qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,700,246 | <p>Let $F=\mathbb{F}_{q}$, where $q$ is an odd prime power. Let $e,f,d$ be a standard basis for the $3$-dimensional orthogonal space $V$, i.e. $(e,e)=(f,f)=(e,d)=(f,d)$ and $(e,f)=(d,d)=1$. I have an element $g\in SO_{3}(q)$ defined by: $g: e\mapsto -e$, $f\mapsto \frac{1}{2}e -f +d$, $d\mapsto e+d$. I would like to determine the spinor norm of this element using Proposition 1.6.11 in the book 'The Maximal Subgroups of the Low-Dimensional Finite Classical Groups' by Bray, Holt and Roney-Dougal. </p>
<p>The proposition is quite long to state so it would be handy if someone who can help already has a copy of the book to refer to. If not, then please let me know and I can post what the proposition says. </p>
<p>I have followed the proposition and have that the matrices
$$A:=I_{3}-g=\left( \begin{array}{ccc}
2 & -\frac{1}{2} & -1 \\
0 & 2 & 0 \\
0 & -1 & 0 \end{array} \right)$$
$$F= \textrm{matrix of invariant symmetric bilinear form =}\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \end{array} \right).$$</p>
<p>If $k$ is the rank of $A$, the proposition says to let $B$ be a $k\times 3$ matrix whose rows form a basis of a complement of the nullspace of $A$. I have that $ker\, A=<(1,0,2)^{T}>$. Now by the way the proposition is stated it seems as if it does not make a difference as to what complement is taken. However, I have tried 3 different complements and I get contradictory answers each time.</p>
<p>Try 1) Orthogonal complement of $Ker\, A$, where $B=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 1 \end{array} \right)$. This gives me $det(BAFB^{T})=-25$.</p>
<p>Try 2) $B=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=-4$.</p>
<p>Try 3) $B=\left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=0$.</p>
<p>The problem is that whatever complement I take the determinants should all be non-zero squares at the same time, but this is not the case. </p>
<p>I am not sure if I have misunderstood how to use the proposition. Any help will be appreciated. Thanks.</p>
| christina_g | 307,603 | <p>It is obvious that if $y \leq 0$ there is no such $r$. If $y>0$ and $x <0$ then $r=0$. Now if $x \geq 0$, then because of the density of rational numbers you can find a $r \in \Bbb Q, \; \sqrt{x} < r < \sqrt{y}$ and you have $x<r^2<y$. I suppose that you do not have to prove the density of $\Bbb Q$ in $\Bbb R$.</p>
|
3,380,081 | <p>Question: Suppose <span class="math-container">$n(S)$</span> is the number of subset of <span class="math-container">$S$</span> and <span class="math-container">$|S|$</span> be the number of elements of <span class="math-container">$S$</span>. If <span class="math-container">$n(A)+n(B)+n(C)=n(A\cup B\cup C)$</span> and <span class="math-container">$|A|=|B|=100$</span>, Find the minimum value of <span class="math-container">$|A\cap B\cap C|$</span>.</p>
<p>Now, I realise PIE is the only way to go, but I don't know how to handle the intersections of the sets taken two at a time.
