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508,550 |
<p>I read in an article the Euclidean distance formula can be estimated with about 6% relative error with the following formula. Would you please <strong>why</strong> this is true and <strong>where</strong> can I find such estimations? Is it possible to <strong>extend</strong> it for higher dimensions?</p>
<p>$d(a,b) = \max(|a_x-b_x|,|a_y-b_y|) + 0.365 \times \min(|a_x-b_x|,|a_y-b_y|) $</p>
<p>The original formula in the article "SOLVING LARGE VEHICLE ROUTING AND SCHEDULING PROBLEMS IN SMALL CORE", by Bordin: </p>
<p><img src="https://i.stack.imgur.com/5Gln7.png" alt="enter image description here"></p>
|
Did
| 6,179 |
<p>Let $u= \max(|a_x-b_x|,|a_y-b_y|)$ and $v=\min(|a_x-b_x|,|a_y-b_y|)$, then algebra shows that
$$
d(a,b)=u+r(u,v)\cdot v,\qquad r(u,v)=\frac{v}{u+\sqrt{u^2+v^2}}.
$$
Why one would think that $r(u,v)\approx0.365$ always, escapes me. In fact, $r(u,v)$ can be anywhere in the interval $[0,\sqrt2-1)\approx[0,0.414)$ and $r(u,v)\approx0.365$ if and only if $v\approx0.8422\cdot u$. </p>
<p>Perhaps vehicle routing involves mainly points $(a,b)$ such that $v\approx0.8422\cdot u$ holds...</p>
|
1,986,247 |
<p>The nth Catalan number is :
$$C_n = \frac {1} {n+1} \times {2n \choose n}$$
The problem 12-4 of CLRS asks to find :
$$C_n = \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) $$
And Stirling's approximation is:
$$n! = \sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ \Theta \left(\frac {1} {n}\right) \right)} $$
So, the nth catalan number becomes :
$$C_n = \frac {2n!}{(n+1)(n!)^2} $$
That, after applying Stirling's approximation becomes:
$$C_n = \left( \frac {1}{1+n} \right) \left( \frac {4^n}{\sqrt{\pi n}} \right) \frac {1}{\left( 1+\Theta \left(1/n\right) \right)}$$
And then, it becomes hopeless. The Asymptotic bound comes in the denominator, not in the numerator.<br>
What should be done now? </p>
<p>Any help appreciated.<br>
Moon </p>
|
epi163sqrt
| 132,007 |
<p>We can use Stirling's approximation formula
\begin{align*}
n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)
\end{align*}
to prove:</p>
<blockquote>
<p>The following is valid</p>
<p>\begin{align*}
C_n=\frac{1}{n+1}\binom{2n}{n}= \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) \tag{1}
\end{align*}</p>
<p>We obtain using (1)
\begin{align*}
\frac{1}{n+1}\binom{2n}{n}&=\frac{1}{n+1}\cdot\frac{(2n)!}{n!n!}\\
&=\frac{1}{n+1}\cdot\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}\left(1+O\left(\frac{1}{n}\right)\right)\\
&\qquad
\cdot \left(\frac{1}{\sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ O \left(\frac {1} {n}\right) \right)}}\right)^2\\
\\
&=\frac{1}{n+1}\cdot\frac{4^n}{\sqrt{\pi n}}\cdot
\frac{\left(1+O\left(\frac{1}{n}\right)\right)}{\left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)}\tag{2}\\
&=\frac{1}{n\left(1+O\left(\frac{1}{n}\right)\right)}\cdot\frac{4^n}{\sqrt{\pi n}}
\left(1+O\left(\frac{1}{n}\right)\right)^3\tag{3}\\
&=\frac{1}{n}\cdot\frac{4^n}{\sqrt{\pi n}}
\left(1+O\left(\frac{1}{n}\right)\right)^4\\
&=\frac{4^n}{\sqrt{\pi}n^{3/2}}
\left(1+O\left(\frac{1}{n}\right)\right)\tag{4}\\
\end{align*}
and the claim follows.</p>
</blockquote>
<p><em>Comment:</em></p>
<ul>
<li><p>In (2) we do some cancellation</p></li>
<li><p>In (3) we use the geometric series expansion
\begin{align*}
\frac{1}{1+O\left(\frac{1}{n}\right)}=1+O\left(\frac{1}{n}\right)
\end{align*}</p></li>
<li><p>In (4) we use
\begin{align*}
\left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)=1+O\left(\frac{1}{n}\right)
\end{align*}</p></li>
</ul>
|
3,909,972 |
<p>I used Photomath and Microsoft Math to compute an equation, but they gave me two different results (-411 and -411/38) Why did that happen and which is the correct answer?</p>
<p><a href="https://i.stack.imgur.com/egt5A.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/egt5A.jpg</a>
<a href="https://i.stack.imgur.com/RIHb0.png" rel="nofollow noreferrer">https://i.stack.imgur.com/RIHb0.png</a></p>
|
Robby the Belgian
| 19,298 |
<p>Note the first step in the second solution. It starts by multiplying both sides by 38.</p>
<p>Anyway, this is not a good way to do arithmetic. You are trying to find a solution that makes <span class="math-container">$\frac{411}{38}$</span> equal to <span class="math-container">$0$</span>, which cannot be done.</p>
<p>Maybe you could try to solve <span class="math-container">$x = \frac{16}{19} - \frac{5 \times 81}{38} - 1$</span>, which should give you the solution <span class="math-container">$x = -\frac{411}{38}$</span>.</p>
<p>Or, depending on the calculator, just type in <span class="math-container">$\frac{16}{19} - \frac{5 \times 81}{38} - 1$</span> (without the <span class="math-container">$=0$</span>). That should work.</p>
|
396,440 |
<p>Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$</p>
<p>I am trying to show that $b$ is the absolute minimum of $f$. </p>
<p>I proceeded as follows:</p>
<p>$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$</p>
<p>Now I will look for critical points by searching for the zeros of this function.</p>
<p>$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$. </p>
<p>Now I can see that $b$ is a critical point. </p>
<p>How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$</p>
<p>But this could not be equal to $b$, where did I go wrong?</p>
|
Community
| -1 |
<p>Note that $$\dfrac{p}{q(p-1)}=1$$ since $\dfrac1p + \dfrac1q = 1$. This is because, we have $\dfrac1q = 1 - \dfrac1p = \dfrac{p-1}p \implies \dfrac{p}{q(p-1)} = 1$.</p>
|
223,582 |
<p>Maps $g$ maps $\left\{1,2,3,4,5\right\}$ onto $\left\{11,12,13,14\right\}$ and $g(1)\neq g(2)$. How many g are there.</p>
<p><strong>My answer</strong>:
I transformed the question to a easy-understand way and find out the solution.
Consider there are five children and four seats. Two of them are willing sitting together but only two of them never seat together.</p>
<p>$$\left(\begin{pmatrix}
5 \\
2
\end{pmatrix}-1\right)*4!=456$$</p>
<p>However the answer is 216. I don't know what's wrong.</p>
<p>Could you please help me find out what's wrong or give a right way to solve the problem?</p>
<p>Thanks!</p>
|
Mick
| 42,351 |
<ol>
<li>From AC, find its midpoint F.</li>
<li>Draw the circle using F as center and FA (or FC) as radius.</li>
<li>The point(s) of intersection of the circles is D. </li>
</ol>
|
2,788,276 |
<p>Let$\ f_n (x)=n^2x(1-x)^n$ I need to prove that$\ f_n→0$ in the interval$\ [0,1]$.</p>
<hr>
<p>Let$\ f_n(x) = nx^n$ prove that$\ f_n→0$ in the interval$\ [0,1)$.</p>
<p>For both of these sequences I tried the following:</p>
<p>By taking the function$\ f(x)=0$ we can see that</p>
<p>$$\lim_{n\rightarrow\infty}f_n(x) = 0$$</p>
<p>for both of the sequences, but I don't know if this is the correct way of solving both problems and I have my doubts if this even means that$\ f_n→0$, I'm thinking that what I did before actually implies that$\ f_n→f$.</p>
<p>Help would be greatly appreciated.</p>
|
Youngsu
| 84,157 |
<p>The existence of $f(X)$ guarantees that every element in $R$ is integral over $\mathbb{Z}/(n)$. In other words, $\mathbb{Z}/(n) \subset R$ is an integral extension. So, $\dim R = \dim \mathbb{Z}/(n) = 0$.</p>
|
3,069,684 |
<p>My question goes like this</p>
<p>If 5a+4b+20c=t, then what is the value of t for which the line ax+by+c-1=0 always passes through a fixed point?</p>
<p>I tried but couldn't solve it so I looked at the solution. The solution says that the equation has 2 independent parameters. I get that. If we choose a and b, c automatically gets defined. Hence c is not really independent.</p>
<p>In the next line it says that it can pass through a fixed point if there is only one independent parameters. This is where I got stuck. I tried a lot but can't understand their logic. </p>
<p>Then they defined a relation between a/(c-1) and b/(c-1) to get the answer which is t=20.</p>
<p>Please help me because I feel understanding this would really clear my concepts and may prove very beneficial later.</p>
|
amd
| 265,466 |
<p>I’ll speak to your first question, which wasn’t really addressed in the other answer. </p>
<p>A line in the plane normally has two degrees of freedom. You’re probably used to specifying them via slope and <span class="math-container">$y$</span>-intercept (with an exception for vertical lines). For lines that go through a fixed point, you can still make an arbitrary choice of slope, but that choice determines the line’s <span class="math-container">$y$</span>-intercept, so you’ve taken away a degree of freedom by requiring that the line pass through a certain point. To put it differently, there is a one-parameter family of lines through any given point. This means that, after substituting for <span class="math-container">$c$</span> in <span class="math-container">$ax+by+c=1$</span>, the remaining unknown coefficients <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are not independent, and this equation tells you how they’re related.</p>
|
3,634,416 |
<p>First of all, English is not my native language, but Chinses is. I tried to spilt the integration interval into 2 pieces: <span class="math-container">$ [0, 1-1/n] $</span> and <span class="math-container">$ [1-1/n, 1] $</span>. In both intervals I use the mean value theorem:
<span class="math-container">$$
\int_{0}^{1-1/n}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\xi_{n}^{n}}\left( 1-\frac{1}{n} \right), \qquad \text{and} \qquad \int_{1-1/n}^{1}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\eta_{n}^{n}}\frac{1}{n},
$$</span>
where <span class="math-container">$ \xi_{n}\in(0, 1-1/n), \eta_{n}\in(1-1/n, 1) $</span>.I found that the latter formula has a limit of <span class="math-container">$ 0 $</span> when <span class="math-container">$ n\to\infty $</span>. However I can't handle the previous formula. Does anyone has some thoughts? </p>
|
Community
| -1 |
<p>Consider <span class="math-container">$\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx$</span> and <span class="math-container">$\int_{1-n^{-1/2}}^1\frac1{1+x^n}\,dx$</span> instead. The second integral is still bounded above by <span class="math-container">$n^{-1/2}$</span>.</p>
<p>For the first integral, <span class="math-container">$$1\ge \int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx=\frac{1}{1+\xi_n^{n}}(1-n^{-1/2})\ge \frac1{1+(1-n^{-1/2})^n}(1-n^{-1/2})$$</span></p>
<p>However, <span class="math-container">$\left(1-n^{-1/2}\right)^n=\left(\left(1-n^{-1/2}\right)^{n^{1/2}}\right)^{n^{1/2}}\stackrel{n\to\infty}{\longrightarrow} \left[\left(e^{-1}\right)^{\infty}\right]=0$</span>, therefore <span class="math-container">$$\liminf_{n\to\infty}\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx\ge \frac1{1+0}(1-0)=1$$</span></p>
|
2,629,133 |
<p>In keno, the casino picks 20 balls from a set of 80 numbered 1 to 80. Before the draw is over, you are allowed to choose 10 balls. What is the probability that 5 of the balls you choose will be in the 20 balls selected by the casino?</p>
<p>My attempt: The total number of combinations for the 20 balls is $80\choose20$. However, I get stuck at the numerator. I thought it will be $\binom{80}{10}\binom{10}5$ but that's wrong. </p>
<p>Thanks.</p>
|
bames
| 464,998 |
<p>Another way to think about this is to realize that the casino must choose $5$ balls from the $10$ that you chose and $15$ balls from the $70$ that you didn't choose. So:
$$P = \frac{\binom{10}{5} \binom{70}{15}}{80\choose 20} \approx 0.0514...$$</p>
|
1,146,050 |
<p>given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.</p>
<p>Need to find the min value of $f(x)$.</p>
<p>I know it can be easily done by polynomial division but my question is if there's another way</p>
<p>(more elegant maybe) to find the min? </p>
<p><strong>About my way</strong>: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$. (long division)</p>
<p>$x_{min}=\frac{-b}{2a}=\frac{1}{2}$. (when $ax^2+bx+c=0$)</p>
<p>So $f(0.5)=0.5^2-0.5+1=\frac{3}{4}$</p>
<p>Thanks. </p>
|
Community
| -1 |
<p>$$f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$$
$$f'(x) = 2x-1=0,x=\frac12$$
$$f''(\frac12)=2\gt0\text{ hence this is a local minimum by the second derivative test}$$</p>
|
2,573,458 |
<p>Given $n$ prime numbers, $p_1, p_2, p_3,\ldots,p_n$, then $p_1p_2p_3\cdots p_n+1$ is not divisible by any of the primes $p_i, i=1,2,3,\ldots,n.$ I dont understand why. Can somebody give me a hint or an Explanation ? Thanks.</p>
|
Sri-Amirthan Theivendran
| 302,692 |
<p>Towards a contradiction, if some prime $p_i$ ($1\leq i\leq n$) divides $p_1p_2p_3\cdots p_n+1$, then because $p_i$ also divides $p_1p_2p_3\cdots p_n$, it follows that $p_i\mid 1$ (the difference), a contradiction. </p>
|
2,877,080 |
<p>Let A denote a commutative ring and let e denote an element of A such that $e^2 = e$.
How to prove that $eA \times (1 - e)A \simeq A$?
I thought that $\phi: A \mapsto eA \times (1 - e)A, \ \phi(a) = (ea, (1-e)a)$ is an isomorphism but I don't know how to prove that $\phi$ is a bijection.</p>
|
gandalf61
| 424,513 |
<p>Remember that your primary objective in this model lecture is to communicate clearly and effectively to your scenario audience (first year analysis students), not to impress you actual (experienced) audience with the breadth and depth of your knowledge.</p>
<p>I imagine your biggest challenge when delivering your model lecture will be time management. The various versions of the MVT, with motivation, background, generalisations, connections, applications etc. is an enormous topic. You could easily fill a two or three hour lecture. But I expect you only have 30 or 45 minutes - and you will need to start from basics.</p>
<p>So focus on getting the essentials in first. Motivation is key - why do we want to prove the MVT ? Why is it not obvious ? Can't we just draw a graph ? Remember that results like the MVT <em>look</em> obvious at first glance to students who are used to working with polynomials and other well behaved, smooth functions. A few well chosen examples where the MVT property does <em>not</em> hold will help with motivation.</p>
<p>Next go through a simple and clear proof. Outline the overall proof strategy. Then walk through the proof step by step. Show how each of the conditions in the theorem statement is used in the proof. Don't forget that your scenario audience may not be completely comfortable with formal proofs, so take time over this.</p>
<p>You can then mention generalisations, applications etc. briefly at the end <em>if you have time</em>. And provide some further reading references/links so that the more able/interested students have something to take away.</p>
|
169,531 |
<p>Let me preface this by saying that I have essentially no background in logic, an I apologize in advance if this question is unintelligent. Perhaps the correct answer to my question is "go look it up in a textbook"; the reasons I haven't done so are that I wouldn't know which textbook to look in and I wouldn't know what I'm looking at even if I did.</p>
<p>Anyway, here's the setup. According to my understanding (i.e. Wikipedia), Godel's first incompleteness theorem says that no formal theory whose axioms form a recursively enumerable set and which contain the axioms for the natural numbers can be both complete and consistent. Let $T$ be such a theory, and assume $T$ is consistent. Then there is a "Godel statement" $G$ in $T$ which is true but cannot be proven in $T$. Form a new theory $T'$ obtained from $T$ by adjoining $G$ as an axiom. Though I don't know how to prove anything it seems reasonably likely to me that $T'$ is still consistent, has recursively enumerable axioms, and contains the axioms for the natural numbers. Thus applying the incompleteness theorem again one deduces that there is a Godel statement $G'$ in $T'$.</p>
<p>My question is: can we necessarily take $G'$ to be a statement in $T$? Posed differently, could there be a consistent formal theory with recursively enumerable axioms which contains arithmetic and which can prove every true <em>arithmetic</em> statement, even though it can't prove all of its own true statements? If this is theoretically possible, are there any known examples or candidates?</p>
<p>Thanks in advance!</p>
|
Thomas Andrews
| 7,933 |
<p>If you had a theory, $T$, with a recursively enumerable axiom set, and it completely resolved all arithmetic questions, then, if the set of arithmetic questions was recursively enumerable in $T$, you could recursively enumerate all the arithmetic statements in $T$ which are provable in $T$, and hence you'd have an r.e. enumeration of a set of complete and consistent arithmentical statements.</p>
<p>Therefore, if such a $T$ existed, the map of statements from Peano arithmetic statements to the equivalent statements in $T$ would have to be non-recursive.</p>
<p>One nice thing about having a recursively enumerable set of axioms is that we can enumerate the set of all proofs in our theory, $T$. If we have a recursively enumerable set of statements in such a theory, then the subset of such statements that can be proven is also recursively enumerable.</p>
|
2,304,318 |
<p>Let $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be two real series, where we have $\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = M >0$. Show that either one of these options happen:</p>
<ol>
<li>Both series converge</li>
<li>Both series diverge</li>
</ol>
<p>I have no clue on how to solve this. Can someone help?</p>
<p><strong>EDIT</strong>: After the comments, I think this can only be shown true if we assume $a_n >0$ and $b_n>0$ for all $n\in\mathbb{N}$</p>
|
Martin Argerami
| 22,857 |
<p>Clue: for $\varepsilon=M/2$ and large enough $n$, $$\frac M2\,b_n\leq a_n\leq \frac{3M}2\,b_n.$$</p>
|
1,513,373 |
<p>Let M be a cardinal with the following properties:<br>
- M is regular<br>
- $\kappa < M \implies 2^\kappa < M$<br>
- $\kappa < M \implies s(\kappa) < M$ where $s(\kappa)$ is the smallest strongly inaccessible cardinal strictly greater than $\kappa$ </p>
<p>My question is: Is M a Mahlo cardinal ? If so, how does the definition above connect to the usual definition in terms of stationary sets ?</p>
<p>Motivation: My intuition about a Mahlo cardinal is that it you cannot reach it by taking unions, power sets or "the next inaccessible cardinal", which is my definition above.
My worry is, I might have arrived at something much smaller than Mahlo.</p>
|
Cosmonut
| 287,070 |
<p>Thanks a lot for your answers, Wojowu and Andreas.
I find I am not able to simply reply with a comment while being a guest, so I am writing it this way.</p>
<p>The mental picture I am getting of Mahlo versus inaccessibles is somewhat like $\aleph_1$ versus the Veblen hierarchy of countable ordinals - you keep recursively strengthening definitions and taking fixed points, but just never get there. Don't know how strong the analogy is, though.</p>
|
2,196,539 |
<p>I'm having a complete mind blank here even though i'm pretty sure the solution is relatively easy.</p>
<p>I need to make X the subject of the following equation:</p>
<p>$$AB - AX = X $$</p>
<p>All i've done so far is:
$$A(B-X) = X$$
$$B-X = A^{-1} X$$</p>
<p>Not sure if thats right?</p>
<p>Thanks in advance.</p>
|
rookie
| 346,911 |
<p>Whatever you have written is correct if inverse of A exists.</p>
<p>Hint for another way of writing an expression for $X$: $AB = (I+A)X.$</p>
|
211,803 |
<p>I ended up with a differential equation that looks like this:
$$\frac{d^2y}{dx^2} + \frac 1 x \frac{dy}{dx} - \frac{ay}{x^2} + \left(b -\frac c x - e x \right )y = 0.$$
I tried with Mathematica. But could not get the sensible answer. May you help me out how to solve it or give me some references that I can go over please? Thanks.</p>
|
doraemonpaul
| 30,938 |
<p>$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{ay}{x^2}+\left(b-\dfrac{c}{x}-ex\right)y=0$</p>
<p>$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)y=0$</p>
<p>Let $y=\dfrac{u}{\sqrt{x}}$ ,</p>
<p>Then $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x\sqrt{x}}$</p>
<p>$\dfrac{d^2y}{dx^2}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}$</p>
<p>$\therefore\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}+\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x^2\sqrt{x}}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)\dfrac{u}{\sqrt{x}}=0$</p>
<p>$\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)\dfrac{u}{\sqrt{x}}=0$</p>
<p>$\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)u=0$</p>
<p>The above ODE is hypergeometric only when the special cases below:</p>
<p>$1$. $e=0$</p>
<p>$2$. $b=0$ and $c=0$</p>
<p>$3$. $c=0$ and $a=\dfrac{1}{4}$</p>
<p>Other than the above special cases the above ODE is not hypergeometric.</p>
<p>Unfortunately it also not belongs to any <a href="http://dlmf.nist.gov/31.12" rel="nofollow">confluent forms of Heun’s equation</a>.</p>
<p>Therefore to solve the above ODE generally is extremely difficult.</p>
<p>One of the main reason is that the coefficient of $u$ has too many terms or contains too high power terms. The similar situation also appear in <a href="http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Titchmarsh" rel="nofollow">Titchmarsh's ODE</a>.</p>
|
211,803 |
<p>I ended up with a differential equation that looks like this:
$$\frac{d^2y}{dx^2} + \frac 1 x \frac{dy}{dx} - \frac{ay}{x^2} + \left(b -\frac c x - e x \right )y = 0.$$
I tried with Mathematica. But could not get the sensible answer. May you help me out how to solve it or give me some references that I can go over please? Thanks.</p>
|
Przemo
| 99,778 |
<p>Consider a slightly more general ODE.
