qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,458,406 | <p>Define <span class="math-container">$f(n)=(2n)^2 + 1,n \in \mathbb{N}$</span></p>
<p>From <span class="math-container">$1$</span> to <span class="math-container">$10^7$</span> there's <span class="math-container">$15$</span> numbers that <span class="math-container">$f(n)$</span> is prime, <span class="math-container">$f(f(n)), f(f(f(n)))$</span> and <span class="math-container">$f(f(f(f(n))))$</span> are also primes.</p>
<p>The <span class="math-container">$15$</span> numbers are:</p>
<pre><code>625678, 704613, 717718, 1182168, 3147353,
4869813, 5339178, 5363578, 5411562, 846777,
7848283, 7970403, 8152962, 9220303, 9727978
</code></pre>
<p>Is that normal <span class="math-container">$4$</span> numbers in this <span class="math-container">$15$</span> numbers end in <span class="math-container">$78$</span>?</p>
| Empy2 | 81,790 | <p>There are <span class="math-container">$40$</span> possible final two digits -those that end <span class="math-container">$2, 3, 7$</span> and <span class="math-container">$8$</span><br>
The chance that one of the <span class="math-container">$40$</span> appears four times is <span class="math-container">$$40{15\choose4}0.025^40.975^{11}=0.016$$</span><br>
So it's a bit surprising, but not if you have done half a dozen of these functions. There are also other things that might have drawn your attention if the numbers had come out differently - for example several close together.</p>
|
535,533 | <p>My confusion is how do we define : $\sin (x)$ for $x\in \mathbb{R}$.</p>
<p>I only know that $\sin(x)$ is defined for degrees and radians..</p>
<p>Suddenly, I have seen what is $\sin (2)$.. </p>
<p>I have no idea how to interpret this when not much information is given what $2$ is... </p>
<p>does this mean $2$ radians or $2$ degrees or some thing else...</p>
<p>I always wanted to clarify this but could not do it... </p>
<p>I guess most of the school students have this confusion.. </p>
<p>please help me to understand this... </p>
<p>Thank you....</p>
| Arthur | 99,272 | <p>You are using radians in your case. The most common definition of the sine is $\sin(x) := \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} x^{2n+1}$ though, which coincides with the sine in radians as you know it.</p>
|
1,857,196 | <p><strong>Question :</strong>
Let $f(x) = \sum^n_{k=0}c_kx^k$ be a polynomial function then prove that if $f(x) = 0$ for $n+1$ distinct real values, then every coefficient $c_k$ in $f(x)$ is $0$ , thus $f(x) = 0$ for all real values of $x$.</p>
<p><strong>What I think :</strong>
My problem is that I have no other original thought as to how come a polynomial function $f(x)$ of degree $n$ have more than $n$ values that satisfy $f(x) = 0$; if we factorize any $n$ degree polynomial we would get something of the form $(x-a_1)(x-a_2)........(x-a_n) = 0$ so we get here $n$ values again if we see graphically, we find that the $f(x)$ curve would meet the $Y$ axis $n$ times.</p>
<p>So does the question mean that $f(x)$ does not exist when it says that all the coefficients in $f(x)$ would be zero if $n +1$ real distinct values satisfy $f(x) = 0$ or does it mean something else and how do I prove it ?</p>
| learning_math | 350,148 | <p>It is actually possible for a polynomial of degree $n$ to have more than $n$ roots. For example, consider the ring $R=\mathbb{Z}/25\mathbb{Z}$. Consider the polynomial $f(x)=x^2\in R[x]$. Then $f$ has $5$ roots, namely, $\overline{0}, \overline{5}, \overline{10}, \overline{15}, \overline{20}$ in $R$.</p>
<p>A polynomial of degree $n$, has $n$ roots if the underlying ring is a field. </p>
|
1,448,213 | <p>In other words, consider $A_n$, the alternating group of the $n$-th symmetrical group $S_n$, is it true that
$$A_n=\{a^2\mid a\in A_n\}$$?
I tested for $S_3$ and it seemed to hold. If it is true, it will be very helpful to me for solving another problem. </p>
| Shervin | 273,520 | <p>Show that $\Vert Lf\Vert\le\alpha$ for all $\Vert f \Vert=1$. </p>
<p>Because </p>
<p>$1\ge \pm f, \forall x \in [0,1]$ </p>
<p>we have that </p>
<p>$1\pm f\ge 0$, </p>
<p>thus, </p>
<p>$L(1\pm f)\ge 0$. </p>
<p>By linearity </p>
<p>$-L1\le Lf\le L1$, </p>
<p>that is </p>
<p>$|Lf|\le L1, \forall x \in [0,1]$, </p>
<p>hence </p>
<p>$\max_{x\in[0,1]}|Lf|= \Vert Lf \Vert \le \max_{x\in [0,1]}L1=\alpha$. </p>
<p>Because boundness gives continuity the conclusion follows.</p>
|
3,806,122 | <p>I tried using Chinese remainder theorem but I kept getting 19 instead of 9.</p>
<p>Here are my steps</p>
<p><span class="math-container">$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$</span></p>
| Stinking Bishop | 700,480 | <p><span class="math-container">$y_1$</span> should've been the inverse of <span class="math-container">$8\pmod{11}$</span>, not of <span class="math-container">$9\pmod{11}$</span>, so <span class="math-container">$y_1=7$</span>.</p>
<p>Similarly, <span class="math-container">$y_2$</span> should've been the inverse of <span class="math-container">$11\pmod 8$</span>, not of <span class="math-container">$1\pmod 8$</span>, so <span class="math-container">$y_2=3$</span>.</p>
<p>Therefore, the result is: <span class="math-container">$9\times\frac{88}{11}\times \color{red}{7}+1\times\frac{88}{8}\times \color{red}{3}=537\equiv 9\pmod{88}$</span></p>
|
135,252 | <p>Evaluate $\displaystyle \lim_{n \to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$. What are the ways of counting such things? My last topic in school was Riemann integral, can I use it here?</p>
| Adam Rubinson | 29,156 | <p>The sum equals (the sum of 1/k from k=1 to 2n) minus (the sum of 1/k from k=n+2 to 2n). Then take the limit.</p>
|
376,861 | <p>A knot can be represented with a <a href="http://katlas.math.toronto.edu/wiki/MorseLink_Presentations" rel="nofollow noreferrer">Morse link presentation</a>, as a combination of cups, caps and crossings (which is not uniquely determined by the knot, of course):</p>
<p><a href="https://i.stack.imgur.com/47mxu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/47mxu.png" alt="enter image description here" /></a></p>
<p>Two Morse link presentations of the same knot can be related by a sequence of the following moves:</p>
<ul>
<li>Swapping the order of two independent operations</li>
</ul>
<p><a href="https://i.stack.imgur.com/OZ2dq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OZ2dq.png" alt="enter image description here" /></a></p>
<ul>
<li>Pulling a cap, a cup or a crossing through two crossings (generalization of Reidemeister type 3)
<a href="https://i.stack.imgur.com/barWn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/barWn.png" alt="enter image description here" /></a></li>
<li>Cancelling out two successive crossings of alternating types (Reidemeister type 2)</li>
</ul>
<p><a href="https://i.stack.imgur.com/6BpPZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6BpPZ.png" alt="enter image description here" /></a></p>
<ul>
<li>Twisting a cap or a cup (Reidemeister type 1)</li>
</ul>
<p><a href="https://i.stack.imgur.com/sy37Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sy37Z.png" alt="enter image description here" /></a></p>
<ul>
<li>Introducing or eliminating a zig-zag</li>
</ul>
<p><a href="https://i.stack.imgur.com/hyBtB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hyBtB.png" alt="enter image description here" /></a></p>
<p>Given Morse presentation of the unknot, it seems to me that one can always simplify it to the trivial unknot diagram without ever introducing a zig-zag (so, never introducing new strands). So the last equality could be used in the simplifying direction only. Is this fact known and if so, how does the proof go? If not, I would be interested in a counter-example.</p>
| M. Ozawa | 46,903 | <p>I understood your question.
I think it is true. First we isotope a Morse position to a bridge position without zig-zag moves. Then we have a bridge position of the trivial knot, which has been proved to be unique up to bridge isotopies by Otal.
Hence we have the trivial knot diagram in a bridge position.</p>
<p><em>Otal, Jean-Pierre</em>, Présentations en ponts du nœud trivial, C. R. Acad. Sci., Paris, Sér. I 294, 553-556 (1982). <a href="https://zbmath.org/?q=an:0498.57001" rel="noreferrer">ZBL0498.57001</a>.</p>
|
4,521,774 | <p>In many posts on MSE, it is discussed that Cauchy sequences can't be defined in General topological spaces and in a typical topology book it is discussed what converging sequences are, but, what I don't understand is, why, on an abstract level, does convergence generalize even without a metric while cauchy-ness doesn't?</p>
<p><strong>My issue with the posts <a href="https://math.stackexchange.com/questions/3839837/cauchy-sequence-is-not-a-topological-notion">(eg)</a> is that they show that Cauchyness is not a topological property using specific sequences. But what I wish to know is the abstract idea behind it.</strong></p>
| Lee Mosher | 26,501 | <p>The first abstract idea to come to terms with is this:</p>
<blockquote>
<p>The following statement is <em><strong>false</strong></em>: For every topological space <span class="math-container">$X$</span>, for any two metrics <span class="math-container">$d,d'$</span> that generate the topology on <span class="math-container">$X$</span>, and for every sequence <span class="math-container">$(x_n)$</span> in <span class="math-container">$X$</span>, the sequence <span class="math-container">$(x_n)$</span> is a Cauchy sequence with respect to <span class="math-container">$d$</span> if and only if it is a Cauchy sequence with respect to <span class="math-container">$d'$</span>.</p>
</blockquote>
<p>In your post you wrote</p>
<blockquote>
<p>"... they show that Cauchyness is not a topological property using specific sequences. But..."</p>
</blockquote>
<p>... But ... an important thing to realize is that those specific sequences amount to a <em>proof</em> that the above statement is false. All one needs for that proof is to give one counterexample, i.e. one example of <span class="math-container">$X,d,d',(x_n)$</span> such that <span class="math-container">$d,d'$</span> generate the topology on <span class="math-container">$X$</span>, and <span class="math-container">$(x_n)$</span> is a Cauchy sequence with respect to <span class="math-container">$d$</span>, and <span class="math-container">$(x_n)$</span> is <em>not</em> a Cauchy sequence with respect to <span class="math-container">$d'$</span>.</p>
<hr />
<p>With that out of the way, your question regarding the abstract idea behind "Cauchyness" is still very interesting. The comments give two distinct answers to this question: that abstract idea can be <a href="https://en.wikipedia.org/wiki/Uniform_space" rel="nofollow noreferrer">uniform structures</a>; or it can be <a href="https://en.wikipedia.org/wiki/Cauchy_space" rel="nofollow noreferrer">Cauchy structures</a>. One thing I'm unsure of is whether these two notions of "Cauchyness" are equivalent, but I believe they are, in fact I think that this equivalence is covered in <a href="https://en.wikipedia.org/wiki/Uniform_space#Uniform_cover_definition" rel="nofollow noreferrer">this part of the first link above</a> (if I'm wrong about this, I hope someone corrects me).</p>
<hr />
<p>Okay, so let me try to answer your question by explaining why a <a href="https://en.wikipedia.org/wiki/Uniform_space" rel="nofollow noreferrer">uniform structure</a> captures the abstract idea behind Cauchyness. Consider a sequence <span class="math-container">$(x_n)$</span> in a topological space <span class="math-container">$X$</span>.</p>
<p>Assuming that the topology on <span class="math-container">$X$</span> is generated by the metric <span class="math-container">$d$</span>, let recall the standard definition: to say that <span class="math-container">$(x_n)$</span> is a Cauchy sequence means that for any <span class="math-container">$\epsilon > 0$</span> there exists an integer <span class="math-container">$N \ge 1$</span> such that for any integers <span class="math-container">$m,n \ge 1$</span> we have <span class="math-container">$|x_n-x_m| < \epsilon$</span>.</p>
<p>But now, let me reword this definition in a vague and intuitive fashion, without referring to the metric nor to any number <span class="math-container">$\epsilon$</span>: Given some <em>uniform measurement of closeness</em> in <span class="math-container">$X$</span> there exists an integer <span class="math-container">$N \ge 1$</span> such that all of the terms of the sequence starting from <span class="math-container">$x_N$</span>, namely all of the terms in the set <span class="math-container">$\{x_N,x_{N+1},x_{N+2},...\}$</span>, are simultaneously close to each other with respect to <em>that given uniform measurement of closeness</em>.</p>
<p>What goes wrong in an ordinary topological space <span class="math-container">$X$</span> is that there is no such thing as a "uniform measurement of closeness". At best, an open neighborhood of a point <span class="math-container">$x \in X$</span> gives a local notion of closeness around <span class="math-container">$x$</span>, in the sense that being an element of that neighborhood is a notion of "being close to" <span class="math-container">$x$</span>. But "closeness to <span class="math-container">$x$</span>" is not (and should not) be transitive! That local notion gives you no way to test whether all the points in an entire infinite set like <span class="math-container">$x_N,x_{N+1},x_{N+2},...$</span> are simultaneously close <em>to each other</em>.</p>
<p>The idea of a uniform structure is that a "uniform measurement of closeness" can be expressed by choosing a subset <span class="math-container">$U \subset X \times X$</span>. Having made a choice of <span class="math-container">$U$</span>, one can then declare that the points of a set such as <span class="math-container">$\{x_N,x_{N+1},x_{N+2},...\}$</span> are all uniformly close to each other if and only if all ordered pairs drawn from that subset are elements of <span class="math-container">$U$</span>. A uniform structure is then defined to be a set <span class="math-container">$\Phi$</span>, each of whose elements <span class="math-container">$U \in \Phi$</span> is a subset <span class="math-container">$U \subset X \times X$</span>, such that <span class="math-container">$\Phi$</span> satisfies a <a href="https://en.wikipedia.org/wiki/Cauchy_space#Definition" rel="nofollow noreferrer">bunch of axioms</a>.</p>
<p>Here's an exercise: show that if <span class="math-container">$d$</span> is a metric on <span class="math-container">$X$</span>, and if for each <span class="math-container">$\epsilon > 0$</span> we define <span class="math-container">$U_\epsilon = \{(x,y) \in X \times X \mid d(x,y) < \epsilon\}$</span>, then <span class="math-container">$\Phi = \{U_\epsilon \mid \epsilon > 0\}$</span> is a <a href="https://en.wikipedia.org/wiki/Cauchy_space#Definition" rel="nofollow noreferrer">uniform structure</a> on <span class="math-container">$X$</span>.</p>
<p>One can then define a sequence <span class="math-container">$(x_n)$</span> in <span class="math-container">$X$</span> to be Cauchy if for every <span class="math-container">$U \in \Phi$</span> there exists an integer <span class="math-container">$N \ge 1$</span> such that every ordered pair drawn from the set of terms <span class="math-container">$\{x_N,x_{N+1},x_{N+2},\ldots\}$</span> is an element of the set <span class="math-container">$U$</span>.</p>
|
2,258,139 | <p>A natural number $n>1$ is called <em>good</em> if$$n \mid 2^n+1.$$ For example, $n=3$ is good, as $3 \mid 2^3+1=9$. Prove that if $N_1$ and $N_2$ are good, then:</p>
<ul>
<li>$\mathrm{lcm}(N_1,N_2)$ and $\gcd(N_1,N_2)$ are good,</li>
<li>$N_1\cdot N_2$ is good. </li>
</ul>
<p>This seems pretty difficult for me. Any hints?</p>
| SiXUlm | 58,484 | <p>Let $n_1,n_2$ be two good numbers. Denote: $k = LCM(n_1,n_2)$ and $d = (n_1,n_2)$. Then we have simple relation: $dk = n_1n_2$.</p>
<p>For the first part, @Aaron has already done. The LCM part can be done more simple using the following property: if $a | x$ and $b | x$ then $LCM(a,b) | x$. In our case, $n_1 | 2^{n_1} + 1 | 2^k + 1$ and $n_2 | 2^{n_2} + 1 | 2^k + 1$ (as $k$ is odd). Thus $k | 2^k + 1$.</p>
<p>For the second part,
$$2^{n_1n_2}+1 = 2^{dk} + 1 = (2^k+1)\bigg((2^k)^{d-1} - ... +1\bigg)$$</p>
<p>The first factor is divisible by $k$ (due to part $1$).</p>
<p>The second factor has $d$ terms. As $d| k|2^k + 1$ (due to part $1$), we have $2^k \equiv -1 \pmod d$. Thus, the second factor sums up to $d$ (modulo $d$) (note that $d-1$ is even). So the whole product is divisible by $dk = n_1n_2$.</p>
|
4,235,480 | <p><span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~~ \leftarrow~~ \text{The set of integers each of which is greater or equal than zero} ~$</span></p>
<p>I want to know or create the alternative(s) of set of <span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~$</span></p>
<p>As I write <span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~$</span> using a pen in a paper, then a symbol of <span class="math-container">$~ \cup ~$</span> may sometimes be seen as letter U and the ambiguity seems happens.</p>
<p>Does anyone know it?</p>
<p><strong>Add</strong></p>
<p>Thought that <span class="math-container">$~ \mathbb Z_{\geq 0} ~$</span> may be the one of the answers .</p>
| David A. Craven | 804,921 | <p>Here is a proof that does not use the sizes of the classes, just that it has four classes. Let <span class="math-container">$x_i$</span> be an element from class <span class="math-container">$C_i$</span>. We start by inserting the trivial character, the action of <span class="math-container">$\chi_2$</span> on <span class="math-container">$x_3^{-1}$</span> (which, since <span class="math-container">$\chi_2$</span> acts as <span class="math-container">$\zeta\not\in\mathbb R$</span>, must not be conjugate to <span class="math-container">$x_3$</span>), then the dual of <span class="math-container">$\chi_2$</span>, to obtain the following partial table.</p>
<p><span class="math-container">$$\begin{matrix}
&x_1&x_2&x_3&x_3^{-1}\\
\chi_1&1&1&1&1
\\\chi_2&1&&\zeta&\zeta^2
\\\chi_3&1&&\zeta^2&\zeta
\\\chi_4&a
\end{matrix}$$</span></p>
<p>Now let's compute the order, which is <span class="math-container">$1+1+1+a^2$</span>. Since the order of <span class="math-container">$x_3$</span> is a multiple of <span class="math-container">$3$</span> (as the image of <span class="math-container">$x_3$</span> under the homomorphism <span class="math-container">$\chi_3$</span> has order <span class="math-container">$3$</span>), <span class="math-container">$3\mid |G|=3+a^2$</span>. Thus <span class="math-container">$a$</span> is a multiple of <span class="math-container">$3$</span>, and in particular, <span class="math-container">$|G/G'|=3$</span>. Thus <span class="math-container">$G'\neq 1$</span>, and some non-trivial class must be in it. This is <span class="math-container">$x_2$</span>.</p>
<p><span class="math-container">$$\begin{matrix}
&x_1&x_2&x_3&x_3^{-1}\\
\chi_1&1&1&1&1
\\\chi_2&1&1&\zeta&\zeta^2
\\\chi_3&1&1&\zeta^2&\zeta
\\\chi_4&a&b
\end{matrix}$$</span></p>
<p>Now by column orthogonality, <span class="math-container">$ab=-3$</span>. But <span class="math-container">$3\mid a$</span>, and <span class="math-container">$b$</span> is an algebraic integer. Thus <span class="math-container">$a=3$</span>, <span class="math-container">$b=-1$</span>. Finally, <span class="math-container">$\chi_4$</span> is clearly real, so its values on <span class="math-container">$x_3$</span> and <span class="math-container">$x_4=x_3^{-1}$</span> are the same.</p>
<p><span class="math-container">$$\begin{matrix}
&x_1&x_2&x_3&x_3^{-1}\\
\chi_1&1&1&1&1
\\\chi_2&1&1&\zeta&\zeta^2
\\\chi_3&1&1&\zeta^2&\zeta
\\\chi_4&3&-1&c&c
\end{matrix}$$</span></p>
<p>Column orthogonality with (say) <span class="math-container">$x_1$</span> gives the result. If you don't want to work with any complex numbers, note that orthogonality of columns 1 and 3 gives <span class="math-container">$\alpha+3c=0$</span> for some <span class="math-container">$\alpha$</span>, and columns 2 and 3 gives <span class="math-container">$\alpha-c=0$</span>. Thus <span class="math-container">$c=0$</span>.</p>
<p><span class="math-container">$$\begin{matrix}
&x_1&x_2&x_3&x_3^{-1}\\
\chi_1&1&1&1&1
\\\chi_2&1&1&\zeta&\zeta^2
\\\chi_3&1&1&\zeta^2&\zeta
\\\chi_4&3&-1&0&0
\end{matrix}$$</span></p>
|
2,571,395 | <p>I recently reached got a nice answer from my <a href="https://math.stackexchange.com/questions/2567486/integrating-int-x-1x-2-sqrt12at-ax-1x-22-dt">previous question</a> but I quickly that the problem would be unreasonable unless $x_1$ is not a variable and always holds some value, preferably 0, which simplifies the problem significantly leading to the above $L=\frac{1}{a}\int_0^{ax_2}\sqrt{1+t^2}dt$. I know that you can take the easy first step to get $La=\int_0^{ax_2}\sqrt{1+t^2}dt$ and then take a derivative to get rid of the integral, but I am unsure what variable(s) it would make sense to take the derivative in respect to, would I take the derivative in terms of $ax_2$? And then if that was the case, how would I take the derivative of $La$ in respect to $ax_2$?</p>
| Jack D'Aurizio | 44,121 | <p>Let us assume to have a function $L$ defined in terms of $a,b>0$ as
$$ L(a,b) = \frac{1}{a}\int_{0}^{ab}\sqrt{1+t^2}\,dt \stackrel{t\mapsto au}{=}\int_{0}^{b}\sqrt{1+a^2 u^2}\,du. \tag{0}$$
Geometric interpretation: $\int_{x_0}^{x_1}\sqrt{1+f'(u)^2}\,du$ is the length of the graph of $f(x)$ over the interval $[x_0,x_1]$, hence $L(a,b)$ is measuring the length of the parabolic arc $y=\frac{a}{2}u^2$ over the interval $[0,b]$. Assume that we want to express $a$ in terms of $b$ and $L$. This means to solve the trascendental equation</p>
<p>$$ ab \sqrt{1+a^2 b^2}+\operatorname{arcsinh}(ab) = 2aL $$
with respect to $a$. Equivalently, we may solve
$$ \sqrt{1+d^2}+\frac{\operatorname{arcsinh}(d)}{d} = \frac{2L}{b}\tag{A} $$
with respect to $d$, then recall $a=\frac{d}{b}$. The last problem is a root finding problem: the function $g(d)=\sqrt{1+d^2}+\frac{\operatorname{arcsinh}(d)}{d}$ behaves like $2+\frac{d^2}{3}$ in a right neighbourhood of the origin and like $d+\frac{1+\log(4d^2)}{2d}$ in a left neighbourhood of $+\infty$. Additionally, it is convex over $\mathbb{R}^+$, and this is a real bless, since it implies that $(A)$ can be efficiently solved through <a href="https://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow noreferrer">Newton's method</a>. A not-so-bad starting point is given by the solution of
$$ \sqrt{1+d^2}+\frac{1}{\sqrt{1+\frac{d^2}{3}}} = \frac{2L}{b}\tag{B} $$
which can be found in explicit terms by locating the roots of a fourth degree polynomial. A better starting point is provided by the solution of
$$ \sqrt{1+d^2}+\frac{1}{\sqrt[3]{1+\frac{d^2}{2}}} = \frac{2L}{b}.\tag{C} $$</p>
<p>A similar problem arising from the study of planetary motion is the notorious <a href="https://en.wikipedia.org/wiki/Kepler_problem" rel="nofollow noreferrer">Kepler's problem</a> (see also <a href="https://www.willbell.com/MATH/mc12.htm" rel="nofollow noreferrer">this introduction</a> to a very nice book on the topic).</p>
<p>Differentiation can also be used to approximate $a(L,b)$, the <a href="https://en.wikipedia.org/wiki/Implicit_function" rel="nofollow noreferrer">implicit function</a> defined by $(0)$.<br> We simply have:
$$ \frac{\partial L}{\partial a} = \int_{0}^{b}\frac{\partial}{\partial a}\sqrt{1+a^2 u^2}\,du = \frac{ab\sqrt{1+a^2 b^2}-\operatorname{arcsinh}(ab)}{2a^2}.$$</p>
<p>Through <a href="https://arxiv.org/pdf/1602.00436" rel="nofollow noreferrer">Huygens' inequality</a> we get the approximation
$$ a\approx \frac{\sqrt{2}}{b^2}\sqrt{L^2-b^2+(L-b)\sqrt{L^2+2b L-2b^2}}.\tag{D}$$</p>
|
1,458,144 | <p><a href="https://i.stack.imgur.com/pJcCp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pJcCp.png" alt="enter image description here"></a></p>
<p>I have this function and I'm trying to write a program to compute it as n approaches 100. The problem is it overflows once it reaches around 50. The hint to solving this question is to rewrite the part inside the parentheses (sqrt(1+2^-n*Xn) - 1) to (a-b)(a+b), but I'm having trouble thinking of how to do that.</p>
| John | 7,163 | <p>Well, $(a+b)(a-b)$ is $a^2 - b^2$, so perhaps you treat the two terms as $a^2$ and $b^2$ to get:</p>
<p>$$x_{n+1} = 2^{n+1}((1+2^{-n}x_n)^{1/4} + 1)((1+2^{-n}x_n)^{1/4} - 1)$$</p>
<p>Maybe this helps?</p>
|
4,286,296 | <p>I am trying to prove the following claim:</p>
<blockquote>
<p>Let <span class="math-container">$ 0\leq n \in \Bbb Z$</span> and suppose that there exists a <span class="math-container">$k \in \Bbb Z$</span> such that <span class="math-container">$n=4k+3$</span>.