Also, I know that this is a duplicate, but the original on wasn't answered fully. I'd request you o kindly provide a solution before mercilessly closing it off.</p>
<p>Plus, I'm not good at even very basic set theory. If you could recommend a short but good book for set-theory, I'd be much obliged.</p>
<p>Thank you all!</p>
| Arthur | 15,500 | <blockquote>
<p>Is it safe to assume that if I keep going infinitely the answer will be exactly 14/3?</p>
</blockquote>
<p>Strictly speaking, no, it is not. The answer could be some number <em>really</em> close to <span class="math-container">$\frac{14}{3}$</span>, and the only way to tell would be to actually do the exact integral. At no point will a finite Riemann sum be able to tell you that the answer is <em>definitely</em> <span class="math-container">$\frac{14}3$</span>.</p>
<p>However, in this case, with a very simple function and very simple bounds... Odds are definitely in your favour. Also, it is entirely correct to use your calculations to conclude that
<span class="math-container">$$
\int_1^4\sqrt x\,dx\approx\frac{14}3
$$</span></p>
|
621,461 | <p>I'm having trouble understanding division when the divisor is greater than the dividend, for ex 1/4.</p>
<p>I think of division as "how many times can the divisor fit into the dividend evenly". </p>
<p>Intuitively, when I see 1/4 in the context of slices of pizza, I think of it as 1 "out of" 4, but I can't seem to grasp it in terms of "how many times does 4 fit into 1" if that makes any sense.</p>
<p>In other words my question could be why do we use division to represent "one out of 4"?.</p>
<p>If it helps you guys understand, this came about as I was trying to find the percentage representation of two populations, as in "there are 1253 A's and 747 B's, what is the proportion of each in % ?".</p>
<p>Conceptually I understand that I need to add up those two populations and then find the proportion they represent of that total. However, when I got to that second part, I couldn't reason through whether I needed to divide the total by a population, or a population by the total in order to find the desired result. </p>
<p>Obviously I eventually found the right way to do it, but it still doesn't make sense to me.</p>
<p>Sorry if it's very vague, this is really bothering me; I can't seem to reconcile those two ways of thinking about division.</p>
| Bill Dubuque | 242 | <p>$X$ equals $\,\dfrac{1}4\,$ of the total $\,T$ means $\,X\, =\, \dfrac{1}4 T\, =\, \dfrac{T}4\ \,$ or $\ \color{#c00}{\dfrac{X}T = \dfrac{1}4}\,\ $ or $\,\ 4X = T$</p>
<p>$X$ equals $\dfrac{A}B$ of the total $\,T$ means $\,X = \dfrac{A}B T = \dfrac{AT}B\ $ or $\ \color{#c00}{\dfrac{X}T = \dfrac{A}B}\ $ or $\ BX = AT$</p>
<p>$X$ equal $A$% of the total $\,T$ is the special case $\,B=100\,$ of the above, i.e. $A$% denotes the fractional proportion $\,A/100\,$ (here $A$ can be any real number). For example, $25$% denotes $25/100 = 1/4$. </p>
<p>To convert a fraction to percent form $\ \dfrac{X}{100} = \dfrac{A}B\ \Rightarrow\ X = \dfrac{100A}B\, $ (usually written in decimal).</p>
<p>Said equivalently, write $\,A/B\,$ in decimal form, then shift the point two places rightward. </p>
<p>Hence the meaning of "percent" boils down to the meaning of fractional proportions, which is simply the ratio of the part $X$ to the whole $T$, expressed as a fraction (see the red equations above). Hopefully this makes it clearer which number goes on the top vs. bottom of the fraction.</p>
|
1,762,001 | <p>I recently watched a <a href="https://www.youtube.com/watch?v=SrU9YDoXE88" rel="noreferrer">video about different infinities</a>. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \ldots, \omega_\omega$, etc..</p>
<p>I can't find myself in all of this. Why there are so many infinities, and why even bother to classify infinity, when infinity is just... infinity? <strong>Why do we use all of these symbols? What does even each type of infinity mean?</strong></p>
| Cody Rudisill | 279,692 | <p>Your question touches on issues both mathematical and philosophical; in fact, you might enjoy an introductory text in philosophy of mathematics and logic, or on the foundations of mathematics.</p>
<p>Truth is, an infinity is not <em>just</em> an infinity. One reason for this peculiarity is due to the seminal work of Georg Cantor.</p>
<p><strong>Cantor proved that the set of real numbers is greater than the set of natural numbers.</strong></p>
<p>Crazy.</p>
<p>Even crazier is how he used grade-school concepts, such as a one-to-one correspondence, to do so.</p>