<span class="math-container">\begin{equation}
\frac{d^2 y(x)}{d x^2} + \frac{1}{x} \frac{d y(x)}{d x} + \left(-\frac{a}{x^2} + b - \frac{c}{x} - e x + e_1 x^2 \right) y(x)=0
\end{equation}</span>
If we write:
<span class="math-container">\begin{equation}
y(x)=x^{\sqrt{a}}\cdot \exp\left( -\frac{\imath}{2 \sqrt{e_1}} x(-e+e_1 x)\right) \cdot v(\frac{(-1)^{3/4}}{\sqrt{2} e_1^{1/4}} x)
\end{equation}</span>
Then the function <span class="math-container">$v(x)$</span> satisfies the biconfluent Heun equation:</p>
<p><span class="math-container">\begin{equation}
\frac{d^2 v(u)}{d u^2} -\left( \frac{\gamma}{u} + \delta + u\right)\frac{d v(u)}{d u} + \frac{\alpha u - q}{u} v(u) = 0
\end{equation}</span>
where
<span class="math-container">\begin{eqnarray}
\delta&=& -1-2\sqrt{a}\\
\gamma&=& \frac{\left(\frac{1}{2}+\frac{i}{2}\right) e}{e_1^{3/4}}\\
\alpha&=& \frac{\imath \left(8 \imath \left(\sqrt{a}+1\right) e_1^{3/2}-4 b e_1+e^2\right)}{8 e_1^{3/2}}\\
q&=&\frac{\left(\frac{1}{4}+\frac{i}{4}\right) \left(2 \sqrt{a} e+2 i c \sqrt{e_1}+e\right)}{e_1^{3/4}}
\end{eqnarray}</span></p>
<pre><code> In[1304]:= Clear[y]; Clear[v]; Clear[m]; Clear[w]; Clear[f];
m[x_] = Exp[-I/(2 Sqrt[e1]) x (-e + e1 x)] x^(Sqrt[ a]);
y[x_] = m[x] w[x];
myeqn = Collect[
Simplify[(y''[x] +
1/x y'[x] + (-a/x^2 + b - c/x - e x + e1 x^2) y[x])], {w[x],
w''[x]}, Expand];
myeqn1 = Collect[Simplify[myeqn/m[x]], {w[x], w'[x], w''[x], x^_},
Simplify];
T = (-1)^(3/4)/(Sqrt[2] e1^(1/4));
f[x_] = T x;
subst = {x :> f[x],
Derivative[1][w][x] :> 1/f'[x] Derivative[1][w][x],
Derivative[2][w][x] :> -f''[x]/(f'[x])^3 Derivative[1][w][x] +
1/(f'[x])^2 Derivative[2][w][x]};
Collect[T^2 (myeqn1 /. subst /. w[f[x]] :> w[x]), {w[x], w'[x],
w''[x], x^_}, Simplify]
</code></pre>
<p><a href="https://i.stack.imgur.com/AmaXR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AmaXR.png" alt="enter image description here"></a></p>
|
1,460,488 |
<p>There are $4$ girls and $3$ boys but there are only $5$ seats. How many ways can you seat the $3$ boys together?</p>
<p>The order of the seat matters, for example:
there's the order
$B_1$ $B_2$ $B_3$ $G_2$ $G_4$
and there's
$B_2$ $B_3$ $B_1$ $G_2$ $G_4$</p>
<p>Here's my answer:
There are $3!$ ways to seat the $3$ boys. The $2$ remaining seats are to be occupied by $2$ out of the $4$ girls, so $^4P_2$. So we now have $3! \cdot $ $^4P_2$. </p>
<p>Lastly, there are $3$ ways to make that arrangement, <BR>
$1)$ two girls on the left, <BR>
$2)$ two girls on the right, and <BR>
$3)$ a girl on both ends.</p>
<p>So my final equation is $3! \cdot $ $^4P_2$ $\cdot 3 = 216$</p>
<p>But then again, that was just a guess, I'm not really sure how to get it. So please confirm if my answer is right, and if it's wrong, please tell me how to get it.</p>
|
N. F. Taussig
| 173,070 |
<p>You are correct. Here is another approach: </p>
<p>We can select the two girls in $\binom{4}{2}$ ways. We treat the three boys as a unit, so we have three objects to permute (the two girls we select and the unit of three boys). We can permute the three objects in $3!$ ways. We can also permute the unit consisting of three boys internally in $3!$ ways. Hence, there are $$\binom{4}{2} \cdot 3! \cdot 3! = 6^3 = 216$$ seating arrangements in which the seats are occupied by the three boys and two of the four girls if the three boys sit together.</p>
|
63,052 |
<p>Suppose I have a square matrix $M$, which you can think of as the weighted adjacency matrix of a graph $G$. I want to order the vertices of $G$ in such a way that the entries of the matrix $M$ are clustered. By this I mean that the weights that are close in value should appear close in $M$.</p>
<p>I know Mathematica has some clustering algorithms implemented. How can I do this with Mathematica?</p>
|
kh40tika
| 6,395 |
<p><a href="https://en.wikipedia.org/wiki/Spectral_clustering" rel="nofollow noreferrer">Spectral clustering</a> might be a good candidate here. Generally, spectral clustering works as following:</p>
<ol>
<li><p>Find a few largest eigenvectors of the adjacency matrix by magnitude, let's say we choose largest M vectors.</p>
</li>
<li><p>Treat each vertex as a N dimensional one-hot unit vector (where N is the number of vertices). Project each vertex into M dimensional "feature" space using the eigenvectors. This can be trivially done by transposing the MxN matrix of eigenvectors. In practice, M can be much smaller than N.</p>
</li>
<li><p>Use general purpose clustering algorithm in the "feature space", such as k-means.</p>
</li>
</ol>
<pre><code>(*sample input*)
nPoints = 2048;
nFeatureDim = 8;
nClusters = 18;
points = Normalize /@ RandomReal[{-1., 1.}, {nPoints, 3}];
adjMatrix = Power[DistanceMatrix[points], 4];
(*cluster on feature space*)
features = Transpose@Eigenvectors[adjMatrix, nFeatureDim];
clusters = FindClusters[features -> Range[nPoints], nClusters, Method ->"KMeans"];
sortedPoints = points[[#]] & /@ clusters;
(*make a render*)
colors = RGBColor /@ Tuples[{{0, 1}, {0, 0.5, 1}, {0, 0.5, 1}}];
vectorRender = Graphics3D[Transpose[{colors, Point /@ sortedPoints}]]
</code></pre>
<p>The above example uses a dense matrix. Finding eigenvectors should work equally well on sparse matrices.</p>
<p><a href="https://i.stack.imgur.com/6sBnq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6sBnq.png" alt="spectral clustering on " /></a></p>
|
944,840 |
<p>For vectors u, w, and v in a vector space V, I am trying to prove:</p>
<p>If $u + w = v + w$ then $u = v$</p>
<p><strong>without</strong> using the additive inverse and only using the 8 axioms which define a vector space. I am coming up short. I don't see how to do this without assuming that if $u + w = v + w$ then I can just add something to both sides as in $(u+w) + w' = (v+w) + w'$.</p>
<p>Thank you.</p>
|
rschwieb
| 29,335 |
<p>Apparently the only approach for a vector space $V$ that avoids additive inverses <em>in $V$</em> is to use $w'=(-1)w$, which would leverage additive inverses in <em>the field</em> and not in the abelian group. Presumably your axioms (whatever they are) include that $0w=0$, $1w=w$ and that scalars distribute, allowing you to continue $$u+0=u+(1-1)w=u+w+(-1)w=v+w+(-1)w=v+(1-1)w=v+0$$.</p>
<p>If this sorcery of scalars is not available to you, and you are only allowed to talk about a binary operation, then it is not always possible. A set with a binary operation $\cdot$ (no inverses assumed) does not have this "cancellation property." </p>
<p>For example, in a boolean ring, the multiplication operation satisfies $x\cdot x =x$ for all $x$, and so you would have, in particular, the equation $x\cdot x = x\cdot 1$, but of course you can choose and example where $x\neq 1$.</p>
|
3,595,451 |
<p><strong>Question:</strong></p>
<p>Let <span class="math-container">$P_{3}(\mathbb{R})$</span> have the standard inner product and <span class="math-container">$U$</span> be the subset spanned by the two vectors (which are polynomials) <span class="math-container">$u_{1}=1+2x-3x^2$</span> and <span class="math-container">$u_{2}=x-x^2+2x^3$</span>. Find the basis for the orthogonal complement <span class="math-container">$U^{⊥}$</span>.</p>
<p>I honestly have no idea how to approach this question. I know what orthogonal complement and a basis are but I don't understand where to begin or even solve this question. Any help would be much appreciated. Thanks in advance. </p>
|
Graham Kemp
| 135,106 |
<blockquote>
<p>How would I represent the distance R from the point (X,Y) and would this be the cdf or pdf?</p>
</blockquote>
<p>By definition of Euclidean Distance, <span class="math-container">$R=\sqrt{X^2+Y^2}$</span>. </p>
<p>So immediately we know the support is <span class="math-container">$\{r: 1\leqslant r^2\leqslant 4\}$</span></p>
<p>Now, since the points are uniformly distributed: the pdf for <span class="math-container">$R$</span>, at any point <span class="math-container">$r$</span> within that support, will be equal to the ratio of the circumference of a circle with radius <span class="math-container">$r$</span> to the area of the annulus. Call this <span class="math-container">$f_R(r)$</span>.</p>
<p>And the CDF will be the integral <span class="math-container">$\displaystyle F_R(r)=\int_1^r f_R(s)\mathrm d s$</span> .</p>
|
3,920,469 |
<p>The topic of <a href="https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers" rel="nofollow noreferrer">odd perfect numbers</a> likely needs no introduction.</p>
<p>The question is as is in the title:</p>
<blockquote>
<p>If <span class="math-container">$p^k m^2$</span> is an odd perfect number with special prime <span class="math-container">$p$</span>, then what is the optimal constant <span class="math-container">$C$</span> such that <span class="math-container">$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$</span></p>
</blockquote>
<p>(Note that the special prime <span class="math-container">$p$</span> satisfies <span class="math-container">$p \equiv k \equiv 1 \pmod 4$</span> and <span class="math-container">$\gcd(p,m)=1$</span>.)</p>
<p><strong>MY OWN PROOF FOR <span class="math-container">$C = 1$</span></strong></p>
<p>It is trivial to show that <span class="math-container">$m^2 - p^k \equiv 0 \pmod 4$</span>. This implies that
<span class="math-container">$$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$</span>
since <span class="math-container">$\sigma(m^2)/p^k$</span> is always odd.</p>
<p>Now, suppose to the contrary that
<span class="math-container">$$\sigma(m^2) > p^k m^2 - p^{2k}.$$</span>
<span class="math-container">$$p^{2k} + \sigma(m^2) > p^k m^2$$</span>
<span class="math-container">$$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$</span>
<span class="math-container">$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$</span>
<span class="math-container">$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$</span>
where we have used the lower bound <span class="math-container">$\sigma(p^k) \geq p^k + 1$</span>.
But from the paper <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html" rel="nofollow noreferrer">Dris (2012)</a>, we have the lower bound
<span class="math-container">$$3p^k \leq \sigma(m^2)$$</span>
while we also have the upper bound
<span class="math-container">$$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$</span></p>
<p>This implies that
<span class="math-container">$$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$</span>
<span class="math-container">$$p^k - 2 < \frac{2}{p^k - 1}$$</span>
<span class="math-container">$$(p^k - 1)(p^k - 2) < 2.$$</span>
This last inequality is a contradiction, as <span class="math-container">$p^k \geq 5$</span>.</p>
<p>We therefore conclude that the inequality
<span class="math-container">$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$</span>
holds when <span class="math-container">$C=1$</span>.</p>
<blockquote>
<p>Does the inequality still hold when the constant <span class="math-container">$C > 1$</span>? If so, then what is the optimal value for <span class="math-container">$C$</span>?</p>
</blockquote>
|
mathlove
| 78,967 |
<p>This is a partial answer.</p>
<blockquote>
<p>Does the inequality still hold when the constant <span class="math-container">$C > 1$</span>?</p>
</blockquote>
<p>Yes, one can get
<span class="math-container">$$\frac{\sigma(m^2)}{p^k} \le \frac{m^2 - p^k}{\color{red}{4/3}}\tag1$$</span></p>
<p>I don't know if this is the optimal value for <span class="math-container">$C$</span>.</p>
<p><em>Proof for <span class="math-container">$(1)$</span></em> :</p>
<p>Note that <span class="math-container">$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}\iff \frac{p^k(m^2 - p^k)}{\sigma(m^2)}\gt C$$</span></p>
<p>Let <span class="math-container">$F:=\dfrac{p^k(m^2 - p^k)}{\sigma(m^2)}$</span>. We can write <span class="math-container">$F$</span> as</p>
<p><span class="math-container">$$F=\frac{p^k(m^2 - p^k)\sigma(p^k)}{\sigma(m^2)\sigma(p^k)}=\frac{p^k(m^2 - p^k)\sigma(p^k)}{2p^km^2}=\frac{m^2\sigma(p^k)-p^{k}\sigma(p^k)}{2m^2}$$</span></p>
<p><span class="math-container">$$=\frac{\sigma(p^k)}{2}-\frac{p^{k}\sigma(p^k)}{2m^2}=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}$$</span></p>
<p>Using <span class="math-container">$3p^k\le \sigma(m^2)$</span>, we get</p>
<p><span class="math-container">$$F=\frac{\sigma(p^k)}{2}-\frac{p^{2k}}{\sigma(m^2)}\ge \frac{\sigma(p^k)}{2}-\frac{p^{k}}{3}=\frac{p^{k+1}-1}{2(p-1)}-\frac{p^{k}}{3}=\frac{p^{k+1}+2p^k-3}{6(p-1)}$$</span></p>
<p>Let <span class="math-container">$g(k):=\dfrac{p^{k+1}+2p^k-3}{6(p-1)}$</span>. Since <span class="math-container">$g(k)$</span> is increasing, we get
<span class="math-container">$$F\ge g(k)\ge g(1)=\frac{p+3}{6}\ge \frac{5+3}{6}=\frac 43$$</span>
to have
<span class="math-container">$$F\ge \frac 43$$</span>
which is equivalent to <span class="math-container">$(1)$</span>.<span class="math-container">$\quad\blacksquare$</span></p>
|
164,137 |
<p>I want to animate a Point moving on an ellipse, but the angle I need to use is a numerical solution from an epression. How can I get mathematica to just take the value from the animated nummerical solution and use it as a variable for my pointfunction?
This is my Code and this way it does not work. I think I have to remove the Set-Brackets around my Value for Psi.</p>
<pre><code>ρ = 1; ε := 0.8 ; T := 10 π
curvE = ParametricPlot[{ρ/(1 - ε^2) Cos[ψ], ρ/Sqrt[1 - ε^2] Sin[ψ]}, {ψ, 0, 2 π}]
Animate[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T} /. ε -> 0.8 /. T -> 10 π, {ψ, 0}], {t, 0, 10 π}]
Animate[Show[curvE, Graphics[{PointSize[Large], Red, Point[
Dynamic[{ρ/(1 - ε^2) Cos[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]],
ρ/Sqrt[1 - ε^2] Sin[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]]}]]}]],
{t, 0, 10 π, AppearanceElements -> None}]
</code></pre>
<p>I would appreciate your help.</p>
|
Coolwater
| 9,754 |
<pre><code>FindRoot[t, {t, 0}]
</code></pre>
<p>Returns <code>{t -> 0.}</code>. You need to extract the numeric value, e.g.</p>
<pre><code>t /. FindRoot[t, {t, 0}]
</code></pre>
<p>which returns <code>0.</code>. Inserting the needed <code>ψ /.</code> in your code makes it work:</p>
<pre><code>Animate[Show[curvE, Graphics[{PointSize[Large], Red,
Point[Dynamic[{ρ/(1 - ε^2) Cos[#], ρ/Sqrt[1 - ε^2] Sin[#]}]] &[
ψ /. FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]]}]], {t, 0, 10 π, AppearanceElements -> None}]
</code></pre>
|
164,137 |
<p>I want to animate a Point moving on an ellipse, but the angle I need to use is a numerical solution from an epression. How can I get mathematica to just take the value from the animated nummerical solution and use it as a variable for my pointfunction?
This is my Code and this way it does not work. I think I have to remove the Set-Brackets around my Value for Psi.</p>
<pre><code>ρ = 1; ε := 0.8 ; T := 10 π
curvE = ParametricPlot[{ρ/(1 - ε^2) Cos[ψ], ρ/Sqrt[1 - ε^2] Sin[ψ]}, {ψ, 0, 2 π}]
Animate[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T} /. ε -> 0.8 /. T -> 10 π, {ψ, 0}], {t, 0, 10 π}]
Animate[Show[curvE, Graphics[{PointSize[Large], Red, Point[
Dynamic[{ρ/(1 - ε^2) Cos[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]],
ρ/Sqrt[1 - ε^2] Sin[FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]]}]]}]],
{t, 0, 10 π, AppearanceElements -> None}]
</code></pre>
<p>I would appreciate your help.</p>
|
Bob Hanlon
| 9,362 |
<pre><code>ρ = 1; ε = 4/5; T = 10 π;
curvE = ParametricPlot[{ρ/(1 - ε^2) Cos[ψ], ρ/Sqrt[1 - ε^2] Sin[ψ]}, {ψ, 0, 2 π}];
sol[t_?NumericQ] := FindRoot[{ψ - ε Sin[ψ] == 2 π t/T}, {ψ, 0}]
Animate[
Show[
curvE,
Graphics[{PointSize[Large], Red,
Point[{ρ/(1 - ε^2) Cos[ψ], ρ/Sqrt[1 - ε^2] Sin[ψ]}]} /. sol[t]]],
{t, 0, 10 π, AppearanceElements -> None}]
</code></pre>
|
4,029,249 |
<blockquote>
<p>Prove that the function <span class="math-container">$f(x)=e^x-(ax^2+bx+c)$</span> has 3 solutions at most .</p>
<p><span class="math-container">$a$</span>,<span class="math-container">$b$</span> and <span class="math-container">$c$</span> are constants.</p>
</blockquote>
<p>This is the information given about the function, I tried a couple of things and I am not sure if what I did is right.</p>
<p>First the function is continuous and differentiable since <span class="math-container">$e^x$</span> is continuous and differentiable and <span class="math-container">$ax^2+bx+c$</span> is a polynomial so it is also continuous and differentiable therefore we can use rolles theorem.</p>
<p>I tried doing a couple of derivatives such as <span class="math-container">$f'(x)=e^x-2ax-b$</span> and <span class="math-container">$f''(x)=e^x-2a$</span> and lastly <span class="math-container">$f'''(x)=e^x$</span></p>
<p>so the third derivative has no solution , the second one is <span class="math-container">$x=ln(2a)$</span> since the second derivative has only 1 solution , then the first derivative has 2 at most and the original has 3 at most.</p>
<p>Is it the right way to solve it? am I missing something? thank you for the help!
By solution I mean f(x)=0</p>
|
Bob Dobbs
| 221,315 |
<p>I like this question but you must correct the statement of the question as we want to solve the equation <span class="math-container">$f(x)=0$</span> where <span class="math-container">$f(x)=e^x-ax^2-bx-c$</span>. And the claim is that it has at most <span class="math-container">$3$</span> roots or in other words <span class="math-container">$3$</span> solutions.</p>
<p>My thinking:</p>
<p><span class="math-container">$f''(x)=e^{x}-2a$</span>. We have two cases:</p>
<p>i) <span class="math-container">$a\leq 0$</span>: Then <span class="math-container">$f''(x)$</span> is always positive. The graph of <span class="math-container">$f(x)$</span> is concave-up shape. So, it can cut the <span class="math-container">$x$</span>-axis at <span class="math-container">$0$</span>,<span class="math-container">$1$</span> or <span class="math-container">$2$</span> points. <span class="math-container">$1$</span> point solution is tangency situation to the <span class="math-container">$x$</span>-axis.</p>
<p>ii) <span class="math-container">$a>0$</span>: Solving <span class="math-container">$f''(x)=0$</span>, we have an inflection point at <span class="math-container">$x=\ln(2a)$</span> for the graph of <span class="math-container">$y=f(x)$</span>. Moreover, till the inflection point the first derivative <span class="math-container">$f'(x)=e^{x}-2ax-b$</span> is decrasing and after then increasing with only one local minumum. So, the graph of <span class="math-container">$y=f'(x)$</span> can cut <span class="math-container">$x$</span>-axis at most <span class="math-container">$2$</span> points. That means the graph of the original function <span class="math-container">$y=f(x)$</span> has at most two local extrema. When it has two, one is local maximum on the left and the other is local minumum on the right side of the graph making an <a href="https://www.wolframalpha.com/input?i=y%3De%5Ex-x%5E2-2x" rel="nofollow noreferrer">snake shape</a>. With this shape, the graph can cut <span class="math-container">$x$</span>-axis at most at <span class="math-container">$3$</span> points. But, in all stiuations, when <span class="math-container">$a>0$</span>, we have at least <span class="math-container">$1$</span> solution.</p>
<p>Pavel Kocourek's answer here, gives a nice Lemma which can make the solution immediate: <a href="https://math.stackexchange.com/questions/4620162/what-is-the-maximum-number-of-points-at-which-an-n-degree-polynomial-can-inter">What is the maximum number of points at which an $n$-degree polynomial can intersect the power function $x^m, m>n$.</a></p>
|
84,204 |
<p>Say I have some object or quantity and an instance or special case of it, how to formally write this down? </p>
<p>I don't (just) mean that $X$ is a set and $x$ an element, i.e. $x\in X$ is not it. I'm dealing with things as general like "<em>the specific group $g$ is a group/is a case of a group</em>". Or "<em>the Integers can be viewed as a restriction from the reals</em>". It should be able to handle such different cases. Are there standard ways to do such a "<em>succession</em>"?</p>
<p>Do I have to intruduce a two valued predicate, which says "<em>is instance</em>" or "<em>is special case of</em>"? Is this even legal/formally right/possible? Do I have to elevate the bigger things to some sort of set or category first? Does this have to do with lattices (since I generate some order)?</p>
|
hmakholm left over Monica
| 14,366 |
<p>If I understand the question correctly, you want to know how to write down a formal formula that says that $g$ is a group?</p>
<p>Usually one would just introduce a predicate specifically for saying this, so your formula would be
$$\mathrm{Group}(g)$$
The definition of this predicate would be as an abbreviation of</p>
<blockquote>
<p>There exist $A$ and $F$ such $g=\langle A,F\rangle$ <em>and</em> $A$ is nonempty <em>and</em> $F:A\times A\to A$ <em>and</em> $F$ is associative <em>and</em> for each $a\in A$ there is a $b\in A$ such that for each $c\in A$ it holds that $F(F(a,b),c)=F(c,F(a,b))=c$ <em>and</em> ...</p>
</blockquote>
|
3,453,483 |
<p>I'm reading Serre's <span class="math-container">$\textit{A course in Arithmetic}$</span> where he defines a Dirichlet series to be an infinite sum of the form
<span class="math-container">$$f(z) = \sum\limits_{n=1}^{\infty} a_ne^{-\lambda_nz}
$$</span>
where <span class="math-container">$\lambda_n$</span> is an increasing sequence of reals diverging to infinity and <span class="math-container">$a_n \in \mathbb{C}$</span>.</p>
<p>One can associate with these series, half-planes <span class="math-container">$H$</span> (including <span class="math-container">$\mathbb{C}$</span> and <span class="math-container">$\varnothing$</span>) on which they converge.