Prove or disprove: <span class="math-container">$\sqrt n \notin \Bbb Q$</span> .</p>
</blockquote>
<p>The problem I am having is that I am trying to assume by contradiction that <span class="math-container">$\sqrt n \in \Bbb Q$</span> and then I say that there are <span class="math-container">$a,b \in \Bbb Z$</span> such that <span class="math-container">$n=\sqrt {4k+3}=\frac ab$</span>. I finally get to a point where <span class="math-container">$k=\frac {a^2-3b^2}{4b^2}$</span>. Yet I can't find any <span class="math-container">$a,b \in \Bbb Z$</span> that will help me show that the claim is false, nor show a contradiction that will cause the claim to be true.
Any help will be welcomed.</p>
| lone student | 460,967 | <blockquote>
<p><strong>Statement:</strong></p>
<p>Let <span class="math-container">$a,b,k\in\mathbb Z^{+}$</span>, where <span class="math-container">$\gcd (a,b)=1$</span> and if <span class="math-container">$4k+3=\frac{a^2}{b^2}$</span>, then <span class="math-container">$b^2=1$</span> or <span class="math-container">$b=1$</span>.</p>
</blockquote>
<p>Thus we have,</p>
<p><span class="math-container">$$\sqrt{4k+3}=a,\thinspace a\in\mathbb Z^{+}$$</span></p>
<p>and</p>
<p><span class="math-container">$$k=\frac{a^2-4+1}{4}=\frac{a^2+1}{4}-1$$</span></p>
<p>This immediately implies,</p>
<p><span class="math-container">$$a=2m-1, \thinspace m\in\mathbb Z^{+}$$</span></p>
<p>This means,</p>
<p><span class="math-container">$$\begin{align}a^2+1&=4(m^2-m)+2\not\equiv 0\thinspace\thinspace\thinspace\text{(mod 4)}.&\end{align}$$</span></p>
<hr />
<p><strong>Conclusion:</strong></p>
<p>We conclude that, there doesn't exist <span class="math-container">$n=4k+3,\thinspace k\in\mathbb Z^{+}$</span>, such that <span class="math-container">$\sqrt n\in\mathbb Q^{+}$</span>.</p>
|
137,435 | <p>Consider the braid group on n strands given in the usual Artin presentation. Then add extra relations: each Artin generator has order d. For example, if d=2, one recovers the symmetric group. I would like to know what the order of the group is for arbitrary n and d. Even knowing the name of such groups would be helpful, though, as my attempts to determine this by searching the literature have so far failed.</p>
| David Mitra | 18,986 | <p>Assuming $A+B=\{a+b\mid a\in A, b\in B\}$:</p>
<p>$A=\{\,1,2,3,\ldots\,\}$ and $B=\{ \,-1 +{1\over2}, -2 +{1\over3} ,-3+{1\over4},\ldots\,\}$. The sum contains $\{\,{1\over2},{1\over3},{1\over4},\ldots\,\}$ but not its limit point $0$.</p>
|
396,297 | <p>Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?</p>
<p>I've calculated that the recurrence relation for this integral is:</p>
<p>$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$</p>
<p>So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.</p>
<p>Can this be useful here?</p>
<p>I would appreciate all your help.</p>
| Community | -1 | <p>Following the Ishan Banerjee comment let $t=x^n$ hence $dx=\frac{1}{n}t^{\frac{1}{n}-1}dt$ and then
$$ (2n+1) \int_0 ^{1} x^n e^x dx=\frac{2n+1}{n}\int_0^1t^{1/n}e^{t^{1/n}}dt\to2e$$
by using the dominated convergence theorem.</p>
|
396,297 | <p>Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?</p>
<p>I've calculated that the recurrence relation for this integral is:</p>
<p>$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$</p>
<p>So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.</p>
<p>Can this be useful here?</p>
<p>I would appreciate all your help.</p>
| Did | 6,179 | <p>Let $I_n=\int\limits_0^1x^n\mathrm e^x\mathrm dx$. By integration by parts, $(n+1)I_n=\left.x^{n+1}\mathrm e^x\right|_0^1-I_{n+1}=\mathrm e-I_{n+1}$. Now, $0\leqslant I_{n+1}\leqslant I_n$ hence $(n+1)I_n\leqslant\mathrm e\leqslant(n+2)I_n$. </p>
<p>This is enough to show that
$$
\left(2-\frac3n\right)\cdot\mathrm e\leqslant(2n+1)I_n\leqslant2\mathrm e,
$$
hence
$(2n+1)I_n\to2\mathrm e$.</p>
|
396,297 | <p>Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?</p>
<p>I've calculated that the recurrence relation for this integral is:</p>
<p>$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$</p>
<p>So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.</p>
<p>Can this be useful here?</p>
<p>I would appreciate all your help.</p>
| Mhenni Benghorbal | 35,472 | <p>A <a href="https://math.stackexchange.com/questions/375950/asymptotic-for-the-integral-involving-exponential/375970#375970">related technique</a>. You can use integration by parts technique by letting $u=e^{x}$ which leads to</p>
<p>$$ I_n = \left( 2\,n+1 \right) \left( {\frac {{{\rm e}}}{n+1}}-{\frac {{
{\rm e}}}{2+3\,n+{n}^{2}}}+\int _{0}^{1}\!{\frac {{x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right) \left( n+1 \right) }}{dx} \right) .$$</p>
<p>$$\implies \lim_{n\to \infty}I_n = 2 \,\rm{e} + 0 + \lim_{n\to \infty } \int _{0}^{1}\!{\frac {(2n+1){x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right) \left( n+1 \right) }}{dx} $$</p>
<p>$$ \implies \lim_{ n\to \infty } = 2 \rm e . $$ </p>
<p>Note that, the change of the limit with integral is due to the uniform convergence of the sequence $$ \frac {(2n+1){x}^{n+2}}{ \left( n+2 \right) \left( n+1 \right) }. $$</p>
<p><a href="https://math.stackexchange.com/questions/370023/how-to-prove-a-sequence-of-a-function-converges-uniformly/370071#370071">Here</a> is a technique for proving uniform convergence. </p>
<p><strong>Added:</strong> Integration by parts,</p>
<p>$$ \int u\, dv = u\,v -\int v \,du. $$</p>
<p>So, in your case $ u = e^{x} $ and $ dv = x^n dx $.</p>
|
239,720 | <p>If $A$ is unital C$^*$-algebra, is it true that the multiplier algebra of $A \otimes \mathcal{K} $ is $ A \otimes \mathcal{B}(\mathcal{H})$? Where $\mathcal{K}$ is C$^*$-algebra of compact operators on the Hilbert space $\mathcal{H}$.</p>
| Ulrich Pennig | 3,995 | <p>The fact stated in the answer by vap is proven in the paper "<a href="http://www.sciencedirect.com/science/article/pii/0022123673900360" rel="noreferrer">Multipliers of C*-algebras</a>" by Akemann, Pedersen and Tomiyama (see Theorem 3.3, I guess). Moreover, they prove in Theorem 3.8 that multiplier algebras are not very well behaved with respect to minimal tensor products:</p>
<blockquote>
<p>Let $A$ and $B$ be $C^*$-algebras and assume that $B$ has a countable
approximate unit, but no unit (think of $\mathcal{K}$ here in your
case) and that $A$ is infinite dimensional. Then<br>
$$
M(A) \otimes M(B) \subsetneq M(A \otimes B)
$$
where the tensor product is the minimal one.</p>
</blockquote>
<p>So, in particular, for any <em>infinite dimensional</em> unital $C^*$-algebra $A$, the tensor product $A \otimes \mathcal{B}(\mathcal{H})$ is always a proper subalgebra of $M(A \otimes \mathcal{K})$.</p>
|
2,379,955 | <p>Assume I want to minimise this:
$$ \min_{x,y} \|A - x y^T\|_F^2$$
then I am finding best rank-1 approximation of A in the squared-error sense and this can be done via the SVD, selecting $x$ and $y$ as left and right singular vectors corresponding to the largest singular value of A.</p>
<p>Now instead, is possible to solve:
$$ \min_{x} \|A - x b^T\|_F^2$$
for $b$ also fixed?</p>
<p>If this is possible, is there also a way to solve:
$$ \min_{x} \|A - x b^T\|_F^2 + \|C - x d^T\|_F^2$$
where I think of $x$ as the best "average" solution between the two parts of the cost function.</p>
<p>I am of course longing for a closed-form solution but a nice iterative optimisation approach to a solution could also be useful.</p>
| mathreadler | 213,607 | <p>You already have solutions with CVX (convex optimization), but in fact you can solve this using simple ordinary linear least squares and representing matrix multiplication with Kronecker products. Let $M_E$ represent multiplication by $E$ (from the right) and $v_A,v_C,v_x$ be the respective vectorization of $A,C,x$ we can rewrite it:</p>
<p>$$\min_{v_x}\left\{\|v_A- M_{b^t}v_x\|_2^2 + \|v_C- M_{d^t}v_x\|_2^2\right\}$$</p>
<p>Which you can expand using $\|Y\|^2 = Y^TY$, then expand, differentiate, sum up and set derivative equal $0$ and solve.</p>
<p>You can read more about the details of the vectorization on <a href="https://en.wikipedia.org/wiki/Kronecker_product" rel="nofollow noreferrer">Wikipedia entry Kronecker Products</a></p>
|
136,264 | <p>I have a question concerning the stability analysis for a kind of differential equation taking the form
$$\dot x=Ax+Bw,$$
where $A\in \mathbb{R}^{n \times n}$, $B\in \mathbb{R}^{n \times m}$ are constant matrices and
$w \in \mathbb{R}^m$ is a normal random variable, i.e., $w\sim \mathcal{N}(0,W)$ with $W$ be a symmetric and positive definite matrix.</p>
<p>Since the zero is not an equilibrium of the system, the Lyapunov analysis does not make sense. When the input-to-state stability analysis is considered, the robust control theory does not apply due to the unboundness of $w$. By resorting to stochastic stability in the sense of mean square or almost surely, the Ito formula seems to be invalid.</p>
<p>HOW to carry out the stability analysis of this kind of systems? Any pointer will be helpful. Thanks!</p>
| 27hel27 | 47,623 | <p>You can also check out <a href="http://zbmath.org/journals/" rel="nofollow">http://zbmath.org/journals/</a> for details on the journal content. It doesn't give you a ranking though but you see at a glance, who published in the journal you are interested in or what topics are represented in the articles.</p>
|
2,872,701 | <blockquote>
<p>Let $K\subset N\subset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.</p>
</blockquote>
<p>It is easy to show that </p>
<blockquote>
<p>If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.</p>
</blockquote>
<p>I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.</p>
<blockquote>
<p>Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/K\oplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/K\oplus N'/K=M/K$.</p>
</blockquote>
<p>I want to say something like $N\oplus N'=M$ which is not true since $N'\cap N\supset K\neq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=N\oplus B$ but $B$ is not a submodule.</p>
| Matthé van der Lee | 75,745 | <p>Taking $K = N$, it would follow that $N$ is always a direct summand of $M$ - which is false in general (unless $R$ is semisimple, i.e. absolutely projective, i.e. all $R$-modules are projective; for example a field).</p>
|
4,466,733 | <p><span class="math-container">$$\frac{df(x)}{dx}=f(x+5)$$</span>
I am unable to solve this kind of integration using high school mathematics.
Please help.</p>
| emacs drives me nuts | 746,312 | <p>Just an ansatz or guess: Because differentiating <span class="math-container">$a^x$</span> gives a multiple of <span class="math-container">$a^x$</span>, and also <span class="math-container">$a^{x+\mathrm{const}}$</span> is a multiple of <span class="math-container">$a^x$</span>: Try <span class="math-container">$f(x)=k\exp(ax)$</span>, then:</p>
<p><span class="math-container">$$\frac d{dx} f(x) = kae^{ax} \stackrel!= f(x+c) = ke^{a(x+c)} = ke^{ca}e^{ax}$$</span>
with <span class="math-container">$c=5$</span>. Dividing out <span class="math-container">$\exp(ax)$</span>:</p>
<p><span class="math-container">$$ ka \stackrel!= k\exp(ca)$$</span>
so <span class="math-container">$k=0$</span> or <span class="math-container">$a = \exp(ca)$</span>. To solve the latter, divide by <span class="math-container">$\exp(ca)$</span> and rewrite as</p>
<p><span class="math-container">$$(-ca)\exp(-ca) = -c$$</span>
and then apply <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert-<em>W</em></a>:</p>
<p><span class="math-container">$$-ca = W(-c) \quad \implies\quad a = -\frac1cW(-c)$$</span></p>
<p>Now for <span class="math-container">$c = 5$</span> there is no (real) solution because the minimum <span class="math-container">$-1/e$</span> of <span class="math-container">$x\mapsto xe^x$</span> is greater than <span class="math-container">$-5$</span>. But I am not familiar with <span class="math-container">$W$</span> and its branches, so there is likely a different branch of <span class="math-container">$W$</span> that gives (complex) solutions.</p>
<p>Or either there is no solution (except the trivial <span class="math-container">$k=0$</span>) or the ansatz was too restricting.</p>
<hr />
<p>That said, if you had used <span class="math-container">$c=-5$</span>, there was the real solution</p>
<p><span class="math-container">$$a = \frac15W(5) \approx1.326725/5 = 0.265345 \approx \ln 1.30388$$</span></p>
<p>so that <span class="math-container">$f(x) = k\cdot 1.30388^x$</span> where you can pick <span class="math-container">$k$</span> as you like.</p>
<hr />
<p>Note: Here is a <a href="https://www.wolframalpha.com/input?i=-lambertw%28-5%29%2F5" rel="nofollow noreferrer">complex solution for <span class="math-container">$c=5$</span></a>.</p>
|
2,548,177 | <p>I'd like to define <code>sumdiv</code> in Maple such that this:</p>
<pre><code>with(numtheory);
f:=x->x^2;
sumdiv(f(d)*mobius(100/d), d=1..100);
</code></pre>
<p>would do a sum on all divisors <code>d</code> of $100$.</p>
<p><strong>How to do such a sum over divisors in Maple?</strong></p>
<p>Here's what I've tried:</p>
<pre><code>isdivisible:=(a,b)->if a mod b = 0 then 1 else 0 fi;
sum(f(d)*mobius(100/d)*isdivisible(100,d), d=1..100);
</code></pre>
<p>but even if <code>isdivisible(100,d)</code> is $0$ (i.e. $d$ not divisible by $100$), it tries to evaluate <code>mobius(100/d)</code> anyway which is impossible, thus an error.</p>
<blockquote>
<p>Error, (in mobius) invalid arguments</p>
</blockquote>
| Brethlosze | 386,077 | <p>In Matlab, <code>D</code> is a vector containing all the divisors, for any <code>n</code>:</p>
<pre><code>n=10;
k=1:n;
D=K(rem(n,k)==0);
s=sum(D)
</code></pre>
<p>Edit: The sum of a function <code>f</code> over the divisors of <code>n</code>. Note the <code>.</code> operator before the <code>^</code> and <code>*</code> operators, for applying them component-wise instead of matrix-wise (default).</p>
<pre><code>n=10;
k=1:n;
D=K(rem(n,k)==0);
f=@(x)(x.^2+2.*x+sin(x)+1);
s=sum(f(D))
</code></pre>
|
2,573,487 | <p>I have given this set</p>
<blockquote>
<p>$$ M = \{ x \in [1,2]\times [3,4] ~|~ x\in\mathbb{Q}^2 \} \subset \mathbb{R}^2 $$</p>
</blockquote>
<p>First I have to identify the boundary $\partial M$ and then tell if it is open or closed.</p>
<p>I think that $$ \partial M = \{ (x,y) ~|~ x\not\in\mathbb{Q}^2, 1\leq x \leq 2, 3\leq y\leq 4 \} $$</p>
<p>is the boundary, but I am not sure about it. Is this correct? If not, what is the boundary?</p>
<p>Also: I am pretty sure that the set is open, because you can for sure find a series (e.g. for Pi) that converges to an irrational number, but with fraction series values in $\mathbb{Q}$.</p>
| José Carlos Santos | 446,262 | <p>Since $\overline M=[1,2]\times[3,4]$ and $\mathring M=\emptyset$, the boundary of $M$ is $\overline M\setminus\mathring M=[1,2]\times[3,4]$, which is a closed set.</p>
|
1,878,975 | <p>X is for continuous random variable and it's nonnegative.
Then this is the formula.</p>
<p>$$E(X)=\int_0^\infty(1-F(x))dx$$</p>
<p>Does anyone know the proof?
I appreciate any help.</p>
| egreg | 62,967 | <p>Consider
$$
u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}>
\sqrt{\frac{1}{2}+\frac{1}{2}}=1
$$
Assuming your substitutions are correct, the integral becomes
$$
\sqrt{2}\int\frac{1}{1-u^2}\,du=
\frac{\sqrt{2}}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\,du=
\frac{\sqrt{2}}{2}\log\left|\frac{1+u}{1-u}\right|+c
$$
You cannot apply the substitution $u=\tanh v$, as $u>1$.</p>
|
2,170,501 | <p>How do you prove the sum of two monotone sequences is also monotone? </p>
<p>Here is my thought process: </p>
<p>Let $a_n$ and $b_n$ be two monotone increasing sequences. Then $\forall n \in N$, $a_n \leq a_{n+1}$ and $b_n \leq b_{n+1}$. Adding both inequalities you get $a_n + b_n \leq a_{n+1} + b_{n+1}$. Therefore in this specific case of both sequences being monotone increasing I have proven their sum is monotone increasing, and so it is also monotone. </p>
<p>But I'm having trouble figuring out how to apply similar logic for an $a_n$ is monotone increasing but $b_n$ is monotone decreasing. I also tried to do proof by contradiction (like supposing the sum is not monotone), but that also lead to nowhere. </p>
<p>How do I prove the other cases?</p>
| Adren | 405,819 | <p>Consider the sequences defined by :</p>
<p>$$a_n=3n+(-1)^n$$</p>
<p>and</p>
<p>$$b_n=-2n$$</p>
<p>It is readily seen that $(a_n)$ is increasing, since for all $n\in\mathbb{N}$ : $a_{n+1}-a_n=3+2(-1)^{n+1}>0.$</p>
<p>Obviously, $(b_n)$ is decreasing.</p>
<p>And finally $(a_n+b_n)$ is known to be non monotonic.</p>
|
233,075 | <p>I am trying to solve the following equation in the Natural Numbers, with the condition <span class="math-container">$a\ge1$</span>, <span class="math-container">$b\ge1$</span>, and <span class="math-container">$r\ge3$</span>:</p>
<p><span class="math-container">$$\frac{a(a + 3)(a(r - 5) + (12 - r))}{9}=\frac{b (9 + b (-14 + r) - r)}{3}\tag1$$</span></p>
<p>The method I know use is, that I solve the equation for <span class="math-container">$b$</span> and I got:</p>
<p><span class="math-container">$$b=\displaystyle\frac{1}{6} \left(\sqrt{3\cdot\frac{4 a (a+3) (r-14) (a (r-5)-r+12)+3 (r-9)^2}{(r-14)^2}}+\frac{15}{r-14}+3\right)\tag2$$</span></p>
<p>Now, I used Mathematica to check when the function under the square root is a perfect square, with the following code:</p>
<pre><code>ParallelTable[
If[IntegerQ[
FullSimplify[
Sqrt[3*((
4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)]]], {a, r}, Nothing], {a, 1,
10^5}, {r, 3, 10^5}] //. {} -> Nothing
</code></pre>
<p>And the solutions I got, I put in equation <span class="math-container">$(1)$</span> to check if I can find a solution to the original problem.</p>
<blockquote>
<p>This method takes a very very long time, but I am not knowing if there is a faster and smarter way to program this. Can you help me with this. Thanks a lot in advance.</p>
</blockquote>
| bbgodfrey | 1,063 | <p>The excellent second solution by Roman, with <code>R</code> slightly modified, produces</p>
<pre><code>R = HornerForm[(a (3 + a) (-12 + 5 a) + 3 (9 - 14 b) b)/
((-1 + a) a (3 + a) - 3 (-1 + b) b)]
With[{s = 10^4}, Do[If[Divisible[a (3 + a) (-12 + 5 a) + 3 (9 - 14 b) b,
(-1 + a) a (3 + a) - 3 (-1 + b) b] && R >= 3, Sow[{a, b, R}]],
{a, s}, {b, s}] // Reap // Last // First]
(* {{3, 6, 24}, {5, 8, 244}, {5, 10, 31}, {5, 14, 19}, {9, 18, 177}, {9, 20, 46},
{12, 30, 45}, {32, 112, 139}, {33, 114, 573}, {35, 126, 220}, {45, 180, 553},
{47, 450, 16}, {48, 204, 129}, {63, 294, 3750}, {77, 396, 3889}, {116, 728, 46750},
{117, 2340, 15}, {159, 1166, 6826}, {240, 2156, 2098129}, {243, 2214, 576},
{357, 3906, 72807}, {372, 4154, 2509849}, {492, 6314, 398389}} *)
</code></pre>
<p>in about 350 seconds. I attempted to find faster approaches using various combinations of <code>Tuples</code>, <code>Table</code>, <code>Cases</code>, and <code>Select</code>, but the best I could do was</p>
<pre><code>Flatten[Table[If[Divisible[a (-36 + a (3 + 5 a)) + (27 - 42 b) b,
a (-3 + a (2 + a)) + (3 - 3 b) b] && R > 2, {a, b, R}, Nothing, Nothing],
{a, 10000}, {b, 10000}], 1]
</code></pre>
<p>which produced the same results in the same amount of time.</p>
<p>The tutorial, <a href="https://reference.wolfram.com/mathematica/tutorial/DiophantineReduce.html" rel="nofollow noreferrer">DiophantineReduce</a> discusses, among many other cases, "Equations with a Linear Variable", which this question is. Applying <code>Reduce</code></p>
<pre><code>Reduce[R == r && a > 0 && b > 0 && r > 2, {a, b, r}, Integers]
</code></pre>
<p>yields a lengthy result in less than a second, a portion of which is, in effect,</p>
<pre><code>(* b > 1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3]) && r == R *)
</code></pre>
<p>(Not coincidentally, <code>1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3])</code> is the value of <code>b</code> for which <code>Denominator[R] == 0</code>.) Employing the inequality in my approach above,</p>
<pre><code>Flatten[Table[If[Divisible[a (-36 + a (3 + 5 a)) + (27 - 42 b) b,
a (-3 + a (2 + a)) + (3 - 3 b) b] && R > 2, {a, b, R}, Nothing, Nothing], {a, 10000},
{b, Ceiling[1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3])], 10000}], 1]
</code></pre>
<p>reproduces the results given at the beginning of this answer in 15 seconds, a significant improvement. Applying this approach to a much larger domain (and using <code>ParallelTable</code> on a six-processor PC) then yields</p>
<pre><code>Flatten[ParallelTable[If[Divisible[a (-36 + a (3 + 5 a)) + (27 - 42 b) b,
a (-3 + a (2 + a)) + (3 - 3 b) b] && R > 2, {a, b, R}, Nothing, Nothing],
{a, 6000}, {b, Ceiling[1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3])],
300000}], 1]
(* {{3, 6, 24}, {5, 8, 244}, {5, 10, 31}, {5, 14, 19}, {9, 18, 177}, {9, 20, 46},
{12, 30, 45}, {32, 112, 139}, {33, 114, 573}, {35, 126, 220}, {45, 180, 553},
{47, 450, 16}, {48, 204, 129}, {63, 294, 3750}, {77, 396, 3889}, {116, 728, 46750},
{117, 2340, 15}, {159, 1166, 6826}, {240, 2156, 2098129}, {243, 2214, 576},
{357, 3906, 72807}, {372, 4154, 2509849}, {492, 6314, 398389}, {768, 12336, 1769},
{1266, 26028, 12553000}, {1545, 43860, 30}, {3792, 138336, 186},
{5973, 266574, 121035}} *)
</code></pre>
<p>in 1070 seconds. Here is a plot of <code>b</code> vs <code>a</code>.</p>
<pre><code>Show[ListLogLogPlot[%[[2, All, ;;2]], PlotRange -> All, ImageSize -> Large, AxesLabel ->
{a, b}, LabelStyle -> {14, Bold, Black}], LogLogPlot[1/2 + Sqrt[3 - 12 a + 8 a^2 +
4 a^3]/(2 Sqrt[3]), {a, 1, 10000}, PlotRange -> All]]
</code></pre>
<p><a href="https://i.stack.imgur.com/tP2Y6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tP2Y6.png" alt="enter image description here" /></a></p>
<p>Evidently, most of the points lie just above the inequality curve. This suggests that most, although not all, solutions can be obtained by searching just above the curve. For instance,</p>
<pre><code>Flatten[ParallelTable[If[Divisible[a (-36 + a (3 + 5 a)) + (27 - 42 b) b,
a (-3 + a (2 + a)) + (3 - 3 b) b] && R > 2, {a, b, R}, Nothing, Nothing],
{a, 1000000}, {b, Ceiling[1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3])],
Ceiling[1/2 + Sqrt[3 - 12 a + 8 a^2 + 4 a^3]/(2 Sqrt[3])] + 100}], 1]
(* {{3, 6, 24}, {5, 8, 244}, {5, 10, 31}, {5, 14, 19}, {9, 18, 177}, {9, 20, 46},
{12, 30, 45}, {32, 112, 139}, {33, 114, 573}, {35, 126, 220}, {45, 180, 553},
{48, 204, 129}, {63, 294, 3750}, {77, 396, 3889}, {116, 728, 46750},
{159, 1166, 6826}, {240, 2156, 2098129}, {243, 2214, 576}, {357, 3906, 72807},
{372, 4154, 2509849}, {492, 6314, 398389}, {768, 12336, 1769},
{1266, 26028, 12553000}, {5973, 266574, 121035}, {12440, 801136, 1730566},
{43329, 5207358, 30979126197}, {44517, 5422980, 3270113811},
{137796, 29532312, 8075577424022}} *)
</code></pre>
<p>in 220 seconds. Plotted as before,</p>
<p><a href="https://i.stack.imgur.com/nA6J9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nA6J9.png" alt="enter image description here" /></a></p>
<p><strong>Addendum: Direct Solution with <code>Reduce</code></strong></p>
<p>Further review of Ref. <a href="https://reference.wolfram.com/mathematica/tutorial/DiophantineReduce.html" rel="nofollow noreferrer">1</a> indicates that <code>Reduce</code> can obtain integer zeros for bounded regions of <code>{a, b}</code>, for instance,</p>
<pre><code>SetSystemOptions["ReduceOptions" -> {"DiscreteSolutionBound" -> Infinity}];
SetSystemOptions["ReduceOptions" -> {"SieveMaxPoints" -> {10^3, 10^6}}];
Values@Solve[{r == R, 1000 >= a > 0, 1000 >= b > 0, r > 2}, {a, b, r},
Integers, Method -> Reduce]
</code></pre>
<p>yields the same sixteen results obtain by Roman in his answer, but over three times more slowly.</p>
|
1,921,879 | <blockquote>
<p>Find all positive integers $n$ for which $\dfrac{x^n + y^n + z^n}2$ is a perfect square, whenever $x$, $y$, and $z$ are integers such that $x + y + z = 0$.</p>
</blockquote>
<p>I don't even know where to start.</p>
| Dietrich Burde | 83,966 | <p>I have no complete answer, but a start (as you wanted to know where to start). For $n=1$ we have that $(x^1+y^1+z^1)/2=0$ is a perfect square for all $x,y,z$ with $x+y+z=0$. So we may assume $n\ge 2$. Now choose, say, $(x,y,z)=(1,1,-2)$. Then $x+y+z=0$ and
$$
\frac{x^n+y^n+z^n}{2}=\frac{2+(-2)^n}{2}.
