qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,143,200 | <p>The jump diffusion model is defined as
$$dS_t = \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\;\;\;\;\;\;\;(1)$$
, where ${V_i}$ is a sequence of iid non-negative random variables and it is independent of $W_t$. In the Merton's jump diffusion model ,
$log(V) \sim N(\mu_J, \sigma^2_J)$ and $ N_t$ is a poisson process with rate $\lambda$.</p>
<p>I was asked to apply Ito lemma to $(d \;logS_t)$ to obtain the following:</p>
<p>$$S_t = S_0 exp \left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t\right) \prod^{N_t}_{j=1}V_j \;\;\;\;\;\;\;(2)$$</p>
<p>I literally do not know how to solve this problem because of the term $d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)$. This is how far I got to:</p>
<p>$$dlogS_t = \frac{1}{S_t} \left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\right) - \frac{1}{2S_t^2} d[S,S]_t$$</p>
<p>What exactly is $d[S,S]_t$ ? I know that
$$d[S,S]_t = \sigma^2 S^2_t dt + ...$$
But what is that "..." ? </p>
| ki3i | 202,257 | <p>I suspect that your SDE defining the $S_t$ process is problematic in the term "$S_t\text d\Big(\sum\limits_{i=1}^{N_t}(V_i-1)\Big)$". Intuitively, this term should capture instantaneous random jumps when they occur in any given realisation of the process. That is, if a jump occurs at time $t$, then the process jumps from $S_{t-}$ to $S_{t-}V_t$, where $S_{t-}$ is the left limit of the process at time $t$. So, it makes more sense for $S_t$ to be the jump-diffusion Levy process (so <em>right-continuous, left-limit</em>) defined by
$$
\text dS_t = \mu S_{t-}\text dt + \sigma S_{t-}\text dW_t + \text d\Big(\sum\limits_{0\leqslant r\leqslant t}S_{r-}(V_r - 1){\bf 1}_{\{N_r-N_{r-}=1\}}\Big),\tag{1}
$$</p>
<p>with jumps determined by a Compound Poisson process independent of $\{W_t\}_{t>0}$. That is, a Poisson process $\{N_t\}_{t>0}$ determines when jumps occur, and the size of such a jump at time $t$ is determined as $S_{t-}(V_t - 1)$ (where $\log V_t\sim\mathcal N(\mu_{J},\sigma^2_{J})\ $ for i.i.d. $V_t$).</p>
<p>Assuming $(1)$ for your SDE and $\mathbb E[\int_{0}^{T}S_{u-}^2\text du] < \infty$, Ito's lemma for jump-diffusion processes is applicable here. Note that, for a $\ C^{1,2}\ $ function $\ f:[0,T]\times\mathbb R\to\mathbb R$, the process $f(t,S_t)$ satisfies</p>
<blockquote>
<p>$$
\begin{eqnarray*}
f(t,S_t) - f(0,S_0) &=& \int_{0}^{t}(\partial_uf + \mu S_{u-}\partial_sf + \frac{1}{2}\sigma^2S_{u-}^2\partial_{ss}f)\text du \\
&& + \sigma \int_{0}^{t}S_{u-}\partial_sf~\text dW_u + \sum_{0\leqslant r\leqslant t}(f(r,S_r)-f(r-,S_{r-})){\bf 1}_{\{N_{r}-N_{r-}=1\}}\,.
\end{eqnarray*}
$$ </p>
</blockquote>
<p>This is simply Ito's lemma for a diffusion process but, in addition, the final term adds the change in the process due to the jumps determined by the compound Poisson process. So, choosing $f(S_t):=\log S_t$ gives us</p>
<p>$$
\begin{eqnarray*}
\log\Big(\frac{S_t}{S_0}\Big) &=& (\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log\Big(\frac{S_{r-}V_r}{S_{r-}}\Big) \\
&=&(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log V_r\,.
\end{eqnarray*}
$$
Therefore, in your notation,</p>
<blockquote>
<p>$$
S_t = S_0 e^{(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t}\prod_{i=1}^{N_t}V_i\,.
$$</p>
</blockquote>
|
4,056,073 | <p>I need help with this task, if anyone had a similar problem it would help me !</p>
<p>The task is: Determine the type of interruption at the point x = 0 for the function</p>
<p><span class="math-container">$$f(x)=2^{-\frac{1}{x^{2}}}$$</span></p>
<p>I did:</p>
<p><span class="math-container">$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$L=R=x=0$$</span></p>
<p>And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?</p>
<p>Thanks in advance !</p>
| Community | -1 | <p>There is no paradox.</p>
<p>You cannot reason with real coefficients, because they convey infinite information and you can add them extra bits at no cost. You have to reason with coefficients having a finite representation, hence a finite number of polynomials. Let <span class="math-container">$p$</span> this number.</p>
<p>The information content of a polynomial chosen among <span class="math-container">$p$</span> is not the total number of bits of the coefficients, but just <span class="math-container">$\lg(p)$</span>, assuming equiprobable polynomials.</p>
<p>There is only one way to produce the roots in sorted order, then this yields <span class="math-container">$\lg(p)$</span> bits of information again. If you enforce a specific order, you increase the amount of information by <span class="math-container">$\lg(n!)$</span>. So supplying the roots as a tuple in fact takes more information than as a set.</p>
|
48,629 | <p>Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$.</p>
<p>On my desk you could find the following books: <strong>Algebraic Geometry</strong> by <em>Hartshorne</em>, <strong>Ideals, Varieties, and Algorithms</strong> by <em>Cox & Little & O'Shea</em>, <strong>Algorithms in Real Algebraic Geometry</strong> by <em>Basu & Pollack & Roy</em> and <strong>A SINGULAR Introduction to Commutative Algebra</strong> by <em>Greuel & Pfister</em>. Unfortunately, neither of them introduced notions and ideas I'm looking for.</p>
<p>If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: <em>is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space?</em></p>
<p>Further questions I have are, for example:</p>
<ol>
<li>Can I define <em>geodesics</em> (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities?</li>
<li>How about <em>curvature</em>? Is it defined for these objects?</li>
<li>Can we talk about <em>convexity</em> of subsets of the algebraic surface?</li>
<li>What other tools and term can be imported from differential/Riemannian geometry?</li>
</ol>
<p>I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help.</p>
| Will Sawin | 18,060 | <p>While a real/complex variety, being a subset of a real/complex vector space, will have a Riemannian/Kahler metric, this metric will not be defined up to isomorphism of the algebraic variety. For instance, if the equation defining it is linear, then the variety is isomorphic to a scaled-up version of itself.</p>
<p>So the induced metric exists, but depends on the embedding (which I guess is usually the case)</p>
<p>Since it is a Riemannian manifold, you can define curvature et. al. using the standard definitions. Some of these properties will be algebraic functioms, defined by polynomial functions or sections of algebraic vector bundles, and some will not. The main problem will be complex conjugation, </p>
<p>Geodesics, having real dimension 1, will not have a complex-algebraic description, and will probably fail to be real-algebraic curves. (Though they might be in some cases - e.g. the sphere.)</p>
|
88,199 | <p>Is there a function that would satisfy the following conditions?:</p>
<p>$\forall x \in X, x = f(f(x))$ and $x \not= f(x)$,</p>
<p>where the set $X$ is the set of all triplets $(x_1,x_2,x_3)$ with $x_i \in \{0,1,\ldots,255\}$.</p>
<p>I would like to find a function that will have as an input RGB color values (triplets) and return the original color after two applications of the function.</p>
| Community | -1 | <p>I think $f(x,y,z) = (x + 1, y, z)$ if $x$ is even and $f(x,y,z) = (x-1, y, z)$ when $x$ is odd does the trick. Suppose $x$ is even. Then $x+1$ is odd so
$$f(f(x,y,z)) = f(x+1,y,z) = (x,yz)$$
and similarly when $x$ is odd. That the function has no fixed points is obvious because $x \neq x+1$ and $x \neq x-1$ for all integers $x$. Moreover, if $x \geq 0$ is odd then $x\geq 1$ and if $x \leq 255$ is even then $x \leq 254$ so $f$ maps your set into itself.</p>
<p>EDIT: ofer's comment gives a function that is faster to compute and less complicated. So you should probably use that.</p>
|
88,199 | <p>Is there a function that would satisfy the following conditions?:</p>
<p>$\forall x \in X, x = f(f(x))$ and $x \not= f(x)$,</p>
<p>where the set $X$ is the set of all triplets $(x_1,x_2,x_3)$ with $x_i \in \{0,1,\ldots,255\}$.</p>
<p>I would like to find a function that will have as an input RGB color values (triplets) and return the original color after two applications of the function.</p>
| Michael Hardy | 11,667 | <p>$$f(x) = 5-x$$</p>
<p>$$f(x) = \frac{16}{x}$$</p>
<p>There are tons, scads, bushels, truckloads, imponderable multitudes, of functions like this. You could fill a fathomless abyss up to <em>here</em> with them.</p>
<p>They're called "involutions".</p>
<p>(Except, technically, the ones that satisfy $f(x)=x$ are also involutions.)</p>
|
1,627,050 | <p>Let $f:[a,b]\rightarrow R$ be differentiable at $c\in [a,b]$. Show that for every $\epsilon >0$, there is a $\delta(\epsilon) >0$ s.t if $0<|x-y|<\delta(\epsilon)$ and $a\leq x \leq c\leq y \leq b$, then\
$$ |{\frac{f(x)-f(y)}{x-y}-f'(c)}|<\epsilon$$
I can only think of using triangle inequality, but it seems does not work. Can anyone help me?</p>
| DanielWainfleet | 254,665 | <p>$$\text {Let } d>0 \text { where } 0<|z-c|<d \implies |\frac {f(z)-f(c)}{z-c}-f'(c)|<e/3.$$ For $c-d<x\leq c\leq y<c+d)$ with $x\ne y$ we have $$f(x)=f(c)+(x-c)(f'(c)+e_x),$$ $$ f(y)=f(c)+(y-c)(f'(c)+e_y),$$ $$\text {where }\; |e_x|<e/3 \;\text { and }\; |e_y|<e/3.$$ $$\text {Therefore }\; f(x)-f(y)=(x-y)f'(c)+(x-c)e_x-(y-c)e_y=$$ $$=(x-y)(f'(c)+(x-c)(e_x-e_y)+e_y(x-y).$$ $$\text {Thus }\; |\frac {f(x)-f(y)}{x-y}-f'(c)|=|(e_x-e_y)\frac {(x-c)}{(x-y)}+e_y|\leq $$ $$\leq (|e_x|+|e_y|)\frac {|x-c|}{|x-y|}+|e_y|\leq (|e_x|+|e_y|)+|e_y|<e$$ $$\text {because }\; |x-c|/|x-y|\leq 1.$$</p>
|
516,544 | <p>The following is an <a href="http://placement.freshersworld.com/placement-papers/IBM/Placement-Paper-Whole-Testpaper-37851" rel="nofollow">aptitude problem (question no: 29-32)</a>, I am trying to solve:- </p>
<blockquote>
<p>Questions 29 - 32:</p>
<p>A, B, C, D, E and F are six positive integers such that</p>
<p>B + C + D + E = 4A</p>
<p>C + F = 3A</p>
<p>C + D + E = 2F</p>
<p>F = 2D</p>
<p>E + F = 2C + 1</p>
<p>If A is a prime number between 12 and 20, then</p>
<ol>
<li>The value of F is</li>
</ol>
<p>(A) 14</p>
<p>(B) 16</p>
<p>(C) 20</p>
<p>(D) 24</p>
<p>(E) 28</p>
<ol>
<li>Which of the following must be true?</li>
</ol>
<p>(A) D is the lowest integer and D = 14</p>
<p>(B) C is the greatest integer and C = 23</p>
<p>(C) B is the lowest integer and B = 12</p>
<p>(D) F is the greatest integer and F = 24</p>
<p>(E) A is the lowest integer and A = 13</p>
</blockquote>
<p>Now there are 5 equations to solve 6 variables. So I am at a loss on how should I start solving the problem? Any help from anybody is appreciated.</p>
| Mufasa | 49,003 | <p>F=2D implies F must be even.<br/>
E+F=2C+1 implies E+F must be odd, hence E must be odd (as F is even).<br/>
Let E=2q+1, hence 2q+1+F=2C+1, hence 2q+2D=2C, hence C=q+D<br/>
C+D+E=2F, hence q+D+D+2q+1=2(2D), hence 3q+1=2D, hence q must be odd.<br/>
Let q=2r+1, hence 6r+4=2D, hence D=3r+2.<br/></p>
<p>Summarising so far, we get:<br/>
D=3r+2<br/>
F=2D=6r+4<br/>
E=2q+1=4r+3<br/>
C=q+D=5r+3<br/></p>
<p>C+F=3A, hence 5r+3+6r+4=3A, hence 11r+7=3A, hence 11(r-1)+18=3A, hence r-1 must be a multiple of 3.<br/>
Let r-1=3s, hence 11(3s+1)+7=3A, hence 33s+18=3A, hence A=11s+6.<br/></p>
<p>Summarising so far, we get:<br/>
D=3r+2=9s+5<br/>
F=6r+4=18s+10<br/>
E=4r+3=12s+7<br/>
C=5r+3=15s+8<br/>
A=11s+6<br/></p>
<p>Finally B+C+D+E=4A, hence B+15s+8+9s+5+12s+7=4(11s+6)<br/></p>
<p>Hopefully you can finish off from here.</p>
|
1,677,035 | <p>I'm new to this website so I apologize in advance if what I'm going to ask isn't meant to be posted here.</p>
<p>A bit of background though: I haven't been to school in 6 years and the last level I've graduated was Grade 7 due to financial problems, as well as my mom frequently being in and out of the hospital. I am now 18 and I wish to go to college as soon as I can, but I need to be caught up on all the math I've missed (I have been studying these past few years with what's available on the internet, but I don't think it's enough).</p>
<p>So my question is, are there any good, easy to understand, high school math books suited for my situation? I learn better with a teacher who can explain the lesson, but since I don't have one I'd prefer books that aren't too difficult, but at the same time provide everything necessary for high school level math and more. I used to be a bright student so I'm sure I can do this on my own with the right material.</p>
<p>Again, if this question isn't meant to be on this site I'd be more than willing to delete it asap! That's all. Thank you for reading. :)</p>
| G987 | 686,625 | <p>Currently, these two websites are down for me. (If anyone can check, please let me know if it isn't.) I will remove the links in a few days if it is still down:
<a href="http://math30.ca" rel="nofollow noreferrer">math30.ca</a>
<a href="http://math20.ca" rel="nofollow noreferrer">math20.ca</a></p>
<p>The Math 30 website corresponds to the Alberta (Canada) Grade 12 Precalculus curriculum, and Math 20 website similarly corresponds to the Grade 11 math curriculum. Both these resources were immensely useful for understanding concepts while I was taking those classes, as they provided animations and interactive slides for demonstrating math concepts. </p>
<p>I did find a pdf version of a part of the booklet available <a href="https://www.ilearnacademy.net/uploads/3/9/2/2/3922443/math30-1_workbook_one.pdf" rel="nofollow noreferrer">here</a>, and a playlist for the <a href="https://www.youtube.com/playlist?list=PLZsJtiruHkwqdczctQQerNF5SaUwwTX3n" rel="nofollow noreferrer">youtube videos</a> are for someone's 30-1 (university-stream math) class, without the interactivity. </p>
<p>In my own searchings in the past, I have also found <a href="https://learn.lboro.ac.uk/archive/olmp/olmp_resources/pages/wbooks_fulllist.html" rel="nofollow noreferrer">HELM booklets from the UK</a> to be a possible source of self-study material, as they cover needed portions for engineering math from the ground up. As Carser has mentioned, <a href="https://www.khanacademy.org/math/math-1-2-3" rel="nofollow noreferrer">Khan Academy high school math</a> offers a lot of videos that can be self-studied, and I found Khan Academy useful up to intermediate-level university math as well (Differential Equations, Vector Calculus and the like). HELM could complement KA well by providing a "workbook" of sorts to practice from.</p>
<p>A huge part of my degree was spent learning about how to study as well, so here are also some tips that seem no-brainer but were good for me: </p>
<ul>
<li>Google the concepts when you don't know them, and watch the YouTube tutorials on the similar problems</li>
<li>Don't be distracted by flashy videos while you're on YouTube</li>
<li>Try reworking the youtube examples after it's finished, and if there's points where you're stuck, go back and revisit the concept</li>
<li>Start from the simpler questions, and work your way up in difficulty. </li>
<li>Don't give up when you encounter difficulties -there will be a lot of those times- but do remember to take breaks when you feel that sufficient progress has been made and something was too difficult to understand in one sitting</li>
<li>Keep up your energy for the long run, as self-studying like this isn't done like a short sprint</li>
<li>Consider enrolling in a class to get support from peers. </li>
</ul>
<p>Hope this helps!</p>
|
1,643,201 | <p>The spectrum-functor
$$
\operatorname{Spec}: \mathbf{cRng}^{op}\to \mathbf{Set}
$$
sends a (commutative unital) ring $R$ to the set $\operatorname{Spec}(R)=\{\mathfrak{p}\mid \mathfrak{p} \mbox{ is a prime ideal of R}\}$ and a morpshim $f:S\to R$ to the map $\operatorname{Spec}(R)\to \operatorname{Spec}(S)$ with $\mathfrak{p}\mapsto f^{-1}(\mathfrak{p})$. Does this functor send pullback squares
\begin{eqnarray}
S\times_R T&\to& T\\
\downarrow && \downarrow\\
S&\to& R
\end{eqnarray}
of (commutative unital) rings to pushout squares
\begin{eqnarray}
\operatorname{Spec}(R)&\to& \operatorname{Spec}(T)\\
\downarrow && \downarrow\\
\operatorname{Spec}(S)&\to& \operatorname{Spec}(S\times_R T)
\end{eqnarray}
of sets? Put in other words, does the functor $\operatorname{Spec}$ from above preserve pushouts?</p>
| Joshua | 430,661 | <p>Find the lower bound and upper bound that is a multiple of 7! so for a three digit integer that is a multiple of 7, 100 is the lowest possible three digit integer and 999 is the highest, however, neither of these are a multiple of 7!
So starting at 100 find a multiple of 7! (105) is the product of (15*7), that is your lower bound. Now find your upper bound,starting at 999; you find, (994) which is a product of (142*7)! Now use a THEOREM (IF m and n are integers, and m <= n,then there are n-m +1 integers from m to n, inclusive). This means take the integer 142 from the product of 142*7 from your upper bound, and subtract that from your lower bound product which is the integer 15 from the product 15*7. You will get your event (E) so E = 142 - 15 + 1 = 128. So E = 128, that is how many three digit integers are a multiple of 7! So now take your E and put that over your sample space (S) where S = all possible outcomes in a random outcomes. so (s) = 9 * 10 * 10 where the first digit of your three digit number is a natural number between 1-9, the second digit can be any natural number from 0 to 9, and the third digit number can also be a natural number between 0 and 9. So you get (S) = 9 * 10 * 10 = 900. So E/S gives you your ratio or probability of number of three digit integers that are a multiple of 7. You can use this for any three digit multiple, or four digit multiple ... (N) multiple where N is a non negative member of Z. </p>
|
1,386,682 | <p>How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?</p>
<p>I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>And that I could not get out, can anyone help me?</p>
| Eclipse Sun | 119,490 | <p>Since the modulus of $\dfrac{\bar z^2}{z}$ is $|z|$, the limit is $0$.</p>
|
1,386,682 | <p>How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?</p>
<p>I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>And that I could not get out, can anyone help me?</p>
| Community | -1 | <p>Let $z=re^{i\theta}$. Then this equals:</p>
<p>$$\lim_{r\to 0} \frac{(re^{-i\theta})^2}{re^{i\theta}}=\lim_{r\to 0} \frac{r^2e^{-2i\theta}}{re^{i\theta}}=\lim_{r\to 0} r e^{-3i\theta}=0$$</p>
|
2,113,777 | <p>I have the following IVP (Initial value problem, Cauchy-Problem), and I do not know how to solve this.</p>
<p>$$y'=e^{-x}-\frac{y}{x} \qquad \qquad y(1)=2$$</p>
<p>I hope you can help me, cause I really do not know how to start.</p>
<p>Thank you! :)</p>
| MatheMagic | 397,530 | <p>\begin{align}
\frac{dy}{dx}+\frac yx&=e^{-x}\\
&\text{IF:}e^{\int\frac1xdx}=x\\
&yx=\int xe^{-x}dx\\
&=-(x+1)e^{-x}+c\\
&\text{putting the initial value $x=1,y=2$}\\
&2=-2e^{-1}+c\\
&c=2+\frac2e\\
&\text{our Ans is: $yx=-(x+1)e^{-x}+2+\frac2e$}
\end{align}</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| Brethlosze | 386,077 | <p>$$
6-2√3 \sim 3√2-2\\
8 \sim 3√2 +2√3 \\
64 \sim 30+12√6\\
34 \sim 12√6\\
17 \sim 6√6\\
289 \sim 36 \cdot 6\\
289 > 216
$$</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| TonyK | 1,508 | <p>$6-2\sqrt 3 \gtrless 3\sqrt 2-2$ </p>
<p>Rearrange: $8 \gtrless 3\sqrt 2 + 2\sqrt 3$ </p>
<p>Square: $64 \gtrless 30+12\sqrt 6$</p>
<p>Rearrange: $34 \gtrless 12\sqrt 6$</p>
<p>Square: $1156 \gtrless 864$</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| Oscar Lanzi | 248,217 | <p>Define</p>
<p>$a=6-2\sqrt 3>0$</p>
<p>$b=3\sqrt 2-2>0$</p>
<p>$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$</p>
<p>$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6)
< 60 < 64$
because $6=2×3 < (5/2)^2$</p>
<p>$a-b > 8-8=0, a>b$</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| Will Jagy | 10,400 | <p>Using simple continued fractions for $\sqrt {12}$
and $\sqrt {18}.$ Worth learning the general technique...Method described by Prof. Lubin at <a href="https://math.stackexchange.com/questions/2215918/continued-fraction-of-sqrt67-4/2216011#2216011">Continued fraction of $\sqrt{67} - 4$</a> </p>
<p>$$ \sqrt { 12} = 3 + \frac{ \sqrt {12} - 3 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{3 } = 2 + \frac{ \sqrt {12} - 3 }{3 } $$
$$ \frac{ 3 }{ \sqrt {12} - 3 } = \frac{ \sqrt {12} + 3 }{1 } = 6 + \frac{ \sqrt {12} - 3 }{1 } $$ </p>
<p>Simple continued fraction tableau:<br>
$$
\begin{array}{cccccccccc}
& & 3 & & 2 & & 6 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 7 }{ 2 } \\
\\
& 1 & & -3 & & 1
\end{array}
$$ </p>
<p>$$
\begin{array}{cccc}
\frac{ 1 }{ 0 } & 1^2 - 12 \cdot 0^2 = 1 & \mbox{digit} & 3 \\
\frac{ 3 }{ 1 } & 3^2 - 12 \cdot 1^2 = -3 & \mbox{digit} & 2 \\
\frac{ 7 }{ 2 } & 7^2 - 12 \cdot 2^2 = 1 & \mbox{digit} & 6 \\
\end{array}
$$ </p>
<p>Continued fraction convergents alternate above and below the irrational number, we get
$$ \frac{ 3 }{ 1 } < \sqrt {12} < \frac{ 7 }{ 2 } $$
Your first number was $6 - \sqrt {12},$
$$ 3 > 6 - \sqrt {12} > \frac{ 5 }{ 2 } $$
$$ \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 $$</p>
<p>Next 18......................========================================</p>
<p>$$ \sqrt { 18} = 4 + \frac{ \sqrt {18} - 4 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{2 } = 4 + \frac{ \sqrt {18} - 4 }{2 } $$
$$ \frac{ 2 }{ \sqrt {18} - 4 } = \frac{ \sqrt {18} + 4 }{1 } = 8 + \frac{ \sqrt {18} - 4 }{1 } $$ </p>
<p>Simple continued fraction tableau:<br>
$$
\begin{array}{cccccccccc}
& & 4 & & 4 & & 8 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 17 }{ 4 } \\
\\
& 1 & & -2 & & 1
\end{array}
$$ </p>
<p>$$
\begin{array}{cccc}
\frac{ 1 }{ 0 } & 1^2 - 18 \cdot 0^2 = 1 & \mbox{digit} & 4 \\
\frac{ 4 }{ 1 } & 4^2 - 18 \cdot 1^2 = -2 & \mbox{digit} & 4 \\
\frac{ 17 }{ 4 } & 17^2 - 18 \cdot 4^2 = 1 & \mbox{digit} & 8 \\
\end{array}
$$ </p>
<p>This time the number is $\sqrt {18} - 2.$</p>
<p>It is enough to use
$$ 2 < \sqrt {18} - 2 < \frac{9}{4} $$</p>
<p>$$ \color{red}{ 2 < \sqrt {18} - 2 < \frac{9}{4} < \frac{ 5 }{ 2 } < 6 - \sqrt {12} < 3 } $$</p>
|
192,125 | <p>Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<</p>
| André Nicolas | 6,312 | <p>We will assume that $x$ ranges over the reals $\ge 4$, to make sure that the square roots are real. Note that
$$\sqrt{x+4}-\sqrt{x-4}=\frac{(\sqrt{x+4}-\sqrt{x-4})(\sqrt{x+4}+\sqrt{x-4})}{\sqrt{x+4}+\sqrt{x-4}} =\frac{8}{\sqrt{x+4}+\sqrt{x-4}} .$$
For $x\ge 4$, $\sqrt{x+4}+\sqrt{x-4}\ge 2\sqrt{2}$. It follows that $\sqrt{x+4}-\sqrt{x-4}\le \dfrac{8}{2\sqrt{2}}=2\sqrt{2}$ for all $x\ge 4$. In particular, $\sqrt{x+4}-\sqrt{x-4}$ cannot be equal to $10$.</p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| lab bhattacharjee | 33,337 | <p>$$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$</p>
<p>$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$</p>
<p>$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ which can be $\cdots,-\frac{3\pi}2,\frac{\pi}2,\frac{5\pi}2,\cdots$</p>
<p>$$\text{Taking the '-' sign,} \sin x=2n\pi- \left(\frac\pi2-\cos x\right)\implies \sin x-\cos x=\frac{(4n-1)\pi}2$$ which can be $\cdots,-\frac{5\pi}2,-\frac{\pi}2,\frac{3\pi}2,\cdots$</p>
<p>Now let, $1=r\cos\theta,1=r\sin\theta$ where $r>0$</p>
<p>So, $(r\cos\theta)^2+(r\sin\theta)^2=1+1=2\implies r^2=2\implies r=\sqrt2$</p>
<p>$\sin x\pm\cos x=r\cos\theta\sin x\pm r\sin\theta\cos x=\sqrt2\sin(x\pm \theta)$</p>
<p>So, $-\sqrt2\le \sin x\pm\cos x\le \sqrt 2 $</p>
<p>Now, $\sqrt 2<1.5<\frac\pi2$ as $3<\pi$</p>
<p>$\implies -\sqrt 2>-\frac\pi2$</p>
<p>$\implies -\frac\pi2<-\sqrt2\le \sin x\pm\cos x\le \sqrt 2<\frac\pi2 $</p>
<p>Hence, there is no real soultion</p>
|
56,481 | <p>I don't know if the question trivial or not (if so I will delete it), but here is what I have:</p>
<p>I know that </p>
<pre><code>x x
(* x^2 *)
</code></pre>
<p>Looking at the full form gives</p>
<pre><code>FullForm[Times[x, x]]
(* Power[x, 2] *)
</code></pre>
<p>It is clear that the <code>Times</code> head has been changed to <code>Power</code> head.