More precisely, if <span class="math-container">$f$</span> converges at <span class="math-container">$z_0$</span>, then it must converge uniformly on compact subsets of the half plane <span class="math-container">$\Re(z)>\Re(z_0)$</span>.
This shows that <span class="math-container">$f$</span> is holomorphic here too.</p>
<blockquote>
<p>Given a holomorphic <span class="math-container">$f$</span> on some half plane <span class="math-container">$H$</span>, is it representable by a Dirichlet series?</p>
</blockquote>
<p>Going through the basic theorems hasn't thrown up any obvious holomorphic functions precluded from having such a representation.
Am I missing something?</p>
|
Henry Crawford
| 274,113 |
<p>Let <span class="math-container">$d(z) = \sum\limits_{n} a_n e^{-\lambda_n z}$</span> be a Dirichlet series converging in some non-empty half plane <span class="math-container">$H$</span>.
Proposition <span class="math-container">$6$</span> on page <span class="math-container">$66$</span> of the book mentioned implies that the sum must converge to <span class="math-container">$d(z)$</span> uniformly on the real line intersected with <span class="math-container">$H$</span>.
You can see this uniform convergence by using Abel's summation lemma.</p>
<p>I claim that the function <span class="math-container">$f(z)=z$</span> cannot be represented by a Dirichlet series on any non-empty <span class="math-container">$H$</span>.
If it were so, one could subtract the terms having <span class="math-container">$\lambda_n<0\ ^*$</span> to get a function which was bounded on the half real axis. This is impossible for a function of the form
<span class="math-container">$$z - (a_1e^{-\lambda_1 z} + \dots + a_me^{-\lambda_m z})
$$</span>
with <span class="math-container">$\lambda_1 < \dots < \lambda_m < 0$</span>. </p>
<p>Additionally, as Conrad points out in the comments, Dirichlet series enjoy some 'almost-periodic' properties on vertical lines not seen for general holomorphic functions (further explanation would be nice at some point).</p>
<p><span class="math-container">$^*$</span> The book actually assumes each <span class="math-container">$\lambda_n\geq 0$</span>, so strictly speaking we do not have to deal with this case.</p>
|
3,258,372 |
<p>I'm doing a practice exam questions and am stuck at this question:</p>
<blockquote>
<p>Are there topological spaces X,Y (each with more than one point), such that [0,1] is homeomorphic to X×Y? What if we replace [0,1] with R?</p>
</blockquote>
<p>I'm not even sure how to start tackle it, any help and clues will be appreciated! My head is leading me to "cut-points" area, but I'm not sure abuot it./</p>
<p>Thanks in advance! </p>
|
José Carlos Santos
| 446,262 |
<p><strong>Hints:</strong> Prove that:</p>
<ul>
<li>If <span class="math-container">$X\times Y$</span> is homeomorphic to <span class="math-container">$[0,1]$</span>, then both <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are connected.</li>
<li>If <span class="math-container">$x_0\in X$</span> and <span class="math-container">$y_0\in Y$</span>, then <span class="math-container">$(X\times Y)\setminus\{(x_0,y_0)\}$</span> is still connected.</li>
</ul>
|
3,863,495 |
<p>This is my solution to an old exam problem that I'd appreciate some feedback on. The problem:</p>
<blockquote>
<p>Let <span class="math-container">$f:[0,\infty)\to\mathbb{R}$</span>, <span class="math-container">$f\geq 0$</span> and <span class="math-container">$\int _0^{\infty} f(x) dx=L<\infty;$</span> that is, <span class="math-container">$f$</span> is Riemann integrable in any finite interval <span class="math-container">$[0,R]$</span> and <span class="math-container">$\lim\limits_{R\to\infty} \int _0^R f(x) dx$</span> exists. Show that
<span class="math-container">$$
\lim_{R\to \infty}\frac{\int_0^R x f(x)dx}{R}=0.
$$</span></p>
</blockquote>
<p>Proof:
Note that if <span class="math-container">$f\equiv 0$</span> there's nothing to prove. Else, we would like to apply L'Hôpital's Rule, but the problem is the integrand is not necessarily continuous, hence the integral is not differentiable in <span class="math-container">$R.$</span> To that end, we use Fubini's Theorem to rewrite as a double-integral:
<span class="math-container">$$
\int _0^R x f(x)\,dx = \int _0^R \int _0^x f(x)\,dydx
$$</span><span class="math-container">$$
= \int _0^R \int _y^R f(x)\; dxdy
$$</span>Define <span class="math-container">$F(x):=\int _0^x f(t)\;dt;$</span> then <span class="math-container">$F$</span> is continuous. We have
<span class="math-container">$$
\int _0^R \int _y^R f(x)\; dxdy=\int _0^R F(R)-F(y)\;dy;
$$</span>then we have
<span class="math-container">$$
\frac{\int_0^R x f(x)dx}{R}=\frac{\int _0^R F(R)-F(y)\;dy}{R} = F(R)-\frac{\int _0^R F(y)\,dy}{R}
$$</span> Here's the tricky part:</p>
<p>Since <span class="math-container">$f$</span> is integrable and non-negative, by assumption <span class="math-container">$F$</span> approaches <span class="math-container">$L$</span>. This allows us to split up the limit:
<span class="math-container">$$
\lim_{R\to\infty}F(R) - \frac{\int _0^R F(y)\;dy}{R} =\lim_{R\to\infty}F(R) - \lim_{R\to\infty}\frac{\int _0^R F(y)\;dy}{R}
$$</span>
<span class="math-container">$$
=L - \lim_{R\to\infty}\frac{\int _0^R F(y)\;dy}{R}
$$</span> In this case, the remaining integral approaches <span class="math-container">$LR$</span>. <em>Now</em> we can apply L'Hôpital's Rule:
<span class="math-container">$$
L-\lim_{R\to\infty} \frac{\int _0^R F(y)\;dy}{R} = L-\lim_{R\to \infty} F(R)=L-L=0. \square
$$</span>Any comments/suggestions?</p>
|
RRL
| 148,510 |
<p>For a very general approach, we have <span class="math-container">$f \in L^1([0,\infty))$</span> since <span class="math-container">$f$</span> is nonnegative and improperly Riemann integrable.</p>
<p>Thus, by the dominated convergence theorem,</p>
<p><span class="math-container">$$\lim_{R \to \infty} \frac{1}{R}\int_0^R x f(x) \, dx = \lim_{R\to \infty} \int_{[0,\infty)}\frac{x f(x)}{R}\mathbf{1}_{[0,R]} \, dx = 0$$</span></p>
|
31,480 |
<p>I'm having difficulty with my math, fractions and up. I used to understand it all, but it's been so long since I've touched the book (I finished it a couple of months ago, picked it up to review everything), I seem to have forgotten it. </p>
<p>The explanations inside of the individual chapters do no good. They never helped me, and I always resorted to having my older brother helping (who is now away at college), and I can't find any resources online that help at all.</p>
<p>Are there any online tutorials / guides that can help me relearn all this fully, all the way from the basics, up to college level? </p>
|
JonnyQuiznos
| 9,968 |
<p>Agreed w/ Khan Academy, but there are videos out there that teach you essential Calculus topics in less than several minutes. </p>
|
116,634 |
<p>Let $G$ be a graph and $e$ be an edge of the graph $G$ such that the subgraph $G\setminus e$
is connected. The subgraph $G\setminus e$ is the subgraph of $G$ obtained by deletion of the edge $e$ of $G$.
Assume that $G$ has $n$ vertices.
Is it true that $\lambda(G)-\lambda(G\setminus e)\geq \frac{1}{n}$?
Here $\lambda(G)$ denotes the maximum eigenvalue of adjacency matrix of $G$.</p>
|
Chris Godsil
| 1,266 |
<p>It's always useful to test these questions on actual examples. The largest eigenvalue of the cycle $C_n$ is 2 and the largest eigenvalue of the path $P_n$ on $n$ vertices is $2\cos(\pi/(n+1))$. When $n=8$, this is 1.879385 and $2-1.8793852=0.120615 <1/8$.</p>
<p>In fact it is not hard to see that for large $n$ the difference $2-2\cos(\pi/(n+1))$ is
of order $\pi^2/(n+1)^2$, and so your lower bound is not even of the right order.</p>
|
116,634 |
<p>Let $G$ be a graph and $e$ be an edge of the graph $G$ such that the subgraph $G\setminus e$
is connected. The subgraph $G\setminus e$ is the subgraph of $G$ obtained by deletion of the edge $e$ of $G$.
Assume that $G$ has $n$ vertices.
Is it true that $\lambda(G)-\lambda(G\setminus e)\geq \frac{1}{n}$?
Here $\lambda(G)$ denotes the maximum eigenvalue of adjacency matrix of $G$.</p>
|
Shahrooz
| 19,885 |
<p>This conjecture is not true in general. For example, let $G$ be a graph that obtained from joining the end vertex of $P_3$, $P_3$ and $P_4$, where it is an star-like tree. This graph has largest eigenvalue equal to $2.02852$. Now join the end vertex of $P_4$ to the end vertex of $P_3$. Therefore we added an edge to the previous graph. The largest eigenvalue of this graph is $2.13578$. Then the difference of this values is $0.10726$. But this graph has $8$ vertices and $1/8=0.125$. It is a counter example. </p>
<p>Also, we can construct a family of star-like tree that does not have this property. Also, I think for any polynomial $f(n)$, we can construct a graph $G$ such that $\lambda(G)-\lambda(G-e)\geq \frac{1}{f(n)}$ is not true. </p>
|
80,456 |
<p>Given an array <code>sel</code> and an index position <code>i0</code>, how can I find the position of the nearest (left or right) nonzero element?
I'm able to do it with a loop and a couple of awful If's, but I was looking for a functional way...</p>
<pre><code> lr=Length[sel];
For[i = 0, i <= lr, i++,
If[1 <= i0 + i <= lr && sel[[i0 + i]] == 1, Print[i0+i]; Break[],
If[1 <= i0 - i <= lr && sel[[i0 - i]] == 1, Print[i0-i];Break[]]]]
</code></pre>
|
kglr
| 125 |
<pre><code>nrstNZP[l_] := With[{nF = Nearest[Flatten@SparseArray[l]["NonzeroPositions"]]},
With[{nrst = nF[#, 2]}, DeleteCases[nrst, #][[1]]] & /@ #] &
</code></pre>
<p>Example:</p>
<pre><code>SeedRandom[1]
sel = RandomInteger[{0, 2}, 20]
(* {1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 1, 2, 0, 0, 1, 1} *)
Flatten[SparseArray[sel]["NonzeroPositions"]]
(* {1, 3, 4, 8, 13, 15, 16, 19, 20} *)
nrstNZP[sel] @ Range[20]
(* {3, 1, 4, 3, 4, 4, 8, 4, 8, 8, 13, 13, 15, 13, 16, 15, 16, 19, 20, 19} *)
</code></pre>
|
1,212,254 |
<p>I am trying to find all pairs of side integers (a, b) for a given hypothenuse number n so
that (a, b, n) is a Pythagorean triple, i.e.,$ a^2 + b^2 = n^2$</p>
<p>The approach i am using is </p>
<ol>
<li>Sorting the array in ascending order </li>
<li>Finding the square of each element in array</li>
</ol>
<p>for(j->0 to (a.length-1))
for (i->j+1 to (a.length-1))
search the a[j]+a[i] ahead in the array from i+1 to the end of array
if found get the triplet according to sqrt(a[j]),sqrt(a[i]) & sqrt(a[j]+a[o])
end
end</p>
<p>Is this approach correct?</p>
|
Tim Raczkowski
| 192,581 |
<p>Hint: Let $N=\{n\in\Bbb N: x^n\notin\alpha\}$. This is non-empty, because $x>1$, so let $n+1=\min N$.</p>
|
1,212,254 |
<p>I am trying to find all pairs of side integers (a, b) for a given hypothenuse number n so
that (a, b, n) is a Pythagorean triple, i.e.,$ a^2 + b^2 = n^2$</p>
<p>The approach i am using is </p>
<ol>
<li>Sorting the array in ascending order </li>
<li>Finding the square of each element in array</li>
</ol>
<p>for(j->0 to (a.length-1))
for (i->j+1 to (a.length-1))
search the a[j]+a[i] ahead in the array from i+1 to the end of array
if found get the triplet according to sqrt(a[j]),sqrt(a[i]) & sqrt(a[j]+a[o])
end
end</p>
<p>Is this approach correct?</p>
|
CIJ
| 159,421 |
<p>Even though I asked (and solved) this question a lot of time ago, here is a solution for future references.</p>
<p>Case I. $\alpha=1^*.$ This case is trivial since $x^{-1}\in\alpha$ and $x^0\notin\alpha.$</p>
<p>Case II. $\alpha\subsetneq1^*.$ Let $a\in\alpha\cap\mathbb Q_{>0}$ and let $m$ and $n$ be positive integers such that $n>a^{-1}$ and $x>1+(1/m).$ Then $$x^{mn}>\left(1+\dfrac{1}{m}\right)^{mn}\geqslant1+n>a^{-1}$$ and it follows that there exists some positive integer $t$ such that $x^{-t}\in\alpha$ and $x^{-t+1}\notin\alpha.$</p>
<p>Case III. $1^*\subsetneq\alpha.$ Let $a\in\mathbb Q\setminus\alpha,$ let $n$ be a positive integer such that $n>a$ and let $m$ be as in the first case. Then $$x^{mn}>\left(1+\dfrac{1}{m}\right)^{mn}\geqslant1+n>a$$ and it follows that there is some positive integer $t$ such that $x^t\notin\alpha$ and $x^{t-1}\in\alpha.$</p>
<p>Note that the solution that Tim suggests is incomplete (in fact, his "hint" was not helpful at all). As an example, one could argue that the floor function exists as follows.</p>
<p>Let $x$ be a real number. Since $\mathbb Z_{>0}$ is not bounded above, there is some $n\in\mathbb Z_{>0}$ such that $n>x.$ Therefore the set of all positive integers $t$ such that $t>x$ has a least element $s.$ It follows that $s-1\leqslant x<s.$ What happens if $x<0$?</p>
|
198,521 |
<p>I am trying to overlap two <code>Graphics</code> objects <code>g1</code> and <code>g2</code> with <code>Show</code>. However, I found that when the coordinates of each object is defined to quite different ranges, I need to "scale" and "shift" the coordinates of one object to get the desired look. </p>
<p>For example,</p>
<pre><code>g1 = Graphics[{GrayLevel[0.8], Rectangle[{-2, -2}, {2, 2}]}];
g2 = Graphics[{{GrayLevel[0.5], Rectangle[{0, 0}, {1, 1}]}}];
Show[g1, AspectRatio -> 1, Axes -> True, ImageSize -> 230]
Show[g2, AspectRatio -> 1, Axes -> True, ImageSize -> 230]
Show[g1, g2, AspectRatio -> 1, Axes -> True, ImageSize -> 230]
</code></pre>
<p>The overlapped version of <code>g1</code> and <code>g2</code> looks like this
<a href="https://i.stack.imgur.com/tVM01.png" rel="noreferrer"><img src="https://i.stack.imgur.com/tVM01.png" alt="enter image description here"></a></p>
<p>However, I would like to scale the coordinates of <code>g2</code> to make <code>g2</code> twice large and also shift its coordinates so the center can "roughly" coincide the center of <code>g1</code>. I say "roughly" because <code>g1</code> and <code>g2</code> may be some graphics not of a regular shape. The desired result will look like <a href="https://i.stack.imgur.com/VO3jM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/VO3jM.png" alt="enter image description here"></a></p>
<p>So how can I manipulate the coordinates of <code>g2</code> to adjust its relative position and size when <code>Show</code> with <code>g1</code>? Please avoid modifying the definition of <code>g1</code> and <code>g2</code> as they can be any <code>Graphics</code> copy-pasted over.</p>
|
C. E.
| 731 |
<p>I'm not saying that this is what you need for your problem, but sometimes when you want to superimpose graphics like this, you're looking for <code>Inset</code>:</p>
<pre><code>Show[
g1,
Graphics@Inset[
Show[g2, PlotRangePadding -> 0],
{0, 0}, {Center, Center}, {2, 2}
],
Axes -> True, ImageSize -> 230
]
</code></pre>
<p><img src="https://i.stack.imgur.com/yCeim.png" alt="Mathematica graphics"></p>
|
495,622 |
<p>Well, it may seem trivial, but I cannot find it on google. Is a constant function continuously differentiable, of all orders?</p>
<p>Thank you.</p>
|
Umberto P.
