
<p>So I thought I'd ask the question here. Can one get any upper bound for the sum of the reciprocals of the twin primes?</p>
| Michael Lugo | 143 | <p>Crandall and Pomerance, "Prime numbers: a computational perspective" (Google books) says that Brun's constant B, the sum of the reciprocals of the twin primes, is known to be between 1.82 and 2.15.</p>
<p><b>edited to add</b>: I'm aware that this isn't much of a citation. It would be nice if someone who has access to this book could give a better citation. I'd do it, but I'm not near a library today.</p>
|
167,013 | <p>I don't understand this behavior: why does <code>Limit[z/(z - a), z -> 0]</code> give zero and not a condition depending on <code>a</code>, provided it has not been defined before? is there a way to make it work properly? (By working properly I mean give the correct result, namely 1 if $a=0$ and 0 otherwise.) </p>
| halirutan | 187 | <p>Obviously, with the assumptions Mathematica uses for <code>a</code>, the value of it does not matter for the residue or the limit. Here are two counter-examples:</p>
<pre><code>Residue[Gamma[z] Gamma[z - 1] Gamma[z - a], {z, 0}]
(* -Gamma[-a] + 2 EulerGamma Gamma[-a] -
Gamma[-a] PolyGamma[0, -a] *)
</code></pre>
<p>and</p>
<pre><code>Limit[z^a, z -> 0]
(* ConditionalExpression[0, a > 0] *)
</code></pre>
|
159,965 | <blockquote>
<p>Find limits of a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by formula $f(x,y,x)=x+y+z$ on set $M=\left\{ (x,y,z)\in\mathbb{R}^3:x^2+y^2\le z\le 1 \right\}$. Does $f$ reaches all its limits?</p>
</blockquote>
<p>To answer the last question I need to know if $M$ is a closed and bounded set. If it is then $f$ reaches all its limits by Bolzano-Weierstrass theorem. But I'm not sure if it is closed (bounded it is I think it just simply follows from inequalities, but probably it isn't precise proof). Previously I had only sets like $\left\{(x;y)\in\mathbb{R}^2: 4x^2+y^2\le 25\right\}$ and a reason for closedness was rather simple. I always used to state given set as an inverse image by continuous function of closed set. But in situation with $M$ I can't find similar explanation.</p>
<p>Finding limits seems to be hard too. I always used partial derivatives to find extremes in the interior of set and parameterization to explore boundary of set.
But here I'm even not sure what will be interior and what will be boundary of $M$.
Can anybody help?</p>
| H. Kabayakawa | 32,428 | <p>$M$ is like a cone with the vertex in $(0,0,0)$ and the base in the circumference $\{x^2+y^2=1, z=1\}$, then $M$ is a compact set in $\mathbb{R}^3$. The function is lineal with gradient $(1,1,1)$ and $M$ is a convex set because the surface $z=x^2+y^2$ is a revolution paraboloid. The paraboloid has normal $(2x,2y,-1)$ in any point $(x,y,z)$. The minimum is in the point with $\displaystyle\frac{2x}{1}=\frac{2y}{1}=\frac{-1}{1}$, that is $\displaystyle (-\frac{1}{2},-\frac{1}{2},\frac{1}{2})$ . The maximum is obtained for $z=1$. The projection of the gradient over the plane $z=1$ is $(1,1,0)$. Then the maximum is obtained for $x=y$, and hence $x=y=\frac{\sqrt 2}{2}$. The maximum value is $f(\frac{\sqrt 2}{2},\frac{\sqrt 2}{2},1)=\sqrt 2 +1$.</p>
|
159,965 | <blockquote>
<p>Find limits of a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by formula $f(x,y,x)=x+y+z$ on set $M=\left\{ (x,y,z)\in\mathbb{R}^3:x^2+y^2\le z\le 1 \right\}$. Does $f$ reaches all its limits?</p>
</blockquote>
<p>To answer the last question I need to know if $M$ is a closed and bounded set. If it is then $f$ reaches all its limits by Bolzano-Weierstrass theorem. But I'm not sure if it is closed (bounded it is I think it just simply follows from inequalities, but probably it isn't precise proof). Previously I had only sets like $\left\{(x;y)\in\mathbb{R}^2: 4x^2+y^2\le 25\right\}$ and a reason for closedness was rather simple. I always used to state given set as an inverse image by continuous function of closed set. But in situation with $M$ I can't find similar explanation.</p>
<p>Finding limits seems to be hard too. I always used partial derivatives to find extremes in the interior of set and parameterization to explore boundary of set.