$$
This can never be a perfect square for all odd $n>1$, because it is negative in this case. Also for even $n$ this is rarely a perfect square, but it can happen, namely for $n=4$. This is clear, because for $n=4$ we have, with $z=-x-y$,
$$
\frac{x^4+y^4+z^4}{2}=\frac{2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4}{2}=(x^2 + xy + y^2)^2,
$$
which is indeed always a perfect square.</p>
|
3,912,734 | <p>My text book in linear algebra - out of the blue - claims that:</p>
<p><span class="math-container">$|\lambda u|=|\lambda||u|$</span></p>
<p>Where u is a vector and <span class="math-container">$\lambda$</span> is a constant.</p>
<p>I would understand if || were used to denote absolute numbers, but in this book, || is used to denote length (so the dot product of the vector v with itself is written as <span class="math-container">$|v|^2$</span>)</p>
<p>I'm positively dumbfounded. How do I prove this and why would av scalar like <span class="math-container">$\lambda$</span> even have a length in the first place?</p>
<p>EDIT:</p>
<p>This is used in a later section to prove that the projection of v on u equals:</p>
<p><span class="math-container">$$|\frac{u \cdotp v}{|u|^2}u|=\frac{|u \cdotp v|}{|u|^2}|u|=\frac{|u \cdotp v|}{|u|}$$</span> so it definitly seems they're referring to lengths, and not absolute numbers. Otherwise I can't really see that working.</p>
| Aryaman Maithani | 427,810 | <p>It is part of the definition of an inner product <span class="math-container">$\langle \;,\; \rangle$</span> on a complex vector space <span class="math-container">$V$</span> that</p>
<ul>
<li><span class="math-container">$\langle \lambda v, w\rangle = \lambda\langle v, w\rangle,$</span> and</li>
<li><span class="math-container">$\langle v, \lambda w\rangle = \bar{\lambda}\langle v, w\rangle,$</span></li>
</ul>
<p>for all <span class="math-container">$\lambda \in \Bbb C$</span> and <span class="math-container">$v, w \in V$</span>. (As usual, <span class="math-container">$\bar{\lambda}$</span> is the complex conjugate of <span class="math-container">$\lambda$</span>.)</p>
<p>After this, a norm is defined as
<span class="math-container">$$\|v\| = \sqrt{\langle v, v\rangle}.$$</span>
This then gives us</p>
<p><span class="math-container">\begin{align}
\|\lambda v\| &= \sqrt{\langle \lambda v, \lambda v\rangle}\\
&= \sqrt{\lambda\langle v, \lambda v\rangle}\\
&= \sqrt{\lambda\bar{\lambda}\langle v, v\rangle}\\
&= \sqrt{|\lambda|^2\langle v, v\rangle}\\
&= |\lambda|\sqrt{\langle v, v\rangle}\\
&= |\lambda|\|v\|.
\end{align}</span></p>
<hr />
<p>In case you're working over <span class="math-container">$\Bbb R$</span>, you just have <span class="math-container">$\lambda$</span> everywhere instead of <span class="math-container">$\bar{\lambda}$</span>.</p>
|
377,925 | <p>This post comes from the suggestion of <a href="https://mathoverflow.net/users/18698/joel-moreira">Joel Moreira</a> in a <a href="https://mathoverflow.net/questions/377706/an-alternative-to-continued-fraction-and-applications#comment958452_377706">comment</a> on <a href="https://mathoverflow.net/q/377706/34538">An alternative to continued fraction and applications</a> (itself inspired by the Numberphile video <a href="https://youtu.be/_gCKX6VMvmU" rel="nofollow noreferrer">2.920050977316</a> and <a href="https://doi.org/10.1080/00029890.2019.1530554" rel="nofollow noreferrer">Fridman, Garbulsky, Glecer, Grime, and Tron Florentin - A prime-representing constant</a>).</p>
<p>Let <span class="math-container">$u_0 \ge 2$</span> be a rational, and <span class="math-container">$u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1)$</span>.<br />
<strong>Question</strong>: Does the sequence <span class="math-container">$(u_n)$</span> reach an integer?</p>
<p><span class="math-container">$\to$</span> see below the application to irrational number theory.</p>
<p><em>Remark</em>: It is true for <span class="math-container">$u_0=\frac{p}{q}$</span> with <span class="math-container">$p \le 40000$</span> (see Appendix).</p>
<p><strong>Proposition</strong>: It is always true for <span class="math-container">$u_0 = \frac{p}{2}$</span>.<br />
<em>Proof by contradiction</em>: Assume that the sequence never reach an integer, then <span class="math-container">$u_n = k_n + \frac{1}{2}$</span> for all <span class="math-container">$n$</span>. Next note that <span class="math-container">$u_{n+1} = k_n + \frac{k_n}{2}$</span>, so <span class="math-container">$k_n$</span> must be odd for all <span class="math-container">$n$</span>. Let write <span class="math-container">$k_n = 2 h_n +1$</span>, then <span class="math-container">$u_n = 2h_n+1+\frac{1}{2}$</span> (with <span class="math-container">$h_n \ge 1$</span>) and <span class="math-container">$u_{n+1} = 3h_n+1+\frac{1}{2}$</span>. It follows that <span class="math-container">$2h_{n+1} = 3h_n$</span>, and so <span class="math-container">$h_n = (\frac{3}{2})^nh_0$</span>, which implies that <span class="math-container">$2^n$</span> divides <span class="math-container">$h_0$</span> for all <span class="math-container">$n$</span>, contradiction. <span class="math-container">$\square$</span></p>
<p>For <span class="math-container">$u_0=\frac{11}{5}$</span>, then <span class="math-container">$$(u_n)= (\frac{11}{5}, \frac{12}{5}, \frac{14}{5}, \frac{18}{5}, \frac{24}{5}, \frac{36}{5}, \frac{42}{5}, \frac{56}{5}, \frac{66}{5}, \frac{78}{5}, 24, \dots).$$</span> Here is a picture of the dynamic:<br />
<a href="https://i.stack.imgur.com/viHif.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/viHif.png" alt="enter image description here" /></a></p>
<p>By regarding (for example) when <span class="math-container">$u_0=\frac{15}{7}$</span> below, we guess that the general proof should be hard. <span class="math-container">$(u_n) = (\frac{15}{7}, \frac{16}{7}, \frac{18}{7}, \frac{22}{7}, \frac{24}{7}, \frac{30}{7}, \frac{36}{7}, \frac{40}{7}, \frac{60}{7}, \frac{88}{7}, \frac{132}{7}, \frac{234}{7}, \frac{330}{7}, \frac{376}{7}, \frac{636}{7}, \frac{1170}{7}, \frac{1336}{7}, \frac{2470}{7}, \frac{4576}{7}, \frac{7836}{7}, \frac{11190}{7}, \frac{17578}{7}, \frac{20088}{7}, \frac{34428}{7}, \frac{44262}{7}, \frac{50584}{7}, \frac{65034}{7}, \frac{102190}{7}, \frac{160578}{7}, 39324, \dots)$</span></p>
<p>For <span class="math-container">$u_0=\frac{10307}{4513}=\frac{k_0}{q}$</span>, the sequence <span class="math-container">$(\frac{k_n}{q})=(u_n)$</span> reaches an integer at <span class="math-container">$n=58254$</span>. The sequence <span class="math-container">$(u_n)$</span> reaches an integer once <span class="math-container">$k_n \text{ mod } q=0$</span>. Below is the picture for <span class="math-container">$(n,k_n \text{ mod } q)$</span>; it looks completely random. The probability for <span class="math-container">$s$</span> random integers between <span class="math-container">$0$</span> and <span class="math-container">$q-1$</span> to never be zero is about <span class="math-container">$e^{-s/q}$</span> when <span class="math-container">$q$</span> is large enough.</p>
<p><a href="https://i.stack.imgur.com/xdKmq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xdKmq.png" alt="enter image description here" /></a></p>
<blockquote>
<p><strong>Application to irrational number theory</strong></p>
<p>According to <a href="https://doi.org/10.1080/00029890.2019.1530554" rel="nofollow noreferrer">the paper</a> mentioned above, there is a bijection between the set of numbers <span class="math-container">$u_0 \ge 2$</span>, and the set of
sequences <span class="math-container">$(a_n)$</span> such that for all <span class="math-container">$n$</span>:</p>
<ul>
<li><span class="math-container">$a_n \in \mathbb{N}_{\ge 2}$</span>,</li>
<li><span class="math-container">$a_n \le a_{n+1} < 2a_n$</span>.</li>
</ul>
<p>The bijection is given by: <span class="math-container">$$u_0 \mapsto (a_n) \text{ with } a_n =
⌊u_n⌋ \text{ and } u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1),$$</span> <span class="math-container">$$(a_n) \mapsto
u_0 = \sum_{n=0}^{\infty}\frac{a_n-1}{\prod_{i=0}^{n-1}a_i}.$$</span> A
positive answer to the question would provide a kind of alternative to
<a href="https://en.m.wikipedia.org/wiki/Continued_fraction" rel="nofollow noreferrer">continued fraction</a>, in the sense of a natural way to represent
the numbers, with a complete characterization of the irrational ones,
which here would be that <span class="math-container">$\lim_{n \to \infty} (a_n)=\infty$</span>.</p>
</blockquote>
<hr />
<p><strong>Appendix</strong></p>
<p>In the following list the datum <span class="math-container">$[r,(p,q)]$</span> means that the sequence <span class="math-container">$(u_n)$</span>, with <span class="math-container">$u_0=\frac{p}{q}$</span>, reaches an integer at <span class="math-container">$n=r$</span>. The list provides the ones with the longest <span class="math-container">$r$</span> according the lexicographic order of <span class="math-container">$(p,q)$</span>.</p>
<p><em>Computation</em></p>
<pre><code>sage: search(40120)
[1, (2, 1)]
[2, (5, 2)]
[3, (7, 2)]
[4, (7, 3)]
[11, (11, 5)]
[30, (15, 7)]
[31, (29, 14)]
[45, (37, 17)]
[53, (39, 17)]
[124, (41, 19)]
[167, (59, 29)]
[168, (117, 58)]
[358, (123, 53)]
[380, (183, 89)]
[381, (201, 89)]
[530, (209, 97)]
[532, (221, 97)]
[622, (285, 131)]
[624, (295, 131)]
[921, (359, 167)]
[1233, (383, 181)]
[1365, (517, 251)]
[1482, (541, 269)]
[2532, (583, 263)]
[3121, (805, 389)]
[3586, (1197, 587)]
[3608, (1237, 607)]
[3860, (1263, 617)]
[4160, (1425, 643)]
[6056, (1487, 743)]
[9658, (1875, 859)]
[9662, (1933, 859)]
[10467, (2519, 1213)]
[10534, (2805, 1289)]
[11843, (2927, 1423)]
[12563, (3169, 1583)]
[13523, (3535, 1637)]
[14004, (3771, 1871)]
[14461, (4147, 2011)]
[17485, (4227, 1709)]
[18193, (4641, 1987)]
[18978, (4711, 2347)]
[22680, (5193, 2377)]
[23742, (5415, 2707)]
[24582, (5711, 2663)]
[27786, (5789, 2837)]
[27869, (6275, 2969)]
[29168, (6523, 3229)]
[32485, (6753, 2917)]
[33819, (7203, 3361)]
[41710, (7801, 3719)]
[49402, (8357, 3863)]
[58254, (10307, 4513)]
[58700, (10957, 4943)]
[81773, (12159, 5659)]
[85815, (16335, 7963)]
[91298, (16543, 7517)]
[91300, (17179, 7517)]
[98102, (19133, 9437)]
[100315, (19587, 8893)]
[100319, (20037, 8893)]
[102230, (20091, 9749)]
[102707, (21289, 10267)]
[103894, (21511, 10151)]
[105508, (22439, 11149)]
[107715, (22565, 10729)]
[142580, (23049, 11257)]
[154265, (24915, 12007)]
[177616, (27461, 13421)]
[178421, (32063, 15377)]
[190758, (34141, 16547)]
[228068, (34783, 15473)]
[228876, (35515, 17477)]
[277844, (40119, 19391)]
</code></pre>
<p><em>Code</em></p>
<pre><code>def Seq(p,q):
x=Rational(p/q)
A=[floor(x)]
while not floor(x)==x:
n=floor(x)
x=Rational(n*(x-n+1))
m=floor(x)
A.append(m)
return A
def search(r):
m=0
for p in range(2,r):
for q in range(1,floor(p/2)+1):
A=Seq(p,q)
l=len(A)
if l>m:
m=l
print([m,(p,q)])
</code></pre>
| katago | 114,101 | <blockquote>
<p><span class="math-container">$u_{0} \in \mathbb{Q} \quad u_{n+1}=\left[u_{n}\right]\left(u_{n}-\left[u_{n}\right]+1\right)$</span>, then <span class="math-container">$\{u_{n}\}_{n=1}^{+\infty}$</span> reach in integer. <span class="math-container">$\quad (*)$</span></p>
</blockquote>
<p>This can be proved if we prove,</p>
<blockquote>
<p><span class="math-container">$p$</span> prime, <span class="math-container">$t\in \mathbb{N}^{*}$</span>
<span class="math-container">$p^{t} u_{0} \in \mathbb{N}^{*}$</span>, <span class="math-container">$\left\{u_{k}\right\}_{n=1}^{+\infty}$</span> reach an integer. we call this as property <span class="math-container">$I(p,t)$</span>.</p>
</blockquote>
<p>It is easy to check if we proved <span class="math-container">$I(p,t)$</span> <span class="math-container">$\forall t\in \mathbb{N}^{*}$</span>. <span class="math-container">$\forall p$</span> prime
, <span class="math-container">$I(p,t)$</span> is true then we proved <span class="math-container">$ (*)$</span>.</p>
<p>Now we focus ourself for a fix problem <span class="math-container">$I(p,t)$</span>, <span class="math-container">$p$</span> prime and we first consider the case <span class="math-container">$t=1$</span>.</p>
<p>We rescaling the sequence, dilate it with <span class="math-container">$p$</span>, and still use <span class="math-container">$u_k$</span> to express the new sequence, and write it in <span class="math-container">$p$</span>-expension.</p>
<p><span class="math-container">$ \quad u_{0} \longrightarrow p u_{0}, u_{0}=\sum_{k=0}^{+\infty} a_{k} p^{k}$</span></p>
<p>then the sequence satisfied,
<span class="math-container">$$u_{n+1}=\left(\sum^{+\infty}_{k=0} a_{k+1}(n) p^{k}\right)\left(a_{0}(n)+p\right) \quad (**)$$</span></p>
<p><span class="math-container">$\Rightarrow \quad a_{k}(n+1)=a_{0}(n) a_{k+1}(n)+a_{k}(n) . \quad \forall k \geqslant 0$</span></p>
<p>Where <span class="math-container">$a_k$</span> is the <span class="math-container">$k$</span>-th digit of <span class="math-container">$p$</span>-expension of <span class="math-container">$u$</span> and <span class="math-container">$a_k(n)$</span> is the <span class="math-container">$k$</span>-digit of <span class="math-container">$p$</span>-expension of <span class="math-container">$u_n$</span>, then for previous <span class="math-container">$a_k$</span> we have <span class="math-container">$a_k=a_k(0)$</span>.</p>
<blockquote>
<p>Remark. And in general, we can not find the formula of general term, (**) is only true for the first several digits of <span class="math-container">$u_k$</span>, how many digits can involve depending on <span class="math-container">$u_0$</span>, because we need to avoid arithmetic carry.</p>
</blockquote>
<p>then <span class="math-container">$II(p, k)$</span> is the following property
<span class="math-container">$$\left\{u_{n}\right\}_{n=1}^{+\infty}, \exists k\in \mathbb{N}^*, a_0(k)=0$$</span></p>
<p>And it is easy to check <span class="math-container">$II(p, k)$</span> is equivalent to <span class="math-container">$I(p, k)$</span>, for all <span class="math-container">$p$</span> prime, <span class="math-container">$k\in \mathbb{N}^*$</span>.</p>
<hr />
<p>For <span class="math-container">$II(2, 1)$</span> we are lucky this case is very special and can be check <span class="math-container">$II(2, 1)$</span> is true by directly check the first <span class="math-container">$k$</span>-digits in the 2-expension of <span class="math-container">$u_0$</span>, i.e. <span class="math-container">$a_i(0)$</span>, for <span class="math-container">$0\leq i\leq k$</span>.</p>
<p>If the first 2 digits of <span class="math-container">$u_0$</span> are 10 , then <span class="math-container">$II(2.1)$</span> is true.<br />
If the first 3 digits of <span class="math-container">$u_0$</span> are 101 , then <span class="math-container">$II(2.1)$</span> is true.<br />
If the first 4 digits of <span class="math-container">$u_0$</span> are 1001 , then <span class="math-container">$II(2.1)$</span> is true.<br />
<span class="math-container">$......$</span><br />
If the first <span class="math-container">$k$</span> digits of <span class="math-container">$u_0$</span> are <span class="math-container">$1\underbrace{00...0}_{k-2}1$</span> , then <span class="math-container">$II(2.1)$</span> is true.
So the only case <span class="math-container">$II(2.1)$</span> false is when <span class="math-container">$u_0=1$</span>, which corresponding to the case for original <span class="math-container">$u_0$</span>, <span class="math-container">$u_0=\frac{1}{2}$</span>.</p>
<blockquote>
<p>Remark. And this argument itself is hopeless to prove <span class="math-container">$II(3, 1)$</span> is true, the new idea will be required for proving this and general <span class="math-container">$II(p, 1)$</span>, the first main obstacle is we can not find a control(if do not lead contradiction already) from first several digits of <span class="math-container">$u_k$</span> to control first several digits of <span class="math-container">$u_{k+1}$</span> by the following strategy,<br />
If the first <span class="math-container">$k$</span>-th digits of <span class="math-container">$u_0$</span> don't fall into some special case, we get contradictions, thus restricting the first several digits of <span class="math-container">$u_0$</span> to a smaller set.
Lose the control makes us fall into endless case checking.<br />
<span class="math-container">$II(2, k)$</span> seems to be more tractable by the same reason as <span class="math-container">$II(2, 1)$</span> but I can not figure out a proof.</p>
</blockquote>
|
84,254 | <p>For me, a simplicial groupoid is a simplicial object in ${\mathbf{Grpd}}$. I am more general than Goerss-Jardine in this definition.</p>
<p>Do you have examples simplicial groupoids that occur in nature? Here's what I have got:</p>
<ol>
<li>Given a simplicial group $G$ acting on a simplicial set $X$, the action groupoid $X//G$ is a simplicial groupoid.</li>
<li>The fundamental groupoid $\Pi X$ of a bisimplicial set or simplicial space.</li>
<li>Given a functor $F:I\to {\mathbf{sSet}}$ from a diagram category $I$ to the category of simplicial sets, one can form what Goerss-Jardine calls the translation category $E_I F$ and what Mac Lane-Moerdijk calls the category of elements $\int_{I^{\mathrm{op}}}\, F$. The nerve of this category calculates the homotopy colimit ${\mathrm{hocolim}}_I \, F$.
In the case where the diagram category is a groupoid, then this translation category/category of element is a simplicial groupoid.</li>
<li>Given a simplicial set $X$, the loop groupoid $GX$ is a simplicial groupoid.</li>
</ol>
| none | 20,161 | <p>I think it is very doable. I took a class like that as an undergrad. The textbook was Enderton's "Introduction to Mathematical Logic". It did enough Hilbert-style proof theory to get up to the incompleteness theorem, then discussed models, interpretations, Tarski's definition of truth, etc. It seemed great at the time and wasn't a terribly hard course. My gripe these many years later is that I wish it had said something about sequent calculus since that would have changed how I thought of logic if I'd known about it at the time.</p>
|
84,254 | <p>For me, a simplicial groupoid is a simplicial object in ${\mathbf{Grpd}}$. I am more general than Goerss-Jardine in this definition.</p>
<p>Do you have examples simplicial groupoids that occur in nature? Here's what I have got:</p>
<ol>
<li>Given a simplicial group $G$ acting on a simplicial set $X$, the action groupoid $X//G$ is a simplicial groupoid.</li>
<li>The fundamental groupoid $\Pi X$ of a bisimplicial set or simplicial space.</li>
<li>Given a functor $F:I\to {\mathbf{sSet}}$ from a diagram category $I$ to the category of simplicial sets, one can form what Goerss-Jardine calls the translation category $E_I F$ and what Mac Lane-Moerdijk calls the category of elements $\int_{I^{\mathrm{op}}}\, F$. The nerve of this category calculates the homotopy colimit ${\mathrm{hocolim}}_I \, F$.