| 67,536 |
<p>Yes. $f'$ and all higher derivatives are identically equal to zero.</p>
|
229,558 |
<p>When we say that a set $S$ is denumerable, that is, there is a bijection $S \to \omega$, do we mean that there <em>exists</em> such a bijection or do we mean that we have one and are talking about a pair $(S,f)$?</p>
<p>I'm asking because it makes a difference to whether I need choice in some proofs or whether I don't. For example, if we prove that a denumerable union of denumerable sets is denumerable we need countable choice to prove it if we assume the former definition and we do not need choice at all if we assume the latter. </p>
|
Ittay Weiss
| 30,953 |
<p>Denumerable means there exists a bijection between the given set and the set $\mathbb {N}$. This indeed, as you point out, creates subtleties for certain proofs. Further to your comment about the countable union of countable sets being countable, it can be shown that without countable choice this result is false. That is, there exists models of ZF where the countable union of countable sets is not countable. </p>
|
2,586,618 |
<p>I'm trying to study for myself a little of Convex Geometry and I have some doubts with respect the proof of the Theorem 1.8.5 of the book Convex Bodies: The Brunn-Minkowski Theory. Before I presented the proof and my doubts, I will put the definitions used in the theorem below.</p>
<p><span class="math-container">$\textbf{Definitions used in the theorem:}$</span></p>
<p>(i) <span class="math-container">$\mathcal{C}^n := \{ A \subset \mathbb{R}^n \ ; \ A \neq \emptyset \ \text{and} \ A \ \text{is compact} \}$</span>.</p>
<p>(ii) <span class="math-container">$B^n = \overline{B(0,1)} := \{ x \in \mathbb{R}^n \ ; \ d(x,0) \leq 1 \}$</span>.</p>
<p>(iii) The Hausdorff distance of the sets <span class="math-container">$K, L \in \mathcal{C}^n$</span> is defined by</p>
<p><span class="math-container">$$\delta (K,L) := \max \left \{ \sup_{x \in K} \inf_{y \in L} |x - y|, \sup_{x \in L} \inf_{y \in K} |x - y| \right \}$$</span></p>
<p>or, equivalently, by</p>
<p><span class="math-container">$$\delta (K,L) := \min \{ \lambda \geq 0 \ ; \ K \subset L + \lambda B^n, L \subset K + \lambda B^n \} $$</span></p>
<blockquote>
<p><span class="math-container">$\textbf{Theorem 1.8.5.}$</span> From each bounded sequence in <span class="math-container">$\mathcal{C}^n$</span> one can select a convergent subsequence.</p>
<p><span class="math-container">$\textbf{Proof:}$</span></p>
<p>Let <span class="math-container">$(K^0_i)_{i \in \mathbb{N}}$</span> be a sequence in <span class="math-container">$\mathcal{C}^n$</span> whose elements are contained in some cube <span class="math-container">$C$</span> of edge length <span class="math-container">$\gamma$</span>. For each <span class="math-container">$m \in \mathbb{N}$</span>, the cube <span class="math-container">$C$</span> can be written as a union of <span class="math-container">$2^{mn}$</span> cubes of length <span class="math-container">$2^{-m}\gamma$</span>. For <span class="math-container">$K \in \mathcal{C}^n$</span>, let <span class="math-container">$A_m(K)$</span> denote the union of all such cubes that meet <span class="math-container">$K$</span>. Since (for each <span class="math-container">$m$</span>) the number of subcubes is finite, the sequence <span class="math-container">$(K^0_i)_{i \in \mathbb{N}}$</span> has a subsequence <span class="math-container">$(K^1_i)_{i \in \mathbb{N}}$</span> such that <span class="math-container">$A_1(K^1_i) =: T_1$</span> is independent of <span class="math-container">$i$</span>. Similarly, there is an union <span class="math-container">$T_2$</span> of subcubes of length <span class="math-container">$2^{-2} \gamma$</span> and a subsequence <span class="math-container">$(K^2_i)_{i \in \mathbb{N}}$</span> of <span class="math-container">$(K^1_i)_{i \in \mathbb{N}}$</span> such that <span class="math-container">$A_2(K^2_i) = T_2$</span>. Continuing in this way, we obtain a sequence <span class="math-container">$(T_m)_{m \in \mathbb{N}}$</span> of union of subcubes (of edge length <span class="math-container">$2^{-m} \gamma$</span> for given <span class="math-container">$m$</span>) and to each <span class="math-container">$m$</span> a sequence <span class="math-container">$(K^m_i)_{i \in \mathbb{N}}$</span> such that</p>
<p><span class="math-container">$$A_m(K^m_i) = T_m \ (1.61)$$</span></p>
<p>and</p>
<p><span class="math-container">$$(K^m_i)_{i \in \mathbb{N}} \ \text{is a subsequence of} \ (K^k_i)_{i \in \mathbb{N}} \ \text{for} \ k < m. (1.62)$$</span></p>
<p>By <span class="math-container">$(1.61)$</span> we have <span class="math-container">$K^m_i \subset K^m_j + \lambda B^n$</span> with <span class="math-container">$\lambda = 2^{-m} \sqrt{n \gamma}$</span>, hence <span class="math-container">$\delta(K^m_i, K^m_j) \leq 2^{-m} \sqrt{n \gamma}$</span> (<span class="math-container">$i,j,m \in \mathbb{N})$</span> and thus, by <span class="math-container">$(1.62)$</span>,</p>
<p><span class="math-container">$$\delta(K^m_i,K^k_j) \leq 2^{-m} \sqrt{n \gamma} \hspace{1cm} \text{for} \ i,j \in \mathbb{N} \ \text{and} \ k \geq m.$$</span></p>
<p>For <span class="math-container">$K_m := K^m_m$</span>, it follows that</p>
<p><span class="math-container">$$\delta(K_m, K_k) \leq 2^{-m} \sqrt{n \gamma} \hspace{1cm} \text{for} \ i,j \in \mathbb{N} \ \text{and} \ k \geq m.$$</span></p>
<p>Thus, <span class="math-container">$(K_m)_{m \in \mathbb{N}}$</span> is a Cauchy sequence and hence convergent because <span class="math-container">$(\mathcal{C}^n, \delta)$</span> is complete. This is the subsequence that proves the assertation. <span class="math-container">$\square$</span></p>
</blockquote>
<p>My doubts are</p>
<ol>
<li><p>When the author states "Since (for each <span class="math-container">$m$</span>) the number of subcubes is finite the sequence <span class="math-container">$(K^0_i)_{i \in \mathbb{N}}$</span> has a subsequence <span class="math-container">$(K^1_i)_{i \in \mathbb{N}}$</span> such that <span class="math-container">$A_1(K^1_i) =: T_1$</span> is independent of <span class="math-container">$i$</span>", I dind't understand why exists the subsequence <span class="math-container">$(K^1_i)_{i \in \mathbb{N}}$</span> and why <span class="math-container">$T_1$</span> is independent of <span class="math-container">$i$</span>.</p>
</li>
<li><p>Why <span class="math-container">$(1.62)$</span> implies that <span class="math-container">$\delta(K^m_i,K^k_j) \leq 2^{-m} \sqrt{n \gamma} \ \text{for} \ i,j \in \mathbb{N} \ \text{and} \ k \geq m$</span>?</p>
</li>
</ol>
<p>Thanks in advance!</p>
|
Community
| -1 |
<p>Your parametrization is correct. You missed the jacobian factor during the change of variable. $\int_{-2}^{2} \int_{0}^{3\sqrt{1-x^2/4}}-2dxdy=-2.\frac{1}{2}$area of allipse=$-6\pi$
($\because \frac{x^2}{a^2}+\frac{y^2}{b^2}$, area enclosed by the closed ellipse=$\pi a b$)</p>
|
1,428,905 |
<p>I have two functions:</p>
<p>$n!$</p>
<p>$2^{n^{2}}$</p>
<p>What is the difference between the growth of these two? My thought is that $2^{n^2}$ grows much faster than $n!$. </p>
|
bof
| 111,012 |
<p>$$\frac{2^{n^2}}{n!}=\frac{2^n}1\cdot\frac{2^n}2\cdot\frac{2^n}3\cdot\cdots\cdot\frac{2^n}n\rightarrow\infty$$</p>
|
10,949 |
<p>Is it known whether every finite abelian group is isomorphic to the ideal class group of the ring of integers in some number field? If so, is it still true if we consider only imaginary quadratic fields?</p>
|
Arturo Magidin
| 742 |
<p>At least as of 1999, this was still an open question, according to the MathReview of a paper by Marc Perret (<em>On the ideal class group problem for global fields</em>, J. Number Theory <strong>77</strong> (1999), no. 1, pages 27-35; MR1695698 (2000d:11135), review by Bruno Anglès). </p>
<p>Luther Claborn proved in 1966 that every abelian group is the ideal class group of <em>some</em> Dedekind domain (<em>Every abelian group is a class group</em>, Pacific J. Math. <strong>18</strong> (1966), pages 219-222, MR0195889 (33 #4085)). </p>
<p>Gary Cornell (<em>Abhyankar's lemma and the class group</em>. Number theory, Carbondale 1979 (Proc. Southern Illinois Conf., Southern Illinois Univ., Carbondale, Ill., 1979), pp. 82–88, Lecture Notes in Math., 751, Springer, Berlin, 1979, MR0564924 (82c:12007)) proved that every finite abelian group is a <em>subgroup</em> of the ideal class group of a cyclotomic extension of $\mathbb{Q}$. There have been some refinements, but I did not find anything in MathSciNet that would suggest the problem has been settled in the interim.</p>
|
1,673,771 |
<p>I was wondering if this proof is valid. </p>
<p>I use $[x]$ to denote the floor of $x$.</p>
<p><strong>Problem</strong> </p>
<p>Prove that</p>
<p>$$[mx] = \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg]$$</p>
<p>where $m \in \mathbb{N}$ and $x \in \mathbb{R}$.</p>
<p><strong>Proof</strong></p>
<p>Let $m \in \mathbb{N}$ and let $x \in \mathbb{R}$ such that $\epsilon = x - [x]$ where $0 \leq \epsilon < 1$.</p>
<p>Partition the interval $[0,1)$ as </p>
<p>$$[0,1) = \bigcup_{k=0}^{m-1} \, \bigg[\frac{k}{m}, \frac{k+1}{m} \bigg)$$</p>
<p>Let $p \in \{1, \dotsc, m\}$ and consider the interval</p>
<p>$$\frac{p-1}{m} \leq \epsilon < \frac{p}{m}$$</p>
<p>Expanding and simplifying the sum $\sum_{k=0}^{m-1} \, \big[x+\frac{k}{m} \big]$ renders</p>
<p>$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= [x] + \bigg([x] + \bigg[\epsilon + \frac{1}{m}\bigg]\bigg) + \cdots + \bigg([x] + \bigg[\epsilon + \frac{m-1}{m} \bigg]\bigg) \\ &= m[x] + \sum_{k=1}^{m-1} \, \bigg[\epsilon + \frac{k}{m} \bigg] \end{align} $$</p>
<p>Since each term in the sum $\sum_{k=1}^{m-1} \, \big[\epsilon + \frac{k}{m} \big]$ either equals 0 or 1 depending on $p$, we observe that there are at most $(m-p)$ zeros and $(p-1)$ ones amongst the $(m-1)$ terms of the given series. Hence,</p>
<p>$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= m[x] + \sum_{k=1}^{m-1} \, \bigg[\epsilon + \frac{k}{m} \bigg] \\\\ &= m[x] + (m-p)\cdot 0 + (p-1) \cdot 1 \\\\ &= m[x] + (p-1) \end{align}$$</p>
<p>Furthermore, $\frac{p-1}{m} \leq \epsilon < \frac{p}{m}$ implies $p-1 \leq m \cdot \epsilon < p$. Thus we can write</p>
<p>$$\begin{align} [mx] &= \big[m([x]+\epsilon)\big] \\\\ &= m[x] + [m \cdot \epsilon] \\\\ &= m[x] + (p-1). \end{align}$$</p>
<p>Since for all intervals $p = 1, \dotsc, m$ we have </p>
<p>$$\begin{align} \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg] &= [mx] \\ &= m[x] + (p-1) \end{align}$$</p>
<p>we conclude that $[mx] = \sum_{k=0}^{m-1} \, \bigg[x+\frac{k}{m} \bigg]$</p>
|
heropup
| 118,193 |
<p>I think it's a good effort but your proof is difficult to follow due to some issues with presentation and definition. For example, the partition of $[0,1) = \bigcup_{k=1}^m [k/m, (k+1)/m)$ is problematic because it obviously fails to contain $0$. Also, if you say "let $p = 1, 2, \ldots, m$ and consider the $p^{\rm th}$ subinterval..." this is not clear, nor is it clear that if $p$ is to take on a sequence of values, that in fact the following inequality is true for at most one <em>particular</em> $p$. At that point, I stopped reading carefully.</p>
<p>A more elegant proof of the stated identity exploits periodicity: consider the function defined for all nonnegative integers $m$ and reals $x$: $$f_m(x) = -\lfloor mx \rfloor + \sum_{k=0}^{m-1} \left\lfloor x + \frac{k}{m} \right\rfloor.$$ Then note that $$\begin{align*} f_m(x+1/m) &= -\lfloor m(x+1/m) \rfloor + \sum_{k=0}^{m-1} \left\lfloor x + \frac{k+1}{m} \right\rfloor \\ &= -\lfloor mx \rfloor - 1+ \sum_{k=1}^m \left\lfloor x + \frac{k}{m} \right\rfloor \\ &= f_m(x) - 1 + \lfloor x+m/m\rfloor - \lfloor x \rfloor \\ &= f_m(x).\end{align*}$$ That is to say, $f_m$ is a periodic function for which $1/m$ is an integer multiple of its period. So it suffices to consider the behavior of $f_m(x)$ on the interval $x \in [0, 1/m)$. But on this interval, $\lfloor mx \rfloor = 0$, and because $0 \le x + k/m < 1$ for all $0 \le k \le m-1$ and $0 \le x < 1/m$, all the terms in the sum are also zero; thus $f_m(x)$ is identically zero on $[0,1/m)$, and by extension, is zero for all $x \in \mathbb R$. This completes the proof.</p>
|
379,893 |
<p>Define the sequence <span class="math-container">$b_1=1$</span> and
<span class="math-container">$$b_n=\sum_{k=1}^{n-1}\binom{n-1}k\binom{n-1}{k-1}b_kb_{n-k}.$$</span></p>
<p>By now, there is enough in the literature that <span class="math-container">$C_n$</span> is odd iff <span class="math-container">$n=2^k-1$</span> for some <span class="math-container">$k$</span> where <span class="math-container">$C_n$</span> are the Catalan numbers: <span class="math-container">$C_0=1$</span> and
<span class="math-container">$$C_{n+1}=\sum_{k=0}^nC_kC_{n-k}.$$</span></p>
<p>In the same spirit, I ask:</p>
<blockquote>
<p><strong>QUESTION.</strong> is it true that <span class="math-container">$b_n$</span> is odd iff <span class="math-container">$n=2^k$</span> for some <span class="math-container">$k$</span>?</p>
</blockquote>
|
David Kern
| 165,619 |
<p>Multicategories and bicategories, to me, are first of all completely
orthogonal generalisations of <em>monoidal</em> categories, with virtual
double categories as a common generalisation of multicategories and
(strict) <span class="math-container">$2$</span>-categories (they are to multicategories as categories are
to monoids, or <span class="math-container">$2$</span>-categories to monoidal categories). As for
generalising simply categories, they are even more different, as
multicategories are still a strictly associative structure while
coherence issues appear for bicategories.</p>
<p>So, to your second highlighted question, I would say that the answer
is no. A way to more precisely understand the difference, and a
positive answer to your first question, lies in your final (bonus)
question.</p>
<p>In addition to the aforementioned algebraic aspect, double categories
have a large conceptual advantage over bicategories for formal
category theory; in short, if you try to treat the objects of an
arbitrary <span class="math-container">$2$</span>-category as abstract categories, you will be lacking a
lot of elements (the Yoneda structure coming from profunctors a.k.a.
bimodules) to speak about limits in them, while double categories (at least
the ones equipping their vertical category with proarrows) will give
you enough. Virtual double categories are just the relevant
generalisation for when bimodules do not compose, and the augmented
version deals with the case lacking identity bimodules (<em>e.g.</em> for
non-locally small categories). This is, in my opinion, the main conceptual
payoff for (augmented) virtual double categories.</p>
<p>To finish, two technical generalisations, the latter of which was your
second-and-a-half question:</p>
<ul>
<li><p>Virtual double categories (though not the augmented ones, as far as
I know), are an example of "generalised multicategories", something
that can be defined relative to any monad acting on a virtual proarrow equipment. For this, see Leinster's book suggested in <a href="https://mathoverflow.net/questions/379853/multicategories-vs-categories#comment964588_379853">varkor's
comment</a>
(chapters 4 and 5) or the more general and recent <a href="http://tac.mta.ca/tac/volumes/24/21/24-21abs.html" rel="noreferrer"><strong>A unified framework for generalized
multicategories</strong> by Cruttwell and
Shulman</a>.</p>
</li>
<li><p>There does exist a higher-categorical version of multicategories and
even virtual double categories, defined (as "generalised non-symmetric
<span class="math-container">$\infty$</span>-operads") and put to great use (precisely to unify
<span class="math-container">$\infty$</span>-multicategories and double <span class="math-container">$\infty$</span>-categories) in <a href="http://dx.doi.org/10.1016/j.aim.2015.02.007" rel="noreferrer">Gepner–Haugseng's <strong>Enriched
<span class="math-container">$\infty$</span>-categories via non-symmetric
<span class="math-container">$\infty$</span>-operads</strong></a>
and follow-ups. A more general form, close to the generalised
multicategories, is offered by Chu–Haugseng's formalism of algebraic
patterns developed in <a href="https://arxiv.org/abs/1907.03977" rel="noreferrer"><strong>Homotopy-coherent algebra via Segal
conditions</strong></a> (see in particular
section 9 with ex. 9.8). For the moment this is only used for algebraic aspects rather than formal (higher) category theory, but I would argue it is definitely worth learning (not <em>over</em> higher category theory, but as an extension of it).</p>
</li>
</ul>
|
1,269,447 |
<p>I consider the space $C^1[a, b]$ of (complex) functions that are at least once differentiable on $[a, b]$. I want to show that</p>
<p>$$||f||_{C^1} := ||f||_\infty + ||f'||_\infty$$</p>
<p>defines a norm on $C^1[a, b]$.</p>
<p>Now it's easy to see that $||f||_{C^1}$ is non-negative, and that it's zero iff f = 0, and it can be easily shown that it's compatible with multiplication of a scalar. But so far I've been struggling with proving and understanding that the triangle inequality is valid indeed. Thanks in advance.</p>
|
minthao_2011
| 40,146 |
<p>Solution.</p>
<p><strong><em>First way</em></strong></p>
<p>From the first equation, we have
$$\begin{cases}
2x^2\leqslant 1,\\
y^2 \leqslant 1
\end{cases}
\Leftrightarrow
\begin{cases}
-\dfrac{1}{\sqrt{2}} \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
Then, the conditions of $x$ and $y$ are
$$\begin{cases}
0 \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
We have $x^2 + y^2 = 1-x^2$. Therefore $x^2 + y^2 \leqslant 1$.
Another way,
$$1-x^2 = y \sqrt{1-x^2} -(1-y)\sqrt{x} \leqslant y \sqrt{1-x^2}.$$
Because
$$ y \sqrt{1-x^2} \leqslant \dfrac{y^2 + 1 - x^2}{2}.$$
Implies
$$1-x^2 \leqslant \dfrac{y^2 + 1 - x^2}{2} \Leftrightarrow x^2 + y^2 \geqslant 1 .$$
From $x^2 + y^2 \leqslant 1$ and $x^2 + y^2 \geqslant 1$, we have $x^2 + y^2 = 1.$
Solve
$$\begin{cases}
x^2 + y^2 = 1,\\
2x^2 + y^2 = 1,\\
0 \leqslant x \leqslant 1,\\
- 1 \leqslant y \leqslant \dfrac{1}{\sqrt{2}}
\end{cases} \Leftrightarrow \begin{cases}
x = 0,\\
y = 1.\end{cases}$$</p>
<p><strong><em>Second way.</em></strong>
We have $2x^2 + y^2 = 1$, therefore $y=\sqrt{1 - 2x^2}$ or $y=-\sqrt{1 - 2x^2}.$</p>
<p>First case, $y=\sqrt{1 - 2x^2}$, subtitution the second equation, we get
$$x^2 + \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2}=1+(1-\sqrt{1 - 2x^2})\sqrt{x}.$$
equavalent to
$$1 - x^2 - \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2} + (1-\sqrt{1 - 2x^2})\sqrt{x}=0$$
or
$$\sqrt{1-x^2} (\sqrt{1-x^2} - \sqrt{1 - 2x^2}) + (1-\sqrt{1 - 2x^2})\sqrt{x}=0.$$
This is equavalent to
$$\dfrac{\sqrt{1-x^2}\cdot(1-x^2-1+2x^2)}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}}+\dfrac{(1-1+2x^2)\sqrt{x}}{1+\sqrt{1 - 2x^2}} = 0$$
or
$$x^2\left (\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}\right )=0.$$
It is easy to see that
$$\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}> 0$$
Thus $x = 0$.</p>
<p>Second cases. $y=-\sqrt{1 - 2x^2}.$
We can check that
$$x^2 + y \sqrt{1-x^2} \leqslant \dfrac{1}{2}$$
and
$$1+(1-y)\sqrt{x} >1.$$
In this case, the given system of equations has no solution. </p>
|
4,616,048 |
<p>How can I solve this system of coupled differential equations?</p>
<p><span class="math-container">$\frac{d^2\rho}{d\lambda^2}=\frac{5\rho}{(5\rho^2+4t^2)^2}$</span> <span class="math-container">$\frac{d^2t}{d\lambda^2}=\frac{4t}{(5\rho^2+4t^2)^2}$</span></p>
<p>Is it something I could input in the Wolfram calculator?</p>
|
Hans Lundmark
| 1,242 |
<p>I hope you'll excuse me if I write <span class="math-container">$x(t)$</span> and <span class="math-container">$y(t)$</span> instead of <span class="math-container">$\rho(\lambda)$</span> and <span class="math-container">$t(\lambda)$</span>.</p>
<p>The system has the form
<span class="math-container">$$
\begin{pmatrix} \ddot x \\ \ddot y \end{pmatrix}
=
\frac{1}{(5x^2+4y^2)^2}
\begin{pmatrix} 5x \\ 4y \end{pmatrix}
=
- \begin{pmatrix} \partial V/\partial x \\ \partial V/\partial y \end{pmatrix}
,
$$</span>
where
<span class="math-container">$$
V(x,y) = -\frac{1/2}{5x^2+4y^2}
.
$$</span>
This means that the total energy (kinetic + potential) <span class="math-container">$E = \frac12 (\dot x^2 + \dot y^2) + V(x,y)$</span> is a constant of motion.</p>
<p>Moreover, in polar coordinates the potential <span class="math-container">$V$</span> takes the form
<span class="math-container">$$
V = - \frac{1/2}{r^2 (4+\cos^2 \phi)}
,
$$</span>
which matches the general form for potentials that are <em>separable</em> in polar coordinates,
<span class="math-container">$$
V = f(r) + \frac{1}{r^2} \cdot g(\phi)
,
$$</span>
with <span class="math-container">$f(r)=0$</span> and <span class="math-container">$g(\phi) = -\dfrac{1/2}{4+\cos^2 \phi}$</span>.</p>
<p>What this means is that you can apply the <em>Hamilton–Jacobi method</em> in polar coordinates, which in principle allows you to integrate the equations explicitly (although in practice it can get a bit messy). Unfortunately this method is quite complicated, so I won't try to describe it here; see some textbook in classical mechanics, such as <a href="https://en.wikipedia.org/wiki/Course_of_Theoretical_Physics" rel="nofollow noreferrer">Landau & Lifshitz</a> (Vol. 1, §48).</p>
|
160,801 |
<p>Here is a vector </p>
<p>$$\begin{pmatrix}i\\7i\\-2\end{pmatrix}$$</p>
<p>Here is a matrix</p>
<p>$$\begin{pmatrix}2& i&0\\-i&1&1\\0 &1&0\end{pmatrix}$$</p>
<p>Is there a simple way to determine whether the vector is an eigenvector of this matrix?</p>
<p>Here is some code for your convenience.</p>
<pre><code>h = {{2, I, 0 },
{-I, 1, 1},
{0, 1, 0}};
y = {I, 7 I, -2};
</code></pre>
|
Henrik Schumacher
| 38,178 |
<p>Yes! We just check whether $h.y = (u + I v) y$ holds for some real $u, v \in \mathbb{R}$.</p>
<pre><code>h = {{2, I, 0}, {-I, 1, 1}, {0, 1, 0}};
y = {I, 7 I, -2};
expr = Norm[h.y - (u + I v) y, 2]^2 // ComplexExpand;
Minimize[expr, {u, v}]
</code></pre>
<blockquote>
<p>{623/6, {u -> 17/18, v -> 0}}</p>
</blockquote>
<p>Answer: Nope, it's not.</p>
|
2,080,042 |
<blockquote>
<p>I am interested of finding examples of non-zero homomorphisms $f:R\to S$ of rings with unity such that $f(1_R)\neq 1_S$.</p>
</blockquote>
<p>I will provide one example and I will be glad if others can also give examples. </p>
|
Mustafa
| 400,050 |
<p>Let be $f:\mathbb Z \to M_2(\mathbb Z); f(a)= \left( \array{a&0\\0&0}\right) $ then $f$ is ring homomorphism and $f(1)=\left(\array{1&0\\0&0}\right) \ne \left(\array{1&0\\0&1}\right)$ . </p>
|
725,602 |
<p>I am trying to prove the 'second' triangle inequality:
$$||x|-|y|| \leq |x-y|$$</p>
<p>My attempt:
$$----------------$$
Proof:
$|x-y|^2 = (x-y)^2 = x^2 - 2xy + y^2 \geq |x|^2 - 2|x||y| + |y|^2 = (||x|-|y||)^2$</p>
<p>Therefore $\rightarrow |x-y| \geq ||x|-|y||$</p>
<p>$$----------------$$</p>
<p>My questions are: Is this an acceptable proof, and are there alternative proofs that are more efficient?</p>
|
user137500
| 137,500 |
<p>Induction will work fine. Assuming it works for $n$, we have $$F_{n+2}F_n-F_{n+1}^2=(F_{n+1}+F_n)F_n-F_{n+1}^2=F_{n+1}(F_n-F_{n+1})+F_n^2=F_{n+1}(-F_{n-1})+F_n^2$$ $$=-\left(F_{n+1}F_{n-1}-F_n^2\right)=-(-1)^n=(-1)^{n+1}$$</p>
|
1,591,371 |
<p>If we start with $n$ elements and at each step split them into $2$ parts randomnly and repeat with both sub-parts until parts of only $1$ element are left, in how many different ways can these elements be separated?
