qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,028,986 | <p>How is this integral
<span class="math-container">$$\dfrac{1}{4} \int_{0}^{4\pi} \left| \cos \theta \right| \; d\theta$$</span>
equal to
<span class="math-container">$$\dfrac{1}{2} \int_{0}^{2\pi} \left| \cos \theta \right| \; d\theta$$</span> </p>
<p>On attempting to solve this integral, found this on a solution manual, I know how to integrate it, but I don't know how are these two equal to one another?</p>
| Community | -1 | <p>You could split the integral: <span class="math-container">$$\int_{0}^{4 \pi} |\cos(\theta)| d\theta = \int_{0}^{2 \pi} |\cos(\theta)| d\theta + \int_{2 \pi}^{4 \pi} |\cos(\theta)| d\theta,$$</span> then consider the change of variable <span class="math-container">$\alpha = \theta - 2 \pi$</span>, so the RHS above becomes <span class="math-container">\begin{align} \int_{0}^{2 \pi} |\cos(\theta)| d\theta + \int_{0}^{2 \pi} |\cos(\alpha +2\pi)| d\alpha \\ = \int_{0}^{2 \pi} |\cos(\theta)| d\theta + \int_{0}^{2 \pi} |\cos(\alpha)| d\alpha,\end{align}</span> since the cosine function is <span class="math-container">$2 \pi$</span> periodic. Who cares whether we use <span class="math-container">$\alpha$</span> or <span class="math-container">$\theta$</span> as the variable of integration in the second term, i.e., <span class="math-container">$$ \int_{0}^{4 \pi} |\cos(\theta)| d\theta = \int_{0}^{2 \pi} |\cos(\theta)| d\theta + \int_{0}^{2 \pi} |\cos(\theta)| d\theta.$$</span> The rest easily follows, regarding your question.</p>
<p>In a similar fashion, as in the suggestion made by Snookie, you can deduce <span class="math-container">$\int_{0}^{2 \pi} |\cos(\theta)| d\theta = 2\int_{0}^{ \pi} |\cos(\theta)| d\theta$</span>, and furthermore <span class="math-container">$\int_0^{\pi} |\cos(\theta)| d \theta = \int_0^{\frac{\pi}{2}} \cos(\theta) d \theta + \int_0^{\frac{\pi}{2}} \sin(\theta) d \theta = 2$</span>. </p>
<p>Therefore, <span class="math-container">$\frac{1}{4} \int_{0}^{4\pi} |\cos(\theta)| d \theta = \frac{1}{4}\times 4 \times 2 = 2$</span>. </p>
|
1,081,383 | <p>Assume that $A$ is an $n\times n$ symmetric positive-definite matrix.</p>
<p>Prove that:</p>
<blockquote>
<p>the element of $A$ with maximum magnitude must lie on the diagonal. </p>
</blockquote>
| megas | 191,170 | <p>Hint:
Assume that this is not the case, <em>i.e.</em>, the entry of maximum magnitude is an off-diagonal entry $A_{ij}=A_{ji}$, with $j \neq i$.
To show that $A$ cannot be PSD, it suffices to construct an $x$ (a certificate)
such that
$$
x^{T}Ax < 0.
$$
How can we find one such $x$?
In order to answer that, you have to make clear what $x^{T}Ax$ really is. How can you write it as a sum? Which entries of $x$ multiply $A_{ij}$?</p>
|
1,081,383 | <p>Assume that $A$ is an $n\times n$ symmetric positive-definite matrix.</p>
<p>Prove that:</p>
<blockquote>
<p>the element of $A$ with maximum magnitude must lie on the diagonal. </p>
</blockquote>
| Alex Silva | 172,564 | <p>Let $a_{max}$ be the element of $\mathbf{A}$ with maximum magnitude. Assume that it is not lie on the diagonal. Thus, there is exist a $2\times 2$ principal minor equal to $ab- a_{max}^2 \leq 0$ for some $a$ and $b$ in the main diagonal. Hence, as the matrix is positive definite, $a_{max}$ should be actually in the main diagonal.</p>
|
2,572,802 | <p>We were asked to find the number of five digit numbers $N=d_1d_2d_3d_4d_5$, where $d_i$ is the $i$th digit of the number and $d_1 < d_2 < d_3 < d_4 < d_5 $. The solution was trivial as for a given selection of five random distinct digits, there is only one way to arrange them in strict increasing order. Now to raise the difficulty level of the question, our teacher changed all the "less than" signs into "less than or equal to" signs, thus yielding the new constraint $d_1 \leq d_2 \leq d_3 \leq d_4 \leq d_5$.</p>
<p>I proceeded as follows .....</p>
<p>Let $x_i = d_{i+1} - d_i$. So, $$ x_1 + x_2 + x_3 + x_4 = d_5 - d_1,$$ or $$d_1 + x_1 + x_2 + x_3 + x_4 = d_5 \leq 9.$$ But $d_1 \geq 1, x_i \geq 0$ for all $i $ belonging to the set ${1,2,3,4} $. So, the problem reduces to find the number of non negative integral solutions of the inequality $$d_1 + x_1 + x _2 + x_3 + x_4 \leq 8,\quad
(*)$$ which comes out to be $^{13} C_ {5}$. </p>
<p>But looking at the simplicity of the answer, I am bound to think that there must exist some elegant approach to this problem. In Combinatorics, every one thinks differently, which is why I ask this question. I just want to explore the different ways one can approach a problem. Please note that I am just a beginner in Combinatorics. I am familiar with Permutations and Combinations, but I do not know a single thing in Graph Theory and all those "advanced" concepts. So please answer accordingly.</p>
| Especially Lime | 341,019 | <p>Here's an essentially equivalent approach that may be what you're looking for. Instead of looking for possible values $d_1,...,d_5\in\{1,...,9\}$ such that $d_1\leq d_2...\leq d_5$, try setting $c_i=d_i+i$ for each $i$. Now $c_1,...,c_5\in\{2,...,14\}$ and $c_1<c_2<...<c_5$, so there are ${}^{13}C_5$ ways to choose the $c_i$. Each of these choices can be converted back to a (different) valid choice for the $d_i$.</p>
|
1,730,352 | <p>I am really struggling to understand what modular forms are and how I should think of them. Unfortunately I often see others being in the same shoes as me when it comes to modular forms, I imagine because the amount of background knowledge needed to fully appreciate and grasp the constructions and methods is rather large, so hopefully with this post some clarity can be offered, also for future readers.. The usual definitions one comes across are often of the form: (here taken from <a href="https://en.wikipedia.org/wiki/Modular_form">wikipedia</a>)</p>
<blockquote>
<p>A modular form is a (complex) analytic function on the upper
half-plane satisfying a certain kind of functional equation with
respect to the group action of the modular group, and also satisfying
a growth condition.</p>
<p>A modular form of weight $k$ for the modular group $$
\text{SL}(2,\mathbb{Z})=\left\{\begin{pmatrix} a & b \\ c & d
\end{pmatrix}| a,b,c,d \in \mathbb{Z} , ad-bc = 1 \right\} $$ is a
complex-valued function $f$ on the upper half-plane $\mathbf{H}=\{z
\in \mathbb{C},\text{Im}(z)>0 \}$, satisfying the following three
conditions:</p>
<ol>
<li>$f$ is a holomorphic function on $\mathbf{H}.$</li>
<li>For any $z \in \mathbf{H}$ and any matrix in $\text{SL}(2,\mathbb{Z})$ as above, we have: $$
f\left(\frac{az+b}{cz+d}\right)=(cz+d)^k f(z) $$</li>
<li>$f$ is required to be holomorphic as $z\to i\infty.$</li>
</ol>
</blockquote>
<p><strong>Questions</strong>:</p>
<ul>
<li>(a): I guess what I'm having least familiarity with is the <em>modular group</em> part. My interpretation of $\text{SL}(2,\mathbb{Z}):$ <em>The set of all $2$ by $2$ matrices, with integer components, having their determinant equal to $1.$</em> But where does the name come from, as in why do we call this set a group and what modular entails?</li>
<li>(b): If I understand correctly, the group operation here is function composition, of type: $\begin{pmatrix}a & b \\ c & d\end{pmatrix}z = \frac{az+b}{cz+d}$ which is also called a linear fractional transformation. How should one interpret the condition $2.$ that $f$ has to satisfy? My observation is that, as a result of the group operation of $\text{SL}$ on a given integer $z,$ the corresponding image is multiplied by a polynomial of order $k$ (which is the weight of the modular form). </li>
<li>(c) The condition $3.$ I interpret as: $f$ should not exhibit any poles in the upper half plane, not even at infinity. About right? </li>
<li>(d) A more general question: Given the definition above, it is tempting to see modular forms as particular classes of functions, much like the Schwartz class of functions, or $L^p$ functions and so on. Is this an acceptable assessment of modular forms?</li>
<li>(e) Last question: It is often said that modular forms have interesting Fourier transforms, as in their Fourier coefficients are often interesting (or known) sequences. Is there an intuitive way of seeing, from the definition of modular forms, the above expectation of their Fourier transforms?</li>
</ul>
| Andrea Mori | 688 | <p>(a) ${\rm SL}_2(\Bbb Z)$ is a group in the sense that is an example of the algebraic structure called <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29">group</a>. :)</p>
<p>(b) That's not the group operation. The group operation in ${\rm SL}_2(\Bbb Z)$ (and in fact in any <a href="https://en.wikipedia.org/wiki/General_linear_group">linear group</a>) is matrix multiplication. What you describe is the <a href="https://en.wikipedia.org/wiki/Group_action">action</a> of the group ${\rm SL}_2(\Bbb Z)$ on the upper halfplane $\cal H$.</p>
<p>(c) Exactly.</p>
<p>(d) Modular forms lift to functions in $L^2({\rm GL}_2(\Bbb Q)\backslash{\rm GL}_2(\Bbb A))$ where $\Bbb A$ is the ring of <a href="https://en.wikipedia.org/wiki/Adele_ring">rational adeles</a>.</p>
<p>(e) I would say no. The importance of their Fourier coefficients becomes evident only after the <a href="https://en.wikipedia.org/wiki/Hecke_operator">Hecke operators</a> are introduced. The analysis of the way Hecke operators act on modular forms is the main introductory link between the theory of modular forms and arithmetic since, for instance, it allows to show that the space of modular forms of fixed weight (which is finite dimensional) has a basis of modular forms with Fourier coefficients in $\Bbb Z$.</p>
|
3,156,570 | <p>I need to evaluate the following limit:
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c}
$$</span>
for different values of the constant <span class="math-container">$c$</span>.</p>
<p><em>What I've tried thus far:</em></p>
<p>We have that
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c} = \lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
Now I know that <span class="math-container">$1 - e^x \to 0$</span> as <span class="math-container">$x\to 0$</span>. If <span class="math-container">$c\geq 0$</span> then <span class="math-container">$x^c\to 0$</span> as <span class="math-container">$x\to 0$</span> so we have that
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)} = \infty
$$</span>
What I'm not so sure about is what happens when <span class="math-container">$c<0$</span>. If <span class="math-container">$c< 0$</span> then we can write <span class="math-container">$x^c$</span> as <span class="math-container">$x^{-\gamma}$</span> where <span class="math-container">$\gamma = \left|c\right|$</span>. I know that <span class="math-container">$\lim_{x\downarrow 0}x^{-\gamma}\to \infty$</span>, but I'm not sure whether <span class="math-container">$x^{-\gamma}\to \infty$</span> quicker than <span class="math-container">$1 - e^x\to 0$</span> as <span class="math-container">$x\to 0$</span>. I need to know because right now I'm not sure what the term <span class="math-container">$x^c(1 - e^x)$</span> does as <span class="math-container">$x\downarrow 0$</span> when <span class="math-container">$c <0$</span>.</p>
<p><strong>Question:</strong> </p>
<p>How do I evaluate
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
when <span class="math-container">$c< 0$</span>?</p>
| Kavi Rama Murthy | 142,385 | <p><span class="math-container">$\frac x {1-e^{x}} \to -1$</span> as <span class="math-container">$x \to 0+$</span> by L'Hopital's Rule. Multiply numerator by <span class="math-container">$x$</span> and change <span class="math-container">$x^{c}$</span> in the denominator to <span class="math-container">$x^{c+1}$</span>. That doesn't change the expression. Is the rest clear now?</p>
|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| alephzero | 223,485 | <p>Absolutely <strong><em>nothing</em></strong> in physics is completely described by a PDE, if you look at a sufficiently small resolution, because space and time are not continuous. (Since the OP has said in a comment that he doesn't know much physics, google for "Planck length" for more information.) </p>
<p>However almost <strong><em>everything</em></strong> in physics is described at a fundamental level by <em>conservation laws</em> which are most naturally expressed mathematically as <em>integral equations</em> not as differential equations.</p>
<p>Integral equations can be converted to differential equations with some loss of generality - i.e. you exclude solutions of the integral equations which are not sufficiently differentiable. But the solutions you might have excluded are interesting and useful from a physicist's point of view, so excluding them simply "because PDEs are easier to work with than integral equations" is throwing the baby out with the bathwater. </p>
<p>Hence, "weak solutions of PDEs" are a thing worth studying. Of course if you want to convert any interesting theorems about weak solutions back into the language of integral equations, feel free to do that - or even better, figure out a way to unify the two subjects using nonstandard analysis, or something similar! (Nonstandard analysis corresponds very well with physicists' idea of "infinitesimal quantities" which can be treated mathematically as if they are numbers even though they are not!)</p>
|
3,854,785 | <p>Considering <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>
to show <span class="math-container">$$8\sin^4(x) = 3 - 4\cos(2x) +\cos(4x)$$</span>
Assuming I did not how to initially do this proof properly, how would I be able to set up a proof that is still valid to show that <span class="math-container">$$1 = 1$$</span></p>
<p>The steps were square both sides of <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>, then you get <span class="math-container">$$2(1 - \cos(2x))^2 = 3 - 4\cos(2x) +\cos(4x)$$</span>
which can lead to <span class="math-container">$$\cos^2(2x) + \sin^2(x) = 1$$</span> which is true
Apparently this is not properly valid. My feedback was to put "RTP" before <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>. How can you make a proof that meets in the middle valid?</p>
| Math Lover | 801,574 | <p><span class="math-container">$3 - 4 \cos(2x) + \cos(4x) = 3 - 4 (1 - 2 \sin^2 x) + (2 \cos^2 2x - 1)$</span></p>
<p><span class="math-container">$ = 8 \sin^2 x - 2 + 2 (1 - 2 \sin^2 x)^2 = 8 \sin^4x$</span></p>
|
2,398,215 | <p>If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.</p>
<p>(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$</p>
<p>(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$</p>
<p>(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.</p>
<p>Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.</p>
<p>Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.</p>
| trying | 309,917 | <p>Here the answer to your first question, where use is made of one of the main corollaries for continuous functions (see $(4)$ below).</p>
<p>Let $f:\mathbb{R}\rightarrow\mathbb{R}$ and $m\in\mathbb{R}$, $|m|\ne1$
$$f \text{ is continuous}\tag{1}$$
$$\forall x\in\mathbb{R},f(x)=f(mx)\tag{2}$$
<strong>Theorem.</strong> $f\text{ is constant}$</p>
<p><strong>Proof.</strong> Let $A_x=\{\xi\in\mathbb{R}|\exists n\in\mathbb{Z},\xi=m^nx\}$ and $\mathcal{A}=\{A_x, x\in\mathbb{R}\}$</p>
<p>Being $(2)$, then:</p>
<p>$$\forall x\in\mathbb{R},f(x)=f(mx)\iff f(A_x)=\{f(x)\}$$</p>
<p>Then there must follow that:</p>
<p>$$\forall x\in\mathbb{R}, \exists \mathcal{B}_x\in\mathcal{A}, f^{-1}(x)=\cup\mathcal{B}_x\tag{3}$$</p>
<p>Being $(1)$, then:</p>
<p>$$f \text{ continuous }\implies\forall x\in\mathbb{R},f^{-1}(x)\text{ is closed}\tag{4}$$</p>
<p>Notice that every subfamily of $\mathcal{A}$ not including $A_0=\{0\}$ has a union that is not closed, while every one including it has a closed union, that is all the subfamilies with closed union are not disjoint.</p>
<p>$$\forall \mathcal{F}\subset\mathcal{A}, \cup\mathcal{F}\text{ is closed}\iff A_0\in\mathcal{F}\tag{5}$$</p>
<p>So from $(3)$, $(4)$, and $(5)$ it is:
$$\forall x, f^{-1}(x)\supset A_0$$
But it is a fact that:
$f^{-1}(x)\cap f^{-1}(y)=\emptyset\iff f(x)\ne f(y)$ so:
$$\forall x, f(x)=f(0)$$</p>
|
2,185,489 | <blockquote>
<p>Let $\xi \in \mathcal{L}^2(\Omega,P)$ be a random variable with finite variance. Show that $$(E(\xi))^2 \leq E(\xi^2)$$ </p>
</blockquote>
<p>Since $$\operatorname{Var}(\xi) := E(\xi^2) - (E(\xi))^2$$ this boils down to showing $$\operatorname{Var}(\xi) \geq 0$$ which is quite restrictive. Since the lecture is not really good, I have not much of an idea what I should use. Has anyone a hint for me?</p>
| Kenny Wong | 301,805 | <p>For an alternative proof, use Cauchy-Schwarz.</p>
<p>$$ E(\xi)^2 = \left( \int_\Omega \xi \right)^2 = \left(\int_\Omega 1.\xi \right)^2 \leq \int_\Omega 1^2 \times \int_\Omega \xi^2 = E(\xi^2).$$</p>
|
2,524,890 | <p>I know that if matrix $a$ is similar to matrix $b$ then $\operatorname{trace} a=\operatorname{trace} b$.</p>
<p>Does it go to the other side?</p>
<p>Thanks.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>we get
$$(b-4)^2=16(1-a)$$ i hope this solves your problem</p>
|
3,535,088 | <p>A function <span class="math-container">$\phi:X\rightarrow Y$</span> between two topological space <span class="math-container">$(X,\tau)$</span> and <span class="math-container">$(Y,\sigma)$</span> is continuous in <span class="math-container">$x\in X$</span> if and only if for any open set <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> such that <span class="math-container">$\phi(x)\in V$</span> there is a open set <span class="math-container">$U$</span> in <span class="math-container">$X$</span> such that <span class="math-container">$x\in U$</span> and <span class="math-container">$\phi(U)\subseteq V$</span>.</p>
<p>Well with this definition we consider the euclidean space <span class="math-container">$\mathbb{R}^2$</span> and the functions <span class="math-container">\begin{align}f&:\mathbb{R}^2\owns(x,y)\rightarrow(x+y)\in\mathbb{R}\\g&:\mathbb{R}^2\owns(x,y)\rightarrow(xy)\in\mathbb{R}\end{align}</span> and we prove to demonstrate that are continuous functions.</p>
<p>Therefore we consider the point <span class="math-container">$(\alpha,\beta)$</span> and its image under <span class="math-container">$f$</span> or <span class="math-container">$g$</span> and we chose a open range <span class="math-container">$(a,b)$</span> that contains <span class="math-container">$f(\alpha,\beta)$</span> and <span class="math-container">$g(\alpha,\beta)$</span>: so it result that <span class="math-container">$a<\alpha+\beta<b$</span> and <span class="math-container">$a<\alpha\beta<b$</span>.</p>
<p>Now let be
<span class="math-container">$$
\varepsilon_f=\inf\Big\{d\big((\alpha,\beta);(x,a-x)\big),d\big((x,y);(x,b-x)\big)\Big\}\\
\varepsilon_g=\inf \biggl\{d\Big((\alpha,\beta);\Big(x,\frac{a}{x}\Big)\Big),d\biggl((x,y);\Big(x,\frac{b}{x}\Big)\Big)\biggl\}.