But here I'm even not sure what will be interior and what will be boundary of $M$.
Can anybody help?</p>
| Christian Blatter | 1,303 | <p>Your set $M$ is defined by inequalities of the form $g_i(x,y,z)\geq0$ with continuous $g_i:{\mathbb R}^3\to{\mathbb R}$, therefore it is closed; and all $(x,y,z)\in M$ satisfy $x^2+y^2\leq 1$ as well as $0\leq z\leq 1$, therefore $M$ is bounded. (In fact $M$ looks like an inverted sugar cone.) It follows that any continuous function $f:M\to{\mathbb R}$ assumes a global minimum as well as a global maximum on $M$. </p>
<p>These two extremal points $(\xi,\eta,\zeta)$ are lying (a) in the interior of $M$, or (b) on the parabolic part $z=x^2+y^2$ of the boundary of $M$, or (c) in the interior of the top surface $z=1$ of $M$, or (d) on the bounding circle $\phi\mapsto(\cos \phi,\sin \phi,1)$ of the top surface.</p>
<p>In the case (a) one would necessarily have $\nabla f(\xi,\eta,\zeta)=0$. As $\nabla f(x,y,z)\equiv(1,1,1)$ there is no extremal point of this kind.</p>
<p>For (b) we look at the pullback $$f_1(x,y):=f(x,y,x^2+y^2)=x+y+x^2+y^2$$ of $f$ to the paraboloid and its two-dimensional gradient $\tilde f_1(x,y)=(1+2x,1+2y)$. This gradient vanishes at $\bigl(-{1\over2},-{1\over2}\bigl)$ which is a point in the interior of the unit circle. The corresponding point $p$ on the boundary $\partial M$ is $p=\bigl(-{1\over2},-{1\over2},{1\over2}\bigr)$, and one has $f(p)=-{1\over2}$.</p>
<p>For (c) we look at the pullback $$f_2(x,y):=f(x,y,1)=x+y+1$$
of $f$ to the top surface $z=1$. As the two-dimensional gradient $\tilde f_2(x,y)\equiv(1,1)$ is nonzero there are no extremal point of this kind.</p>
<p>For (d) we look at the pullback
$$f_3(\phi):=f(\cos\phi,\sin\phi, 1)=\cos\phi+\sin\phi+1=\sqrt{2}\sin\bigl(\phi+{\pi\over4}\bigr)+1$$
of $f$ to the boundary circle of the top surface. The function $f_3$ assumes its minimum $1-\sqrt{2}>-{1\over2}$ for $\phi=-{3\pi\over4}$ and its maximum $1+\sqrt{2}$ for $\phi={\pi\over4}$. The latter $\phi$-value corresponds to the point $q=\bigl({1\over\sqrt{2}},{1\over\sqrt{2}},1\bigr)\in\partial M$.</p>
<p>All in all we have found that $f$ takes on $M$ the minimal value $-{1\over2}$ at the point $p$ and the maximal value $1+\sqrt{2}$ at the point $q$.</p>
|
3,608,114 | <p>Consider the example where I have a matrix <span class="math-container">$\mathbf{D}$</span> in <span class="math-container">$-1/1$</span> coding with <span class="math-container">$5$</span> columns,</p>
<p><span class="math-container">$$D = \begin{bmatrix}-1&-1&-1&1&1\\1&-1&-1&-1&1\\-1&1&-1&-1&-1\\1&1&-1&1&-1\\
-1&-1&1&1&-1\\1&-1&1&-1&-1\\-1&1&1&-1&1\\1&1&1&1&1\end{bmatrix}$$</span></p>
<p>We see that the fourth and fifth columns are combinations of the first three columns so that if we label the columns <span class="math-container">$a,b,c,d,e$</span> we can say that <span class="math-container">$d=a*b$</span> and <span class="math-container">$e=b*c$</span>. We can separate the columns in two groups: <span class="math-container">$a,b,c$</span> are <strong>basic columns</strong> and <span class="math-container">$e,d$</span> are <strong>added columns</strong>. Furthermore we can call the relations that define <span class="math-container">$e$</span> and <span class="math-container">$d$</span> as a combination of <span class="math-container">$a,b,c$</span> <strong>defining contrasts</strong>.</p>
<p>My question is this:
Is there a way to determine more generally (for a larger matrix or a matrix with different defining contrasts) which columns are combinations of the others and</p>