In the case where the diagram category is a groupoid, then this translation category/category of element is a simplicial groupoid.</li>
<li>Given a simplicial set $X$, the loop groupoid $GX$ is a simplicial groupoid.</li>
</ol>
| user729424 | 5,698 | <p>Kenneth Kunen recently wrote a wonderful introduction to Mathematical Logic, called "The Foundations of Mathematics" (ISBN: 978-1-904987-14-7), published in 2009. The book's only prerequisite is the mathematical maturity that an Introduction to Analysis course would provide, so it sounds like your students would be prepared. The book provides a brief introduction to axiomatic set theory, model theory, and computability theory; and it culminates with a proof of Godel's incompleteness theorems and Tarski's theorem on the non-definability of truth. There are also a couple brief discussions of the philosophy of mathematics; these are given from the perspective of the working mathematician, and they are used to motivate the material. And they are very helpful. In fact, the most salient thing about this book is that it is exceptionally clear, well-written, and easy to learn from. (Kunen also wrote "Set Theory: An Introduction To Independence Proofs" which is also exceptionally clear, well-written, and easy to learn from). Your students will be grateful for the fact that this book is available (new) on amazon.com for less than $25.</p>
|
924,555 | <p>My homework question:</p>
<blockquote>
<p>From the order axioms for $\mathbb{R}$, show that $0 < 1$. [<em>Hint:</em> From the field axioms, $0 \not=1$. By the trichotomy property, either $0<1$ or $4<0$. Assuming $1 < 0$, get $0 < -1$. Now use Exercise 4.]</p>
</blockquote>
<p>Exercise 4 from my textbook problems states:</p>
<p>"From the order axioms for $\mathbb{R}$, show that the set of positive real numbers, {$x \in \mathbb{R} : x > 0$}, is closed under addition and multiplication."</p>
<p>How am I expected to use Exercise 4 as directed?</p>
| orangeskid | 168,051 | <p>How about proving that $x^2 >0$ whenever $x \ne 0$ and then noticing that $1 = 1^2$.</p>
<p>So, let $x \ne 0$. Then $x > 0$ or $x < 0$. If $x>0$ then $x^2 = x\cdot x > 0$. If $ x < 0$ then adding $-x$ to both terms we get $0 < -x$ and therefore $0 < (-x)(-x) = x^2$. </p>
<p>One should also check the sign rule $(-a)(-b) = a b$ in an introductory course.</p>
|
225,253 | <p>Simple question, I just cannot find something that explains it right out and to the point without giving a huge confusing explanation. The question that I am struggling with is to determine a limit of a function if it exists.</p>
<blockquote>
<p>Find: <span class="math-container">$$\lim_{x\to2}{f(x)},$$</span></p>
<p>If</p>
<p><span class="math-container">$$f(x)=\begin{cases}x^{2} & \text{if } x<2 \\ 3 & \text{if } x=2 \\ 3x-2 & \text{if } x>2\end{cases}$$</span></p>
</blockquote>
<p>Now i worked out the limits from both sides and they both equal 4. But it does say that the limit at that point equals 3. Does this mean that the limit doesn't exist? Or does this just mean that the double sided limit exists at 4, but is discontinuous and equals 3 at that point?</p>
<p>Thanks in advance!</p>
| Sidd Singal | 37,043 | <p>The answer is $4$. The limit asks for the value of a function as it <em>approaches</em> some $x$ value, not the exact value. There can be a hole at $x=2$ and your answer would still be valid. </p>
<p>For a more technical answer, take the following definition of a limit: $$\forall \varepsilon \gt 0 \: \exists \delta\gt 0:\forall x(0\lt|x-c|\lt\delta \Rightarrow|f(x)-L|\lt\varepsilon)$$</p>
<p>Then notice how the difference between any $x$ value you choose and the $c$ you are approaching must always be greater than $0$.</p>
|
2,153,421 | <p>I want to find curvatures and torsions for the following curves but get stuck with their natural parametrizations ($s$ is natural if $|\dot{\gamma}(s)| = 1$). Can anyone help me?</p>
<p>(a) $e^t(\cos t,\sin t,1)$</p>
<p>(b) $(t^3+t,t^3-t,\sqrt{3}t^2)$</p>
<p>(c) $3x^2+15y^2=1, z=xy$</p>
<p><strong>Update</strong>:</p>
<p>Here are my attempts on solving (a):</p>
<p>$\dot{\gamma}(t) = (\dot{t}e^t \cos t - \dot{t} e^t \sin t, \dot{t} e^t \sin t + \dot{t}e^t \cos t, \dot{t} e^t)$ which gives $|\dot{\gamma}(t)| = \sqrt{2}\dot{t}e^t=1$ and the solution for this ODE is $t = \ln\frac{\tau}{2}$.</p>
<p>But if I substitute $t$ with $t=\ln\frac{\tau}{2}$ the result will be $\dot{\gamma}(\tau) = (\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2} - \frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2},\frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2}+\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2},\frac{1}{2})$ and $|\dot{\gamma}(\tau)| = \sqrt{\frac{3}{2}}$. So for $|\dot{\gamma}(\tau)| = 1$ we should take $t = \ln\frac{\tau}{3}$. Where is my mistake?</p>
<p>Any help and hints will be very appreciative.</p>
| Ng Chung Tak | 299,599 | <p>$(a)$
\begin{align*}
\mathbf{\dot{r}}(t) &= e^{t}(\cos t-\sin t, \sin t+\cos t,1) \\
|\mathbf{\dot{r}}(t)| &= e^{t}\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2+1} \\
&= e^{t}\sqrt{3} \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \sqrt{3}(e^{t}-1) \\
t &= \ln \left( 1+\frac{s}{\sqrt{3}} \right)
\end{align*}</p>
<p>$(b)$
\begin{align*}
\mathbf{\dot{r}}(t) &= (3t^2+1, 3t^2-1,2t\sqrt{3}) \\
|\mathbf{\dot{r}}(t)| &= \sqrt{(3t^2+1)^2+(3t^2-1)^2+12t^2} \\
&= \sqrt{2(9t^4+6t^2+1)} \\
&= \sqrt{2}(3t^2+1) \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \sqrt{2}(t^3+t) \\
t &= \sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}+\frac{s}{2\sqrt{2}}}-
\sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}-\frac{s}{2\sqrt{2}}} \\
\end{align*}</p>
<p>$(c)$
\begin{align*}
\mathbf{r}(t) &=
\left(
\frac{\cos t}{\sqrt{3}},
\frac{\sin t}{\sqrt{15}},
\frac{\cos t \sin t}{3\sqrt{5}}
\right) \\
\mathbf{\dot{r}}(t) &=
\left(
-\frac{\sin t}{\sqrt{3}},
\frac{\cos t}{\sqrt{15}},
\frac{\cos^2 t-\sin^2 t}{3\sqrt{5}}
\right) \\
|\mathbf{\dot{r}}(t)| &=
\sqrt{\frac{15\sin^2 t+3\cos^2 t+(\cos^2 t-\sin^2 t)^2}{45}} \\
&= \frac{3-\cos 2t}{3\sqrt{5}} \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \frac{6t-\sin 2t}{6\sqrt{5}}
\end{align*}
there is no close form for $t$ in terms of $s$.</p>
<blockquote>
<p>You still can find $\kappa$ and $\tau$ by keeping $t$ parametrization
\begin{align*}
\kappa &=
\frac{|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|}
{|\mathbf{\dot{r}}|^3} \\
\tau &=
\frac{\mathbf{\dot{r}} \cdot \mathbf{\ddot{r}} \times \mathbf{\dddot{r}}}
{|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|^2}
\end{align*}</p>
</blockquote>
|
388,766 | <p>I need to show that every element in $\Bbb Z/p\Bbb Z$ can be written as a sum of two squares. The case $p=2$ is trivial and $0$ is always $0^2 + 0^2$. So all I have to do is show that every element of $(\Bbb Z/p\Bbb Z)^\times$ (the group of units) can be expressed as a sum of two squares. The question hints that I should consider the set of elements of $(\Bbb Z/p\Bbb Z)^\times$ expressible as a sum of two squares.</p>
<p>Associativity is trivial, the identity element $1 = 1^2 + 0^2$ exists.
Closure holds by the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$.
I don't know how to show that inverses exist.</p>
| André Nicolas | 6,312 | <p>The following is another version of the argument by Jyrki Lahtonen. Beside the numbers $1,2,\dots, p-1$ write QR for quadratic residue and NR for quadratic non-residue. So we write QR beside $1$. </p>
<p>Note that we write each of QR and NR $\frac{p-1}{2}$ times. So at some time, QR is followed by NR. This means there is a NR $c$ which is congruent to $a^2+1^2$ modulo $p$. But as $x$ ranges over the numbers from $1$ to $p-1$, $x^2 c$ ranges over the non-residues. So the non-residues can be written as $(xa)^2+x^2$. </p>
|
2,340,487 | <p>I was trying to compute this limit:
$$\lim_{x \to 0}\lim_{y \to 0} (x+y)\sin{\frac{x}{y}}$$</p>
<p>And this is my solution:
$$\lim_{x \to 0}\lim_{y \to 0}|(x+y)\sin{\frac{x}{y}}|\leq\lim_{x \to 0}\lim_{y \to 0} |(x+y)|=0$$</p>
<p>So I got the limit 0.</p>
<p>The answer was different. I have no idea what is wrong with my solution?</p>
| robjohn | 13,854 | <p>Note that
$$
\lim_{y\to0}(x+y)\sin\left(\frac xy\right)
$$
is indeterminate for each non-zero $x$. Thus, there is no way to compute
$$
\lim_{x\to0}\lim_{y\to0}(x+y)\sin\left(\frac xy\right)
$$
Whereas,
$$
\lim_{x\to0}(x+y)\sin\left(\frac xy\right)=0
$$
for each non-zero $y$, so
$$
\lim_{y\to0}\lim_{x\to0}(x+y)\sin\left(\frac xy\right)=0
$$</p>
<hr>
<p>Note that
$$
\lim_{\substack{(x,y)\to(0,0)\\y\ne0}}(x+y)\sin\left(\frac xy\right)=0
$$
Thus, on any path toward $(0,0)$ where $y\ne0$, the limit will be $0$. However, because
$$
\lim\limits_{y\to0}(x+y)\sin\left(\frac xy\right)
$$
does not exist for any $x\ne0$, the function cannot be extended continuously to the line $y=0$ and this messes up
$$
\lim_{x\to0}\lim_{y\to0}(x+y)\sin\left(\frac xy\right)
$$</p>
|
4,458,863 | <p>Let <span class="math-container">$z_1,\;z_2,\;z_3\;$</span> be complex number such that <span class="math-container">$|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2\;\;$</span>. If <span class="math-container">$|z_1-z_3|=|z_1-z_2|\; \;$</span> and <span class="math-container">$z_2 \neq z_3.\; \; $</span> Then Find value of <span class="math-container">$|z_1+z_2||z_1+z_3|$</span>.</p>
<p><strong>My Thinking:</strong></p>
<p>All I could think is that <span class="math-container">$z_1, z_2, z_3 \;$</span> lies on a circle of radius <span class="math-container">$2$</span> with origin as center. Can anyone help me in how to process further.</p>
| Claude Leibovici | 82,404 | <p>In the same spirit as @Dan, at least for small <span class="math-container">$x$</span>, use
<span class="math-container">$$\sqrt{\cosh (x)}=1+\sum_{n=1}^\infty \frac {(-1)^{n+1}}{2^n \,(2n)!}\,a_n\,x^{2n}$$</span> the first <span class="math-container">$a_n$</span> being
<span class="math-container">$$\{1,1,19,559,29161,2368081,276580459,43947282079,\cdots\}$$</span> which correspond to sequence <span class="math-container">$A008990$</span> in <span class="math-container">$OEIS$</span> (use their absolute values).</p>
|
3,355,544 | <p><a href="https://i.stack.imgur.com/oYf7f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oYf7f.jpg" alt="enter image description here"></a></p>
<p>Given the following equations:</p>
<p><span class="math-container">$$x=t^2$$</span></p>
<p><span class="math-container">$$y=t^2$$</span></p>
<p><span class="math-container">$$z=2t+3$$</span></p>
<p>Show that the graph of the curve <span class="math-container">$(t^2,t^2,2t+3)$</span> lives in a plane and that, within that plane, the graph is a parabola.</p>
<p>My attempt, so far:</p>
<p>From the first two equations <span class="math-container">$x=y$</span>, meaning the curve belongs to the <span class="math-container">$x-y=0$</span> plane.</p>
<p>Plugging the first or second equation into the third one:</p>
<p><span class="math-container">$$((z-3)/2)^2=x,\>\>\> ((z-3)/2)^2=y$$</span></p>
<p>This tells me it is parabola in the zx and zy planes respectively , I guess ??</p>
<p>A matlab plot shows it is , in fact, a parabola in the <span class="math-container">$x-y=0$</span> plane. What I am doing wrong and how do I show it?</p>
| Quanto | 686,284 | <p>Rotate the <span class="math-container">$xyz$</span>-coordinates by 45 degrees around the <span class="math-container">$z$</span> axis to the <span class="math-container">$uvz$</span>-coordinates, which is equivalent to </p>
<p><span class="math-container">$$x=u+v, \>\>\> y=u-v$$</span></p>
<p>Rewrite the curve in the <span class="math-container">$uv$</span>-coordinates</p>
<p><span class="math-container">$$u+v=t^2, \>\>\>u-v=t^2$$</span></p>
<p>Then, we have <span class="math-container">$u=t^2$</span> and <span class="math-container">$v=0$</span>. So, the curve lies in the plane <span class="math-container">$v=0$</span>, with the <span class="math-container">$uz$</span> coordinates given by,</p>
<p><span class="math-container">$$u = t^2$$</span>
<span class="math-container">$$z=2t+3$$</span></p>
<p>Eliminate <span class="math-container">$t$</span> to get the standard parabola equation in the plane <span class="math-container">$v=0$</span> (or, the plane <span class="math-container">$x=y$</span> in the original coordinates)</p>
<p><span class="math-container">$$(z-3)^2=4u$$</span></p>
|
544,779 | <p>So here are the contextual statements:
1) Maya either listens to music or does her homework. If she listens to music she feels happy.If she does her homework she feels unhappy. Therefore she will not do her homework while listening to music.</p>
<p>Let P be the statement "Maya listens to Music".
Q "Maya does homework".
R "Maya feels happy".
So am I right to write it as ((P=>Q)∨(Q=>¬R) ) => ( ¬Q∧¬P ) </p>
<p>2) If I drink coffee, then I will get my assignment done on time. If I do not drink coffee, then I will feel sleepy. If I feel sleepy, then I will make mistakes. Therefore, if I will not get the assignment done on time, then I will make mistakes.</p>
<p>Let P be the statement "I drink coffee".
Q "I get the assignment done on time".
R "I will feel sleepy".
S "I will make mistakes".</p>
<p>( (P=>Q)∧(¬P=>R)∧(R=>S) ) =>( ¬Q=>S ) . Is it right?</p>
<p>In order to examine whether the arguments are right do I really need to do the truth tables..? It will be a huge one for the second statement.</p>
| Community | -1 | <p>In defining the topology on the one point compactification, one needs to claim that $X\setminus K \cup \{\infty\}$ be open. But if $X$ is not Hausdorff, $K$ might not be closed. Thus it might be more natural to assume that $X$ is Hausdorff. </p>
<p>There are a lot of compactification, as long as $f(X)$ is dense in $Y$. For example, you can do either a one point or two point compactification of $(0, 1)$, to make it into a circle or a close interval. You can also compactify the open unit disc by one point compactification, which becomes a sphere, or you can compactify by adding the boundary circle. </p>
<p>Which one you choose depends on what you want to achieve. </p>
|
2,269,917 | <p>I've come across this question in a university exam paper. It's causing me a huge headache due to the fact that it goes from one vector space to another ($\def\R{\Bbb R}\R^3 \to \R^2$), otherwise it would be fairly standard. If anyone could shed some light on what I'm missing it would be much appreciated.</p>
<p>A4. Let $T : \R^3 → \R^2$ be given by
$T((x, y, z)) = (2x + 3y − 5z, x − 2y + z)$.</p>
<p>(i) Find a basis for the kernel $\ker T$ of $T$.</p>
<p>(ii) Find a basis for the image $\operatorname{im}T$ of $T$.</p>
| SEWillB | 441,390 | <p>I am assuming since you are a university student that you know the rank-nullity theorem- and hopefully it's proof! We take a basis for the kernel of the linear map, call the map $T$ between vector spaces $V$ and $W$;<br>
Say $\{v_1,...,v_m\} $ is a basis for $ker(T)$<br>
Extend to a basis for $V$, say $\{v_1,...,v_m, u_1, ..., u_n\}$ so $dim(V)=n+m$<br>
Then $\{T(u_1),...,T(u_n)\}$ is a basis for the image of $T$.<br>
Applying that to this case, we first need to find a basis for $ker(T)$, do this however you wish (row-reduction of matrix defining T or just solving the simultaneous equations you need to - both are equivalent.)<br>
You should find that the vector $[1, 1, 1]$ is a basis for the kernel of T -(If I've done this correctly).<br>
Extend this in any way you see fit to a basis for $\Bbb{R}^3$ say, add in $[1, 0, 0]$ and $[0, 1, 0]$. Apply $T$ to these and you'll have you're basis for $Im(T)$<br>
As a check you can check that $T([1, 0, 0]) = [2, 1]$ and $T([0, 1, 0]) = [3, -2]$ are linearly independent...<br>
Hope this helps!</p>
|
649,379 | <p>I'm on the final part of my project, where I have to prove the Noether-Lasker Theorem (or copy out the following proof and "fill in the gaps"). I'm looking for an explanation of what's going on at a macro-level. I think I could follow the proof, but I don't understand how it proves what it says it proves. I've already proved that every ideal is the intersection of finitely many primary ideals, and somehow the following proves that every ideal is the reduced intersection whose radicals are unique. I feel like if I understood the "macro logic" then I'd be away. Here's the link:</p>
<p><a href="https://dl.dropboxusercontent.com/u/17606191/noether.gif" rel="nofollow">https://dl.dropboxusercontent.com/u/17606191/noether.gif</a></p>
<p>Thanks for any replies!</p>
| rschwieb | 29,335 | <p>After proving an ideal can be expressed at least one way as an intersection of primary ideals, the next logical step is to make that expression as "tight" as possible by eliminating redundancy.</p>
<p>The same idea applies to generating sets of vector spaces. After you find one generating set, then you can remove redundant elements to get down to a minimal generating set, and that set is a basis. Linear independence is basically the quality of a set of vectors to generate their span with no redundancy.</p>
<p>Anyhow, it's obvious if your decomposition of an ideal $I$ has the same ideal in it twice, then you would just remove one of those copies because you would still get $I$, but you'd be writing one less ideal in your decomposition.</p>
<p>Actually you can do better than that. If there are primary ideals $Q,Q'$ in the decomposition with the same radical $P$, then $Q\cap Q'=Q''$ is also primary. So why not delete $Q\cap Q'$ and replace it with $Q''$? You'd be writing out one less ideal, and you'd still be getting $I$.</p>
<p>Once you get down to "minimal" (or "optimal") decompositions, then you can start asking about what properties these decompositions share. All the bases of a vector space, for example, must have the same cardinality. One thing that the minimal primary decompositions share is that they all share the same associated primes.</p>
|
3,202,955 | <blockquote>
<p><strong>Note:</strong> Please do not give a solution; I would prefer guidance to help me complete the question myself. Thank you.</p>
</blockquote>
<hr>
<p>I am having trouble understanding and finding the continuous and residual spectrum. I am working through the following problem:</p>
<p>Let <span class="math-container">$\alpha = (\alpha_{i})\in\ell^{\infty}$</span> and let <span class="math-container">$T_{\alpha}:\ell^{2}\rightarrow\ell^{2}$</span> with <span class="math-container">$T_{\alpha}x=(\alpha_{1}x_{1},\alpha_{2}x_{2},\ldots)$</span>.</p>
<p>(i) Compute the spectrum, <span class="math-container">$\sigma(T_{\alpha})$</span>, of <span class="math-container">$T_{\alpha}$</span>.</p>
<p>(ii) Identify the point spectrum, <span class="math-container">$\sigma_{p}(T_{\alpha})$</span>, the continuous spectrum, <span class="math-container">$\sigma_{c}(T_{\alpha})$</span>, and the residual spectrum, <span class="math-container">$\sigma_{r}(T_{\alpha})$</span>, of <span class="math-container">$T_{\alpha}$</span>.</p>
<p><strong>My Solution</strong></p>
<p>(i) Suppose <span class="math-container">$\lambda\in\rho(T_{\alpha})$</span> (where <span class="math-container">$\rho(T_{\alpha})$</span> is the resolvent set of <span class="math-container">$T_{\alpha}$</span>) then <span class="math-container">$\lambda I-T_{\alpha}$</span> is bijective. Hence, for every <span class="math-container">$y\in\ell^{2}$</span> <span class="math-container">$\exists ! x\in\ell^{2}$</span> such that,
<span class="math-container">\begin{align}
(\lambda I-T_{\alpha})x = y\implies x = (\lambda I-T_{\alpha})^{-1}y,
\end{align}</span>
and
<span class="math-container">\begin{align}
x_{n} = \frac{y_{n}}{\lambda - \alpha_{n}},
\end{align}</span>
for each element, <span class="math-container">$n\in\mathbb{N}$</span>. Since <span class="math-container">$x\in\ell^{2}$</span>,
<span class="math-container">\begin{align}
|x|_{2}^{2} = \sum_{n=1}^{\infty}|x_{n}|^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}<\infty.
\end{align}</span>
Therefore, if <span class="math-container">$\lambda\in\overline{\{\alpha_{n}\}}$</span> then <span class="math-container">$|x|_{2} = \infty$</span>. Hence <span class="math-container">$\sigma(T_{\alpha}) = \{\lambda\in\mathbb{C}|\lambda\in\overline{\{\alpha_{n}\}},n\in\mathbb{N}\}$</span>.</p>
<p>(ii) Using the definition of point spectrum,
<span class="math-container">\begin{align}
\sigma_{p}(T_{\alpha}):=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha} \text{is not injective}\},
\end{align}</span>
then there exists <span class="math-container">$x_{1},x_{2}\in\ell^{2}$</span>, with <span class="math-container">$x_{1}\neq x_{2}$</span>, such that <span class="math-container">$(\lambda I-T_{\alpha})x_{1} = (\lambda I-T_{\alpha})x_{2}$</span>. This occurs when <span class="math-container">$|x|_{2}=\infty$</span>, that is, <span class="math-container">$\lambda = \alpha_{n}$</span> for some <span class="math-container">$n\in\mathbb{N}$</span>. Hence, <span class="math-container">$\sigma_{p}(T_{\alpha})=\{\lambda\in\mathbb{C}|\lambda = \alpha_{n}\text{ for some }n\in\mathbb{N}\}$</span>.</p>
<p>At this stage I do not know how to continue. Particularly, I don't understand how to use the definitions of continuous and residual spectrum:
<span class="math-container">\begin{align}
\sigma_{c}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2},(\lambda I-T_{\alpha})^{-1}\text{ is unbounded}\},\\
\sigma_{r}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}\}
\end{align}</span></p>
<p><strong>Finding continuous and residual spectrum</strong></p>
<p>Assume <span class="math-container">$\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}$</span> then <span class="math-container">$\text{im}(\lambda I-T_{\alpha})^{\perp}\neq\{0\}$</span>. Hence there is <span class="math-container">$y\in\text{im}(\lambda I-T_{\alpha})^{\perp}$</span>, <span class="math-container">$y\neq 0$</span>, such that,
<span class="math-container">\begin{align}
\sum_{n=1}^{\infty}(\lambda x_{n}-\alpha_{n}x_{n})y_{n} = 0,
\end{align}</span>
for all <span class="math-container">$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$</span>. This implies <span class="math-container">$\lambda=\alpha_{n}$</span> for each <span class="math-container">$n^{\text{th}}$</span> element of <span class="math-container">$x$</span>, which is impossible. Otherwise, <span class="math-container">$x=0$</span>, however this contradicts our assumption that holds for all <span class="math-container">$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$</span>. Hence by contradiction, <span class="math-container">$\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2}$</span> and <span class="math-container">$\text{im}(\lambda I-T_{\alpha})^{\perp}=\{0\}$</span>. Therefore, <span class="math-container">$\sigma_{r}(T_{\alpha})=\emptyset$</span>.</p>
<p>By above <span class="math-container">$(\lambda I-T_{\alpha})$</span> has trivial kernel and hence injective. Given <span class="math-container">$\lambda\in\sigma_{c}(T_{\alpha})$</span> we need <span class="math-container">$(\lambda I-T_{\alpha})^{-1}$</span> unbounded. From above this implies,
<span class="math-container">\begin{align}
|x|_{2}^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}\rightarrow\infty.
\end{align}</span>
Hence, for all <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$\exists N\in\mathbb{N}$</span> such that for <span class="math-container">$n>N$</span>,
<span class="math-container">\begin{align}
|\lambda-\alpha_{n}|<\epsilon.
\end{align}</span>
Now <span class="math-container">$(\alpha_{n})\in\ell^{\infty}$</span> so they are uniformly bounded. Taking <span class="math-container">$\lambda=\sup_{n\in\mathbb{N}}|\alpha_{n}|$</span> we satisfy the criteria and hence <span class="math-container">$(\lambda I-T_{\alpha})^{-1}$</span> is unbounded.</p>
<p>Therefore,
<span class="math-container">\begin{align}
\sigma_{p}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda=\alpha_{n}\text{ for some }n\in\mathbb{N}\},\\
\sigma_{c}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda = \sup_{n\in\mathbb{N}}|\alpha_{n}|\},\\
\sigma_{r}(T_{\alpha}) &= \emptyset.