$$</span>
If we consider the balls <span class="math-container">$B_f:=B\big((\alpha,\beta),\varepsilon_f\big)$</span> and <span class="math-container">$B_g:=B\big((\alpha,\beta),\varepsilon_g\big)$</span> that contain <span class="math-container">$(\alpha,\beta)$</span> is true that <span class="math-container">$f(B_f)\subseteq(a,b)$</span> and <span class="math-container">$g(B_g)\subseteq(a,b)$</span> and that if <span class="math-container">$(x,y)\in B_f$</span> or <span class="math-container">$(x,y)\in B_g$</span> it result that <span class="math-container">$f(x,y)\in(a,b)$</span> or <span class="math-container">$g(x,y)\in(a,b)$</span>? Then if this is true will we have proved the continuity of <span class="math-container">$f$</span> and <span class="math-container">$g$</span>?</p>
<p>Could someone help me, please?</p>
| Lorenzo Cecchi | 523,193 | <p>Yes, it is true. Since <span class="math-container">$v_n\to v$</span> weakly <span class="math-container">$L^\infty$</span>, you have that <span class="math-container">$\langle u_n v_n,x\rangle=\langle u_n,v_n x\rangle$</span> (I'm assuming you are working with real valued functions) and <span class="math-container">$v_n x\to v x$</span> weakly <span class="math-container">$L^2$</span>, and therefore conclusion follows from the fact that a strong convergence and a weak one allow to have pass to limit in the scalar product.</p>
|
3,293,082 | <p>I know this question was asked on this site, but I didn't understand the answer. Could someone give me the simplest explanation of this? (High school level explanation)</p>
| fleablood | 280,126 | <p>They are completely different.</p>
<p><span class="math-container">$\sec x = \frac 1{\cos x} = 1\div \cos x$</span>. This is the multiplicative reciprocal, which is sometimes call the <em>multiplicative</em> inverse.</p>
<p><span class="math-container">$\arccos x$</span> is the function so that if <span class="math-container">$x = \cos y$</span> then <span class="math-container">$\arccos x$</span> is "going backwards" to get <span class="math-container">$y$</span> for which <span class="math-container">$x$</span> is the <span class="math-container">$\cos y$</span>. so <span class="math-container">$\arccos x$</span> is defined as: If there are any <span class="math-container">$w$</span> so that <span class="math-container">$\cos w = x$</span> then one of those <span class="math-container">$w$</span> will be between <span class="math-container">$0$</span> and <span class="math-container">$\pi$</span>; we define <span class="math-container">$\arccos x$</span> to be that <span class="math-container">$w$</span>.</p>
<p>This is called the <em>functional</em> inverse.</p>
<p>That is it.</p>
<p>THE END.</p>
<p>...........................</p>
<p>Still reading? Well... this is probably why you got confused:</p>
<p>A <em>multiplicative</em> inverse of a value <span class="math-container">$K$</span> is a value <span class="math-container">$m$</span> so that <span class="math-container">$m \times K = 1$</span>. In other words <span class="math-container">$m = \frac 1K$</span>. We write the multiplicative inverse of <span class="math-container">$K$</span> as <span class="math-container">$\frac 1K$</span> but we also say <span class="math-container">$K^{-1}$</span> meaning <span class="math-container">$K$</span> raised to the negative <span class="math-container">$1$</span> power. (Remember <span class="math-container">$a^{-m} = \frac 1{a^m}$</span>.) This will cause trouble later.</p>
<p>So the <span class="math-container">$\sec x = $</span> the <em>multiplicative</em> inverse of <span class="math-container">$\cos x$</span> means <span class="math-container">$\sec x = \frac 1{\cos x}$</span>. Now we can write <span class="math-container">$\sec x = (\cos x)^{-1}$</span> <em>!!!!!IF!!!!!!</em> we mean "the value of <span class="math-container">$\cos x$</span> raised to the negative <span class="math-container">$1$</span> power.</p>
<p>The <em>functional</em> inverse of a function <span class="math-container">$f(x)$</span> is a function <span class="math-container">$g(x)$</span> so that <span class="math-container">$g(f(x)) = x$</span>. And we often write the inverse function of <span class="math-container">$f(x)$</span> as <span class="math-container">$f^{-1}(x)$</span> but notice <strong><em>!!!!!! THIS IS VITALLY IMPORTANT AND YOU WILL !!!!DIE!!!! HORRIBLY !!!!!! IF YOU MISUNDERSTAND IT!!!!!!</em></strong> this does <em>NOT</em> mean <span class="math-container">$(f(x))^{-1}$</span> which is "the value of <span class="math-container">$f(x)$</span> raised to the negative <span class="math-container">$1$</span> power"; this, <span class="math-container">$f^{-1}(x)$</span>, means "the function that 'reverses' <span class="math-container">$f$</span> and gets us back to where we started".</p>
<p>So <span class="math-container">$\arccos x$</span> is the <em>functional</em> inverse of <span class="math-container">$\cos x$</span>. That is to say, <span class="math-container">$\arccos x$</span> is the function where <span class="math-container">$\arccos (\cos x) = x$</span>. It "undoes" the <span class="math-container">$x \mapsto \cos x$</span> to get us back to <span class="math-container">$x$</span>. We can write <span class="math-container">$\arccos x = \cos^{-1}(x)$</span> <em>!!!!!!IF!!!!!!</em> we mean "the function that 'reverses' <span class="math-container">$\cos$</span>" and we <strong><em>!!!!!ABSOLUTELY UNDER RISK OF !!!!HORRIBLE!!!! DEATH!!!! !N!E!V!E!R! CONFUSE IT WITH !!!!!!</em></strong> <span class="math-container">$(\cos x)^{-1}$</span> meaning "raising <span class="math-container">$\cos x$</span> to the negative <span class="math-container">$1$</span> power.</p>
<p>It is VERY unfortunate that we use two very similar looking notation for two <em>completely</em> different concepts. It causes a <em>lot</em> of confusion in students.</p>
<p>(Actually <em>logically</em> the concepts are similar. <span class="math-container">$K^{-1} = \frac 1K$</span> undoes a <em>multiplication</em> whereas <span class="math-container">$f^{-1}(x)$</span> undoes a <em>function</em>. They both undo and get you back to where you started but one undoes multiplication. They other undoes a function.)</p>
<p>(I don't know. Maybe I shouldn't have mentioned that and maybe that's just more confusing. Forget I said anything.)</p>
|
2,000,940 | <p>Suppose $A \subseteq \mathbb{R}$ is measurable and $f\colon A \to \mathbb{R}$ is Lipschitz on the set $A$, i.e there is some $K\ge 0$ such that $\lvert f(x)-f(y)\rvert \le K \lvert x-y\rvert$ for $x,y \in A$. </p>
<p>I'm trying to prove that
$$
m^\ast(f(E)) \le K\,m^\ast(E)\textrm{ for every set }E \subseteq A.
$$</p>
<p>I've tried covering the set $E$ by intervals, yet the function $f$ may not have been defined outside the set $A$. Also, approaching the set $E$ from inside can encounter problem when $E$ is non-measurable.</p>
| ChristophorusX | 239,293 | <p>We do not know if the function can be extended to $\mathbb{R}$ by continuous extension theorem, so we handle it using the method we use to prove Growth Lemma.</p>
<p>For $\forall E \subseteq A$, we can cover $E$ with $\{I_k\}$, s.t.
$$
E \subseteq \cup_k I_k\quad \textrm{and}\quad \sum_k m(I_k) \le m^\ast (E) + \varepsilon
$$
Since $A$ is measurable, we have $\{I^\prime_k\}$ with $I_k^\prime = \ A \cap I_k$ measurable for every $k$.
$$
E \subseteq \cup_k I_k^\prime\quad \textrm{and}\quad \sum_k m(I^\prime_k) \le m^\ast (E) + \varepsilon
$$
For $\forall x, y \in I_k^\prime$,</p>
<p>$\because x, y \in A$ and being Lipschitz on the set $A$.</p>
<p>$\therefore |f(x) - f(y)| \le K |x - y|$.</p>
<p>$\Rightarrow \textrm{diam}(f(I_k^\prime)) = \textrm{sup}\{|f(x) - f(y)|:x,y \in I_k^\prime\} \le K m(I_k^\prime)$</p>
<p>Since $f$ maps $A$ to $\mathbb{R}$, and interval $Q_k$ with $m(Q_k) = {\rm diam}(f(I_k^\prime))$ can cover $f(I_k^\prime)$.</p>
<p>$\Rightarrow m^\ast (f(I_k^\prime)) \le m(Q_k) = {\rm diam}(f(I_k^\prime))$</p>
<p>Hence, $m^\ast (f(I_k^\prime)) \le K m(I_k^\prime)$. Then we can conclude that
\begin{align*}
m^\ast (f(E)) & \le m^\ast(\cup_k f(I_k^\prime))\\ & \le \sum_k m^\ast (f(I_k^\prime))\\ & \le K \sum_k m(I_k^\prime)\\ & \le K \sum_k m(I_k)\\ &\le K (m^\ast (E) + \varepsilon)
\end{align*}</p>
<p>Since $\varepsilon$ is arbitrary, the result follows.</p>
|
125,592 | <p>I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:</p>
<p>$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$</p>
<p>where $u(t)$ is Heavside function</p>
<p>well I applied the formula that says that the convolution of this two signal is equal to</p>
<p>$$Y(f)=X(f)W(f)$$</p>
<p>where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal</p>
<p>well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation:
$\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$</p>
<p>right or not?</p>
<p>Edit</p>
<p>If something is not clear please advice me</p>
| joriki | 6,622 | <p>Funny enough, someone just posted a question on the <a href="https://math.stackexchange.com/questions/125539/power-reduction-formula">Power-reduction formula</a> two hours ago. Using that, you readily get the result </p>
<p>$$\frac{2\pi}{2^n}\binom{n}{n/2}\;.$$</p>
|
454,040 | <p>I need to know whether There exists any continuous onto map from $(0,1)\to (0,1]$</p>
<p>could any one give me any hint?</p>
| martini | 15,379 | <p><strong>Hint</strong>: $(0,1) = (0,\frac 12] \cup [\frac 12, 1)$. Can you map each part onto $(0,1]$?</p>
|
454,040 | <p>I need to know whether There exists any continuous onto map from $(0,1)\to (0,1]$</p>
<p>could any one give me any hint?</p>
| Balbichi | 24,690 | <p>From The Hint of Martini the Map $f(x)=2x; x\in (0,{1\over 2}]$ and $f(x)=1;x\in [{1\over 2},1)$ will work</p>
|
2,396,073 | <p>Let $\omega_1$ be the first uncountable ordinal. In some book, the set $\Omega_0:=[1,\omega_1)=[1,\omega_1]\backslash\{\omega_1\}$ is called the set of countable ordinals. Why? It is obvious that it is an uncountable set, because $[1,\omega_1]$ is uncountable. The most possible reason I think is that for any $x\prec \omega_1$, the set $[1,x)$ is countable. </p>
| Henno Brandsma | 4,280 | <p>Every ordinal is a well-ordered set, and the first one that is uncountable is by definition $\omega_1$. So this set which is uncountable because there are uncountably many different ways to well-order a countable set, is thus called "the set of all countable ordinals". By definition, all $\alpha < \omega_1$ are countable sets, as otherwise they'd contradict the minimality of $\omega_1$, analogous, to $\omega$ being the set of finite ordinals (despite not being finite itself).</p>
<p>$[0,\omega_1]$, including the maximal element $\omega_1$, is the unique compactification of $\omega_1$, also denoted $\omega_1 + 1$. It has one element more than $\omega_1$ so is also uncountable. By the usual definition, every ordinal is the set of all strictly smaller ordinals.</p>
|
2,114,619 | <p>An intruder has a cluster of 64 machines, each of which can try 10^6 passwords per second. How long does it take him to try all legal passwords if the requirements for the password are as follows:</p>
<ul>
<li>passwords can be 6, 7, or 8 characters long</li>
<li>each character is either a lower-case letter or a digit</li>
<li>at least one character has to be a digit</li>
</ul>
<p>My approach is that, for password of length k, there are 36^k possibilities - 26^k forbidden possibilities (because can't be all letters):
∑(k=6 to 8)(36^k−26^k) = 2,684,483,063,360 so
2,684,483,063,360 passwords × (1 second / 64*10^6 passwords) × (1 minute/ 60 seconds) × (1 hour / 60 minutes) = 11.651 hours</p>
<p>Am I approaching this the right way? Is my solution correct?</p>
| Stella Biderman | 123,230 | <p>The methodology is correct, and if the calculator says that the end number is what you say it is, then it's right too :P</p>
|
259,308 | <p>The output of <code>ListPointPlot3D</code> is shown below:
<a href="https://i.stack.imgur.com/ypt73.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ypt73.png" alt="enter image description here" /></a>
I only want to connect the dots in such a way that it forms a ring-like mesh. However, when I use <code>ListPlot3D</code>, the output is like this:
<a href="https://i.stack.imgur.com/zmaI1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zmaI1.png" alt="enter image description here" /></a>
Which is not what I want.</p>
<p>May I know how I should modify the code to achieve the intended result? Thank you.</p>
<p>Here is the data I used (with fewer data points):</p>
<pre><code>{{15.,-4.33154,0.015},{15.4114,-3.89839,0.0154114},{15.7792,-3.46523,0.0157792},{16.1037,-3.03208,0.0161037},{16.3847,-2.59893,0.0163847},{16.6224,-2.16577,0.0166224},{16.8169,-1.73262,0.0168169},{16.968,-1.29946,0.0169681},{17.076,-0.866309,0.017076},{17.1408,-0.433154,0.0171408},{17.1624,-6.01122*10^-16,0.0171624},{17.1408,0.433154,0.0171408},{17.076,0.866309,0.017076},{16.968,1.29946,0.0169681},{16.8169,1.73262,0.0168169},{16.6224,2.16577,0.0166224},{16.3847,2.59893,0.0163847},{16.1037,3.03208,0.0161037},{15.7792,3.46523,0.0157792},{15.4114,3.89839,0.0154114},{14.698,-4.79395,2.99474},{15.1194,-4.31455,3.08061},{15.4564,-3.83516,3.14927},{15.7246,-3.35576,3.2039},{15.9361,-2.87637,3.24701},{16.1009,-2.39697,3.28058},{16.2264,-1.91758,3.30616},{16.3185,-1.43818,3.32492},{16.3815,-0.95879,3.33775},{16.4181,-0.479395,3.34521},{16.4301,-6.65294*10^-16,3.34766},{16.4181,0.479395,3.34521},{16.3815,0.95879,3.33775},{16.3185,1.43818,3.32492},{16.2264,1.91758,3.30616},{16.1009,2.39697,3.28058},{15.9361,2.87637,3.24701},{15.7246,3.35576,3.2039},{15.4564,3.83516,3.14927},{15.1194,4.31455,3.08061},{13.8101,-5.25915,5.85509},{14.2225,-4.73323,6.02996},{14.5189,-4.20732,6.15563},{14.7316,-3.6814,6.24579},{14.8837,-3.15549,6.31026},{14.9917,-2.62957,6.35607},{15.0675,-2.10366,6.3882},{15.1193,-1.57774,6.41015},{15.1527,-1.05183,6.42432},{15.1714,-0.525915,6.43225},{15.1774,-7.29853*10^-16,6.4348},{15.1714,0.525915,6.43225},{15.1527,1.05183,6.42432},{15.1193,1.57774,6.41015},{15.0675,2.10366,6.3882},{14.9917,2.62957,6.35607},{14.8837,3.15549,6.31026},{14.7316,3.6814,6.24579},{14.5189,4.20732,6.15563},{14.2225,4.73323,6.02996},{12.3716,-5.68073,8.48201},{12.7528,-5.11265,8.74341},{13.0014,-4.54458,8.91387},{13.1635,-3.97651,9.02496},{13.2689,-3.40844,9.09726},{13.3374,-2.84036,9.14419},{13.3815,-2.27229,9.17442},{13.4094,-1.70422,9.19355},{13.4263,-1.13615,9.20515},{13.4353,-0.568073,9.21134},{13.4382,-7.88359*10^-16,9.21329},{13.4353,0.568073,9.21134},{13.4263,1.13615,9.20515},{13.4094,1.70422,9.19355},{13.3815,2.27229,9.17442},{13.3374,2.84036,9.14419},{13.2689,3.40844,9.09726},{13.1635,3.97651,9.02496},{13.0014,4.54458,8.91387},{12.7528,5.11265,8.74341},{10.4398,-6.03187,10.7708},{10.7685,-5.42868,11.1099},{10.9656,-4.8255,11.3133},{11.0838,-4.22231,11.4352},{11.1547,-3.61912,11.5083},{11.1971,-3.01594,11.552},{11.2223,-2.41275,11.5781},{11.2372,-1.80956,11.5935},{11.2457,-1.20637,11.6022},{11.25,-0.603187,11.6066},{11.2513,-8.3709*10^-16,11.608},{11.25,0.603187,11.6066},{11.2457,1.20637,11.6022},{11.2372,1.80956,11.5935},{11.2223,2.41275,11.5781},{11.1971,3.01594,11.552},{11.1547,3.61912,11.5083},{11.0838,4.22231,11.4352},{10.9656,4.8255,11.3133},{10.7685,5.42868,11.1099},{8.09191,-6.3005,12.6302},{8.35003,-5.67045,13.033},{8.49481,-5.0404,13.259},{8.57602,-4.41035,13.3858},{8.62155,-3.7803,13.4569},{8.64707,-3.15025,13.4967},{8.66132,-2.5202,13.5189},{8.66922,-1.89015,13.5313},{8.67348,-1.2601,13.5379},{8.67557,-0.63005,13.5412},{8.67619,-8.74371*10^-16,13.5421},{8.67557,0.63005,13.5412},{8.67348,1.2601,13.5379},{8.66922,1.89015,13.5313},{8.66132,2.5202,13.5189},{8.64707,3.15025,13.4967},{8.62155,3.7803,13.4569},{8.57602,4.41035,13.3858},{8.49481,5.0404,13.259},{8.35003,5.67045,13.033},{5.42138,-6.48148,13.986},{5.5955,-5.83333,14.4352},{5.68862,-5.18519,14.6754},{5.73842,-4.53704,14.8039},{5.76505,-3.88889,14.8726},{5.77928,-3.24074,14.9093},{5.78687,-2.59259,14.9289},{5.79089,-1.94444,14.9393},{5.79297,-1.2963,14.9446},{5.79396,-0.648148,14.9472},{5.79425,-8.99486*10^-16,14.9479},{5.79396,0.648148,14.9472},{5.79297,1.2963,14.9446},{5.79089,1.94444,14.9393},{5.78687,2.59259,14.9289},{5.77928,3.24074,14.9093},{5.76505,3.88889,14.8726},{5.73842,4.53704,14.8039},{5.68862,5.18519,14.6754},{5.5955,5.83333,14.4352},{2.53472,-6.57254,14.7843},{2.6163,-5.91529,15.2601},{2.65878,-5.25803,15.5079},{2.6809,-4.60078,15.6369},{2.69241,-3.94352,15.704},{2.69841,-3.28627,15.739},{2.70152,-2.62902,15.7572},{2.70313,-1.97176,15.7666},{2.70395,-1.31451,15.7713},{2.70433,-0.657254,15.7735},{2.70444,-9.12123*10^-16,15.7742},{2.70433,0.657254,15.7735},{2.70395,1.31451,15.7713},{2.70313,1.97176,15.7666},{2.70152,2.62902,15.7572},{2.69841,3.28627,15.739},{2.69241,3.94352,15.704},{2.6809,4.60078,15.6369},{2.65878,5.25803,15.5079},{2.6163,5.91529,15.2601},{-0.452986,-6.5727,14.9932},{-0.467535,-5.91543,15.4747},{-0.475076,-5.25816,15.7243},{-0.478985,-4.60089,15.8537},{-0.481011,-3.94362,15.9207},{-0.48206,-3.28635,15.9555},{-0.482603,-2.62908,15.9734},{-0.482883,-1.97181,15.9827},{-0.483024,-1.31454,15.9873},{-0.483089,-0.65727,15.9895},{-0.483108,-9.12146*10^-16,15.9901},{-0.483089,0.65727,15.9895},{-0.483024,1.31454,15.9873},{-0.482883,1.97181,15.9827},{-0.482603,2.62908,15.9734},{-0.48206,3.28635,15.9555},{-0.481011,3.94362,15.9207},{-0.478985,4.60089,15.8537},{-0.475076,5.25816,15.7243},{-0.467535,5.91543,15.4747},{-3.42264,-6.48177,14.6043},{-3.5319,-5.8336,15.0705},{-3.58953,-5.18542,15.3164},{-3.61993,-4.53724,15.4462},{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</code></pre>
| kglr | 125 | <p><strong>Update 2:</strong> Additional methods using <code>Partition[pts, 20]</code> with <code>BSplineSurface</code> and <code>BSplineFunction</code>:</p>
<pre><code>array = Partition[pts, 20];
</code></pre>
<p>We can get the surface (without lines connecting the points) using <code>array</code> with <a href="https://reference.wolfram.com/language/ref/BSplineSurface.html" rel="nofollow noreferrer"><code>BSplineSurface</code></a> as follows:</p>
<pre><code>bSS = BSplineSurface[array, SplineClosed -> {True, False}, SplineDegree -> {1, 1}]
Graphics3D[{Directive[FaceForm[Opacity[.75, LightBlue]], EdgeForm[Gray]], bSS,
Red, Point @ pts
Gray, Line @ array, Line @ Transpose[Append[First @ #] @ # & @ array]},
ImageSize -> 600, Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/ytZDL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ytZDL.png" alt="enter image description here" /></a></p>
<p>To get the surface with mesh lines we can use <a href="https://reference.wolfram.com/language/ref/BSplineFunction.html" rel="nofollow noreferrer"><code>BSplineFunction</code></a> + <a href="https://reference.wolfram.com/language/ref/ParametricPlot3D.html" rel="nofollow noreferrer"><code>ParametricPlot3D</code></a>:</p>
<pre><code>bSF = BSplineFunction[array,
SplineClosed -> {True, False}, SplineDegree -> {1, 1}];
Show[ParametricPlot3D[bSF[u, v], {u, 0, 1}, {v, 0, 1},
Mesh -> Full,
PlotPoints -> {1, 0} + Most @ Dimensions[array],
MaxRecursion -> 0,
PlotStyle -> Directive[FaceForm[Opacity[.75, LightBlue]], EdgeForm[Gray]],
ImageSize -> 600, Boxed -> False, Axes -> False],
Graphics3D[{Red, AbsolutePointSize[4], Point @ pts}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/oPXNJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oPXNJ.png" alt="enter image description here" /></a></p>
<p><strong>Original answer:</strong></p>
<pre><code>NearestNeighborGraph[ScalingTransform[{1, 5, 1}] @ pts, 4,
VertexCoordinates -> pts, ImageSize -> Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/8l2U1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8l2U1.png" alt="enter image description here" /></a></p>
<p><strong>Note:</strong> the magic factor <code>{1,5,1}</code> is obtained using</p>
<pre><code>Round[1/Normalize[Max /@ Abs[Differences /@ Sort /@ Transpose[pts]], Max]]
</code></pre>
<blockquote>
<pre><code>{1, 5, 1}
</code></pre>
</blockquote>
<p><strong>Update 1:</strong> Realized based on N.J.Evans's answer that the input data is more regular than I thought.</p>
<p>So we get a cleaner result using cycles in a periodic <code>GridGraph</code> to identify indices of polygon coordinates:</p>
<pre><code>edges = Join[EdgeList @ GridGraph[{20, 32}], Thread[Range[20] <-> Range[621, 640]]];
cycles = VertexList /@ FindCycle[edges, {4}, All];
Graphics3D[GraphicsComplex[pts,
{Opacity[.5], RandomColor[], Polygon @ #} & /@ cycles],
ImageSize -> Large, Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/UtrUf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UtrUf.png" alt="enter image description here" /></a></p>
<p>Alternatively, we can directly construct the polygon coordinates for use with <code>Graphics3D</code> + <code>GraphicsComplex</code>:</p>
<pre><code>indexlist = Map[(a |-> (# - 1) 20 + #2 & @@ Mod[a + #, {32, 20}, {1, 1}]) /@
{{0, 0}, {1, 0}, {1, 1}, {0, 1}} &] @ Tuples[{Range[32], Range[19]}];
Graphics3D[GraphicsComplex[pts,
{Opacity[.5], LightBlue, Polygon @ indexlist}],
Boxed -> False, ImageSize -> Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/SSY9A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SSY9A.png" alt="enter image description here" /></a></p>
|
1,854,823 | <p>How do I express the hyperplane $x+y=1$ as the span of two vectors or more?</p>
<p>P. S. We have a 3D space.</p>
| DonAntonio | 31,254 | <p>Using only basic analytic geometry: find three different non-collinear points on the plane, for example</p>
<p>$$A=(1,0,1)\;,\;\;B=(1,0,0)\;,\;\;C=(0,1,0)$$</p>
<p>and now construct the directed vectors</p>
<p>$$\vec{AB}=B-A=(0,0,-1)\;,\;\;\vec{AC}=C-A=(-1,1,-1)$$</p>
<p>and then the plane is</p>
<p>$$\pi:\;A+r\vec{AB}+s\vec{AC}=(1,0,1)+r(0,0,-1)+s(-1,1,-1)\;,\;\;r,t\in\Bbb R$$</p>
<p><strong>Check</strong>: take the vectorial product of the direction vectors to get a perpendicular vector to the plane:</p>
<p>$$\vec{AB}\times\vec{AC}=\det\begin{vmatrix}i&j&k\\0&0&-1\\-1&1&-1\end{vmatrix}=(1,1,0)$$</p>
<p>and thus our plane is $\;x+y+d=0\;$ , and to find $\;d\;$ we can substitute any point on the plane here, say $\;A\;$ , to obtain</p>
<p>$$0+0+d=0\implies d=0$$</p>
<p>and the wanted plane is</p>
<p>$$x+y-1=0$$</p>
<p>which, of course, it is the same as you give. This way is just a standard form to check that what we got at the beginning is correct.</p>
<p>Anyway, you can look at your plane as the translation of a <em>subspace</em>, so;</p>
<p>$$\pi:\;\;\text{Span}\,\left\{\;(0,0,-1)\,,\,\,(-1,1,-1)\;\right\}+(1,0,1)$$</p>
|
2,397,564 | <p>Question:</p>
<p>Prove that if $ \ A\cup B \subseteq C \cup D,\ A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D$.</p>
<p>My attempt:</p>
<p>Let $ \ x\in B \implies x \in A \cup B \implies x \in C \cup D \because A\cup B \subseteq C \cup D$.</p>
<p>Now, $ x \in C \lor x\in D$. If $\ x \in C \implies x \in A \because C \subseteq A$. But that's not possible $\because x \notin A \cap B$, in particular $ x \notin A$. So we must have $ x \in D$.</p>
<p>I found this proof a little challenging. Not quite sure if this is the correct way to prove it. Is my logic correct?</p>
| Aryabhata | 1,102 | <p>It is correct. Well done.</p>
<p>Perhaps a bit of suggested modification in the "But that's not possible part" to make it clearer.</p>
<p>Since we started with $x \in B$, if $x \in A$, then $x \in A \cap B$ which is not possible, as $A \cap B = \phi$.</p>
|
138,520 | <p>I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$</p>
<p>So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$
$$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$</p>
<p>then substituting these into the differential equation I get:</p>
<p>$$(1-2x+x^2)\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>$$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>relabeling the indexes: </p>
<p>$$\sum_{n=-2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=-1}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>and then cancelling the $n=-2$ and $n=-1$ terms:</p>
<p>$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need</p>
<p>$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$</p>
<p>I'm not sure where I have gone wrong?</p>
<p>Thanks very much for any help</p>
| Community | -1 | <p>You are right till the last step.</p>
<p>You have $$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>which gives us
$$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$
and not
$$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - (2n^2+n) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$
as you have written.