| asd.123 | 613,151 | <p>At the comments @Rodrigo de Azevedo said you can apply <a href="https://en.m.wikipedia.org/wiki/Gaussian_elimination" rel="nofollow noreferrer">Gaussian elimination</a> which is quite intuitive method to apply but I personally suggest that if you want to find the linear dependence relation between rows or columns of a square matrix, taking the <a href="https://en.m.wikipedia.org/wiki/Determinant" rel="nofollow noreferrer">determinant</a> of the matrix then if it is 0 then you may want to apply gaussian elimination. It will reduce the work.</p>
|
64,881 | <p>I am having trouble with this problem from my latest homework.</p>
<p>Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.</p>
<p>Any and all help would be appreciated.</p>
| Bruno Joyal | 12,507 | <p>Since $x$ and $y$ are positive, we can write them as $x=u^2$, $y=v^2$. Then</p>
<p>$$(u-v)^2 \geq 0 \Rightarrow u^2 + v^2 \geq 2uv$$</p>
<p>which is precisely it.</p>
|
69,961 | <p>I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s.</p>
<p>$$S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}$$</p>
<p>I want to check whether that would be:
0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on.</p>
<p>Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set?</p>
| André Caldas | 17,092 | <p>First, it will be easier to determine the set
$$I = \{3k+5j | k,j \in \mathbb{Z}\}.$$</p>
<p>This set is an <a href="http://en.wikipedia.org/wiki/Ideal_%28ring_theory%29" rel="nofollow" title="ideal on the Wikipedia">ideal</a> over $\mathbb{Z}$.
That is, if $a,b \in I$ then $a+b \in I$ ($I+I \subset I$),
and if
$a \in I$ and $z \in \mathbb{Z}$, then $za \in I$
($\mathbb{Z}I \subset I$).</p>
<p>The <a href="http://en.wikipedia.org/wiki/Euclidean_algorithm" rel="nofollow" title="Euclidean algorithm on Wikipedia">Euclidean algorithm</a> allows on to prove that all ideals
over $\mathbb{Z}$ are in fact the set of multiples of a sole number.
Take $a \in I$, the smallest positive element of $I$,
and let $b \in I$ be any other element.
The Euclidean algorithm shows that there is a $q \in \mathbb{Z}$,
and an integer $0 \leq r < a$ such that
$$b = qa + r.$$
But since $I$ is an ideal, $r = b - qa \in I + \mathbb{Z}I \subset I$.
From the minimality of $a$, we conclude that $r = 0$.
That is, $I = a\mathbb{Z}$.</p>
<p>It is evident that all elements of
$I$, including $a$, are multiples of $M = \mathrm{GCD}(3,5) = 1$.
And again, the <a href="http://en.wikipedia.org/wiki/Euclidean_algorithm" rel="nofollow" title="Euclidean algorithm on Wikipedia">Euclidean algorithm</a> shows that
$M \in I$.
That is, $a = M$.
In the case of $3$ and $5$, $a = 1$.
That is, $I = \mathbb{Z}$.</p>
<p>So, what we know is that every integer can be written as
$3k + 5j$ for $k,j \in \mathbb{Z}$.
Notice that we can add and subtract multiples of $15$:
$3k + 5j = 3(k+n5) - 5(j-n3)$.
Show that if $3k + 5j \geq 15$, then you can add and subtract
a multiple of $15$ in order to end up with $j,k \geq 0$.
That is, any integer greater then or equal to $15$ is in $S$.
Now, you just have to check the integers between $0$ and $15$.</p>
|
4,627,334 | <p>To my understanding that a primitive triple <span class="math-container">$x$</span> and <span class="math-container">$y$</span> can be written as <span class="math-container">$x = q^2 - p^2$</span> while <span class="math-container">$y=2pq$</span> for relatively prime opposite parity <span class="math-container">$q > p$</span> then the area can be calculated as: <span class="math-container">$pq(q^2 - p^2) = pq(q+p)(q-p)$</span>.