I made a mistake we don't split them in half we split them in a random place.</p>
|
true blue anil
| 22,388 |
<p>Imagine a log of wood which you want to cut into $n$ parts.</p>
<p>A little thought will show that you need to make $(n-1)$ cuts.</p>
<p>These cuts can be in any order, thus # of ways to do it = $(n-1)!$</p>
<p>e.g. if $n = 4,$ you need $(4-1)! = 3\times2\times1 = 6$</p>
|
2,128,380 |
<p>Okay, I must admit that I am lost on how to do this. I have looked up videos and tutorials about this, and they helped a little. The main thing is that my professor asked for us to solve this without using the "determinant method." I have just started linear algebra, so I am still trying to understand determinants and the like. I am just wondering how it is possible to prove the cross product of two vectors with another method? Any help would be great!</p>
<p>Prove the following without using the determinant method:</p>
<p>$A \times B = (A_yB_z - B_yA_z)\hat{i} - (A_xB_z - B_xA_z)\hat{j} + (A_xB_y - B_xA_y)\hat{k}$</p>
|
Reinhard Meier
| 407,833 |
<p>The proof can be given using the distributive property of the cross product and the fact that $c(v\times w) = (cv)\times w = v\times (cw)$ for vectors $v$ and $w$ and a scalar $c$:
$$
A\times B = (A_x \hat i + A_y \hat j + A_z \hat k)\times (B_x \hat i + B_y \hat j + B_z \hat k)\\
= A_xB_x(\hat i\times\hat i)+A_xB_y(\hat i\times \hat j)+A_xB_z(\hat i\times \hat k)\\
+ A_yB_x(\hat j\times\hat i)+A_yB_y(\hat j\times \hat j)+A_yB_z(\hat j\times \hat k)\\
+ A_zB_x(\hat k\times\hat i)+A_zB_y(\hat k\times \hat j)+A_zB_z(\hat k\times \hat k)\\
$$
Now we only need to use the obvious properties of $\hat i$,$\hat j$ and $\hat k$:
$$
(\hat i\times\hat i) = (\hat j\times \hat j) = (\hat k\times \hat k) = 0 \\
(\hat i\times \hat j) = \hat k,\;(\hat j\times \hat k) = \hat i,\;(\hat k\times \hat i) = \hat j\\
(\hat j\times \hat i) = -\hat k,\;(\hat k\times \hat j) = -\hat i,\;(\hat i\times \hat k) = -\hat j\\
$$</p>
|
125,165 |
<p>Hi friends,</p>
<p>I have some questions concerning the critical values of motives, in the sense of Deligne. I will only look at motives of the form $h^i(X)$ where $X$ is a smooth projective algebraic variety over $\mathbb{Q}$. If I understand correctly, the notion of critical value depends only on the Hodge numbers. </p>
<p>Introduce the notations:</p>
<p>$$
\Gamma_\mathbb{C}(s):=2(2\pi)^{-s}\Gamma(s), \quad \Gamma_\mathbb{R}(s):=\pi^{-s/2}\Gamma(\frac{s}{2})
$$ where $\Gamma$ is the usual Gamma function. Then one define the $L$-factor at infinity as follows. </p>
<p>Let us first consider the case where $i$ is odd. Then:</p>
<p>$$
L_\infty(h^i(X), s)=\prod_{p < q} \Gamma_{\mathbb{C}}(s-p)^{h^{p, q}}
$$ where $p+q=k$ and $h^{p, q}$ is the corresponding Hodge number. </p>
<p><strong>Example</strong>: If $i=3$ and $h^{3, 0}=0$ (for instance $X$ is an hypersurface in $\mathbb{P}^4$), then </p>
<p>$L_\infty(h^3(X), s)=\Gamma_{\mathbb{C}}(s-1)^{h^{2, 1}}$</p>
<p>When $i$ is even, the definition is more involved, as one has also to consider the action of complex conjugation on $H^{p, p}$, which decomposes this space as $H^{p^+}\oplus H^{p^{-}}$. Let</p>
<p>$$
h^{p^{\pm}}:=\dim_{\mathbb{C}} H^{p^{\pm}}.
$$</p>
<p>Then
$$
L_\infty(h^i(X), s)=\prod_{p < q} \Gamma_{\mathbb{C}}(s-p)^{h^{p, q}}\cdot \Gamma_{\mathbb{R}}(s-\frac{i}{2})^{h^{i/2+}}
\Gamma_{\mathbb{R}}(s-\frac{i}{2}+1)^{h^{i/2-}}
$$</p>
<p>After having introduced this: an integer $n$ is said to be critical for $M=h^i(X)$ if $L_\infty(M, s)$ has no pole at $s=n$. </p>
<p>In his paper Deligne also asks that $L_\infty(\hat{M}, 1-s)$ has no pole at $n$. Is that necessary or it is a consequence of the former provided one has a functional equation?</p>
<p>Anyway, I would like to know if for my example the critical integers are all integers $n \geq 2$. </p>
<p><strong>Question 2</strong> What about a $K3$ surface? Can one determine the critical integers for the transcedental lattice? In that case there is a $\Gamma(s)$ coming from $h^{2,0}=1$. What about the real Gamma factor?</p>
<p>Thanks for your help!</p>
|
David Loeffler
| 2,481 |
<p>In your example (with Hodge structure of weight 3 concentrated in bidegrees (2, 1) and (1, 2)) there is only one critical value, at s = 2. At all other integer values of $s$, either $L_\infty(M, s)$ or $L_\infty(M^\vee, 1-s)$ will have a pole.</p>
|
365,986 |
<p>If $A$ is an $n \times n$ matrix with $\DeclareMathOperator{\rank}{rank}$ $\rank(A) < n$, then I need to show that $\det(A) = 0$.</p>
<p>Now I understand why this is - if $\rank(A) < n$ then when converted to reduced row echelon form, there will be a row/column of zeroes, thus $\det(A) = 0$</p>
<p>However, I have been told to use the fact that the determinant is multilinear and alternating and subsequently deduce that if $\det(A)$ is non-zero, $A$ is invertible. </p>
<p>How do I use the properties of the determinant to prove these claims? </p>
|
Abel
| 71,157 |
<p>Here goes the sketch of a proof.</p>
<p>Let $A$ be your matrix. Do Gaussian elimination on $A$ so that you end up with an upper triangular matrix $U$.
During this process, you either</p>
<p>1) Switch rows, which switches the sign of the determinant.</p>
<p>2) Add a multiple of a row to another row, which preserves the determinant.</p>
<p>Thus, $\det(U) = \pm\det(A)$. Hence if $\det(A)\neq 0$, then $\det(U)\neq 0$. Since $U$ is upper triangular, this means that the diagonal of $U$ does not contain any zeroes.</p>
<p>Hence $U^{-1}$ exists and doing the 'reverse' of the Gaussian elimination done for $A$ gives $A^{-1}$ from $U^{-1}$.</p>
<p>You do need to work out some messy details, but I'm sure it can be made to work.</p>
|
1,129,712 |
<p>So I'm complete stuck with something. I know it the following statements are true (or at least the seem to be from the results that I got from messing around with it a bit on MATLAB), but I don't understand why they are true or how to show so.
Let $A$ be and $m$X$n$ matrix. Show that:</p>
<p>a) if $x \in N(A^TA)$ then $Ax$ is in both $R(A)$ and $N(A^T)$. </p>
<p>For this one I messed around with it with my own examples and I got $Ax=0$, therefore satisfing the statement, but I don't understand what's actually going on. </p>
<p>b) $N(A^TA)=N(A)$</p>
<p>again, makes sense when I see the results in MATLAB, but don't undestand why it works.</p>
<p>c) $A$ and $A^TA$ have the same rank</p>
<p>d) If $A$ has linearly independent columns, the $A^TA$ is nonsingular.</p>
<p>For the last two I have no idea on how to even start showing the relationship. I feel like I'm missing some crucial relationship between $A$ and $A^TA$, but I'm just not seeing it. </p>
<p>I would greatly appreciate any help of sugestions on how to show that these statements are true. </p>
<p>Thank you very much. </p>
|
Pablo Repetto
| 79,624 |
<p>Let us show that (c) and (d) are both consecuences of (b):</p>
<p>Let $A\in R^{nxm}$ and $nul(A) = nul(A^TA)$</p>
<blockquote>
<p>$ \dim nul(A)
= \dim nul(A^TA) $</p>
<p>$ n-rank(A) = n-rank(A^TA)$</p>
<p>$rank(A)=rank(A^TA)$</p>
</blockquote>
<p>and, if $A$ has linearly independent columns, $nul(A)=\{0_{R^m}\} = nul(A^TA)$; and thus is proven that $\det(A^TA) \neq 0$</p>
<p>As for the proof of (b), and a more rigorous proof of $rank(A) = rank(A^TA)$, I'll post it after dinner.</p>
<h3>EDIT:</h3>
<p>$\dim col(A) + \dim nul(A) = n$</p>
<p>This property is what I had left unproven in my proof that (c) is consequence of (b), so there:</p>
<p>Let $A =
\left( \begin{array}{ccc}
- & v_1 & -\\
& \ldots &\\
- & v_n & -
\end{array} \right)$ then:</p>
<blockquote>
<p>$Ax = 0$ iff $v_ix = 0, \forall i$</p>
<p>$x \perp v_i, \forall i$</p>
</blockquote>
<p>$x$ is perpendicular to every column of $A^T$, so it follows that</p>
<blockquote>
<p>$x \in nul(A)$ iff $x \in (col(A^T))^\perp$</p>
</blockquote>
<p>And so:</p>
<blockquote>
<p>$\dim nul(A) = \dim col(A^T)^\perp$</p>
<p>$ = n- \dim col(A^T) = n-rank(A^T) = n-rank(A)$</p>
</blockquote>
<h3>EDIT:</h3>
<p>Proof of (b):</p>
<p>Let $A \in R^{nxm}$:</p>
<blockquote>
<p>$nul(A^TA) = \{ x \in R^m | A^TAx = 0_{R^m} \}$</p>
<p>$= \{ x \in R^m | Ax = 0_{R^n} \quad$ or $\quad Ax \in nul(A^T) \}$</p>
<p>but we have proven that $x \in nul(A^T) \quad$ iff $\quad x \in (col(A))^\perp$</p>
<p>There can be no $Ax \neq 0_{R^m}$ such that $A^TAx = 0_{R^n} \quad$ !</p>
<p>and thus: $\quad nul(A^TA) = \{ x \in R^m | Ax = 0_{R^n} \} = nul(A)$</p>
</blockquote>
|
1,293,207 |
<p>A ray of light travels from the point $A$ to the point $B$ across the border between two materials. At the first material the speed is $v_1$ and at the second it is $v_2$. Show that the journey is achieved at the least possible time when Snell's law: $$\frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}$$ holds. </p>
<p>To show that do we have to use Lagrange multipliers method?? </p>
<p>But which function do we want to minimize??</p>
|
Hagen von Eitzen
| 39,174 |
<p>The total travel time, which is the sum of travel times from $A$ to the surface and fromthe surface to $B$. The two part times are obtained as length divided by speed, the two lengths depend on the angles.</p>
|
1,556,645 |
<p>I am new to the axiom of choice, and currently working my way through some exercises. I am struggling with the following exercise:</p>
<p><strong>Exercise -</strong>
Prove the Axiom of Choice (every surjective $f: X \to Y$ has a section) in the following two special cases:</p>
<ol>
<li>Y is finite</li>
<li>X is countable</li>
</ol>
<p>(A section has been previously defined as a function $s: Y \to X$ such that $f(s(y)) = y$ for all $y \in Y$)</p>
<p><strong>My confusion -</strong>
Now from what I understand, in (1) you use surjectivity of $f$ to pick an $x \in X$ such that $f(x) = y_0$ proving the case $|Y| = 1$, and then use induction to proof it for general $|Y| = N$.</p>
<p>I am a bit confused at (2) though. Would it be legal to take the previous argument and take the limit $N \to \infty$? I tried to google it but I got more confused after reading <a href="https://math.stackexchange.com/questions/241995/countable-infinity-and-the-axiom-of-choice">this question</a> and seeing other references to the axiom of countable choice, suggesting that this result cannot be proven.</p>
<p>By the way, I am doing 'naive' set theory here; ZF/ZFC axiom systems and those kind of things have not been discussed in the course.</p>
|
Clément Guérin
| 224,918 |
<p>The countable axiom of choice refers to the case where $Y$ is countable, so it is kind of different.</p>
<p>Let us see how we could do this. Since $X$ is countable there is a bijection $\phi$ with $\mathbb{N}$. Now you have a well defined order on $X$ :</p>
<p>$$x\leq y \text{ by definition if } \phi(x)\leq \phi(y)\text{ in }\mathbb{N} $$</p>
<p>I claim that this order on $Y$ verifies that the property " any non-empty set has a least element" is true (I don't remember right now the name of this property...). </p>
<p>Another way to state this is that in any non-empty $C$ set of $Y$ you can canonically $\textit{ choose }$ an element in $C$ namely $min(C)$. Apply this to construct the section for $f$. </p>
|
148,185 |
<p>Let $ X = \mathbb R^3 \setminus A$, where $A$ is a circle. I'd like to calculate $\pi_1(X)$, using van Kampen. I don't know how to approach this at all - I can't see an open/NDR pair $C,D$ such that $X = C \cup D$ and $C \cap D$ is path connected on which to use van Kampen. </p>
<p>Any help would be appreciated. Thanks</p>
|
Lilith
| 114,997 |
<p>You should definitely check out the Hatcher's Algebraic Topology book page 46. </p>
<p>It was very hard for me to imagine at first but $\mathbb{R}^3 - S^1$ deformation retracts onto $S^1 \vee S^2$ so just choose $S^1$ and $S^2$ for $C$ and $D$ respectively, since the space is formed as wedge product of two spaces, the intersection is going to be a point only (by definition) whose fundamental group is trivial for sure. Similarly $\pi_1(S^2)$ is also trivial then $\pi_1(\mathbb{R}^3 - S^1)$ is isomorphic to the fundamental group of the circle which is $\mathbb{Z}$. </p>
|
23,471 |
<p>I'm trying to find an explanation for the different sizes I'm seeing for fonts added to graphics in different ways, and haven't yet located an easy to understand explanation. Here's a minimal example:</p>
<pre><code>Graphics[
{LightGray,
Rectangle[{0, 0}, {72, 72}],
Red,
Style[
Text["Hig", {0, 0}, {-1, -1}], 72,
FontFamily -> "Times New Roman"],
Black,
First[
First[
ImportString[
ExportString[
Style["Hig",
FontFamily -> "Times New Roman",
FontSize -> 72], "PDF"],If[$VersionNumber>=13,{"PDF","PageGraphics"},"PDF"],
"TextMode" -> "Outlines"]]]
},
PlotRange -> {{0, 100}, {0, 100}},
Axes -> True,
Ticks -> {Table[x, {x, 0, 100, 10}], Table[x, {x, 0, 100, 10}]},
Epilog -> {Text["72", {200, 75}], Line[{{0, 72}, {200, 72}}]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/l2wT7.png" alt="fonts" /></p>
<p>The red text doesn't change size when you resize the graphic, although everything else changes. The black text resizes along with everything else. But neither text seems to be the result of specifying 72.</p>
<h2>Update</h2>
<p>After changing the screen resolution to Automatic following @Sjoerd's suggestion, I can see how the red text is basically displaying at a fixed size that's independent of <em>Mathematica</em>. In this picture, the left image shows 72 dpi, the red font is using 70ish pixels of vertical space. The right image is at 'Automatic' (so presumably 133.51 for my machine, according to <a href="http://members.ping.de/%7Esven/dpi.html" rel="nofollow noreferrer">this site</a>, and similarly uses up 70ish pixels.</p>
<p><img src="https://i.stack.imgur.com/Qjbuv.png" alt="resolutions" /></p>
<p>I'm still puzzled by the size of the black font, which doesn't seem to be related to the specified font size or to the screen resolution. Perhaps the PDF translation introduces another scaling factor.</p>
|
geordie
| 4,626 |
<p>The <code>FontSize</code> refers to <a href="http://en.wikipedia.org/wiki/Point_%28typography%29" rel="noreferrer">printers points</a>, which by convention are 1/72 of an inch. So the red 'Hig' is fixed to an external font size and will not scale with the graphic. However, you are right to say that this font size doesn't seem to be strictly 72 points... If you try:</p>
<pre><code>Export["hig.pdf", Style["Hig", FontFamily -> "Times New Roman", FontSize -> 72]]
</code></pre>
<p>a quick comparison with 72pt text generated in Illustrator shows that the fontsize is ~51 pts or ~30% too small (the Mathematica 72pt text is on the right): </p>
<p><img src="https://i.stack.imgur.com/8GYO7.jpg" alt="illustrator 72pt vs mma 72pt"></p>
<p>I have no clue as to why Mathematica text is too small....</p>
<p>Your second block of text is converted to a scalable graphic by the <code>ImportString[
ExportString[Style[...], "PDF"],"TextMode" -> "Outlines"]]]</code> transform. Enclosing this routine in <code>FullForm</code> demonstrates that the text has been converted into a <code>FilledCurve</code>comprised of nested lists (i.e. is no longer treated as text but rather as a scalable graphic element). As a result, it is scaled with the units of the graphic. </p>
<pre><code>FullForm[ImportString[
ExportString[
Style["Hig", FontFamily -> "Times New Roman", FontSize -> 72],
"PDF"], "TextMode" -> "Outlines"]]
</code></pre>
<blockquote>
<p>List[Graphics[
List[Thickness[0.0128601],
Style[List[
FilledCurve[
List[List[List[0, 2, 0], List[0, 1, 0], List[1, 3, 3],
List[1, 3, 3], List[1, 3, 3], List[0, 1, 0], List[0, 1, 0],
List[0, 1, 0], List[0, 1, 0], List[0, 1, 0], List[1, 3, 3],
List[1, 3, 3], List[1, 3, 3],<<.......>>List[59.7569, 11.0567]]]]], Thickness[0.0128601]]],
Rule[ImageSize, List[78., 56.]],
Rule[PlotRange, List[List[0., 77.76], List[0., 56.]]],
Rule[AspectRatio, Automatic]]]</p>
</blockquote>
<p>with,</p>
<pre><code>FullForm[Style["Hig", FontFamily -> "Times New Roman", FontSize -> 72]]
</code></pre>
<blockquote>
<p>Style["Hig", Rule[FontFamily, "Times New Roman"], Rule[FontSize, 72]]</p>
</blockquote>
<p>Interestingly, even when scaled, the size disparity persists, as can be seen if you resize your graphic such that the red and black text overlap, then the grey square is approximately one inch wide...</p>
|
2,604,093 |
<p>I would like to study the convergence of the series:</p>
<p>$$\sum_{n=1}^\infty \frac{\log n}{n^2}$$</p>
<p>I could compare the generic element $\frac{\log n}{n^2}$ with $\frac{1}{n^2}$ and say that
$$\frac{1}{n^2}<\frac{\log n}{n^2}$$ and $\frac{1}{n^2}$ converges but nothing more about.</p>