Hence, setting the coefficients of $x^n$ to zero, we get that
$$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)=0$$
Factorizing $n+1$ out, we get what you need i.e.
$$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)= (n+1) \left((n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n \right)$$
Hence, we get that $$(n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n = 0$$</p>
|
2,448,696 | <p>Show that $\frac{1}{n}<\ln n$, for all $n>1$ where n is a positive integer</p>
<p>I've tried using induction by multiplying both sides by $\ln k+1$ and $\frac{1}{k+1}$ but but all it does is makes it more complicated, I've tried using the fact that $k>1$ and $k+1>2$ during the inductive $k+1$ step, but I'm still stuck. </p>
<p>Looking for clues for this question.</p>
| Randall | 464,495 | <p>Note that the statement is true if $n=2$. Now consider $f(x) = \ln x - \frac{1}{x}$. Since $f'(x) = \frac{1}{x} + \frac{1}{x^2}=\frac{x+1}{x^2}$ is clearly positive for $x \geq 2$, $f$ is increasing over the same span. Thus the gap between $\frac{1}{n}$ and $\ln n$ is only getting wider, so we have $\frac{1}{n} < \ln(n)$ for $n \geq 2$. </p>
|
3,715,715 | <p>Let’s say I have a set <span class="math-container">$X$</span> and a set <span class="math-container">$Y$</span>, and <span class="math-container">$X \subseteq Y$</span>. Is it possible to state that <span class="math-container">$|X| \leq |Y|$</span> (<span class="math-container">$|X|$</span> cardinality of <span class="math-container">$X$</span>)? How can I demonstrate that?</p>
| TransfiniteGuy | 474,380 | <p>If we talk about cardinality of two sets, say <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, the way to show that <span class="math-container">$|A|\leq |B|$</span> is giving an injective function <span class="math-container">$f: A \longrightarrow B$</span>. </p>
<p>Now, in your problem, since <span class="math-container">$A \subseteq B$</span>, the function <span class="math-container">$f: A \longrightarrow B$</span> given by <span class="math-container">$f(x)= x$</span> is an injective function (If A and B are explicit sets, you can find many other injections). </p>
<p>As an important mark, <span class="math-container">$f$</span> is not the indentical function, instead of that and according the hypotesis of <span class="math-container">$A \subseteq B$</span>, <span class="math-container">$f$</span> would be called "The Inclusion of A in B"</p>
|
2,391,453 | <p>Let $A\subseteq[0,1]^2$ be a Lebesgue-measurable set. For $x\in [0,1]$, we define $A_x$ as $\{y:(x,y)\in A\}$.<br><br> <strong>Prove that $A_x$ is measurable for almost all $x\in[0,1]$.</strong> <br><br> I know that the "almost every'' is indeed needed, since we could have $A=V\times\{0\}$ for $V$ the Vitali set -- then $A$ has outer measure zero, so it is measurable, but $A_0$ is not measurable.<br><br> <a href="https://math.stackexchange.com/questions/1153161/every-section-of-a-measurable-set-is-measurable-in-the-product-sigma-algebra">This question</a> is similar, but it asks about product measure, while the Lebesgue measure on $[0,1]^2$ is not just the product measure, but completion of it.</p>
| Matematleta | 138,929 | <p>I think the following argument works:</p>
<p>Let $(I\times I,\mathcal L, \lambda_2)$ be the completion of $(I\times I,\mathscr B([0,1])\times \mathscr B([0,1]), \lambda\times \lambda ),$ set $f=1_A$ and choose a $\mathscr B([0,1])\times \mathscr B([0,1])$- measurable function $g$ such that $g=f\ \text{a.e.}\ \lambda_2.$ Note that the sections $g_x$ are $\mathscr B([0,1])$-measurable.</p>
<p>Then, $h=f-g=0\ $a.e.-$\lambda_2$ and therefore, there is a $B\in \mathscr B([0,1])\times \mathscr B([0,1])$ such that $\left \{ h\neq 0 \right \}\subseteq B$ and $\lambda_2(B)=\lambda \times \lambda (B)=0.$</p>
<p>We have now $0=\lambda \times \lambda (B)=\int_I \lambda (B_x)d\lambda$ so $\lambda(B_x)=0$ for almost all $x\in I.$ But then, $\lambda (\left \{ h\neq 0 \right \}_x)\le \lambda (B_x)=0\Rightarrow A_x=f_x=g_x$ for almost every $x\in I.$</p>
|
2,391,453 | <p>Let $A\subseteq[0,1]^2$ be a Lebesgue-measurable set. For $x\in [0,1]$, we define $A_x$ as $\{y:(x,y)\in A\}$.<br><br> <strong>Prove that $A_x$ is measurable for almost all $x\in[0,1]$.</strong> <br><br> I know that the "almost every'' is indeed needed, since we could have $A=V\times\{0\}$ for $V$ the Vitali set -- then $A$ has outer measure zero, so it is measurable, but $A_0$ is not measurable.<br><br> <a href="https://math.stackexchange.com/questions/1153161/every-section-of-a-measurable-set-is-measurable-in-the-product-sigma-algebra">This question</a> is similar, but it asks about product measure, while the Lebesgue measure on $[0,1]^2$ is not just the product measure, but completion of it.</p>
| JS_ | 354,831 | <p>This is just a more detailed and extended version of the accepted answer. I just want to make sure everything is understood properly.</p>
<p>Let $\lambda_d$ denote the $d$-dimensional Lebesgue measure and $\mathcal{L}_d$ the set of Lebesgue-measurable subsets of $\mathbb{R}^d$. Moreover, let $\mathcal{L}_d\times\mathcal{L}_{d'}$ denote the product $\sigma$-algebra, i.e., the smallest $\sigma$-algebra containing all cartesian products $X\times Y$ for $X\in\mathcal{L}_d$ and $Y\in\mathcal{L}_{d'}$. (That is, $\mathcal{L}_d\times\mathcal{L}_{d'}$ is not just the cartesian product of the two sets.)</p>
<p>Note that $\mathcal{L}_2$ is the <strong>completion</strong> of the product $\sigma$-algebra $\mathcal{L}_1\times\mathcal{L}_1$, i.e., every set $X\in\mathcal{L}_2$ can be expressed as $X=Y\Delta Z$ for $Y\in\mathcal{L}_1\times\mathcal{L}_1$ and $Z$ a null-set.</p>
<p>We will be using the standard Fubini's theorem for product measure spaces that claims that for $X\in\mathcal{L}_d\times\mathcal{L}_{d'}$, we have $X_x\in\mathcal{L}_{d'}$ for <strong>every</strong> $x\in\mathbb{R}^d$.</p>
<p>Let $A$ be given and decompose $A=B\Delta C$ for $B\in\mathcal{L}_1\times\mathcal{L}_1$ and $C$ a null-set. </p>
<p><strong>Claim.</strong> There exists $D\in\mathcal{L}_1\times\mathcal{L}_1$ such that $C\subseteq D$ and $\lambda_2(D)=0$.</p>
<p><strong>Proof of claim.</strong> Since $C$ is null-set, for every $\varepsilon>0$ there exists a countable collection $O_1^\varepsilon, O_2^\varepsilon,\ldots$ of open boxes such that $\sum_{n=1}^\infty |O_n^\varepsilon|\leq\varepsilon$ and $C\subseteq\bigcup_{n=1}^\infty O_n^\varepsilon$. Then we can set $$D=\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty O_n^{1/m},$$ which is an intersection and union of open boxes, hence Borel, hence from $\mathcal{L}_1\times\mathcal{L}_1$.</p>
<p>So by Fubini we have $$0=\int_{[0,1]^2}1_D(x,y)dxy=\int_{[0,1}\left(\int_{[0,1]}1_D(x,y)dy\right)dx=\int_{[0,1]}\lambda_1(D_x)dx.$$</p>
<p>Since $\int f=0$ implies that $f\equiv 0$ almost everywhere, it follows that $\lambda_1(D_x)=0$ for almost every $x\in[0,1]$. Moreover, since $C\subseteq D$, we also have that $\lambda_1(C_x)=0$ for almost every $x\in[0,1]$.</p>
<p>Finally, for almost every $x\in[0,1]$ we have $A_x=B_x\Delta C_x$ with $B_x\in\mathcal{L}_1$ (by Fubini) and $\lambda_1(C_x)=0$, it follows that $A_x$ is Lebesgue measurable (since $\mathcal{L}_1$ is itself a complete measure space).</p>
|
1,163,033 | <p>I want to calculate $ 8^{-1} \bmod 77 $ </p>
<p>I can deduce $ 8^{-1} \bmod 77$ to $ 8^{59} \bmod 77 $ using Euler's Theorem.</p>
<p>But how to move further now. Should i calculate $ 8^{59} $ and then divide it by $ 77 $ or is there any other theorem i can use ? </p>
| Mark Bennet | 2,906 | <p>Your question is equivalent to solving $8x+77y=1$ for integers $x$ and $y$. This can be solved using the division algorithm, or alternatively (equivalently) solving modulo $8$ - which is small enough to do by trial and error or observation.</p>
<p>So noting that $77\equiv 5 \bmod 8$ and $5\times 5 \equiv 1 \bmod 8$ choose $y=5$ and then solve for $x$. Then adjust the answer by an appropriate multiple of $77$.</p>
|
623,428 | <blockquote>
<p>Suppose $$
Y = X^TAX,
$$ where $Y$ and $A$ are both known $n\times n$, real, symmetric matrices. The unknown matrix $X$ is restricted to $n\times n$.</p>
</blockquote>
<p>I think there should be at least one real valued solution for $X$. How do I solve for $X$? </p>
| Doubt | 61,559 | <p>Consider the $2\times2$ case with
\begin{align}
X =
\begin{bmatrix}
x_1 & x_3\\ x_2 & x_4
\end{bmatrix},\quad A=
\begin{bmatrix}
a_1 & a_3\\ a_2 & a_4
\end{bmatrix},\quad Y=
\begin{bmatrix}
y_1 & y_3\\ y_2 & y_4
\end{bmatrix}.
\end{align}
Then
\begin{align}
\begin{bmatrix}
y_1 & y_3\\ y_2 & y_4
\end{bmatrix}=
\begin{bmatrix}
x_1(a_1x_1+a_3x_2)+x_2(a_2x_1+a_4x_2) & x_1(a_1x_3+a_3x_4)+x_2(a_2x_3+a_4x_4)\\ x_3(a_1x_1+a_3x_2)+x_4(a_2x_1+a_4x_2) & x_3(a_1x_3+a_3x_4)+x_4(a_2x_3+a_4x_4)
\end{bmatrix}
\end{align}
This may be generalized to the $n\times n$ case, but if $n$ is small and this is a practical application, perhaps entering these nonlinear equations into a computer algebra system will suffice.</p>
|
2,526,716 | <p>Define, by structural induction, a function $f : A^* \to A^*$ that removes all occurrences of the letter $a$. For instance, we should have
$f(abcbab) = bcbb$ and $f(bc) = bc$.</p>
<p>I came up with this:</p>
<p>$f(\lambda) = \lambda$ (empty word)</p>
<p>$f(xw) = xf(w)$ if $x$ is not $a$</p>
<p>$f(w)$, otherwise.</p>
<p>But I have no idea if this is sufficient and if its a structural induction. </p>
<p>Thanks in advance</p>
| Michael Rozenberg | 190,319 | <p>It's the distance between the tangent line $yy_1=p(x+x_1)$ to the parabola $y^2=x$ and $y=4x+4$, which parallel to the tangent line.</p>
<p>Since $p=\frac{1}{2}$, we obtain $\frac{1}{2y_1}=4$, which gives $y_1=\frac{1}{8}$ and $\left(\frac{1}{64},\frac{1}{8}\right)$ is a touching point. </p>
<p>Id est, for the distance we obtain:
$$\frac{\left|4\cdot\frac{1}{64}-\frac{1}{8}+4\right|}{\sqrt{4^2+(-1)^2}}$$</p>
|
1,591,311 | <p>I've been thinking about this problem which I think is interesting, but can't solve it.</p>
<p>There are $n$ distinguishable items, and $b$ distinguishable bins. Each bin has to include at least one item. But, once some set of items are placed in a bin, they become indistinguishable. How many ways are there to place the items into the bins?</p>
<p>(1) The condition that each bin has to include at least one item can be resolved by simply tweaking the problem a bit: suppose there are $n-b$ items, and proceed. So this is not a big hurdle. (Or, it can be, depending on how we handle the second condition below.)</p>
<p>(2) The second condition that the items in a bin are indistinguishable is a bit tricky. Suppose we have 3 items, and 2 bins. The items are numbered as 1, 2 and 3. The bins are denoted as A and B.</p>
<p>The second condition says that, we have to consider the following as identical: A - 1, B - 2, 3. vs. A - 1, B - 3, 2.</p>
<p>However, we have to consider the following as distinct: A - 1, B - 2, 3 vs. A - 2, 3, B - 1.</p>
<p>How can I compute the total number of ways to place the items into the bins?</p>
<p>Thanks.</p>
| Eric Thoma | 35,667 | <p>First lets not have the restriction that all the bins are nonempty. There are $b$ choices of bins for each ball, and so there are $b^n$ different configurations.</p>
<p>How many configurations are we counting that we should not? Lets leave one bin empty. There are $b$ choices for the empty bin, and $(b-1)^n$ ways to fill the rest.</p>
<p>Now we are overcounting configurations with two empty bins. This is the principle of inclusion exclusion.</p>
<p>We see the total number of configurations is
$$
\sum_{i=0}^{b-1} (-1)^{i} {b \choose i}(b-i)^n.
$$</p>
<p>Disclaimer: I haven't numerically checked this so I would not be surprised if I am in error.</p>
|
1,591,311 | <p>I've been thinking about this problem which I think is interesting, but can't solve it.</p>
<p>There are $n$ distinguishable items, and $b$ distinguishable bins. Each bin has to include at least one item. But, once some set of items are placed in a bin, they become indistinguishable. How many ways are there to place the items into the bins?</p>
<p>(1) The condition that each bin has to include at least one item can be resolved by simply tweaking the problem a bit: suppose there are $n-b$ items, and proceed. So this is not a big hurdle. (Or, it can be, depending on how we handle the second condition below.)</p>
<p>(2) The second condition that the items in a bin are indistinguishable is a bit tricky. Suppose we have 3 items, and 2 bins. The items are numbered as 1, 2 and 3. The bins are denoted as A and B.</p>
<p>The second condition says that, we have to consider the following as identical: A - 1, B - 2, 3. vs. A - 1, B - 3, 2.</p>
<p>However, we have to consider the following as distinct: A - 1, B - 2, 3 vs. A - 2, 3, B - 1.</p>
<p>How can I compute the total number of ways to place the items into the bins?</p>
<p>Thanks.</p>
| Marc van Leeuwen | 18,880 | <p>Your second condition is implicit in the usual notion of bin in combinatorics: there is no imposed ordering among the items inside the bin (which is unrelated to the possibility to distinguish the items among each other). So you are just asking for the number of surjective maps from the set of your $n$ items to the set of $b$ bins.</p>
<p>This is <a href="https://en.wikipedia.org/wiki/Twelvefold_way#case_s" rel="nofollow">one of the problems</a> of the <a href="https://en.wikipedia.org/wiki/Twelvefold_way" rel="nofollow">twelvefold way</a>. Its answer is given by $b!{n\atopwithdelims\{\}b}$, where $n\atopwithdelims\{\}b$ is a <a href="https://en.wikipedia.org/wiki/Stirling_number_of_the_second_kind" rel="nofollow">Stirling number of the second kind</a>. The factor $b!$ can be understood by the fact that their nonempty contents allow distinguishing the $b$ bins even if we forget their labels; the action of the group of permutations of the bins partitions the set of solutions (surjective maps) into orbits of $b!$ each.</p>
|
26,152 | <p>In my textbook, they said:</p>
<p>$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$</p>
<p>The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$</p>
<p>And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:</p>
<p>Let $y = 2x^{3} + 7x - 4$, we have:<br>
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$</p>
<p>What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this? </p>
<p>Thanks, </p>
| Alex J. | 8,057 | <p>$$\begin{aligned}
2x^3+7x-4 \equiv 0 ( mod 5)\\
2x^3+2x-4 \equiv 0 ( mod 5)\\
2(x-1)(x^2+x+2) \equiv 0 ( mod 5)
\end{aligned}$$</p>
<p>Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$.