<p>But none of this really answers <em>Why?</em>. Maybe I can give you something of an answer. Mathematics is a language rigorously defined. It's one thing to throw around a natural-language word like infinity, and totally different to empower that word with a thoroughly defined meaning in a formal language.</p>
<p>Put more poetically, <strong>Cantor provided a definition of infinity by showing that infinite sets of things can be classed and organized like books on a shelf.</strong></p>
|
3,792,683 | <p>Where <span class="math-container">$\alpha$</span> is a real constant, consider the sequence {<span class="math-container">$z_n$</span>} defined by <span class="math-container">$z_n=\frac{1}{n^\alpha}$</span>. For which value of <span class="math-container">$\alpha$</span> is {<span class="math-container">$z_n$</span>} a bounded sequence?</p>
<p>How do I start with this kind of question? I think that <span class="math-container">$\forall\space \alpha\in\Bbb{R}_{\geq0}$</span> the sequence is convergent and therefore bounded, but how do I write it out?</p>
| Quanto | 686,284 | <p>Integrate by parts directly</p>
<p><span class="math-container">$$\begin{array}\
\int^{\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx
&= \int^{\infty}_0 \frac{e^{-x^2}}x d\left( \frac{x^2}{x^2+\frac{1}{2}} \right)=
2\int^{\infty}_0e^{-x^2}dx= \sqrt{\pi} \end{array}$$</span></p>
|
2,021,126 | <p>Let $f :[a , \infty)\to \mathbb{R}$ a positive and Monotonic function such that $\int_a^\infty f$ converge
<br>
prove:
$\lim_{x\to\infty}f(x)=0$</p>
| Gono | 384,471 | <p>That's an easy calculation… assume $$\lim_{x\to\infty}f(x)\not=0$$ then because f is monotone and positive there exists an $\varepsilon > 0$ s.t. $f>\varepsilon$ so $$\int_a^\infty f \ge \int_a^\infty \varepsilon = \infty$$</p>
|
2,021,126 | <p>Let $f :[a , \infty)\to \mathbb{R}$ a positive and Monotonic function such that $\int_a^\infty f$ converge
<br>
prove:
$\lim_{x\to\infty}f(x)=0$</p>
| Patrick Stevens | 259,262 | <p>Alternative answer: if $f$ is constant then the result is trivial; wlog $f$ is decreasing, since if it is increasing then the integral can't converge.</p>
<p>By the integral test for convergence, $\sum_{i=a}^{\infty} f(i)$ converges. Therefore $f(i) \to 0$ as $i \to \infty$ over the integers.</p>
<p>But $f$ is monotone, so $f(\lfloor x \rfloor) \geq f(x) \geq f(\lceil x \rceil)$; so we're done by the squeeze theorem.</p>
|
3,568,230 | <p>My question is: why, in general we cannot write down an formula for the <span class="math-container">$n-$</span>th term, <span class="math-container">$S_{n}$</span>, of the sequence of partial sums?</p>
<p>I will explain better in the following but the question is basically that one above.</p>
<p>Suppose then you have an <em>infinite sequence</em> in your pocket, <span class="math-container">$\{a_{1},a_{2},a_{3},...\}$</span>, or,</p>
<p><span class="math-container">$$\{a_{1},a_{2},a_{3},...\} \equiv \{a_{n}\}_{n=0}^{\infty} \tag{1}$$</span></p>
<p><span class="math-container">$(1)$</span> then is a fundamental object because then you can "sum up" all the terms of this particular sequence, just like: <span class="math-container">$a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot$</span> to define another object. Well, doing that procedure you construct that object, called <em>infinite series of the infinite sequence <span class="math-container">$\{a_{n}\}_{n=0}^{\infty}$</span></em> </p>
<p><span class="math-container">$$a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot \equiv \sum^{\infty}_{n=0}a_{n} \tag{2}$$</span></p>
<p>The next procedure you might like to do is then question yourself if a infinite series have some value <span class="math-container">$s \in \mathbb{K}$</span> (<span class="math-container">$\mathbb{K}$</span> a field) indeed. The procedure to answer that question is then firstly construct another infinite sequence called <em>the sequence of partial sums of the series</em>:</p>
<p><span class="math-container">$$\{S_{0},S_{1},S_{2},S_{3},...,S_{k},...\} \equiv \{S_{n}\}_{n=0}^{\infty} \tag{3} $$</span></p>
<p>Which is:</p>
<p><span class="math-container">$$\begin{cases} S_{0} = \sum^{0}_{n=0}a_{n} = a_{0}\\S_{1} = \sum^{1}_{n=0}a_{n} = a_{0} + a_{1} \\ S_{1} = \sum^{2}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} \\ S_{3} = \sum^{3}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3} \\\vdots\\ S_{k} = \sum^{k}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3}+\cdot \cdot \cdot+a_{k}\\ \vdots \end{cases} $$</span></p>
<p>and then calculate the limit of this sequence <span class="math-container">$(3)$</span>, like:</p>
<p><span class="math-container">$$ \lim_{n\to \infty} \sum^{n}_{j=0}a_{j} \equiv \lim_{n\to \infty} S_{n} \tag{4} $$</span></p>