|
N. S.
| 9,176 |
<p><strong>Hint</strong> Apply the integral test, and integrate by parts.</p>
|
216,031 |
<p>Using image analysis, I have found the positions of a circular ring and imported them as <code>xx</code> and <code>yy</code> coordinates. I am using <code>ListInterpolation</code> to interpolate the data:</p>
<pre><code>xi = ListInterpolation[xx, {0, 1}, InterpolationOrder -> 4, PeriodicInterpolation -> True, Method -> "Spline"];
yi = ListInterpolation[yy, {0, 1}, InterpolationOrder -> 4, PeriodicInterpolation -> True, Method -> "Spline"];
</code></pre>
<p>I plot the results as:</p>
<pre><code>splinePlot = ParametricPlot[{xi[s], yi[s]} , {s, 0, 1}, PlotStyle -> {Red}]
</code></pre>
<p>and the result looks like: </p>
<p><a href="https://i.stack.imgur.com/c2wRi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c2wRi.png" alt="interpolation of data"></a></p>
<p>I am trying to study this shape as it deforms, and I will need to look at the derivatives of this interpretation (notably, second derivatives). I know that there are physical constraints that will not let the local curvature at any point be <strong>larger</strong> than, for example <code>1/10</code> in the units show (so, a radius of curvature <code>10</code>). <strong>Is there a way that I can constrain the interpolation so that the local curvature never exceeds a given value?</strong></p>
<p>Here is the data (Dropbox Link): <a href="https://www.dropbox.com/s/g9vajch0obbcplk/testShape.csv?dl=0" rel="nofollow noreferrer">https://www.dropbox.com/s/g9vajch0obbcplk/testShape.csv?dl=0</a></p>
|
Cesareo
| 62,129 |
<p>After normalizing the data scaling conveniently we can proceed approximating the data points by a conic (ellipse). This can be done as</p>
<pre><code>f[p_List] := c1 p[[1]]^2 + c2 p[[2]]^2 + c3 p[[1]] p[[2]] + c4 p[[1]] + c5 p[[2]] + c6
data0 = Import["testShape.csv"];
factor = 0.01;
data = data0*factor;
xmax = Max[Transpose[data][[1]]];
xmin = Min[Transpose[data][[1]]];
ymax = Max[Transpose[data][[2]]];
ymin = Min[Transpose[data][[2]]];
vars = {c1, c2, c3, c4, c5, c6};
restr = {c1 > 0.5, c2 > 0.5, c1 c2 > 0.25 c3^2};
obj = Sum[f[data[[k]]]^2, {k, 1, Length[data]}];
sol = NMinimize[{obj, restr}, vars]
conic = f[{x, y}] /. sol[[2]]
gr1 = ContourPlot[conic == 0, {x, xmin, xmax}, {y, ymin-0.1, ymax}, ContourStyle -> Black];
gr2 = ListPlot[data, PlotStyle -> Red];
Show[gr2, gr1, AspectRatio -> 1]
</code></pre>
<p><a href="https://i.stack.imgur.com/Q2Iqn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q2Iqn.jpg" alt="enter image description here"></a></p>
<p>NOTE</p>
<p>The scaling can be done with the values</p>
<pre><code>xmax = Max[Transpose[data][[1]]]
xmin = Min[Transpose[data][[1]]]
ymax = Max[Transpose[data][[2]]]
ymin = Min[Transpose[data][[2]]]
</code></pre>
<p>In the present case we choose a multiplicative factor of <span class="math-container">$0.01$</span> over <strong>data</strong></p>
|
1,811,028 |
<p>I want to know which of the following methods is right for graphing $f(ax-b)$ from $f(x)$ and why:</p>
<p>Method 1. First Horizontally Translate it ($f(x)$) by $b$, then Horizontally Stretch/Compress it by $a$.</p>
<p>Method 2. First Horizontally Stretch/Compress it by $a$ then Horizontally Translate it by $b/a$.</p>
<p>Method 3. First Horizontally Translate it by $b/a$ then Horizontally Stretch/Compress it by $a$.</p>
|
Dietrich Burde
| 83,966 |
<p>A connected compact Lie group $G$ has a reductive Lie algebra $L=[L,L]\oplus Z(L)$, where $[L,L]$ is semisimple. Furthermore it has finite center $Z(G)$ if and only if $[G,G]=G$, i.e., if and only if $[L,L]=L$. It follows that a connected compact Lie group $G$ satisfies $[L,L]=L$ if and only if $Z(L)=0$, so that $L=[L,L]$ is semisimple, or equivalently $G$ is semisimple. </p>
|
2,278,018 |
<p>Let $\;\;\displaystyle \sum_{n=1}^\infty U_n\;$ be a divergent series of positive real numbers.</p>
<p>Then, show that the series $\;\displaystyle\sum_{n=1}^\infty \dfrac{U_n}{1+U_n}\;$ is divergent.</p>
<p>Is there is any easy method to prove it? </p>
|
Rigel
| 11,776 |
<p>Assume, for the sake of contradiction, that $\sum_n \frac{U_n}{1+U_n}$ is convergent.</p>
<p>Then $\dfrac{U_n}{1+U_n} \to 0$, hence
$$U_n = \frac{\frac{U_n}{1+U_n}}{1-\frac{U_n}{1+U_n}}\to 0$$
so that there exists $N\in\mathbb{N}$ such that $0\leq U_n \leq 1$ for every $n\geq N$.</p>
<p>Since
$$
\frac{U_n}{1+U_n} \geq \frac{U_n}{2}\qquad \forall n\geq N,
$$
by comparison we conclude that $\sum \dfrac{U_n}{1+U_n}$ diverges, reaching a contradiction.</p>
|
552,474 |
<p>If there are,
Are there unity <strong>(but not division)</strong> rings of this kind?
Are there non-unity rings of this kind?</p>
<p>Sorry, I forgot writting the non division condition.</p>
|
André Nicolas
| 6,312 |
<p>Take the "polynomials" with integer coefficients in two non-commuting variables $x$ and $y$. If you don't want a unit, use even integers only.</p>
<p>A related example replaces integer coefficients by coefficients in $\mathbb{Z}_2$. </p>
|
4,429,162 |
<p><a href="https://i.stack.imgur.com/9nrUn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9nrUn.png" alt="enter image description here" /></a></p>
<p>This is from Rambo's Math subject GRE book.</p>
<p>One solution to this problem is to note that the equation of the circle is <span class="math-container">$x^2+(y-k)^2=1$</span>. By taking the derivative of this and solving for <span class="math-container">$y'$</span>, and then setting this equal to the derivative of the parabola, <span class="math-container">$y'=2x$</span>, we can solve for <span class="math-container">$x$</span> and <span class="math-container">$k$</span> and get all the information we need to set up the necessary integeral:</p>
<p><span class="math-container">$2\int_0^{\frac{\sqrt{3}}{2}}\frac{5}{4}-\sqrt{1-x^2}-x^2dx=\frac{3\sqrt{3}}{4}-\frac{\pi}{3}$</span></p>
<p>Where the expression involving a square root can is evaluated with trig substitution and power reduction identities.</p>
<p>The author of the text mentions that this problem also be solved by using some trigonometry and the fact that:</p>
<p><span class="math-container">$2\int_0^{\frac{\sqrt{3}}{2}}x^2dx=\frac{\sqrt{3}}{2}$</span></p>
<p>However I can't figure out what he means by this. Can anyone show me how to solve this problem using less calculus and more trigonometry? Also, if anyone has any other alternative methods, I'd greatly appreciate it. Thanks!</p>
|
Hussain-Alqatari
| 609,371 |
<p>This solution is not by me, but available in Charles Rambo book,</p>
<p><a href="https://i.stack.imgur.com/DxvRJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DxvRJ.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/hjqH0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hjqH0.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/2eQ6t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2eQ6t.png" alt="enter image description here" /></a></p>
|
2,023,130 |
<p>Suppose I have directed graph $G=(V,E)$ s.t at least one of the following statements is always true:</p>
<ol>
<li>for every $v$ in $V$, v doesn't have any incoming edges.</li>
<li>for every $v$ in $V$, v doesn't have any outgoing edges.</li>
</ol>
<p>How can I prove that G is bipartite? I tried to think of a proof but coldn't figure it out.</p>
|
Brian M. Scott
| 12,042 |
<p>HINT: Let $V_0$ be the set of vertices with no incoming edge and $V_1$ the set of vertices with no outgoing edge.</p>
|
813,825 |
<p>In strong Induction for the induction hypothesis you assume for all K, p(k) for k
<p>If for example I am working with trees and not natural numbers can I still use this style of proof?</p>
<p>For example if I want my induction hypothesis to be that p(k) for k < n where n is a node in the tree and everything smaller than/bellow it (The nodes children,k) is assumed to be true.</p>
<p>In the proof would I have to define what the < operator does for two nodes?</p>
|
Basil
| 36,042 |
<p>Sure you can. As André Nicolas says, this is called <em>structural induction</em>, and it applies to data types that are inductively defined, that is, the structure, or rather, the <em>construction</em> of their elements is given (a) by some <em>base cases</em>, and (b) by some <em>step cases</em> involving already given elements.</p>
<p><em>Binary trees</em> for example are given (a) by one base case, the tree consisting of one <em>leaf node</em> (no children), and (b) one step case declaring that if you have already constructed two binary trees then you can join their roots to a common <em>root node</em> (parent node) and obtain a new binary tree.</p>
<p>In order to show that every binary tree has some property using induction, you would have to follow the construction cases: (a) show that the one-node tree has the property; (b) assume that two already constructed trees have the property (this would be the <em>induction hypothesis</em>), and show that the new tree that you get by joining them via a common root node must also have the property. You can also do the "strong" (also called <em>course-of-values</em>) version (b'): assume that the two already constructed trees, <em>as well as their own subtrees</em> have the property, and then show that the new tree must also have it.</p>
<p>A conceptual subtlety here is that you perform the induction not on <em>the nodes</em> of the tree, but on the different possible ways these can be pieced together to form the tree (that is to say: on <em>the structure</em> of the tree); the different types of nodes, namely leaf nodes and root nodes, are just your induction pivots. To this point, notice that structural induction generalizes the usual induction on natural numbers: just think of the data type of unary trees labelled by the characters $0$, $S$, and defined by (a) $0$ is a unary tree, and (b) if $m$ is an already constructed unary tree, then $Sm$ is also a unary tree. Here, the zero symbol $0$ plays the role of the one-node base case, and the successor symbol $S$ the (joining) root node of the step case. Indeed, when we perform the usual induction we induct not on nodes (which could only mean node labels) $0$ and $S$, but on numbers $0$, $1$, $2$, ..., that is on "unary trees" $0$, $S0$, $SS0$, ..., where $0$ for zero and $S$ for the successor $+1$ serve as the induction pivots.</p>
<blockquote>
<p>In the proof would I have to define what the < operator does for two nodes?</p>
</blockquote>
<p>Well, the symbol "<" in this case would refer to the subtree relation, right?: $t<s$ would mean $t$ is a (proper) subtree of $s$. This has to be defined before you prove anything using it — in this case, you would want to use it if you performed strong induction. Note that definitions in inductively defined data types are wont to be <em>recursive</em>, to follow the inductive pattern as well: (a) define what the subtrees of the one-node tree are; (b) given the subtrees of two already constructed trees, define what the subtrees of the new tree are.</p>
|
1,473,418 |
<p>I have difficulties with this question : </p>
<p>Given the ODE named (1) : $$x'=y+\sin (x^2y)$$ $$y'=x+\sin(xy^2)$$</p>
<p>and the :</p>
<p><strong>Definition.</strong> A <em>Petal</em> is a solution $(x(t),y(t))$ that verifies $\displaystyle \lim _{t \to \pm \infty} (x(t),y(t)) =(0,0)$.</p>
<p>How can I show that the exists at most two distinct <em>petals</em> ? </p>
<p>First I remarked that the question has sense since : </p>
<p>$$\|(x',y')\| \leqslant \|(x,y)\| + C$$ where $C$ is a wisely chosen constant. It follows that the solution $(x(t),y(t))$ is defined on $\mathbb{R}$.</p>
<p>Next, the linearization in $(0,0)$ shows that $(0,0)$ is an hyperbolic point. </p>
<p>But how can I say more ? </p>
<p>Thank you</p>
|
Archis Welankar
| 275,884 |
<p>))U take two case for divisible by 5 case I last digit is 0 so last place. Fixed there 3 places can be filled in 6p3 ways then case 2 last place is 5 so 5 .5.4 solutions and more 29 solutions for at least 1 2 sixes</p>
|
2,983,877 |
<p>I once saw a function for generating successively more-precise square root approximations, <span class="math-container">$f(x) = \frac{1}{2} ({x + \frac{S}{x}})$</span> where S is the square for which we are trying to calculate <span class="math-container">$\sqrt S$</span>. And the function works really well, generating an approximation of <span class="math-container">$\sqrt 2 \approx f^3(1) = \frac{577}{408} \approx 1.414215$</span>.</p>
<p>This fascinated me, so I tried extending the same logic to further radicals, starting with cube roots.</p>
<p>My first guess was <span class="math-container">$f_2(x) = \frac{1}{3} ({x + \frac{S}{x}})$</span>, but when I tried an approximation for <span class="math-container">$\sqrt[3] 3$</span>, I got <span class="math-container">$\sqrt[3] 3 \approx f_2^2(1) = \frac{43}{36} \approx 1.194444$</span>, which is a far cry from <span class="math-container">$\sqrt[3] 3 \approx Google(\sqrt[3] 3) \approx 1.44225$</span>.</p>
<p>How can I extend this logic for <span class="math-container">$n^{a\over b}$</span> where <span class="math-container">$ b > 2$</span>? Was I accurate all-along and just needed more iterations? Or is the presence of <span class="math-container">$\frac{1}{2}$</span> in <span class="math-container">$f(x) $</span> and in <span class="math-container">$n^\frac{1}{2}$</span> a coincidence?</p>
<p>Disclaimer: I am not educated in calculus.</p>
|
Hussain-Alqatari
| 609,371 |
<p>Let me tell you how to differentiate a polynomial, multiply the exponent by the coefficint and reduce <span class="math-container">$1$</span> from the exponent.</p>
<p>For example, if <span class="math-container">$f(x)=5x^3+7$</span>, then <span class="math-container">$f'(x)=15x^2$</span> [we multiplied <span class="math-container">$3$</span> by <span class="math-container">$5$</span> to get <span class="math-container">$15$</span>, and we reduced <span class="math-container">$1$</span> from the exponent, it was <span class="math-container">$3$</span>, then it became <span class="math-container">$2%$</span>]. The <span class="math-container">$7$</span> just cancelled because it is a constant.</p>
<p>Another example, if <span class="math-container">$f(x)=x^5-8$</span>, then <span class="math-container">$f'(x)=5x^4$</span> [we multipled 5 by the coefficient <span class="math-container">$1$</span> to get <span class="math-container">$5$</span>, and we reduced 1 from the exponent, it was <span class="math-container">$5$</span>, then it became <span class="math-container">$4$</span>]. The <span class="math-container">$8$</span> is constant, so it is cancelled.</p>
<hr>
<p>Now you want to approximate <span class="math-container">$^3\sqrt{3}$</span></p>
<p>This means you want to find a number, if you cube it you get <span class="math-container">$3$</span>, therefore you want to solve the equation;</p>
<p><span class="math-container">$x^3=3$</span> , moving all terms to the left we get <span class="math-container">$x^3-3=0$</span>, denoting the left hand side by <span class="math-container">$f(x)$</span></p>
<p>You need to approximate the root of the equation <span class="math-container">$x^3-3=0$</span></p>
<p>Let <span class="math-container">$f(x)=x^3-3$</span>, therefore <span class="math-container">$f'(x)=3x^2$</span> [as you know now]</p>
<p>Newtons methods for approximating root is:</p>
<p><span class="math-container">$x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}$</span>, where <span class="math-container">$x_0$</span> is the initial guess,</p>
<p>Let <span class="math-container">$x_0=1.5$</span></p>
<p><span class="math-container">$x_1=1.5-\frac{f(1.5)}{f'(1.5)}=1.5-\frac{1.5^3-3}{3\times1.5^2}=1.5-\frac{3.375-3}{3\times2.25}=1.5-\frac{0.375}{6.75}=1.44444$</span></p>
<p>Now you have <span class="math-container">$x_1=1.44444$</span>, you can calculate <span class="math-container">$x_2$</span> in the same way;</p>
<p><span class="math-container">$x_1=1.44444-\frac{f(1.44444)}{f'(1.44444)}=1.44444-\frac{1.44444^3-3}{3\times1.44444^2}=1.44444-\frac{3.01369-3}{3\times2.08641}=1.44444-\frac{0.01369}{6.25923}=1.44225$</span></p>
<p>Here it is a good approximation. if you see that it is not a good approximation, just find <span class="math-container">$x_3$</span> or <span class="math-container">$x_4$</span> or until you reach a good approximation.</p>
<hr>
<p>Suppose you want to approximate <span class="math-container">$^7\sqrt{5}$</span> (the seventh root of five)</p>
<p>Then you want to find a number, if you raise it to the power <span class="math-container">$7$</span> you get <span class="math-container">$5$</span>,</p>
<p>this means you want to solve the equation <span class="math-container">$x^7=5$</span> , moving terms to the left we get <span class="math-container">$x^7-5=0$</span></p>
<p>Now <span class="math-container">$f(x)=x^7-5$</span> and <span class="math-container">$f'(x)=7x^6$</span></p>
<p>Let the initial guess, <span class="math-container">$x_0=1.2$</span></p>
<p>So <span class="math-container">$x_1=1.2-\frac{f(1.2)}{f'(1.2)}=1.2-\frac{1.2^7-5}{7\times1.2^6}=1.26778$</span></p>
<p>Again, <span class="math-container">$x_2=1.26778-\frac{f(1.26778)}{f'(1.26778)}=1.26778-\frac{1.26778^7-5}{7\times1.26778^6}=1.2585$</span></p>
<p>Again, find <span class="math-container">$x_3,x_4,...$</span> unit you get a continent accuracy</p>
|
2,078,796 |
<p>In a contest problem book, I found a reference to Newton's little formula that may be used to find the <em>nth</em> term of a numeric sequence. Specifically, it is a formula that is based on the differences between consecutive terms that is computed at each level until the differences match. </p>
<p>An example application of this formula for computing the <em>nth</em> term of the series (15, 55, 123, 225, 367, 555, 795, ....) involves computing the differences as shown below: </p>
<pre>
1) 1st Level difference is (40, 68, 102, 142, 188, 240)
2) 2nd Level difference is (28, 34, 40, 46, 52)
3) 3rd Level difference is (6, 6, 6, 6, 6)
</pre>
<p>Now the nth term is $$15{n-1\choose 0} + 40{n-1\choose 1} + 28{n-1\choose 2} + 6{n-1\choose 3}$$ where the constant multipliers are the first term of the differences at each level in addition to the first term of the sequence itself.</p>
<p>I was not able to find any reference to this formula or a proof of it after searching on the web. Any explanation of this method is appreciated.</p>
|
Daniel McLaury
| 3,296 |
<p>Starting with the second differences we have</p>
<ul>
<li>$28 = 28$,</li>
<li>$34 = 28 + 6$,</li>
<li>$40 = 28 + 6 + 6$</li>
<li>$46 = 28 + 6 + 6 + 6$$</li>
</ul>
<p>Now for the first differences we have</p>
<ul>
<li>$40 = 40$</li>
<li>$68 = 40 + 28$</li>
<li>$102 = 40 + 28 + 34 = 40 + 28 + (28 + 6)$</li>
<li>$142 = 40 + 28 + 34 + 50 = 40 + 28 + (28 + 6) + (28 + 6 + 6)$</li>
</ul>
<p>So the original sequence is</p>
<ul>
<li>$15 = 15$</li>
<li>$55 = 15 + 40$</li>
<li>$123 = 15 + 40 + 68 = 15 + 40 + [40 + 28]$</li>
<li>$225 = 15 + 40 + 68 + 102 = 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)]$</li>
<li>\begin{align}
367 &= 15 + 40 + 68 + 102 + 142 \\
&= 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)] + [40 + 28 + (28 + 6) + (28 + 6 + 6)]
\end{align}</li>
</ul>
<p>The patterns should be clear. So just count how many 15's, 40's, 28's, and 6's there are in each term.</p>
|
1,472,314 |
<p>How many even numbers less than 600 can be made from the digits: 3,3,4,8,9 with each only being used once. I can't figure out what to do for the 3rd case where 3 digits are needed</p>
|
user247327
| 247,327 |
<p>That looks straight forward. In order to be even, the number must end in 4 or 8. If it ends in 4 then the first 4 digits must be 3, 3, 8, 9. If there were 4 distinct digits there would be 4!= 24 different orders but because two of those digits are "3", each of those orders is exactly the same as another with just the "3"s swapped so there are really 4!/2= 24/2= 12 such numbers. If the number ends in 8 then the first four digits must be 3, 3, 4, 9 and exactly the same argument applies.</p>
|
1,472,314 |
<p>How many even numbers less than 600 can be made from the digits: 3,3,4,8,9 with each only being used once. I can't figure out what to do for the 3rd case where 3 digits are needed</p>
|
N. F. Taussig
| 173,070 |
<p>We wish to find how many even numbers less than $600$ can be formed from the digits $3, 3, 4, 8, 9$ if each digit is used at most once. </p>
<p>Since the number is even, the units digit of each number must be $4$ or $8$.</p>
<p><strong>One-digit numbers:</strong> The only possibilities are $4$ or $8$, giving us two possibilities in this case, as you found.</p>
<p><strong>Two-digit numbers:</strong> If the units digit is $4$, then the tens digit can be $3$, $8$, or $9$. If the units digit is $8$, then the tens digit can be $3$, $4$, or $9$. Hence, there are six possibilities, as you found.</p>
<p><strong>Three-digit even numbers:</strong> If the units digit of the even number less than $600$ is $4$, the hundreds digit must be $3$. This leaves us with three choices for the tens digit, namely $3$, $8$, or $9$. Hence, we can form three three-digit even numbers less than $600$ with units digit $4$ by using the digits $3, 3, 4, 8, 9$ at most once. They are $334$, $384$, $394$.</p>
<p>If the units digit of the three digit even number less than $600$ is $8$, we have two possibilities for the hundreds digit, namely $3$ or $4$. If the hundreds digit is $3$, we have three possibilities for the tens digit, namely $3$, $4$, or $9$. If the hundreds digit is $4$, we have two possibilities for the tens digit, namely $3$ or $9$. Thus, we can form five three-digit even numbers less than $600$ with units digit $8$. They are $338$, $348$, $398$, $438$, $498$. </p>
<p>Hence, there are a total of eight three-digit even numbers less than $600$ that can be formed with the digits $3, 3, 4, 8, 9$ if each digit is used at most once.</p>
<p>In all, we can form $2 + 6 + 8 = 16$ even numbers less than $600$ using the digits $3, 3, 4, 8, 9$ at most once.</p>
|
488,258 |
<p>What are the last two digits of $11^{25}$ to be solved by binomial theorem like $(1+10)^{25}$?