You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.</p>
|
750,417 | <p><strong>Question:</strong></p>
<blockquote>
<p>Initially we have a list of numbers $1,2,3,\cdots,2013$.an operation is defined that taking two numbers $a, b$ out from the list, but add $a+b$ into it instead, what is the minimum number of operations required that the sums of any number of numbers in the list can never be $2014$</p>
</blockquote>
<p><strong>This is Math competition in jiangxi province in 2014 the first problem</strong></p>
<p>my idea: if we change this digital: such this:</p>
<p>Initially we have a list of numbers $1,2,3,4,5$.an operation is defined that taking two numbers $a,b$ out from the list, but add $a+b$ into it instead, what is the minimum number of operations required that the sums of any number of numbers in the list can never be $6$</p>
<p>if we an operation is defined that taking two numbers $1,2$then add $1+2=3$,so must operation is defined that taking two numbers $3,3$ out from the list.and $3+3$.so we at lest operation two.</p>
<p><img src="https://i.stack.imgur.com/kQQBs.jpg" alt="enter image description here">
It is said this reslut is $1007$
Thank you for you help </p>
| achille hui | 59,379 | <p>The minimum number of operations is either $503$ or $504$. I cannot figure out which is the answer.</p>
<p>Among the list of $2013$ numbers $( 1, 2, \ldots, 2013 )$, there are $1006$ pairs that sum to $2014$: $$( 1, 2013), ( 2, 2012 ), \ldots, ( 1006, 1008 )\tag{*1}$$</p>
<p>In order for the final list and the sum of elements in the list doesn't contain $2014$, we must break all these pairs. Since each operation can break at most $2$ pairs, we need at least $503 = 1006/2$ operations to break all these pairs.</p>
<p>On the other hand, the original list contains $504$ pairs that sum to $1009$.</p>
<p>$$( 1, 1008 ), ( 2, 1007 ), \ldots, (7,1002),\ldots, ( 504, 505 )\tag{*2}$$</p>
<p>We can use $504$ operations to transform the list to one with $505$ copies of $1009$ together with extra $1004$ numbers greater than $1009$.</p>
<p>$$(\;\underbrace{1009, 1009, \ldots, 1009}_{505\text{ copies}}, 1010, 1011, \ldots, 2013\;)$$</p>
<p>Individually, each elements of this list are smaller than $2014$ and yet the sum of any two of them is at least $2018$. This implies at most $504$ operations is enough to meet the requirement.</p>
|
3,335,081 | <p>Is a temperature change in Celsius larger than a temperature change in Fahrenheit?</p>
<p><strong>The teacher offers this second way of thinking about the question.</strong></p>
<blockquote>
<p>If the temperature increases by 1 degree Celsius, does it also increase by 1 degree Fahrenheit? Or is one temperature change larger than the other?</p>
</blockquote>
<p><strong>My Attempt:</strong>
<span class="math-container">$$T_F = \frac95T_C + 32 \implies (T_F + T_{\triangle F}) = \frac95(T_C+T_{\triangle C}) + 32$$</span>
Where <span class="math-container">$T_{\triangle F}$</span> is some change on the Fahranheit scale and <span class="math-container">$T_{\triangle C}$</span> is the respective change on the Celsius scale. (I am confident this is true as long as we assume the changes in temperature are respective to each other.)</p>
<p>Thus, <span class="math-container">$$T_{\triangle F} = \frac95T_C + 32 - T_F + \frac95T_{\triangle C} \implies T_{\triangle F} = T_F - T_F + \frac95T_{\triangle C}$$</span>
<span class="math-container">$$\implies T_{\triangle C} = \frac59T_{\triangle F} \checkmark$$</span></p>
<p>For some degree change on the Fahrenheit scale, <span class="math-container">$T_{\triangle F}$</span>, the resulting change on the Celsius scale, <span class="math-container">$T_{\triangle C}$</span>, is <span class="math-container">$\frac59$</span> that change. Therefore, a temperature change on the Celsius scale is larger than a temperature change on the Fahrenheit scale.</p>
<p><span class="math-container">$$$$</span>
<span class="math-container">$$$$</span></p>
<p><strong>Question:</strong> Does this show what I want? And does it answer the original question? Thanks in advance! This one is making me over think a bit I think.</p>
| Tanner Swett | 13,524 | <p>What you have so far is correct, but in my opinion, you haven't <em>quite</em> answered the question that was asked.</p>
<p>Before I get to that, though, I have just a minor comment. You write that</p>
<p><span class="math-container">$$T_{\triangle F} = \frac95T_C + 32 - T_F + \frac95T_{\triangle C}.$$</span></p>
<p>This is correct, but in my opinion, you should write one more intermediate step:</p>
<p><span class="math-container">$$T_{\triangle F} = \frac95 \left (T_C + T_{\triangle C} \right) + 32 - T_F = \frac95T_C + 32 - T_F + \frac95T_{\triangle C}.$$</span></p>
<p>I say this simply because when I read through your steps, it took me a minute to figure out how you got from the one expression to the next.</p>
<hr>
<p>Now, the original question was:</p>
<blockquote>
<p>If the temperature increases by 1 degree Celsius, does it also increase by 1 degree Fahrenheit? Or is one temperature change larger than the other?</p>
</blockquote>
<p>You've shown that when a temperature change is measured in each of the two scales, the number of degrees Fahrenheit is larger than the number of degrees Celsius. Equal temperature changes, different numbers of degrees.</p>
<p>But that's not <em>quite</em> what the question was asking. It's asking whether a change of 1 degree Celsius is larger, smaller, or the same as a change of 1 degree Fahrenheit. Equal numbers of degrees, different temperature changes.</p>
<p>So, you should take a change of 1 degree Fahrenheit, convert that to Celsius, and use the result to show that this is smaller than a change of 1 degree Celsius. Alternatively, take a change of 1 degree Celsius, convert that to Fahrenheit, and use the result to show that this is larger than a change of 1 degree Fahrenheit.</p>
<p>Great work so far!</p>
|
3,365,361 | <p>Suppose that <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> is analytic at <span class="math-container">$x=0$</span>, and <span class="math-container">$T(x)$</span> its Taylor series at <span class="math-container">$x=0$</span>, with radius of convergence <span class="math-container">$R>0$</span>. Is it true that <span class="math-container">$f(x)=T(x)$</span> whenever <span class="math-container">$|x|<R$</span> ?</p>
| DonAntonio | 31,254 | <p>Because <span class="math-container">$\;\operatorname{tr}(AB)=\operatorname{tr}(BA)\;$</span> , so</p>
<p><span class="math-container">$$\operatorname{tr}(Q\Lambda^kQ^{-1})=\operatorname{tr}\left(Q(\Lambda^kQ^{-1})\right)=\operatorname{tr}((\Lambda^kQ^{-1})Q)=\operatorname{tr}(\Lambda^k(Q^{-1}Q))=\operatorname{tr}(\Lambda^kI)=\operatorname{tr}(\Lambda^k)$$</span></p>
|
1,622,779 | <p>I have been asked to calculate $\frac{1+i \tan \alpha}{1-i \tan \alpha}$, where $\alpha \in \mathbb{R}$.</p>
<p>So, I multiplied top and bottom by the complex conjugate of the denominator:</p>
<p>$\frac{(1+i \tan \alpha)(1+ i \tan \alpha)}{(1-i \tan \alpha)(1+ i \tan \alpha) } = \frac{1 + 2 i \tan \alpha + i^{2} \tan^{2} \alpha}{1-i^{2} \tan^{2} \alpha} = \frac{1 + 2i \tan \alpha - \tan ^{2} \alpha}{1 + \tan^{2} \alpha} = \frac{(1-\tan ^{2} \alpha)+2 i \tan \alpha}{\sec ^{2} \alpha} =\displaystyle \frac{1-\tan^{2} \alpha }{\sec ^{2} \alpha} + 2\sin\alpha\cos\alpha i$</p>
<p>Is this all that there is to do? Have I simplified it enough?</p>
| Community | -1 | <p>I'll continue from where you left to get a simpler form :</p>
<ul>
<li>The real part :</li>
</ul>
<p>$$\frac{1-\tan \alpha^2}{1+ \tan \alpha^2}=\frac{\cos^2 \alpha-\sin^2 \alpha}{\sin^2 \alpha+ \cos^2 \alpha}=\cos2\alpha$$</p>
<ul>
<li>The imaginary part :</li>
</ul>
<p>$$2\sin \alpha \cos \alpha=\sin 2\alpha$$</p>
<p>A simpler form is thus :</p>
<p>$$\frac{1+i\tan \alpha}{1-i \tan \alpha}=\cos2 \alpha+i \sin2\alpha$$</p>
|
792,356 | <p>There is a dark night and there is a very old bridge above a canyon. The bridge is very weak and only 2 men can stand on it at the same time. Also they need an oil lamp to see holes in the bridge to avoid falling into the canyon.</p>
<p>Six man try to go through that bridge. They need 1,3,4,6,8,9(first man, second man etc.) minutes to pass the bridge.</p>
<p>What is the fastest way for those six men to pass this bridge?</p>
| MJD | 25,554 | <p>A minimal solution is:</p>
<ul>
<li>1 and 6 cross the bridge; 1 comes back (7)</li>
<li>1 and 3 cross the bridge, 1 comes back (4)</li>
<li>8 and 9 cross the bridge; 3 comes back (12)</li>
<li>1 and 3 cross the bridge, 1 comes back (4)</li>
<li>1 and 4 cross the bridge (4)</li>
</ul>
<p>Total is $7+4+12+4+4 = 31$. There are minor variations that are still minimal. For example, 3 can return in step 2 instead of in step 3.</p>
<p>I have nothing useful to say about how to find the solution; I programmed the computer to do an exhaustive search. I only posted this because the other guy was getting upvotes for his wrong answer.</p>
<p>[ By request, <a href="https://github.com/mjdominus/perl-misc/blob/master/search/bridge" rel="nofollow">my search program is available</a> ]</p>
|
239,136 | <p>I was given this question and I'm not really sure how to approach this...</p>
<p>Assume $(r,s) = 1$. Prove that If $G = \langle x\rangle$ has order $rs$, then $x = yz$, where $y$ has order $r$, $z$ has order $s$, and $y$ and $z$ commute; also prove that the factors $y$ and $z$ are unique.</p>
| Hagen von Eitzen | 39,174 | <p>From $(r,s)=1$ we find integers $n,m$ with $nr+ms=1$.
Let $y=x^{ms}$, $z=x^{nr}$.
Then $yz=x^{ms+nr}=x$.
The fact that $x$ and $y$ commute is trivial because the cyclic group $G$ is abelian.
Also, we have $y^r=(x^m)^{rs}=1$, $z^s=(x^n)^{rs}=1$, hence the orders are at least divisors of $r$ and $s$, respectively.
If the actual orders are $r'|r$ and $s'|s$, then $x^{r's'}=y^{r's'}z^{r's'}=1$, hence $r's'$ is a positive multiple of $rs$, hence at least $rs$. We conclude that $r'=r$, $s'=s$.
Finally, assume we have another solution $x=y'z'$ with the required properties.
Then $z^r=y^rz^r=x^r = y'^rz'^r=z'^r$ implies $z=z^{nr+ms}={z^r}^n{z^s}^m={z^r}^n={z'^r}^n={z'^r}^n{z'^s}^m=z'^{nr+ms}=z'$ and similarly $y=y'$.</p>
|
732,996 | <p><img src="https://i.stack.imgur.com/kXJEt.png" alt="enter image description here"></p>
<p>Hi! I am working on some ratio and root test online homework problems for my calc2 class and I am not sure how to completely solve this problem. I guessed on the second part that it converges, but Im not sure how to solve of the value that it converges to. If someone could possibly help me with this problem it would be greatly appreciated. </p>
| Cookie | 111,793 | <p>We have $a_n = \frac{1}{(2n)!}$ and $a_{n+1} = \frac{1}{(2n+2)!}$</p>
<p>\begin{align}\lim_{n \rightarrow \infty} \Big|\frac{a_{n+1}}{a_n}\Big| &= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)!}{(2n+2)!}\Bigg| \\
&= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)(2n-1)(2n-2)\cdots}{(2n+2)(2n+1)(2n)(2n-1)(2n-2)\cdots}\Bigg| \\
&= \lim_{n \rightarrow \infty} \Bigg|\frac{1}{(2n+2)(2n+1)}\Bigg| \\
&=0 < 1
\end{align}</p>
<p>Hence, by the Ratio Test, the series $\sum_{n=1}^{\infty} \frac{1}{(2n)!}$ is convergent.</p>
|
605,772 | <p>Solving $x^2-a=0$ with Newton's method, you can derive the sequence $x_{n+1}=(x_n + a/x_n)/2$ by taking the first order approximation of the polynomial equation, and then use that as the update. I can successfully prove that the error of this method converges quadratically. However, I can't seem to prove this for the residual, and this is likely a simple problem in arithmetic:</p>
<p>\begin{align*}
|x_{n+1}^2 - {a}| &= \left|\frac{1}{4}\Big(x_n+\frac{a}{x_n}\Big)^2 - {a}\right| \\
&= \left|\frac{1}{4}\Big(x_n^2+2a +\frac{a^2}{x_n^2}\Big) - {a}\right| \\
&= \left|\big(\frac{1}{2}x_n\big)^2-\frac{1}{2}a +\big(\frac{a}{2x_n}\big)^2\right| \\
&= \frac{1}{4}\left|x_n^2-2a +\big(\frac{a}{x_n}\big)^2\right| \\
&= \frac{1}{4}\left|\big(x_n+\frac{a}{x_n}\big)^2-2a +\big(\frac{a}{x_n}\big)^2\right|
\end{align*}</p>
<p>I get stuck here, as well as trying other expansions/factorizations. Is there a way to have this simplify?</p>
| MoonKnight | 115,071 | <p>1) It is symmetry when $x>0$ and $x<0$, so we can only prove $x>0$. And the other half part of the proof is very similar</p>
<p>2) $\forall x_0>0$, it is easy to see that $x_1=(x_0+a/x_0)/2>\sqrt{a}$</p>
<p>3) If $x_n^2>a$, then follow your derivation until the second last line</p>
<p>$$
x_{n+1}^2-a = \frac{1}{4}\left(x_n-\frac{a}{x_n}\right)^2 = \frac{x_n^2-a}{4x_n^2} ( x_n^2-a)
$$</p>
<p>$$
\text{So, }0<x_{n+1}^2-a<\frac{1}{4}(x_n^2-a)
$$
So the residual converges since the second step.</p>
|
135,426 | <p>$$\frac{d}{dq}\int_{s_{1}-z-q}^{z-s_{1}} \varphi(w) \, dw$$</p>
<p>(if it helps, in my setting $\varphi$ is the CDF of some arbitrary uniform distribution). So I want to get a nice expression for this integral and it seems to suggest FTC, but I tried a change of variable and ended up with a $q$ inside the integrand which was not nice. Any help much appreciated.</p>
| The Chaz 2.0 | 7,850 | <p>For a similar approach to Jyrki's, we use the result that the <a href="http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Alternating_harmonic_series" rel="nofollow">alternating harmonic series</a> converges to $\ln 2$. Let's subtract one from both sides of your inequality, then change all the signs to get
$$1 - \dfrac{1}{2} +... + \dfrac{1}{2009} - \dfrac{1}{2010} > \dfrac{5}{8}$$and denote the left hand side of this new inequality by $(\#)$. Then
$$ (\#) + \sum_{n = 2011}^{\infty} \dfrac{(-1)^{n+1}}{n} = \ln 2 \approx 0.6931$$</p>
<p>So $$( \#) \approx 0.6931 - \sum_{n=2011}^{\infty}\dfrac{(-1)^{n+1}}{n}$$</p>
<p>Which is greater than $\dfrac{5}{8} = 0.625$ if we can bound the tail of this infinite sum (as in the comments to Jyrki's answer). I'll leave that as an exercise :)</p>
|
3,498,199 | <p>Suppose if a matrix is given as</p>
<p><span class="math-container">$$ \begin{bmatrix}
4 & 6\\
2 & 9
\end{bmatrix}$$</span></p>
<p>We have to find its eigenvalues and eigenvectors.</p>
<p>Can we first apply elementary row operation . Then find eigenvalues.</p>
<p>Is their any relation on the matrix if it is diagonalized or not.</p>
| Aryaman Maithani | 427,810 | <p>No, elementary row operations need not preserve eigenvalues and/or eigenvectors.</p>
<p>Examples.</p>
<p><span class="math-container">$$\begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix} \overset{R_1 \mapsto R_1 + R_2}{\longrightarrow} \begin{bmatrix}1 & 1 \\0 & 1\end{bmatrix}$$</span></p>
<p>In this case, the eigenvalue remains the same but the eigenvectors don't. This can be easily observed by the fact that the left matrix is diagonalisable but the right one isn't.</p>
<p><span class="math-container">$$\begin{bmatrix}1 & 1 \\0 & 1\end{bmatrix} \overset{R_2 \mapsto R_1 + R_2}{\longrightarrow} \begin{bmatrix}1 & 1 \\1 & 2\end{bmatrix}$$</span></p>
<p>In this case, the eigenvalues don't remain the same.</p>
|
221,351 | <p>I asked the following question (<a href="https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con">https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con</a>) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.</p>
<p>The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?</p>
<p>Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.</p>
<p>If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.</p>
<p>Thanks for any assistance.</p>
| David E Speyer | 297 | <p>This argument is fairly standard, but it is quicker to repeat it than to find a reference: Let $G$ be a finite subgroup of $GL_n(\mathbb{Q})$. Set $\Lambda = \sum_{g \in G} g \cdot \mathbb{Z}^n \subset \mathbb{Q}^n$. Then $\Lambda$ is a finitely generated torsion free abelian group, hence isomorphic to $\mathbb{Z}^r$ for some $r$. Since $\mathbb{Z}^n \subseteq \Lambda \subset \mathbb{Q}^n$, we have $r=n$. Clearly, $g \Lambda = \Lambda$ for all $g \in G$.</p>
<p>Let $h \in GL_n(\mathbb{Q})$ take the standard basis to a basis of $\Lambda$, so $h \mathbb{Z}^n = \Lambda$ and $\mathbb{Z}^n = h^{-1} \Lambda$. Then $h^{-1} g h$ takes $h^{-1} \Lambda = \mathbb{Z}^n$ to itself for all $g \in G$, so $h^{-1} G h \subset GL_n(\mathbb{Z})$. </p>
|
4,337,820 | <p>We know that during projection 3D space points <span class="math-container">$(x, y, z)$</span> projects to projection plane which has 2D points <span class="math-container">$(x, y).$</span> But during matrix calculation we use homogenous coordinates is of the form <span class="math-container">$(x, y, 1).$</span>
And we know that projective plane is of the form <span class="math-container">$(x, y, 1).$</span></p>
<p>My question is that homogenous coordinates and projective plane points both are same thing, I mean "is all projective plane points are said homogenous coordinates"?</p>
<p>My second question is what is the difference between projection plane and projective plane? I mean "is projective plane is real plane or just imagination plane"? And we know that projection plane is real thing where we do all projection. But both have similar names. What is the relationship between these?</p>
| bubba | 31,744 | <p>Short answer: the two concepts “projective plane” and “projection plane” are different things, though they are loosely related.</p>
<p>Longer answer …</p>
<p>The “projective plane”, often denoted by <span class="math-container">$P^2$</span>, is an abstract mathematical concept. It’s used in a field of mathematics called “projective geometry”. As the other answer explained, the basic idea is to represent each 2D point by a 3D line passing through the origin. The benefit is that this allows you to represent 2D points that are “at infinity”. You can use this technique with <strong>any</strong> plane.</p>
<p>The “projection plane” is a <strong>specific</strong> plane that’s used in 3D computer graphics. The points of a 3D object are projected onto the projection plane to produce a 2D image. Quite often, the projection plane has equation <span class="math-container">$z=1$</span> in some coordinate system.</p>
<p>People often use 4D (homogeneous) coordinates and <span class="math-container">$4\times 4$</span> matrices to represent the 3D-to-2D projection in computer graphics. This approach is not much related to the projective plane <span class="math-container">$P^2$</span>, but it is somewhat related to projective 3-space, <span class="math-container">$P^3$</span>.</p>
<p>Similarly, if you use 3D (homogeneous) coordinates to represent points in any plane, you are effectively working with the projective plane, <span class="math-container">$P^2$</span>. But note that this is true of <strong>any</strong> plane. In particular, it’s true of the projection plane that you use in computer graphics, so this is the connection between “projection plane” and “projective plane”.</p>
<p>The main reason homogeneous coordinates are used in computer graphics is so that perspective projection can be represented by a matrix multiplication. But you don’t have to use matrices and homogeneous coordinates if you don’t want to —- the whole projection calculation can be done just using ordinary 3D coordinates. And this approach doesn’t involve <span class="math-container">$P^2$</span> or <span class="math-container">$P^3$</span> or any other concept from projective geometry.</p>
|
2,697,069 | <p>Two series of functions are given in which I cannot figure out how to find $M_n$ of the second problem. $$1.\space \sum_{n=1}^{\infty} \frac{1}{1+x^n}, x\in[k,\infty)\\ 2. \space \sum_{n=1}^{\infty} (\cos x)^n, x\in(0,\pi)$$.. </p>
<p>I have determined the $M_n$ for problem no. $1.$ [$\space|\sum_{n=1}^{\infty} \frac{1}{1+x^n}|<|\sum_{n=1}^{\infty} \frac{1}{1+k^n}|<\sum_{n=1}^{\infty} \frac{1}{k^n}$] </p>