<p>Now, if the limit <span class="math-container">$(4)$</span> has a value <span class="math-container">$s = L$</span> then the can say that the <em>Sum of the Series</em> is that limit:</p>
<p><span class="math-container">$$ \sum^{\infty}_{n=0}a_{n} = s \tag{5}$$</span></p>
<p><span class="math-container">$$ * * * $$</span></p>
<p>Now, if we do not have a proper expression for <span class="math-container">$S_{n} = \sum^{k}_{n=0}a_{n}$</span>, then the whole "direct limit calculus" do not work and then we need other methods for search the value (more generally the convergence) of a series (e.g. integral test). The thing is, I do not see (understand) why we cannot in general write down a formula for <span class="math-container">$S_{n}$</span> and some times we can. For instance, I do not see why in one hand we can write down a formula for geometric series but on the other hand we cannot for harmonic series, for me the <span class="math-container">$S_{n}$</span> term, of the harmonic series, to plug up in the limit is given by:</p>
<p><span class="math-container">$$ S_{n} = \sum^{n}_{k=0}\frac{1}{n} = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \equiv \Big( 1+ \frac{1}{2} + \frac{1}{3} + ... \Big) + \frac{1}{n} = $$</span></p>
<p><span class="math-container">$$= C + \frac{1}{n} $$</span></p>
<p><span class="math-container">$C$</span> a constant since it's a finite sum. Then,</p>
<p><span class="math-container">$$\lim_{n\to \infty} C + \frac{1}{n} = C$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\sum^{\infty}_{n=0}\frac{1}{n} = C$$</span></p>
<p>I know that what I wrote above isn't write, but I simply do not understand why. There's a subtle thing that I do not understand. Anyway, the question is posted above.</p>
<p>Thank You. </p>
| hamam_Abdallah | 369,188 | <p>You had to write <span class="math-container">$C_n$</span> instead of <span class="math-container">$ C$</span>.
In fact we have
<span class="math-container">$$C_n=S_{n-1} \text{ and } \; S_n=C_n+\frac 1n$$</span>
and all we can say is</p>
<p>If <span class="math-container">$(C_n)$</span> converges then the series is convergent.</p>
|
3,711,744 | <p>Let <span class="math-container">$C_1\geq C_2\geq\dots\geq C_n$</span> be a fixed set of positive numbers. Maximize the linear function <span class="math-container">$L(x_1, x_2, \dots, x_n)=\sum^n_1C_jx_j$</span> in the closed set described by the inequalities <span class="math-container">$0\leq x_j\leq 1, \sum^n_1 x_j\leq A$</span><br>
I think this question can be solved by simplex method, but it happens in a calculus book, so can I solve it using some pure calculus method? I tried to find the critical points, but it equals <span class="math-container">$[c_1, c_2, \dots, c_n]$</span>, and can not equal <span class="math-container">$0$</span>.</p>
| Sergio | 731,870 | <p>Let <span class="math-container">$A,B$</span> be matrices associated to <span class="math-container">$U,T$</span>. You could look for two matrices such that <span class="math-container">$AB=0$</span>, <span class="math-container">$BA\ne AB$</span>, e.g. you could look for two diagonalizable, but not simultaneously diagonalizable, matrices (see <a href="https://math.stackexchange.com/questions/56307/simultaneous-diagonalization-of-commuting-linear-transformations">Simultaneous diagonalization of commuting linear transformations</a>).</p>
<p>Example: <span class="math-container">$U(x,y)=(-2x+y,-2x+y)$</span>, <span class="math-container">$T=(x+2y,2x+4y)$</span>, i.e.
<span class="math-container">$$A=\begin{bmatrix} -2 & 1 \\ -2 & 1 \end{bmatrix},\quad
B=\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$</span>
and <span class="math-container">$AB=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$</span>, but <span class="math-container">$BA=\begin{bmatrix} -6 & 3 \\ -12 & 6 \end{bmatrix}$</span>. So <span class="math-container">$UT(x,y)=AB(x,y)=(0,0)$</span>, but <span class="math-container">$TU(x,y)=BA(x,y)=(-6x+3y, -12x+6y)\ne (0,0)$</span> for all <span class="math-container">$x\ne y/2$</span>.</p>
|
3,920,812 | <p>Please excuse if the formatting of this post is wrong.</p>
<p>There's a question that asks for the 2nd derivative of <span class="math-container">$y-2x-3xy=2$</span></p>
<p>From what I know, I have to use implicit differentiation, using which I get: <span class="math-container">$$\frac{12+18y}{(1-3x)^{2}}$$</span>
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)?
By doing that I got: <span class="math-container">$$\frac{48}{(1-3x)^{3}}$$</span>
I'm not sure if this is a correct answer as I am new to differentiation.