I just wonder how does this happen. Can I control this behavior to stop this conversion between heads?</p>
| Mr.Wizard | 121 | <p>We can use this function to see that the conversion is not made during parsing:</p>
<pre><code>parseString[s_String, prep : (True | False) : True] :=
FrontEndExecute[FrontEnd`UndocumentedTestFEParserPacket[s, prep]]
parseString["x x"]
</code></pre>
<blockquote>
<pre><code>{BoxData[RowBox[{"x", "x"}]], StandardForm}
</code></pre>
</blockquote>
<p>We can use this to see that the conversion does not take place while converting boxes to <a href="http://reference.wolfram.com/mathematica/ref/StandardForm.html" rel="nofollow noreferrer"><code>StandardForm</code></a>:</p>
<pre><code>ToExpression[RowBox[{"x", "x"}], StandardForm, HoldComplete] // FullForm
</code></pre>
<blockquote>
<pre><code>HoldComplete[Times[x,x]]
</code></pre>
</blockquote>
<p>We can see that the rule is applied as part of the normal evaluation sequence:</p>
<pre><code>Times[x, x] // TracePrint
</code></pre>
<blockquote>
<p>x x</p>
<p>Times</p>
<p>x</p>
<p>x</p>
<p>(x^2)</p>
<p>Power</p>
<p>x</p>
<p>2</p>
</blockquote>
<p>You cannot generally "stop this conversion" but you can temporarily disable the rules for <code>Times</code> using <a href="http://reference.wolfram.com/mathematica/ref/Block.html" rel="nofollow noreferrer"><code>Block</code></a>:</p>
<pre><code>Block[{Times},
ToString[x x]
]
</code></pre>
<blockquote>
<pre><code>"Times[x, x]"
</code></pre>
</blockquote>
<p>You can also <a href="http://reference.wolfram.com/mathematica/ref/Hold.html" rel="nofollow noreferrer"><code>Hold</code></a> or <a href="http://reference.wolfram.com/mathematica/ref/HoldComplete.html" rel="nofollow noreferrer"><code>HoldComplete</code></a> expressions and apply your own rules.</p>
<p>Neither of these methods lets <code>Times</code> otherwise behave as desired. I have <a href="https://mathematica.stackexchange.com/q/40927/121">lamented this</a> problem myself with regard to <code>Subtract</code> and <code>Divide</code> as the "equivalent" forms they are converted into are neither equivalent nor as fast.</p>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| Syed | 81,355 | <p>Consider using <code>NestList</code> to see how the solution evolves with every step. The square root of <code>num</code> is being calculated starting at various initial values as shown on <a href="https://math.stackexchange.com/questions/861509/what-is-the-sum-that-the-square-root-button-on-calculator-does-so-i-can-do-it-wi">this</a> page.</p>
<pre><code>nsteps = 10;
num = 6;
evols = NestList[(Numerator@# + num Denominator@#)/(
Numerator@# + Denominator@#) &, #, nsteps] & /@ Range[1, 3, 0.5]
ListLinePlot[evols
, PlotRange -> All
, InterpolationOrder -> 2
]
</code></pre>
<p><a href="https://i.stack.imgur.com/JQx5d.png" rel="noreferrer"><img src="https://i.stack.imgur.com/JQx5d.png" alt="enter image description here" /></a></p>
<hr />
<p><em>EDIT</em></p>
<p>To get the numbers from the first answer on the linked page, start at <code>1/2</code> and use <code>2</code> as the number whose square root is needed.</p>
<pre><code>NestList[(Numerator@# + 2 Denominator@#)/(
Numerator@# + Denominator@#) &, 1/2, 5]
</code></pre>
<p><span class="math-container">$$\left\{\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},\frac{65}{46},\frac{157}{111}\right\}$$</span></p>
|
3,561,664 | <p>I did part of this question but am stuck and don't know how to continue</p>
<p>I let <span class="math-container">$x= 2k +1$</span></p>
<p>Also noticed that <span class="math-container">$x^3+x = x(x^2+1)$</span></p>
<p>therefore
<span class="math-container">$4m+2 = 2k+1((2k+1)^2+1)$</span></p>
<p>I simplified this and ended up with</p>
<p><span class="math-container">$4m+2 = 8k^3+12k^2+8k+2$</span></p>
<p>I don't know how to continue from and prove that <span class="math-container">$x^3+x$</span> has remainder 2 when divided by 4</p>
| Toby Mak | 285,313 | <p>Consider <span class="math-container">$f(x) =x^3+x-2$</span> instead. This can be factorised as <span class="math-container">$(x-1)(x^2+x+2)$</span> instead using the remainder theorem (<span class="math-container">$1^3+1-2 = 0$</span>).</p>
<p>If <span class="math-container">$x = 1 \pmod 2$</span>, then <span class="math-container">$x -1 \equiv 0 \pmod 2$</span> and <span class="math-container">$x^2+x+2 \equiv 1^2+1+2 \equiv 4 \equiv 0 \pmod 2$</span>. Therefore <span class="math-container">$x^3+x-2 \equiv 0 \pmod 4$</span>, so <span class="math-container">$x^3+x \equiv 2 \pmod 4$</span>.</p>
|
1,386,683 | <p>I posted early but got a very tough response.</p>
<p>Point $A = 2 + 0i$ and point $B = 2 + i2\sqrt{3}$ find the point $C$ $60$ degrees ($\pm$) such that Triangle $ABC$ is equilateral. </p>
<p>Okay, so I'll begin by converting into polar form:</p>
<p>$A = 2e^{2\pi i}$ and $B = 4e^{\frac{\pi}{3}i}$</p>
<p>$\overline{AB} = \sqrt{13}$</p>
<p>How should I find a point with length $\overline{BC} = \overline{AC} = \sqrt{13}$ and the sufficient angle?</p>
| mvw | 86,776 | <p>How about
$$
\lim_{z\to 0} \frac{\bar{z}^2}{z} =
\lim_{z\to 0} \frac{\bar{z}^2z^2}{z^3} =
\lim_{z\to 0} \frac{\lvert z\rvert^4}{z^3} =
\lim_{z\to 0} \frac{\lvert z\rvert^4}{\lvert z \rvert^3 e^{3i\phi(z)}}
= 0
$$</p>
|
170,240 | <p>I have the following function of ω</p>
<pre><code>f[ω_] := (2 Sqrt[Γ] (4*g2^2 + (κ1 - 2*I*ω) (κ2 - 2*I*ω)))/(4*g2^2 (Γ -
2*I*ω) + (4*
g1^2 + (Γ - 2*I*ω) (κ1 -
2*I*ω)) (κ2 - 2*I*ω))
</code></pre>
<p>And I wish to obtain the poles for the denominator of the function:</p>
<pre><code>wroots1 = x /. Solve[(Denominator[f[ω]] /. {ω -> x}) == 0, x]
</code></pre>
<p>The result is of the following:</p>
<pre><code>{-(1/6) I (Γ + κ1 + κ2) + (I (-16 (\
Γ + κ1 + κ2)^2 +
48 (4 g1^2 +
4 g2^2 + Γ κ1 + Γ...}
</code></pre>
<p>Basically a really long ugly solution. My goal is to obtain the conjugate of wroots1, however, when I do it straightforwardly:</p>
<pre><code>wroots2 =
Simplify[Conjugate[wroots1],
Assumptions -> {Γ ∈
Reals, κ1 ∈ Reals, κ2 ∈ Reals,
g1 ∈ Reals,
g2 ∈ Reals, ω ∈ Reals,
4 (-16 (Γ + κ1 + κ2)^2 +
48 (4 g1^2 +
4 g2^2 + Γ κ1 + (Γ + \
κ1) κ2))^3 +
4096 (-36 g1^2 (Γ + κ1 -
2 κ2) + (2 Γ - κ1 - \
κ2) (36 g2^2 + (Γ + κ1 -
2 κ2) (Γ -
2 κ1 + κ2)))^2 > 0}];
</code></pre>
<p>I am returned with:</p>
<pre><code>{1/48 I (8 (Γ + κ1 + κ2) +
8 2^(1/3) (-12 g1^2 -
12 g2^2 + (Γ + κ1 + κ2)^2 -
3 (κ1 κ2 + Γ (κ1 + \
κ2))) Conjugate[
1/(-36 g1^2 Γ + 72 g2^2 Γ +
2 Γ^3 - 36 g1^2 κ1 -
36 g2^2 κ1 - 3 Γ^2 κ1 -
3 Γ κ1^2 + 2 κ1^3 +
72 g1^2 κ2 - 36 g2^2 κ2 -
3 Γ^2 κ2 +
12 Γ κ1 κ2 -
3 κ1^2 κ2 - 3 Γ κ2^2 -
3 κ1 κ2^2 + 2 κ2^3...}
</code></pre>
<p>Clearly, Conjugate refuses to take the conjugate of said function. I tried doing</p>
<pre><code>wroots2 = wroots1 /. {I -> -I}
</code></pre>
<p>But that doesn't work as well. I'm at lost at what to do here and I could use any help I can get. Thank you very much in advance.</p>
| ulvi | 1,714 | <p>You can use <code>ComplexExpand</code>, but the cube root term (what I define as <code>cRoot</code> below) interferes with its logic.
The following could be a work-around (suppressing all output):</p>
<p><code>wroots2 =</code>
<code>FullSimplify[Conjugate[wroots1],
Assumptions -> {Γ ∈
Reals, κ1 ∈ Reals, κ2 ∈ Reals,
g1 ∈ Reals,
g2 ∈ Reals, ω ∈ Reals,
4 (-16 (Γ + κ1 + κ2)^2 +
48 (4 g1^2 +
4 g2^2 + Γ κ1 + (Γ + \
κ1) κ2))^3 +
4096 (-36 g1^2 (Γ + κ1 -
2 κ2) + (2 Γ - κ1 - \
κ2) (36 g2^2 + (Γ + κ1 -
2 κ2) (Γ -
2 κ1 + κ2)))^2 > 0}];</code></p>
<p><code>wroots2 = wroots2 /. (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \[Kappa]2) \
(36 g2^2 + (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) (\[CapitalGamma] -
2 \[Kappa]1 + \[Kappa]2)) +
1/64 \[Sqrt](4 (-16 (\[CapitalGamma] + \[Kappa]1 + \[Kappa]2)^2 \
+ 48 (4 g1^2 +
4 g2^2 + \[CapitalGamma] \[Kappa]1 + (\[CapitalGamma] \
+ \[Kappa]1) \[Kappa]2))^3 +
4096 (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \
\[Kappa]2) (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) (\[CapitalGamma] -
2 \[Kappa]1 + \[Kappa]2)))^2))^(1/3) -> cRoot</code></p>
<p><code>wroots2 = wroots2 /.
1/(-36 g1^2 (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \[Kappa]2) \
(36 g2^2 + (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) (\[CapitalGamma] -
2 \[Kappa]1 + \[Kappa]2)) +
1/64 \[Sqrt](4 (-16 (\[CapitalGamma] + \[Kappa]1 + \[Kappa]2)^2 \
+ 48 (4 g1^2 +
4 g2^2 + \[CapitalGamma] \[Kappa]1 + (\[CapitalGamma] \
+ \[Kappa]1) \[Kappa]2))^3 +
4096 (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \
\[Kappa]2) (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 -
2 \[Kappa]2) (\[CapitalGamma] -
2 \[Kappa]1 + \[Kappa]2)))^2))^(1/3) -> (1/cRoot)</code></p>
<p><code>ComplexExpand[wroots2]</code></p>
<p>You can then substitute the expression for <code>cRoot</code> back into the answer.</p>
|
2,668,616 | <p>I am fairly new at MAPLE and I'm having some trouble solving this ODE. </p>
<p>$$(t+1)\frac{dy}{dt}-2(t^2+t)y=\frac{e^{t^2}}{t+1}$$</p>
<p>My initial value problem is $$t>-1, y(0)=5$$</p>
<p>I put the equation in standard form and typed into maple</p>
<p><a href="https://i.stack.imgur.com/rctFl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rctFl.png" alt="What I entered in MAPLE"></a></p>
<p>However I am aware that when my initial value is $y(0)=5$ my equation becomes $4=0$ which is not possible. So I am confused on whether or not it is possible to find a general solution. </p>
| acer | 12,448 | <p>The image of your Maple code shows an italic <code>e</code>, which indicates that you may have tried to just type the letter "e" as the base of the natural logarithm. That would be committing a common usage mistake.</p>
<p>If entered correctly the terms like <code>exp(t^2)</code> would get displayed in 2D Output as an upright Roman "e" rather than with an italic "e" as the base.</p>
<p>In 1D plaintext Maple code the input would look like as follows (where I also show 1D output just for simplicity of my inlining the code here).</p>
<pre><code>restart;
eq := diff(y(t),t)*(t+1) - 2*(t^2+t)*y(t) = exp(t^2)/(t+1);
2
/d \ 2 exp(t )
eq := |-- y(t)| (t + 1) - 2 (t + t) y(t) = -------
\dt / t + 1
dsolve({eq, y(0)=5});
/ 1 \ 2
y(t) = |- ----- + 6| exp(t )
\ t + 1 /
normal(%);
2
(5 + 6 t) exp(t )
y(t) = -----------------
t + 1
</code></pre>
<p>To enter <code>exp(t^2)</code> while in (default) fancy 2D Input mode in a Maple Document, you have three choices:
1) Just type it in as <code>exp(t^2)</code>
2) Use the Expression palette in the GUI's left panel which has an item for inserting a typeset 2D e^blah .
3) Type the letters <code>exp</code> and then the Esc key, to get a popup of command-completion suggestions, one of which is in the typeset e^blah form.</p>
<p>The keyboard acceleration to invoke command-completion varies by platform, and on Linux the Esc key does it. See the Help system for details, or look <a href="https://www.maplesoft.com/support/help/Maple/view.aspx?path=worksheet%2Fdocumenting%2F2DMathShortcutKeys" rel="nofollow noreferrer">here</a>.</p>
|
2,736,426 | <p>Let's imagine a point in 3D coordinate such that its distance to the origin is <span class="math-container">$1 \text{ unit}$</span>.</p>
<p>The coordinates of that point have been given as <span class="math-container">$x = a$</span>, <span class="math-container">$y = b$</span>, and <span class="math-container">$z = c$</span>.</p>
<p>How can we calculate the angles made by the vector with each of the axes?</p>
| Hope | 426,849 | <p>I used a way finding the angle between two vectors.</p>
<pre><code>angle = aCos ( (V1.V2) / (|V1|.|V2|) )
</code></pre>
<p>When I want to calculate the angle of a point to x axis.</p>
<p>x Axis is</p>
<pre><code>V1 = 1.i + 0.j + 0.k
</code></pre>
<p>The point is </p>
<pre><code>V2 = a.i + b.j + c.k
</code></pre>
<p>This way I calculated correct solution.</p>
|
2,371,108 | <p>Cubic equations of the form $ax^3+bx^2+cx+d$ can be solved in various ways. Some are easy to easy to factor in a pair, for some the roots can be found out by trial-and-error, some are one-of-a-kind, some can be reduced to a quadratic equation. A compilation of all possible ways to solve cubic equations would be very helpful for students and learners. </p>
| J.G. | 56,861 | <p>As with quadratics, first we divide through to make the polynomial monic and then shift the unknown by a constant so the second highest power has zero coefficient, giving $x^3+px+q=0$. We now use Cardano's method. There exist complex numbers $u,\,v$ with $u+v=x,\,uv=-\frac{p}{3}$ (since if only you knew $x$ we'd just have to solve $t^2-xt-\frac{p}{3}=0$) so $u^3+v^3=x(x^2+p)=-q$ and $u^3v^3=-\frac{p^3}{27}$. Solving a quadratic gives $u^3,\,v^3$, so $u=u_0\omega^n,\,v=v_0\omega^{-n}$ say with $\omega=\exp\frac{2\pi i}{3},\,n\in\{ 0,\,1,\,2\}$. Summing gives three values for $x=u+v$.</p>
|
1,462,379 | <p>I have been given the task to compute $\int_{-1}^1 \sqrt{1-x^2} dx$ by means of calculus. We got the hint to substitute $x=\sin u$, but that only seems to make things more complicated:</p>
<p>$$\int_{-1}^1 \sqrt{1-x^2}dx = \int_{\arcsin-1}^{\arcsin1} \sqrt{1-\sin^2u} \frac{d\arcsin u}{du} du = \int_{\arcsin-1}^{\arcsin1} \sqrt{\frac{(1-\sin u)(1+\sin u)}{(1-u)(1+u)}} du.$$</p>
<p>I know that the answer should be $\frac\pi2$, because this is the area under one half of the unit circle, but how to arrive there by means of calculus is completely unclear to me. Could someone point me in the right direction?</p>
| Surb | 154,545 | <p><strong>Hint</strong></p>
<p>$$\int_{-1}^1\sqrt{1-x^2}dx=\int_{\arcsin(-1)}^{\arcsin(1)}\sqrt{1-\sin^2(x)}\cos(x)dx=...$$</p>
|
3,382,464 | <p>Let <span class="math-container">$g$</span> be a <strong>smooth</strong> Riemannian metric on the closed <span class="math-container">$n$</span>-dimensional unit disk <span class="math-container">$\mathbb{D}^n$</span>. Let <span class="math-container">$f$</span> be a harmonic function w.r.t <span class="math-container">$g$</span>.</p>
<blockquote>
<p>Is it true that <span class="math-container">$f$</span> must be real-analytic?</p>
</blockquote>
<p>I <em>think</em> that this is true if we assume that <span class="math-container">$g$</span> is real-analytic, but I am not sure. Is it true in that case? I would like to find a reference.</p>
<p>This should be related to whether or not the Riemannian laplacian <span class="math-container">$\Delta_g$</span> is "analytically hypoelliptic".</p>
| YiFan | 496,634 | <p>Per the suggestion of fleablood, I'm going to assume the question asks you to find the integer <span class="math-container">$n$</span> so that <span class="math-container">$n<a^6<n+1$</span>. You already obtained
<span class="math-container">$$a^7=2a^2-a+2,$$</span>
so we divide through on both sides by <span class="math-container">$a$</span> to get
<span class="math-container">$$a^6=2a-1+\frac2a.$$</span>
Since <span class="math-container">$a$</span> is positive, we may use the AM-GM inequality to get:
<span class="math-container">$$a^6=\left(2a+\frac2a\right)-1>2\sqrt{4}-1=3.$$</span>
Note that the inequality is strict because <span class="math-container">$a\neq 1$</span>. To show that <span class="math-container">$a^6<4$</span> is a bit harder. Note that the function <span class="math-container">$f(a)=a^5-a^3+a$</span> is a monotonically increasing function (if you know calculus, prove this by showing the derivative is everywhere positive). Furthermore you can check that <span class="math-container">$f(\sqrt[6]4)=f(\sqrt[3]2)$</span> is greater than <span class="math-container">$2$</span>, so that the value of <span class="math-container">$a$</span> satisfying <span class="math-container">$f(a)=2$</span> must be less than <span class="math-container">$\sqrt[6]4$</span>. In other words, <span class="math-container">$a^6<4$</span>.</p>
|
3,438,653 | <p>I have this thing written on my notes: let <span class="math-container">${x}, {y}\in\mathbb R^n$</span> be two distinct points, then the set <span class="math-container">$$\{ \lambda x + (1-\lambda){y}\;\lvert\; \lambda \in [0,1] \}$$</span> contains all the points on the line segment that connects <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.