Am I missing something obvious that helps prove that a Pythagorean Triple triangle cannot have an odd area?</p>
| Joseph Harrison | 1,131,061 | <p>As you say, we can write <span class="math-container">$x = q^2 - p^2$</span> and <span class="math-container">$y = 2pq$</span> where <span class="math-container">$p, q$</span> are relatively prime with different parity. Now the area of the triangle is half of <span class="math-container">$xy$</span>. If this area is odd, then <span class="math-container">$xy \equiv 2 \; (\textrm{mod } 4)$</span> and
<span class="math-container">\begin{align}
2pq(q^2 - p^2) \equiv 2 \; (\textrm{mod } 4)
\end{align}</span>
which implies <span class="math-container">$pq(q^2 - p^2) \equiv 1 \; (\textrm{mod } 2)$</span>. This is impossible because one of <span class="math-container">$p$</span> or <span class="math-container">$q$</span> must be even.</p>
<p>Let me know if I have made any errors :)</p>
|
90,940 | <p>It seems known that the category of hypergraphs is a topos.
I am looking for any reference here, or just a statement of this in the literature, but can't find anything. There is one paper </p>
<blockquote>
<p>A category-theoretical approach to hypergraphs,
W. Dörfler and D. A. Waller, ARCHIV DER MATHEMATIK, Volume 34, Number 1, 185-192, <a href="https://doi.org/10.1007/BF01224952" rel="nofollow noreferrer">DOI:10.1007/BF01224952</a>, 1980</p>
</blockquote>
<p>which might contain information about that, but I don't have access to this paper (and it might take some time to get a copy, likely a paper-copy).</p>
<p>By a hypergraph I mean here a triple $(V,E,h)$, where $V$, $E$ are arbitrary sets, while $h$ is a map from $E$ to the set of finite subsets of $V$ (so $V$ is the set of vertices, $E$ the set of hyperedge-labels, and $h$ yields the hyperedge of a hyperedge-label). Morphisms are pairs $a: V \rightarrow V'$, $b: E \rightarrow E'$, which fulfill the usual commutativity condition.</p>
| YKY | 11,555 | <p>According to this presentation:</p>
<blockquote>
<p>Will Grilliette and Lucas Rusnak, <em>Natural Generalizations of Graphs Part II: Commas, Topoi, & Homomorphisms</em>,
Discrete Mathematics Seminar, Texas State University (2017) DOI:<a href="https://doi.org/10.13140/RG.2.2.13627.92961" rel="nofollow noreferrer">10.13140/RG.2.2.13627.92961</a></p>
</blockquote>
<p>Quote:</p>
<blockquote>
<p>The classical categories H of hypergraphs and M of (undirected) multigraphs arise naturally as a comma category using the power-set functor P. However, P is well-known not to preserve limit processes, and both H and M fail to be Cartesian closed as a result, among other issues. On the other hand, the category Q of quivers arises equivalently as both a comma category and a functor category. Consequently, Q can be represented as a topos of presheaves, inheriting a significant amount of structure immediately. </p>
</blockquote>
<p>note: quiver = directed multigraph.</p>
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2,262,011 | <p>This might be a somewhat stupid question, but I've been wondering if it is possible to define some other topology on $\mathrm{Spec} (A)$ other than Zariski topology in a way that it has some interesting properties as well.</p>
<p>First of all, I am new as this is my first encounter with anything close or related to algebraic geometry, so be easy on me =).</p>
<p>And second, what I'd like to know is if, for example, there is a topology on $\mathrm{Spec} (A)$ such that, say, $\mathrm{Spec} (A)$ is Hausdorff or has any other nice properties (connectedness, compactness, etc...), or why if such a topology exists, isn't as interesting as Zariski topology. </p>
<p><strong>Note:</strong> I am aware of a "similar" question <a href="https://math.stackexchange.com/questions/161884/why-zariski-topology">here</a>. However, I'm not interested that much in why Zariski topology is important since I think I understand how it arises naturally.</p>
| Tsemo Aristide | 280,301 | <p>There is a generalization of the notion of topology called a Grothendieck topology. Grothendieck has defined many examples of Grothendieck topolgies in the category of schemes for example the Etale topology which is one notion used in the proof of Weil conjectures.</p>
<p><a href="https://en.wikipedia.org/wiki/Grothendieck_topology" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Grothendieck_topology</a></p>
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