\end{align}</span></p>
| Disintegrating By Parts | 112,478 | <p>A quick overview might be helpful, even though it does not really fit your requirement for an answer.</p>
<p><span class="math-container">$T_{\alpha}$</span> is a bounded normal operator. That rules out all but continuous and point spectrum. Every <span class="math-container">$\alpha_j$</span> is in the point spectrum of <span class="math-container">$T_{\alpha}$</span>, which is easily demonstrated by showing <span class="math-container">$T_{\alpha}e_n = \alpha_n e_n$</span> where
<span class="math-container">$$
e_n=\{0,0,\cdots,1,0,0,\cdots\},
$$</span>
and where <span class="math-container">$1$</span> is in the <span class="math-container">$n$</span>-th position. Every cluster point of the set <span class="math-container">$\{ \alpha_n \}$</span> that is not in the set of eigenvalues is in the continuous spectrum, which is the approximate point spectrum.</p>
|
13,030 | <p>At work, we were discussing when is it the best time to change to winter tires for bikes and/or cars.</p>
<p>Using <code>WeatherData[]</code> and <code>DateListPlot[]</code>, it was fairly straightforward for me to create the diagram below:</p>
<p><img src="https://i.stack.imgur.com/Y5wNT.png" alt="Mean temperature per day in Stockholm"> </p>
<p><em>Fig 1 Mean temperature per day in Stockholm, red is negative and a risk without winter tires.</em></p>
<p>The code for this is</p>
<pre><code>cityTemp = WeatherData["Stockholm", "MeanTemperature", {{1977, 1, 1}, {2011, 12, 31}, "Day"}];
iceRiskDays = Select[cityTemp, Last[#] < 0 &];
yearStrip[ dataItem_] := {{0, Part[First[dataItem], 2], Last[First[dataItem]]}, Last[dataItem]}
DateListPlot[{yearStrip[#] & /@ cityTemp, yearStrip[#] & /@ iceRiskDays} ]
</code></pre>
<p>My question is: <strong>How do I calculate for each day the proportion of values for that day that are below 0 Celsius?</strong> (e.g. for a date at the end of November, the proportion is likely to be bigger than 0.5)?</p>
<p>My attempts to do this ended with trying to create separate lists for each date, but I felt that this was a less elegant way and also creates less "fit" with DateListPlot.</p>
| J. M.'s persistent exhaustion | 50 | <p>If my understanding of the question is correct, this does the job:</p>
<pre><code>cityTemp = WeatherData["Stockholm", "MeanTemperature",
{{1977, 7, 1}, {2011, 12, 31}, "Day"}];
(* proportions; I ignore February 29 for this *)
props = Array[(Count[#, {_List, _?Negative}]/Length[#]) &[Cases[cityTemp,
{Prepend[Rest[DatePlus[{2011, 1, 1}, #]], yr_], temp_}, Infinity]] &, 365, 0];
Take[N[props], 5] (* sample *)
{0.705882, 0.588235, 0.705882, 0.676471, 0.705882}
</code></pre>
<p>To plot these proportions, you can do this:</p>
<pre><code>DateListPlot[props, {{2011, 1, 1}, {2011, 12, 31}, "Day"},
FrameLabel -> {None, "Proportion"}, Joined -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/XxPol.png" alt="plot of proportion of cold days"></p>
<p>Pick the day of the year with the most number of "cold" instances:</p>
<pre><code>Pick[Array[Rest[DatePlus[{2011, 1, 1}, #]] &, 365, 0], props, Max[props]]
{{2, 18}}
</code></pre>
<p>So, at least for Stockholm, February 18 is often a cold day.</p>
|
1,449,776 | <p>I have always known that $a^n=a*a*a*.....$(n times)</p>
<p>Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?</p>
<p>I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$. </p>
<p>But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?</p>
<p>Any help is highly appreciated.</p>
| Russ H | 273,811 | <p>You are correct in that $a$ is repeated zero times.</p>
<p>$a^n = a*a*a*... = 1 * a*a*a*...$</p>
<p>And so $a^0 = 1$ when $a$ is repeated zero times</p>
|
1,217,557 | <p>I was tasked with drawing the contour lines of $ z = \sqrt{xy} $, which I find a bit problematic since I can see no way in which one can plot (by hand, and not with wolfram and others....) the $ z = \sqrt{xy} $ graph in $R^2( x-, y- $ projection} to begin with for this surface...</p>
<p>How can one draw this contour graph manually? </p>
| MissMonicaE | 227,754 | <p>You're right, but I often see people using "linear" to mean "polynomial of degree 1" outside of linear algebra contexts, especially in introductory e.g. calculus courses. In fact, I say this to my calc tutoring students <em>all the time</em> and I didn't realize it's actually wrong until now.</p>
|
2,852,550 | <p>I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(\bar A,d)$ is complete. But are there spaces s.t. indeed for all $\varepsilon>0$, there is $N$ s.t. $d(x_n,x_{m})<\varepsilon$ for all $n\geq N$, and all $r\geq 0$, but the ball also move to $+\infty $ ? </p>
| Florian R | 511,528 | <p>No, that cannot happen. Suppose that $(x_n)_{n \geq 0}$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N \geq 0$ such that for all $n,m \geq N$ we have $d(x_n,x_m) \leq 1$. Thus, all the points $\{x_n ~|~ n \geq N\}$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,\dots,x_{N-1}$, this sequence cannot become unbounded.</p>
|
374,881 | <p>I'd like to know how I can recursively (iteratively) compute variance, so that I may calculate the standard deviation of a very large dataset in javascript. The input is a sorted array of positive integers.</p>
| Did | 6,179 | <p>Recall that, for every $n\geqslant1$,
$$
\bar x_n=\frac1n\sum_{k=1}^nx_k,
$$
and
$$
\bar\sigma^2_n=\frac1n\sum_{k=1}^n(x_k-\bar x_n)^2=\frac1n\sum_{k=1}^nx_k^2-(\bar x_n)^2.
$$
Hence simple algebraic manipulations starting from the identities
$$
(n+1)\bar x_{n+1}=n\bar x_n+x_{n+1},
$$
and
$$
(n+1)(\bar\sigma^2_{n+1}+(\bar x_{n+1})^2)=n(\bar\sigma^2_n+(\bar x_n)^2)+x_{n+1}^2,
$$
lead to
$$
\bar x_{n+1}=\bar x_n+\frac{x_{n+1}-\bar x_n}{n+1},
$$
and
$$
\bar\sigma^2_{n+1}=\bar\sigma^2_n+(\bar x_n)^2-(\bar x_{n+1})^2+\frac{x_{n+1}^2-\bar\sigma^2_n-(\bar x_n)^2}{n+1}.
$$
Thus, $(n,\bar x_n,x_{n+1})$ yield $\bar x_{n+1}$ and $(n,\bar\sigma^2_n,\bar x_n,\bar x_{n+1},x_{n+1})$ yield $\bar\sigma^2_{n+1}$.</p>
|
688,782 | <p>$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$</p>
<p>I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit. </p>
| Cameron Williams | 22,551 | <p>Hint: try defining $b_n = \ln(a_n)$ (which is well-defined) and see what limit this goes to. Then use a certain exponential function to see what $\lim_n a_n$ is.</p>
|
19,996 | <p>In 1556, Tartaglia claimed that the sums<br>
1 + 2 + 4<br>
1 + 2 + 4 + 8<br>
1 + 2 + 4 + 8 + 16<br>
are alternative prime and composite. Show that his conjecture is false. </p>
<p>With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:<br>
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?</p>
| Gottfried Helms | 1,714 | <p>Although Ross's proof is much shorter and elegant I show my exercising because it uses a slightly different logic. </p>
<p>Assume, at some <em>n</em> we have $\small 2^n-1 = p_0 $ where $\small p_0 $ is prime. We use only the fact, that <em>n</em> must then be odd (otherwise it would factor into $\small (2^{n/2}-1)(2^{n/2}+1) $), so we write explicitely $\small n=2m + 1$. </p>
<p>With this Tartaglia's claim means, that </p>
<ol>
<li><p>$\small 2^{2m+1} -1 =p_0 \in \mathbb P $ (we observe that this is true for instance for <em>m=1</em>, so: $\small 2^3-1 = 7 \in \mathbb P $ )</p></li>
<li><p>$\small 2^{2m+1+2k} -1 =p_k \in \mathbb P $ for all positive integer <em>k</em> </p></li>
</ol>
<p><hr>
To arrive at a contradiction we rewrite the expressions </p>
<p>$\small \begin{array} {rclllllll}
2^{2m+1} &=& p_0+1 \\
2^{2m+1+2k} &=& p_k+1 & =& 2^{2m+1}*4^k &=& 4^k(p_0+1) \\
&\to& \\
p_k&=&4^k p_0 + 4^k-1
\end{array}
$ </p>
<p>The little fermat says, that for each odd prime <em>p</em> there is some <em>k</em> such that $\small 4^k-1 $ is divisible by <em>p</em> namely $\small k=p-1 $ so we'lll have $\small 4^{p_0-1}-1 = p_0 \cdot x $ where <em>x</em> is some positive integer. From this follows that there is some <em>k</em> such that $\small p_k \notin \mathbb P $: </p>
<p>$ \small p_{p_0-1} = 4^{p_0-1} p_0 + 4^{p_0-1}-1 =4^{p_0-1} p_0 + p_0 \cdot x = p_0 \cdot (4^{p_0-1} + x ) \notin \mathbb P $ </p>
|
1,841,882 | <p>By Cayley's theorem, we know that for any finite group $G$, there exists $N \in \mathbb{N}$ such that $G$ is isomorphic to a subgroup of $S_N$, the symmetric group on $N$ letters. Can we prove that for every finite group $G$ there is some symmetric group $S_N$ such that $G$ is isomorphic to a $normal$ subgroup of $S_N$?</p>
| wnoise | 7,882 | <p>In general (i.e. for $N \neq 4$), the only normal subgroups of $S_N$ are $S_N$ itself, $A_N$, and $1.$ Therefore no, because most $G$ will not map to one of these. ($S_4$ has an additional normal subgroup, the Klein $4$-group hiding in it.)</p>
|
2,521,710 | <p>I am trying to do a proof for convergence. But I am stuck in my proof not getting any further... What is missing to finish that proof?</p>
<p>$$a_n = \frac{1}{(n+1)^2}$$
Show that: $$\lim_{n \to \infty}a_n=0$$</p>
<p>Let $e > 0$ and $\forall n \ge n_0 = \lceil \frac{1}{\sqrt{\epsilon}}\rceil+1 \in \mathbb Z^+:$</p>
<p>$\begin{align}
|a_n-0| &\equiv |a_n| \\
&\equiv | \frac{1}{(n+1)^2}|\\
&\equiv|\frac{1}{(n+1)} \cdot \frac{1}{(n+1)}| \\
&\equiv |\frac{1}{n+1}| \cdot |\frac{1}{n+1}| \\
&< |\frac{1}{n}| \cdot |\frac{1}{n}| \\
&(\text{because } n \in \mathbb N) \\
&= \frac{1}{n^2} < \epsilon \text{ (by definition of convergence) }
\end{align}$</p>
<p>$\begin{align}
n \ge n_0 &= \lceil \frac{1}{\sqrt{\epsilon}}\rceil +1 \\
&> \lceil \frac{1}{\sqrt{\epsilon}} \rceil \\
&\ge \frac{1}{\sqrt{\epsilon}}
\end{align}$</p>
<p>thus</p>
<p>$\begin{align}
n &> \frac{1}{\sqrt{\epsilon}} \\
&\equiv \frac{1}{n} > \sqrt{\epsilon} \\
&\equiv \frac{1}{n^2} > \epsilon
\end{align}$</p>
<p>But the line that should follow: </p>
<p>$\begin{align}
\frac{1}{n^2} &< \epsilon
\equiv \frac{1}{n^2} < \frac{1}{n^2}
\end{align}$</p>
<p>which is wrong.. </p>
| copper.hat | 27,978 | <p>You are working too hard.</p>
<p>Look for a simpler upper bound before computing an $\epsilon$.</p>
<p>Note that if $n \ge 1$ then $0 \le a_n = {1 \over (1+n)^2} \le {1 \over 1+n} \le { 1\over n}$.</p>
<p>Choose $n \ge \max(1, {1 \over \epsilon})$.</p>
|
2,521,710 | <p>I am trying to do a proof for convergence. But I am stuck in my proof not getting any further... What is missing to finish that proof?</p>
<p>$$a_n = \frac{1}{(n+1)^2}$$
Show that: $$\lim_{n \to \infty}a_n=0$$</p>
<p>Let $e > 0$ and $\forall n \ge n_0 = \lceil \frac{1}{\sqrt{\epsilon}}\rceil+1 \in \mathbb Z^+:$</p>
<p>$\begin{align}
|a_n-0| &\equiv |a_n| \\
&\equiv | \frac{1}{(n+1)^2}|\\
&\equiv|\frac{1}{(n+1)} \cdot \frac{1}{(n+1)}| \\
&\equiv |\frac{1}{n+1}| \cdot |\frac{1}{n+1}| \\
&< |\frac{1}{n}| \cdot |\frac{1}{n}| \\
&(\text{because } n \in \mathbb N) \\
&= \frac{1}{n^2} < \epsilon \text{ (by definition of convergence) }
\end{align}$</p>
<p>$\begin{align}
n \ge n_0 &= \lceil \frac{1}{\sqrt{\epsilon}}\rceil +1 \\
&> \lceil \frac{1}{\sqrt{\epsilon}} \rceil \\
&\ge \frac{1}{\sqrt{\epsilon}}
\end{align}$</p>
<p>thus</p>
<p>$\begin{align}
n &> \frac{1}{\sqrt{\epsilon}} \\
&\equiv \frac{1}{n} > \sqrt{\epsilon} \\
&\equiv \frac{1}{n^2} > \epsilon
\end{align}$</p>
<p>But the line that should follow: </p>
<p>$\begin{align}
\frac{1}{n^2} &< \epsilon
\equiv \frac{1}{n^2} < \frac{1}{n^2}
\end{align}$</p>
<p>which is wrong.. </p>
| Peter Szilas | 408,605 | <p>And one more:</p>
<p>$|a_n| = |\dfrac{1}{(n+1)^2}| \lt |\dfrac{1}{n^2}| \le |\dfrac{1}{n}|.$</p>
<p>Let $\epsilon \gt 0$ be given.</p>
<p>There is a $n_0$ such that $n_0 \gt 1/\epsilon.$
(Archimedes).</p>
<p>For $n \ge n_0 :$</p>
<p>$|a_n| \lt |\dfrac{1}{n}| \le \dfrac{1}{n_0} \lt \epsilon$.</p>
|
2,774,792 | <p>How would you go about proving the recursion
$$T(n) = T\left(\frac n4\right) + T\left(\frac{3n}4\right) + n$$is $\mathcal O(n\log n)$ using induction?</p>
<p>Thanks!</p>
| drhab | 75,923 | <p>Your answer is wrong.</p>
<p>The probability of "exactly $0$ the same" is indeed $\frac36\frac36\frac36=\frac{27}{216}$ as you suggest. </p>
<p>But the probability of "exactly $1$ the same" equals:$$\left[\frac16\frac16\frac16+3\frac16\frac16\frac36+3\frac16\frac36\frac36\right]3=\frac{111}{216}$$</p>
<p>If e.g. the first roll was $(1,2,3)$ then between the brackets $[$ and $]$ you find the probability that number $1$ appears in the second roll and the numbers $2$ and $3$ do not. The first term of the three gives the probability on roll $(1,1,1)$, the second term gives the probability of a roll with exactly twice a $1$ in it and the third term gives the probability of a roll with once a $1$ in it.</p>
<p>The final result is: $$1-\frac{27}{216}-\frac{111}{216}=\frac{78}{216}\approx0.3611$$</p>
|
191,307 | <pre><code>roots = NSolve[Sin[2z] == Cos[Sin[z]] + 1 && -2π <= Re[z] <= 2π && -2π <= Im[z] <= 2π, z];
w = z\.roots;
ListPlot[{Re[w], Im[w]}, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
<p>In this code, <code>Ticks</code> command is not working properly, that is, real axis range is not showing between <span class="math-container">$-2\pi$</span> to <span class="math-container">$2\pi$</span>, where as imaginary axis is showing properly.</p>
<p>Also, is there any command for complex list plot and complex plot? I mean a command such that the following command works </p>
<pre><code>ComplexListPlot[w, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
<p>instead of </p>
<pre><code>ListPlot[{Re[w], Im[w]}, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
| kglr | 125 | <p>Using Chip Hurst's example data:</p>
<pre><code>SeedRandom[1234];
pts = RandomReal[{0, 1}, {1000, 3}];
</code></pre>
<h3>SmoothKernelDistribution + DensityPlot3D</h3>
<pre><code>pdf[x_, y_, z_] := PDF[SmoothKernelDistribution[pts, MaxExtraBandwidths -> 0,
MaxMixtureKernels -> All], {x, y, z}]
DensityPlot3D[Evaluate[pdf[x, y, z]], {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
ColorFunction -> (Directive[Opacity[#], Blend[{{0, White}, {0.5, Blue}, {1, Red}}, #]] &),
PlotLegends -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/ubniZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ubniZ.png" alt="enter image description here"></a></p>
<pre><code>DensityPlot3D[Evaluate[pdf[x, y, z]], {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
OpacityFunction -> Function[f, f/20],
ColorFunction -> (Blend[{{0, White}, {0.5, Blue}, {1, Red}}, #] &),
PlotLegends -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/Jn5OV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jn5OV.png" alt="enter image description here"></a></p>
<h3>HistogramList + ListDensityPlot3D</h3>
<pre><code>ListDensityPlot3D[HistogramList[pts, 10][[2]], DataRange -> {{0, 1}, {0, 1}, {0, 1}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/mWLAK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mWLAK.png" alt="enter image description here"></a></p>
|
275,310 | <p>I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated</p>
| Dmitri Zaitsev | 59,225 | <p>An affine function between vector spaces is <em>linear</em> if and only if it fixes the origin.</p>
<p>In the simplest case of scalar functions in one variable,
<em>linear</em> functions are of the form $f(x)=ax$ and <em>affine</em> are $f(x)=ax +b$, where $a$ and $b$ are arbitrary constants.</p>
<p>More generally, <em>linear</em> functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ are $f(v)=Av$, and <em>affine</em> functions are $f(v)=Av +b$, where $A$ is arbitrary $m\times n$ matrix and $b$ arbitrary $m$-vector. Further, $\mathbb{R}$ can be replaced by any field.</p>
<p>More abstractly, a function is <em>linear</em> if and only if it preserves the linear (aka vector space) structure, and is <em>affine</em> if and only if it preserves the affine structure. A <em>vector space</em> structure consists of the operations of vector addition and multiplication by scalar, which are preserved by linear functions:</p>
<p>$$f(v_1+v_2) = f(v_1) + f(v_2), \quad f(kv) = k f(v).$$</p>
<p>An <em>affine structure</em> on a set $S$ consists of a transitive free action $(v,s)\to v + s$ by a vector space $V$ (called the associated vector space). Informally, $v+s\in S$ is the translation of the point $s\in S$ by the vector $v\in V$, and for any pair of points $s_1, s_2\in S$, there is an unique translation vector $v\in V$ with $v + s_1=s_2$, also written as $v = \overrightarrow{s_1 s_2}$. Then a function between two affine spaces $S$ and $T$ (i.e. sets with affine structures) is <em>affine</em> if and only if it preserves the structure, that is</p>
<p>$$f(v + s) = h(v) + f(s)$$</p>
<p>where $h$ is a linear function between the corresponding associated vector spaces.</p>
|
118,701 | <p>I have two vectors of 134 elements each ($mu$, and $gt$). $mu$ contains Integers, and $gt$ contains machine precision Reals. I execute the following simple expression multiple times without changing either mu or gt:
$$
(mu/2*gt).gt
$$
I will get one of two different results: $88474.52216839303$ or $88474.52216839301$ (they differ in the last digit). What's up with that? I have never seen Mathematica behave non-deterministically on simple mathematical expressions before (I'm using version 9.0.1). Could it be performing the dot product or the multiplication in parallel or something? I really need deterministic results (at least on the same computer). I'm using Windows 7 on a Dell laptop.</p>
<p><strong>Update 2.</strong>
More fun (but apparently only when n = 134).</p>
<pre><code>n = 134;
</code></pre>
<p>Manually evaluate the following expression over and over again in a notebook. I get a different output for Hash[a.a] about every 5 or 6 evaluations (I realize that I'm using RandomReal, but I'm resetting the seed every time so $a$ always receives the same value).</p>
<pre><code>SeedRandom[1]
a = RandomReal[{0, 1}, n];
Hash[a]
Hash[a.a]
</code></pre>
<p><strong>Update 1.</strong>
Here's the output of two supposedly identical computations:</p>
<pre><code>InputForm[(mu/2 gt).gt]
88474.52216839301
InputForm@Sum[(mu[[i]]/2 gt[[i]]) gt[[i]], {i, Length@mu}]
88474.52216839303
</code></pre>
<p>Sometimes the first expression comes out identical to the second expression.</p>
<p>The two vectors are:</p>
<pre><code>InputForm[mu];
{1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000,
1000, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10};
InputForm[gt]
{-0.004653655618380714, -0.009080939192239956, -0.031034220835509885,
0.06650237015337201, -0.009447749078904444, 0.03336916426793746,
-0.4631539984577655, -6.265611574451846, 0.5050454470606143,
-0.19087962150762205, -4.3828201756709735, 0.49876272299955815,
0.3861679596859371, -6.1635962937354645, -0.15452549964472695,
0.3473530604725559, -4.199256945736951, -0.040039918046105516,
0.004474673856913913, -0.08203350223995054, 0.00755080417433085,
0.19742429370731282, -0.058492983677093546, -0.07179532877025352,
-0.06654467292071248, 0.053278032391796154, 0.021824097049037344,
-0.15494663566691225, 0.0490691048534343, 0.06064592126063756,
-0.07745582064698114, -0.00746967588173042, 0.03559995124065775,
-0.013295758759910827, -0.0021771809232924225, 0.014751625280327196,
-0.012490576622492892, -0.004059719351428424, 0.02260878725186556,
0.00906537308368245, 0.00005640969962929232, 0.013322793809346312,
0.01939688322252025, 0.0012921413237499124, 0.005353410478874665,
0.013744515181619552, -0.0008260952031691714, 0.0019523689330869926,
0.0036484573285596087, 0.0002894048077433703, 0.0009607045972313589,
-0.0005226991236028443, -0.002322046639359741, -0.024682514037441098,
0.027220500404569047, 0.0545967335062756, 0.05642841641401902,
0.004367474285221214, -0.0005337849541497026, -0.0002226867162167549,
0.00019518810603202322, 4.710483244170917*^-6,
-0.00016670488036184576, -0.00011284285936618362,
-0.00008617428801798452, -0.00009997965667755927,
-0.00011813749998897688, -0.00011200191449290103,
-0.00003504307476243085, -0.00007170970464401294,
8.314045093271716*^-6, -0.000012108637423846602,
-0.00008318583029878757, 6.872184070694232*^-6,
0.000039598867001222615, -0.00030723457677017674,
0.000021126291712914025, 0.00017867733864129764, -0.00132238312878738,
0.00008352828516677846, 0.00016343481838493323,
-0.0009495187710002947, 0.000014501494743246468,
0.00010295794136475078, -0.0007041704456466069,
-0.000038074551413580515, 0.000024071562009758685,
-0.000368866036143653, -0.000035157347339220824,
-7.738001399044264*^-6, -0.00012744527523419168,
-0.00001767740166705445, -0.00006515075020033605,
-0.0005012315885292151, -0.00009246376382040004,
0.000035397133843528654, 0.00010514304758398063, 0.000164070573737746,
0.00033034132631488825, 0.0002563940375876836, 0.0001581254676494522,
0.00010340418143852661, 0.0001271214003914325, 0.000144619937452181,
-6.445829617805765*^-8, -7.918633438308738*^-8,
-5.331850464748234*^-10, -5.395049641407357*^-11,
-9.692675239369703*^-10, -2.3443641234029144*^-10,
-2.741707881994002*^-9, -1.8602797897764294*^-9,
1.852265547443208*^-10, -8.268466237349803*^-9,
-2.873207919632894*^-10, -8.965460213556384*^-9,
-2.7681125512775745*^-9, -6.542104162408522*^-9,
-4.0174693373664825*^-9, -3.665420456657473*^-9,
-2.843221227293432*^-9, -1.5467319689177169*^-9, 0.12984908695319053,
0.3495235744679681, 0.31365238179137667, 0.5566921439809294,
0.09068947055843457, 78.52618761194245, 1.462920938774925,
0.13819238950337862, 0.24562317270333922, 0.28816819236661184,
0.02950510971255533, 0.19955099011468305}
</code></pre>
| Daniel Lichtblau | 51 | <p>It's not a bug and it's not so uncommon. For an explanation have a look <a href="http://blog.nag.com/2011/02/wandering-precision.html" rel="nofollow noreferrer">here</a>. This and some related issues also appear in <a href="http://forums.wolfram.com/mathgroup/archive/2011/Mar/msg00974.html" rel="nofollow noreferrer">this MathGroup thread</a>.</p>
<p>Also relevant: <a href="https://stackoverflow.com/questions/5185281/what-is-causing-this-strange-mathematica-result">1</a> <a href="http://www.yqcomputer.com/1164_13938_1.htm" rel="nofollow noreferrer">2</a>.</p>
|
510,732 | <p>I am trying to think of a case where this is not true:</p>
<p>$f(n) = O(g(n))$ and $f(n) \neq \Omega(g(n))$, does $f(n) = o(g(n))$?</p>
<p>I suspect that it has to do with the varying $c$ and $n_{0}$ constants but am not completely sure. </p>
<p>Thanks!</p>
| Jonas Meyer | 1,424 | <p>Consider $f(n) = n+(-1)^nn$ and $g(n) = n$.</p>
|
1,530,874 | <p>Is there a case where a function $f$ that is not differentiable at $0$ and a function $g$ that is differentiable at $0$ where $f+g$ is differentiable at $0$?</p>
| C. Falcon | 285,416 | <p>One has: $$f=f+g-g.$$
Then, if $f+g$ and $g$ are differentiable at the origin, $f$ is differentiable at the origin with: $$f'(0)=(f+g)'(0)-g'(0).$$
Therefore, the answer to your question is no.</p>
|
3,580,293 | <blockquote>
<p>The value of <span class="math-container">$$\lim\limits_{x \rightarrow \infty} \left(5^x + 5^{3x}\right)^{\frac{1}{x}}$$</span> is...</p>
</blockquote>
<p>My approach :</p>
<blockquote>
<p><span class="math-container">$$\lim\limits_{x \rightarrow \infty} \left(5^x + 5^{3x}\right)^{\frac{1}{x}}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} \left(5^x\left(1 + 5^{2x}\right)\right)^\frac{1}{x}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} \left(5^x\right)^\frac{1}{x}\left(1 + 5^{2x}\right)^\frac{1}{x}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} 5\left(1 + 5^{2x}\right)^\frac{1}{x}$$</span></p>