I can't seem to understand why it is so, and why we need to require that <span class="math-container">$\lambda \in [0,1]$</span>.
Thanks for any clarification.</p>
| herb steinberg | 501,262 | <p>Any point on the line passing through <span class="math-container">$x$</span> and <span class="math-container">$y$</span> is expressed as a linear combination of these points. The requirement on <span class="math-container">$\lambda$</span> insures that the points are in between <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, rather than outside the segment.</p>
|
2,386,689 | <p>I know I can find solution for These equations using subtraction</p>
<p>$3x+2y=14.....(\text{i})$</p>
<p>$4x+3y=20.....(\text{ii})$</p>
<p>My question is can I divide $(\text{i})$ by $(\text{ii})$ to get values of $x$ and $y$ ?</p>
| hamam_Abdallah | 369,188 | <p>Your equations can be written as
$$3x=14-2y $$
$$4x=20-3y $$</p>
<p>$x=0$ yields to a contradiction.
we can assume $x\ne 0$.</p>
<p>by division, we get
$$\frac {3x}{4x}=\frac {14-2y}{20-3y}=\frac {3}{4} $$</p>
<p>thus
$$4 (14-2y)=3 (20-3y )$$
or
$$y=60-56$$
You can finish.</p>
|
3,996,218 | <p>Well I wanted to know whether or not <span class="math-container">$y = x^2 + x + 7$</span> is a quadratic equation since the general form is <span class="math-container">$ax^2 + bx + c = 0$</span> here the equation
<span class="math-container">$y=x^2+x+7$</span>. Isn't equal to zero so I'm a bit confused</p>
| Teshan | 505,730 | <p>It could be taken as both a relationship between <span class="math-container">$y$</span> and <span class="math-container">$x$</span> or as a quadratic equation, depending on the context of the problem.</p>
<p>Most of the time the form <span class="math-container">$y=ax^2+bx+c$</span> is used to indicate the relationship between <span class="math-container">$y$</span> and <span class="math-container">$x$</span>. This is called is a quadratic function. But if it is given that <span class="math-container">$y$</span> is a constant, then you could simply rewrite it as <span class="math-container">$ax^2+bx+(c-y)=0$</span> and this is a quadratic equation.</p>
<p>The difference between these two is that <span class="math-container">$y$</span> in a <strong>quadratic function</strong> could have any value depending on the value of <span class="math-container">$x$</span>. Whereas <span class="math-container">$y$</span> in a <strong>quadratic equation</strong> is independent of <span class="math-container">$x$</span>.</p>
<p>Hope that clears it up.</p>
|
92,983 | <p><strong>Does every polyhedron in $\mathbb{R}^3$ with $n$ triangular facets have a <em>topological</em> triangulation with complexity $O(n)$?</strong></p>
<p>Suppose $P$ is a non-convex polyhedron in $\mathbb{R}^3$ with $n$ triangular facets, possibly with positive genus. A <em>topological</em> triangulation of $P$ is a simplicial complex whose underlying space is the closure of the interior of $P$, such that every facet of $P$ is a cell in the complex. These boundary facets are true geometric triangles, but interior simplices may be arbitrarily bent and twisted. In the more standard <em>geometric</em> triangulations, every simplex is the convex hull of its vertices.</p>
<p>Results of <a href="http://www.cs.princeton.edu/~chazelle/pubs/BoundsSizeTetrahedral.pdf" rel="nofollow">Chazelle and Shouraboura</a> imply that every polyhedron has a geometric triangulation with complexity $O(n^2)$. Moreover, a classical construction of <a href="http://www.cs.princeton.edu/~chazelle/pubs/ConvexPartitionPolyhedra.pdf" rel="nofollow">Chazelle</a> implies that the $O(n^2)$ bound is is optimal in the worst case, even when the genus is zero.</p>
<p>But we can get tighter bounds for topological triangulations, at least for genus-zero polyhedra. If $P$ has genus zero, <a href="http://en.wikipedia.org/wiki/Steinitz%27s_theorem" rel="nofollow">Steinitz's theorem</a> implies that there is a <em>convex</em> polyhedron $Q$ that is combinatorially equivalent to $P$. <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085499/" rel="nofollow">Alexander's extension of the Schönflies theorem</a> implies that the interiors of $P$ and $Q$ are both homeomorphic to open balls. Thus, applying a suitable homeomorphism to a minimal <em>geometric</em> triangulation of $Q$ gives us a <em>topological</em> triangulation of $P$ with complexity $O(n)$. (Alternatively, we can triangulate $P$ by joining an arbitrary interior point to every facet.)</p>
<p>What makes the question tricky for higher-genus polyhedra is the possibility of knottedness; the topology of the interior of $P$ is not determined by its genus. Intuitively, the question is how knotted the interior of a polyhedron can be, as a function of the number of facets.</p>
<p>The following question may be equivalent: Let $K$ be a closed polygonal chain (or "stick knot") in $S^3$ with $n$ edges. Is there a <em>topological</em> triangulation of $S^3$ with complexity $O(n)$ that includes $K$ in its 1-skeleton? Again, if we insist on <em>geometric</em> triangulations, $\Theta(n^2)$ tetrahedra are always sufficient and sometimes necessary, even if $K$ is unknotted.</p>
<p><strong>Added for bounty (Apr 13):</strong> Partial results, subquadratic upper bounds, or references that imply this problem is open (or the crossing-number problem in my comment on the first answer) would be welcome.</p>
| Sam Nead | 1,650 | <p>The question in the comment was as follows.</p>
<blockquote>
<p>Is there an infinite family of (hyperbolic) stick knots whose crossing numbers are quadratic in the number of edges?</p>
</blockquote>
<p>The answer is yes. Suppose that $T$ is the $(p,q)$-torus knot, with $2 \leq p < q \leq 2p$. Wikipedia tells us that $T$ has crossing number $(q−1)p$ and has stick number $2q$. Taking $p$ and $q$ close together gives a non-hyperbolic example. Now change a few crossings randomly - that changes the stick number by a small constant and makes the knot hyperbolic, almost surely. If you'd rather, you can instead use <em>twisted torus knots</em>. See the paper "The simplest hyperbolic knots" or the paper "The next simplest hyperbolic knots". </p>
<p>However, neither of these obviously give $O(n^2)$ lower bounds for your original question -- these knots typically have very small hyperbolic volume and small triangulations; hence the name of the two papers I cited. </p>
<p>EDIT -
My previous sentence can be made even more concrete. Fixing $n$ there are torus knots with stick number $O(n)$, with crossing number $O(n^2)$ and whose complements only require $O(\log(n))$ tetrahedra to triangulate. See the last paragraph of Agol's answer to this question: <a href="https://mathoverflow.net/questions/46149/lower-bound-on-number-of-tetrahedra-needed-to-triangulate-a-knot-complement">Lower bound on number of tetrahedra needed to triangulate a knot complement</a></p>
<p>Of course, these minimal triangulation do not have the desired boundary patterns. But it does indicate that the statement "large crossing number implies large triangulation" fails rather badly. </p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Robin Chapman | 4,213 | <p>One big book on distributions is the first volume
of Hormander's <em>The Analysis of Linear Partial Differential Operators</em>.
This may not be the easiest book to read, but it is comprehensive
and a definitive reference.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| John D. Cook | 136 | <p>Robert Adams' <em>Sobolev Spaces</em>. Maybe not the best first book, but a very good second book.</p>
|
618,339 | <p>Could somebody please check my solution?</p>
<p>I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges or diverges.</p>
<p>Using the Comparison test:</p>
<p>Let $a_n = \frac{(1+\frac{1}{n})^n}{n^2},~ b_n=\frac{1}{n^2}$</p>
<p>Since $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}= \lim\limits_{n \rightarrow \infty} \frac{(1+\frac{1}{n})^n}{n^2} \frac{n^2}{1} = \lim\limits_{n \rightarrow \infty} (1+\frac{1}{n})^n = e$. </p>
<p>Since $0<e<\infty$, $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges.</p>
| mathematics2x2life | 79,043 | <p>Your solution is correct. However, the simpler trick implied by the problem is to notice the numerator. Recall that
$$
\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e
$$
Since all the terms are positive, we have
$$
0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{e}{n^2}= e \sum_{n=1}^\infty \frac{1}{n^2}
$$
By the $p$-test, the series on the right converges. Therefore, your original series converges. </p>
<p><strong>EDIT</strong> As suggested by user21820, it might be unclear what I was saying. The limit just produces me a number, $e$ to use to bound the original series. The solution 'works' because $\left(1+\frac{1}{n}\right)^n<e$ for any positive integer $n$. This is usually discussed when one learns the definition of $e$. So then each term on the right sum is larger than that of corresponding ones in the original. The use of $e$ was arbitrary (but suggested by the numerator). We could have used 'any' number, $\pi, e^2,4,10,\sqrt{17}$, so long as $\left(1+\frac{1}{n}\right)^n$ was smaller than our choice of number, say $x$, for all positive integer $n$ (or at least all but finitely many of them). Then we would have written
$$
0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{x}{n^2}= x \sum_{n=1}^\infty \frac{1}{n^2}
$$
for whatever larger number we chose. </p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Sage Stark | 745,622 | <p>First, by Fermat's Last Theorem, this equation has no integral solutions, and thus no prime solutions, for all <span class="math-container">$n\geq 3$</span>, so we only need to focus on the <span class="math-container">$n=2$</span> case.</p>
<p><strong>Case 1: <span class="math-container">$x, y, z>2$</span></strong></p>
<p>Because <span class="math-container">$x, y,$</span> and <span class="math-container">$z$</span> are and all primes greater than <span class="math-container">$2$</span>, they are trivially odd. Hence, <span class="math-container">$x^2$</span>, <span class="math-container">$y^2$</span>, and <span class="math-container">$z^2$</span> are all odd. Adding <span class="math-container">$x^2$</span> and <span class="math-container">$y^2$</span> then yields an even number, but <span class="math-container">$x^2 +y^2 =z^2$</span>, a contradiction.</p>
<p><strong>Case 2:</strong> <span class="math-container">$x$</span> or <span class="math-container">$y=2$</span></p>
<p>WLOG, let <span class="math-container">$x=2$</span>. Then, the equation yields <span class="math-container">$$2^2 +y^2=z^2.$$</span> Rearranging, <span class="math-container">$$y^2=z^2-2^2$$</span> <span class="math-container">$$\therefore y^2=(z-2)(z+2).$$</span> Because the prime factorization of <span class="math-container">$y^2$</span> is unique, the values <span class="math-container">$z-2$</span> and <span class="math-container">$z+2$</span> must be the products of prime factors (and possibly <span class="math-container">$1$</span>) of <span class="math-container">$y^2$</span>. However, because <span class="math-container">$y$</span> is the only prime factor of <span class="math-container">$y^2$</span>, <span class="math-container">$y=z-2=z+2$</span>, a contradiction.</p>
<p><strong>Case 3:</strong> <span class="math-container">$z=2$</span></p>
<p>The only lattice points that the curve <span class="math-container">$x^2 +y^2 =2^2$</span> passes through are <span class="math-container">$(2,0)$</span>, <span class="math-container">$(0,2)$</span>, <span class="math-container">$(-2,0)$</span>, and <span class="math-container">$(0,-2)$</span>. The number <span class="math-container">$0$</span> is composite, so the equation does not have any solutions. <span class="math-container">$\blacksquare$</span></p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Wolfgang Kais | 640,973 | <p>If <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> are primes with <span class="math-container">$x^n+y^n=z^n$</span> for an integer <span class="math-container">$n\ge 2$</span>, then they can't all be odd, so at least one of them must be even. Since <span class="math-container">$2$</span> is the only even prime, at least one of them must be <span class="math-container">$2$</span>. Since <span class="math-container">$2$</span> also is the smallest prime and <span class="math-container">$z$</span> must be greater than <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, <span class="math-container">$z$</span> can't be <span class="math-container">$2$</span>, so w.l.o.g. <span class="math-container">$x=2$</span>.</p>
<p>Now, since <span class="math-container">$z>y$</span> and <span class="math-container">$z\equiv{y}\pmod2$</span>, we have <span class="math-container">$z \ge y+2$</span> and so finally we get a contradiction (because <span class="math-container">$n\ge2$</span>):
<span class="math-container">$$
2^n=z^n-y^n\ge (y+2)^n-y^n=2^n+\sum_{k=1}^{n-1}\binom{n}{k}2^ky^{n-k}>2^n
$$</span></p>
|
3,837,548 | <p>The triple integral
<span class="math-container">$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=\zeta(3) \dots (1)$$</span>
is not separable in <span class="math-container">$x,y,z$</span> and the integral representation of reciprocal: <span class="math-container">$\frac{1}{1-xyz}=\int_{0}^{1} t^{-xyz} dt$</span> also doesn't help separation of the integrand. I want help to prove (1).</p>
| Angina Seng | 436,618 | <p>Expand out as a geometric progression. You get
<span class="math-container">$$\sum_{n=0}^\infty\int_0^1\int_0^1\int_0^1x^n y^nz^n\,dx\,dy\,dz.$$</span>
Now do the integral.</p>
|
3,837,548 | <p>The triple integral
<span class="math-container">$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=\zeta(3) \dots (1)$$</span>
is not separable in <span class="math-container">$x,y,z$</span> and the integral representation of reciprocal: <span class="math-container">$\frac{1}{1-xyz}=\int_{0}^{1} t^{-xyz} dt$</span> also doesn't help separation of the integrand. I want help to prove (1).</p>
| Z Ahmed | 671,540 | <p><span class="math-container">$$I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=
\int_{0}^{1} \int_{0}^{1} \frac{ln(1-yz)}{yz} dy dz =\int_{0}^{1} \int_{0}^{z} \frac{\ln(1-u)}{u} \frac{du dz}{z}, u=yz$$</span>
<span class="math-container">$$ \implies I=\int_{0}^{1}\int_{0}^{z}[1+u/2+u^2/3+u^3/4+....]\frac{ du dz}{z}= \int_{0}^{1} \sum_{k=1}^{\infty}\frac{z^{k-1}}{k^2} dz= \sum_{k=1}^{\infty} \frac{1}{k^3}=\zeta(3).$$</span></p>
|
4,101,974 | <p>I'm trying to understand what does this matrix operator norm means and what it does to matrix A. <span class="math-container">$${{\left\| A \right\|}_{1,\,\infty }}:={{\max }_{{{\left\| x \right\|}_{\infty }}=1}}{{\left\| Ax \right\|}_{1}}$$</span> Can somebody help with the explanation and maybe an example?</p>
| Lee Mosher | 26,501 | <p>Let me reword that sufficient condition in order to make it more explicit:</p>
<blockquote>
<p>... it suffices to know that the representation of <span class="math-container">$1$</span> by the empty word <em>is the</em> unique <em>representation of <span class="math-container">$1$</span> by a reduced word</em>.</p>
</blockquote>
|
935,454 | <p>Suppose we have the integral operator $T$ defined by</p>
<p>$$Tf(y) = \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}f(xy)\,dx,$$</p>
<p>where $f$ is assumed to be continuous and of polynomial growth at most (just to guarantee the integral is well-defined). If we are to inspect that the kernel of the operator, we would want to solve</p>
<p>$$0 = \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}f(xy)\,dx.$$</p>
<p>If $f$ were odd, this would be trivially zero so we would like to consider the case that $f$ is even. My hunch is that $f$ should be identically zero but I haven't been able to convincingly prove it to myself. The reason that I feel like it should be the zero function is that by scaling the Gaussian, we can make it arbitrarily close to $0$ or $1$. I'll sketch some thoughts of mine.</p>
<p>If $y=0$, then we would have that $0 = \sqrt{2\pi}f(0)$, forcing $f(0)=0$. Making use of even-ness and supposing instead that $y\neq 0$, we can make a change of variable to get</p>
<p>$$0 = \int_0^{\infty} e^{-\frac{x^2}{2y^2}}f(x)\,dx.$$</p>
<p>Since $f$ has polynomial growth at most, given a fixed $y$, for any $\varepsilon > 0$, there exists $R_y$ such that</p>
<p>$$\left|e^{-\frac{x^2}{2y^2}}f(x)\right| \le \frac{\varepsilon}{1+t^2}$$</p>
<p>for all $x > R_y$. Thus we can focus instead on the integral from $0$ to $R_y$ since the tail effectively integrates to zero:</p>
<p>$$0 = \int_0^{R_y}e^{-\frac{x^2}{2y^2}}f(x)\,dx.$$</p>
<p>Since $[0,R_y]$ is compact and $f$ is a continuous function, it can be approximated uniformly by (even) polynomials with constant term $0$ by Stone-Weierstrass, i.e.</p>
<p>$$f(x) = \lim_n p_n(x),$$</p>
<p>where $p_n(x) = \sum\limits_{m=1}^n a_{m,y}x^{2m}$. Here the coefficients are tacitly dependent upon the upper bound (so I've made it explicit to prevent any confusion). From here, we have</p>
<p>$$0 = \int_0^{R_y}e^{-\frac{x^2}{2y^2}}\lim_n p_n(x)\,dx.$$</p>
<p>However since the convergence is uniform, we can commute limit and integral to get that</p>
<p>$$0 = \lim_n\sum_{m=0}^n a_{m,y}\int_0^{R_y}e^{-\frac{x^2}{2y^2}}x^{2m}\,dx.$$</p>
<p>Making use of a change of variable, this gives</p>
<p>$$0 = \lim_n\sum_{m=0}^n a_{m,y}y^{2m+1}\int_0^{\frac{R_y}{y}} e^{-\frac{x^2}{2}}x^{2m}\,dx.$$</p>
<p>I would like to be able to say that the coefficients must be zero but this is pretty messy at this point. Does anyone have any clue as to how to proceed? Or is there a better way to do this? (I would like to avoid Fourier transform-based or Weierstrass transform-based arguments.)</p>
| Jack D'Aurizio | 44,121 | <p>If you are able to prove that, given your constraints, $Tf$ is an analytic function, we have:
$$(Tf)^{(2n+1)}(0) = 0,\qquad (Tf)^{(2n)}(0) = 2^{n+1/2}\cdot\Gamma(n+1/2)\cdot f^{(2n)}(0)$$
hence $Tf\equiv 0$ implies that all the derivatives of $f$ of even order in the origin must vanish. Assuming that $f$ can be well-approximated by analytic functions over larger and larger neighbourhoods of the origin, we have that $Tf\equiv 0$ implies that $f$ is an odd function. We cannot hope in more than this since when $f(x)=x$, $Tf\equiv 0$.</p>
|
3,491,816 | <p>Find the min and max values of the function <span class="math-container">$$f(x,y)=10y^2-4x^2$$</span> with the constraint <span class="math-container">$$g(x,y)=x^4+y^4=1$$</span>
I have done the following working;
<span class="math-container">$$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$</span><span class="math-container">$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$</span><span class="math-container">$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$</span><span class="math-container">$$\lambda x^3=2\\\lambda y^3=5$$</span> My question is how can I find the value of lambda for the constraint to hold.</p>
| Martund | 609,343 | <p>There are errors in both of your equations.
<span class="math-container">$$x(2+\lambda x^2)=0$$</span>
<span class="math-container">$$y(5-\lambda y^2)=0$$</span>
Now, if <span class="math-container">$\lambda\ge 0$</span>, from the first equation we get <span class="math-container">$x=0$</span>. If <span class="math-container">$\lambda\le 0$</span>, from the second equation we get, <span class="math-container">$y=0$</span>. Hence, at optimum points, atleast one of the coordinate is <span class="math-container">$0$</span>. So, required optimum points are <span class="math-container">$(\pm1,0)$</span> and <span class="math-container">$(0,\pm1)$</span> on putting in the constraint.</p>
|
1,392,209 | <blockquote>
<p>Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$</p>
</blockquote>
<p>My attempt </p>
<p>So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$</p>
<p>$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$</p>
<p>Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$</p>
<p>$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$</p>
<p>Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$</p>
<p>Thanks for any help. </p>
| Paramanand Singh | 72,031 | <p>It is done much simpler in the following manner with just one application of LHR.
\begin{align}
L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{1}{\tan^{2}x}\right)\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{2}\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot\frac{x^{2}}{\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \frac{\tan x + x}{x}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \lim_{x \to 0}\left(\frac{\tan x}{x} + 1\right)\notag\\
&= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (via LHR)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\
&= \frac{2}{3}\notag
\end{align}</p>
|
3,652,518 | <p>I am no mathematician but have studied mathematics some 20 years ago. So I know basics of number theory but have lost the skills to solve problems. </p>
<p>I was wondering if the equation <span class="math-container">$3ax^2 + (3a^2+6ac)x-c^3=0$</span> in which <span class="math-container">$a$</span> and <span class="math-container">$c$</span> are arbitrary positive integers and <span class="math-container">$c$</span> is divisible by 6 has an integer solution. </p>
<p>So we need to show there is either no <span class="math-container">$a$</span>, <span class="math-container">$c$</span> and <span class="math-container">$x$</span> all positive that satisfy the equation, or one counter example exist. </p>
<p>Rational root theorem didn't help me. </p>
<p>Thanks. </p>
| achille hui | 59,379 | <p>Multiply <span class="math-container">$x + x^2 + \cdots + x^n$</span> by <span class="math-container">$1-x$</span> and rearrange terms, you get
<span class="math-container">$$\begin{array}{c}
x &+& \color{red}{x^2} &+& \color{green}{x^3} &+& \cdots &+&\color{blue}{x^n}\\
&-& \color{red}{x^2} &-& \color{green}{x^3} &-& \cdots &-&\color{blue}{x^n} &-& x^{n+1}
\end{array}$$</span>
Notice the massive cancellation of terms, the result simplifies to</p>
<p><span class="math-container">$$(1-x)(x + x^2 + \cdots + x^n) = x - x^{n+1} = x(1-x^n)$$</span>
Divide both sides by <span class="math-container">$1-x$</span>, you get what you want to show.</p>
|
1,680,862 | <p>I am attempting to find the expected value and variance of the random variable $X$ analytically (in addition to a decimal answer). $X$ is the random variable <code>expression(100)[-1]</code> where <code>expression</code> is defined by:</p>
<pre><code>def meander(n):
x = [0]
for t in range(n):
x.append(x[-1] + 3*random.random())
return x
</code></pre>
<p>For those that do not understand the Python, $X$ is essentially the sum of a sequence of $100$ values with value of <code>3*random.random()</code>, where <code>random.random()</code> is uniformly distributed on $[0,1)$.</p>
<p>I am almost certain that I will need to apply the concepts of:
$$\text{mean}\left(\bar{X}\right)=E\left(\frac{1}{n}\left(X_1+X_2+...+X_n\right)\right)=E\left(X\right)$$
$$\text{and}$$
$$\text{var}\left(\bar{X}\right)=var\left(\frac{1}{n}\left(X_1+X_2+...+X_n\right)\right)=\frac{1}{n}\text{var}\left(X\right)$$
$$\text{where, }\bar{X}=\frac{1}{n}\left(X_1+X_2+...+X_n\right)$$</p>
<p>I am having difficulty understand how I should be plugging in this equation and representing it symbolically, let alone calculating it. I created a simulation in order to better understand the distribution of the data (in addition to getting an estimate of the expected value) and it seems to be a Gaussian distribution (<a href="https://i.stack.imgur.com/1NmfF.png" rel="nofollow noreferrer">histogram of distribution after 100,000 trials</a>). The simulation suggests an estimated expected value of $150.038527551$.</p>
<p>These solutions will culminate in the usage of the Central Limit Theorem in finding an analytical expression that approximates the pdf of $X$.</p>
<p>Any guidance or help to point me in the right direction would be very much appreciated!</p>
| Umang Gupta | 188,747 | <p>Clement C has already explained the theoritical part. Here is my programmatic explanation. </p>
<p>Let the variables be $X = \sum_1^k x_k$ where $x_k$ is $U(0,1)$</p>
<p>You can calculate mean & variance in two ways :
One is straight forward. Compute X and use following formulae.