I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?</p>
<p>Thank you.</p>
| Raffaele | 83,382 | <p>Sure you can in this case
<span class="math-container">$$y=\frac{2 (x+1)}{1-3 x}$$</span>
and then
<span class="math-container">$$y'=\frac{8}{(1-3 x)^2}$$</span>
and finally
<span class="math-container">$$y''=\frac{48}{(1-3 x)^3}$$</span></p>
|
3,920,812 | <p>Please excuse if the formatting of this post is wrong.</p>
<p>There's a question that asks for the 2nd derivative of <span class="math-container">$y-2x-3xy=2$</span></p>
<p>From what I know, I have to use implicit differentiation, using which I get: <span class="math-container">$$\frac{12+18y}{(1-3x)^{2}}$$</span>
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)?
By doing that I got: <span class="math-container">$$\frac{48}{(1-3x)^{3}}$$</span>
I'm not sure if this is a correct answer as I am new to differentiation.
I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?</p>
<p>Thank you.</p>
| Community | -1 | <p>We differentiate once,</p>
<p><span class="math-container">$$y'-2-3y-3xy'=0$$</span>
and twice,</p>
<p><span class="math-container">$$y''-3y'-3y'-3xy''=0.$$</span></p>
<p>Now we can eliminate <span class="math-container">$y'$</span>, using</p>
<p><span class="math-container">$$(1-3x)y'=3y+2$$</span></p>
<p>and</p>
<p><span class="math-container">$$(1-3x)y''=6y'=6\frac{3y+2}{1-3x}.$$</span></p>
<p>You can also eliminate <span class="math-container">$y$</span> using the initial equation.</p>
|
1,102,709 | <p>I'd like to know if there is an explicit atlas for the manifold $\mathbb{R}P^3$ which is defined as the quotient of the three-sphere by the antipodal mapping.</p>
<p>Thanks.</p>
| Julián Aguirre | 4,791 | <p>What you have proved is that for a fixed $t\in\mathbb{R}$, given $\epsilon>0$ there is a polynomial $p$ such that $|f(t)-p(t)|<\epsilon$. But the Weierstrass approximation theorem is about uniform approximation; $|f(t)-p(t)|<\epsilon$ must hold for all $t$ on a given set.</p>
<p>It is true that given an interval $[-i,i]$ and $\epsilon>0$ there is a polynomial $p$ such that $|f(t)-p(t)|<\epsilon$ for all $t\in[-i,i]$. But $p$ depends on $\epsilon$ <strong>and</strong> $i$.</p>
<p>You can see that the theorem does not hold on unbounded sets by considering for example $f(t)=e^t$. Suppose that there is a polynomial $p$ such that $|e^t-p(t)|<1$ for all $t\in\mathbb{R}$. Let $n$ be the degree of $p$. Then we would have $e^t\le p(t)+1$ and
$$
\lim_{t\to+\infty}\frac{e^t}{t^{n+1}}=0.
$$</p>
|
238,392 | <p>Do you know interesting examples of purely geometric or topological results which can be proved using group theory? To make precise what I have in mind, let us consider the two following examples:</p>
<blockquote>
<p>There does not exist any Riemannian metric on the torus whose sectional curvature is $<0$.</p>
</blockquote>
<p>This is a consequence of Milnor's article <em>A note on curvature and fundamental group</em>, where he proves that the fundamental group of a negatively-curved Riemannian manifold has exponential growth. On the other hand, the fundamental group of the torus, namely $\mathbb{Z}^2$, has quadratic growth.</p>
<blockquote>
<p>Any compact Riemannian manifold whose sectional curvature is $\equiv 0$ has a torus as a finite cover.</p>
</blockquote>
<p>This is a consequence of Bieberbach theorem. </p>
<p>More recently, showing that quasiconvex subgroups of hyperbolic cubulable groups are separable was the key point in the proof of the virtual Haken's conjecture. However, this is more technical.</p>
| HJRW | 1,463 | <p>I'm still a little uncertain about this question, but I'll try to say something about the Virtual Haken conjecture (discussed above) and in the process explain why I think it's a good example.</p>
<p>The Virtual Haken conjecture (now Agol's theorem) can be stated as follows --</p>
<blockquote>
<p>Every hyperbolic 3-manifold has a finite-sheeted cover that contains an embedded, incompressible surface.</p>
</blockquote>
<p>-- a thoroughly topological statement (though more group-theoretic statements can be given). </p>