</blockquote>
<p>But I don't know how to proceed. Can anyone help? Thanks!</p>
| Axion004 | 258,202 | <p>From your last step</p>
<p><span class="math-container">$$\left(1 + 5^{2x}\right)^{1/x}=\Big[5^{2x}(1+5^{-2x})\Big]^{1/x}=\Big(5^{2}\Big)\Big(1+5^{-2x}\Big)^{1/x}$$</span></p>
<p>then take the limit as <span class="math-container">$x\to\infty$</span> and the above expression will approach <span class="math-container">$25$</span> giving you the desired result. </p>
|
281,717 | <p>Suppose that $\beta \mathbb{R}$ is Stone–Čech compactification of $\mathbb{R}$. What is the closure of $\mathbb{Q}$? </p>
| Rustyn | 53,783 | <p>It means your first assumption. </p>
<p>Setting the derivative equal to $0$, we obtain:<br></p>
<p>$3x^2 - 6x = 0 \Rightarrow$<br>
$x(3x - 6)=0 \Rightarrow$ <br></p>
<p>$x=0,$ or $x=2$</p>
<p>$f(2) = 8$, $f(0) = 12$</p>
<p>Now we test end points,</p>
<p>$f(4) = 64 - 48 + 12 = 28$ <br>
$f(-2)= -8 -12 + 12 = -8$</p>
<p>Hence $f(4)=28$ = $\max \{ f(x): x \in [-2,4]\}$ </p>
|
3,929,703 | <p>so, we know how to solve if the question was only <span class="math-container">$5$</span> different rings in <span class="math-container">$4$</span> different fingers, which is <span class="math-container">$4^5$</span>. but what if internal order of rings within the finger matters, is this counted in this answer or we have a different answer?</p>
| user2661923 | 464,411 | <p>For me, this is simply a take off my shoes and count problem.</p>
<p>Type of distribution : <br>
Explanation <br>
Number of ways <br></p>
<p><span class="math-container">$5,0,0,0 :$</span><br>
<span class="math-container">$4$</span> ways of choosing which finger, <span class="math-container">$5!$</span> ways of permuting the rings. <br>
<span class="math-container">$T_1 = 4 \times 5!.$</span></p>
<p><span class="math-container">$4,1,0,0 :$</span><br>
<span class="math-container">$4$</span> ways of choosing the heavy finger, <span class="math-container">$3$</span> ways of choosing the 2nd finger, <span class="math-container">$5$</span> ways of choosing which ring goes on the 2nd finger, <span class="math-container">$4!$</span> ways of permuting the rings on the heavy finger. <br>
<span class="math-container">$T_2 = 4 \times 3 \times 5 \times 4!.$</span></p>
<p><span class="math-container">$3,2,0,0 :$</span><br>
<span class="math-container">$4$</span> ways of choosing the heavy finger, 3 ways of choosing the 2nd finger, <span class="math-container">$\binom{5}{3}$</span> ways of selecting which rings go on the heavy finger, <span class="math-container">$3!$</span> ways of permuting the rings on the heavy finger, <span class="math-container">$2!$</span> ways of permuting the rings on the 2nd finger. <br>
<span class="math-container">$T_3 = 4 \times 3 \times \binom{5}{3} \times 3! \times 2!.$</span></p>
<p><span class="math-container">$3,1,1,0 :$</span><br>
<span class="math-container">$4$</span> ways of choosing the heavy finger, <span class="math-container">$3$</span> ways of choosing the no-ring-finger, <span class="math-container">$\binom{5}{3}$</span> ways of selecting the rings on the heavy finger, <span class="math-container">$2$</span> ways of selecting the ring on the 1st light finger, <span class="math-container">$3!$</span> ways of permuting the rings on the heavy finger. <br>
<span class="math-container">$T_4 = 4 \times 3 \times \binom{5}{3} \times 2 \times 3!.$</span></p>
<p><span class="math-container">$2,2,1,0 :$</span><br>
<span class="math-container">$4$</span> ways of choosing the no-ring-finger, <span class="math-container">$3$</span> ways of choosing the 1-ring finger, <span class="math-container">$5$</span> ways of selecting the ring for the 1-ring finger, <span class="math-container">$\binom{4}{2}$</span> ways of distributing the rings on the 2-ring fingers, <span class="math-container">$(2! \times 2!)$</span> ways of permuting the rings on the 2-ring fingers. <br>
<span class="math-container">$T_5 = 4 \times 3 \times 5 \times \binom{4}{2} \times 2! \times 2!.$</span></p>
<p><span class="math-container">$2,1,1,1 :$</span><br>
<span class="math-container">$4$</span> ways of choosing the 2-rings-finger, <span class="math-container">$\binom{5}{2}$</span> ways of selecting which rings go on the 2-rings finger, <span class="math-container">$2!$</span> ways of permuting these 2 rings, <span class="math-container">$3!$</span> ways of distributing the remaining rings among the 3 light fingers.<br>
<span class="math-container">$T_6 = 4 \times \binom{5}{2} \times 2! \times 3!.$</span></p>
<p>Final answer:
<span class="math-container">$$T_1 + T_2 + T_3 + T_4 + T_5 + T_6.$$</span></p>
<p><strong>Edit</strong><br>
I got a total of <span class="math-container">$6720$</span> ways.</p>
|
881,572 | <p>A relation $R$ on the set of real numbers can be thought of as a subset of the $xy$ plane. Moreover an equivalence relation on $S$ is determined by the subset $R$ of the set $S \times S$ consisting of those ordered pairs $(a,b)$ such that $ a \sim b$. </p>
<p>With this notation explain the geometric meaning of the reflexive and symmetric properties.</p>
<p>Since reflexivity implies the presence of all ordered pairs of the type $(a,a)$, may be the geometric meaning is the straight line passing through $(0,0)$ and $(a,a)$ which is nothing but the line $y=x$.</p>
<p>For symmetric presence of $(a,b)$ implies the presence of $(b,a)$. Is its geometric meaning the straight line joining $(a,b)$ and $(b,a)$??</p>
<p>Thanks for the help!!</p>
| dioid | 165,958 | <p>Your description of reflexivity is correct.</p>
<p>For symmetry it means that the subset $R$ is "symmetric" around the line $y = x$, this means that for any point $(a, b)\in R$ its mirror point $(b, a)\in R$ (it's the point you get by doing reflection in the line $y=x$), i.e. either none of the two points $(a, b)$ and $(b, a)$ is included in $R$, or both of them are included in $R$.</p>
<p>Thus the "graph" of $R$ is the same as its mirror image when doing reflection in $y=x$.</p>
|
881,572 | <p>A relation $R$ on the set of real numbers can be thought of as a subset of the $xy$ plane. Moreover an equivalence relation on $S$ is determined by the subset $R$ of the set $S \times S$ consisting of those ordered pairs $(a,b)$ such that $ a \sim b$. </p>
<p>With this notation explain the geometric meaning of the reflexive and symmetric properties.</p>
<p>Since reflexivity implies the presence of all ordered pairs of the type $(a,a)$, may be the geometric meaning is the straight line passing through $(0,0)$ and $(a,a)$ which is nothing but the line $y=x$.</p>
<p>For symmetric presence of $(a,b)$ implies the presence of $(b,a)$. Is its geometric meaning the straight line joining $(a,b)$ and $(b,a)$??</p>
<p>Thanks for the help!!</p>
| M. Vinay | 152,030 | <p>First, you have to define clearly <em>how</em> you are representing a relation between two real numbers graphically.</p>
<p>For example, if $a \sim b$, are you plotting the point $(a, b)$ on the graph? Then it would make sense to plot each point $(a, a)$ for every $a \in S$ (but it will not make up the entire line $y = x$ unless $S = \mathbb R$). If this is your representation, then the relation is symmetric if whenever $(a, b)$ is plotted, so is $(b, a)$. Similarly, for transitivity, whenever $(a, b)$ and $(b, c)$ are plotted, so is $(a, c)$.</p>
<p>Example:
<img src="https://i.stack.imgur.com/FaXRz.png" alt="Graph of Equivalence Relation"><br>
Each color is a partition (and the ones with more than one element are shown along with their bounding squares).</p>
<p>Joining $(a, b)$ and $(b, a)$ by a straight line (segment) does not make much sense, because the two <em>points</em> are not related, the two <em>elements</em> $a$ and $b$ are. Joining elements by lines works when you have a single point representing every element (instead of a pair of elements). Then you can join the elements by lines or rather, by arrows. This will give you a "graph" (in the <a href="https://en.wikipedia.org/wiki/Graph_theory" rel="nofollow noreferrer">graph theoretic</a> sense). So, for an equivalence relation on $S$, you will have a <a href="https://en.wikipedia.org/wiki/Complete_graph" rel="nofollow noreferrer">complete graph</a> with $S$ as the vertex set.</p>
|
3,742,517 | <p>Let <span class="math-container">$f(x)=x^3-1$</span>. To approximate the root <span class="math-container">$x^{\star}=1$</span>, we consider the sequence <span class="math-container">$(x_n)$</span> that we get if we apply Newton's method with <span class="math-container">$x_0>0$</span>. Show that the sequence converges to <span class="math-container">$1$</span>.</p>
<p>I have done the following:</p>
<p>I used <span class="math-container">$x_0=0,5$</span> and applied the method and in that way we see that the sequence converges to <span class="math-container">$1$</span>.</p>
<p>Is that correct?</p>
<p>Now I think that we could also do the following:</p>
<p>From Newton's method we get <span class="math-container">$x_{n+1}=x_n-\frac{x_n^3-1}{3x_n^2}=\frac{2x_n^3+1}{3x_n^2}$</span> and we have to show that this sequence converges to <span class="math-container">$1$</span>, or not?</p>
| Marjan van den Akker | 805,370 | <p>I think that indeed the small value of epsilon is the problem. My suggestion is to change the objective to <span class="math-container">$M \sum x_i + \sum y_i$</span>, where <span class="math-container">$M$</span> is a big number (at least big enough to prioritize <span class="math-container">$\sum x_i$</span> over <span class="math-container">$\sum y_i$</span>. This has been applied in examples where <span class="math-container">$x_i >0$</span> means infeasibilty e.g. because <span class="math-container">$x_i$</span> signals that something in unassigned.</p>
|
28,955 | <p>I need to crack a stream cipher with a repeating key.</p>
<p>The length of the key is definitely 16. Each key can be any of the characters numbered 32-126 in ASCII.</p>
<p>The algorithm goes like this:</p>
<p>Let's say you have a plain text:</p>
<p>"Welcome to Q&A for people studying math at any level and professionals in related fields."</p>
<p>Let's say that the password is:</p>
<p>"0123456789abcdef"</p>
<p>Then, to encrypt the plaintext, just XOR them together. If the key isn't long enough, just repeat it. e.g., </p>
<p>Welcome to Q&A for people studying math at any level and professionals in related fields.</p>
<pre><code> XOR
</code></pre>
<p>0123456789abcdef0123456789abcdef0123456789abcdef0123456789abcdef0123456789ab</p>
<p>I have 2 English messages encrypted with the above algorithm and with the same key.
I know about the communicative property of xor and that it can be exploited for the example above.
I've read that this is a pretty weak cipher and it has been cracked. However, I have no idea how to do it. So, where can I find a cryptanalysis tool to do it for me?</p>
| Alon Amit | 308 | <p>Take the two encrypted messages and XOR them with each other. You'll get the XOR of the two original, unencrypted messages since the identical keys cancel out. Deciphering this just requires patience and a good understanding of the encoding (what exactly is being XORed - the ASCII values of the letters? Some other binary encoding? Are spaces retained?). </p>
<p>You can try some common words and letter combinations and look for places where XORing the word with the ciphertext yields an English-looking string (which would then be the corresponding word or letter combination at the other text). Also, the ciphertext will likely have numerous 0 values corresponding to places where both messages have the same letter. Such matches are more likely to be a pair of E's or a pair of Spaces than a pair of Q's or X's. </p>
<p>Having said all that, I don't know where you can find a tool to do it for you, and if you find one it'll be very sensitive to the particular encoding. It's much more fun to do this yourself.</p>
<p>EDIT: in the comments you're mentioning that it's indeed ASCII encoding. This makes the task much easier since many 8-bit sequences don't correspond to any legal ASCII character at all. Make a table of all the possible XORs of legal ASCII characters in your plain text and this will tell you, for each position in your text, what are the possible pairs of letters from the two messages.</p>
|
28,955 | <p>I need to crack a stream cipher with a repeating key.</p>
<p>The length of the key is definitely 16. Each key can be any of the characters numbered 32-126 in ASCII.</p>
<p>The algorithm goes like this:</p>
<p>Let's say you have a plain text:</p>
<p>"Welcome to Q&A for people studying math at any level and professionals in related fields."</p>
<p>Let's say that the password is:</p>
<p>"0123456789abcdef"</p>
<p>Then, to encrypt the plaintext, just XOR them together. If the key isn't long enough, just repeat it. e.g., </p>
<p>Welcome to Q&A for people studying math at any level and professionals in related fields.</p>
<pre><code> XOR
</code></pre>
<p>0123456789abcdef0123456789abcdef0123456789abcdef0123456789abcdef0123456789ab</p>
<p>I have 2 English messages encrypted with the above algorithm and with the same key.
I know about the communicative property of xor and that it can be exploited for the example above.
I've read that this is a pretty weak cipher and it has been cracked. However, I have no idea how to do it. So, where can I find a cryptanalysis tool to do it for me?</p>
| castarco | 4,976 | <p>If you want to decrypt these texts, a good method is the old "Kasysky's method":</p>
<p>First, you have to know the frequencies of the characters in your plain text (if you know the language, it's easy). Then, search repeated characters and measure the space between them. If you make the greatest common divisor of the distances, it's probably that the GCD is the length of the key (name it L).</p>
<p>Lastly, when you know the length of the key, you can divide the ciphered message into chunks and measure the frequencies of characters (join the characters spaced by L positions). It's probably that the frequencies of characters in the ciphered text are similar to the frequencies of characters in the language that are written.</p>
|
942,651 | <p>Let $n,r,k$ be non negative integers such that $r,k\leq n$.</p>
<p>If $r^{n-r}=k^{n-k}$, when is this true other than $r=k$ ?</p>
<p>For example it is holds for $n=6,r=2,k=4$.</p>
| user153012 | 153,012 | <p>We have that $n,r,k$ be non negative integers such that $r,k\leq n$.</p>
<p>First of all $(r,k,n)=(0,0,n)$, where $n \neq 0$ is a trivial solution for $r^{n-r} = k^{n-k}$.</p>
<p>For further solutions, we have</p>
<p>$$\begin{align}
r^{n-r} & = k^{n-k} \\
(n-r) \cdot \ln r & = (n-k) \cdot \ln k \\
n & = {\frac{r\ln r - k\ln k}{\ln r - \ln k}}.
\end{align}$$</p>
<p>Now to provide, that $n$ is an integer let $a$ be a nonnegative integer, and take $k=r^a$. If $a=1$, then $r=k$ and then if $r=k=0$, then $n\neq r$, but else $n$ could be anything. If $a \neq 1$, then</p>
<p>$$n = \frac{r-ar^a}{1-a},$$</p>
<p>which is an integer for appropriate $(r,a)$ pairs. I don't know a condition for this.</p>
<p>So I think there are infinitly many solutions in the form
$$(r,k,n) = \left(r,r^a,\frac{r-ar^a}{1-a} \right).$$</p>
<p>If you can find such condition for $n$, that when is it integer, then you could find all solutions.</p>
<p>Examples: $(2,4,6), (2,8,11), (3,9,15), (3,27,39), (3,81,107), (3,243,303) (4,16,28), (4,64,94), (4,256,340), (4,1024,1279), (5,25,45), \ \dots$</p>
|
32,849 | <p>I am trying to simulate a signal that randomly increases its phase, so far I have tried two thing but neither worked. I usually use matlab but I want to learn some <em>Mathematica</em> so I thought I would try this in <em>Mathematica</em>.</p>
<p>My first try was</p>
<pre><code>times = Table[i, {i, 0, 2, 0.05}];
function[source_, t_][fiIN_] :=
With[{fi = fiIN + 0.01*RandomReal[]},source*(1 + Sin[2 Pi*t + fi])];
</code></pre>
<p>Where I wanted to feed <code>fi</code> back into <code>fiIN</code> for each subsequent <code>t</code> value (in the list <code>times</code>). I did not know how to makes this work though so I went on with my second try:</p>
<pre><code>fiupdate[fi_] := fiupdate[fi - 1] + 0.01*RandomReal[];
fiupdate[1] = 0;
fitimes = Range[Length[times]];
</code></pre>
<p>However this function does not remember the value of <code>RandomReal[]</code> for the earlier steps, so <code>fiupdate[10]</code> could be smaller than <code>[9]</code> or <code>[8]</code>. Also when using this function I get an error:</p>
<pre><code>fis = fiupdate[fitimes];
$RecursionLimit::reclim: Recursion depth of 1024 exceeded
</code></pre>
<p>Im not sure how to make this work. Any help is appreciated. Thank you!</p>
| PlatoManiac | 240 | <p>One way will be to use <code>Nest</code> or something like that...</p>
<pre><code>tmax = 20;
times = Table[i, {i, 0, tmax, 0.05}];
fi = With[{dist = .8}, (*here you can control how big jump is possible between phases*)
NestList[RandomReal[{#, # + dist}] &,RandomReal[],-1 + Length@times]];
phaseIncrese =Transpose@{times, (1 + Sin[2 Pi*#1 + #2]) & @@@ (Transpose[{times,fi}])};
With[{(f = Interpolation@phaseIncrese)},
Plot[{1 + Sin[2 Pi*t], f[t]}, {t, 1, 4}, Frame -> True]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/O93Nd.png" alt="enter image description here"></p>
|
32,849 | <p>I am trying to simulate a signal that randomly increases its phase, so far I have tried two thing but neither worked. I usually use matlab but I want to learn some <em>Mathematica</em> so I thought I would try this in <em>Mathematica</em>.</p>
<p>My first try was</p>
<pre><code>times = Table[i, {i, 0, 2, 0.05}];
function[source_, t_][fiIN_] :=
With[{fi = fiIN + 0.01*RandomReal[]},source*(1 + Sin[2 Pi*t + fi])];
</code></pre>
<p>Where I wanted to feed <code>fi</code> back into <code>fiIN</code> for each subsequent <code>t</code> value (in the list <code>times</code>). I did not know how to makes this work though so I went on with my second try:</p>
<pre><code>fiupdate[fi_] := fiupdate[fi - 1] + 0.01*RandomReal[];
fiupdate[1] = 0;
fitimes = Range[Length[times]];
</code></pre>
<p>However this function does not remember the value of <code>RandomReal[]</code> for the earlier steps, so <code>fiupdate[10]</code> could be smaller than <code>[9]</code> or <code>[8]</code>. Also when using this function I get an error:</p>
<pre><code>fis = fiupdate[fitimes];
$RecursionLimit::reclim: Recursion depth of 1024 exceeded
</code></pre>
<p>Im not sure how to make this work. Any help is appreciated. Thank you!</p>
| ybeltukov | 4,678 | <p>You can use <code>Sin[t + phase[t]]</code> where <code>phase[t]</code> is <code>Interpolation</code> of a random process.</p>
<p>This random process can be implemented by <code>Accumulate</code></p>
<pre><code>tmin = 0;
tmax = 10;
dt = 0.1;
phase = Interpolation[
Transpose@{ Range[tmin, tmax, dt],
Accumulate@RandomReal[1.0, Floor[(tmax - tmin)/dt + 1]]},
InterpolationOrder -> 1];
Plot[phase[t], {t, tmin, tmax}]
</code></pre>
<p><img src="https://i.stack.imgur.com/XpsnH.png" alt="enter image description here"></p>
<p>or by <code>PoissonProcess</code></p>
<pre><code>phase = Interpolation[
RandomFunction[
PoissonProcess[1/dt], {0, Floor[(tmax - tmin)/dt + 1]}]["Path"],
InterpolationOrder -> 1];
Plot[phase[t], {t, tmin, tmax}]
</code></pre>
<p><img src="https://i.stack.imgur.com/pY63d.png" alt="enter image description here"></p>
|
43,743 | <p>An alternative title is: When can I homotope a continuous map to a smooth immersion?</p>
<p>I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated.</p>
<p>The set-up is the following:
Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$. In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion. </p>
<p>I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion.</p>
<p>Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.</p>
<p>References would be greatly appreciated.</p>
<p>Thanks!</p>
<p>Edit:</p>
<p>As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves. This is where my intuition says that there should be such a smooth immersion.</p>
| Johannes Ebert | 9,928 | <p>There is a general strategy for these kind of problems, which sometimes helps (the ''h-principle''): separate the homotopical and smooth aspects of the problem. Setup: $f:N \to M$ a map of smooth manifolds, $dim (N) < dim (M)$, $f|_{\partial N}$ is an immersion. </p>
<p>Step 1: if your $f$ is going to be homotopic to an immersion $g$, then there should exist a bundle monomorphism $\phi:TN \to f^* TM$, extending the derivative of $f|_{\partial}$ to all of $N$. Otherwise, the hypothetical immersion won't exist, because there is no candidate for its derivative. Constructing this monomorphism is a purely bundle theoretic problem. Whether this can be successfully attacked depends very much on the concrete situation -- maybe it is trivial, maybe it is very hard.
In any case -- if there is no candidate for an immersion in sight -- it is be easier to construct a bundle map than an immersion.</p>
<p>Step 2: if you succeed in solving the homotopy problem, you are done! This is the content of the Smale-Hirsch-Theorem: once $\phi$ exists, $f$ will be homotopic to an immersion $g$ and the derivative is homotopic (as a bundle mono) to $\phi$. There is a relative statement as well; you have to extend the immersion $f_{\partial M}$ to an immersion of a collar of $\partial M$ in $M$. This is not trivial since the collar has one dimension
more. If $dim(N)=dim(M)$, then the result still holds (in your situation, where $N$ is a connected cobordism with nonempty boundary), but you cannot fix $f$ on the boundary.</p>
<p>Your intuition in the low-dimensional case seems to be true: if $N$ is a surface and $M$ a domain in $R^3$, then both are parallelizable.</p>
<p>References: Eliashberg-Michachev: Introduction to the h-principle, Haefliger: Lectures on the theorem of Gromov, Ponaru: Homotopy theory and differentiable singularities. Adachi: Immersions and embeddings.</p>
|
1,134,510 | <p>Regarding My Background I have covered stuff like </p>
<p>1.Single Variable Calculus</p>
<p>2.Multivariable Calculus (Multiple Integration,Vector Calculus etc) (Thomas Finney)</p>
<p>3.Basic Linear Algebra Course (Containing Vector spaces,Linear Transformation)</p>
<p>4.Ordinary Differential Equation</p>
<p>5.Real Analysis (Sequences And series)</p>
<p>I am interested In Number theory and i am big fan of Ramanujan .I have not been through rigorous proofs before . But i want to dig deep into number theory especially area where Ramanujan was working .Anyone researcher,Professor can advice me about which preparations are needed for going into number theory and which books should i need .I will be highly obliged .I am asking this for self study or you can say pursuing research at home .</p>
<p>Note - All stuff i have covered is with help of youtube videos and self study .Thanks </p>
| Community | -1 | <p>Another way to think is -
No of ways of selecting one alphabet out of 26 is $\binom {26}{1}$.
Then the number of ways of arranging it in 3 places out of 7 is $\binom {7}{3}$.