$$mean(X) = \sum_1^nX$$
$$var(X) = \sum_1^nX^2/n -mean(x)^2 $$</p>
<p>However since x is independent, we can use following formulate
$$ mean (X) = mean(\sum_1^k x_k) = \sum_1^k(mean(x_k)) $$
$$ var(X) = var(\sum_1^k x_k) = \sum_1^k(var(x_k)) $$</p>
<p><code>
import random
import math
def meander(n):</p>
<h1>x is 2-D array of values</h1>
<h1>x[i][j]; j corresponds to k & i is the ith sample</h1>
<pre><code> x = [ [ 3*random.random() for i in range(0,100)] for i in range(0,n)]
</code></pre>
<h1>evaluate X variables by sum over j</h1>
<pre><code> X = [ sum(i) for i in x]
</code></pre>
<h1>calculate mean & variance of X</h1>
<pre><code> X_ = sum(X)/n
var_X_ = sum([i*i for i in X])/n - X_*X_
</code></pre>
<h1>by formulae</h1>
<h1>calculate mean of each X_k ; i.e sum over i</h1>
<pre><code> mean_xk = []
var_xk = []
for i in range(0,100):
s = 0
s_2 = 0
for j in range(0,n):
s = s+x[j][i]
s_2 = s_2+x[j][i]*x[j][i]
mean_k = s/n
mean_xk.append(mean_k)
sum_of_square = s_2
var_xk.append( sum_of_square/n - mean_k*mean_k)
formula_X_ = sum(mean_xk)
formula_var_X_ = sum(var_xk)
return [X_, var_X_, formula_X_, formula_var_X_]
</code></pre>
<p>print meander(100000)
</code></p>
|
947,254 | <p>The problem is part (b):</p>
<p><b>1.4.7.</b> A pair of dice is cast until either the sum of seven or eigh appears.</p>
<p> <b>(a)</b> Show that the probability of a seven before an eight is 6/11.</p>
<p> <b>(b)</b> Next, this pair of dice is cast until a seven appears twice or until each of a six and eight has appeared at least once. Show that the probability of the six and eight occurring before two sevens is 0.546.</p>
<p>I would like to try to solve this problem using Markov chains, but I'm encountering a dilemma. To calculate the probability, I would need to multiply down the branches that lead to a terminating state, and then sum all of those branches. But I have loops in my diagram, so I'm not sure how to account for the fact that I could remain in a state for an indefinite number of rolls:</p>
<p>[I only drew the branches corresponding to rolling a 6, but there are of course the two other branches (and sub-branches) for rolling a 7 or 8.]</p>
<p><img src="https://i.stack.imgur.com/Cy6S6.jpg" alt="enter image description here"></p>
<p>If that's hard to read, here <a href="https://i.imgur.com/T8BE5ix.jpg" rel="nofollow noreferrer">is a higher resolution</a>. This is my chain of reasoning: We start out in a state of not having a 6, 7, or 8 yet. We could stay here indefinitely. Rolling a 6 takes us to the next state. We could also stay here indefinitely, or roll an 8 and get an accept state. Or we could roll a 7. At that state, we could roll another 7 and get an accept state or roll and 8 or indefinitely roll a 6 (or any other number). All of those probabilities are noted in the transitions.</p>
<p>How do I account for these possibilities?</p>
| awkward | 76,172 | <p>Others have already given some excellent answers to this problem, and the original post was over two years ago. Nevertheless, I would like to show how the problem can be solved by using an exponential generating function.</p>
<p>We know the last number rolled must be a 6 or an 8, and by symmetry we know these two cases are equally likely, so let's suppose the last number is an 8 in order to simplify the problem a bit. Then an acceptable sequence of rolls starts with <span class="math-container">$n >0$</span> rolls consisting of no 8's, at least one 6, and at most one 7, followed by a final 8. Let <span class="math-container">$a_n$</span> be the probability of rolling the initial sequence (not including the final 8), for <span class="math-container">$n \ge 0$</span>. Any acceptable initial sequence is the "labeled product" of </p>
<ul>
<li>any number of rolls which are not 6's, 7's, or 8's</li>
<li>at least one roll of 6</li>
<li>at most one roll of 7</li>
</ul>
<p>so the exponential generating function for <span class="math-container">$a_n$</span> is
<span class="math-container">$$\begin{align}
f(x) &= \sum_{n=0}^{\infty} \frac{1}{n!} a_n x^n \\
&= \left(1 + qx + \frac{q^2}{2!} x^2 + \frac{q^3}{3!} x^3 + \dots \right) \left( p_6 x + \frac{p_6^2}{2!} x^2 + \frac{p_6^3}{3!} x^3 + \dots \right) (1 + p_7x) \\
&= e^{qx} (e^{p_6 x} -1) (1+ p_7 x)
\end {align}$$</span>
where <span class="math-container">$q$</span> is the probability of rolling anything but a 6, 7, or 8, and <span class="math-container">$p_k$</span> is the probability of rolling a <span class="math-container">$k$</span>. The numerical values are <span class="math-container">$q = 20/36$</span>, <span class="math-container">$p_6 = 5/36$</span>, and <span class="math-container">$p_7 = 6/36$</span>. </p>
<p>The probability of rolling an acceptable initial sequence of length <span class="math-container">$n$</span> followed by an 8 is <span class="math-container">$a_n p_8$</span>, so the total probability of any acceptable initial sequence followed by an 8 is
<span class="math-container">$$p = \sum_{n=0}^{\infty} a_n p_8$$</span>
We can use the following trick to extract this sum from <span class="math-container">$f(x)$</span>. Since
<span class="math-container">$$n! = \int_0^{\infty} e^{-x} x^n \; dx$$</span>
we have
<span class="math-container">$$p = \sum_{n=0}^{\infty} a_n p_8 = \int_0^{\infty} f(x) \; e^{-x} \; p_8 \; dx = \int_0^{\infty} e^{qx} (e^{p_6 x} -1) (1+ p_7 x) \; e^{-x} \; p_8 \; dx $$</span>
Evaluating the integral yields
<span class="math-container">$$p = \frac{4225}{15488}$$</span>
which is the probability of an acceptable sequence of rolls ending in 8.</p>
<p>The answer to the original problem, where the final roll may be either a 6 or an 8, is then
<span class="math-container">$$2p = \frac{4225}{7744}$$</span></p>
|
804,483 | <p>The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?</p>
<p>$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$</p>
| xpaul | 66,420 | <p>For $I_2$, we can use a well-known result:
$$ \int_{-\infty }^{\infty } \frac{\sinh (ax)}{\sinh(bx)}dx=\frac{\pi}{b}\tan\frac{a\pi}{2b}. $$
Note $\sinh(ix)=\sin(x), \tanh(ix)=\tan(x)$. Thus
$$ \int_{-\infty }^{\infty } \frac{\sin (ax)}{\sinh(bx)}dx=\int_{-\infty }^{\infty } \frac{\sinh (iax)}{\sinh(bx)}dx=\frac{\pi}{b}\tanh\frac{a\pi}{2b}. $$
For $I_2$, define
$$ I_2(a)=\int_{-\infty }^{\infty } \frac{\sin^2(ax)}{x\sinh (x)}dx. $$
Then $I_2(0)=0$ and $I_2(1)=I_2$. Now
\begin{eqnarray}
I_2'(a)&=&\int_{-\infty }^{\infty } \frac{2\sin(ax)\cos(ax)}{\sinh (x)}dx\\
&=&\int_{-\infty }^{\infty } \frac{\sin(2ax)}{\sinh (x)}dx\\
&=&\pi\tanh(a\pi).
\end{eqnarray}
So
$$ I_2(1)=\int_0^1\pi\tan(a\pi)da=\ln(\cosh(a\pi)). $$</p>
|
3,689,513 | <p>I've been stuck on one of my homework numbers.
The number precise that the following equation is a non-linear equation of order 1 with x>0.</p>
<p><span class="math-container">$$y' + {{y\ln(y)}\over x}= xy$$</span></p>
<p>So far, I tried 2 different methods to solve them. As suggested by internet (link below): Bernoulli, and as an inexact one. However, I get stuck every time. </p>
<p><a href="https://www.symbolab.com/solver/ordinary-differential-equation-calculator/y'%2B%5Cfrac%7By%5Ccdot%20ln%5Cleft(y%5Cright)%7D%7Bx%7D%20%3D%20xy" rel="nofollow noreferrer">Bernoulli sample</a></p>
<p>**As a small side note, can someone explain to me the following from the link: how, by passing ln(y) to the right side, we get y^1+1 ? </p>
<p>For the Bernoulli equation, when I try to solve it manually, I get in a kind of infinite integral, no matter how many time I integrate.</p>
<p>Am I missing something?</p>
| user577215664 | 475,762 | <p><span class="math-container">$$y' + {{y\ln(y)}\over x}= xy$$</span>
<span class="math-container">$$\dfrac {y'}{y}+\dfrac {\ln y}{x}=x$$</span>
<span class="math-container">$$(\ln y)'+\dfrac {\ln y}{x}=x$$</span>
It's a linear first order DE. Substitute <span class="math-container">$u=\ln y$</span>
<span class="math-container">$$u'+\dfrac {u}{x}=x$$</span>
<span class="math-container">$$xu'+u=x^2$$</span>
<span class="math-container">$$(xu)'=x^2$$</span>
Integrate.
<span class="math-container">$$\ln y =\dfrac {x^2}3+\dfrac {c_1}{x}$$</span>
This answer dosent agree with Symbolab link but it agrees with <a href="https://www.wolframalpha.com/input/?i=y%27%20%2B%20%7B%7By%5Cln%28y%29%7D%2F%20x%7D%3D%20xy" rel="nofollow noreferrer">Wolfram Alpha's answer</a>
<span class="math-container">$$ y(x) =\exp {\left (\dfrac {x^2}3+\dfrac {c_1}{x}\right)}$$</span></p>
|
878,939 | <p>I have found the eigen vaues, I also know that you can find the eigenvectors through a Gausian Jordan.
-- x1, gauss jordan gives me rows(1 -1/3 ,, 0 0 ), so [a, b] = [1,3]
For vector x2, GJ gives (1 -2/5 ,, 0 0 ), I would assume [a,b] = [2,5], but why did they choose to go with [-2,-5]. I don't get it?</p>
<p>A bigger picture is on this webpage if needed;</p>
<p><a href="http://oi59.tinypic.com/2v7unw1.jpg" rel="nofollow noreferrer">http://oi59.tinypic.com/2v7unw1.jpg</a></p>
<p><img src="https://i.stack.imgur.com/5h7Ix.jpg" alt="enter image description here"></p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ By the Division with Remainder Algorithm </p>
<p>$\qquad\qquad\qquad\qquad n\, =\, 4q + r,\ \ \ 0 \le r < 4$</p>
<p>Now notice that $\ 2\mid n\iff 2\mid 4q+r\iff 2\mid r$</p>
<p>That is: $\,n\,$ is even iff its remainder mod $4$ is even iff $\,r \in \{0,2\}$</p>
<p><strong>Remark</strong> $ $ We began by writing $\,n\,$ in the form $\, 4q+r\,$ (vs. $\,5q+r),\,$ because that form is a <em>slight</em> <em>generalization</em> of the form that we seek, viz. $\,4q+r,\ r\in \{0,2\},\,$ already having the same modulus/quotient $(= 4)$ as the sought form, so leaving only further analysis of the remainder $\,r,\,$ which succumbs to a simple <em>parity</em> argument.</p>
|
3,417,001 | <blockquote>
<p>I am wondering if the ring of polynomials <span class="math-container">$F[x]$</span> with coefficients in the field <span class="math-container">$F$</span> is ever isomorphic to <span class="math-container">$\mathbb{Z}$</span> for some field <span class="math-container">$F$</span>. </p>
</blockquote>
<p>For all the fields I've examined, such as <span class="math-container">$F = \mathbb{C}$</span> or <span class="math-container">$F = \mathbb{R}$</span>, it's true that both <span class="math-container">$\mathbb{C}[x]$</span> and <span class="math-container">$\mathbb{R}[x]$</span> cannot be isomorphic to <span class="math-container">$\mathbb{Z}$</span>. Is it true that <span class="math-container">$F[x]$</span> is not isomorphic to <span class="math-container">$\mathbb{Z}$</span> for any field <span class="math-container">$F$</span> ? If so, is there a nice way to prove this in general? </p>
<p>Thanks! </p>
| Mark | 470,733 | <p>If there was such an isomorphism of rings then the additive groups would be isomorphic. However, the additive group of <span class="math-container">$\mathbb{Z}$</span> is cyclic. Now, can the additive group of <span class="math-container">$F[x]$</span> be cyclic? </p>
|
127,808 | <p>I have <a href="https://math.stackexchange.com/questions/356925/a-basis-of-the-symmetric-power-consisting-of-powers">asked this question on math.se</a>, but did not get an answer - I was quite surprised because I thought that lots of people must have though about this before:</p>
<p>Let $V$ be a complex vector space with basis $x_1,\ldots,x_n\in V$. Denote by $v_1\odot\cdots\odot v_k$ the image of $v_1\otimes\cdots\otimes v_k$ in the symmetric power $\newcommand{\Sym}{\mathrm{Sym}}\Sym^k(V)$. It is well-known that the Elements $v^{\odot k}$ for $v\in V$ generate this space (see, for instance, <a href="https://math.stackexchange.com/questions/137912/can-e-n-always-be-written-as-a-linear-combination-of-n-th-powers-of-linear-p/138411#138411">this answer on math.se</a>), so they must contain a basis. </p>
<p>In other words, let $N=\binom{n+k-1}k$, then there must be $v_1,\ldots,v_N\in V$ with
$$\mathrm{Sym}^k V = \mathbb Cv_1^{\odot k} \oplus \cdots \oplus \mathbb C v_N^{\odot k}.$$
I am looking for an explicit description of such a basis. Is such a description known? Is there maybe even a <em>"nice"</em> or somewhat <em>"natural"</em> choice for the $v_i$ as linear combinations of the $x_i$?</p>
| Peter Michor | 26,935 | <p>Can you work with polarization?
<BR>
$2 x_1x_2 = (x_1-x_2)^2 -x_1^2-x_2^2$.
<BR>
If $k=2$ then $(x_i+x_j)^2$ for $i\le j$ is already a basis.
<BR>
In general,
$(x_{i_1}+\dots +x_{i_k})^k$ for $1\le i_1\le\dots\le i_k\le n$ might do the job. </p>
|
4,408,772 | <p>I'm read about a Lienard System in Perko books, but I don't understand how this applies
<a href="https://i.stack.imgur.com/PFVeR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PFVeR.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/vgttf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vgttf.png" alt="enter image description here" /></a></p>
<p>I have managed to understand the proof, the shape of the trajectory is intuitively clear, doing an analysis of the vector field <span class="math-container">$F(x,y)=(y-F(x),-g(x))$</span>, b ut how do I formally justify it? Any suggestion? If anyone knows a book or article where this system and its generalizations are treated in detail, I would be very grateful if you could cite it.</p>
<p>Postscript: The result mentioned in Perko is as follows (Ordinary differential equations by Philip Hartman):
<a href="https://i.stack.imgur.com/HyaYl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HyaYl.png" alt="enter image description here" /></a></p>
| Futurologist | 357,211 | <p>To be honest, the theorem you quote is a bit out of context for my taste. For that reason, maybe one can just cook up their own argument (or maybe that argument is analogous to the theorem, I am not sure). So here is what I think, by assessing the situation quickly. I am going to assume we are dealing with the following system of ODEs:
<span class="math-container">\begin{align}
&\frac{dx}{dt} \,=\, y \, -\, F(x)\\
&\\
&\frac{dy}{dt} \,=\, -\, g(x)
\end{align}</span>
where <span class="math-container">$F(x)$</span> and <span class="math-container">$g(x)$</span> are continuously differentiable functions with enough continuous derivatives with respect to <span class="math-container">$x$</span>. I am assuming the graph of the function <span class="math-container">$y = F(x)$</span> is as depicted on your diagram, and I am going to assume that
<span class="math-container">\begin{align}
&g(x) < 0 \,\,\text{ for }\,\, x < 0\\
&g(x) = 0 \,\,\text{ for }\,\, x = 0\\
&g(x) > 0 \,\,\text{ for }\,\, x > 0\\
\end{align}</span>
So basically, to analyse this situation, I would consider the following four curves :
<span class="math-container">\begin{align}
L \, &=\, \big\{\, (x, y) \in \mathbb{R}^2 \,\, :\,\, x = 0\,\,\text{ and }\,\, y > 0\, \big\}\\
FG_{+} \, &= \, \big\{\, (x, y)\in \mathbb{R}^2\, \,:\, \, y = F(x) + g(x) \text{ where } x > 0\, \big\}\\
R \, &= \, \big\{\, (x, y)\in \mathbb{R}^2 \, \, :\,\, y = F(x) \text{ where } x > 0\, \big\}\\
FG_{-} \, &= \, \big\{\, (x, y)\in \mathbb{R}^2\, \,:\, \, y = F(x) - g(x) \text{ where } x > 0\, \big\}
\end{align}</span>
With these four curves as boundaries, define the two bigger domains
<span class="math-container">\begin{align}
&D_{+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, > \, F(x) \,\text{ and }\, x > 0\, \big\}\\
&D_{-} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, < \, F(x) \,\text{ and }\, x > 0\, \big\}
\end{align}</span>
and the four subdomains
<span class="math-container">\begin{align}
&D_{x+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, > \, F(x) + g(x) \text{ and } x > 0\, \big\}\\
&D_{y+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, F(x) + g(x)
\, >\, y > F(x)\, \big\}\\
&D_{y-} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, F(x) \, >\, y \,>\, F(x) - g(x) \text{ and } x > 0\, \big\}\\
&D_{x-} \,=\, \big\{ \, (x, y) \in \mathbb{R}^2\, \,:\, \, F(x) - g(x)\, >\, y \text{ and } x > 0\, \big\}
\end{align}</span>
where <span class="math-container">$D_{x+} \cup D_{y+} \subset D_+$</span> and <span class="math-container">$D_{x-} \cup D_{y-} \subset D_{-} \,$</span>.
Assume that for some interval <span class="math-container">$t \in (\alpha, \beta)$</span> a solution <span class="math-container">$\big(\,x(t), \, y(t)\,\big)$</span> of the original system of differential equations lies in the domain <span class="math-container">$D_{+}$</span>. Then, since <span class="math-container">$x > 0$</span>
<span class="math-container">$$\frac{dx}{dt} \, = \, y - F(x) \, > \, F(x) + g(x) - F(x) \, =\, g(x)
\, >\, 0$$</span> which means that <span class="math-container">$x=x(t)$</span> is a monotonously increasing function, and as such it has a well defined inverse <span class="math-container">$t = t(x)$</span> and therefore, <span class="math-container">$y=y(t)$</span> can be reparametrized as <span class="math-container">$y = y\big(t(x)\big) = y(x)$</span> and so the solution trajectory of the original system can be reparametrized as <span class="math-container">$\big(x, \, y(x)\big)$</span>, where, by the chain rule, <span class="math-container">$y=y(x)$</span> should satisfy the differential equation
<span class="math-container">$$\frac{dy}{dx} \,=\, \frac{dy}{dt} \frac{dt}{dx} \, =\, \frac{\,\,\frac{dy}{dt}\,\,}{\frac{dx}{dt}} \,=\, \frac{-\,g(x)}{y - F(x)} \, =\, \frac{g(x)}{F(x)-y} $$</span> or in short
<span class="math-container">$$\frac{dy}{dx} \, =\, \frac{g(x)}{F(x)-y} $$</span>
One can check that in <span class="math-container">$ D_{+}$</span> one has <span class="math-container">$\frac{dy}{dx}\,=\,\frac{g(x)}{F(x)-y} \, < \, 0$</span>, because <span class="math-container">$g(x) > 0$</span> and <span class="math-container">$y > F(x)$</span> implies <span class="math-container">$F(x) - y < 0$</span>, which means that <span class="math-container">$y=y(x)$</span> is a monotonously decreasing function.