<p>From problem no. $2.$, since $-1\leq \cos x\leq1$, therefore for higher $n$ the values of $\cos x$ will lie between $[-1,1]$ and in $(0,\pi)$ $\cos x$ is decreasing. But is it correct to choose $n$ as $M_n$, so that $$|f_n(x)|=|(\cos x)^n|<n,$$ where $n$ is decreasing. </p>
<p>I am not sure what the $M_n$ should be. Any help or suggestion please? Any help is greatly appreciated.</p>
| Botond | 281,471 | <p>Hint:
From $2x=5y$ you have that $y=\frac{2}{5}x$, and from $\frac{y}{3}=\frac{z}{4}$ you have that $\frac{y}{z}=\frac{3}{4}$</p>
|
2,697,069 | <p>Two series of functions are given in which I cannot figure out how to find $M_n$ of the second problem. $$1.\space \sum_{n=1}^{\infty} \frac{1}{1+x^n}, x\in[k,\infty)\\ 2. \space \sum_{n=1}^{\infty} (\cos x)^n, x\in(0,\pi)$$.. </p>
<p>I have determined the $M_n$ for problem no. $1.$ [$\space|\sum_{n=1}^{\infty} \frac{1}{1+x^n}|<|\sum_{n=1}^{\infty} \frac{1}{1+k^n}|<\sum_{n=1}^{\infty} \frac{1}{k^n}$] </p>
<p>From problem no. $2.$, since $-1\leq \cos x\leq1$, therefore for higher $n$ the values of $\cos x$ will lie between $[-1,1]$ and in $(0,\pi)$ $\cos x$ is decreasing. But is it correct to choose $n$ as $M_n$, so that $$|f_n(x)|=|(\cos x)^n|<n,$$ where $n$ is decreasing. </p>
<p>I am not sure what the $M_n$ should be. Any help or suggestion please? Any help is greatly appreciated.</p>
| Peter Szilas | 408,605 | <p>An option:</p>
<p>1)$2x=5y,$ divide both sides by by $2z:$</p>
<p>1') $x/z = (5/2)(y/z).$</p>
<p>2)$y/3=z/4$, then</p>
<p>2')$ y/z = 3/4$ (why?)</p>
<p>Substuting 2') into the RHS of 1'):</p>
<p>$x/z= (5/2)(3/4)=15/8.$</p>
|
2,416,424 | <p>It is known that the collection of finite mixtures of Gaussian Distributions over $\mathbb{R}$ is dense in $\mathcal{P}(\mathbb{R})$ (the space of probability distributions) under convergence in distribution metric.</p>
<p>I'm interested to know the following:</p>
<p>Let $P_X$ be a random variable with finite $p$ th moment i.e. $\mathbb{E}_{P_X}[|X|^p]<\infty$, and $P_{X_n}\stackrel{d}{\to} P_X$ where $P_{X_n}$ are mixtures of Gaussian distributions. Then, suppose $X_n \sim P_{X_n}$ and $X\sim P_X$, does it follow that
$$\mathbb{E}[|X_n|^p] \to \mathbb{E}[|X|^p]$$? $\quad (*)$</p>
<p><strong>My attempt:</strong></p>
<p>I've been able to show that
$$\liminf_{n \to \infty} \mathbb{E}[|X_n|^p] \geq \mathbb{E}[|X|^p]$$
In fact this didn't even use the mixtures part. However I'm having difficulty showing the upper bound. Here are some steps:
\begin{eqnarray}
\mathbb{E}[|X_n|^p] &=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + \mathbb{E}[|X_n|^p1_{\{|X_n|>A\}}] \\
&=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + A^pPr[|X_n|\geq A] + \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}]
\end{eqnarray}
Let $f(x)=x1_{0\leq x\le A} + A1_{x\geq A}$ which is a continuous and bounded function. Then the first two terms of RHS are equal to $\mathbb{E}[f(|X_n|^p)]$, which by definition of weak convergence, will converge to $\mathbb{E}[f(|X|^p)]$ as $n \to \infty$. Now we may use MCT as $A\to \infty$ to get $\mathbb{E}[|X|^p]$. Hence in a nutshell, we need to show:
\begin{equation}
\limsup_{A\to\infty} \limsup_{n \to \infty} \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}] \leq 0
\end{equation}
However I do not know how to proceed from here.</p>
<p>I'd like to point out that I don't know the answer to the question I asked in $(*)$ but my guess is that this is true. The reason is because I read a similar result in Soren Asmussen's book. The result states that for distributions over non-negative reals, we have phase type distributions not only being weak dense but also the moments converging. But in the provided proof, they said "It can be easily shown that the moments converge".</p>
<p><strong>Update:</strong>
Sincerest apologies for I forgot one very essential condition. The mixtures are not any mixtures but specific ones. Namely, I'm looking at distributions $P_X$ such that $P_{X_n}$ has the form $\sum_{k=1}^n \alpha_k \mathcal{N}(\mu_k;P_k)$ ($\mathcal{N}(\mu_k;P_k)$ stands for Gaussian distribution with mean $\mu_k$ and variance $P_k$.) where given $P<\infty$, $\sum_{k=1}^n \alpha_k =1$, $\alpha_k >0$ and $\sum_{k=1}^n\alpha_k P_k = P$. </p>
<p><strong>Update 2:</strong> The reference I mentioned earlier is Soren Asmussen "Applied probability and queues", 2nd edition, page 84.</p>
<p><strong>Update 3:</strong> Looks like I misinterpreted Asmussen. What he wanted to say was that for any distribution $P_X$ with finite $p$th moment, <strong>there exists</strong> a sequence of phase type distributions $P_X^k$ such that $P_X^k\stackrel{d}{\to}P$ and $E_{P_X^K}[|X|^p] \to E_{P_X}[|X|^p]$. This doesn't mean any weak converging mixture will have moment convergence as the answer points out.</p>
| kingW3 | 130,953 | <p>The equations $2^x=16$ and $-2^x=-16$ are equivalent however the functions $f(x)=-2^x+16$ and $f(x)=2^x-16$ are not. </p>
<p>You can take a look at the <a href="https://www.desmos.com/calculator/iesqpyxuqg" rel="nofollow noreferrer">graph</a> here, they are symmetric around the $x$ axis which means that when one is <em>negative</em> the other one is <em>positive</em>.</p>
<p>So if you want to solve $$-2^x+16>0$$ you are solving
$$2^x<16$$
which gives $x<4$ while if you wanted to solve $2^x-16>0$ you'd get $x>4$.</p>
<p>Getting back to the inequality which is true when either both $-2^x+16,2^x-8$ are positive or both are negative. They are both positive when $x<4$ and $x>3$ and they are both negative when $x>4$ and $x<3$ which is impossible so the only solution is $x\in (3,4)$</p>
|
3,180,914 | <p>Let <span class="math-container">$G$</span> be a cyclic group of order <span class="math-container">$n$</span>. Let <span class="math-container">$G_k$</span> the subgroup
<span class="math-container">$$G_k=\left\{x^k: x\in G\right\}.$$</span>
Is it true that <span class="math-container">$[G:G_k]\in\{1,k\}$</span>?</p>
<p>If <span class="math-container">$n=p-1$</span> and <span class="math-container">$k=2$</span> this is true and I used many times in some number theory exercises. How much can I generalize this thing?</p>
<p>What if <span class="math-container">$G$</span> is any (maybe abelian) group?</p>
| ΑΘΩ | 623,462 | <p>If <span class="math-container">$G$</span> is cyclic of order <span class="math-container">$n$</span>, say <span class="math-container">$G=\langle a \rangle$</span> then the subgroup of <span class="math-container">$m$</span>-powers is also cyclic generated by <span class="math-container">$a^m$</span> and it can be shown that the order of <span class="math-container">$a^m$</span> is <span class="math-container">$\frac{n}{(m,n)}$</span>: if <span class="math-container">$a^{km}=1_G$</span> then <span class="math-container">$n|km$</span>; after simplifying this divisibility relation by <span class="math-container">$(m,n)$</span>, since <span class="math-container">$\left(\frac{n}{(m,n)}, \frac{m}{(m,n)}\right)=1$</span> we infer right away that <span class="math-container">$\frac{n}{(m,n)}|k$</span>. </p>
<p>Hence, <span class="math-container">$|G:\langle a^m \rangle|=(m,n)$</span>.</p>
<p>In general, for arbitrary <span class="math-container">$n \in \mathbb{N}^{*}$</span> and arbitrary abelian group <span class="math-container">$(G, +)$</span> the index <span class="math-container">$|G:nG|$</span> can be arbitrarily large: consider for instance a direct sum <span class="math-container">$G=(\mathbb{Z}/2n\mathbb{Z})^{(T)}$</span>, for arbitrary set <span class="math-container">$T$</span>. You will have
<span class="math-container">$$G/nG \approx (\mathbb{Z}/2\mathbb{Z})^{(T)}$$</span>
and thus <span class="math-container">$|G:nG|=|T|$</span>, when <span class="math-container">$T$</span> is infinite.</p>
|
54,496 | <p>If there a group G acting on a variety V.
The action is algebraic.
What is the definition of algebro-geometric quotient of this action?</p>
<p>I hope you can give a very basic explanation.</p>
<p>Thanks.</p>
| JBorger | 1,114 | <p>The are several possible meanings. Which one it is would surely depend on the context. </p>
<p>The straight-up meaning is the one that works in any category. If $G$ acts on $X$ then a quotient is a universal object $X/G$ with a $G$-equivariant map $X\to X/G$, where $X/G$ has the trivial $G$ action. Here, 'universal' means that if $Y$ is any other such object, then there is a unique map $X/G\to Y$ commuting with the two maps from $X$. Then $X/G$ is unique up to unique isomorphism. </p>
<p>Unfortunately, if you're working in the category of schemes, such quotients sometimes don't exist. If you work in the slightly larger category of algebraic spaces (IMHO 'schemes done right'), they are more likely to exist, and there are even nice theorems (M Artin) to this effect. (There are also some not-so-nice theorems about scheme quotients in SGA 3.) You can also work in some big ambient topos, such as the category of sheaves of sets on the category of affine schemes, equipped with your favorite Grothendieck topology. In this big category, quotients always exist, and they have all the nice formal properties you could ask for (i.e. they are universal and effective, in the language of category theory).</p>
<p>As Steven Landsburg points out, you can also ask for the quotient in the stack-theoretic sense.</p>
<p>There are also various kinds of quotients that come up in geometric invariant theory, which are important if you're interested in projective algebraic geometry, but I'm embarrassed to admit that I never got to the bottom of what's going on there.</p>
<p>Finally, there might even be a theory of quotients in Weil's foundations for algebraic geometry, which I hear that some people working in algebraic groups still use.</p>
|
1,406,796 | <p>Can someone please show me how they would work it out as I have never come across this before.</p>
<p>$$(x^2-5x+5)^{x^2-36} =1$$</p>
| Hirshy | 247,843 | <p>We have that $a^b=1$ if and only if $a=1$ and arbitrary $b$ or $b=0$ and arbitrary $a$ (one might have to talk about the case $a=b=0$). Having this you should be able to get two seperate equations from $$(x^2-5x+5)^{x^2-36}=1$$ which can be easily solvedfor $x$.</p>
<p>Edit: I should drink a cup of coffee first after getting up...one also has to consider $a=-1$ and $b=2k$ for $k\in\mathbb N$, e.g. $(-1)^2=(-1)^4=(-1)^6=\dots =1$.</p>
|
1,808,258 | <p>I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix
$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$
is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix
$$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$
necessarily orthogonal?
At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$:
\begin{align*}
&\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\
&\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\
&\implies v \cdot v' = 0.
\end{align*}
But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.</p>
| ajotatxe | 132,456 | <p>The third point implies that the parametrization goes over the circle at a constant speed $1$, which is, of course, false in general.</p>
<p>Take for example $v(t)=\langle \sin t^2,\cos t^2\rangle $ for $t\in[0,\sqrt{2\pi})$.</p>
|
3,182,802 | <p>Show that if <span class="math-container">$ \sigma $</span> is a solution to the equation <span class="math-container">$ x^2 + x + 1 = 0 $</span> then the following equality occurs:</p>
<p><span class="math-container">$$ (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma) \geq 0 $$</span></p>
<p>I looked at the solution in my textbook and it says I should multiply the parentheses and take into account that <span class="math-container">$ \sigma + \sigma^2 + 1 = 0 $</span>.
I tried factoring the rest but I just can't seem to manage to solve it?</p>
<p>Maybe I messed up at multiplying the parentheses? Here's what I got:</p>
<p><span class="math-container">$$ a^2 + ab\sigma^2 + ac\sigma + ab\sigma + b^2\sigma^3 + bc\sigma^2 + ac\sigma^2 + bc\sigma^4 + c^2\sigma^3 $$</span> </p>
| Travis Willse | 155,629 | <p><strong>Hint</strong> Rearranging the defining equation gives <span class="math-container">$\require{cancel} \sigma^2 = -\sigma - 1$</span>. So, for example, the second term, <span class="math-container">$a b \sigma^2$</span> becomes <span class="math-container">$-a b \sigma - a b$</span>.</p>
<p>Alternatively, we can avoid tedious computation by showing that <span class="math-container">$$(\sigma^2)^2 + (\sigma^2) + 1 = 0 ,$$</span>
which implies that <span class="math-container">$\sigma^2$</span> is also a root of <span class="math-container">$x^2 + x + 1$</span>, and hence that <span class="math-container">$\sigma^2 = \bar \sigma$</span>. Here (provided <span class="math-container">$a, b, c$</span> are all real---the claim isn't true for general complex parameters) we can rewrite the inequality as <span class="math-container">$$z \bar z \geq 0 , \qquad \textrm{where} \qquad z := a + b \sigma + c \sigma^2 .$$</span></p>
|
1,304,344 | <p>How do I find the following:</p>
<p>$$(0.5)!(-0.5)!$$</p>
<p>Can someone help me step by step here?</p>
| Zelos Malum | 197,853 | <p>Gamma function the rescue! It is generalized of factorial to all non-negative even values
$$(0.5)!=\Gamma(-0.5)=-2\sqrt{\pi}$$
$$(-0.5)!=\Gamma(-1.5)=\frac{4}{3}\sqrt{\pi}$$</p>
|
382,293 | <p>Not sure where to go with this one. Clearly will have to use the axiom of choice at some point. I haven't been able to think of a good example for the set $A.$ Once we've got that, it'd be a matter of showing that a representation $($as a sum, $q+a)$ exists for each real number $($which should be the case by construction of $A)$, and then subsequently that this representation is unique.</p>
| Brian M. Scott | 12,042 | <p>HINT: You need $A$ to have the property that if $a,b\in A$ and $a\ne b$, then $(a+\Bbb Q)\cap(b+\Bbb Q)=\varnothing$. Suppose that $x\in(a+\Bbb Q)\cap(b+\Bbb Q)$; then there are $p,q\in\Bbb Q$ such that $a+p=x=b+q$ and hence $b=a+(p-q)\in a+\Bbb Q$ Conversely, if $b\in a+\Bbb Q$, then it’s clear that $(a+\Bbb Q)\cap(b+\Bbb Q)\ne\varnothing$. Thus, you need to be sure that if $a,b\in A$ with $a\ne b$, then $b\notin a+\Bbb Q$.</p>
<p>Now note that $\{x+\Bbb Q:x\in\Bbb R\}$ is a partition of $\Bbb R$, so you can choose ... what?</p>
<p>(You can do this from a group-theoretic point of view, but it’s not necessary.)</p>
|
158,720 | <p>By induction I can prove :
$$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!} = \frac{(D+M)!}{D!M!} $$</p>
<p>However, I couldn't derive the right hand side directly.</p>
<p>It would be of great help if anyone can solve it!!</p>
| Norbert | 19,538 | <p>Here is a combinatorial proof.</p>
<p>Assume you have $D+1$ types of cakes, and you allowed to choose $M$ cakes. Of course you can take several cakes of the same type. The amount of possible choices is <a href="http://en.wikipedia.org/wiki/Combination" rel="nofollow">combination with repetitions</a>:
$$
{(D+1)+M - 1 \choose D}=\frac{(M+D)!}{M!D!}\tag{1}
$$
How can we classify all possible choices? We say our choice belongs to the $t$-th class if we choose $M-t$ cakes of the first type. How many choices of $t$-th type are there? Well this is amount of possible choices of remaining $t$ cakes from remaining $D$ types of cakes, which is equal to
$$
{D+t-1 \choose D-1}=\frac{(D+t-1)!}{(D-1)!t!}
$$
Since we are allowed to take only $M$ cakes, there are $M+1$ classes - $t$ varies from $0$ to $M$. Now we see that total amount of choices is sum of amount of choices over all classes:
$$
\sum\limits_{t=0}^M{D+t-1\choose D-1}=\sum\limits_{t=0}^M\frac{(D+t-1)!}{(D-1)!t!}
$$
On the other hand this amount is equals to $(1)$.</p>
|
205,080 | <p>I have a problem that I cannot figure out how to do. The problem is:<br>
Suppose $s(x)=\frac{x+2}{x^2+5}$. What is the range of $s$?<br><br>
I know that the range is equivalent to the domain of $s^{-1}(x)$ but that is only true for one-to-one functions. I have tried to find the inverse of function s but I got stuck trying to isolate y. Here is what I have done so far:<br>
$y=\frac{x+2}{x^2+5}$<br><br>
$x=\frac{y+2}{y^2+5}$<br><br>
$x(y^2+5)=y+2$<br>
$xy^2+5x=y+2$<br>
$xy^2-y=2-5x$<br>
$y(xy-1)=2-5x$<br></p>
<p>This is the step I got stuck on, usually I would just divide by the parenthesis to isolate y but since y is squared, I cannot do that. Is this the right approach to finding the range of the function? If not how would I approach this problem?</p>
| Community | -1 | <p><strong>Hint:</strong> one way would be to sketch it, notice the minimum and maximum (call them $m$ and $M$), and show that $y>M$ and $y<m$ lead to contradictions (show also that $y=M$ and $y=m$ for certain values of $x$)</p>
|
205,080 | <p>I have a problem that I cannot figure out how to do. The problem is:<br>
Suppose $s(x)=\frac{x+2}{x^2+5}$. What is the range of $s$?<br><br>
I know that the range is equivalent to the domain of $s^{-1}(x)$ but that is only true for one-to-one functions. I have tried to find the inverse of function s but I got stuck trying to isolate y. Here is what I have done so far:<br>
$y=\frac{x+2}{x^2+5}$<br><br>
$x=\frac{y+2}{y^2+5}$<br><br>
$x(y^2+5)=y+2$<br>
$xy^2+5x=y+2$<br>
$xy^2-y=2-5x$<br>
$y(xy-1)=2-5x$<br></p>
<p>This is the step I got stuck on, usually I would just divide by the parenthesis to isolate y but since y is squared, I cannot do that. Is this the right approach to finding the range of the function? If not how would I approach this problem?</p>
| Matthew Conroy | 2,937 | <p>To find the range, we want to find all $y$ for which there exists an $x$ such that
$$ y = \frac{x+2}{x^2+5}.$$
We can solve this equation for $x$:
$$ y x^2 + 5y = x+2$$
$$ 0 = y x^2 -x + 5y-2$$
If $y \neq 0$, this is a quadratic equation in $x$, so we can solve it with the quadratic formula:
$$
x = \frac{ 1 \pm \sqrt{ 1 - 4y(5y-2)}}{2y}.$$
So, for a given $y$, $y$ is in the range if this expression yields a real number. That is, if
$$ 1 - 4y(5y-2) = -20y^2 +8y +1 \ge 0$$
If you study this quadratic, you will find that it has roots at $y=1/2$ and $y=-1/10$, and between these roots it is positive, while outside these roots it is negative. Hence, there exists an $x$ such that $s(x)=y$ only if
$$
-\frac{1}{10} \le y \le \frac{1}{2}.
$$
Thus, this is the range of $s$.</p>
<p>(Note we excluded $y=0$ earlier, but we know $y=0$ is in our range since $s(-2)=0$.)</p>
|
2,473,780 | <p>So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:</p>
<ol>
<li><p>Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$
Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?</p></li>
<li><p>One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?</p></li>
</ol>
<p>We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$</p>
| Bernard | 202,857 | <ol>
<li>This is a standard result from high school</li>
<li>If you nevertheless want to deduce it from the limit of $\dfrac{\ln x}x$, use the properties of logarithm:
$$\frac{\ln x}{\sqrt x}=\frac{2\ln(\sqrt x)}{\sqrt x}\xrightarrow[\sqrt x\to\infty]{}2\cdot 0=0$$</li>
</ol>
|
148,257 | <p>Let $f:\mathbb R\to\mathbb R$ be continuous. Suppose $(x_n)_n$ and $(y_n)_n$ are sequences in $\mathbb R$ such that the sequence $(x_n-y_n)_n$ converges to $0$. Does this mean that the sequence $(f(x_n)-f(y_n))_n$ converges to $0$?</p>
<p>I feel like it is true, since the definition of continuity states that $f$ preserves limits of convergent sequences, but I do not know how to prove it.</p>
| copper.hat | 27,978 | <p>Try $f(x) = e^{x^2}$, with $y_n = n$, $x_n = n+\frac{1}{n}$. Then $(x_n-y_n) \to 0$, but $f(x_n)-f(y_n) = e^{n^2}(e^{2+\frac{1}{n^2}}-1)$, which is unbounded.</p>
|
2,375,736 | <p>If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?</p>
| Donald Splutterwit | 404,247 | <p>Hint :
let $\alpha$ and $\beta$ be the roots of $ax^2+bx+c=0$ & let $\gamma$ and $\delta$ be the roots of $a_1 x^2+b_1 x+c_1 =0$.
The ratio of their roots are equal if
\begin{eqnarray*}
\frac{\alpha}{\beta} = \frac{\gamma}{\delta}.