Then we are left with 25 alphabets and we have to select one. This can be done in $\binom{25}{1}$ ways. And we have only one way of arranging it. And therefore our final answer is $$\binom{26}{1}\binom{7}{3}\binom{25}{1}$$.</p>
|
304 | <p>Per <a href="http://blog.stackoverflow.com/2010/07/moderator-pro-tempore/">this post on the SO/SE blog</a> (which, curiously, does not include math.SE in its graphic list), it looks like the admins will choose moderators pro tempore at about 7 days into the public beta. In the roughly 24 hours that we've been in public beta, I've wondered several times: should we push to have the admins choose moderators pro tempore sooner (e.g. <em>now</em>)?</p>
<p><em>edit</em>: to try to get a bit more clarity in response to this question, I've created CW answers "YES" and "NO" below--please up/down vote those as you see fit.</p>
<p><em>edit2</em>: see also: <a href="https://math.meta.stackexchange.com/questions/150/elect-our-provisional-moderators">elect our (Provisional) Moderators</a></p>
| Isaac | 72 | <p><strong>NO</strong> we do not need moderators pro tempore now</p>
|
130,914 | <p>I dont know how to proceed with solving $$\sum_{i=1}^{n}i^{k}(n+1-i).$$ Please give advise.</p>
| bspk | 28,987 | <p>$$\sum_{i=1}^{n}i^{k}(n+1-i)$$ </p>
<p>is same as </p>
<p>$$\sum_{i=1}^{n}i(n+1-i)^{k}$$</p>
<p>which looks like some combination of Eulerian number.</p>
|
431,690 | <p>As far as I know, for any $A$:
$$\mathbf{x}^{T}A\mathbf{y}=0;\forall\mathbf{x},\mathbf{y}\in R^n\Rightarrow A=0$$</p>
<p>Does it mean that
$$\mathbf{x}^{T}A\mathbf{x}=0;\forall\mathbf{x}\in R^n\Rightarrow A=0$$</p>
<p>The condition of the first claim $\forall\mathbf{x},\mathbf{y}\in R^n$ implies that we could take $y=x$, and, therefore the second claim should hold. Correct?</p>
| Doctor Dan | 84,274 | <p>$x^T A x = 0 \; \forall x => (x, Ax) = 0 \; \forall x$, hence $A$ can be an orthogonal matrix. This is what A in the @Daniel Fisher's counterexample is.</p>
|
4,194,611 | <p>Let <span class="math-container">$A \in \mathbb{R}^{m \times n}$</span>.</p>
<p><span class="math-container">$\forall i \in \mathbb{N} \; [ x_i \in \mathbb{R}^n \; \mbox{and} \; x_i \geq \textbf{0}] $</span></p>
<p>Assume that the sequence <span class="math-container">$Ax_1, Ax_2, ...$</span> converges to <span class="math-container">$p$</span>. Show that <span class="math-container">$ \exists x \in \mathbb{R}^n, x \geq \textbf{0}$</span>, such that <span class="math-container">$p=Ax$</span>.</p>
<p>I am trying to prove this statement to prove that a cone formed by finitely many vectors in <span class="math-container">$\mathbb{R}^n$</span> is closed.</p>
<p>Edit (additional context): I got the question while reading the proof of Farkas's Lemma here: <a href="https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf" rel="nofollow noreferrer">https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf</a>. <br />
The problem statement is assumed in this proof.</p>
| Mark Saving | 798,694 | <p>Come up with a basis <span class="math-container">$b_1, b_2, ..., b_k$</span> for the image of <span class="math-container">$A$</span>. Given some <span class="math-container">$y$</span> in the image of <span class="math-container">$A$</span>, there are unique <span class="math-container">$w_1, w_2, ..., w_k$</span> such that <span class="math-container">$y = \sum\limits_{j = 1}^k w_j b_j$</span>. So take these to be functions <span class="math-container">$y \mapsto w_j(y)$</span>. Note that these functions are continuous.</p>
<p>Then we have</p>
<p><span class="math-container">$\begin{equation}
\begin{split}
p &= \lim\limits_{i \to \infty} A x_i \\
&= \lim\limits_{i \to \infty} \sum\limits_{j = 1}^k w_j(A x_i) b_j \\
&= \sum\limits_{j = 1}^k \lim\limits_{i \to \infty} w_j(A x_i) b_j \\
&= \sum\limits_{j = 1}^k \lim\limits_{i \to \infty} w_j(p) b_j
\end{split}
\end{equation}$</span></p>
<p>Since <span class="math-container">$p$</span> can be written as a linear combination of the <span class="math-container">$b_k$</span>, we see that <span class="math-container">$p$</span> is in the image of <span class="math-container">$A$</span>. This is exactly what we sought to prove.</p>
|
4,194,611 | <p>Let <span class="math-container">$A \in \mathbb{R}^{m \times n}$</span>.</p>
<p><span class="math-container">$\forall i \in \mathbb{N} \; [ x_i \in \mathbb{R}^n \; \mbox{and} \; x_i \geq \textbf{0}] $</span></p>
<p>Assume that the sequence <span class="math-container">$Ax_1, Ax_2, ...$</span> converges to <span class="math-container">$p$</span>. Show that <span class="math-container">$ \exists x \in \mathbb{R}^n, x \geq \textbf{0}$</span>, such that <span class="math-container">$p=Ax$</span>.</p>
<p>I am trying to prove this statement to prove that a cone formed by finitely many vectors in <span class="math-container">$\mathbb{R}^n$</span> is closed.</p>
<p>Edit (additional context): I got the question while reading the proof of Farkas's Lemma here: <a href="https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf" rel="nofollow noreferrer">https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec07.pdf</a>. <br />
The problem statement is assumed in this proof.</p>
| Mathews Boban | 948,760 | <p><strong>Claim 1</strong>: Let <span class="math-container">$\{ v_1, .. v_k\}$</span> be a linearly independent subset of <span class="math-container">$\mathbb{R}^n$</span>. Let <span class="math-container">$\{c_{i1}, c_{i2},\; ..,\; c_{ik}\}_{i = 1}^{\infty}$</span> be reals.
<span class="math-container">$$ (c_{i1}v_1 + c_{i2}v_2 + .. + c_{ik}v_k)_{i=1}^{\infty} \; \text{converges to} \; \vec{0} \implies [\; (c_{i1})_{i=1}^{\infty} \rightarrow \; 0 \; \text{and}\; (c_{i2})_{i=1}^{\infty} \rightarrow \; 0 \; \text{and} \; ... \; (c_{ik})_{i=1}^{\infty} \rightarrow \; 0 ] $$</span></p>
<p><strong>Proof of claim 1</strong>:
The cases <span class="math-container">$k=1$</span> and <span class="math-container">$k=2$</span> are slightly easier. I will prove the claim for the <span class="math-container">$k=3$</span>. For <span class="math-container">$k \geq 4$</span>, the proof is essentially the same.</p>
<p>Observe that
<span class="math-container">$$ (c_{i1}v_1 + c_{i2}v_2 + c_{i3}v_3)_{i=1}^{\infty} \;\text{converges to} \; \vec{0} \iff ({|| c_{i1}v_1 + c_{i2}v_2 + c_{i3}v_3 ||}^2)_{i=1}^{\infty}\; \text{converges to } 0 $$</span>
And
<span class="math-container">$$ {|| (c_{i1}v_1 + c_{i2}v_2) + c_{i3}v_3 ||}^2 = ( ||c_{i1}v_1 + c_{i2}v_2|| + \frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3}v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||} ||c_{i3}v_3||)^2 + c_{i3}^{2} ({ 1 - (\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3} v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||}) }^{2} )||v_3 ||^2 $$</span></p>
<p>Note that
<span class="math-container">$$ (a_n^2 + b_n^2)_n \rightarrow 0 \implies [\;(a_n)_n \rightarrow 0 \; \text{and} \; (b_n)_n \rightarrow 0\;] $$</span></p>
<p>Hence,
<span class="math-container">$$ (\; c_{i3}^{2} ({1 - (\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3} v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||}) }^{2} )||v_3 ||^2 \;)_i \rightarrow 0 $$</span></p>
<p>And
<span class="math-container">$$ (\; c_{i3}^{2} ({1 - (\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3} v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||}) }^{2} ) \;)_i \rightarrow 0 $$</span></p>
<p>Now we show that there exists a <span class="math-container">$t > 0$</span> that depends only on <span class="math-container">$v_1,v_2,v_3$</span> such that <span class="math-container">$ ({1 - (\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3} v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||}) }^{2} ) \geq t $</span></p>
<p>This would imply that <span class="math-container">$(c_{i3}^2)_i \rightarrow 0$</span>, which in turn implies <span class="math-container">$(c_{i3})_i \rightarrow 0$</span></p>
<p>Let <span class="math-container">$k_1v_1 + k_2v_2$</span> be the projection of <span class="math-container">$v_3$</span> onto the subspace spanned by <span class="math-container">$\{v_1,v_2\}$</span></p>
<p><span class="math-container">$$\begin{aligned} {|\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; c_{i3} v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||c_{i3}v_3||}| } &= |\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; v_3 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||v_3||}|
\\ &= |\frac{ \langle c_{i1}v_1 + c_{i2}v_2, \; k_1v_1 + k_2v_2 \rangle}{|| c_{i1}v_1 + c_{i2}v_2|| * ||v_3||}|
\\&=
\frac{|| k_1v_1 + k_2v_2 ||}{||v_3||}*|\langle \frac{ c_{i1}v_1 + c_{i2}v_2 }{ ||c_{i1}v_1 + c_{i2}v_2 || }, \; \frac{k_1v_1 + k_2v_2}{||k_1v_1 + k_2v_2||} \rangle|
\\ &\leq \frac{|| k_1v_1 + k_2v_2 ||}{||v_3||}
\\ &< 1 \end{aligned}$$</span></p>
<p>Last inequality follows since <span class="math-container">$v_3$</span> must have a non-zero component that lies outside <span class="math-container">$span\{v_1,v_2\}$</span></p>
<p>Define the <span class="math-container">$t$</span> that we wanted as <span class="math-container">$ (1 - {(\frac{|| k_1v_1 + k_2v_2 ||}{||v_3||})}^2 ) $</span>.</p>
<p>We just showed that <span class="math-container">$(c_{i3})_i \rightarrow 0$</span></p>
<p>Using the same line of arguments, we can prove that <span class="math-container">$(c_{i1})_i \rightarrow 0$</span> and <span class="math-container">$(c_{i2})_i \rightarrow 0$</span></p>
<p>Proof of claim 1 (for <span class="math-container">$k = 3$</span> case) ends
<span class="math-container">$$\tag* {$\blacksquare$}$$</span></p>
<p>Now let us see why claim 1 is useful.</p>
<p>Let the column vectors of <span class="math-container">$A$</span> be <span class="math-container">$v_1,..,v_n$</span>. There is a linearly independent subset of <span class="math-container">$\{ v_1,..,v_n \}$</span> that spans <span class="math-container">$span\{ v_1,..,v_n \} = Image(A)$</span>. Wlog, assume that <span class="math-container">$\{v_1,..,v_k\}$</span> is such a linearly independent set.</p>
<p>Hence, there exists <span class="math-container">$\{d_{i1}, d_{i2}, \; .., \; d_{ik}\}_{i=1}^{\infty}$</span> with each <span class="math-container">$d_{ij} \geq 0 $</span> such that for each <span class="math-container">$i \in \mathbb{N}$</span> :</p>
<p><span class="math-container">$$Ax_i = d_{i1}v_1 + d_{i2}v_2 + .. + d_{ik}v_k $$</span></p>
<p>We know that <span class="math-container">$Ax_1, Ax_2, ..$</span> converges to <span class="math-container">$p$</span>. It can be shown that <span class="math-container">$p \in Image(A)$</span> from this statement.</p>
<p>Let <span class="math-container">$p = b_1v_1+ b_2v_2 + .. b_kv_k$</span></p>
<p>Hence,
<span class="math-container">$$ (d_{i1}v_1 + d_{i2}v_2 + .. + d_{ik}v_k)_{i=1}^{\infty} \; \rightarrow \; b_1v_1+ b_2v_2 + .. b_kv_k $$</span></p>
<p>This implies,
<span class="math-container">$$ ( \; (d_{i1} - b_1) v_1 + (d_{i2}- b_2) v_2 + .. + (d_{ik} - b_k)v_k \;)_{i=1}^{\infty} \; \rightarrow \; \vec{0} $$</span></p>
<p>Applying claim 1, we get
<span class="math-container">$$ (d_{i1})_{i} \rightarrow b_1 \; \text{and} \; (d_{i2})_{i} \rightarrow b_2 \; .. \; \text{and} \; (d_{ik})_{i} \rightarrow b_k $$</span></p>
<p>Recall that <span class="math-container">$d_{ij}$</span>'s are all <span class="math-container">$\geq 0$</span>. Hence <span class="math-container">$b_1, .. b_k$</span> are <span class="math-container">$\geq 0$</span>. Now, the proof is complete.</p>
|
367,669 | <p><img src="https://i.stack.imgur.com/zQFyC.jpg" alt="enter image description here"></p>
<p>This is probably a very simple questions but I am not clear on Möbius transformations and how to solve this problem. I'd appreciate if somebody can point me towards a method to do these sort of questions or a webpage that explains what I need to solve this problem.</p>
| Zen | 72,576 | <p>How about this:</p>
<ol>
<li>{x<sub>1</sub>: x<sub>1</sub> is prime OR 1}$\to 1$.</li>
<li>{x<sub>2</sub>: $x_2=2\cdot p$, for prime p>2}$\to 2$.</li>
<li>{x<sub>3</sub>: $x_3=3\cdot p$, for prime p>3}$\to 3$.</li>
<li>{x<sub>4</sub>: $x_4=4\cdot p$, for prime p>4}$\to 4$
etc.</li>
</ol>
<p>Sorry, I was a little sloppy.</p>
|
3,716,619 | <p>Evaluating
<span class="math-container">$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$</span>
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.</p>
| Community | -1 | <p>Well, you've got your answer, and it's a good one, I'd use series expansions <em>always</em> in such case, but then, the answerer couldn't know you've ever heard of those expansions, and some of your comments show you aren't too familiar with them. That's why SE encourages sharing information about your mathematical background, btw. Most people ignore that. But then, you risk to get an answer like the following, without any l'Hospitals, based only on elementary principles:</p>
<p>"From the elementary identity
<span class="math-container">$$\frac1{\sin^2x}-\frac1{x^2}=\sum^\infty_{k=1}3^{-2k}\,\frac{\frac83-\frac{16}9\sin^2\frac{x}{3^k}}{\left(1-\frac43\sin^2 \frac{x}{3^k}\right)^2},$$</span> letting <span class="math-container">$x\to0,$</span> we get <span class="math-container">$$\frac1{\sin^2x}-\frac1{x^2}\to\sum^\infty_{k=1}3^{-2k}\cdot\frac83=\frac13,$$</span> and the result we're looking for follows after replacing <span class="math-container">$x\to\pi x.$</span>"</p>
<p>The joke is: that identity is an elementary consequence of the triplication formula <span class="math-container">$$\sin3y=3\sin y-4\sin^3y$$</span> and the limit <span class="math-container">$\sin y/y\to1$</span> as <span class="math-container">$y\to0,$</span> indeed.</p>
<p>Of course, such an answer isn't helpful, not just because it is rather obscure, but also because the method is applicable only in exceptional cases.</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| Anurag A | 68,092 | <p><strong>Hint</strong></p>
<p>Try the function $f(x)=\arctan{x} - x$. It's derivative is $\frac{-x^2}{1+x^2}$. Now use the monotonicity to get the required inequality.</p>
|
83,565 | <p>I am learning Mathematica on the fly, one of my tasks is to find the variance of white noise. I followed the tutorial for finding white noise by using the code:</p>
<pre><code>WN = WhiteNoiseProcess[NormalDistribution[0, 10]];
data = RandomFunction[WN, {0, 10000}];
</code></pre>
<p>I know I can use the following code to find the variance:<code>Variance[data]</code> </p>
<p>However, I would like to find it by using the formula for variance. I checked the reference built into Mathematica and it says I can simply use: </p>
<pre><code>Total[(list-Mean[list])^2]/(Length[list]-1)
</code></pre>
<p>I input data for the list:</p>
<pre><code>Total[(data-Mean[data])^2]/(Length[data]-1)
</code></pre>
<p>When I do this, I don't get the same output as when I use the <code>Variance[data]</code> code, but instead get:
<img src="https://i.stack.imgur.com/ZcpyR.png" alt="enter image description here"></p>
<p>So, I am curious what I am doing wrong? I'm sure it's something simple I am not doing, but after spending a couple of hours wrestling with this, I am breaking down to ask. Sorry if this is a dumb question. Thank you in advance for your time. </p>
| ubpdqn | 1,997 | <p>It is valuable to look at the properties of these complex objects, e.g. in your example: </p>
<pre><code>data["Properties"]
</code></pre>
<p>To do your own variance:</p>
<pre><code>val = First@data["ValueList"];
Variance[val]
Total[(val - Mean[val])^2]/(Length[val] - 1)
</code></pre>
<p>You can compare results of <code>Variance</code> and your mimic.</p>
|
3,059,833 | <blockquote>
<p>If the equation <span class="math-container">$2^{2x} + a*2^{x+1} + a + 1=0$</span> has roots of opposite sign then the exhaustive values of a are?</p>
</blockquote>
<p>I tried taking <span class="math-container">$2^x = t$</span>. But then didn't know what to do.</p>
<p>The equation became, <span class="math-container">$t^2 + 2at + a +1 =0$</span>. But then, what conditions should I impose?</p>
<p>Since one root is negative and the other positive, the only conclusion that I could draw was that <span class="math-container">$x<0$</span> for the first condition and <span class="math-container">$x>0$</span> for the second condition. But, I don't know how do I proceed from here?</p>
<p>Any help would be appreciated.</p>
| Michael Rozenberg | 190,319 | <p>The hint:</p>
<p>Solve the following system.
<span class="math-container">$$1^2+2a\cdot1+a+1<0$$</span> and
<span class="math-container">$$a+1>0.$$</span>
The first inequality says that <span class="math-container">$1$</span> is placed between <span class="math-container">$2^{x_1}$</span> and <span class="math-container">$2^{x_2}.$</span></p>
<p>The second inequality says that the smaller root of the quadratic equation is positive.</p>
|
3,059,833 | <blockquote>
<p>If the equation <span class="math-container">$2^{2x} + a*2^{x+1} + a + 1=0$</span> has roots of opposite sign then the exhaustive values of a are?</p>
</blockquote>
<p>I tried taking <span class="math-container">$2^x = t$</span>. But then didn't know what to do.</p>
<p>The equation became, <span class="math-container">$t^2 + 2at + a +1 =0$</span>. But then, what conditions should I impose?</p>
<p>Since one root is negative and the other positive, the only conclusion that I could draw was that <span class="math-container">$x<0$</span> for the first condition and <span class="math-container">$x>0$</span> for the second condition. But, I don't know how do I proceed from here?</p>
<p>Any help would be appreciated.</p>
| zipirovich | 127,842 | <p>So far, so good! Now: if <span class="math-container">$x>0$</span>, then <span class="math-container">$t=2^x>1$</span>; and if <span class="math-container">$x<0$</span>, then <span class="math-container">$t=2^x<1$</span>. So for your new equation <span class="math-container">$t^2+2at+a+1=0$</span> you want it to have two roots <span class="math-container">$t$</span> of which one is greater and one is less than <span class="math-container">$1$</span>. Comparing roots with <span class="math-container">$1$</span> seems a bit difficult; comparing with <span class="math-container">$0$</span> would be easier… So let's make another change of variables: <span class="math-container">$t$</span> is greater than or less than <span class="math-container">$1$</span> if and only if <span class="math-container">$y=t-1$</span> is greater than or less than <span class="math-container">$0$</span> respectively. Substituting <span class="math-container">$y=t-1$</span>, i.e. <span class="math-container">$t=y+1$</span>, the quadratic equations transforms into
<span class="math-container">$$\begin{split}
t^2+2at+a+1=0 &\iff (y+1)^2+2a(y+1)+a+1=0\\
&\iff y^2+2(a+1)y+(3a+2)=0.
\end{split}$$</span></p>
<p>Now, we have to satisfy two conditions:</p>
<ul>
<li>A quadratic equation (with real coefficients and the leading term of <span class="math-container">$1$</span>, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need <span class="math-container">$3a+2<0$</span>.</li>
<li>But also don't forget that <span class="math-container">$t=2^x$</span> can't be negative! In other words, <span class="math-container">$t>0$</span>, and therefore <span class="math-container">$y=t-1>-1$</span>. For a parabola <span class="math-container">$f(y)=y^2+2(a+1)y+(3a+2)$</span> opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than <span class="math-container">$-1$</span>, or in other words, that <span class="math-container">$-1$</span> lies to the left of both roots. This conditions will be satisfied if <span class="math-container">$f(-1)>0$</span>, i.e. <span class="math-container">$(-1)^2+2(a+1)\cdot(-1)+(3a+2)>0$</span>, which simplifies to <span class="math-container">$a+1>0$</span>.</li>
</ul>
|
1,452,121 | <p>I very well know that every open ball is an open set. and that every open set need not be an open ball. But illustrate me some counter example.</p>
| mkausp | 243,800 | <p>The most simple examples of open sets, which are not balls, in every metric space are $\emptyset$ and the space itself, which are open.</p>
<p>Even if you consider those sets to be balls with radius $0$ or $\infty$, respectively, you can take the union of two or more open balls, which is not necessarily an open ball anymore (see <a href="https://math.stackexchange.com/users/131740/user46944">user46944</a>'s answer for the one-dimensional case).</p>
<p>What you can actually show instead is that every open set in a metric space can be written as a union of open balls (see <a href="https://math.stackexchange.com/questions/135740/the-union-of-open-balls">this quesion</a>). In $\mathbb{R}^n$ you can furthermore show that a set is open if and only if it is a <strong>countable</strong> union of open balls (see <a href="https://math.stackexchange.com/questions/299260/union-of-a-countable-collection-of-open-balls">this question</a>).</p>
|
3,401,630 | <p>I am trying to prove the inequality
<span class="math-container">$$\frac{1}{n}-\frac{1}{(n+1)^2}>\frac{1}{n+1}\quad \forall \ n>1.$$</span> How would I go about doing this? I've tried solving it on my own but my final answer is <span class="math-container">$1>0$</span>. </p>
| Allawonder | 145,126 | <p>Your inequality is equivalent to <span class="math-container">$$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{n+1}>0.$$</span> Simplifying LHS gives <span class="math-container">$$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{n+1}{(n+1)^2}=\frac1n-\frac{n+2}{(n+1)^2}=\frac{(n+1)^2-n(n+2)}{n(n+1)^2}.$$</span> Since the denominator is positive, it only suffices to show that the numerator is positive too. Pursuing this by expanding gives <span class="math-container">$$(n+1)^2-n(n+2)=n^2+2n+1-n^2-2n=1>0,$$</span> as desired.</p>
|
3,689,096 | <p>This was the Question:- Find all positive integers <span class="math-container">$n$</span> such that <span class="math-container">$\varphi(n)$</span> divides <span class="math-container">$n^2 + 3$</span></p>
<p>What I tried:-</p>
<p>I knew the solution and explanation of all positive integers <span class="math-container">$n$</span> such that <span class="math-container">$\varphi(n)\mid n$</span> .
<a href="https://math.stackexchange.com/q/135756/11619">That answer</a> was when <span class="math-container">$n = 1$</span>, or <span class="math-container">$n$</span> is of the form of <span class="math-container">$2^a$</span> or <span class="math-container">$2^a3^b$</span> . </p>
<p>I tried to relate this fact with this problem in many ways, but couldn't get to a possible solution.</p>
<p>Any hints or suggestions will be greatly appreciated</p>
| SB1729 | 466,737 | <p>First, we observe that <span class="math-container">$n$</span> can't be even. If it were even, <span class="math-container">$n^2+3$</span> would be odd and hence <span class="math-container">$\varphi(n)$</span> couldn't divide <span class="math-container">$n^2+3$</span> as <span class="math-container">$\varphi(n)$</span> is always even unless <span class="math-container">$n=1,2$</span>. Since <span class="math-container">$\varphi(n)=1$</span> for <span class="math-container">$n=1,2$</span>, these are two trivial solutions.</p>
<p>Let <span class="math-container">$n=3^ap_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$</span>, where <span class="math-container">$p_j$</span>'s are primes bigger than <span class="math-container">$3$</span> and <span class="math-container">$a_j\geq1$</span>.