Hence, for <span class="math-container">$x > 0$</span> the solution trajectory <span class="math-container">$(x, \,y(x))$</span> has the property <span class="math-container">$y(x) < y(0) = y_0$</span>. The initial condition <span class="math-container">$y_0$</span> can be taken to be the <span class="math-container">$y$</span>-coordinate of the point <span class="math-container">$P_0 = (0, y_0)$</span> on the picture.</p>
<p>Next, we will apply the following theorem from the theory of systems of ordinary differential equations:</p>
<p><strong>Theorem.</strong> <em>Let <span class="math-container">$f \, : \, U \, \to \, \mathbb{R}^n$</span> be a Lipschitz continuous function (this includes continuously differentiable functions) defined on an open set <span class="math-container">$U \, \subset \, \mathbb{R} \times \mathbb{R}^n$</span>. Furthermore, let <span class="math-container">$D \subset U$</span> be a bounded connected open subset of <span class="math-container">$U$</span> and let <span class="math-container">$(x_0, \,y_0) \, \in \, {D}$</span>. Then there exists a unique (local) solution <span class="math-container">$y = y(x)$</span> of the initial value problem
<span class="math-container">\begin{align}
&\frac{dy}{dx} \, =\, f(x,\, y)\\
&y(x_0) \, =\, y_0
\end{align}</span>
Furthermore, the solution <span class="math-container">$y = y(x)$</span> can be extended over a closed interval <span class="math-container">$x \in [a, b]$</span>, so that <span class="math-container">$\big(x,\, y(x)\big) \in D$</span> for all <span class="math-container">$x \in (a, b)$</span>, while <span class="math-container">$y(a) \in \partial D$</span> and <span class="math-container">$y(b) \in \partial D$</span>.</em></p>
<p>By applying the theorem above to the equation <span class="math-container">$$\frac{dy}{dx} \, =\, \frac{g(x)}{F(x)-y} $$</span> defined in the open set <span class="math-container">$U = \big\{\,(x,y) \in \mathbb{R}^2 \,\, :\,\, F(x) \neq y\,\big\}$</span> and restricted to the open bounded domain <span class="math-container">$D \,=\, D_{x+} \, \cap\, \big\{\,(x,y) \in \mathbb{R}^2 \,\, : \,\, y < y_0\,\big\} $</span>, we can conclude that any solution in <span class="math-container">$D$</span>, being decreasing, cannot go over the boundary <span class="math-container">$y = y_0$</span> and since it is extended forward, the only boundary component of <span class="math-container">$D$</span> it can intersect, as per the theorem, is the curve <span class="math-container">$FG_{+}$</span>.</p>
<p>Absolutely analogously, in the domain <span class="math-container">$D_{+}$</span>, any solution trajectory of the original system of differential equations can be reparametrized as <span class="math-container">$\big(x(y), y\big)$</span> where <span class="math-container">$x = x(y)$</span> is the solution to the differential equation
<span class="math-container">$$\frac{dx}{dy} \, =\, \frac{F(x)-y}{g(x)} $$</span>
Again, applying the theorem above, but this time for the latter differential equation defined in the open set <span class="math-container">$U = \big\{\,(x,y) \in \mathbb{R}^2 \,\, :\,\, g(x) \neq 0\,\big\}$</span> and restricted to the bounded open set <span class="math-container">$D = D_{y+} \cap\, \big\{\,(x,y) \in \mathbb{R}^2 \,\, : \,\, y < y_0\,\big\} $</span> we can conclude that any solution trajectory (x(y), y) in <span class="math-container">$D$</span> cannot go over the boundary <span class="math-container">$y = y_0$</span>, because the same trajectory can be reparametrized as <span class="math-container">$(x, y(x))$</span> with <span class="math-container">$\frac{dy}{dx} = \frac{g(x)}{F(x) - y}$</span> and we already know that such solution <span class="math-container">$y=y(x)$</span> is decreasing and cannot reach <span class="math-container">$y=y_0$</span>. Therefore, according to the theorem above, the only boundary component of <span class="math-container">$D$</span> the solution trajectory can intersect is the curve <span class="math-container">$R$</span>. This point of intersection is denoted by <span class="math-container">$P_2$</span> on your picture.</p>
<p>The reason for this ping-pong between the two representations
<span class="math-container">$$\frac{dy}{dx} \,=\, \frac{g(x)}{F(x) - y} \, \text{ and } \, \frac{dx}{dy} \,=\, \frac{F(x) - y}{g(x)}$$</span> of a solution trajectory of the original system is because on the curve <span class="math-container">$L$</span> at the point <span class="math-container">$P_0$</span> we have <span class="math-container">$\frac{dy}{dx} = 0$</span> and <span class="math-container">$\frac{dx}{dy} = \infty$</span>, while on the curve <span class="math-container">$R$</span> at the point <span class="math-container">$P_2$</span> we have <span class="math-container">$\frac{dy}{dx} = \infty$</span> and <span class="math-container">$\frac{dx}{dy} = 0$</span>.</p>
<p>Following absolutely the same method for the other domain <span class="math-container">$D_{-}$</span>, which splits into the subdomains <span class="math-container">$D_{y-}$</span> and <span class="math-container">$D_{x-}$</span>, ping-ponging between the two differential equation representation of a solution trajectory of the original system, you can complete the argument that the solution trajectory reaches a point <span class="math-container">$P_4$</span> on the negative half of the vertical axis <span class="math-container">$x=0$</span>.</p>
|
46,837 | <p>I am looking for jokes which involve some serious mathematics. Sometimes, a totally absurd argument is surprisingly convincing and this makes you laugh. I am looking for jokes which make you laugh and think at the same time. </p>
<p>I know that a similar <a href="https://mathoverflow.net/questions/1083/do-good-math-jokes-exist-closed">question</a> was closed almost a year ago, but this went too much in the direction "<span class="math-container">$e^x$</span> was walking down the street ...". There is also the community wiki <a href="https://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples">Jokes in the sense of Littlewood</a>, but that is more about notational curiosities. In order to motivate you, let me give an example:</p>
<blockquote>
<p>The real numbers are countable. Indeed, let <span class="math-container">$r_1,r_2,r_3,\dots$</span> be a list of real numbers and suppose that there is a real number missing. Just add it to the list.</p>
</blockquote>
<p>If moderators or audience decide to close this question as off-topic or duplicate, I can fully understand. I just thought it could be interesting and entertaining to have this question open for at least some time.</p>
<hr>
<p><strong>Added by joro Sat Apr 27 08:59:45 UTC 2019</strong> There is <a href="https://chat.stackexchange.com/rooms/92902/jokes">chat room</a> about general jokes and it appears close resistant.</p>
| Michael Hardy | 6,316 | <p>Cosgrove's writings in the <em>Mathematical Intelligencer</em> about 20 or 30(?) years ago had lots of puns, many of which would be understood only by mathematicians. E.g. someone was even worse than an unprincipled infiltrator: he was a non-principal ultrafilter. The biographies of Victoria Cross (famous for Cross products and Cross-ratios, and also particular kinds of word puzzles and a certain style of country running), Montmorency Royce Sebastian Carlow (whose "methods" you've heard of), and Karl-Heinz Normal (Normal subgroups, the Normal distribution,....) were of that sort.</p>
|
2,113,413 | <p>A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore.
<img src="https://i.stack.imgur.com/iHWdl.jpg" alt="enter image description here"></p>
<p>My Attempt,
We know, Speed $=\frac {dist. }{time }$
$=\frac {CD}{10}$.</p>
<p>Also, $Tan 60=\frac {AB}{BC}$
$\sqrt {3}=\frac {AB}{BC}$
$AB=\sqrt {3} BC$.</p>
<p>Now, what do I have to do further? </p>
| Community | -1 | <p>If the side length of <span class="math-container">$CB $</span> is <span class="math-container">$x $</span>, then in <span class="math-container">$\triangle ABC $</span>, we have length of <span class="math-container">$AB =\sqrt {3}x $</span>. Now in <span class="math-container">$\triangle ABD $</span>, we have <span class="math-container">$BD =\sqrt {3}x \cot 30 =3x $</span>. Thus <span class="math-container">$CD=2x $</span>.</p>
<p>If the boat covers a distance of <span class="math-container">$2x $</span> in ten minutes, it can cover a distance of <span class="math-container">$x $</span> in what time?? Hope it helps. <img src="https://i.stack.imgur.com/SybNv.jpg" alt="enter image description here"></p>
|
2,667,230 | <p>Let (X, d) be a complete metric space. Let$ f : X → X$ be a function such that for all distinct$ x, y ∈ X$ ,</p>
<p>$ d(f^ k (x), f^ k (y)) < c · d(x, y)$, for some real number $c < 1$ and an integer $k > 1$. Show that f has a unique fixed point. </p>
<p>my attempt : i take $f(x) = x $and $f(y) = y$ .now $f^k = f(f(...(x),,))))= x $ similarly $f^k =f(f(,,,,,(y))..) = y $ </p>
<p>here im getting $ d(f^ k (x), f^ k (y)) =c · d(x, y)$ $=0$.By fixed point theorem $f(x) =f(y) = x= y$ so im getting$ $$ d(f^ k (x), f^ k (y)) = 0$</p>
<p>im getting uniques fixed points</p>
<p>Is my proof is coorect or not correct . Pliz tell me </p>
<p>if not correct . PLiz help me</p>
| G Cab | 317,234 | <p>Supposing "chain" means string, i.e. word, then we are dealing with binary words of lenght $n$, where</p>
<p><em>a) all the ones (including the last) shall be followd by at least $d$ two's</em></p>
<p>In this case we can figure out that each one be followed by a "bumper" string of $d$ two's.<br>
Assuming that the word contains $q$ ones, that is the same as deleting $qd$ two's from the total length
and distributing the $q$ ones into $n-qd$ places.<br>
That can be done in
$$ \bbox[lightyellow] {
N\left( {n,d,q} \right) = \left( \matrix{
n - qd \cr
q \cr} \right) = \left( \matrix{
n - qd \cr
n - q\left( {d + 1} \right) \cr} \right)
} \tag{a.1} $$
where the second writing shall be preferred, because it ensures that $N(n,d,q)$ be null when
$n<q(d+1)$, taking for the binomial the <a href="https://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_and_connection_to_the_binomial_series" rel="nofollow noreferrer">definition through falling factorial</a>. </p>
<p>Then the answer to your question is
$$ \bbox[lightyellow] {
N\left( {n,d} \right) = \sum\limits_{\left( {0\, \le } \right)\,q\,\left( { \le \left\lfloor {n/\left( {d + 1} \right)} \right\rfloor \, \le \,n} \right)\;} {\left( \matrix{
n - qd \cr
n - q\left( {d + 1} \right) \cr} \right)} \quad \left| {\;0 \le n,d \in Z} \right.
} \tag{a.2} $$
where the bounds for $q$ can be generally extended to $0,\,cdots,\, n$, since the actual bounds are intrinsic to the binomial.</p>
<p>In the hypothesis (<em>user</em>'s comment) that the first one is not preceded by any two, then the position of the first $d+1$ block of letters would be fixed, and the number of ways of obtaining such a configuration would just become $N(n-(d+1),\, d,\, q-1)$.</p>
<p><em>b) all the ones (except the last) shall be followed by at least $d$ two's, i.e. the ones are separated by at least $d$ two's.</em></p>
<p>In this alternative interpretation, it also means that each $1$ occupies $d+1$ consecutive places, except the last, which
can be followed by whichever number of $2$'s.<br>
Let's place the last one at position $j$, with $1 \le j \le n$.<br>
Then we are left with $j-1$ positions where to place $q-1$ ones occupying $d+1$ places.<br>
That is the same as telling that we are disposing $q-1$ identical objects into a total of $j-1-(q-1)d$ cells.</p>
<p>So the total number of ways of arranging $q$ ones will be
$$
\eqalign{
& N\left( {n,d,q} \right) = \sum\limits_{1\, \le \,j\, \le \,n} {\left( \matrix{
j - 1 - \left( {q - 1} \right)d \cr
q - 1 \cr} \right)} = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{
k - \left( {q - 1} \right)d \cr
k - \left( {q - 1} \right)\left( {d + 1} \right) \cr} \right)} = \cr
& = \sum\limits_{(0\, \le) \,k\, (\le \,n - 1)} {\left( \matrix{
n - 1 - k \cr
n - 1 - k \cr} \right)\left( \matrix{
k - \left( {q - 1} \right)d \cr
k - \left( {q - 1} \right)\left( {d + 1} \right) \cr} \right)} = \left( \matrix{
n - \left( {q - 1} \right)d \cr
n - 1 - \left( {q - 1} \right)\left( {d + 1} \right) \cr} \right) \cr}
$$
where the steps are:<br>
-1) change index range<br>
-2) bring index range inside the sum, as an additional binomial, so as to let $k$ free of bounds<br>
-3) apply the "double convolution" formula<br>
$$
\eqalign{
& \sum\limits_k {\left( \matrix{
a - k \cr
m - k \cr} \right)\left( \matrix{
b + k \cr
n + k \cr} \right)} = \sum\limits_k {\left( { - 1} \right)^{\,m - k} \left( \matrix{
m - a - 1 \cr
m - k \cr} \right)\left( { - 1} \right)^{\,n + k} \left( \matrix{
n - b - 1 \cr
n + k \cr} \right)} = \cr
& = \left( { - 1} \right)^{\,m + n} \left( \matrix{
m + n - a - b - 2 \cr
n + m \cr} \right) = \left( \matrix{
a + b + 1 \cr
n + m \cr} \right) \cr}
$$</p>
<p>Now let's check the validity of the above for low values of $q,\, n,\, d$ :<br>
- $0$ ones can be placed in $1$ way, checks with ${{n+d} \choose {n+d}}$;<br>
- $1$ one can be placed in $n$ ways, checks with ${{n} \choose {n-1}}$;<br>
- $2$ ones can be placed in ${{n-d} \choose {2}}$, checks with ${{n-d} \choose {n-2-d}}$;<br>
- for $n=q=0$ we get $1$, the empty string;<br>
- for $d=0$ we get ${{n} \choose {n-q}}$ as it shall be. </p>
<p>Thus we can conclude that
$$ \bbox[lightyellow] {
N\left( {n,d,q} \right) = \left( \matrix{
n - \left( {q - 1} \right)d \cr
n - 1 - \left( {q - 1} \right)\left( {d + 1} \right) \cr} \right)\quad \left| {\;0 \le n,q,d \in \mathbb Z} \right.
} \tag{b.1} $$
and of course the answer to your question will be the sum of $N(n,d,q)$ over $q$
$$ \bbox[lightyellow] {
N\left( {n,d} \right) = \sum\limits_{\left( {0\, \le } \right)\,q\,\left( { \le \,n} \right)} {\left( \matrix{
n - \left( {q - 1} \right)d \cr
n - 1 - \left( {q - 1} \right)\left( {d + 1} \right) \cr} \right)} \quad \left| {\;0 \le n,d \in \mathbb Z} \right.
} \tag{b.2} $$
where the bounds for $q$ can be generally extended to $0,\,\cdots,\, n$, since the actual ones are intrinsic to the binomial.</p>
<p><strong>Example</strong></p>
<p>$n=5,\; d=2,\; q=2 \quad \Rightarrow \quad N(n,d,q)=3$<br>
$$(1,\, 0,\, 0,\, 1,\, 0) ,\; (1,\, 0,\, 0,\, 0,\, 1) ,\; (0,\, 1,\, 0,\, 0,\, 1) $$</p>
<p>$n=6,\; d=2,\; q=2 \quad \Rightarrow \quad N(n,d,q)=6$<br>
$$(1,\, 0,\, 0,\, 1,\, 0,\, 0) ,\; (1,\, 0,\, 0,\, 0,\, 1,\, 0) ,\; (1,\, 0,\, 0,\, 0,\, 0,\, 1) ,\; (0,\, 1,\, 0,\, 0,\, 1,\, 0) ,\; (0,\, 1,\, 0,\, 0,\, 0,\, 1) ,\; (0,\, 0,\, 1,\, 0,\, 0,\, 1) $$</p>
|
345,310 | <p>This is computed based on the following recursive formula <span class="math-container">$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$</span> where: <span class="math-container">$n$</span> is the inital state, State <span class="math-container">$0$</span> is absorbing, <span class="math-container">$\lambda_n$</span> and <span class="math-container">$\mu_n$</span> are the up and down rates respectively and <span class="math-container">$$\sum_{n=0}^\infty\prod_{j=1}^n\frac{\mu_j}{\lambda_j}$$</span>diverges (to make extinction certain). To get the recursion started, we need <span class="math-container">$w_0=0$</span> and <span class="math-container">$$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$</span>The derivation of the last formula can be found in S. Karlin's classic book "A first course in stochastic processes". The last step of his proof requires showing that <span class="math-container">$$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$</span>To prove that is, according to Karlin, "more involved but still possible" (but he does not do it). How does one prove that the last limit must equal to <span class="math-container">$0$</span>?</p>
| mike | 143,907 | <p>Here is a suggestion: <span class="math-container">$\omega_n - \omega_{n+1}$</span> is the time to move from state n+1 to n, and it can't be much worse that the return time to state n+1 (proof: every time you return you have a fixed probability of succeeding in getting to state n on next trial, so no worse than a geometric number of return times). It ought to be easy to write down the expected return time explicitly, as it is the reciprocal of the probability under the stationary distribution of being in that state, which is easily got for a B&D process. Whether this is then tractable, I do not know.</p>
|
17,134 | <p>On a very regular basis we see new users that are not accustomed with the use of MathJaX on MSE. Sometimes even some users that aren't that new to the site. Most of us, when this happens, kindly bring to this users attention that there is a <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">MathJax basic tutorial and quick reference</a> and give a proper link. </p>
<p>Now this is not a bad habit at all, because you can't expect new users to magically know this. My idea however was that we could promote this a bit more actively. </p>
<p>For instance: a new user is taken by the hand when he/she is taking the <a href="https://math.stackexchange.com/tour">tour</a>. Would it not be to this new user's benefit to point out in the <a href="https://math.stackexchange.com/tour">tour</a>, that we write our maths in MathJaX. For example place a link to <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">-MathJax basic tutorial and quick reference-</a> or even a link to a -MathJaX tour page- (which would probably have the same structure as the normal <a href="https://math.stackexchange.com/tour">tour</a> page). And perhaps even -but this is really going ahead of things- award a badge for reading this page too.</p>
<p>This was a thought of mine that I wanted to share. I would like to know how the community feels about this. Or perhaps someone could make clear why this would not be helpfull.</p>
| copper.hat | 27,978 | <p>Yes, it should. It took me 5 mins. to locate a pointer to a tutorial (in fact, the one in the question) to add as a comment in response to a new user's MathJax-free question. And I am moderately familiar with the site.</p>
<p>Either the tour or the 'Ask Question' page should have a suitable pointer.</p>
|
151,864 | <p>I would like to generate a random password of a defined length which can easily be typed in with a standard keyboard.</p>
<p>As a start I tried the following:</p>
<pre><code>SeedRandom["pass"];
StringJoin[RandomChoice[CharacterRange[33, 126], 10]
(* "=IP@7mbYcB" *)
</code></pre>
<p>Do you know other solutions?</p>
| Mr.Wizard | 121 | <p>A shorter formulation equivalent to your own code is:</p>
<pre><code>FromCharacterCode @ RandomInteger[{33, 126}, 10]
</code></pre>
<blockquote>
<pre><code>"+(pCT4W#;T"
</code></pre>
</blockquote>
<p>However quite a few places only accept alphanumeric passwords, and not all keyboards have the same easily accessible character sets. If we assume that your given example is sufficiently secure you need 94^10 ~= 5*10^19 unique passwords. This is easily accomplished by adding a single alphanumeric character as 62^11 ~= 5*10^19. Therefore I propose:</p>
<pre><code>rnd =
FromCharacterCode @
RandomChoice[Join @@ Range[{48, 97, 65}, {57, 122, 90}], #] &;
rnd[11]
</code></pre>
<blockquote>
<pre><code>"liLC2RoA3cR"
</code></pre>
</blockquote>
<p>Or five passwords at once:</p>
<pre><code>rnd[{5, 11}]
</code></pre>
<blockquote>
<pre><code>{"suPwm0c7FxS", "CV3khXaowWS", "Lac9z1IVCwc", "gptfkp2GMwH", "HDhRuFPLxte"}
</code></pre>
</blockquote>
|
771,959 | <p>Let $p$ be prime, $n \in \mathbb{N}$ and $p \nmid n$. </p>
<p>$\Phi_n$ is the $n$-th cyclotomic polynomial.</p>
<p>How can I find the maximum $n \in \mathbb{N}$ (with $p \nmid n)$ so that $\Phi_n$ splits into linear factors over $\mathbb{Z}/(p)$.</p>
| user88595 | 88,595 | <p>The answer by Kaj Hansen is very good. Here's a slightly different approach with simple equalities like we all know them :
$$a\equiv b \mod n \Longleftrightarrow a = b + kn\Longrightarrow ra = rb + (rk)n$$</p>
<p>Writing it back under modulo form you obtain $ra \equiv rb \mod n$</p>
|
2,138,009 | <p>Let $f(z)=(1+i)z+1$. Then $f(z)=\sqrt 2 e^{i\pi/4}z+1$ and thus $f=t\circ h\circ r$ where $t$ is the translation of vector $1$, $r$ the rotation of center $0$ and angle $\pi/4$ and $h$ the homothetic of parameter $\sqrt 2$. I found a fix point $z=\frac{-1}{2}+\frac{i}{2}$.</p>
<p>1) What if $f\circ f\circ f\circ f\circ f$ ? It looks to be an translation in the direction of $i$ and an homothetic of parameter 4, but is there an easy way to prove it (with out calculation) ?</p>
<p>2) Let $z_0=i$, $z_1=\sqrt 3+2i$ and $z_2=\sqrt 3-2i$, and consider the triangle $T=\Delta (z_0,z_1, z_3)$. Let $g_n=\underbrace{f\circ ...\circ f}_{n\ times}$. What is the smallest $n$ s.t. $$Area(g_n(T))\geq 100 Area(T) \ \ ?$$</p>
<p>First, $Area(T)= 2\sqrt 3$. I think that $g_n$ is the same triangle with length of side bigger of $n\sqrt 2$, and thus, I would says that $$Area(g_n(T))=\frac{n4\sqrt 2\cdot n\sqrt 2\sqrt 3}{2}=4\sqrt 3n^2.$$
Therefore, $$4\sqrt 3n^2\geq 200\sqrt 3\implies n^2\geq 50\implies n\geq 8.$$
We conclude that the smallest $n$ is $n=8$. Is it correct ? </p>
| Surb | 154,545 | <p>1) The linear part of $f^4$ is $-4z$, therefore, $f^4$ will be of the form $$f^4(z)=-4z+\tau,$$
therefore, it will be an homothetic transformation with scale factor $4$ composed by a point reflexion (you can also say that it's an homothetic transformation with scale factor $-4$) and a translation of vector $\tau$. To find $\tau$, you unfortunately need to do some calculation (i.e. compute explicitly $f^4$).</p>
<p>2) Yes, correct.</p>
|
1,920,994 | <p>My calculus teacher gave us this interesting problem: Calculate</p>
<p>$$ \int_{0}^{1}F(x)\,dx,\ $$ where $$F(x) = \int_{1}^{x}e^{-t^2}\,dt $$</p>
<p>The only thing I can think of is using the Taylor series for $e^{-t^2}$ and go from there, but since we've never talked about uniform convergence and term by term integration, I suppose that there is an easier way to do this.</p>
| Math1000 | 38,584 | <p>To rigorously justify the use of <em>Tonelli's</em> theorem, let $g:\mathbb R^2\to\mathbb R$ be defined by $g(x,t) = e^{-t^2}\mathsf 1_E$, where $$E = \{(x,t)\in\mathbb R^2 : 0 < x < t < 1\}. $$
Since $g(x,t)\geqslant 0$ for all $x,t$ and $g$ is measurable as the product of measurable functions, it follows that $$\int_{\mathbb R^2} g(x,t)\,\mathsf d(x\times t) = \int_{\mathbb R}\int_{\mathbb R} g(x,t)\,\mathsf dt\mathsf dx = \int_{\mathbb R}\int_{\mathbb R} g(x,t)\,\mathsf dx\mathsf dt. $$</p>
<p>We can also use <em>Fubini's</em> theorem. Since $0\leqslant g(x,t)\leqslant 1$ for all $x,t$ and $m(E)<1$, we have $$\int_{\mathbb R^2} |g(x,t)|\,\mathsf d(x\times t) <1<\infty,$$ so that $g\in L^1(\mathbb R^2)$ and again we conclude that the integral of $g$ is finite and equal to both of the iterated integrals.</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| bluemaster | 460,565 | <p>Clearly impossible in base 10. Can it be done in another number base?</p>
<p>In base 5: $11+11+3=30$</p>
<p>Actually there are many other possibilities!</p>
<p>Another possibility: fill the boxes with $\binom{5}{3}$,15 and 5.</p>
<p>It holds that $\binom{5}{3}+15+5=30$ (base 10). Notice that $\binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| Eric Towers | 123,905 | <ul>
<li><p>Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 \text{.}$$</p></li>
<li><p>Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \\ (15+15,15+15) = 30 \text{.}$$ Here, the parens represent the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor" rel="nofollow noreferrer">GCD</a>.</p></li>
</ul>
<p><strong>Edit:</strong></p>
<p>Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 \text{.} $$</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| Andrew | 259,577 | <p>What about
6+9+15=30?</p>
<p>they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)</p>
|
1,415,505 | <p>I was trying to find out how to prove </p>
<p>$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ 0.91 \ \sin^2 \ A \ \ . $$
When I put this equation into my calculator both sides appear to be exactly the same, but I have no idea how to prove it.</p>
| colormegone | 71,645 | <p>[I thought of another way since I left my comment above.]</p>
<p>Applying the "product-to-sum" formula for two sine factors,</p>
<p>$$ \sin \ \alpha \ \cdot \ \sin \ \beta \ = \ \frac{1}{2} \ [ \ \cos(\alpha \ - \ \beta) \ - \ \cos(\alpha \ + \ \beta) \ ] \ \ , $$</p>
<p>with $ \ \alpha \ \ \text{and} \ \ \beta \ $ being sums and differences of angles, we can write</p>
<p>$$ \sin (\theta \ + \ \phi) \ \cdot \ \sin \ (\theta \ - \ \phi) \ = \ \frac{1}{2} \ [ \ \cos \ 2 \phi \ - \ \cos \ 2 \theta \ ] \ \ . $$</p>
<p>So the left-hand side of your equation becomes</p>
<p>$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ \frac{1}{2} \ [ \ \cos ( 2 \ \cdot \arcsin \ [0.3 \ \sin \ A]) \ - \ \cos \ 2 A \ ] \ \ . $$</p>
<p>We're looking ultimately for an expression which has a factor of $ \ \sin^2 A \ $ , so we'll use the "double-angle" formula for cosine, $ \ \cos \ 2 \theta \ = \ \cos^2 \ \theta \ - \ \sin^2 \ \theta \ = \ 1 \ - \ 2 \ \sin^2 \ \theta \ $ . The left-hand side becomes</p>
<p>$$ = \ \frac{1}{2} \ [ \ ( 1 \ - \ 2 \ \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A] \ ) \ ) \ - \ ( 1 \ - \ 2 \ \sin^2 A ) \ ] $$</p>
<p>$$ = \ \frac{1}{2} \ [ \ 2 \ \sin^2 A \ - \ 2 \ \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A] \ ) \ ] \ = \ \sin^2 A \ - \ \sin^2 ( \ \arcsin \ [0.3 \ \sin \ A] \ ) \ \ . $$</p>
<p>Now we need to consider the angle in the second term. [Since arcsine has as its domain $ \ -\frac{\pi}{2} \ \le \ \theta \ \le \ \frac{\pi}{2} \ $ , we would have to deal with negative as well as positive angles. But we can restrict the discussion to positive angles, since the argument for negative angles is similar.] Suppose angle $ \ A \ $ has a sine value of $ \frac{y}{1} \ $ ; then the angle in the second term has a sine value of $ \ 0.3 \ y \ $ . That angle has measure $ \ \arcsin \ (0.3 \ y ) \ $ , but we don't care what that number is, since we are going to take the sine of that angle and square it. [You don't indicate at level you're familiar with working with trigonometric functions; if you are used to the inverse functions, this is something you already know and I apologize for the explanantion.]</p>
<p>Hence, the expression reduces to</p>
<p>$$ \sin^2 A \ - \ ( \ 0.3 \ \sin \ A \ )^2 \ = \ \sin^2 A \ - \ 0.09 \ \sin^2 \ A \ = \ 0.91 \ \sin^2 \ A \ \ . $$</p>
|
539,027 | <p>How many ways are there to distribute $18$ different toys among $4$ children?</p>
<ol>
<li><p>without restrictions</p></li>
<li><p>if $2$ children get $7$ toys each and $2$ children get $2$ toys each.</p></li>
</ol>
<p>For $1$ since toys are different, then there are $4^{18}$ ways to distribute .</p>
<p>for $2$, Im kind of stuck. I know I have two do it in $2$ cases. Can someone help me?</p>
| André Nicolas | 6,312 | <p>Choose the $2$ children who will get $7$ toys. Then choose the $7$ toys the older of these will get. Then choose the $7$ for the younger one. Then choose the $2$ toys for the older of the remaining children. That gives a total of $\binom{4}{2}\binom{18}{7}\binom{11}{7}\binom{4}{2}$. </p>
|
154,722 | <p>Let $A = \pmatrix{1 & 0 \\ \alpha & 1} $ and $ B = \pmatrix{1 & 1 \\ 0 & 1}$, where $\alpha \in \mathbb{C}$ is a complex parameter.</p>
<p>Now consider the family of representations $r_{\alpha}$ of the free group on two generators $F_2 = \langle a,b\rangle$ in $\mathrm{SL}(2, \mathbb{C})$ setting $r_{\alpha}(a) = A$ and $r_{\alpha}(b) = B$. One can see that when $\alpha$ is transcendental over $\mathbb{Q}$, the representation $r_{\alpha}$ is faithful (see T. Church & A. Pixton "Separating twists and the Magnus representation of the Torelli group" Lemma 5.1). </p>
<p>The question I am interested in is the following : when is (or is not) $r_{\alpha}(F_2)$ a discrete subgroup of $\mathrm{SL}(2, \mathbb{C})$ ? </p>
<p>I suppose this is a difficult question of dynamics, I am curious if anyone has ever studied similar questions. </p>
| Danny Ruberman | 3,460 | <p>If you want to know if your group is not discrete, apply Jorgensen's inequality (Jørgensen, Troels (1976), "On discrete groups of Möbius transformations", American Journal of Mathematics 98 (3): 739–749). It says that if $A$ and $B$ generate a non-elementary discrete subgroup of $SL_2(C)$, then
$$
|\text{Tr}(A)^2 -4| + |\text{Tr}(ABA^{-1}B^{-1})-2|\ge 1.