If there is any other way to solve this it would help if that is shown too.</p>
|
Harish Kayarohanam
| 30,423 |
<p>A short cut for finding last two digits of any power of a number that ends in 1 .</p>
<blockquote>
<p>Last two digits of</p>
<p><span class="math-container">$$(\ldots a1)^{\displaystyle\ldots x}$$</span></p>
<p><strong>TENS DIGIT</strong>: unit's digit of (<span class="math-container">$x$</span> <span class="math-container">$\times$</span> a)</p>
<p><strong>UNITS DIGIT</strong> : <span class="math-container">$1$</span></p>
</blockquote>
<p>so here <span class="math-container">$$11^{25}$$</span> last two digits are units digit of <span class="math-container">$(1*5)$</span> and <span class="math-container">$1$</span>
so <span class="math-container">$$51$$</span></p>
|
3,196,797 |
<p>Suppose <span class="math-container">$gcd(m,n)=1$</span>, and let <span class="math-container">$F :Z_n→Z_n$</span> be defined by <span class="math-container">$F([a])=m[a]$</span>. Prove that <span class="math-container">$F$</span> is an automorphism of the additive group <span class="math-container">$Z_n$</span>. I find it is diffcult to prove <span class="math-container">$F$</span> is injective and surjective. Could you please to help me proof it with all the details. I type it roughly, and i am sorry and sincerely looking for a result.</p>
|
Andreas Caranti
| 58,401 |
<p>To show that the map is injective and surjective is equivalent to showing that the map has a two-sided inverse.</p>
<p>The extended Euclidean algorithm yields that there are numbers <span class="math-container">$m', n'$</span> such that
<span class="math-container">$$
m m' + n n '= 1.
$$</span>
Consider the map <span class="math-container">$G : Z_{n} \to Z_{n}$</span> given by <span class="math-container">$G([b]) = m' [b]$</span>. Then for all <span class="math-container">$a$</span> one has
<span class="math-container">$$
G \circ F([a]) = G(F([a])) = G(m [a]) = m'm [a] = (1 - n' n) [a]
= [a],
$$</span>
as <span class="math-container">$n [a] = [0]$</span>. Similarly <span class="math-container">$F \circ G([b]) = [b]$</span> for all <span class="math-container">$b$</span>.</p>
|
120,667 |
<p>Let $V, W$ be two finite-dimensional vector spaces, $f: V\rightarrow W$ a linear map, and $U \subseteq W$ a vector subspace. I'm trying to show that $(f^{-1}(U))^0 = f^*(U^0)$, i.e. that the annihilator of the inverse image of $U$ is the image of the annihilator under the the dual $f^*$ of $f$. $(f^{-1}(U))^0 \supseteq f^*(U^0)$ is easy to prove, but I'm having troubles with the other direction…</p>
<p>(the annihilator for $Y \subseteq X$ vector spaces is defined here as $Y^0 := \{x^* \in X^*\ |\ \forall y \in Y: x^*(y) =0 \}$, with $X^*$ the dual space of $X$; for inner product spaces, $X^0 \cong X^\perp$).</p>
|
Arturo Magidin
| 742 |
<p>I claim that $(f^*(U^0))^0\subseteq f^{-1}(U)$.</p>
<p>Let $\mathbf{v}\in (f^*(U^0))^0$. Then for every $\mathbf{x}\in f^*(U^0)$, we have $\langle \mathbf{v},\mathbf{x}\rangle = 0$. Therefore, for every $\mathbf{w}\in U^0$,
$$
\langle f(\mathbf{v}),\mathbf{w}\rangle = \langle \mathbf{v},f^*(\mathbf{w})\rangle
= 0,$$
(since $f^{*}(\mathbf{w})\in f^*(U^0)$). As this holds for all $\mathbf{w}\in U^0$, it follows that $f(\mathbf{v})\in (U^0)^0 = U$. Thus, $\mathbf{v}\in f^{-1}(U)$. This proves the inclusion.</p>
<p>Therefore,
$$(f^*(U^0))^0 \subseteq f^{-1}(U),$$
hence
$$f^*(U^0) = ((f^*(U^0))^0)^0 \supseteq (f^{-1}(U))^0,$$
which shows that $(f^{-1}(U))^0\subseteq f^*(U^0)$, as required.</p>
|
935,331 |
<p>Previously, to integrate functions like $x(x^2+1)^7$ I used integration by parts. Today we were introduced to a new formula in class: $$\int f'(x)f(x)^n dx = \frac{1}{n+1} {f(x)}^{n+1} +c$$
I was wondering how and why this works. Any help would be appreciated. </p>
|
Brightsun
| 118,300 |
<p>The reason the formula holds is that for the chain rule:
$$
\left(\frac{1}{n+1}f^{n+1}(x)\right)' =\frac{1}{n+1}(n+1)f^n(x) f'(x) = f^n(x) f'(x)
$$
and this shows your identity by definition of indefinite integral as anti-derivative.</p>
<p>However, applying integration by parts to the same problem we have:
$$
\int f'(x) f^n(x)dx = f(x)f^n(x)-\int f(x)nf^{n-1}(x)f'(x)dx
$$
which is an identity that can be recast precisely as:
$$
\int f'(x) f^n(x) dx = \frac{1}{n+1}f^{n+1}(x).
$$
We can observe that this is a general fact, since the integration by parts rule is just an application of the chain rule itself:
$$
\left(f(x)g(x)\right)' = f'(x)g(x) + f(x)g'(x)\\
f'(x)g(x) = \left(f(x)g(x)\right)'- f(x)g'(x)\\
\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx.
$$
The difference in the two methods above amounts exactly to the difference between the first line and the last one.</p>
|
119,506 |
<p>Let $\kappa$ be a singular cardinal, and let $\langle \kappa_i \mid i<\mathrm{cf}(\kappa) \rangle$ be an increasing sequence of regular cardinals cofinal in $\kappa$. Recall that a scale on $\Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$ is a sequence $\langle f_\alpha \mid \alpha < \kappa^+ \rangle$ such that:</p>
<ol>
<li>For every $\alpha < \kappa^+$, $f_\alpha \in \Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$.</li>
<li>For every $\alpha < \beta < \kappa^+$, there is $i < \mathrm{cf}(\kappa)$ such that $f_\alpha <_i f_\beta$, i.e. for every $j\geq i$, $f_\alpha(j) < f_\beta(j)$.</li>
<li>For every $g\in \Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$, there is $\alpha < \kappa^+$ and $i < \mathrm{cf}(\kappa)$ such that $g <_i f_\alpha$.</li>
</ol>
<p>Question: Is it consistent that there is a scale on $\Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$ such that, for every $\beta < \kappa^+$ and every $i<\mathrm{cf}(\kappa)$,
$\left|{\{\alpha < \beta \mid f_\alpha <_i f_\beta\}}\right| < \kappa$ ?</p>
<p>My intuition is that the answer should be no, but I haven't been able to find a proof.</p>
|
Eran
| 10,708 |
<p>Shelah's Dichotmoy theorem (see <a href="http://math.cmu.edu/~sunger/PCFtalk1.pdf" rel="nofollow">link text</a>) says more or less that both options are valid. Every increasing sequence can either:</p>
<p>a) Have an exact upper bound (in which case your condition fails.)</p>
<p>b) Or have an "interleaved cofinal sequence" i.e. another (very small length) sequence is "interleaved" with the original one, in which case your condition must hold.</p>
|
987,620 |
<p>$P$ and $Q$ are two distinct prime numbers. How can I prove that $\sqrt{PQ}$ is an irrational number?</p>
|
Alexander Berliner
| 634,134 |
<p><strong>Proof:</strong> Assume, to the contrary, that <span class="math-container">$\sqrt{pq}$</span> is rational. Then <span class="math-container">$\sqrt{pq}=\frac{x}{y}$</span> for two integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span> and we further assume that <span class="math-container">$gcd(x,y)=1$</span>. Observe that <span class="math-container">$pqy^2=(qy^2)p=x^2$</span>. Since <span class="math-container">$qy^2$</span> is an integer, <span class="math-container">$p\mid x^2$</span> and by Euclid's Lemma, <span class="math-container">$p\mid x$</span>. Thus <span class="math-container">$x=pd$</span> for some integer <span class="math-container">$d$</span> and so <span class="math-container">$pqy^2=x^2=(pd)^2$</span> and so <span class="math-container">$qy^2=p(d^2)$</span>. Hence, <span class="math-container">$p\mid qy^2$</span>. Since <span class="math-container">$p,q$</span> are distinct primes, <span class="math-container">$p \neq q$</span> and so <span class="math-container">$p \nmid q$</span>, which means <span class="math-container">$p\mid y^2$</span>. Thus, <span class="math-container">$p\mid y$</span>. This contradicts our assumption that <span class="math-container">$gcd(x,y)=1$</span>.</p>
|
33,330 |
<p>Let $p$ be a complex number. Let $ z_0 = p $ and, for $ n \geq 1 $, define $z_{n+1} = \frac{1}{2} ( z_n - \frac{1}{z_n}) $ if $z_n \neq 0 $. Prove the following:</p>
<p>i) If $ \{ z_n \} $ converges to a limit $a$, then $a^2 + 1 = 0 $</p>
<p>ii) If $ p $ is real, then $ \{ z_n \} $, if defined, does not converge</p>
<p>iii) If $ p = iq $, where $ q \in \mathbb{R} \backslash \{0\} $, then $ \{ z_n \} $ converges.</p>
<p>I have been able to do the first two parts of this (the second is because the sequence would be real, but would have to have a complex limit). I am stuck on the third part, though. Any help would be greatly appreciated.</p>
<p>Thanks</p>
|
Samrat Mukhopadhyay
| 83,973 |
<p>From the recursion relation we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{z_{n}+i}{z_{n}-i}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{p+i}{p-i}\right)^{2^{n+1}}$$ Rearranging, we get, $$z_{n+1}=i\frac{1+w^{2^{n+1}}}{1-w^{2^{n+1}}}$$ where $$w=\frac{p-i}{p+i}$$ When $p=iq$ $$ w=\frac{q-1}{q+1}$$ Clearly, $|w|<1$, since $q\ne 0$. Hence $$\lim_{n\rightarrow \infty}w^{2^{n+1}}=0.
$$ Thus $$\lim_{n\rightarrow \infty}z_n=1 \hspace{0.6cm}\Box$$</p>
|
302,243 |
<p>Let $f:[0,1]\to\mathbb{R}$ be a Lipschitz function, and $\pi f$ be its piecewise linear interpolant on an equispaced grid with $n$ points.</p>
<p>It should be true (if I am not making mistakes with the constant) that
$$
\int_0^1 |f - \pi f| \leq \frac{1}{4n} \operatorname{Lip}(f).
$$</p>
<p>Do you have a reference that I can cite for this result, without having to re-prove it? All the references I have found by looking around assume better regularity.</p>
|
Will Sawin
| 18,060 |
<p>Let $E$ be the elliptic curve. Let $E_1$ and $E_2$ be two different double covers of $E$, with $E = E_1 /x_1$ and $E=E_2/x_2$ for two-torsion points $x_1,x_2$. </p>
<p>Let $M$ be the minimal resolution of singularities of $ E_1 \times E_2 /\langle (a,b) \to (-a,-b), (a,b) \to (a+ x_1, x_2-b) \rangle $.</p>
<p>Then $M$ is an Enriques surface, with fundamental group $\mathbb Z/2$. We can embed $E$ into $M$ either horizontally or vertically. In the horizontal case the inverse image in the double cover is $E_1$, and in the verticle case the inverse image is $E_2$.</p>
<p>By combining these, we get an embedding of $E$ into $M \times M$, whose inverse image in the universal cover is $E$, mapping to $E$ by the multiplication by two map. </p>
<p>On the other hand the Picard group of $M \times M$ is $\mathbb Z/2 \times \mathbb Z/2$ (as long as it is defined using rational Chern classes, and not integral Chern classes, for which the statement is trivially false as $\mathcal O_X$ on $M$ is a counterxample). Pulled back to $E$, these line bundles are all the two-torsion vector bundles.</p>
<p>So the question becomes:</p>
<blockquote>
<p>Is any coherent sheaf on $E$ which is isomorphic to its own tensor product with any two-torsion line bundle equal to a pushforward for $E$ under the multiplication-by-two map?</p>
</blockquote>
<p>Then the answer is "no", because indecomposable vector bundles on $E$ are classified by their determinant, so every indecomposable rank two bundle has this property, but none is a pushforward under multiplication by two as that would force it to have rank a multiple of four.</p>
<p>However, it may still be true if the fundamental group is cyclic.</p>
|
615,093 |
<p>How to prove the following sequence converges to $0.5$ ?
$$a_n=\int_0^1{nx^{n-1}\over 1+x}dx$$
What I have tried:
I calculated the integral $$a_n=1-n\left(-1\right)^n\left[\ln2-\sum_{i=1}^n {\left(-1\right)^{i+1}\over i}\right]$$
I also noticed ${1\over2}<a_n<1$ $\forall n \in \mathbb{N}$.</p>
<p>Then I wrote a C program and verified that $a_n\to 0.5$ (I didn't know the answer before) by calculating $a_n$ upto $n=9990002$ (starting from $n=2$ and each time increasing $n$ by $10^4$). I can't think of how to prove $\{a_n\}$ is monotone decreasing, which is clear from direct calculation.</p>
|
Tim Ratigan
| 79,602 |
<p>EDIT: I feel kind of stupid for not thinking of the easier ways in other posts, but I think this method is kind of cool.</p>
<p>I apologize in advance, this is a lot of math and few words.</p>
<p>$$\begin{align} \int_0^1\frac{nx^{n-1}}{1+x}\text dx&=\int_0^1nx^{n-1}\sum_{k=0}^\infty(-x)^k\text dx\\
&=\sum_{k=0}^\infty \int_0^1nx^{n-1+k}(-1)^k\text dx\\
&=\sum_{k=0}^\infty\frac{(-1)^kn}{n+k} \end{align}$$</p>
<p>Now, you want $$\begin{align}\lim_{n\to\infty}n\sum_{k=0}^\infty \frac{(-1)^k}{n+k}&=\lim_{n\to\infty}n\left(\sum_{k=0}^\infty \frac{1}{n+2k}-\frac{1}{n+2k+1}\right)\\
&=\lim_{n\to\infty}n\sum_{k=0}^\infty \frac1{(n+2k)^2+n+2k}\\
&=\lim_{n\to\infty}n\sum_{k=0}^\infty\frac{1}{(n+2k)^2}\tag 1\\
&=\lim_{n\to\infty}\frac1n\sum_{k=0}^\infty \frac{1}{(1+2\frac kn)^2}\\
&=\int_0^\infty \frac{1}{(1+2x)^2}\text dx\tag 2\\
&=\frac12\int_0^\infty \frac1{(1+x)^2}\text dx\\
&=\frac12\left.\left(-\frac1{x+1}\right)\right|_0^\infty\\
&=\frac12\end{align}$$</p>
<p>$(1)$ is obtained by realizing that $n+2k$ is negligible in comparison to $(n+2k)^2$ as $n$ approaches $\infty$</p>
<p>$(2)$ uses the well-known identity $$\lim_{n\to\infty}\frac1n\sum_{k=an}^{bn}f\left(\frac kn\right)=\int_a^bf(x)\text dx$$</p>
|
615,093 |
<p>How to prove the following sequence converges to $0.5$ ?
$$a_n=\int_0^1{nx^{n-1}\over 1+x}dx$$
What I have tried:
I calculated the integral $$a_n=1-n\left(-1\right)^n\left[\ln2-\sum_{i=1}^n {\left(-1\right)^{i+1}\over i}\right]$$
I also noticed ${1\over2}<a_n<1$ $\forall n \in \mathbb{N}$.</p>
<p>Then I wrote a C program and verified that $a_n\to 0.5$ (I didn't know the answer before) by calculating $a_n$ upto $n=9990002$ (starting from $n=2$ and each time increasing $n$ by $10^4$). I can't think of how to prove $\{a_n\}$ is monotone decreasing, which is clear from direct calculation.</p>
|
Avi Steiner
| 13,487 |
<p>Male the u-substitution $u=x^n$, them apply the dominated convergence theorem. </p>
|
615,093 |
<p>How to prove the following sequence converges to $0.5$ ?
$$a_n=\int_0^1{nx^{n-1}\over 1+x}dx$$
What I have tried:
I calculated the integral $$a_n=1-n\left(-1\right)^n\left[\ln2-\sum_{i=1}^n {\left(-1\right)^{i+1}\over i}\right]$$
I also noticed ${1\over2}<a_n<1$ $\forall n \in \mathbb{N}$.</p>
<p>Then I wrote a C program and verified that $a_n\to 0.5$ (I didn't know the answer before) by calculating $a_n$ upto $n=9990002$ (starting from $n=2$ and each time increasing $n$ by $10^4$). I can't think of how to prove $\{a_n\}$ is monotone decreasing, which is clear from direct calculation.</p>
|
iballa
| 116,491 |
<p>Thinking about the graph of $x^n$ on $[0,1]$ we observe that it stays near $0$ and then sharply jumps to $1$. As such, it makes sense to break up the integral into $[0,c)$ and $[c,1]$ (for some $c$ to be chosen later).</p>
<p>$$
a_n = \int_0^c{\frac{n x^{n-1}}{x+1}dx} + \int_c^1{\frac{n x^{n-1}}{x+1}dx} \leq \int_0^c{n x^{n-1}dx} + \int_c^1{\frac{n x^{n-1}}{c+1}dx}\\
= c^n + \frac{1 - c^n}{c+1}.
$$</p>
<p>Now observe that for any fixed $c < 1$, $c^n + \frac{1 - c^n}{c+1} \rightarrow 1/(c+1)$ as $n \rightarrow \infty$. Thus we have $\limsup_{n \rightarrow \infty}{a_n} \leq 1/(c+1)$, and now letting $c \rightarrow 1$ from below we conclude $\limsup_{n \rightarrow \infty}{a_n} \leq 1/2$.</p>
<p>On the other hand $a_n = \int_0^1{\frac{n x^{n-1}}{x+1}dx} \geq \int_0^1{\frac{n x^{n-1}}{2}dx} = 1/2$ for all $n$, so that $\liminf_{n \rightarrow \infty}{a_n} \geq 1/2$.</p>
<p>Thus $$\lim_{n \rightarrow \infty}{a_n} = \frac{1}{2}.$$</p>
|
834,508 |
<p>Show that for any natural number $n$, between $n^2$ and $(n+1)^2$ one can find three distinct natural numbers $a,b,c$ such that $a^2+b^2$ is divisible by $c$.</p>
<p>A friend and I found a general case that always work with a computer problem, I would like to see a different solution, or a solution that tells the motivation of how to find that general case without a computer.</p>
|
Simply Beautiful Art
| 272,831 |
<p>A simple straightforward algorithm starts from <span class="math-container">$0$</span> and repeatedly adds numbers onto pre-made numbers until a streak is found. For example, consider <span class="math-container">$3$</span> and <span class="math-container">$5$</span>.</p>
<p>We start off with <span class="math-container">$\{0\}$</span>.</p>
<p>Taking the minimum of <span class="math-container">$0+3$</span> and <span class="math-container">$0+5$</span> gives us <span class="math-container">$\{0,3\}$</span>.</p>
<p>Taking the minimum of <span class="math-container">$0+5,3+3,$</span> and <span class="math-container">$3+5$</span>, we get <span class="math-container">$\{0,3,5\}$</span>.</p>
<p>The next few steps are:</p>
<p><span class="math-container">$\{0,3,5,6\}$</span></p>
<p><span class="math-container">$\{0,3,5,6,8\}$</span></p>
<p><span class="math-container">$\{0,3,5,6,8,9\}$</span></p>
<p><span class="math-container">$\{0,3,5,6,\color{red}{8,9,10}\}$</span></p>
<p>From which we can see the Frobenius number of <span class="math-container">$\{3,5\}$</span> is <span class="math-container">$7$</span>.</p>
<hr>
<p>As it would turn out, an asymptotically faster algorithm would be to rewrite this into a linear recurrence:</p>
<p><span class="math-container">$$a_n=\begin{cases}0,&n<0\\1,&n=0\\a_{n-3}+a_{n-5},&n>0\end{cases}$$</span></p>
<p>We seek to find the largest <span class="math-container">$n$</span> s.t. <span class="math-container">$a_n=0$</span>. The computation of linear recurrences is well studied and you can compute <span class="math-container">$(a_n,a_{n-1},a_{n-2},a_{n-3},a_{n-5})$</span> in <span class="math-container">$\mathcal O(\log(n))$</span> steps by rewriting it as:</p>
<p><span class="math-container">$$\begin{bmatrix}a_n\\a_{n-1}\\a_{n-2}\\a_{n-3}\\a_{n-4}\end{bmatrix}=A^n\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}\tag{$n\ge0$}\\A=\begin{bmatrix}0&0&1&0&1\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{bmatrix}$$</span></p>
<p>and applying exponentiation by squaring. Once a streak is found, we work backwards with a binary search, using our previously computed <span class="math-container">$A^k$</span> and their inverses as needed. Note that this isn't super helpful on very small values, but can be more useful for finding the Frobenius number of something like <span class="math-container">$\{2000,3001,4567\}$</span>.</p>
|
1,762,522 |
<p>This question was asked by one of my Professor during the class. I'm getting intuition that these functions should be one-one (I'm wrong maybe). But, I'm unable to classify all such functions.</p>
<p>Please help in this!!</p>
|
Siddharth Bhat
| 261,373 |
<p>What have you tried so far? </p>
<p>And, as for being one-to-one, can you think of examples where a function maps, say, every open interval to $R$? what would such a function look like? is it one-one? onto?</p>
|
1,762,522 |
<p>This question was asked by one of my Professor during the class. I'm getting intuition that these functions should be one-one (I'm wrong maybe). But, I'm unable to classify all such functions.</p>
<p>Please help in this!!</p>
|
Brian M. Scott
| 12,042 |
<p>HINT: Not all bijections work, but you should be able to prove that the monotone (increasing or decreasing) bijections do. You should also be able to prove that these are precisely the continuous bijections. Going a bit further, try to prove that if $f:\Bbb R\to\Bbb R$ has an absolute maximum or minimum on some open interval $(a,b)$, then $f$ does <em>not</em> have the desired property, and use this to show that the continuous bijections are the <em>only</em> continuous functions that work.</p>
<p>However, you won’t be able to characterize these functions completely, because (assuming the axiom of choice) there are wildly discontinuous functions, impossible to visualize, that take open intervals to open intervals — that in fact map every open interval onto a single open interval. The construction of such a function requires a bit of set-theoretic background that you quite likely don’t have, but I’m going to include it for the sake of completeness.</p>
<p>Let $\{\langle a_\xi,b_\xi,c_\xi\rangle:\xi<2^\omega\}$ be an enumeration of $\{\langle a,b,c\rangle\in\Bbb R^3:a<b\}$, and let $\{u_\xi:\xi<2^\omega\}$ be an enumeration of the open interval $(0,1)$. Suppose that $\eta<2^\omega$, and we’ve chosen distinct $x_\xi\in(a_\xi,b_\xi)$ for $\xi<\eta$. Let</p>
<p>$$\alpha=\min\big\{\beta<2^\omega:u_\beta\in(a_\eta,b_\eta)\setminus\{x_\xi:\xi<\eta\}\big\}\;,$$</p>
<p>and set $x_\eta=u_\alpha$; this is always possible, since $|\{x_\xi:\xi<\eta\}|=|\eta|<2^\omega=|(a_\eta,b_\eta)|$. Thus, the recursion goes through to $2^\omega$, and for each $\xi<2^\omega$ we have a distinct $x_\xi\in(a_\xi,b_\xi)$.</p>
<p>Now let $X=\{x_\xi:\xi<2^\omega\}$ and define</p>
<p>$$f:\Bbb R\to(0,1):x\mapsto\begin{cases}
c_\xi,&\text{if }x=x_\xi\\
1/2,&\text{if }x\in\Bbb R\setminus X\;.