\end{eqnarray*}</p>
<p>Further hint : $\color{red}{\alpha+\beta=-\frac{b}{a}}$ & $\alpha \beta=\frac{c}{a}$
\begin{eqnarray*}
\frac{b^2}{ac} = \color{red}{\frac{b^2}{a^2}} \frac{a}{c} = \frac{\color{red}{(\alpha+\beta)^2}}{\alpha \beta}=\frac{\alpha}{\beta}+2+\frac{\beta}{\alpha} = \cdots
\end{eqnarray*}</p>
|
2,375,736 | <p>If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?</p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>Note that , if the roots of the equation $ax^2+bx+c=0$ are $x_1$ and $x_2$, than
$$
\frac{b}{ac}=\frac{\left(\frac{x_1}{x_2}+1\right)^2}{\frac{x_1}{x_2}}
$$</p>
|
3,784,471 | <p>To solve this exercise,</p>
<p><span class="math-container">$$|\arccos(\cos(x))|<\pi/4$$</span></p>
<p>I have thought to apply this condition,
<span class="math-container">$$|f(x)|<k, \quad k\in \Bbb R^+, \iff -k<f(x)<k$$</span></p>
<p>Hence,</p>
<p><span class="math-container">$$-\frac \pi4<\arccos(\cos(x))<\frac \pi4$$</span>
Being <span class="math-container">$\arccos\colon [-1,1]\to [0,\pi]$</span>, I can have</p>
<p><span class="math-container">$$\cos\left(-\frac \pi4\right)<\cos(\arccos(\cos(x)))<\cos\left(\frac \pi4\right) \iff \frac{\sqrt2}2<\cos(x)<\frac{\sqrt2}2$$</span></p>
<p>false for all <span class="math-container">$x\in \mathbb{R}$</span>. Are they correct my steps?</p>
| tkf | 117,974 | <p>Consider the algebra over <span class="math-container">$\mathbb{F}_3$</span> with basis (as a vector space over <span class="math-container">$\mathbb{F}_3$</span>) the set <span class="math-container">$\{1,x,x^2\}$</span> and multiplication given by:
<span class="math-container">\begin{eqnarray*}
x(x^2)&=&x+2,\\
(x^2)x&=&1+x+x^2,\\
(x^2)(x^2)&=&x.
\end{eqnarray*}</span></p>
<p>By construction it is finite, has a two-sided identity <span class="math-container">$1$</span> and multiplication distributes over addition. The first two equations demonstrate that it is non-associative. Also left or right multiplication by any fixed non-zero element is bijective (see proof below).</p>
<p>I fixed the first equation and did a computer search through the <span class="math-container">$676$</span> possibilities for the other two. Of these <span class="math-container">$14$</span> came out as having the left and right cancellation property. One of these was of course <span class="math-container">$\mathbb{F}_{27}$</span>. The other <span class="math-container">$13$</span> are non-associative, and of them the above algebra seemed like the nicest.</p>
<p>Proof of left and right cancellation:</p>
<hr />
<p>It suffices to prove that left multiplication by any non-zero element is injective, as then it must also be surjective and the algebra will not contain non-zero zero-divisors. Thus right multiplication by any non-zero element would also be injective, hence bijective.</p>
<p>Both <span class="math-container">$x^3-x^2-x-1$</span> and <span class="math-container">$x^3-x-2$</span> are irreducible over <span class="math-container">$\mathbb{F}_3$</span> as they have no roots in <span class="math-container">$\mathbb{F}_3$</span>.</p>
<p>Left multiplication by a non-zero <span class="math-container">$\mathbb{F}_3$</span>-linear combination <span class="math-container">$\alpha(x)$</span> of <span class="math-container">$1$</span> and <span class="math-container">$x$</span> is the same map as left multiplication by <span class="math-container">$\alpha(x)$</span> in <span class="math-container">$\mathbb{F}_3[x]/(x^3-x-2)\cong\mathbb{F}_{27}$</span> - hence bijective.</p>
<p>Similarly left multiplication by a non-zero <span class="math-container">$\mathbb{F}_3$</span>-linear combination <span class="math-container">$\alpha(y)$</span> of <span class="math-container">$1$</span> and <span class="math-container">$y=x^2$</span> is the same map as left multiplication by <span class="math-container">$\alpha(y)$</span> in <span class="math-container">$\mathbb{F}_3[y]/(y^3-y^2-y-1)\cong\mathbb{F}_{27}$</span> - hence bijective.</p>
<p>Thus without loss of generality, if there is a non-zero left zero-divisor, there will be one of the form <span class="math-container">$\lambda+x\pm x^2$</span>, for some <span class="math-container">$\lambda\in \mathbb{F}_3$</span>. Thus it suffices to check that the matrices representing left multiplication by <span class="math-container">$x\pm x^2$</span> have no eigenvalues in <span class="math-container">$\mathbb{F}_3$</span>. The characteristic polynomials of these matrices are:</p>
<p><span class="math-container">$$
\left| \begin{array}{ccc} t&2 &1 \\
2&t+2&1\\
2&1&t
\end{array}\right|
= t^3-t^2-t-1
,\qquad
\left| \begin{array}{ccc} t&1 &1 \\
2&t+1&0\\
1&0&t
\end{array}\right|
= t^3+t^2+2
.$$</span></p>
<p>Neither of these cubics have roots in <span class="math-container">$\mathbb{F}_3$</span>.</p>
<hr />
|
3,784,471 | <p>To solve this exercise,</p>
<p><span class="math-container">$$|\arccos(\cos(x))|<\pi/4$$</span></p>
<p>I have thought to apply this condition,
<span class="math-container">$$|f(x)|<k, \quad k\in \Bbb R^+, \iff -k<f(x)<k$$</span></p>
<p>Hence,</p>
<p><span class="math-container">$$-\frac \pi4<\arccos(\cos(x))<\frac \pi4$$</span>
Being <span class="math-container">$\arccos\colon [-1,1]\to [0,\pi]$</span>, I can have</p>
<p><span class="math-container">$$\cos\left(-\frac \pi4\right)<\cos(\arccos(\cos(x)))<\cos\left(\frac \pi4\right) \iff \frac{\sqrt2}2<\cos(x)<\frac{\sqrt2}2$$</span></p>
<p>false for all <span class="math-container">$x\in \mathbb{R}$</span>. Are they correct my steps?</p>
| tkf | 117,974 | <p>We provide a family of examples of finite <a href="https://en.wikipedia.org/wiki/Division_algebra" rel="nofollow noreferrer">division algebras</a>, which additionally have a right identity. This does not answer the revised version of the question which asks for a finite division algebra (other than a finite field) which has a two-sided identity.</p>
<p>On any finite field <span class="math-container">$\mathbb{F}_q$</span> with <span class="math-container">$q=p^r$</span>, and <span class="math-container">$p$</span> prime, <span class="math-container">$r>1$</span>, we can define <span class="math-container">$a\star b= ab^p$</span>. This is non-commutative (<span class="math-container">$a\star b\neq b\star a \iff a^{-1}b\notin \mathbb{F}_p$</span>), but has the two-sided cancellation property: <span class="math-container">$$a\star b=0\implies a=0\,\, {\rm or}\,\, b=0,$$</span> and has a right identity.</p>
|
775,265 | <p>Please help me get the answer to this question.</p>
<p>Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)</p>
| Mr.Fry | 68,477 | <p>There is no problem with "reo's" work, he/she gave you a good sketch of the background work. Now we will proceed in the forward argument which somehow where we "magically" know what $\delta$ has to be. </p>
<p>Proof: Given $\epsilon >0$ choose $\delta $= min{$1,\frac{\epsilon}{\sqrt{2}}$}. Then, $|f(x)-f(4)| = \sqrt{2}|\frac{x-4}{\sqrt{x-3}+1}|$. </p>
<p>Suppose $|x-4|<\delta \Rightarrow \sqrt{2}|\frac{x-4}{\sqrt{x-3}+1}|< \sqrt{2}\delta = \sqrt{2}(\frac{\epsilon}{\sqrt{2}})=\epsilon$.</p>
<p>$\textbf{Additional}$: Note: We can also have, $|x-4|< \delta \leq 2 \Rightarrow 2<x<6 \Rightarrow \frac{|x-4|}{|\sqrt{x-3}+1|}<\frac{\delta}{2}$. Thus, picking $\delta$=min{$2\epsilon,1$} works as well. We typically want to bound $x-a$ so that the denominator doesn't blow up.</p>
|
1,022,523 | <blockquote>
<p>Make a series expansion of $f(z)=\dfrac{1}{z^2+z-6}$ valid in the region $2<|z|<3$.</p>
</blockquote>
<p>By partial fractions,</p>
<p>$$f(z) = \frac{1}{(z-2)(z+3)} = \frac{1}{5(z-2)}-\frac{1}{5(z+3)}.$$</p>
<p>From here, how are these fractions expanded into a geometric series?</p>
| Jo Wehler | 169,961 | <p>The correspondence between Grothendieck and Serre during the period 1955-1987 illustrates how Grothendieck's mathematical work shifted from functional analysis to algebraic geometry. The correspondence edited by the American Mathematical Society has the special flavour of being bilingual.</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/B0088OUXXY" rel="nofollow">http://www.amazon.com/Grothendieck-Serre-Correspondence-Pierre-Colmez-Grothendieck/dp/B0088OUXXY/ref=sr_1_1?s=books&ie=UTF8&qid=1416203841&sr=1-1</a></p>
|
2,747,753 | <p>Let $x\in\mathbb{R}$. Demonstrate that if the numbers $a = x^3–x$ and $b = x^2 +1$ are rational, then $x$ is rational.</p>
| lulu | 252,071 | <p>$x^2+1\in \mathbb Q \implies x^2\in \mathbb Q$</p>
<p>From this, we deduce that $x^3=rx$ for some $r\in \mathbb Q$. It is easy to see that $r=1\implies x=0,\pm 1$ which are all rational. Assume, then, that $r\neq 1$.</p>
<p>But this implies that $x^3-x=x(r-1)\in \mathbb Q\implies x\in \mathbb Q$ as desired.</p>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| rm -rf | 5 | <p>You can use <code>Subtract</code> or <code>Differences</code> to get the differences in the index and use <code>FreeQ</code> to test if every element is <code>True</code>. This removes the need for the ugly <code>#2[[1]]-#2[[2]]</code> and also the <code>Flatten</code> and <code>And@@</code> at the end.</p>
<pre><code>triDiagonalQ[mat_?MatrixQ] :=FreeQ[MapIndexed[(#1 == 0 || Abs[Subtract @@ #2] <= 1) &,
mat, {2}], False]
</code></pre>
<p>The pattern test <code>_?MatrixQ</code> ensures that you only operate on matrices.</p>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| Leonid Shifrin | 81 | <p>You can use macros / code generation to be able to use the syntax you like. Here is one possibility:</p>
<pre><code>ClearAll[withLiteralIndices];
SetAttributes[withLiteralIndices, HoldAll];
withLiteralIndices[code_, inds : {__Symbol}] :=
Block[inds,
Unevaluated[code] /.
MapIndexed[
Function[{i, pos}, pos /. {p_} :> (i :> #2[[p]])],
inds
]
]
</code></pre>
<p>Now, you can pretty much <em>literally</em> use the code you would like to use:</p>
<pre><code>withLiteralIndices[
triDiagonalQ[mat_] :=
MapIndexed[#1 == 0 || Abs[i - j] <= 1 &, mat, {2}] // Flatten // And @@ # &,
{i, j}
]
</code></pre>
<p>When you look at the resulting definition, you can see that this is entirely equivalent to hand-written code using slots:</p>
<pre><code>?triDiagonalQ
(*
Global`triDiagonalQ
triDiagonalQ[mat_]:=
(And@@#1&)[Flatten[MapIndexed[#1==0||Abs[#2[[1]]-#2[[2]]]<=1&,mat,{2}]]]
*)
</code></pre>
<p><strong>EDIT</strong></p>
<p>Here is a simpler and perhaps more elegant version of the macro, which uses the <a href="https://mathematica.stackexchange.com/questions/1929/injecting-a-sequence-of-expressions-into-a-held-expression/1937#1937">injector pattern</a> more explicitly:</p>
<pre><code>ClearAll[withLiteralIndices];
SetAttributes[withLiteralIndices, HoldAll];
withLiteralIndices[code_, inds : {__Symbol}] :=
Block[inds,
Unevaluated[code] /.
Replace[
Range[Length[inds]],
p_ :> (inds[[p]] :> #2[[p]]),
{1}
]
]
</code></pre>
|
876,310 | <p>So I <em>think</em> I understand what differentials are, but let me know if I'm wrong.</p>
<p>So let's take $y=f(x)$ such that $f: [a,b] \subset \Bbb R \to \Bbb R$. Instead of defining the derivative of $f$ in terms of the differentials $\text{dy}$ and $\text{dx}$, we take the derivative $f'(x)$ as our "primitive". Then to define the differentials we do as follows:</p>
<p>We find some $x_0 \in [a,b]$ where there is some neighborhood of $x_0$, $N(x_0)$, such that all $f(x)$ in $\{f(x) \in \Bbb R \mid x \in N(x_0)\}$ are differentiable. Then we choose another point in $N(x_0)$, let's call it $x_1$, such that $x_1 \ne x_0$. Then let $dx = \Delta x = x_1 - x_0$. Now this $\Delta x$ doesn't actually have to be very small like we're taught in Calculus 1 (in particular it's not infinitesimal, it's finite). In fact, as long as $f(x)$ is differentiable for all $x \in [-10^{10}, 10^{10}]$ we could choose $x_0 = -10^{10}$ and $x_1 = 10^{10}$.</p>
<p>Then we know that $\Delta y = f'(x_0) \Delta x + \epsilon(\Delta x)$, where $\epsilon(\Delta x)$ is some nonlinear function of $\Delta x$. If $f(x)$ is smooth, we know that $\epsilon(\Delta x)$ is equal to the sum of powers of $\Delta x$ with some coefficients, by Taylor's theorem. But of course, $\epsilon(\Delta x)$ won't be so easy to describe if $f(x)$ is only once differentiable. So we define $dy$ as $dy = f'(x_0) dx$: that is, $dy$ is the <em>linear part</em> of $\Delta y$. This has the very useful property that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x_0)$. This is then <em>not a definition</em> of the derivative, but a consequence of our definitions.</p>
<p>It can be seen from this $dy$ really depends on what we choose as $dx$, but $f'$ is independent of both. </p>
<p>This definition can be extended to functions of multiple variables, like $z = f(x, y)$ as well, by letting $\Delta x = dx,\ \Delta y=dy$ and defining $dz$ as $dz = \frac{\partial f(x_0, y_0)}{\partial x}dx + \frac{\partial f(x_0, y_0)}{\partial y} dy$. So $dz$ is the linear part of $\Delta z$. Does all of the above look correct?</p>
<p>If so, then where I'm having a problem is: <br>1) how then <em>do</em> we define the derivative of $f(x)$ if not by $f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$? <br>2) how do we apply this definition of $dx$ to $\int_a^b f(x)dx$? It seems like the inherit arbitrariness of $dx$ is really going to get in the way of a good definition of the integral.</p>
| user121955 | 121,955 | <p>Also, the definition of a derivative that I learned was $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$...basically rise over run as run ($\mathrm{d}x$) approaches $0$ (thus the tangent line concept).</p>
|
881,282 | <p>Same as above, how to simplify it. I am to calculate its $n$th derivative w.r.t x where t is const, but I can't simplify it. Any help would be appreciated. Thank you.</p>
| 2'5 9'2 | 11,123 | <p>One solution is "clear" at $\frac{2}{\sqrt{3}}$. I was motivated to look for something like this by trying to write $1$ as the sum of two simple fractions, and the presence of $3$ and $4$. </p>
<p>It's also halfway between the two vertical asymptotes of $\frac1{x^2}+\frac{1}{\left(4-\sqrt{3}x\right)^2}$, and in a sketch of that function (which clearly is never negative) it was natural to want to see what the output was at that midpoint.</p>
|
3,596,566 | <p>I've been going through <a href="https://math.stackexchange.com/a/27177/288450">this proof</a>.</p>
<p>And I'm wondering what allows me to change the order of the integral and the infinite sum.</p>
<p><span class="math-container">$$\int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} dx = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$</span></p>
<p>I know the result that for a function series that converges uniformly on <span class="math-container">$[a, b]$</span> the following holds:</p>
<p><span class="math-container">$$ \int^{b}_{a} \sum_{n=0}^{\infty} f_n(x) \ dx = \sum_{n=0}^{\infty} \int_a^b f_n(x) \ dx$$</span></p>
<p>But here the limits of integration are <span class="math-container">$-\infty$</span> and <span class="math-container">$+\infty$</span>, and I'm not sure what to make of it.</p>
| AlanD | 356,933 | <p>The series can be written as
<span class="math-container">$$
\sum_{n=0}^\infty \frac{(2tx)^n}{n!}e^{-x^2}\leq \sum_{n=0}^\infty \frac{(2|t||x|)^n}{n!}e^{-x^2}=e^{-(x^2-2|t||x|)}=e^{-[(|x|-|t|)^2-|t|^2]}=e^{t^2}e^{-(|x|-|t|)^2}
$$</span>
Obviously, the right hand term is integrable (it is essentially a Gaussian distribution centered at <span class="math-container">$|t|$</span>). Call it <span class="math-container">$f(x;t)$</span>. Now let the partial sum be denoted as <span class="math-container">$g_N(x;t)=\sum_{n=0}^N \frac{(2tx)^n}{n!}e^{-x^2}$</span>. Clearly, <span class="math-container">$|g_N(x;t)|\leq \sum_{n=0}^N \frac{(2|t||x|)^n}{n!}e^{-x^2}\leq f(x;t)$</span> (since all terms in sum are positive). By dominated convergence theorem,
<span class="math-container">$$
\lim_{N\to\infty}\int_{-\infty}^\infty g_N(x;t)\,dx=\int_{-\infty}^\infty\lim_{N\to\infty}g_N(x;t)\,dx.