$$
For your group, this works out to $|\alpha|^2 \geq 1$. So for $|\alpha|<1$, your group is either elementary (eg if $\alpha=0$) or non-discrete. </p>
<p>In the case at hand, your matrices are parabolic, so you should be able to get the same result by an elementary analysis of fixed points on the sphere at infinity; this was actually Jorgensen's starting point (per the intro to his paper).</p>
|
1,105,971 | <p>I've got two independent bernoulli distributed random variables $X$ and $Y$ with parameter $\frac{1}{2}$. Based on those I define two new random variables </p>
<p>$X' = X + Y , E(X') = 1$</p>
<p>$Y' = |X - Y|, E(Y') = \sum_{x=0}^1\sum_{y=0}^1|x-y|*P(X=x)*P(Y=y) = \frac{1}{2}$ </p>
<p><strong>How can I calculate E(X'Y')?</strong> As X' and Y' and not independent (e.g. it is impossible for Y' to assume 1 if X' is 0) I must not use the sum of all possible outcomes multiplied with their likelihood as I did for $E(Y')$ but I cannot find another formula to calculate the expected value.</p>
| Ian | 83,396 | <p>For general dependent variables with finite variance, $E((X'-E(X'))(Y'-E(Y')))$ could be anything between $-\sqrt{E((X'-E(X'))^2) E((Y'-E(Y'))^2)}$ and $\sqrt{E(X'^2) E(Y'^2)}$. The fact that this must hold follows from the Cauchy-Schwarz inequality, while the fact that any value can be attained can be proven by construction.</p>
<p>So there is no general solution; you must find the joint distribution function and calculate the expectation directly. In this particular case you have a discrete variable that takes on at most $4$ values (one for each possible pair $(X,Y)$). So this is not too hard to do (tau_cetian has already done it).</p>
|
3,679,806 | <p>this is Probability Density Function(pdf)
if <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent, how to prove <span class="math-container">$X^2$</span> and <span class="math-container">$Y^2$</span> are also independent</p>
| Jonathan Hole | 661,524 | <p>That <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent just means that the sigma algebras they generate are independent. But the sigma algebra <span class="math-container">$X^2$</span> generates is containedin the sigma algebra generates by <span class="math-container">$X$</span> and the same goes for <span class="math-container">$Y^2$</span>.</p>
|
368,461 | <p>Let <span class="math-container">$G=(V,E)$</span> be a finite simple graph. We say a map <span class="math-container">$p:V\to [n]:=\{1,\ldots,n\}$</span> is a <em>pseudo-coloring</em> if for all <span class="math-container">$a\neq b\in[n]$</span> there is <span class="math-container">$v\in\psi^{-1}(\{a\})$</span> and <span class="math-container">$w\in\psi^{-1}(\{b\})$</span> such that <span class="math-container">$\{v,w\}\in E$</span>. We denote the maximal number <span class="math-container">$m$</span> such that there is a pseudo-coloring <span class="math-container">$p:V\to [m]$</span> by <span class="math-container">$\psi(V)$</span>.</p>
<p>An easy argument shows that every coloring in the traditional sense is also a pseudo-coloring, which implies that <span class="math-container">$\psi(G) \geq \chi(G)$</span>.</p>
<p>Let <span class="math-container">$G, H$</span> be finite simple undirected graphs.
Do we necessarily have <span class="math-container">$$\psi(G\times H) \leq \min\{\psi(G),\psi(H)\}?$$</span></p>
<p>(By <span class="math-container">$G\times H$</span> we denote the categorical product, sometimes also referred to as the <a href="https://en.wikipedia.org/wiki/Tensor_product_of_graphs" rel="nofollow noreferrer">tensor product of graphs</a>.)</p>
| bof | 43,266 | <p>Here is a very simple example for <span class="math-container">$\psi(G\times H)\gt\max\{\psi(G),\psi(H)\}$</span>.</p>
<p>The graph <span class="math-container">$G=H=K_3$</span> has <a href="https://en.wikipedia.org/wiki/Complete_coloring" rel="nofollow noreferrer">achromatic number</a> and pseudo-chromatic number equal to <span class="math-container">$3$</span>. The tensor product <span class="math-container">$K_3\times K_3$</span> has achromatic number and pseudo-chromatic number <span class="math-container">$5$</span>, as shown by the following coloring: assuming <span class="math-container">$V(K_3)=[3]$</span>, define <span class="math-container">$p(1,1)=p(1,2)=1$</span>, <span class="math-container">$p(1,3)=p(2,3)=2$</span>, <span class="math-container">$p(3,3)=p(3,2)=3$</span>, <span class="math-container">$p(3,1)=p(2,1)=4$</span>, <span class="math-container">$p(2,2)=5$</span>.</p>
|
4,545,364 | <blockquote>
<p>Solve the quartic polynomial :
<span class="math-container">$$x^4+x^3-2x+1=0$$</span>
where <span class="math-container">$x\in\Bbb C$</span>.</p>
<p>Algebraic, trigonometric and all possible methods are allowed.</p>
</blockquote>
<hr />
<p>I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.</p>
<p>I realized there is no any rational root, by the rational root theorem.</p>
<p>The harder part is, WolframAlpha says the factorisation over <span class="math-container">$\Bbb Q$</span> is impossible.</p>
<p>Another solution method can be considered as the quasi-symmetric equations approach. (divide by <span class="math-container">$x^2$</span>).</p>
<p><span class="math-container">$$x^2+\frac 1{x^2}+x-\frac 2x=0$$</span></p>
<p>But the substitution <span class="math-container">$z=x+\frac 1x$</span> doesn't make any sense.</p>
<p>I want to ask the question here to find possible smarter ways to solve the quartic.</p>
| Yuri Negometyanov | 297,350 | <p>HINT.</p>
<p><span class="math-container">$$4(x^4+x^3-2x+1)=(2x^2+x-1)^2+3(x-1)^2$$</span>
<span class="math-container">$$=\big(2x^2+(1+i\sqrt3\,)(x-1)\big)\big(2x^2+(1-i\sqrt3)(x-1)\big),$$</span></p>
<p>DETAILS.</p>
<ol>
<li><p>The first quadratic term can be chosen in the form of <span class="math-container">$(2x^2+x+a)^2,$</span> which provides the coefficients near <span class="math-container">$x^4$</span> and <span class="math-container">$x^3.$</span></p>
</li>
<li><p>The difference between the left part and the first quadratic term should be a full square.</p>
</li>
</ol>
|
2,179,253 | <p>$$n{n-1 \choose 2}={n \choose 2}{(n-2)}$$
Give a conceptual
explanation of why this formula is true.</p>
| PSPACEhard | 140,280 | <p>Suppose from $n$ students, you want to select $1$ student to clean the blackboard and $2$ students to clean the ground. There are $n$ ways to choose a student to clean the blackboard, and from the remaining students there are totally $\binom{n-1}{2}$ ways to select students to clean the ground. Therefore, the total # of ways
is
$$
n\binom{n-1}{2}
$$
Alternatively, you can first select $2$ students to clean the ground, which is $\binom{n}{2}$, and from the remaining $n - 2$ students, you select one to clean the blackboard. If you count this way, the total # of ways is $\binom{n}{2}(n-2)$ and it should be equal to $n\binom{n-1}{2}$ because you count the same thing.</p>
|
1,537,676 | <p>$f(x,y)=\begin{cases}
|\frac{y}{x^2}|exp(-|\frac{y}{x^2}|) & , x\ne0\\
0 & , x=0
\end{cases}$</p>
<p>I need to show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ the limit does not exist in $(0,0)^T$. I tried to prove it with the sequence criteria but I could not find a good sequence. Polar form doesn't get me further. </p>
| Michael Medvinsky | 269,041 | <p>On $y=x^2$ you have
$$\lim\limits_{(x,y)\to(0,0)}\left|\frac{y}{x^2}\right|\exp\left(-\left|\frac{y}{x^2}\right|\right)
=\lim\limits_{(x,y)\to(0,0)}\exp(-1)=\frac{1}{e}$$</p>
<p>Denote
$$f(x,y)=\left|\frac{y}{x^2}\right|\exp(-\left|\frac{y}{x^2}\right|)
$$
on $y=0$
we have $f(x,0)=0$
and therefore
$$\lim\limits_{(x,y)\to(0,0)}f(x,y)=
\lim\limits_{x\to0}f(x,0)=0$$</p>
<p>thus, the limit doesn't exists because otherwise it would be the same on every path\curve.</p>
|
1,112,081 | <p>Does $\int_0^\infty e^{-x}\sqrt{x}dx$ converge? Thanks in advance.</p>
| idm | 167,226 | <p>set $u=\sqrt x$ then
$$...=\int_0^\infty 2u^2e^{-u^2}du=-\int_0^\infty u(-2ue^{-u^2})du.$$</p>
<p>Set $f'(u)=-2ue^{-u^2}$ and $g(u)=u$, and by part you'll get
$$-\int_0^\infty u(-2ue^{-u^2})du=\underbrace{-[ue^{-u^2}]_0^\infty}_{=0} +\int_0^{\infty }e^{-u^2}du$$</p>
<p>Now use the fact that $\int_{-\infty }^\infty e^{-x^2}dx=\sqrt \pi$ and that $x\mapsto e^{-x^2}$
to conclude. But a way to prove is either to use polar coordinate on
$$\int_0^\infty \int_0^\infty e^{-(x^2+y^2)}dxdy$$
and take the square root to conclude, or you know that
$$\int_{-\infty }^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=1$$
(cf. normal law) then you make the substitution $u=\frac{1}{\sqrt 2}x$ and remark that the function $u\mapsto e^{-u^2}$ is even to conclude.</p>
|
1,890,040 | <p>This is related to <a href="https://math.stackexchange.com/questions/1888881/expanding-a-potential-function-via-the-generating-function-for-legendre-polynomi">this previous question of mine</a> where (with lots of help) I show that $$\sum_{l=0}^\infty \frac{R^l}{a^{l+1}}P_l(\cos\theta)=\frac{1}{\sqrt{a^2-2aR\cos\theta+R^2}}\tag{1}$$ by using the Legendre generating function.</p>
<p>The generating function for Legendre Polynomials is:</p>
<p>$$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{2}$$
or
$$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{3}$$</p>
<p>I need to show that $$\sum_{l=0}^\infty\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r(R^2/a)\cos\theta+r^2}}\tag{4}$$ I note the striking similarity between the RHS's of $(1)$ and $(4)$ within the square root of the denominator.</p>
<hr>
<p><code>It is not necessary to read the following if you understand how to prove that the sum can be written as above; But here is some context anyway:</code>
<a href="https://i.stack.imgur.com/DIR6f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DIR6f.png" alt="Diagram"></a>
<a href="https://i.stack.imgur.com/PkIFq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkIFq.png" alt="Potential function 1"></a>
<a href="https://i.stack.imgur.com/CtYlI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CtYlI.png" alt="Potential function 2"></a> </p>
<hr>
<p><code>So here is my attempt:</code></p>
<p>From the RHS of equation $(4)$:</p>
<p>$$\displaystyle\begin{align}\frac{R}{a\sqrt{r^2-2r(R^2/a)\cos\theta+(R^2/a)^2}}&=\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}\end{align}$$</p>
<p>Changing variables to match equation $(2)$: Let $h=R^2/ar$ and $x=\cos\theta$</p>
<p>Then
$$\displaystyle\begin{align}\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}&=\frac{R}{ar\sqrt{1-2hx+h^2}}\\&=\frac{R}{ar}(1-2xh+h^2)^{-1/2}\\&=\frac{R}{ar}\Phi(x,h)\\&=\frac{R}{ar}\sum_{l=0}^\infty h^l P_l(x)\\&=\frac{R}{ar}\sum_{l=0}^\infty \frac{R^{2l} P_l(\cos\theta)}{(ar)^l}\\&=\sum_{l=0}^\infty \frac{R^{2l+1} P_l(\cos\theta)}{a^{l+1}r^{l+1}}\\&=\sum_{l=0}^\infty \frac{R^{2l+1}r^{-l-1} P_l(\cos\theta)}{a^{l+1}}\quad\fbox{}\end{align}$$</p>
<hr>
<p>You may be wondering what I'm asking at this point; so here is the question:</p>
<p>In the last line of the extract the textbook says "by summing the series"; and this seems to suggest that one must arrive at the RHS <em>from</em> the LHS (instead of going from RHS to LHS as I did). So is my interpretation correct (LHS $\to$ RHS) and if so <em>can</em> it be done in this order?</p>
<p>Many thanks.</p>
| hmakholm left over Monica | 14,366 | <p>Here's a graphical count, based on intersecting the configuration with spheres of different sizes -- though in order to make things visible actually I'm normalizing the size of the spheres, such that the images here actually shows a sphere intersected by the planes of a dodecahedron of increasing size.</p>
<p>The 12 planes of the dodecahedron come in 6 parallel pairs, and in the figures, the regions of each sphere are color-coded according to (and labeled with) the number of pairs where the region is <em>between</em> the two parallel planes.</p>
<p><img src="https://i.stack.imgur.com/SaQpk.png" alt="sphere with bands of width 0.08D">
<img src="https://i.stack.imgur.com/M4gPR.png" alt="sphere with bands of width 0.25D"></p>
<p>The first image shows a sphere far from the dodecahedron. Each of the 6 pairs of planes become a narrow band around the sphere, and the bands create the graph of an <a href="https://en.wikipedia.org/wiki/Icosidodecahedron" rel="nofollow noreferrer">icosidodecahedron</a>. Each face, edge, and vertex of the icosidodecahedron corresponds to an unbounded region of space -- there are $12$ of type <code>0</code>, $20$ of type <code>0'</code>, $60$ of type <code>1</code>, and $30$ of type <code>2</code>.</p>
<p>As we zoom in towards the dodecahedron (or the dodecahedron become larger), the edges of each band move away from each other. The first change that happens is that the the <code>0'</code> regions disappear and are replaced by $20$ cells of type <code>3</code>. These new cells are the triangular spires of the <a href="https://en.wikipedia.org/wiki/Great_stellated_dodecahedron" rel="nofollow noreferrer">great stellated dodecahedron</a>.</p>
<p><img src="https://i.stack.imgur.com/pIcCU.png" alt="sphere with bands of width 0.4D">
<img src="https://i.stack.imgur.com/N8Yq6.png" alt="sphere with bands of width 0.5D"></p>
<p>Nothing new happens in the first of these pictures, except that the bands keep growing wider.</p>
<p>In the second picture the bands have just met in groups of five, and the last of the <code>0</code> and <code>1</code> regions disappear. Instead we get cells that are between <code>4</code> and <code>5</code> pairs of parallel planes.</p>
<p><img src="https://i.stack.imgur.com/FvXyc.png" alt="sphere with bands of width 0.57D">
<img src="https://i.stack.imgur.com/nSwuh.png" alt="sphere with bands of width 0.77D"></p>
<p>The new yellow regions of type <code>4</code> merge, and we can see there are $30$ such cells, just as there were thirty <code>2</code>s before, corresponding to the edges of either an icosahedron or a dodecahedron. In fact each of these cells is an irregular tetrahedron that shares one edge with the central dodecahedron, and the opposite edge with a <a href="https://en.wikipedia.org/wiki/Great_dodecahedron" rel="nofollow noreferrer">great dodecahedron</a> (whose edge figure looks like an icosahedron).</p>
<p>Removing the yellow tedrahedra from the great dodecahedron leaves a <a href="https://en.wikipedia.org/wiki/Small_stellated_dodecahedron" rel="nofollow noreferrer">small stellated dodecahedron</a>, whose pentagonal spires are the $12$ magenta <code>5</code> cells that came into being together with the <code>4</code>s.</p>
<p><img src="https://i.stack.imgur.com/wL6kP.png" alt="sphere with bands of width 0.83D">
<img src="https://i.stack.imgur.com/5G6fG.png" alt="sphere with bands of width 0.9D"></p>
<p>Finally a single red <code>6</code> cell appears: the original dodecahedron in the middle.</p>
<p>So how many regions were that? Summing everything we get
$$ \underbrace{(12+20)+60+30}_{\text{unbounded}}+\underbrace{20+30+12+1}_{\text{bounded}} = 185 $$</p>
|
1,890,040 | <p>This is related to <a href="https://math.stackexchange.com/questions/1888881/expanding-a-potential-function-via-the-generating-function-for-legendre-polynomi">this previous question of mine</a> where (with lots of help) I show that $$\sum_{l=0}^\infty \frac{R^l}{a^{l+1}}P_l(\cos\theta)=\frac{1}{\sqrt{a^2-2aR\cos\theta+R^2}}\tag{1}$$ by using the Legendre generating function.</p>
<p>The generating function for Legendre Polynomials is:</p>
<p>$$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{2}$$
or
$$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{3}$$</p>
<p>I need to show that $$\sum_{l=0}^\infty\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r(R^2/a)\cos\theta+r^2}}\tag{4}$$ I note the striking similarity between the RHS's of $(1)$ and $(4)$ within the square root of the denominator.</p>
<hr>
<p><code>It is not necessary to read the following if you understand how to prove that the sum can be written as above; But here is some context anyway:</code>
<a href="https://i.stack.imgur.com/DIR6f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DIR6f.png" alt="Diagram"></a>
<a href="https://i.stack.imgur.com/PkIFq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkIFq.png" alt="Potential function 1"></a>
<a href="https://i.stack.imgur.com/CtYlI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CtYlI.png" alt="Potential function 2"></a> </p>
<hr>
<p><code>So here is my attempt:</code></p>
<p>From the RHS of equation $(4)$:</p>
<p>$$\displaystyle\begin{align}\frac{R}{a\sqrt{r^2-2r(R^2/a)\cos\theta+(R^2/a)^2}}&=\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}\end{align}$$</p>
<p>Changing variables to match equation $(2)$: Let $h=R^2/ar$ and $x=\cos\theta$</p>
<p>Then
$$\displaystyle\begin{align}\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}&=\frac{R}{ar\sqrt{1-2hx+h^2}}\\&=\frac{R}{ar}(1-2xh+h^2)^{-1/2}\\&=\frac{R}{ar}\Phi(x,h)\\&=\frac{R}{ar}\sum_{l=0}^\infty h^l P_l(x)\\&=\frac{R}{ar}\sum_{l=0}^\infty \frac{R^{2l} P_l(\cos\theta)}{(ar)^l}\\&=\sum_{l=0}^\infty \frac{R^{2l+1} P_l(\cos\theta)}{a^{l+1}r^{l+1}}\\&=\sum_{l=0}^\infty \frac{R^{2l+1}r^{-l-1} P_l(\cos\theta)}{a^{l+1}}\quad\fbox{}\end{align}$$</p>
<hr>
<p>You may be wondering what I'm asking at this point; so here is the question:</p>
<p>In the last line of the extract the textbook says "by summing the series"; and this seems to suggest that one must arrive at the RHS <em>from</em> the LHS (instead of going from RHS to LHS as I did). So is my interpretation correct (LHS $\to$ RHS) and if so <em>can</em> it be done in this order?</p>
<p>Many thanks.</p>
| Nate | 91,364 | <p>There is a nice combinatorial formula for the number of regions in a hyperplane arrangement in $\mathbb{R}^d$.</p>
<p>$$r(\mathcal{A}) = (-1)^d\chi_\mathcal{A}(-1)$$</p>
<p>Where $r(\mathcal{A})$ denotes the number of regions and $\chi_\mathcal{A}(t)$ is the characteristic polynomial of the underlying poset of flats.</p>
<p>In this case we get that $\chi_\mathcal{A}(t) = t^3-12t^2+60t-112$, so $$r(\mathcal{A}) = (-1)^3(-1-12-60-112)= 185$$</p>
<p>This setup also lets you read off the number of bounded regions by the formula:</p>
<p>$$b(\mathcal{A}) = \pm\chi_\mathcal{A}(1)$$</p>
<p>So we get 63 bounded regions. </p>
<p>For a reference on all this I'll recommend Richard Stanley's lecture notes on hyperplane arrangements available here: <a href="http://www.cis.upenn.edu/~cis610/sp06stanley.pdf" rel="nofollow">http://www.cis.upenn.edu/~cis610/sp06stanley.pdf</a></p>
|
3,294,369 | <p>I am reading a paper which asserts that the value of the function <span class="math-container">$f(x) = x^{\frac{1}{x}}$</span> at <span class="math-container">$x=0$</span> is equal to <span class="math-container">$0$</span>. I can believe that this is so if I write <span class="math-container">$f(x) = e^{{\frac{1}{x}}{{\rm ln}(x)}}$</span>, but I want a rigorous proof of this fact! Is this function even defined at <span class="math-container">$x=0?$</span></p>
| Robert Israel | 8,508 | <p>The limit of <span class="math-container">$x^{1/x}$</span> as <span class="math-container">$x \to 0+$</span> is indeed <span class="math-container">$0$</span>, as you see by writing <span class="math-container">$f(x) = \exp(\ln(x)/x)$</span>. The limit as <span class="math-container">$x \to 0-$</span>, on the other hand, is certainly not <span class="math-container">$0$</span> (not even defined, if you want to stick to real numbers). <strong>At</strong> <span class="math-container">$x=0$</span> the expression is undefined because <span class="math-container">$1/0$</span> is undefined. </p>
<p>You can, if you wish, define the function <span class="math-container">$f(x)$</span> to be <span class="math-container">$0$</span> at <span class="math-container">$x=0$</span>. You can even say that is the "usual convention". But it's not something that can be proven: there is nothing there to prove.</p>
|
510,130 | <p>Let $(r_i)_{i=1}^m$ be a sequence
of positive reals such that
$\sum_i r_i < 1$
and let $t$ be a positive real.