\end{cases}$$</p>
<p>Let $(a,b)$ be any open interval in $\Bbb R$ and $c$ any real number in $(0,1)$; then $\langle a,b,c\rangle=\langle a_\xi,b_\xi,c_\xi\rangle$ for some $\xi<2^\omega$, we chose $x_\xi$ so that $x_\xi\in(a_\xi,b_\xi)=(a,b)$, and we defined $f$ so that $f(x_\xi)=c_\xi=c$. Thus, $f[(a,b)]=(0,1)$: $f$ maps every open interval of $\Bbb R$ onto the interval $(0,1)$.</p>
<p>Note that we can replace $(0,1)$ in this argument with any other open interval of $\Bbb R$, including $\Bbb R$ itself.</p>
|
1,117,458 |
<p>How do I integrate an expression of the form
$$
\frac{f'(x)}{[f(x)]^n}
$$
with respect to $x$?</p>
<p>Could I use some kind of recognition method, thus avoiding partial fractions?</p>
<p>For example:
$$
\frac{(2x+1)}{(x^2+x-1)^2}
$$</p>
|
lab bhattacharjee
| 33,337 |
<p>$$S=\int\frac{f'(x)}{f^n(x)}dx=\int\dfrac{d[f(x)]}{f^n(x)\cdot dx}dx=\int f^{-n}(x)\ d[f(x)$$</p>
<p>For $-n+1\ne0\iff n\ne1$</p>
<p>$$S=\frac{f^{-n+1}(x)}{-n+1}+K$$</p>
<p>For $n=1,$</p>
<p>$$S=\int\frac{d[f(x)}{f(x)}=\ln|f(x)|+C$$</p>
|
1,117,458 |
<p>How do I integrate an expression of the form
$$
\frac{f'(x)}{[f(x)]^n}
$$
with respect to $x$?</p>
<p>Could I use some kind of recognition method, thus avoiding partial fractions?</p>
<p>For example:
$$
\frac{(2x+1)}{(x^2+x-1)^2}
$$</p>
|
Martingalo
| 127,445 |
<p>Chain rule. You know that
$$\frac{d}{dx}[f(x)]^n = n [f(x)]^{n-1} f'(x).$$</p>
<p>Then $$\int [f(x)]^m f'(x) dx = \frac{[f(x)]^{m+1}}{m+1} + C$$
Just set $m =-n$. For the case $n=1$ you get the logarithm, that is
$$\int \frac{f'(x)}{f(x)} dx = \log f(x) + C$$</p>
<p>In Your example the numerator is the derivative of the denominator, so youcan Write</p>
<p>$$\int (x^2 + x -1)^{-2} (2x+1) dx = \frac{(x^2 + x -1)^{-2+1}}{-2+1} + C = \frac{-1}{x^2 + x -1}+ C$$</p>
|
2,530,820 |
<p>Let $1\leq p<\infty$ and $q$ be the conjugate exponent of $p$. Suppose that $\lbrace a^n \rbrace_{n=1}^{\infty} \subset \ell^q$ in a sequence in $\ell^q$ such that $f_{a^n}(x) \mapsto 0~( n \mapsto \infty)$ for all $x \in \ell^p$ where $f_{a^n}(x)=\sum_{i=1}^{\infty} a_i^{(n)}x_i$. Show that the sequence $\lbrace a^n \rbrace_{n=1}^{\infty}$ is bounded, i.e., there exists $M\geq0$ such that $\lVert a^n\rVert_q\leq M$ for all $n$. </p>
<p>I get across with a theorem in Bryan P. Rynne Book which shows that $f_{a^n}$ defines a linear functional $f_{a^n} \in (\ell^q)'$ with $\lVert f_{a^n}\rVert=\lVert a^n\rVert_q$. I think it will help but I am still confused how to prove the statement above</p>
|
Matematleta
| 138,929 |
<p>If $\lbrace a^n \rbrace_{n=1}^{\infty} \subset l^q$ and $x\in l^p$, consider $J(a^n)(x)=x(a^n)=\sum_{i=1}^{\infty} a_i^{(n)}x_i.$ Since the series converges to $0,$ each element of $\left \{ J(a^n) \right \}$ is pointwise bounded, so by Banach-Steinhaus, there is an $M<\infty $ such that $\|J(a^n)\|<M$ for all $n\in \mathbb N$. But since $y\mapsto J(y)$ is an isometry, it follows that $\|a^n\|<M$ as well. </p>
|
15,480 |
<p>Say I have two lists,</p>
<pre><code>list1 = {a, b, c}
list2 = {x, y, z}
</code></pre>
<p>and I want to map a function f over them to produce</p>
<pre><code>{f[a,x], f[a,y], f[a,z], f[b,x], f[b,y], f[b,z], f[c,x], f[c,y], f[c,d]}
</code></pre>
<p>I would assume I map the function over the first list to produce a "list of functions", which then run over the 2nd list, something like:</p>
<pre><code>Map[Map[f[#1, #2]&,list1]&, list2]
</code></pre>
<p>but I can't figure out how to leave #2 "empty" until the 2nd map kicks in. How can I separate them to generate all combinations of arguments?</p>
|
MinHsuan Peng
| 1,376 |
<p><code>Distribute</code> is also handy.</p>
<p>Assuming <code>f</code> is not <code>Listable</code>:</p>
<pre><code>In[39]:= Distribute[f[{a, b, c}, {x, y, z}], List]
Out[39]= {f[a, x], f[a, y], f[a, z], f[b, x], f[b, y], f[b, z],
f[c, x], f[c, y], f[c, z]}
</code></pre>
|
2,831,199 |
<p>What is the probability of getting $6$ $K$ times in a row when rolling a dice N times?</p>
<p>I thought it's $(1/6)^k*(5/6)^{n-k}$ and that times $N-K+1$ since there are $N-K+1$ ways to place an array of consecutive elements to $N$ places.</p>
|
Michael Burr
| 86,421 |
<p>This is really just a long comment, but I thought that it would be a mess to write it as a comment:</p>
<p>Please clarify your question with, at least, the following data (some of the answers are in the comments):</p>
<ol>
<li><p>Does the run of $6$'s need to be the only $6$'s in the string? If $k=2$, would $6656$ be allowable?</p></li>
<li><p>Can the run of $6$'s be longer than $k$? If $k=2$, would $6665$ be counted?</p></li>
<li><p>Can you have more than one run of $6$'s in the string? If $k=2$, would $66566$ be allowable?</p></li>
</ol>
|
3,523,205 |
<p>The given series of function is as follow</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^\infty x^{n-1}(1-x)^{2}$$</span>
prove that given series is uniformaly convergent on <span class="math-container">$[0,1]$</span></p>
</blockquote>
<p><strong>The solution i tried</strong>-The given series form an <span class="math-container">$G.P$</span> with ratio <span class="math-container">$x \leq1$</span> </p>
<p>i.e</p>
<blockquote>
<p><span class="math-container">$$(1-x)^2+x(1-x)^2+x^2(1-x)^2+...$$</span></p>
</blockquote>
<p>Now if i form partial sum of <span class="math-container">$n$</span> terms it will be
<span class="math-container">$$s_n=(1-x)^2 \frac{1-x^n}{1-x}$$</span>
<span class="math-container">$$\lim_{n\to \infty}s_n=\frac{(1-x)^2}{1-x}$$</span></p>
<p>after that we get <span class="math-container">$$s=(1-x)$$</span></p>
<p>now what can i say about convergence </p>
<p>Because we know that if the series of partial sum is uniform convergence then series is uniform convergence ,but here <span class="math-container">$s$</span> is something polynomial type .</p>
<p>Please Help</p>
|
Peter Szilas
| 408,605 |
<p>Option:</p>
<p>Weierstrass M test .</p>
<p><span class="math-container">$f_n (x)=x^{n-1}(1-x)^2$</span>;</p>
<p><span class="math-container">$f_n'(x)=$</span></p>
<p><span class="math-container">$(n-1)x^{n-2}(1-x)^2-2x^{n-1}(1-x)=0;$</span></p>
<p><span class="math-container">$(n-1)(1-x)-2x=0$</span>;</p>
<p><span class="math-container">$x(n+1)=n-1$</span>;</p>
<p><span class="math-container">$x=\frac{n-1}{n+1}$</span>;</p>
<p>This is the maximum of <span class="math-container">$f_n$</span> on <span class="math-container">$[0,1]$</span>.(Why?)</p>
<p>(Recall : A continuous function on a compact interval attains its maximum, <span class="math-container">$f_n(x) \ge 0$</span> on <span class="math-container">$[0,1]$</span>)</p>
<p><span class="math-container">$f_n(\frac{n-1}{n+1})=$</span></p>
<p><span class="math-container">$(\frac{(1-1/n)^{n-1}}{(1+1/n)^{n-1}})(\frac{2}{n+1})^2 <4/n^2.$</span></p>
<p>Weierstrass M test : <span class="math-container">$\sum f_n(x)$</span> is uniformly convergent on <span class="math-container">$[0,1]$</span>.</p>
<p><a href="https://en.m.wikipedia.org/wiki/Weierstrass_M-test" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Weierstrass_M-test</a></p>
|
1,439,850 |
<p>So the problem states that the centre of the circle is in the first quadrant and that circle passes through $x$ axis, $y$ axis and the following line: $3x-4y=12$. I have only one question. The answer denotes $r$ as the radius of the circle and then assumes that centre is at $(r,r)$ because of the fact that the circle passes through $x$ and $y$ axis. I was thinking that this single fact does not permit one to assume that centre must be at $(r,r)$, simply because the centre may be positioned in such a manner that the distance to $y$ and $x$ axis is not the same and not necessarily $r$. Is my thinking correct? If not, why? </p>
|
MrYouMath
| 262,304 |
<p>Rewrite </p>
<p>$$y'=-3\frac{y}{x}+\frac{1}{x^2}$$</p>
<p>Solving the homogenous equation is pretty eays (e.g. using method of separation) $$y_h=c_1x^{-3}$$</p>
<p>Then you guess (is faster for easy ODEs) a particular solution of the form $y_p=\frac{k}{x}$. Plug this into the equation and find k. </p>
<p><strong>Note</strong>: You can also get the particular solution using the general formula for a linear ODE or variation of constants.</p>
<p>As always you general solution is
$$y=y_h+y_p=\frac{c}{x^3}+\frac{k}{x}$$</p>
|
1,439,850 |
<p>So the problem states that the centre of the circle is in the first quadrant and that circle passes through $x$ axis, $y$ axis and the following line: $3x-4y=12$. I have only one question. The answer denotes $r$ as the radius of the circle and then assumes that centre is at $(r,r)$ because of the fact that the circle passes through $x$ and $y$ axis. I was thinking that this single fact does not permit one to assume that centre must be at $(r,r)$, simply because the centre may be positioned in such a manner that the distance to $y$ and $x$ axis is not the same and not necessarily $r$. Is my thinking correct? If not, why? </p>
|
Kwin van der Veen
| 76,466 |
<p>When you have a first order non-linear and/or non-autonomous differential equation you could always try testing if it is an <a href="http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html" rel="nofollow">exact differential equation</a></p>
<p>$$
p(x,y) + q(x,y) \frac{dy}{dx} = 0,
$$</p>
<p>where</p>
<p>$$
p(x,y) = 3xy - 1,
$$</p>
<p>$$
q(x,y) = x^2.
$$</p>
<p>In this case $\frac{\partial p}{\partial y} \neq \frac{\partial q}{\partial x}$, however if you multiply this equation by some other function $\mu(x,y)$ the new $p(x,y)$ and $q(x,y)$ might satisfy this relation.</p>
<p>Since all terms in the existing differential equation only contain powers of $x$ and $y$, a good guess would be</p>
<p>$$
\mu(x,y) = x^a y^b.
$$</p>
<p>The values for $a$ and $b$ can be found by equating</p>
<p>$$
\frac{\partial \mu(x,y) p(x,y)}{\partial y} = \frac{\partial \mu(x,y) q(x,y)}{\partial x},
$$</p>
<p>which is equal to $\frac{\partial}{\partial x} \frac{\partial}{\partial y} \Psi(x,y)$, such that differential equation is equivalent to</p>
<p>$$
\frac{\partial}{\partial x} \Psi(x,y) + \frac{\partial}{\partial y} \Psi(x,y) \frac{dy}{dx} = \frac{d}{dx} \Psi(x,y) = 0,
$$</p>
<p>thus $\Psi(x,y)$ has to be equal to a constant, represented by $c$.</p>
<p>The solution to the differential equation can then be found by integrating to find $\Psi(x,y)$</p>
<p>$$
\Psi(x,y) = \int{\mu(x,y) p(x,y) dx} + f(y),
$$</p>
<p>$$
\Psi(x,y) = \int{\mu(x,y) q(x,y) dy} + g(x).
$$</p>
<p>Both integrals will contain the same terms, which both depend on $x$ and $y$, such that $g(x)$ will be equal to the terms of the first integral, which only depends on $x$ and $f(y)$ will be equal to the terms of the second integral, which only depends on $y$.</p>
|
1,821,186 |
<p>Why is the solution of $|1+3x|<6x$ only $x>1/3$? After applying the properties of modulus, I get $-6x<1+3x<6x$. And after solving each inequality, I get $x>-1/9$ and $x>1/3$, but why is $x>-1/9$ rejected? </p>
|
cQQkie
| 70,348 |
<p>Well, take a dense subset of $D\subset X$ and then the metric guarantees you a countable neighbourhood basis at every point of $D$. A countable union of countable sets is again countable. Check that this is enough.</p>
|
744,034 |
<p>How do I show that for all integers $n$, $n^3+(n+1)^3+(n+2)^3$ is a multiple of $9$?
Do I use induction for showing this? If not what do I use and how? And is this question asking me to prove it or show it? How do I show it? </p>
|
Community
| -1 |
<p>$$n^3+(n+1)^3+(n+2)^3=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8=\\
3n^3+9n^2+15n+9=3(n^3+3n^2+5n+3)$$
Thus, this amounts to showing that $n^3+3n^2+5n+3$ is divisible by $3$ for all $n$. Obviousl, the terms $3n^2,3$ are divisible by $3$. Thus, we need to check that $\forall n,n^3+5n$ is divisible by $3$. Let's proceed by induction, letting the statement $\forall n,n^3+5n$ is divisible by $3$ be the induction hypothesis.</p>
<p>It's true for $n=1$. Suppose it's true for $n\geq1$. Then, $(n+1)^3+5(n+1)=(n^3+5n)+(3n^2+3n+6)$. Use now the induction hypothesis ($\forall n,n^3+5n$ is divisible by $3$).</p>
|
2,343,958 |
<p>I am interested in a mathematical approach to quantum information theory. I have observed that several probabilists have been working in this area. What can be a suitable background and good book for this subject?</p>
|
user 1987
| 243,227 |
<p>I myself started learning quantum information theory with Quantum Computing by J. Gruska and Quantum Computation and Information Theory by Nielsen-Chuang. I think the former is more mathematical. It covers very well Hilbert spaces and related notions. I like very much the treatment of observalbles and measurements. On the other hand Nielsen-Chuang, I think, has a very good computational approach. </p>
<p>In overall, if you look for an $\textbf{introduction}$, it will be difficult to find a lot of abstract math. I work in mathematical aspects of quantum error correction and the books I mentioned worked well for the basics. Afterwards most of literature was articles and lecture notes.</p>
|
2,717,821 |
<p>Since we have 4 digits there is a total of 10000 Password combinations possible.</p>
<p>Now after each trial the chance for a successful guess increases by a slight percentage because we just tried one password and now we remove that password from the "guessing set". That being said I am struggling with the actual calculation.</p>
<p>I first calculate the probability of me NOT guessing the password and then subtract that from 1.</p>
<p>\begin{align}
1-\frac{9999}{10000} \cdot \frac{9998}{10000} \cdot \frac{9997}{10000} = 0.059\%
\end{align}</p>
|
Bram28
| 256,001 |
<p>Since you're trying $3$ different PINS, the chances of one of them being correct is $$\frac{3}{10000}$$</p>
|
2,717,821 |
<p>Since we have 4 digits there is a total of 10000 Password combinations possible.</p>
<p>Now after each trial the chance for a successful guess increases by a slight percentage because we just tried one password and now we remove that password from the "guessing set". That being said I am struggling with the actual calculation.</p>
<p>I first calculate the probability of me NOT guessing the password and then subtract that from 1.</p>
<p>\begin{align}
1-\frac{9999}{10000} \cdot \frac{9998}{10000} \cdot \frac{9997}{10000} = 0.059\%
\end{align}</p>
|
Ross Millikan
| 1,827 |
<p>The simple approach is that there are $10000$ possible PINs and you have tried $3$ of them, so your chance of finding the right one is $\frac 3{10000}$. In your calculation, the denominators should decrease $10000,9999,9998$, so you will get the same result.$$1-\frac {9999}{10000}\cdot \frac {9998}{9999}\cdot\frac{9997}{9998}=1-\frac {9997}{10000}=\frac 3{10000}$$</p>
|
58,870 |
<p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
|
R. Andrew Hicks
| 13,754 |
<p>I don't believe either of those books covers distributions and the theorem of Frobenius. Connections to partial differential equations in general I think are good topics. </p>
<p>Guillemin and Pollack is a book I like a lot, but chapters 2 & 3 (transversality and intersection) always seemed a bit specialized for a first course. Although, the title is, after all, "Differential Topology". My experience is that people tend to cover just chapters 1 & 4. </p>
<p>The definition of a manifold in G&P is as a subset of $\mathbb{R}^n$ (as in Milnor). As I recall the the definition of diffeomorphism is such that a cube and a sphere are considered not to be diffeomorphic. This is because G&P define a map at point of a manifold to be smooth if it can be extended to a map on an open set of the ambient space that is smooth in the sense that it is a map from an open set in $\mathbb{R}^n$ to $\mathbb{R}^m$. I never understood, or saw, how this approach can be used to think about different differentiable structures on manifolds. Since there is only one differential structure on $S^2$, the definition I mention above of diffeomorphism seems to at odds with the general one, given for example in Spivak volume 1. (If anyone could explain this to me I'd be grateful. As a student I found this confusing and still do.) </p>
<p>What I am getting at in the above paragraph is that an additional topic might be the general definition of differentiable manifold. It's nice have projective spaces and Grassmanians at least in ones collection of examples.</p>
|
58,870 |
<p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
|
Anton Petrunin
| 1,441 |
<p>If this is the first course in Differential geometry,
you should not go further than Gauss--Bonnet for surfaces.
I would not even consider anything with dimension >2.
By the way here is our <a href="https://arxiv.org/abs/2012.11814" rel="nofollow noreferrer">textbook</a> on the subject.
If they like Differential geometry, they could take another course.</p>
<p>If you cover more, then it is easy to produce lammers.
If you skip these topics, then (most likely) your student will have no idea what is differential geometry at the end of the course.</p>
|
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