<p>For me, modern (geometric) group theory enters the picture in a truly astonishing way via the following result, Wise's <em>Malnormal Special Quotient Theorem</em> (MSQT).</p>
<blockquote>
<p>Let $G$ be a hyperbolic, virtually special group, and $H$ a malnormal, quasiconvex subgroup. Then, for all sufficiently deep finite-index subgroups $K\lhd H$, the quotient $G/\langle\langle K\rangle\rangle$ is hyperbolic and virtually special.</p>
</blockquote>
<p>I won't explain the definitions here, but rather point out that this tells us that we can kill large subgroups and stay in the (very well behaved) world of hyperbolic, virtually special groups. <em>Note that this is not true of manifolds.</em> You can't crush an immersed surface in a hyperbolic manifold and get a new manifold.</p>
<p>Although the proofs of the MSQT can (and usually are) phrased in an entirely topological/geometric manner, the key point here is that they concern geometric complexes (more precisely, CAT(0) cube complexes), not manifolds. The transition from manifolds to more general geometric complexes is surely the hallmark of modern infinite group theory.</p>
<p>Agol's proof, still the only proof we know, makes essential use of the MSQT. In this sense, the Virtual Haken conjecture is truly a theorem of geometric group theory rather than topology, in the sense that we don't know how to keep the proof purely in the world of topology (ie manifolds) -- you have to pass to the world of CAT(0) cube complexes (ie group theory).</p>
|
401,389 | <p>I worked through some examples of Bayes' Theorem and now was reading the proof.</p>
<p>Bayes' Theorem states the following:</p>
<blockquote>
<p>Suppose that the sample space S is partitioned into disjoint subsets <span class="math-container">$B_1, B_2,...,B_n$</span>. That is, <span class="math-container">$S = B_1 \cup B_2 \cup \cdots \cup B_n$</span>, <span class="math-container">$\Pr(B_i) > 0$</span> <span class="math-container">$\forall i=1,2,...,n$</span> and <span class="math-container">$B_i \cap B_j = \varnothing$</span> <span class="math-container">$\forall i\ne j$</span>. Then for an event A,</p>
<p><span class="math-container">$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$</span></p>
</blockquote>
<p><img src="https://i.stack.imgur.com/nnbZU.png" alt="enter image description here" /></p>
<p>The numerator is just from definition of conditional probability in multiplicative form.</p>
<p>For the denominator, I read the following:</p>
<p><span class="math-container">$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$</span></p>
<p>Now this is what I don't understand:</p>
<blockquote>
<p><strong>The sets <span class="math-container">$A \cup B_i$</span> are disjoint because the sets <span class="math-container">$B_1, B_2, ..., B_n$</span> form a partition.<span class="math-container">$\tag{$\clubsuit$}$</span></strong></p>
</blockquote>
<p>I don't see how that is inferred or why that is the case. What does B forming a partition have anything to do with it being disjoint with A. Can someone please explain this conceptually or via an example?</p>
<p>I worked one example where you had 3 coolers and in each cooler you had either root beer or soda. So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda. But I don't see why these would be disjoint. If anything, I would say they weren't disjoint because each cooler contains both types of drinks.</p>
<p>Thank you in advance! :)</p>
| Abhra Abir Kundu | 48,639 | <p>You can easily show this using set theoretic arguement. </p>
<p>$A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)$</p>
|
3,387,458 | <p>Show that a bounded sequence having one limit point is convergent. </p>
<p>The converse holds true. The fact that a convergent seq is bounded has been shown in Baby Rudin. The fact that it will have only one limit point can be found <a href="https://math.stackexchange.com/questions/3386703/prove-that-a-convergent-sequence-has-only-one-limit-point-proof-verification/3386711?noredirect=1#comment6967723_3386711">here</a>.</p>