If for some <span class="math-container">$j$</span> we have <span class="math-container">$a_j\geq2$</span>, then <span class="math-container">$p_j\mid \varphi(n)\implies p_j\mid n^2+3\implies p_j\mid3$</span>. A contradiction! Hence <span class="math-container">$a_j=1$</span> for all <span class="math-container">$1\leq j\leq k$</span>. Now if <span class="math-container">$a>0$</span>, <span class="math-container">$v_3(n^2+3)=1$</span>. Since <span class="math-container">$v_3(\varphi(n))\geq(a-1)$</span>, we must have <span class="math-container">$a=2$</span> and <span class="math-container">$3\nmid p_j-1$</span> for all <span class="math-container">$1\leq j\leq k$</span>.</p>
<p><strong>case <span class="math-container">$1$</span>:</strong>
Let <span class="math-container">$a=0$</span>. Then <span class="math-container">$n=p_1p_2\ldots p_k$</span> and <span class="math-container">$\varphi(n)=(p_1-1)(p_2-1)\ldots(p_k-1)$</span>. clearly <span class="math-container">$v_2(\varphi(n))\geq2^k$</span> and hence <span class="math-container">$v_2(n^2+3)\geq2^k$</span>. Since <span class="math-container">$n$</span> is odd, <span class="math-container">$n^2\equiv1\pmod{8}\implies n^2+3\equiv4\pmod{8}\implies v_2(n^2+3)=2$</span>. This means <span class="math-container">$k\leq2$</span>.</p>
<p>For <span class="math-container">$k=1$</span>, we have <span class="math-container">$n$</span> is a prime <span class="math-container">$p$</span> and <span class="math-container">$(p-1)\mid (p^2+3)$</span>. Now <span class="math-container">$(p-1)\mid(p-1)^2=p^2-2p+1\implies (p-1)\mid((p^2+3)-(p^2-2p+1))=2(p+1)$</span>. Since <span class="math-container">$p$</span> is odd, <span class="math-container">$(p-1)=2$</span> or <span class="math-container">$(p-1)=4$</span>. since we have assumed that <span class="math-container">$p>3$</span>, in this case the only solution is <span class="math-container">$p=5$</span>.</p>
<p>For <span class="math-container">$k=2$</span> we have the situation <span class="math-container">$n=pq$</span> for two distinct primes bigger than <span class="math-container">$3$</span>. We have to solve the congruence <span class="math-container">$(pq)^2+3\equiv0\pmod{(p-1)(q-1)}$</span></p>
<p><span class="math-container">$(p-1)(q-1)\mid((pq)^2-p^2q-pq^2+pq)\implies (p-1)(q-1)\mid(p^2q+q^2p+3-pq)$</span></p>
<p><span class="math-container">$(p-1)(q-1)\mid(p^2q-p^2-pq+p)$</span> and <span class="math-container">$(p-1)(q-1)\mid(q^2p-q^2-pq+q)$</span>. Therefore,</p>
<p><span class="math-container">$(p-1)(q-1)\mid(p^2q+q^2p-p^2-q^2-2pq+p+q-(p^2q+q^2p+3-pq))\implies (p-1)(q-1)\mid(p^2+q^2+pq-p-q+3)\implies (p-1)(q-1)\mid(p^2+q^2+2)$</span></p>
<p><span class="math-container">$(p-1)\mid(p^2+q^2+2)\implies(p-1)\mid(p^2+q^2+2-p^2+2p-1)=(q^2+2p+1)\implies (p-1)\mid(q^2+2p-2p+3)=(q^2+3)$</span>. Similarly we get <span class="math-container">$(q-1)\mid(p^2+3)$</span>. Let <span class="math-container">$\mathrm{WLOG}$</span> <span class="math-container">$p<q$</span>. If <span class="math-container">$p=3$</span> then we can deduce that <span class="math-container">$q=7$</span>. Therefore <span class="math-container">$n=21$</span> is a solution. Let <span class="math-container">$p>q>3$</span>. Since <span class="math-container">$(pq)^2+3\equiv4\pmod{8}$</span> and <span class="math-container">$(p-1)(q-1)\mid((pq)^2+3)$</span> we get <span class="math-container">$v_2(p-1)=v_2(q-1)=1$</span>. Any odd prime dividing <span class="math-container">$p-1$</span> or <span class="math-container">$q-1$</span> divides <span class="math-container">$(pq)^2+3$</span> and hence <span class="math-container">$-3$</span> is a quadratic residue modulo those primes. Therefore they are either <span class="math-container">$3$</span> or of the form <span class="math-container">$6l+1$</span>. If <span class="math-container">$(q-1)/2\equiv1\pmod{6}$</span> then <span class="math-container">$q-1\equiv2\pmod{6}$</span>, which implies <span class="math-container">$3\mid pq$</span>. A contradiction! Therefore <span class="math-container">$p=3,q=7$</span> is the only solution in this case.</p>
<p><strong>case <span class="math-container">$2$</span>:</strong> Let <span class="math-container">$a=1$</span>. In this case <span class="math-container">$n=3$</span> is a solution as <span class="math-container">$\varphi(3)=2\mid3^2+3=12$</span>.</p>
<p>We investigate now the other possibilities. For <span class="math-container">$a=1$</span>, if <span class="math-container">$n\neq3$</span>, then <span class="math-container">$n$</span> can be of the form <span class="math-container">$3p$</span> for some odd prime <span class="math-container">$p>3$</span>. Otherwise <span class="math-container">$v_2(\varphi(n))>2$</span> which can't be possible as we have shown before.</p>
<p>In this case, the situation is, <span class="math-container">$\varphi(3p)=2(p-1)\mid(9p^2+3)$</span>. We have <span class="math-container">$(p-1)\mid(9p^2+3)\implies (p-1)\mid(9p^2+3-9p^2+9p)=(9p+3)\implies (p-1)\mid12$</span>. Then <span class="math-container">$p$</span> can be <span class="math-container">$7$</span> or <span class="math-container">$13$</span>. For <span class="math-container">$p=13$</span>, <span class="math-container">$v_2(\varphi(n))=3$</span> which is not possible. So in this case only possible solution is <span class="math-container">$n=3\cdot7=21$</span></p>
<p><strong>last case:</strong> For <span class="math-container">$a=2$</span>, we have <span class="math-container">$9$</span> is a solution. By similar arguments as above, we can show that there can't be any other solutions.</p>
<p><strong>Hence only possible solutions are <span class="math-container">$n=1,2,3,5,9,21$</span></strong></p>
<p><strong>DONE!</strong></p>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| Victor Protsak | 5,740 | <p>This is a frivolous item solely to demonstrate the pitfalls of running MathSciNet searches and working with large datasets:</p>
<p>Type "Wang and Zhang" in the author field and get a list of 2417 items. Li and Wang are close contenders with 2300 total. I wouldn't venture a guess how many collaborations that represents!</p>
|
3,968,508 | <p>All I have got so far is that <span class="math-container">$a$</span> must divide <span class="math-container">$2002$</span>... Could anyone share some ideas? I mean is there any method rather than trial and error?</p>
| QED | 864,951 | <p>This isn't an answer, but it's an idea on how you could further narrow down your options</p>
<p>You can try taking (mod a) of either side to get <span class="math-container">$-1\equiv2001($</span>mod <span class="math-container">$a)$</span> (I'm assuming that's how you got a divides 2002).</p>
<p>Taking (mod 3) of either side shows that a isn't congruent to 0 or 2 (mod 3), so it must be congruent to 1 (mod 3), and n must be even.</p>
<p>Finally you can take the (mod a+1) of either side to get <span class="math-container">$-1^{n+1}\equiv 2001($</span>mod <span class="math-container">$a+1) \Rightarrow (a+1)|2002$</span> since n is even.</p>
<p>This reduces the possible values of a to 1 and 13, and a=1 obviously won't have any solutions.</p>
|
3,968,508 | <p>All I have got so far is that <span class="math-container">$a$</span> must divide <span class="math-container">$2002$</span>... Could anyone share some ideas? I mean is there any method rather than trial and error?</p>
| quasi | 400,434 | <p>At most one of <span class="math-container">$a,a+1$</span> is a multiple of <span class="math-container">$3$</span>.</p>
<p>
If exactly one of <span class="math-container">$a,a+1$</span> is a multiple of <span class="math-container">$3$</span>, then
<span class="math-container">$$
a^{n+1}-(a+1)^n
$$</span>
would not be a multiple of <span class="math-container">$3$</span>, contradiction, since <span class="math-container">$2001$</span> is a multiple of <span class="math-container">$3$</span>,
<p>
Thus neither of <span class="math-container">$a,a+1$</span> is a multiple of <span class="math-container">$3$</span>, hence <span class="math-container">$a\equiv 1\;(\text{mod}\;3)$</span>,
<p>
If <span class="math-container">$n$</span> is odd, then
<p><span class="math-container">\begin{align*}
&
a^{n+1}-(a+1)^n=2001
\\[4pt]
\implies\;&
a^{n+1}-(a+1)^n\equiv 2001\;(\text{mod}\;3)
\\[4pt]
\implies\;&
1^{n+1}-2^n\equiv 0\;(\text{mod}\;3)
\\[4pt]
\implies\;&
1-2\equiv 0\;(\text{mod}\;3)
\end{align*}</span>
contradiction.</p>
<p>
Hence <span class="math-container">$n$</span> is even.<span class="math-container">$\;$</span>Then
<p><span class="math-container">\begin{align*}
&
a^{n+1}-(a+1)^n=2001
\\[4pt]
\implies\;&
a^{n+1}-(a+1)^n\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
(-1)^{n+1}-0^n\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
-1\equiv 2001\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
2002\equiv 0\;(\text{mod}\;(a+1))
\\[4pt]
\implies\;&
(a+1){\,\mid\,}2002
\\[4pt]
\end{align*}</span></p>
<p>and as you noted, we also have <span class="math-container">$a{\,\mid\,}2002$</span>, thus both <span class="math-container">$a$</span> and <span class="math-container">$a+1$</span> are divisors of <span class="math-container">$2002$</span>.</p>
<p>
Hence since <span class="math-container">$\gcd(a,a+1)=1$</span>, it follows that <span class="math-container">$\bigl(a(a+1)\bigr){\,\mid\,}2002$</span>.
<p>
Then <span class="math-container">$a(a+1)\le 2002$</span>, so <span class="math-container">$a < 45$</span>.
<p>
From the prime factorization of <span class="math-container">$2002=(2)(7)(11)(13)$</span>, we get that the only positive integer divisors of <span class="math-container">$2002$</span> which are less than <span class="math-container">$45$</span> are <span class="math-container">$1,2,7,11,13,14,22,26$</span>, and of those potential values of <span class="math-container">$a$</span>, only <span class="math-container">$a=1$</span> and <span class="math-container">$a=13$</span> are such that <span class="math-container">$a+1$</span> is also a divisor of <span class="math-container">$2002$</span>.
<p>
But if <span class="math-container">$a=1$</span>, then <span class="math-container">$a^{n+1}-(a+1)^n=1-2^n < 0$</span>, contradiction.
<p>
It remains to analyze the case <span class="math-container">$a=13$</span>.
<p>
Suppose <span class="math-container">$13^{n+1}-14^n=2001$</span>.
<p>
If <span class="math-container">$n > 2$</span>, then
<p><span class="math-container">\begin{align*}
&
13^{n+1}-14^n=2001
\\[4pt]
\implies\;&
13^{n+1}-14^n\equiv 2001\;(\text{mod}\;8)
\\[4pt]
\implies\;&
13^{n+1}\equiv 1\;(\text{mod}\;8)
\\[4pt]
\implies\;&
(13^n)(13)\equiv 1\;(\text{mod}\;8)
\\[4pt]
\implies\;&
(1)(5)\equiv 1\;(\text{mod}\;8)
\\[4pt]
\end{align*}</span>
contradiction.</p>
<p>
Hence since <span class="math-container">$n$</span> is even, the only remaining possibility is <span class="math-container">$n=2$</span>.
<p>
By direct evaluation, it can be verified that <span class="math-container">$13^3-14^2=2001$</span>.
<p>
Therefore <span class="math-container">$a=13, n=2$</span> is the only solution in positive integers to the given equation.
|
85,343 | <p>I am looking for a reference to study classical (i.e., not quantized) Yang-Mills theory. </p>
<p>Most of the sources I find focus on mathematical aspects of the theory, like Bleecker's book <em>Gauge theory and variational principles</em>, or Baez & Muniain's <em>Gauge fields, knots and gravity</em>.</p>
<p>But I am more interested something similar to the standard development of electromagnetism, as can be found, for example, in Landau & Lifchitz's course on theoretical physics. To be more exact, I would like to learn about the field equations (Yang-Mills and Einstein equations), but also about the corresponding Lorentz Law, the energy of the Yang-Mills field,... and the analogous concepts of what is made for the electromagnetism.</p>
<p>That is, I look for a rigorous exposition where, at the same time, I could learn whether it is possible to prove, at the classical level, the quick decrease of the strong interaction, and things like that.</p>
| Liviu Nicolaescu | 20,302 | <p>Have you tried the book "the Geometry of physics" by Th. Frankel?</p>
|
85,343 | <p>I am looking for a reference to study classical (i.e., not quantized) Yang-Mills theory. </p>
<p>Most of the sources I find focus on mathematical aspects of the theory, like Bleecker's book <em>Gauge theory and variational principles</em>, or Baez & Muniain's <em>Gauge fields, knots and gravity</em>.</p>
<p>But I am more interested something similar to the standard development of electromagnetism, as can be found, for example, in Landau & Lifchitz's course on theoretical physics. To be more exact, I would like to learn about the field equations (Yang-Mills and Einstein equations), but also about the corresponding Lorentz Law, the energy of the Yang-Mills field,... and the analogous concepts of what is made for the electromagnetism.</p>
<p>That is, I look for a rigorous exposition where, at the same time, I could learn whether it is possible to prove, at the classical level, the quick decrease of the strong interaction, and things like that.</p>
| Qfwfq | 4,721 | <p>Maybe you can have a look to Nakahara's <em>Geometry, Topology and Physics</em>, or is it too elementary for your purposes?</p>
|
1,808,881 | <p>I am struggling with finding all roots of unity in $\mathbb{Q}(i)$. I know that if $a+bi$ is a root of unity in $\mathbb{Q}(i)$, then $a^2+b^2=1$, and I know how to find all $a, b \in \mathbb{Q}$ that satisfy that equation. However, I do not know how to filter out the roots of unity. I think there should be a somewhat easier way, for example looking at cyclotomic polynomials. Any help would be very nice.</p>
| Pipicito | 93,689 | <p>Let $\xi \in \mathbb{Q}(i)$ be a root of unity. We know that there exists $n\in \mathbb{N}$ such that $\xi$ is $n$-primitive. So, let's use the suggestive notation $\xi_n = \xi$. Now the cyclotomic extension $\mathbb{Q}(\xi_n) / \mathbb{Q}$ is a subextension of $\mathbb{Q}(i) / \mathbb{Q}$. By multiplicativity of the degree you have $$\phi(n) = [\mathbb{Q}(\xi_n) : \mathbb{Q}] \; | \; [\mathbb{Q}(i) : \mathbb{Q}] = 2$$ You have to look for the $n\in \mathbb{N}$ with $\phi(n)| 2$.</p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| Ross Millikan | 1,827 | <p>There is an infinite family of solutions coming from the idea $(2+\sqrt 3)^3=26+15\sqrt 3$. We can form $(2+\sqrt 3)^{3n}$ and find another solution. The next one is $(2+\sqrt 3)^6=1351+780 \sqrt 3$ and $(1351+780\sqrt 3)^{(1/3)}+(1351-780\sqrt 3)^{(1/3)}=14$ There is a recurrence, if $(a,b)$ is a solution, the next is $(26a+45b,15a+26b)$ and so we get triplets $(1351,780,14),(70226,40545,52),(3650401,2107560,194),(189750626,109552575,724)
9863382151,5694626340,2702)$</p>
<p>and on. I have not shown that these are all the solutions.</p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| Ian Mateus | 17,751 | <p>The solutions are of the form $\displaystyle(p, q)= \left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)$, for any rational parameter $t$. To prove it, we start with $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n\tag{$\left(p,q,n,r\right)\in\mathbb{N}^{4}$}$$
and cube both sides using the identity $(a+b)^3=a^3+3ab(a+b)+b^3$ to, then, get $$\left(\frac{n^3-2p}{3n}\right)^3=p^2-rq^2,$$ which is a nicer form to work with. Keeping $n$ and $r$ fixed, we see that for every $p={1,2,3,\ldots}$ there is a solution $(p,q)$, where $\displaystyle q^2=\frac{1}{r}\left(p^2-\left(\frac{n^3-2p}{3n}\right)^3\right)$. When is this number a perfect square? <a href="http://www.wolframalpha.com/input/?i=%5Cfrac%7Bp%5E2-%5Cleft%28%5Cfrac%7Bn%5E3-2p%7D%7B3n%7D%5Cright%29%5E3%7D%7Br%7D">Wolfram</a> says it equals $$q^2 =\frac{(8p-n^3) (n^3+p)^2}{(3n)^2\cdot 3nr},$$ which reduces the question to when $\displaystyle \frac{8p-n^3}{3nr}$ is a perfect square, and you get solutions of the form $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right).$ Note that when $r=3$, this simplifies further to when $\displaystyle \frac{8p}{n}-n^2$ is a perfect square.</p>
<hr>
<p>Now, we note that if $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right) \in\mathbb{Q}^2$, $\displaystyle\sqrt{\frac{8p-n^3}{3nr}}$ must be rational as well. Call this rational number $t$, our parameter. Then $8p=3t^2nr+n^3$. Substitute back to get $$(p,q)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right).$$ This generates expressions like <a href="http://www.wolframalpha.com/input/?i=%5Cleft%28p%2bq%5Csqrt%7Br%7D%5Cright%29%5E%7B1/3%7D%2b%5Cleft%28p-q%5Csqrt%7Br%7D%5Cright%29%5E%7B1/3%7D%20where%20r=11,%20p=2589437/8%20and%20q=56351/4&a=%5E_Real">$$\left(\frac{2589437}{8}+\frac{56351}{4}\sqrt{11}\right)^{1/3}+\left(\frac{2589437}{8}-\frac{56351}{4}\sqrt{11}\right)^{1/3}=137$$</a></p>
<p><a href="http://www.wolframalpha.com/input/?i=%5Cleft%28p%2bq%5Csqrt%7Br%7D%5Cright%29%5E%7B1/3%7D%2b%5Cleft%28p-q%5Csqrt%7Br%7D%5Cright%29%5E%7B1/3%7D%20where%20r=3,%20p=11155/4%20and%20q=6069/4&a=%5E_Real">$$\left(\frac{11155}{4}+\frac{6069}{4}\sqrt{3}\right)^{1/3}+\left(\frac{11155}{4}-\frac{6069}{4}\sqrt{3}\right)^{1/3}=23$$</a></p>
<p>for whichever $r$ you want, the first using $(r,t,n)=(11,2,137)$ and the second $(r,t,n)=(3,7,23)$.</p>
|
234,409 | <p>I'm trying to obtain the coordinates of the border of the continents. I need this information to be ordered such that when I do, for example,</p>
<pre><code>ListLinePlot[data]
</code></pre>
<p>It does not yield a messed up image, as happens for disordered points. Initially I was trying by highlighting points on images of maps, and the detecting the points. This approach was really problematic. However, I find that Mathematica has a built in functionality which gives me just what I need, but for countries. That is for example,</p>
<pre><code>CountryData["Iran", "FullCoordinates"][[1]]
</code></pre>
<p>I'm exploring if this can be done for continents as entities, and I have found that the information for a continent as a polygon can be obtained as,</p>
<pre><code>africa = Entity["GeographicRegion", "Africa"]["Polygon"]
</code></pre>
<p>Is there any way by which I can obtain a list of points corresponding to the coordinates of this continent?</p>
| flinty | 72,682 | <p>I'm assuming you want points from the Mercator projection because you're trying to plot in 2D with <code>ListLinePlot</code>? Otherwise you'd be asking for the polygon wrapped on the sphere.</p>
<pre><code>(* get the GeoPosition points from the polygon *)
points=Flatten[First@First[Entity["GeographicRegion","Africa"]["Polygon"]],1];
(* convert to Mercator projection and plot *)
ListLinePlot[
GeoGridPosition[First@Entity["GeographicRegion","Africa"]["Polygon"],"Mercator"][[1]]
, AspectRatio->1]
</code></pre>
<p>Bear in mind Africa has loads of islands, so if you just want the main continent border then you need to select the first set of points:</p>
<pre><code>ListLinePlot[
GeoGridPosition[
First@Entity["GeographicRegion","Africa"]["Polygon"],
"Mercator"][[1,1]]
, AspectRatio->1, PlotRange->All
]
</code></pre>
<p><a href="https://i.stack.imgur.com/qU5wE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qU5wE.png" alt="enter image description here" /></a></p>
|
685,918 | <p>I'm doing exercises in Real Analysis of Folland and got stuck on this problem. I don't know how to calculate limit with the variable on the upper bound of the integral. Hope some one can help me solve this. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Show that $\lim\limits_{k\rightarrow\infty}\int_0^kx^n(1-k^{-1}x)^k~dx=n!$</p>
</blockquote>
</blockquote>
<p>Thanks so much for your consideration.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Let
$$
f_k(x)=\left\{\begin{array}{ccc}
\left(1-\frac{x}{k}\right)^{k}x^n & \text{if} & x\in [0,k], \\ \\
0 & \text{if} & x>k.
\end{array}
\right.
$$
Then
$$
0\le f_k(x)\le \mathrm{e}^{-x}x^n,
$$
for all $k$ and $x\ge 0$, and $\lim_{k\to\infty}f_k(x)=\mathrm{e}^{-x}x^n$, for all $x\ge 0$.</p>
<p><a href="http://en.wikipedia.org/wiki/Dominated_convergence_theorem" rel="nofollow">Lebesgue Dominated Convergence Theorem</a> is applicable as $\mathrm{e}^{-x}x^n\in L^1(0,\infty)$, and implies that
$$
\lim_{k\to\infty}\int_0^k \left(1-\frac{x}{k}\right)^k x^n\,dx=
\lim_{k\to\infty}\int_0^\infty f_k(x)\,dx=\int_0^\infty \mathrm{e}^{-x}\,x^n\,dx=\Gamma(n+1)=n!
$$</p>
|
1,744,698 | <p>How to show that the characteristic polynomials of matrices A and B are $\lambda^{n-1}(\lambda ^2-\lambda -n)=0$ and $\lambda^{n-1}(\lambda^2+\lambda-n)=0$ respectively by applying elementary row or column operations.</p>
<p>$A=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$</p>
<p>$B=\begin{bmatrix}
-1 & 1 & 1 & \cdots & 1 \\
1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 \\
\end{bmatrix}$</p>
<p>Where $A$ and $B$ are symmetric matrices of order $n+1$.</p>
| openspace | 243,510 | <p>About first matrix. </p>
<p>Lets take the bottom line. As we know :
$$\det{A} = \sum{(-1)^{i+j}\cdot a_{i,j}\cdot M_{i,j}},$$ so we got:
$$S_{n+1} = a_{n+1,n+1}\cdot (-1)^{2n+2}\cdot S_{n} + (-1)^{n+2}a_{n+1,1}S'_{n},$$ where $S'_{n}= (-1)^{1 + n} \lambda^{n-1}(-1)^{n-1}$, because of :</p>
<p>$S'_{n} = \begin{bmatrix}
1 & 1 & \cdots & 1 & 1\\
-\lambda & 0 & \cdots & 0 & 0 \\
0 & -\lambda & \cdots & 0 & 0\\
\vdots & \vdots & \cdots &\ddots & \vdots \\
0 & 0 & \cdots & -\lambda & 0 \\
\end{bmatrix}$</p>
|
542,951 | <p>I am trying show that the function $f:[0,1]\to \mathbb{R}$ defined by $f(x)=\sin \dfrac{1}{x}$ if $x\neq 0$ and $f(0)=0$ possesses IVP. Though it looks easy, but I am not getting any clue how to start with. Any help would be appreciated.</p>
| Dan Rust | 29,059 | <p>Show that there exists a subset $A$ of $(0,1]$ such that $f(A)=f([0,1])$ and such that $f|_A$ ($f$ restricted to the domain $A$) is a continuous function. You may then apply the intermediate value theorem to $f|_A$.</p>
<p>Note that the above proves that $f$ has the property that you mentioned in the comments, but this is not what one would usually call the IVP. The usual intermediate value property is that for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$. We call functions which satisfy the IVP <em>Darboux functions</em>. This question highlights the fact that the set of Darboux functions $[0,1]\rightarrow\mathbb{R}$ is a proper superset of the set of continuous functions $[0,1]\rightarrow \mathbb{R}$.</p>
<p>Assuming $a<b$, the above proves the IVP for $a=0$, $b=1$. For $b<1$, you will need to show that an $A$ exists as above, but such that $A\subset (0,b)$. It also remains to show that the IVP holds for $a\neq 0$ but this case is handled rather easily by simply restricting $f$ to the interval $[a,b]$ and noting that $f|_{[a,b]}$ is continuous. </p>
|
1,851,084 | <p>I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.</p>
| BDS | 59,605 | <p>Consider the Jordan Canonical Form for $A$; that is, $A = PJP^{-1}$ for some invertible $P$ and block diagonal matrix $J$ whose blocks are either diagonal or Jordan (same entry on the diagonal, have $1$s on the the diagonal above the main diagonal, and $0$s elsewhere).</p>
<p>Then, the equation reduces to $J^2 + J = I$. Looking at the diagonal entries on both sides, this reduces to solving $r^2 + r = 1$, which has solutions $r = \frac{-1 \pm \sqrt{5}}{2}$. In other words, $J$ has diagonal entries are $\frac{-1 \pm \sqrt{5}}{2}$. However, we want $|A| = 1$, which is equivalent to $|J| = 1$. This is only possible if $n$ is even and $J$ has an equal even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal, due to $(\frac{-1 + \sqrt{5}}{2})(\frac{-1 -\sqrt{5}}{2}) = -1$ and no positive integral power of $\frac{-1 \pm \sqrt{5}}{2}$ equaling $\pm 1$.</p>
<p>As a summary, there are no solutions when $n$ is not a multiple of $4$.
Otherwise, when $n$ is not a multiple of $4$, there are infinitely many solutions. To illustrate this, let $J$ be diagonal with an equal and even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal. Letting $P$ vary (as there are infinitely many non-invertible matrices that do not commute with $J$) will give infinitely many possibilities for $A$ when $n$ is a multiple of $4$. </p>
|
377,266 | <p>My question is very direct:</p>
<blockquote>
<p>What are the motivations for the name "jet"(subjet, superjet) in the context of viscosity solutions for second order fully nonlinear elliptic PDE?</p>
</blockquote>
<p>The definition of which can be seen in Crandall, Ishii, Lions:</p>
<p><em>Crandall, Michael G.; Ishii, Hitoshi; Lions, Pierre-Louis</em>, <a href="http://dx.doi.org/10.1090/S0273-0979-1992-00266-5" rel="nofollow noreferrer"><strong>User’s guide to viscosity solutions of second order partial differential equations</strong></a>, Bull. Am. Math. Soc., New Ser. 27, No. 1, 1-67 (1992). <a href="https://zbmath.org/?q=an:0755.35015" rel="nofollow noreferrer">ZBL0755.35015</a>.</p>
<p>see also <a href="https://arxiv.org/pdf/math/9207212.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/math/9207212.pdf</a>.</p>
| asymptotic | 85,716 | <p>As I understand it, the word "jet" is meant to evoke the idea of a "spray" of curves through a point, or more accurately, their equivalence classes up to kth order contact</p>
|
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