Consider the sequence $T(n)$
defined by $T(0) = t$,
$T(n) = \sum_i T(\lfloor r_i n \rfloor) $
for $n \ge 1$.</p>
<p>Show that
$T(n) = o(n)$,
that is,
$\lim_{n \to \infty} \dfrac{T(n)}{n}
= 0
$.</p>
<p>Note:
This is a variation on
<a href="https://math.stackexchange.com/questions/506489/if-tn-un-sum-i-t-lfloor-r-i-n-rfloor-show-that-tn-thetan">If $T(n) = un + \sum_i T(\lfloor r_i n \rfloor) $, show that $T(n) = \Theta(n)$</a>.
It is gotten by setting $u=0$ there.</p>
<p>I am close to a solution,
and hope to have one in a few days.
If I find one,
I will post it.</p>
<p>Note:
It is easy to prove that
$T(n) = O(n)$.
The problem is showing that
$T(n)/n \to 0$.</p>
| marty cohen | 13,079 | <p>Here is what I have
been able to come up with.</p>
<p>I will show that $T(n)/n$ can be made arbitrarily small,
although the number of steps
seems to be exponential
in the reciprocal of the bound.
More specifically,
for a real $0 < c < 1$,
there are
explicitly computable constants
$A$ and $B$
which both depend on the recurrence
and $B$ also depending on
initial values of $T(n)$
such than
$T(n)/n < c$
for $n > (1/c)^A/B$.</p>
<p>Let
$r = \sum_i r_i$
and
$s = 1-r$
so that
$0 < s < 1$.
Let $r_0 = \min_i r_i$.</p>
<p>I first show that if
$T(k) \le bk$
for
$r_0n \le k < n$
then $T(n) \le bnr$
for all subsequent $n$.</p>
<p>Suppose that
$T(k) \le bk$
for
$r_0n \le k < n
$.
We want to show that
$T(n) \le bn
$. </p>
<p>$\begin{align}
T(n)
&=\sum_i T(\lfloor r_i n \rfloor)\\
&\le \sum_i b r_i n\\
&= n\sum_i b r_i \\
&= nb r\\
\end{align}
$ </p>
<p>This shows that the bound on
$T(n)/n$
gets reduced by a factor
of $r$.
The next step is
to see when the bound
gets be reduced by
a factor of $r^2$.</p>
<p>Let $n_0 = n$.
We want to find an
$n_1$ such that
$r_0 n_1 = n_0$,
so that the interval
from $r_0 n_1$ to $n_1-1$
can be the basis for
another reduction in the bound
by a factor of $r$.</p>
<p>This obviously is satisfied by
$n_1 = n_0/r_0$.
Similarly, setting
$n_2 = n_1/r_0$
allows the bound to be reduced
by a factor of $r^2$.</p>
<p>By induction,
if $n_k = n_{k-1}/r_0
= n_0/r_0^k$,
$T(n)/n < b r^k$
for $n > n_k$.</p>
<p>Let $s = 1/r_0$,
so $s > 1$.
restating the preceding results
in terms of $s$,
if $n_k
= n_{k-1}s
= n_0 s^k$,
$T(n)/n < b r^k$ for
$n > n_k$.</p>
<p>This shows that
the bound is reduced
by a factor of $r^k$
in $n_0 s^k
$
steps.</p>
<p>If the bound,
starting at $b$,
is to be less than $c$,
where $0 < c < 1$,
$b r^k < c$
or
$k > \dfrac{\log (c/b)}{\log r}$
(since $r < 1$).</p>
<p>The number of steps
to have the bound reduced to
$c$
is thus</p>
<p>$\begin{align}
n_0 s^{\frac{\log (c/b)}{\log r}}
&=n_0 \exp(\log s(\log (c)-\log(b))/(\log r)\\
&=n_0 \exp\left(\frac{\log s\log c}{\log r}-\frac{\log s\log b}{\log r}\right)\\
&=n_0 \exp\left(\frac{\log s\log c}{\log r}\right)/K\\
&=n_0 \exp\left(\frac{\log s\log (1/c)}{\log (1/r)}\right)/K\\
&=n_0 (1/c)^{(\log s)/(\log (1/r))}/K\\
\end{align}
$</p>
<p>where
$K
= \exp\left(\frac{\log s\log b}{\log r}\right)
= b^{\log s/\log r}
$.</p>
<p>In these formulas,
$c$ is the value the
bound on $T(n)/n$
is to be reduced to
(e.g., $c=0.01$),
$r$ and $s$ depend
on the recurrence,
and $b$ is the initial bound on
$T(n)/n$.</p>
<p>Therefore,
$T(n)/n$ can be made
arbitrarily small,
although the number of steps
seems to be
exponential in the
reciprocal of the bound.</p>
|
2,636,712 | <p>I have question about my proof. I could not tell whether it is sufficient enough since my professor approached it differently. </p>
<p><strong>The problem:</strong></p>
<blockquote>
<p>Let $z \in \mathbb{C}^{*}$. If $|z| \neq 1$, prove that the order of $z$ is infinite. </p>
</blockquote>
<p><strong>My proof:</strong> (by contradiction)</p>
<p>Let $z = r\cos(\theta)+r\sin(\theta)=r\operatorname{cis}(\theta)$, where $r> 0, \theta \in [0, 2\pi]$. Since $|z| \neq 1$, then $r \neq 1$. </p>
<p>Suppose that the order of $z$ is finite i.e. $\exists m \in \mathbb{Z}_{+} s.t. z^{m}=1$. Then, observe that: </p>
<p>$z^{m}=r^{m}\operatorname{cis}(m\theta)$, so</p>
<p>$|z^{m}|=|1| \implies \sqrt{r^{2m}\operatorname{cis}^{2}(m\theta)}=1 \implies r^{m}=1$.</p>
<p>However, since $m$ is the least positive integer that $z^{m}=1$ i.e. $m> 0$. Then, $r$ has to be $1$. Yet, this contradicts the assumption that $r\neq 1$. </p>
<p>Hence, we proved that the order of $z$ can’t be finite. </p>
<p><strong>My question:</strong> </p>
<p>Is this proof complete? I did not do the way my professor talked about, which using induction and state that $z \neq 1$ is equivalent to $z \in \mathbb{Q}^{*}$. </p>
<p>Any suggestion or different approach is highly appreciated.</p>
| Patrick Stevens | 259,262 | <p>The proof looks fine, but I always prefer to do things without contradiction where possible.</p>
<p>Let $z \in \mathbb{C}^*$, and consider the sequence of reals whose $n$th term is $|z^n| = |z|^n$.</p>
<p>If $|z| > 1$, then this sequence is strictly increasing, and hence in particular the sequence $z^n$ cannot repeat.</p>
<p>If $|z| < 1$, then this sequence is strictly decreasing, so $z^n$ does not repeat.</p>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| NiloCK | 308 | <p>Most experienced mathematicians will immediately identify $x = 0$ as a solution to this equation, and then do the same division (factoring) as your student in order to obtain the other solution at $x=2$. The only difference is that the experienced mathematician has already done due diligence with respect to the possibility of erasing a root/dividing by zero.</p>
<p>I am all for dividing both sides by $6x$ here, with the caveat that when dividing by an unknown quantity you are responsible for first checking whether that unknown quantity is (or might be) zero. Your student is most of the way there - all that's left is to push the conclusion C from information A and B which they already seem to have.</p>
<p>A) I don't know what $x$ is (or $6x$, as is the case here)<br>
B) I'm 'not allowed' to divide by zero</p>
<hr>
<p>C) I don't know whether or not I'm allowed to divide by $x$ (because it might be zero)</p>
<p>There are useful things to think about in this neighborhood as well. A similar situation where the variable-including-divisor being equal to zero represents the only solution, then we end up producing an absurdity. Eg,
$$6x = 4x$$
$$\frac{6x}{x} = \frac{4x}{x}$$
$$6 = 4$$
If we trust our teacher, or the textbook, to have provided us with a logically sound equation in the first place, then the absurdity reached here tells us that we've broken some rules in our working with it. Since the only thing we did was to divide by $x$, we can guess that dividing by $x$ introduced the absurdity. Since we know that dividing both sides of an equation by a quantity is fair game for any value other zero, we now know that $x=0$, etc...</p>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| user21820 | 1,550 | <p>Firstly, division must obey rules regardless of what you are dividing. If the division is performed over real numbers, then you have to respect the rules for division of real numbers, that is no division by <span class="math-container">$0$</span>. In other structures <span class="math-container">$0$</span> is not the only thing that you cannot "divide".</p>
<p>Secondly, when <span class="math-container">$x$</span> is a variable it should be made crystal clear what it refers to. If it can refer to any arbitrary object, then division by <span class="math-container">$x$</span> is meaningless. If it can refer to a real number, whether known or unknown, whether zero or not, then division by <span class="math-container">$x$</span> must again obey real number rules.</p>
<p>Thirdly, the rules must not be given empty and without meaning. For real numbers, multiplication corresponds to scaling or stretching something, and division is to undo that. If you multiply by zero, you are scaling something until it is gone; zero size. You can never recover the original size, so division by zero is not allowed.</p>
<p>Fourthly, and only after the first three points are made absolutely clear, then you can give some examples to show that indeed division by zero can lead to nonsense.</p>
<blockquote>
<p>Fail to solve <span class="math-container">$x = 2x$</span> by dividing by <span class="math-container">$x$</span>!</p>
<p>Fail to solve <span class="math-container">$x^2 = 0$</span> by dividing by <span class="math-container">$x$</span>, twice!</p>
</blockquote>
|
1,818,764 | <p>I think everything I have done is kosher, but unless I am missing an identity it is a different answer than the online quiz and wolfram alpha give.</p>
<p>I tried to use the trig substitution
$$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$</p>
<p>Which yields
$$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{4\sin^2(\theta)}{2\sqrt{1-\sin^2{\theta}}}2\cos(\theta)d\theta=4\int \sin^2(\theta)d\theta\\
=2\int (1-\cos(2\theta))d\theta=2\theta-\sin(2\theta)$$ </p>
<p>By the half angle formula. Then since $x=2\sin(\theta)\Rightarrow \theta=\arcsin(x/2)$ this gives a final answer of </p>
<p>$$\int\frac{x^2}{\sqrt{4-x^2}}dx=2\arcsin(x/2)-\sin(2\arcsin(x/2))+c $$</p>
<p>Is this right? If not, where did I go wrong?</p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
\int \frac{x^{2}}{\sqrt{4-x^{2}}} d x &=-\int x d \sqrt{4-x^{2}} \\
&=-x \sqrt{4-x^{2}}+\int \sqrt{4-x^{2}} d x \\
&=-x \sqrt{4-x^{2}}+\frac{x \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)+C\\&= -\frac{x \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)+C
\end{aligned}
$$</span></p>
|
2,638,679 | <p><a href="https://i.stack.imgur.com/S4p0Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S4p0Y.jpg" alt="enter image description here"></a></p>
<p>Due apologies for this rustic image. But while drawing this lattice arrangement about the "square numbers" , I discovered a pattern here wherein if I add the alternate red dots (as depicted in the image above) to the square number, I get the next square number. For instance, $4 + 5(red\ dot) = 9$ , $9+7(red\ dot)=16$, $16+9(red\ dot)=25$, $25+11(red\ dot)=36$, $36+13 (red\ dot)=49$.</p>
<p>The red dotted numbers themselves have a pattern as is obvious from the image. Is there any mathematical explanation to this pattern.</p>
| Joca Ramiro | 264,635 | <p>The eigenvalues $\lambda_1$ and $\lambda_2$ are the roots of the characteristic polynomial $\det(A-\lambda 1)=a\lambda^2+b\lambda+c$, where $a,b,c$ are real numbers. It is easy to see that $a=1$. Being the polynomial second degree, the sum of its roots is $\lambda_1+\lambda_2=-b/a=3$ and their product $\lambda_1\cdot\lambda_2=c/a=3$. Then the characteristic polynomial is $\lambda^2-3\lambda+3$. The roots can be computed by Bhaskara and are equal to $\frac{3\pm i\sqrt{3}}{2}$. The trace of $A^{-1}$ is the sum of inverse eigenvalues for $A$:
$$
\frac{2}{3+i\sqrt{3}}+\frac{2}{3-i\sqrt{3}}=1.
$$</p>
|
2,638,679 | <p><a href="https://i.stack.imgur.com/S4p0Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S4p0Y.jpg" alt="enter image description here"></a></p>
<p>Due apologies for this rustic image. But while drawing this lattice arrangement about the "square numbers" , I discovered a pattern here wherein if I add the alternate red dots (as depicted in the image above) to the square number, I get the next square number. For instance, $4 + 5(red\ dot) = 9$ , $9+7(red\ dot)=16$, $16+9(red\ dot)=25$, $25+11(red\ dot)=36$, $36+13 (red\ dot)=49$.</p>
<p>The red dotted numbers themselves have a pattern as is obvious from the image. Is there any mathematical explanation to this pattern.</p>
| Widawensen | 334,463 | <p>Every matrix satisfies its own characteristic equation so </p>
<p>$A$ satisfies </p>
<ul>
<li><p>$A^2-\operatorname{tr}(A)A+\det(A)I=0 $</p>
<p>and $A^{-1}$ satisfies </p></li>
<li><p>$A^{-2}-\operatorname{tr}(A^{-1})A^{-1}+\det(A^{-1})I=0$.</p></li>
</ul>
<p>With multiplying both sides of the first equation by $A^{-2}$ and dividing by $\det(A)$ we can easily obtain form of the second one.</p>
<p><strong>Edit</strong><br>
I.e. </p>
<p>$$A^{-2}A^2-\operatorname{tr}(A)A^{-2}A+\det(A)IA^{-2}=0$$</p>
<p>$$I -\operatorname{tr}(A)A^{-1}+\det(A)A^{-2}=0$$</p>
<p>After changing the order of summands and dividing by $\det(A)$</p>
<p>$$A^{-2}- \frac{1}{\det(A)}\operatorname{tr}(A)A^{-1}+\frac{1}{\det(A)}I=0.$$</p>
<p>Comparing the last equation with the second one $\bullet$ we finally obtain </p>
<p>$$\operatorname{tr}(A^{-1})=\frac{\operatorname{tr}(A)}{\det(A)}$$</p>
|
1,747,696 | <p>First of all: beginner here, sorry if this is trivial.</p>
<p>We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .</p>
<p>My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ?
The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on.