<p>The similar question seems to have been asked before on MSE except that there is no complete or well-detailed solution.</p>
| copper.hat | 27,978 | <p>Suppose <span class="math-container">$x_n \in \mathbb{R}$</span> is bounded and has exactly one limit point. Then
<span class="math-container">$x_n$</span> converges.</p>
<p>Suppose <span class="math-container">$x_{n_k} \to x$</span>, and suppose <span class="math-container">$|x_n| \le B$</span>.</p>
<p>Let <span class="math-container">$\epsilon>0$</span>, then <span class="math-container">$C_\epsilon=[-B,x-\epsilon] \cup [x+\epsilon,B]$</span> is a compact set, hence at most a finite number of <span class="math-container">$x_n \in C_\epsilon$</span> (otherwise there would be another limit point).</p>
<p>In particular, for all <span class="math-container">$\epsilon>0$</span> there is some <span class="math-container">$N$</span> such that for <span class="math-container">$n \ge N$</span>, we have <span class="math-container">$|x-x_n| < \epsilon.$</span> Hence <span class="math-container">$x_n \to x$</span>.</p>
|
1,943,840 | <p>I'm trying to find out for square matrices with $n \geq 2$ :
$$ \det(A-B) = \det(A)-\det(B).$$ </p>
<p>I know that $\det(AB) = \det(A)\det(B)$, but I'm unable to find proof on why a subtraction (or addition) is not equal. Thanks.</p>
| M.Hannan | 372,918 | <p>This is true if $A=B$. </p>
<p>In that case $\det(A-A) = 0$</p>
<p>and $\det(A) - \det(A) = 0$</p>
<p>so $\det(A-A) = \det(A) - \det(A)$</p>
|
367,254 | <p>Solve the following quadratic congreunce</p>
<p>$x^2+ 7x + 10 \equiv 0$ (mod $11$).</p>
<p>I want to know a general and easy method how to solve this kind of questions.</p>
| shobon | 73,393 | <p>Use the quadratic formula $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$</p>
<p>so $b^2-4ac = 7^2 - 40 = 9 = 3^2$</p>
<p>so $\pm \sqrt{b^2-4ac} = \pm 3$</p>
<p>and $-b \pm 3 = -10,-4$</p>
<p>and so the answers are -5 and -2.</p>
|
367,254 | <p>Solve the following quadratic congreunce</p>
<p>$x^2+ 7x + 10 \equiv 0$ (mod $11$).</p>
<p>I want to know a general and easy method how to solve this kind of questions.</p>
| Warren Moore | 63,412 | <p>For $ax^2+bx+c\equiv 0$ (mod $p$), if $p$ is an odd prime not dividing $a$, then this is the same as solving a simpler question about quadratic residues $y^2\equiv b^2-4ac$ (mod $p$), and $y\equiv 2ax+b$ (mod $p$).</p>
|
476,895 | <p>Show that for $u,v \in \mathbb{C}$ with $|u|<1, |v|<1$, and $\bar{u}v\neq u\bar{v}$, we always have
$$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|>\left|u\bar{v}-\bar{u}v\right|.$$</p>
| achille hui | 59,379 | <p>Divide both sides of the inequality we want to prove by $(1+|u|^2)(1+|v|^2)$, we find
it is equivalent to following statement
$$| y - x | \stackrel{?}{>} |x\bar{y} - y\bar{x}|
\quad\text{ where }\quad x = \frac{u}{1+|u|^2}\quad\text{ and }\quad y = \frac{v}{1+|v|^2}
$$
Notice
$$\begin{align}
& \bar{u} v \ne u\bar{v} \implies x \ne y \implies |x-y| > 0\\
\text{ and }\quad & \;|v| < 1\; \implies |y| = \frac{|v|}{1+|v|^2} = \frac{1}{|v|+|v|^{-1}} < \frac12
\end{align}$$</p>
<p>We have
$$|x\bar{y} - y\bar{x}| = |(x - y)\bar{y} + y(\bar{y} - \bar{x})|
\le |x-y||\bar{y}| + |y||\bar{y}-\bar{x}| = 2|x-y||y| < |x-y|$$
This means the inequality we want to prove is indeed true.</p>
|
476,895 | <p>Show that for $u,v \in \mathbb{C}$ with $|u|<1, |v|<1$, and $\bar{u}v\neq u\bar{v}$, we always have
$$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|>\left|u\bar{v}-\bar{u}v\right|.$$</p>
| TheSimpliFire | 471,884 | <p>I have answered here as <a href="https://math.stackexchange.com/q/3896729/471884">this question</a> which shows context has been closed as a duplicate.</p>
<hr />