$$</span>
For the left hand side, partial sums can always be pulled out of integrals. The conclusion then holds.</p>
|
3,970,641 | <p>I have 927 unique sequences of the numbers 1, 2 and 3, all of which sum to 12 and represent every possible one-octave scale on the piano, with the numbers representing the intervals between notes in half-steps (i.e., adjacent keys). For example, the <a href="https://en.wikipedia.org/wiki/Major_scale" rel="nofollow noreferrer">Major Scale</a> is { 2,2,1,2,2,2,1 }, while the hexatonic (6-note) <a href="https://en.wikipedia.org/wiki/Blues_scale" rel="nofollow noreferrer">Blues Scale</a> is { 3, 2, 1, 1, 3, 2 }. It can help mentally to add a zero at the beginning to represent the root note.</p>
<p>I'm trying to decide how to categorize them, and believe Shannon entropy is a natural place to begin, with one problem: The default Shannon Equation returns the same figure agnostic of the order of the sequence (as we would expect), whereas this is a case where the order is supremely relevant. (Music theory and information theory have a lot in common, though this appears to be largely unexplored.)</p>
<p>Let's take a simple example: Both of these are valid scales, though neither has a name, to my knowledge:</p>
<pre><code>{ 1,2,1,2,1,2,1,2 }
{ 2,2,1,2,1,1,1,2 }
</code></pre>
<p>Since there are equal numbers of 1s and 2s in each set, we don't need a calculator to see that the Shannon Entropy for each is 4:</p>
<p><span class="math-container">$$-\sum_{i=1}^8 \frac{1}{2} \log_{2}\frac{1}{2}$$</span></p>
<p>However, I believe--I may be very wrong here--that there is significantly more information contained in the second sequence, given that the first can be communicated as "{1,2}, four times" while the second is considerably more arbitrary. (I believe this is debate in bioinformatics, where one can locate long spans of simple repeating sequences of nucleotides that some consider "junk DNA," though others dispute that adjective.)</p>
<p><strong>How does one account for order when measuring surprisal?</strong> If I were to take, say, <em>The Great Gatsby</em> and sort the ~261,000 characters (as in letters, etc., not fictional people!) in ASCII order, I don't think it would contain the same amount of information as the original text.</p>
<p>One strategy I tried was a rolling measure of probability, where the algorithm is initially unaware of the total frequency of each number but accumulates the odds as it goes. (I suppose this is like measuring the accumulating surprisal of a die you cannot see and don't know how many sides it has.) This did not produce very intuitive or useful results.</p>
<p>Another thought I had was to divide each sequence into "words" that are repeating sub-sequences of at least two digits, then measure the entropy of each word and ... I'm not sure what to do with them. A sum of <span class="math-container">$H[{1,2}]$</span> four times still adds to 4. I'm a bit out of my depth here both in reducing sequences to the most efficient subsequences and knowing what to do with them.</p>
<p>I've read Shannon's 1948 paper twice, including the discussion of the relative odds of letters in a given language, which seems relevant but is, in fact, not useful in this case, where every possible ordered combination is present and the intervals are added independently of those preceding them, with the edge case that they cannot exceed a sum of 12.</p>
<p><strong>Which is all to say, how does one quantify the entropy of an <em>ordered</em> sequence, when the order is of extreme relevance but each item is independent of the others?</strong> Below I'll include how I generated the 927 sequences for anyone interested, but it's ancillary to the question, I think.</p>
<p>***</p>
<p>First, I wrote a simple recursive algorithm that begins with an empty set and calls itself three times after adding 1, 2 or 3, terminating when the sum is 12 and tossing any that go over. Happy to provide code or pseudo-code, but I think it's intuitive.</p>
<p>To check myself, I generated the all the <em>combinations</em> of [1,2,3] that add to 12--there are 19--using a <a href="https://algorithmist.com/wiki/Coin_change" rel="nofollow noreferrer">Coin-Change algorithm</a>, and then calculated the unique, distinct <em>permutations</em> of each of them (treating identical digits as indistinguishable). The sum of the permutations is also 927, which range from 1 for the sequence of twelve 1s (the Chromatic Scale) or four 3s (a sparse diminished scale) to 140 possible scales for the intervals { 1, 1, 1, 2, 2, 2, 3 }, which <em>almost</em> subsumes the Harmonic Minor Scale.</p>
<h3>One more thought</h3>
<p>Regarding the above example of four 1s and four 2s: If I was flipping an unbiased coin 12 times and coding Heads as "1" and Tails as "2", I would be more surprised, in a general sense of the word, to get alternating results than to get the second sequence with small clusters. So it's possible that "surprisal" is the wrong frame of mind here versus information.</p>
| Linguosaurus Rex | 1,149,751 | <p>Very interesting question! I've also been thinking about entropy in sequences, but with regards to language. I'm surprised I haven't been able to find much about the entropy of sequences (still looking), but I think I've been able to work some parts of it out.</p>
<p>As the first commenter mentioned, Shannon's entropy <span class="math-container">$H(X)$</span> assumes that <span class="math-container">$X$</span> is i.i.d. (independent and identically-distributed). That means in the limit there's no correlation between outcomes of <span class="math-container">$X$</span>, and that the probability distribution of <span class="math-container">$X$</span> remains the same for each trial.</p>
<p>This works great for coin flips, but not when outcomes are correlated. For that, we need the concept of <a href="https://en.wikipedia.org/wiki/Conditional_entropy" rel="nofollow noreferrer">conditional entropy</a>. We can use the conditional entropy between two correlated random variables, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>, to calculate their joint entropy, <span class="math-container">$H(X,Y)$</span>:</p>
<p><span class="math-container">$$H(X,Y) = H(X) + H(Y|X)$$</span></p>
<p>or alternatively,</p>
<p><span class="math-container">$$H(X,Y) = H(Y) + H(X|Y)$$</span></p>
<p>For sequences, not only are random variables correlated, but they're also ordered. This order contains much of the sequence's information; as you said, sorting the characters in The Great Gatsby into ASCII order would change the information in the novel. To calculate the entropy in a sequence, its items must be generated in this order. Changing this order changes the entropy, because it changes the sequence.</p>
<p>For a sequence of outcomes generated by a sequence of random variables <span class="math-container">$X_0,\dots,X_{n-1}$</span>, its entropy would then be</p>
<p><span class="math-container">$$H(X_0,\dots,X_{n-1}) = H(X_0) + \sum_{i=1}^{n-1} H(X_i|X_0,\dots,X_{i-1})$$</span></p>
<p>or alternatively, backwards:</p>
<p><span class="math-container">$$H(X_{n-1},\dots,X_0) = H(X_{n-1}) + \sum_{i=n-2}^{0} H(X_i|X_{i+1},\dots,X_{n-1})$$</span></p>
<p>(In practice, this can be hard to calculate for longer sequences, if there aren't very many of them. If most of the correlations are very local, we can assume that each conditional entropy is only conditional on the last <span class="math-container">$k$</span> outcomes, for some constant positive integer <span class="math-container">$k$</span>. When <span class="math-container">$k=1$</span>, we get the <a href="https://en.wikipedia.org/wiki/Markov_property" rel="nofollow noreferrer">Markov assumption</a>).</p>
<p>Since items in a sequence are generated in order, one after another, it also makes sense to talk about the entropy of a particular sequence:</p>
<p><span class="math-container">$$H(\{X_0,\dots,X_{n-1}\}=\{x_0,\dots,x_{n-1}\})$$</span>
<span class="math-container">$$= H(X_0) + \sum_{i=1}^{n-1} H(X_i|X_0=x_0,\dots,X_{i-1}=x_{i-1})$$</span></p>
<p>To make this more concrete, we can imagine a situation where one person has to indicate to another one of the scales you generated. First, person A randomly picks one of the 927 scales, and gives it person B. Then B has to send a message to person C, to tell C which scale they've been given by A. B and C can agree on a protocol beforehand, but can't communicate after B gets a scale from A. The entropy of a scale would then be the amount of information or uncertainty contained in the message that describes it.</p>
<p>So for the first scale, we only have to know the first two intervals to know the rest. The message "12" would be enough to specify it; the transmitter and receiver could agree beforehand that, for example, if the intervals didn't add up to 12, the receiver would just repeat the written intervals until they did. Since <span class="math-container">$H(X_0) = -log_2 (\frac{1}{3})$</span>, and since <span class="math-container">$H(X_1|X_0)$</span> is the same, the joint entropy <span class="math-container">$H(\{X_0,\dots,X_7\}=\{1,2,1,2,1,2,1,2\})$</span> is then <span class="math-container">$-2log_2 \frac{1}{3} \approx 3.17$</span> bits, or <span class="math-container">$0.396$</span> bits/interval.</p>
<p>For the second scale, however, there's no pattern that can shorten the message, so all intervals are independent and must be specified. With the exception of the final 2, which otherwise could only have been a 1, all the other intervals could have been 1, 2, or 3. So <span class="math-container">$H(\{X_0,\dots,X_7\}=\{2,2,1,2,1,1,1,2\})$</span> would then be <span class="math-container">$-7log_2 \frac{1}{3} - log_2 \frac{1}{2} \approx 12.09$</span> bits, or <span class="math-container">$1.51$</span> bits/interval.</p>
|
2,339,101 | <blockquote>
<p>There are six socks in a drawer. The socks are of two colors: black and white. If you draw two socks randomly, the probability that you get white socks is $\frac{2}{3}$. What is the probability of getting black socks, when two socks are drawn at a time?</p>
</blockquote>
<p>There is no detail about the number of white and black socks. Can anyone help me with this?</p>
| jvdhooft | 437,988 | <p>If there are $k$ white socks, the probability of getting two white socks equals:</p>
<p>$$\frac{k}{6} \cdot \frac{k-1}{5} = \frac{k(k-1)}{30} = \frac{2}{3} \iff k(k-1) = 20 \iff k=5$$</p>
<p>Since there are five white socks, it is impossible to draw more than one black sock. As such, the probability of getting two black socks is 0. The probability of getting one black sock and one white sock equals:</p>
<p>$$\frac{{5 \choose 1}{1 \choose 1}}{6 \choose 2} = \frac{5}{15} = \frac{1}{3}$$</p>
<p>Indeed, $\frac{2}{3} + \frac{1}{3} + \frac{0}{3} = 1.$</p>
|
3,014,766 | <p>I am supposed to find the derivative of <span class="math-container">$ 2^{\frac{x}{\ln x}} $</span>. My answer is <span class="math-container">$$ 2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot \frac{\ln x-x\cdot \frac{1}{x}}{\ln^{2}x}\cdot \frac{1}{x} .$$</span> Is it correct? Thanks. </p>
| Robert Z | 299,698 | <p>You are almost correct. You have just an <strong>extra</strong> factor <span class="math-container">$1/x$</span> at the end. The correct derivative is
<span class="math-container">$$D(2^{\frac{x}{\ln x}})=2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot D\left(\frac{x}{\ln x}\right)=2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot \frac{\ln x-x\cdot \frac{1}{x}}{\ln^{2}x}.$$</span></p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| orangeskid | 168,051 | <p>It is enough to assume
<span class="math-container">$$|f'(x)|\le f^2(x)$$</span></p>
<p>To show that if <span class="math-container">$\{x \ | \ f(x) = 0\}$</span> is nonvoid then it equals <span class="math-container">$\mathbb{R}$</span>. </p>
<p>Clearly it is a closed set. Let us show that it is also open.</p>
<p>Take <span class="math-container">$x_0$</span>, <span class="math-container">$f(x_0) = 0$</span>. There exists <span class="math-container">$0< \delta < 1$</span> so that
<span class="math-container">$$\sup_{x \in (x_0-\delta, x_0 + \delta)} |f(x)| < 1$$</span>
Denote that supremum by <span class="math-container">$M$</span>. Note that <span class="math-container">$0\le M < 1$</span>.</p>
<p>For <span class="math-container">$x \in (x_0-\delta, x_0 + \delta)$</span> we have</p>
<p><span class="math-container">$$|f(x)| = |f(x) - f(x_0) | = |\int_{x_0}^{x} f'(t) dt| \le
|x-x_0| \cdot \sup |f'(t)| \le \delta \cdot M^2$$</span>
so
<span class="math-container">$$M \le \delta M^2\le \delta M$$</span>
We conclude <span class="math-container">$M=0$</span>, so <span class="math-container">$f=0$</span> on <span class="math-container">$(x_0-\delta,x_0+\delta)$</span></p>
<p>Note: we can see how to generalize this to functions satisfying
<span class="math-container">$$|f'(x)|\le F(f(x))$$</span>
where <span class="math-container">$F$</span> is a positive function so that <span class="math-container">$F(t) \le K\cdot |t|$</span> around <span class="math-container">$0$</span>.</p>
|
352,983 | <p>How to find this expression $(1000!\mod 3^{300})$?</p>
| lab bhattacharjee | 33,337 | <p>Using <a href="https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n">this</a>, the highest power of $3$ in $1000!>300$</p>
<p>So, the remainder $=0$</p>
|
1,195,625 | <p>We have $X = R^n$ and the discrete metric:</p>
<p>$d(x,y) = 0$, if $x=y$ and $d(x,y) = 1$ in all other cases.</p>
<p>Is this space separable or not? I tried to prove, that the answer for that is no.</p>
<p>Let us have a random $x=(x_1, x_2, ..., x_n)$ vector from $R^n$. If $X$ is separable, then such $q$ exists, that $q=(q_1, q_2, ..., q_n)$ and $d(x_i-q_i) < ε$ for all $i$ in ${1,...n}$</p>
<p>The problem with that is, for example if I choose $ε = 1/3$ , then I can't find such $q$ vector, because in discrete metric, we either have 1 as distance, or 0. If some $q_i$ is not equal to its pair, $x_i$, then $d(x_i-q_i)= 1 < 1/3$ is impossible.</p>
<p>That would mean that all $x_i$ and $q_i$ are equal, which makes $x$ equal to $q$, but that's impossible too.</p>
<p>Is that proving good or what you think? Thanks! :)</p>
| Casteels | 92,730 | <p>Let $n=2k$. Then $T_2(n)$ is just the complete bipartite graph $K_{k,k}$. </p>
<p>Let $G=(V,E)$ be a triangle-free graph on $n$ vertices with $\delta(G)=k$. Let $v\in V$ have degree $\delta(G)$. Let $S$ be the neighbours of $v$, and let $T=V\setminus(S\cup v)$. So $|S|=k$ and $|T|=k-1$</p>
<p>Since $G$ is triangle-free, no two vertices in $S$ share an edge, and so the only way for a vertex $s\in S$ to have degree at least $k$ is if $s$ is adjacent to every vertex in $T$.</p>
<p>But now you have $T_2(n)$ as a spanning subgraph of $G$ (one side $S$, the other $T\cup v$), and of course we cannot add any more edges without creating a triangle. Therefore, $G$ must be $T_2(n)$.</p>
|
3,753,060 | <blockquote>
<p>If <span class="math-container">$\int f(x)dx =g(x)$</span> then <span class="math-container">$\int f^{-1}(x)dx $</span> is equal to</p>
<p>(1) <span class="math-container">$g^{-1}(x)$</span></p>
<p>(2) <span class="math-container">$xf^{-1}(x)-g(f^{-1}(x))$</span></p>
<p>(3) <span class="math-container">$xf^{-1}(x)-g^{-1}(x)$</span></p>
<p>(4) <span class="math-container">$f^{-1}(x)$</span></p>
</blockquote>
<p>My approach is as follows:
Let <span class="math-container">$f(x)=y$</span>, therefore <span class="math-container">$f^{-1}(y)=x$</span>, <span class="math-container">$\int f^{-1}(f(x))dx =g(f(x))$</span></p>
<p>On differentiating we get <span class="math-container">$x=g'(f(x))f'(x)$</span></p>
<p>After this step, I am not able to proceed.</p>
| Kavi Rama Murthy | 142,385 | <p>Ignoring the constant of integration the answer is (2):<span class="math-container">$$\int f^{-1}(x)dx=\int yf'(y)dy=yf(y)-\int f(y)dy$$</span> (where I have used integration by parts). Hence <span class="math-container">$$\int f^{-1}(x)dx=f^{-1}(x)x-g(y)=xf^{-1}(x)-g(f^{-1}(x))$$</span>.</p>
|
265,189 | <p>Integrate, $$\int_{0}^{\frac{\pi}{2}}\sin (\tan\theta) \mathrm{d\theta}$$</p>
| Mhenni Benghorbal | 35,472 | <p>First, make the change of variables $ x = \arctan(t) $ to transform the integral t0</p>
<p>$$\int_{0}^{\frac{\pi}{2}}\sin (\tan\theta) \mathrm{d\theta} = \int_{0}^{\infty}\frac{\sin(t)}{t^2+1} {dt}
\\
= -\frac{1}{2}\,{{\rm e}^{-1}}{ \operatorname {E_1} } \left( -1 \right) +\frac{1}{2}\,{{\rm e}}\,{\operatorname{E_1}} \left( 1 \right) - \frac{1}{2}\,i\pi \,{{\rm e}^{-1}}$$</p>
<p>where $\operatorname{E}_a(z)$ is the exponential integral</p>
<blockquote>
<p>$$ \operatorname{E}_a \left( z \right) =\int _{1}^{\infty }\!{{\rm e}^{-{t}z}}{{ t}}^{-a }{d{ t}}.$$</p>
</blockquote>
<p>To see the details of evaluation of the last integral see <a href="https://math.stackexchange.com/questions/264929/evaluate-int-0-infty-frac-alpha-sin-x-alpha2x2-mathrmdx-space/264936#264936">here</a>.</p>
|
2,820,796 | <p>In How many ways can a 25 Identical books can be placed in 5 identical boxes. </p>
<p>I know the process by counting but that is too lengthy .
I want different approach by which I can easily calculate required number in Exam hall in few minutes. </p>
<p>Process of Counting :
This problem can be taken partitions of 25 into 5 parts.</p>
<p>25 = 25+0+0+0+0</p>
<p>25 = 24 +1 + 0 + 0 +0</p>
<p>25 = 23+ 1 +1 +0 + 0
... ....
Like this way many combinations are made.: about 377 </p>
<p>How can we calculate it without this process of manual counting. </p>
| Boyku | 567,523 | <p>Number of partitions of m into b positive parts (blocks) is described here: <a href="https://oeis.org/A008284" rel="nofollow noreferrer">https://oeis.org/A008284</a> in the OEIS article. </p>
<p>$P(m,b) = P(m-1, b-1) + P(m-k,b)$</p>
<p>The recurrence formula means that given a $P(m,b)$ partition of m into b blocks, two situations may occur:</p>
<p>1) a singleton is in the partition; we remove it and obtain $ P(m-1, b-1)$ cases</p>
<p>2) there are no singletons in the partition; we remove one piece from each block and we get the $P(m-k, b)$ cases.</p>
<p>for example, for P(7,3)</p>
<p>5+1+1 <----> 5+1</p>
<p>4+2+1 <----> 4+2</p>
<p>3+3+1 <----> 3+3</p>
<p>3+2+2 <----> 2+1+1</p>
<p>After completing the table given by Andrew Wood to the 25th row we get:</p>
<p>P(25,1) + P(25,2) + P(25,3)+ P(25,4)+ P(25,5) = $1+12+52+120+192 = 377$ </p>
|
11,266 | <p>I have a list of time durations, which are strings of the form: <code>"hh:mm:ss"</code>. Here's a sample for you to play with:</p>
<pre><code>durations = {"00:09:54", "00:31:24", "00:40:07", "00:11:58", "00:13:51", "01:02:32"}
</code></pre>
<p>I want to convert all of these into numbers in seconds, so that I can actually do useful things with this data!</p>
<p>How would I go about doing this? There must be some way to get <em>Mathematica</em> to extract the relevant things from each string.</p>
<p>Thanks.</p>
<p>Edit: I'm surprised that there doesn't seem to be a question related to this already - but it may be that I was using the wrong search terms.</p>
| J. M.'s persistent exhaustion | 50 | <p>One method:</p>
<pre><code>3600 FromDMS[ToExpression[StringSplit[#, ":"]]] & /@ durations
{594, 1884, 2407, 718, 831, 3752}
</code></pre>
|
34,874 | <p>If you visit this <a href="http://www.springerlink.com/content/ug8h1563j3484211/" rel="nofollow">link</a>, you'll see at the top of the PDF view. Basic properties of finite abelian groups:</p>
<p>Every quotient group of a finite abelian group is isomorphic to a subgroup.</p>
<p>If the above statement true, it would make some proofs in Serge Lang's Algebra easier, particularly in the p-Sylow groups section.</p>
<p>I know that there is a correspondence between subgroups of G/N and subgroups of G containing N, but the corresponding groups are not necessarily isomorphic or are they?</p>
| Keivan Karai | 3,635 | <p>The quotients of an abelian group are in bijection with the subgroups of its Pontryagin dual. Now, every finite abelian group is isomorphic to its dual.</p>
|
2,555,399 | <p>The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$</p>
<p>I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?</p>
| Bernard | 202,857 | <p>Expand $(1+u)^{\tfrac12}$ up to order $3$:
$$(1+u)^{\tfrac12}=1+\frac12 u-\frac18u^2+\frac1{16}u^3+o(u^3),$$
and compose with $u=-2x+3x^2-4x^3$:</p>
<ul>
<li><p>$u^2=4x^2-12x^3+o(x^3)$,</p></li>
<li><p>$u^3=u^2\cdot u=-8x^3+o(x^3)$.</p></li>
</ul>
<p>One finally obtains$$1-x+x^2-x^3+o(x^3).$$</p>
|
1,136,192 | <p>I need to solve this integral but I have no idea about how to procede, this is the integral:</p>
<p>$$\int \frac{x-1}{x+4x^3}\mathrm dx$$</p>
<p>This is how I solve the first part:</p>
<p>$$\int \frac{x}{x+4x^3}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>$$\int \frac{1}{1+4x^2}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>So I solved the first integral:</p>
<p>$$\int \frac{1}{1 + (2x)^2}\mathrm dx = \frac{1}{2}\arctan(2x) + C$$</p>
<p>But how can I solve the second?</p>
<p>$$- \int \frac{1}{x(1+4x^2)}\mathrm dx$$</p>
| idm | 167,226 | <p><strong>Hint:</strong></p>
<p>$$\ln(n+1)\leq n+1\implies \frac{1}{\ln(n+1)}\geq\frac{1}{n+1}$$</p>
|
244,241 | <p>How can I find minimum distance between cone and a point ?</p>
<p><strong>Cone properties :</strong><br/>
position - $(0,0,z)$<br/>
radius - $R$<br/>
height - $h$</p>
<p><strong>Point properties:</strong><br/>
position - $(0,0,z_1)$</p>
| WimC | 25,313 | <p>This was already mentioned by Rahul but I think it deserves an answer in its own right. Digital signal processing of 1d (sound) and 2d (images) real data would take incredible amounts of time and would be much harder to understand if it weren't for the discrete Fourier transform and its fast implementations. This field is very real and complex numbers play a major role in it.</p>
|
1,344,464 | <p>Consider <span class="math-container">$$f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^{3n+1}$$</span></p>
<p>It's a power series with a radius, <span class="math-container">$R=1$</span>. at <span class="math-container">$x=1$</span> it converges. Hence, by Abel's thorem:</p>
<p><span class="math-container">$$\lim_{x\to 1^-} f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$</span></p>
<p>Evaluating the derivative</p>
<p><span class="math-container">$$f'(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} (3n+1)x^{3n+1} = \ldots = \frac{1}{1+x^3}$$</span></p>
<p>Now, consider this claim: "Since <span class="math-container">$f(0) = 0$</span>: <span class="math-container">$$f(x) = \int_0^x \frac{1}{1+t^3} \ dt$$</span></p>
<p>I am familiar with the fundamental theorem of calculus, yet not sure why this claim true, More presicely; Why is the integral starts at <span class="math-container">$0$</span>?</p>
<p>Thanks.</p>
| Ian | 83,396 | <p>The fundamental theorem tells you that</p>
<p>$$f(x)=f(a)+\int_a^x f'(t) dt.$$</p>
<p>It's convenient to choose $a$ such that $f(a)=0$, because then</p>
<p>$$f(x)=f(a)+\int_a^x f'(t) dt = \int_a^x f'(t) dt.$$</p>
<p>Since your function is a power series with no constant term, it's not hard to see that you can use $a=0$ for this purpose.</p>
|
4,034,709 | <p>What will be the operator norm of the matrix <span class="math-container">$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$</span> where <span class="math-container">$a,b,c,d \in \Bbb C\ $</span>?</p>
<p>According to the definition of the operator norm it turns out that <span class="math-container">$$\|A\|^2 = \sup \left \{ \left (|a|^2 + |c|^2 \right ) |z|^2 + \left (|b|^2 + |d|^2 \right ) |w|^2 + 2\ \mathfrak R\ (a \overline b z \overline w) + 2\ \mathfrak R\ (c \overline d z \overline w) \ \bigg |\ |z|^2 + |w|^2 = 1,\ z,w \in \Bbb C \right \}.$$</span></p>
<p>Is there any way to simplify the above expression?
Any help will be highly appreciated.</p>
<p>Thanks for your time.</p>
| Yalikesifulei | 521,468 | <p>Let <span class="math-container">$A$</span> be <span class="math-container">$n\times n$</span> matrix, <span class="math-container">$\|{A}\| = \sup\left\{ \|Ax\| : \|x\| = 1\right\}$</span>. For any <span class="math-container">$x$</span> with <span class="math-container">$\| x\| = 1$</span> we have
<span class="math-container">$\|Ax\|^2 = (Ax, Ax) = (A^* A x, x)$</span>, where <span class="math-container">$A^*$</span> is conjugate transpose matrix for A.