In this case, this is also true but I can't seem to find a way to generalize it.</p>
| gnasher729 | 137,175 | <p>There's a simple rule for linear progressions: The sum equals the number of items, times the average of the first and last item. So in this case n (11n + 11) / 2. </p>
<p>If you added 7, 17, 27, 37 and so on, the first item would be 7, the last would be 10n - 3, the average (10n - 3 + 7) / 2 = 5n + 2, so the sum would be n (5n + 2). </p>
|
4,226,455 | <blockquote>
<p>Let <span class="math-container">$X$</span> be a set and <span class="math-container">$Y$</span> a topological space. What is the topology on <span class="math-container">$X$</span> induced by constant maps <span class="math-container">$f:X \to Y$</span>?</p>
</blockquote>
<p>The induced topology is <span class="math-container">$\tau_X = \{f^{-1}[V] : V \in \tau_Y\}$</span> and <span class="math-container">$f^{-1}[V] = \{x \in X : f(x) \in V \}$</span></p>
<p>So if <span class="math-container">$f(x ) = c(x)$</span>, then I initially considered that <span class="math-container">$f^{-1}[V]$</span> would be just the singletons <span class="math-container">$\{x\}$</span>, but I don’t see why there couldn’t be a open set where the there could be multiple <span class="math-container">$f(x) \in V$</span> and thus the preimage wouldn’t be just the singleton?</p>
| mohottnad | 955,538 | <p>Regarding your claim:</p>
<blockquote>
<p>My book says that the proposition "∀x∈D,P(x)" can be vacuously true, because it can be turned into the form "P→Q"</p>
</blockquote>
<p>You seem to conflate the truth value of the material conditional P→Q with quantified formula ∀xP(x) where x ranges over D. P→Q may be an unquantified material conditional which may be vacuously true if the antecedent P is false. Or ∀x(P(x)→Q(x)) may be the quantified version what your teacher talked about. It has nothing to do with the truth value of the formula ∀xP(x) where x ranges over D.</p>
<p>As for the definition of atomic well formed formula, see reference <a href="https://en.wikipedia.org/wiki/First-order_logic#Evaluation_of_truth_values" rel="nofollow noreferrer">here</a>:</p>
<blockquote>
<p>Next, each formula is assigned a truth value. The inductive definition used to make this assignment is called the T-schema.</p>
</blockquote>
<blockquote>
<p>Atomic formulas (1). A formula P(t1,... ,tn)...</p>
</blockquote>
<blockquote>
<p>Atomic formulas (2). A formula t1=t2 is assigned true if t1 and t2 evaluate to the same object of the domain of discourse (see the section on equality below).</p>
</blockquote>
<p>So there're two types of atomic formula (wff) in the first order logic with equality.</p>
|
837,570 | <p>Prove that $\arctan{x}=\frac{1}{x^2}$ has only one solution on the set of real numbers.</p>
<p>I need some help with it, would greatly appreciate it.</p>
| JimmyK4542 | 155,509 | <p>For $x > 0$, $\arctan x$ is strictly increasing while $\dfrac{1}{x^2}$ is strictly decreasing. </p>
<p>For $x < 0$, $\arctan x < 0 < \dfrac{1}{x^2}$. </p>
<p>What does this tell you about the number of positive and negative solutions?</p>
|
4,255,587 | <p>There was a quiz posted on F*cebook by someone. Here's the problem.</p>
<p><a href="https://i.stack.imgur.com/fGu2W.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fGu2W.jpg" alt="enter image description here" /></a></p>
<p>And here's my attempt:</p>
<p><a href="https://i.stack.imgur.com/EMP6g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EMP6g.jpg" alt="enter image description here" /></a></p>
<p>First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green square. The degree has been symbolized as alpha (<span class="math-container">$\alpha$</span>).</p>
<p><span class="math-container">$$\begin{align}
\alpha &= \tan^{-1}\left(\frac{16}{23}\right)\\
\end{align}$$</span></p>
<p>That means the degree between the line (almost diagonally to the green square) and the side line of the green square is <span class="math-container">$90-\alpha$</span>. Let's say the side green square is <span class="math-container">$x$</span></p>
<p>We know <span class="math-container">$\cot(x)=\tan(90^{0} - x)$</span>, so</p>
<p><span class="math-container">$$\begin{align}
\tan(90^{0} - \alpha) = \cot(\alpha) &= \frac{16}{x}\\
x &= 16 \tan(\alpha)\\
&= 16 \tan\left(\tan^{-1}\left(\frac{16}{23}\right)\right)\\
&= \frac{16^2}{23} = \frac{256}{23}
\end{align}$$</span></p>
<p>Now, I assume (I can't tell reason, it's just my guesswork) that <span class="math-container">$x$</span> I've found earlier has the same length as the blue side one. I also assume those two blue are right triangles.</p>
<p>Finally, the last calculation is:</p>
<p><span class="math-container">$$\begin{align}
& \left(\frac{1}{2}\cdot \frac{256}{23}\cdot 16\right) + \left(\frac{1}{2}\cdot 7 \left( \frac{256}{23} + 9\right)\right)\\
&= 159.5
\end{align}$$</span></p>
<p>The poster gave me cry emoji and didn't say anything, I conclude my answer is incorrect. Where's the mistake?</p>
| Michael Rozenberg | 190,319 | <p>Let <span class="math-container">$\Delta ABC$</span> be our triangle, <span class="math-container">$AB=BC$</span>, <span class="math-container">$PQRL$</span> be a square,</p>
<p>where <span class="math-container">$P\in AB$</span>, <span class="math-container">$Q\in BC$</span>, <span class="math-container">$R\in QC$</span>, <span class="math-container">$L\in AR$</span>, <span class="math-container">$PQ=16.$</span></p>
<p>Also, let <span class="math-container">$MRNK$</span> be a square, <span class="math-container">$M\in LR$</span>, <span class="math-container">$N\in RC$</span> and <span class="math-container">$K\in AC$</span>.</p>
<p>Let <span class="math-container">$AL=x$</span>.</p>
<p>Thus, since <span class="math-container">$\Delta ALP\sim\Delta PQB,$</span> we obtain <span class="math-container">$$\frac{BQ}{16}=\frac{16}{x}$$</span> or <span class="math-container">$$BQ=\frac{256}{x}.$$</span></p>
<p>Also, since <span class="math-container">$\Delta KNC\sim\Delta AMK,$</span> we obtain:
<span class="math-container">$$\frac{NC}{7}=\frac{7}{x+9}$$</span> or
<span class="math-container">$$NC=\frac{49}{x+7}.$$</span>
Thus, <span class="math-container">$$BC=\frac{256}{x}+23+\frac{49}{x+9}$$</span> and from <span class="math-container">$\Delta ARB$</span> we obtain:
<span class="math-container">$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+23+\frac{49}{x+9}\right)^2$$</span> or
<span class="math-container">$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+16+\frac{7(x+16)}{x+9}\right)^2$$</span> or
<span class="math-container">$$1+\frac{256}{x^2}=\left(\frac{16}{x}+\frac{7}{x+9}\right)^2$$</span> or
<span class="math-container">$$(x-12)(x^2+30x+168)=0,$$</span> which gives <span class="math-container">$x=12.$</span></p>
<p>Can you end it now?</p>
<p>I got <span class="math-container">$169.5.$</span></p>
|
2,934,973 | <p><a href="https://i.stack.imgur.com/XQ80d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XQ80d.png" alt="enter image description here"></a> </p>
<p>Let <span class="math-container">$\theta = \angle BAC$</span>. Then we can write <span class="math-container">$\cos \theta = \dfrac{x}{\sqrt{2}}$</span>
Find x.</p>
<p>Currently, that is all I have. I think I should use Law of cosine. Where should I progress?</p>
| clathratus | 583,016 | <p>Assuming that </p>
<p><span class="math-container">$A=(1,1), B=(2,3), C=(4,0)$</span>,</p>
<p>we know that <span class="math-container">$\vec{AB}=(2-1)\vec{x}+(3-1)\vec{y}=\vec{x}+2\vec{y}$</span>, and <span class="math-container">$\vec{AC}=(4-1)\vec{x}+(0-1)\vec{y}=3\vec{x}-\vec{y}$</span>.</p>
<p>Thus, <span class="math-container">$||\vec{AB}||=\sqrt{1^2+2^2}=\sqrt{5}$</span>, and <span class="math-container">$||\vec{AC}||=\sqrt{3^2+(-1)^2}=\sqrt{10}$</span>. Therefore, <span class="math-container">$||\vec{AB}||||\vec{AC}||=\sqrt{5}\sqrt{10}=5\sqrt{2}$</span>.</p>
<p>And also: <span class="math-container">$\vec{AB}\cdot\vec{AC}=(1\cdot3)+(-1\cdot2)=3-2=1$</span>.</p>
<p>Therefore, using a rule about dot products, <span class="math-container">$\cos\theta=\frac{\vec{AB} \cdot\vec{AC}}{||\vec{AB}||||\vec{AC}||}$</span></p>
<p>Plugging in, we get <span class="math-container">$\cos\theta=\frac{1}{5\sqrt{2}}$</span>. Therefore, <span class="math-container">$x=\frac{1}{5}$</span></p>
|
965,851 | <p>I have a computer problem that I was able to reduce to an equation in quadratic form, and thus I can solve the problem, but it's a little messy. I was just wondering if anybody sees any tricks to simplify it?</p>
<p>$$\sin^2\beta ⋅ d^4 + c^2\left(\cos^2\beta⋅\cos^2\alpha-\frac{\cos^2\beta}{2}-\frac12\right)d^2 + sin^2\beta⋅\frac{c^4}{16} = 0$$</p>
<p>Obviously I am using the quadratic formula to solve for $d^2$.</p>
<p>$\beta$, $\alpha$, and $c$ are known. </p>
<p>That middle term is so ugly. Perhaps this could even be simplified and solved without the quadratic formula? </p>
<p>By dividing through with $\sin^2\beta$ I came up with (assuming I did it correctly, I didn't double check it):</p>
<p>$$d^4 + \left[\cot^2\beta\cos^2\alpha - \frac{\csc^2\beta}{2}-\frac{\cot^2\beta}{2}\right]c^2d^2 + \frac{c^4}{16} = 0$$</p>
<p>Which is a little less ugly. Am I missing a cool trick to simplify this?</p>
| Aaron Maroja | 143,413 | <p>Try this, first suppose $x > 0$, then you take $x<0$. </p>
<p>$$\begin{align}x^2 = 2^x &\Rightarrow (x^2)^{\frac{1}{2}} = (2^x)^\frac{1}{2} \Rightarrow x= 2^\frac{x}{2} \\ & \Rightarrow x \ e^{-x\frac{\ln\ 2}{2}} = 1 \Rightarrow -x \frac{\ln\ 2}{2}\ e^{-x\frac{\ln\ 2}{2}} = -\frac{\ln \ 2}{2} \\ &\Rightarrow -x \frac{\ln\ 2}{2} = W(-\frac{\ln\ 2}{2}) \Rightarrow x = -\frac{2\ W(-\frac{\ln \ 2}{2})}{\ln\ 2}\end{align}$$</p>
<p>Which gives us </p>
<p>$x = -\frac{2\ W(\frac{-\ln \ 2}{2})}{\ln\ 2} = 2$ , in case $x > 0$ </p>
<p>Similarly we may find </p>
<p>$x = -\frac{2\ W(\frac{\ln \ 2}{2})}{\ln\ 2} \approx -0,76666$, in case $x < 0$ </p>
<p>Where $W$ is the <a href="http://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow">Lambert's funtion</a>.</p>
|
965,851 | <p>I have a computer problem that I was able to reduce to an equation in quadratic form, and thus I can solve the problem, but it's a little messy. I was just wondering if anybody sees any tricks to simplify it?</p>
<p>$$\sin^2\beta ⋅ d^4 + c^2\left(\cos^2\beta⋅\cos^2\alpha-\frac{\cos^2\beta}{2}-\frac12\right)d^2 + sin^2\beta⋅\frac{c^4}{16} = 0$$</p>
<p>Obviously I am using the quadratic formula to solve for $d^2$.</p>
<p>$\beta$, $\alpha$, and $c$ are known. </p>
<p>That middle term is so ugly. Perhaps this could even be simplified and solved without the quadratic formula? </p>
<p>By dividing through with $\sin^2\beta$ I came up with (assuming I did it correctly, I didn't double check it):</p>
<p>$$d^4 + \left[\cot^2\beta\cos^2\alpha - \frac{\csc^2\beta}{2}-\frac{\cot^2\beta}{2}\right]c^2d^2 + \frac{c^4}{16} = 0$$</p>
<p>Which is a little less ugly. Am I missing a cool trick to simplify this?</p>
| fleablood | 280,126 | <p>I wrote about this in another post.</p>
<p>The other solution is 4.</p>
<p>For $x^a = a^x; a > 0$ in general... Well, notice the shapes of the graphs for $x \ge 0$ are such that they intersect twice if $a \ne e$ and intersect once if $a = 1$ (at $x = e$).</p>
<p>In my other post I went into great detail about taking the first and second derivatives to show both $a^x$ and $x^a$ are concave and that they can only have at most two intersections and as $a^0 = 1 > 0^a = 0$ there must have at least one. But I'll leave that as an excercise to the reader this time.</p>
<p>But obviously $x = a \implies a^a = a^a$ is a solution. The other ... hmm... I forget and I'm too lazy to figure it out a second time. But the solutions are "paired" (that is if $x = b$ solves $a^x = x^a$ then $x = a$ solves $b^x = x^b$) and one solution is less than e and the other is greater than e. I came up with a formula relating the solutions.</p>
<p>2 and 4 solves both $x^2 = 2^x$ and $x^4 = 4^x$ were the slickest solutions but every other positive value for $a$ and a $a$ and $a'$ solution.</p>
<p>If $a$ is even or an even rational then there is an $x = -1/a$ solution. But if a is odd then there is no negative solution as for $x < 0$ then $a^x > 0$ but $x^a < 0$. If a is irrational then $x^a$ is undefined for $x < 0$. These are the only solutions as $a^x$ is increasing but $x^a$ is decreasing when $x < 0$ and $a$ is even.</p>
|
75,791 | <p>When will a probabilistic process obtained by an "abstraction" from a deterministic discrete process satisfy the Markov property?</p>
<p>Example #1) Suppose we have some recurrence, e.g., $a_t=a^2_{t-1}$, $t>0$. It's a deterministic process. However, if we make an "abstraction" by just considering the one particular digit of each $a_t$, we have a probabilistic process. We wonder whether it satisfy a Markov property or not?</p>
<p>Example #2) Suppose we have a finite state automaton. Now we make an "abstraction" by grouping the states into sets of states and obtaining a probabilistic finite state automaton. We consider this automaton in time and we wonder whether it satisfies the Markov property or not.</p>
<p>The particular examples are not important, of interest are some general conditions when a deterministic process becomes a Markov process after an "abstraction" of the kind above (in any context). I'm looking for any references on this matter. </p>
<hr>
<p>Edit: as pointed out in the comments below, the examples #1 and #2 were not well specified. Now there's a distribution on the starting state $a_0$ and $a_t=f(a_{t-1})$ is a deterministic function. Then, $a_t$, $t\geq 0$ is a degenerate Markov chain. Now the question is whether grouping some of the states of such a chain can yield a Markov chain (i.e., pointers to literature where the conditions would be discussed is required). </p>
<p>A more general problem seems to be "Given any Markov chain (i.e., not a degenerate one), group the states into sets: what are the conditions under the resulting process satisfies the Markov assumption?"</p>
| Craig | 15,279 | <p>The simple answer is: every discrete deterministic process is already a Markov process. The "probabilities" in question are just restricted to be 0 or 1.</p>
|
1,837,356 | <p>I'm reading <a href="http://www.careerbless.com/aptitude/qa/permutations_combinations_imp8.php" rel="nofollow">this passage</a> and wondering why</p>
<p>Number of ways in which
k identical balls can be distributed into
n distinct boxes =</p>
<p>$$\binom {k+n-1}{n-1}$$</p>
<p>could someone explain it to me please?</p>
| JMoravitz | 179,297 | <p>The proof technique is what is often referred to as <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="nofollow noreferrer">stars and bars</a>.</p>
<p>Imagine if you will you have $n$ distinct boxes and $k$ identical balls.</p>
<p>We arrange a sequence of $k$ $\star$'s and $n-1$ $|$'s to denote how the balls are distributed. The number of stars to the left of the leftmost bar will correspond to the number of balls in the first box. The number of stars between the $(i-1)$'st and $i$'th bar will correspond to the number of balls in the $i$'th box. And the number of stars after the $(n-1)$'st bar will correspond to the number of balls in the $n\text{'th}$ box.</p>
<p>E.g.: with $n=4$ and $k=5$ the sequences below correspond to the following arrangements:</p>
<p>$\star|\star\star|\star|\star$ corresponds to $(1,2,1,1)$</p>
<p>$\star\star\star\star|||\star$ corresponds to $(4,0,0,1)$</p>
<p>Recognize now that there is a <a href="https://en.wikipedia.org/wiki/Bijection" rel="nofollow noreferrer">bijection</a> between the ways to arrange the balls in the boxes and the ways to arrange the stars and bars in a line. If we know how to count the number of possibilities for one of these problems, then it must be the same answer for the other as well.</p>
<p>The number of ways to arrange $n-1$ $|$'s and $k$ $\star$'s will be $\binom{n+k-1}{k} = \binom{n+k-1}{n-1}$.</p>
<p>This is a common enough problem, that it even receives its own special notation. $\left(\!\!\binom{n}{k}\!\!\right)$. <a href="http://mathworld.wolfram.com/Multichoose.html" rel="nofollow noreferrer">Multichoose @ WolframMathworld</a></p>
<hr>
<p>As an aside, from the conversation <a href="https://math.stackexchange.com/questions/1836185/how-many-ways-are-there-to-choose-5-ice-cream-cones-if-there-are-10-flavors/1836211?noredirect=1#comment3754025_1836211">here</a>, some people are taught how to answer the question where every box must contain at least one ball first, and use that result to derive the result above.</p>
<p>With $n$ distinct boxes, and $n+k$ identical balls, where each box must contain at least one ball, imagine laying $n+k$ $\star$'s in a line. We place dividers between the balls to designate as above where one box ends and another box begins. Due to the constraint that every box must have at least one ball, we must pick $n-1$ spaces from the $n+k-1$ available spaces in which to place dividers.</p>
<p>There are then $\binom{n+k-1}{n-1}$ solutions with all positive integers. Removing one ball from every box then makes it so there are $k$ balls total to place and it is now allowed to be non-negative as opposed to strictly positive, yielding the same result as the first method of explaining.</p>
|
825,703 | <p>I have been working with vector spaces for a while and I now take for granted what the vector space does. I feel like I dont really understand why multiplication and addition must be defined on a vector space. For example, it feels like adding two vectors and having their sum contained within the space is just a name for a vector space and I dont get what necessarily happens IF the two vector's sum arent in the space. In other words, I dont know why must addition and multiplication must be defined on a vector space, is it to take advantage of nice properties? Thanks!</p>
| Ittay Weiss | 30,953 | <p>Imagine the plane $\mathbb R^2$ as a vector space. Now truncate it on all sides so you are just left with a little rectangle about the origin. Now, you this is your world and you take two vectors in it and add them, you may land inside the rectangle or you may not. You don't want to have to worry about such issues, and thus simply make the assumption that your vector space is not truncated. That you can always add vectors and stay firmly in the space. Same goes for scalar product.</p>
|
825,703 | <p>I have been working with vector spaces for a while and I now take for granted what the vector space does. I feel like I dont really understand why multiplication and addition must be defined on a vector space. For example, it feels like adding two vectors and having their sum contained within the space is just a name for a vector space and I dont get what necessarily happens IF the two vector's sum arent in the space. In other words, I dont know why must addition and multiplication must be defined on a vector space, is it to take advantage of nice properties? Thanks!</p>
| Empy2 | 81,790 | <p>A lot of the calculus problems we can solve are linear. For these problems, if we know two solutions, then the sum is another solution. So is a scalar multiple of either. So the set of solutions to a large set of problems is a vector space. It makes sense to understand vector spaces.</p>
<p>Many things aren't vector spaces : if you take two vectors on the surface of a sphere, their sum is usually not on the sphere. </p>
|
2,245,631 | <blockquote>
<p>$x+x\sqrt{(2x+2)}=3$</p>
</blockquote>
<p>I must solve this, but I always get to a point where I don't know what to do. The answer is 1.</p>
<p>Here is what I did: </p>
<p>$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$</p>
<p>Then I got:
$-2x^{3}-x^{2}-6x+9=0$ </p>
| Jack | 138,160 | <p>As in the comment $x-1$ is a factor of $f(x) = -2x^3 -x^2-6x+9$. After an easy long division we get $f(x) = (x-1)(-2x^2-3x-9)$. From this we can use the quadratic equation to see the other two roots are not real.</p>
<p>We can do the long division in the following way:</p>
<p>We need to divide $-2x^3-x^2-6x+9$ by $x-1$ so we can ask what times $x-1$ gives the first term $-2x^3$? This is clearly $\boxed{-2x^2}$. But this also gives $+2x^2$ that we didn't want. So we need to take this away from what is left: we now have $-x^2-6x+9 - (+2x^2) = -3x^2-6x +9$. Again we look for a term that when multiplied with $x-1$ gives the highest order term ($-3x^2$). This is $\boxed{-3x}$ but this gives an extra $+3x$. Now we are left with $-6x+9 - (+3x) = -9x+9$. Here we see that this is $\boxed{-9}$ times $(x-1)$. Adding together the terms that we used along the way, we have $-2x^2-3x-9$.</p>
|
23,846 | <p>I'm stuck with this algebra question.</p>
<p>I try to prove that the exterior algebra $R$ over $k^d$, that is, the $k$-algebra that is generated by $x_1,\ldots,x_d$ and $x_ix_j=- x_jx_i$ for each $i,j$, has just one simple module which is not faithful.</p>
<p>I think the only simple module is $k$, but I am not really sure if my idea does work or not.</p>
<p>Can I use the fact $(x_i)^2=0$ for all $i$, then $k$ has cyclic subrings? If yes, then HOW?</p>
<p>Also, one more question: if $k$ is finitely generated, is that enough to say that $R$ is Artinian?
Thank you</p>
| rschwieb | 29,335 | <p>Assuming that $k$ is a field for the same reasons Mariano gave, the short answer is:</p>
<blockquote>
<p>The exterior algebra is <strong><em>local</em></strong>, and so any simple module over it looks like $R/M$ where $M$ is the unique maximal right ideal.</p>
</blockquote>
<p>This is detailed in <a href="https://math.stackexchange.com/a/135957/29335">a similar question here</a>.</p>
<p>This isoclass of module is clearly not faithful since $M$ annihilates it.</p>
|
1,685,895 | <blockquote>
<blockquote>
<p>Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.</p>
</blockquote>
</blockquote>
<hr>
<p>My attempt:</p>
<p>$\displaystyle \binom{n}{7}=\binom{n}{8} $</p>
<p>$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$</p>
<p>$$ \frac{6}{7!} = \frac{n-7}{40320} $$</p>
<p>$$ n-7 = 48 $$</p>
<p>$$ n=55 $$</p>
| Upstart | 312,594 | <p>The coefficient of $x^7$ will be ${n\choose 7}{2^{n-7}\over 3^7}$ and the coefficient of $x^8$ will be ${n\choose 8}{2^{n-8}\over 3^8}$ and not what you have got .you just missed out the factor of 3 that plays a role as a coefficient.So equating it you get
$${n\choose 7}{2^{n-7}\over 3^7}={n\choose 8}{2^{n-8}\over 3^8}$$
and get the answer as $55$</p>
|
155,024 | <p>I am trying to plot a polynomial inside <code>Manipulate</code> with indexed coefficients.</p>
<p>Surprisingly, only an empty plot is generated.</p>
<p>Any help welcome.</p>
<pre><code>Block[{n = 3},
Manipulate @@ {
Column[{
Sum[a[i] x^i, {i, 0, n}],
Plot[Evaluate@Sum[a[i] x^i, {i, 0, n}], {x, -5, 5}]
}],
Sequence @@ Table[{{a[i], 0}, -2, 2}, {i, 0, n}]
}]
</code></pre>
<p>Regards
Robert</p>
| rhermans | 10,397 | <p>Best so far is <code>BoolEval[Rest[y] == Most[y]]</code></p>
<p><img src="https://i.stack.imgur.com/ensyM.png" alt="Mathematica graphics"></p>
<hr>
<pre><code>data = Union @ RandomReal[E + {0, 10^-7}, 10^6];
</code></pre>
<hr>
<pre><code>pwb1[y_] := Boole[SameQ @@@ Transpose[{Rest[y], Most[y]}]]
pwb1[data]; // RepeatedTiming
(* {0.730, Null} *)
</code></pre>
<hr>
<pre><code>pwb2[y_] := Inner[Boole @* SameQ, Rest[y], Most[y], List]
pwb2[data]; // RepeatedTiming
(* {0.9117, Null} *)
</code></pre>
<hr>
<pre><code>pwb3[y_] := Boole @ Inner[SameQ, Rest[y], Most[y], List]
pwb3[data]; // RepeatedTiming
(* {0.66, Null} *)
</code></pre>
<hr>
<pre><code>pwb4[y_] := Boole[Developer`PartitionMap[SameQ, y, 2, 1]]
pwb4[data]; // RepeatedTiming
(* {0.55, Null} *)
</code></pre>
<hr>
<pre><code>pwb5[y_] := Unitize @ Differences[y]
pwb5[data]; // RepeatedTiming
(* {0.024, Null} *)
</code></pre>
<hr>
<pre><code>pwb6[y_] := 1 - Unitize @ Differences[y]
pwb6[data]; // RepeatedTiming
(* {0.0269, Null} *)
</code></pre>
<hr>
<pre><code>pwb7[y_] := 1 - Unitize[Rest[y] - Most[y]]
pwb7[data]; // RepeatedTiming
(* {0.0203, Null} *)
</code></pre>
<hr>
<pre><code>pwb8[y_] := 1 - Unitize[Rest[y] - Most[y], 1*^-17]
pwb8[data]; // RepeatedTiming
(* {0.0233, Null} *)
</code></pre>
<hr>
<pre><code>pwb9[y_] := ListConvolve[{1, 1}, y, {-1, 1}, {}, Times, SameQ]
pwb9[data]; // RepeatedTiming
(* {0.712, Null} *)
</code></pre>
<hr>
<pre><code>pwb10[y_] := BoolEval[Rest[y] == Most[y]]
pwb10[data]; // RepeatedTiming
(* {0.00671, Null} *)
</code></pre>
<hr>
<pre><code>pwb11[y_] := Subtract[1, Unitize[Subtract[Rest[y], Most[y]]]]
pwb11[data]; // RepeatedTiming
(* {0.0165, Null} *)
</code></pre>
|
1,180,199 | <p>I am not quite sure how to deal with discrete IVP</p>
<p>Find self-similar solution
\begin{equation}
u_t=u u_x\qquad -\infty <x <\infty,\ t>0
\end{equation}</p>
<p>satisfying initial conditions</p>
<p>\begin{equation}
u|_{t=0}=\left \{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{equation}</p>
<p>Here is my attempt</p>
<p>Characteristic equation:
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
\frac{dx}{dt} &=\ -u = -f(C)\\
x &=\ C - tf(C)\\
\end{align}</p>
<p>Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) = u|_{t=0}=\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
\begin{align}
x = C+t \left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
Where the characteristic lines are never across each other</p>
| EditPiAf | 418,542 | <p>This is a Riemann problem for the PDE <span class="math-container">$u_t - u u_x = 0$</span>, which cannot be solved by applying the method of characteristics alone: weak solutions must be considered. This PDE becomes equal to the inviscid Burgers' equation <span class="math-container">$v_t + vv_y = 0$</span> under the transformation <span class="math-container">$y := -x$</span>. The initial data transforms as <span class="math-container">$v(y,0) = 1$</span> if <span class="math-container">$y < 0$</span> and <span class="math-container">$v(y,0) = -1$</span> if <span class="math-container">$y \geq 0$</span>. The Lax entropy conditions states that the solution is a shock wave, which speed <span class="math-container">$s = \frac{1}{2} (1 -1) = 0$</span> is given by the Rankine-Hugoniot condition. Finally, a stationary shock solution is obtained:
<span class="math-container">$$
u(x,t) = v(-x,t) = \left\lbrace
\begin{aligned}
&{-1} && \text{if}\quad x\leq 0\\
&1 && \text{if}\quad x> 0
\end{aligned}\right.
$$</span></p>
|
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