<span class="math-container">$A^* A$</span> is self-adjoint, so <span class="math-container">$A^* A = U D U^*$</span> for some unitary matrix <span class="math-container">$U$</span> and <span class="math-container">$D = \mathrm{diag}(\lambda_1, ..., \lambda_n)$</span>, where <span class="math-container">$\lambda_k$</span> are non-negative eigenvalues. Hence, <span class="math-container">$(A^* A x, x) = (U D U^* x, x) = (D U^* x, U^*x)$</span>. Let <span class="math-container">$y = U^* x$</span>. As <span class="math-container">$U$</span> is unitary, <span class="math-container">$\| U^* x\| = \|x\|$</span>, so <span class="math-container">$\|Ax\|^2 = (Dy, y)$</span> and <span class="math-container">$\|{A}\| = \sup\left\{ \sqrt{(Dy, y)} : \|y\| = 1\right\}$</span>.
<span class="math-container">$$ (Dy, y) = \lambda_1|y_1|^2 + \lambda_2|y_2|^2 + ... + \lambda_n|y_n|^2 \leq
\underset{1\leq k \leq n}{\max} \lambda_k \cdot \sum_{j=1}^n |y_j|^2 =
\underset{1\leq k \leq n}{\max} \lambda_k \cdot ||y||^2 = \underset{1\leq k \leq n}{\max} \lambda_k$$</span>
Moreover, for <span class="math-container">$y = (0, ..., 0, 1, 0, ..., 0)$</span> with <span class="math-container">$1$</span> at <span class="math-container">$k$</span>-th place (<span class="math-container">$k$</span> is index of largest eigenvalue) there is an equality.</p>
<p>So, <span class="math-container">$\|A\| = \sqrt{\underset{1\leq k \leq n}{\max} \lambda_k}$</span>, where <span class="math-container">$\lambda_k$</span> are eigenvalues of <span class="math-container">$A^* A$</span>.</p>
|
45,911 | <p>I've been wondering for some time now about the difference between a point and a vector. In high school, it was very important to distinguish them from each other, and we used the notation $(x,y,z)$ for points and $[x,y,z]$ for vectors. We always had to translate the point $P=(a,b,c)$ to the vector $\overrightarrow{OP} =[a,b,c]$ before we started calculating with them. </p>
<p>Now, after I started at the university, people don't seem to care anymore. My professors either say that they're the same, or that they're almost the same, and the books I have seem to share that view. The book I use for my calculus course (Colley's Vector Calculus) says, among other things, the following:</p>
<blockquote>
<p>[...] we adopt the point of view that a vector field assigns to each <em>point</em> $\textbf{x}$ in X a <em>vector</em> $\textbf{F}(\textbf{x})$ in $\mathbb{R}^n$, represented by an arrow whose tail is at the point $\textbf{x}$.</p>
</blockquote>
<p>So it seems like a point is also a vector.</p>
<p>My question is this: Do mathematicians distinguish between points and vectors, and if they do, in what circumstances?</p>
| Zhen Lin | 5,191 | <p>It's a good habit to distinguish the coordinates of a point from vectors. As everyone else has pointed out, Euclidean space is special — but I'll add that on top of that, cartesian coordinates on Euclidean space is special. If you use, for example, polar coordinates on Euclidean space, you'll find that you can't subtract the coordinate components of different points to obtain the components of the displacement vector. For instance, the displacement vector <em>from</em> the point $(r, \theta)$ <em>to</em> the origin is $r \mathbf{e}_r$, not $r \mathbf{e}_r + \theta \mathbf{e}_\theta$, where $\mathbf{e}_r$ and $\mathbf{e}_\theta$ are the normalised coordinate basis. It only gets worse when you start working with general curved spaces. </p>
<p>(A pet peeve of mine is when people talk about position vectors, especially in the context of non-affine space.)</p>
|
2,054,676 | <p>I know that m is even and m/2 is odd, but I don't know where/how I can use this. Also, 3y^2 is odd and the sum is odd when x^2 is even. I'm trying to prove that its always odd, but I'm stuck.
Can someone please help?
Thanks</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ {\rm mod}\,\ 2\!:\ x\equiv y\ \Rightarrow\ {\rm mod}\,\ 4\!:\ x^2\equiv y^2\equiv -3y^2$</p>
|
617,389 | <p>How to sketch $y = \frac1{\sqrt{x-1}}$</p>
<p>My way:(which does not work here)</p>
<p>I normally solve these problems by squaring and converting them to equations of 2 degree curves(such as parabola, hyperbola, etc.) which I can easily plot. But this seems to go 3 degree as $xy^2$ term is coming.</p>
<p>Please help me to solve this.</p>
<p>Note: Please don't say to use a graph plotter and see for myself since in the exam if this question comes I won't have the graph plotter with me. </p>
| Clive Newstead | 19,542 | <p>Consider some basic properties of the function, which you can work out either by inspection or by considering derivatives:</p>
<ul>
<li>It is only defined for $x \ge 1$;</li>
<li>It has no roots, stationary points, inflection points, etc.;</li>
<li>It is always decreasing and convex;</li>
<li>It tends to $0$ as $x \to \infty$;</li>
<li>It tends to $\infty$ as $x \to 1^+$.</li>
</ul>
<p>Just this information is enough for you to give a rough sketch of the function.</p>
<p>If you want to make it more accurate then you could consider some points which the function passes through, e.g. $(2,1)$ and $(5,2)$.</p>
|
3,358,592 | <p>In the following quote, what does the notation <span class="math-container">$\{a_n\}$</span> mean?</p>
<blockquote>
<p>Дана последовательность Фибоначчи <span class="math-container">$\{a_n\}$</span>.</p>
</blockquote>
<p><strong>Translation:</strong> "You are given the Fibonacci sequence <span class="math-container">$\{a_n\}$</span>."</p>
| Bill | 735,723 | <p>I apologize for the late response. I recently came across this post and as it so happens, I have been experimenting with such functions recently and figured I would add my two-cents.</p>
<p>For my current research, I needed a smooth(ish) approximation to the Heaviside step function defined on a compact interval. I really wanted to use a similar approach to the one discussed in chapter 13 of the book "An Introduction to Manifolds" by Loring W. Tu. This approach uses the function <span class="math-container">$$f(x)=\left\{\begin{array}{cc}e^{-\frac{1}{x}},&x>0,\\0,&x\le0\end{array}\right.$$</span> to construct a <span class="math-container">$C^\infty$</span> bump function. When I tried implementing this numerically I ran into some problems due to the very fast nature of how this function approaches <span class="math-container">$0$</span> (overflow errors, division by zero, etc. ).</p>
<p>This led me to an approach that is extremely similar to that used in the book but instead of using the function <span class="math-container">$e^{-\frac{1}{x}}$</span>, I needed to find a function with the same basic ''shape'' and behavior as the exponential but which was simpler to compute and used only addition/subtraction, multiplication/division (and was perhaps not <span class="math-container">$C^\infty$</span> but just <span class="math-container">$C^1$</span> or <span class="math-container">$C^2$</span>). I thought for a while and finally decided to use the function</p>
<p><span class="math-container">$$f(x)=\left\{\begin{array}{cc}1-\frac{1}{1+x^p},&x>0,\\0,&x\le0\end{array}\right.$$</span> where <span class="math-container">$p$</span> is a positive integer greater than or equal to 2.</p>
<p>To construct the Heaviside step function approximation from this you can simply follow a slightly altered version of the procedure in the book by Tu</p>
<p>Define <span class="math-container">$g(x)=\frac{f(x)}{f(x)+f(1-x)}$</span> and let <span class="math-container">$\epsilon$</span> be the desired width of the smoothed interval around the corner point (in the plots below I use <span class="math-container">$\epsilon=.1$</span>)</p>
<p>The approximation to the Heaviside-step function is then given by <span class="math-container">$$H_\epsilon(x)=g\left(\frac{x-\frac{\epsilon}{2}}{\epsilon}\right)$$</span></p>
<p>Here is a comparison what Tu's Heaviside step function (red) and my Heaviside step function (purple) approximation look like for <span class="math-container">$p=2$</span>
<a href="https://i.stack.imgur.com/gzmy6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gzmy6.png" alt="enter image description here" /></a></p>
<p>They look very similar and get sharper as you increase the value of <span class="math-container">$p$</span>. As an added benefit, the function <span class="math-container">$f$</span> gains more degrees of smoothness as <span class="math-container">$p$</span> is increased.</p>
<p>When I apply this approximation to your desired function <span class="math-container">$x(H(x)-H(x-1))+H(x-1)$</span> for <span class="math-container">$\epsilon=.1$</span> and <span class="math-container">$p=4$</span> the result is visually indistinguishable from the desired result. Actual function (blue), My approximation (red)
<a href="https://i.stack.imgur.com/5ZPzv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5ZPzv.png" alt="enter image description here" /></a></p>
<p>I hope this helps you out! Please let me know if you have any questions regarding my approach </p>
|
306,848 | <p>The question is in the title. Suppose that $X$ and $Y$ are two pointed connected CW-complexes. I was wondering if there exists a spectral sequence computing the homology of the function space $$H_{\ast}(map_{\ast}(X,Y);k) $$
where $k$ is a fixed field. Could we say something interesting in the case when $H_{\ast}(X;k)$ is trivial. </p>
<p>For simplicity we can assume that $X$ and $Y$ are simply connected. </p>
| Gregory Arone | 6,668 | <p>An alternative spectral sequence for $H_∗(map_∗(X,Y);k)$ can be constructed using the approach in <a href="https://www.ams.org/journals/tran/1999-351-03/S0002-9947-99-02405-8/S0002-9947-99-02405-8.pdf" rel="nofollow noreferrer">this paper</a> of mine (apols for self-promotion). Actually, this spectral sequence is discussed already in an <a href="http://www.ams.org/journals/tran/1991-326-01/S0002-9947-1991-1010881-8/S0002-9947-1991-1010881-8.pdf" rel="nofollow noreferrer">earlier paper of Bendersky and Gitler</a>. This spectral sequence converges in the same cases as the Anderson spectral sequence that is alluded to in John Klein's answer. An explicit comparison of the two spectral sequences <a href="https://link.springer.com/content/pdf/10.2478%2Fs11533-011-0084-1.pdf" rel="nofollow noreferrer">was given by Podkorytov</a>. See also <a href="https://projecteuclid.org/euclid.agt/1513882737" rel="nofollow noreferrer">the paper of Ahearn and Kuhn</a> that gives a nice introduction to the spectral sequence, and studies its structure in detail.</p>
<p>I think that the approach in this paper can answer the question on what happens when $H_*(X;k)=0$. Indeed, the spectral sequence is based on the following model for the chains on $map_*(X, Y)$. Let $C_*(Y)$ denote the (reduced) complex of singular chains on $Y$ and $C^*(X)$ denote the reduced singular cochains on $X$. Let $\mathcal E$ be the category of non-empty finite sets and surjective functions, and let $\mathcal E^t$ be the "twisted arrow category" of $\mathcal E^t$. An object of $\mathcal E^t$ is a surjection of sets $i\twoheadrightarrow j$ and a morphism from $i\twoheadrightarrow j$ to $s\twoheadrightarrow t$ consists of sujections $i\twoheadrightarrow s$ and $t\twoheadrightarrow j$ that make the evident square commute (note the twist in the directions of the arrows).</p>
<p>There is an evident contravariant functor from $\mathcal E^t$ to chain complexes that sends $i\twoheadrightarrow j$ to $C^*(X^{\wedge j}) \otimes C_*(Y^{\wedge i}) \otimes k$. The point is that the homotopy limit of this functor is quasi-isomorphic to $C_*(map_*(X,Y);k)$ when $X$ is a finite q-dimensional complex and $Y$ is $q-1$-connected. Now, suppose $H_*(X;k)=0$. Then $C^*(X^{\wedge j})\otimes k$ is an acyclic complex for all $j$, and therefore the homotopy limit is acyclic as well. So we may conclude the following</p>
<p><strong>Claim</strong>: Suppose $X$ is a finite $q$-dimensional complex and $Y$ is a $q-1$-connected space. Suppose furthermore that $H_*(X;k)$ is trivial. Then $H_*(map_*(X,Y);k)$ is trivial.</p>
<p>Remark 1: it is not necessary to assume that $X$ is simply connected.</p>
<p>Remark 2: The conclusion should hold whenever either one of the spectral sequences converges. The condition that $X$ is q-dimensional and $Y$ is $q-1$-connected is the standard condition that guarantees convergence. But when $k={\mathbb Z}/p$ there also are results about "exotic convergence" of these types of spectral sequences that may be relevant.</p>
|
285,114 | <blockquote>
<p>Find the solution of the differential equation
$$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$</p>
</blockquote>
<p>Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ \implies v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)} \end{align}$$ I can't seperate $x$ and $v$.</p>
| Kns | 27,579 | <p>Your differential equation can be written as
$$ (x^{3}+x y^{2}-10 x)dx+(x^{2}y+y^{3}+5y)dy=0$$
So it is of the form $M(x,y)dx+N(x,y)dy=0$
Now, $$M_{y}=2xy$$ and $$N_{x}=2xy.$$
$\therefore$ The given differential equation is exact.
So its general solution is
$$\int M dx+\int (\text{Terms in $N$ which does not contain $x$})dy=C$$
$$\Rightarrow \int (x^{3}+x y^{2}-10 x)dx+\int 0 dy=C$$
$$\Rightarrow \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=c$$
Now $y(0)=1$, we have $C=0$.</p>
<p>$\therefore$ The solution of given differential equation is
$\displaystyle \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=0$.</p>
|
3,897,361 | <p>Find the GS of the following system of DE's where the independent variable is <span class="math-container">$t$</span> and <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are the dependent variables</p>
<p><span class="math-container">\begin{cases}
x' = x-3y\\
y'=3x+7y
\end{cases}</span></p>
<p>I know using eigenvalues and eigenvectors or operators is one way to do this. But I wish to double check my answer using a substitution method.</p>
<p>So my work:</p>
<p>The second DE <span class="math-container">$y'=3x+7y$</span> can be rewritten as <span class="math-container">$x = \cfrac{y'}{3}-\cfrac 73y$</span></p>
<p>then <span class="math-container">$x' = \cfrac{y''}{3}-\cfrac73y'$</span></p>
<p>When we plug these values of <span class="math-container">$x$</span> and <span class="math-container">$x'$</span> into the first DE (<span class="math-container">$x' = x -3y)$</span>, we get with some rearranging</p>
<p><span class="math-container">$\cfrac{y''}{3}-\cfrac83y'+\cfrac{16}{3}y = 0$</span></p>
<p>Which has a characteristic equation of</p>
<p><span class="math-container">$\cfrac{r^2}{3}-\cfrac83r+\cfrac{16}{3} = 0$</span></p>
<p>with roots <span class="math-container">$r_1=4$</span> and <span class="math-container">$r_2 = 4$</span></p>
<p>Then the solution for <span class="math-container">$y$</span> is <span class="math-container">$y$</span> = <span class="math-container">$C_1e^{4t}+C_2te^{4t}$</span></p>
<p>Then we back sub to solve for <span class="math-container">$x$</span> using <span class="math-container">$x = \cfrac{y'}{3}$$-\cfrac73y$</span> with the solution of y we just found.</p>
<p>We get <span class="math-container">$x =-C_1e^{4t}-C_2te^{4t} + \cfrac{C_2}{3}e^{4t} =-C_1e^{4t}-C_2te^{4t} + C_3e^{4t}$</span></p>
<p>so the GS to the homo system is</p>
<p><span class="math-container">\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}</span></p>
<p>If this solution is right, then I'm confident that I understand how substitution method works for solving DE systems. (Also it would boost my confidence in using the operator method to solve this as I got the same answer as this using the operator method). I'm a little thrown off on the roots being the same but I still think my methodology is still sound. I would appreciate it if someone could tell me if I've got this right cause then I know I completely understand how to solve a system of DE's.</p>
<p>If more work is necessary to show please let me know.</p>
| user577215664 | 475,762 | <p><span class="math-container">$$\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}$$</span>
You should only end with two constants not three.
<span class="math-container">$$y=C_1e^{4t}+C_2te^{4t}$$</span>
<span class="math-container">$$y'=4C_1e^{4t}+C_2e^{4t}+4tC_2e^{4t}$$</span>
<span class="math-container">$$y'=4C_1e^{4t}+C_2e^{4t}(1+4t)$$</span>
And since:
<span class="math-container">$$x = \cfrac{y'}{3}-\dfrac 73 y$$</span>
You should only have <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> constant. You applied the method of substitution correctly.</p>
|
392,442 | <p>What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed <em>Eureka</em>?</p>
| Caleb Stanford | 68,107 | <p>There are in fact many 'formula's which always generate prime numbers. Among the simplest ones listed by <a href="http://en.wikipedia.org/wiki/Formula_for_primes" rel="noreferrer">Wikipedia</a> are:</p>
<ul>
<li>There is some real number $A$ such that $\left\lfloor A^{3^n}\right\rfloor$ is always prime.</li>
<li>There is some real number $\alpha$ such that $\left\lfloor 2^{2^{\cdots^{\alpha}}} \right\rfloor$ is always prime.</li>
<li>The formula $2 + [2(n!) \bmod (n+1)]$ always gives a prime, where here '$\bmod$' (nonstandardly) denotes the remainder.</li>
</ul>
<p>The third in particular generates every prime.</p>
<p>These formulas are mathematically valid, fun to prove, and highly interesting. However, they are actually completely useless, and not just because we don't know the values of $A$ and $\alpha$. So (as some have suggested in the comments) a better question is how <em>fast</em> one can generate primes.</p>
<p>The ability (or inability) to generate or check for primes in a certain amount of time is fundamentally important to cryptographic systems such as RSA. However, the "practical" applications of prime numbers (to fields like physics, chemistry, etc.) are, as far as I understand, very few -- cryptography is <em>the</em> major application.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Unknown | 5,627 | <p>Wiki also says: </p>
<blockquote>
<p>A famous network of conditional proofs is the NP-complete class of complexity theory</p>
</blockquote>
|
3,522,752 | <p>Solve the following equation:
<span class="math-container">$$y=x+a\tan^{-1}p$$</span>
<span class="math-container">$$\text{where p}=\frac{dy}{dx}$$</span>
Differentiating both side w.r.t. x,
<span class="math-container">$$\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}$$</span>
I have tried till this...but what to do next?..please help..</p>
| Oliver Jones | 55,622 | <p>The equation can be written as</p>
<p><span class="math-container">$$\frac{dy}{dx}=\tan \frac{1}{a}(y-x).$$</span></p>
<p>Now make the substitution <span class="math-container">$u=y-x$</span> to get </p>
<p><span class="math-container">$$
\frac{du}{dx}+1=\tan\frac{u}{a}.
$$</span></p>
<p>This equation is now separable.</p>
|
1,032,714 | <p>'Let $X$ be a topological space and let $(U_i)_{i \in I}$ be a cover of $X$ by connected subspaces $U_i$. Supposed for all $i,j \in I$ there exists some $n \geq 0$ and $k_0,...,k_n \in I$ such that $k_0 = i, k_n = j$ and $$U_{k_0} \cap U_{k_1} \neq \emptyset, U_{k_1} \cap U_{k_2} \neq \emptyset, ..., U_{k_{n-1}} \cap U_{n} \neq \emptyset$$
Show that $X$ is connected.</p>
<p>My intuition tells me the proof is along the lines of since each $U_i$ is connected, there are no subsets $A,B \subseteq U_i$ for $i \in I$ such that $A$ and $B$ are disjoint (unless they are the empty set.) Also, since the intersection of $U_{k_i} \cap X/U_{k_i}$ is non empty (as there always exists at least some other subcover it intersects with) there are no two nonempty subsets $A,B \subset X$ such that $A \cap B = X$. Thus $X$ is connected.</p>
<p>I was looking for some help formulating this into a nice mathematical proof. Thanks.</p>
| Jimmy R. | 128,037 | <p>No, the first formula is a special case of the second. Let $A=\{ω\}\in B$ and $B=B$, then if you substitute in the second formula, you obtain that $$Pr(ω\mid B)=Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{P(\{ω\}\cap B)}{Pr(B)}\overset{ω\in B}=\frac{Pr(\{ω\})}{Pr(B)}$$ as the first formula states.</p>
|
1,456,407 | <p><a href="https://i.stack.imgur.com/oy6T7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oy6T7.jpg" alt="enter image description here"></a></p>
<p>We need to find the area of the shaded region , where curves are in polar forms as $r = 2 \sin\theta$ and $r=1$.</p>
<p>I formulated the double integral as follows : </p>
<p>We find the area in the first quadrant and then multiply it by $2$ , </p>
<p>Area of the circle $r=1$ in the first quadrant is $\frac{\pi}{4}$ , we need to subtract the area of the curve $r = 2\sin\theta$ from this , thus , area is given by : </p>
<p>$[\dfrac{\pi}{4} - \int^{\dfrac{\pi}{6}}_{0}\int^{2\sin\theta}_{0}r.dr.d\theta]\times2$ </p>
<p>Is this correct ?
The solution says , " first consider $0< \theta < \dfrac{\pi}{6}$ and then $\dfrac{\pi}{6}< \theta < \dfrac{\pi}{2}$ etc etc..... "</p>
| lhf | 589 | <p>By the binomial theorem, $42^n = (43-1)^n=43a+(-1)^n$.</p>
<p>If $n$ is even, then $42^n-1$ is a multiple of $43$.</p>
<p>On the other hand, $42^n = (41+1)^n=41b+1$, and so $42^n-1$ is always a multiple of $41$. Thus, $42^n-1$ is not prime if $n>1$, regardless of the parity of $n$.</p>
|
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