<p>Let <span class="math-container">$u=re^{i\theta}$</span> and <span class="math-container">$v=\rho e^{i\phi}$</span>. Then the inequality is equivalent to <span class="math-container">$$\left|(1+r^2)\rho e^{i\phi}-(1+\rho^2)re^{i\theta}\right|>r\rho\left|e^{i(\theta-\phi)}-e^{-i(\theta-\phi)}\right|=2r\rho|\sin(\theta-\phi)|$$</span> and dividing both sides by <span class="math-container">$r\rho$</span> yields <span class="math-container">$$\left|r+\frac 1r-\left(\rho+\frac1\rho\right)e^{i(\theta-\phi)}\right|>2|\sin(\theta-\phi)|.$$</span> Denoting <span class="math-container">$s=r+1/r$</span> and <span class="math-container">$t=\rho+1/\rho$</span>, we have <span class="math-container">$s,t>2$</span>. Therefore, it suffices to show that <span class="math-container">$$(s-t\cos(\theta-\phi))^2+t^2\sin^2(\theta-\phi)^2>4\sin^2(\theta-\phi)$$</span> on squaring both sides. Denoting <span class="math-container">$\varphi=\theta-\phi$</span>, the inequality simplifies to <span class="math-container">$$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)>0.$$</span> When <span class="math-container">$\cos\varphi\le0$</span> the LHS is minimised at <span class="math-container">$s=2$</span> so that <span class="math-container">$$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge4-4t\cos\varphi+(t^2-4+4\cos^2\varphi)=(t-2\cos\varphi)^2>0.$$</span> When <span class="math-container">$\cos\varphi>0$</span> the LHS is minimised at <span class="math-container">$s=t\cos\varphi$</span> so that <span class="math-container">$$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge-t^2\cos^2\varphi+t^2-4\sin^2\varphi=(t^2-4)\sin^2\varphi>0$$</span> which proves <span class="math-container">$\vert (1+\vert u \vert^2)v - (1+\vert v \vert^2)u \vert > \vert u\overline{v}-\overline{u}v \vert$</span>.</p>
|
3,602,323 | <p>Let <span class="math-container">$ m $</span>, <span class="math-container">$ m+1 $</span>, <span class="math-container">$ m+2 $</span>, <span class="math-container">$ \dots $</span>, <span class="math-container">$ m+p-1 $</span> be an integers and let <span class="math-container">$ p $</span> be an odd prime. I want to show that
<span class="math-container">$$
m + (m+1)^{p-2} + (m+2)^{p-2} + \cdots + (m+p-1)^{p-2} \equiv 0\pmod p.
$$</span>
This comes down to showing that
<span class="math-container">$$
1 + 2^{p-2} + 3^{p-2} + \cdots + (p-1)^{p-2} \equiv 0\pmod p,
$$</span>
because by the pigeonhole principle, without loss of generality we may assume that <span class="math-container">$ m \equiv 0 \pmod p$</span>, and then <span class="math-container">$ m+i \equiv i \pmod p$</span> for <span class="math-container">$ i = 1,2,\dots,p-1 $</span></p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>Work in the field <span class="math-container">$\,\mathbf F_p=\mathbf Z/p\mathbf Z$</span> and observe that by Fermat, for any <span class="math-container">$k\in[1..\,p-1]$</span>, one has
<span class="math-container">$$k^{p-2}=k^{-1}.$$</span>
Now, in this field, the map
<span class="math-container">\begin{align}
\mathbf F_p^{\times}&\longrightarrow \mathbf F_p^{\times} \\
k&\longmapsto k^{-1}
\end{align}</span>
is an automorphism of the (multiplicative) group of non-zero elements. </p>
<p>Therefore the given sum is just, modulo <span class="math-container">$p$</span>, the sum <span class="math-container">$\;1+2+\dots+(p-1)$</span> in another order. Can you finish the proof?</p>
|
614,962 | <blockquote>
<p>We have a continuous function <span class="math-container">$f:(a,b)\to \mathbb R$</span></p>
<p>Prove that: <span class="math-container">$\forall n: x_1...x_n\in(a,b):\exists x\in(a,b)$</span> such that:</p>
<p><span class="math-container">$$f(x)=\frac1n ( f(x_1)+...+f(x_n) ) $$</span></p>
</blockquote>
<p>Experience tells me that it may be possible with induction but I have no clue on how to begin, I don't even see how is that possible.</p>
<p>Help please ?</p>
| Suraj M S | 85,213 | <p>$$\lim_{x\to 0} \frac{x}{\sin (2x)\cos (5x)}$$
apply the formula $$\sin A\cos B=\frac{1}{2}(\sin (A+B)+\cos (A-B))$$
$$\lim_{x\to 0} \frac{2x}{\sin (7x)-\sin (3x)}$$
further use L hospital to get the limit as $\frac{1}{2}$.</p>
|
282,050 | <p>I have equation $y = -x^2 + 2x + 7$.
How can I change it to canonical form, which looks like $y^2 = 2px$ ?
($p$ will be parameter)</p>
<p>What i ve tried so far:
$$\begin{align}
y &= -x^2 + 2x + 7\\
y &= -(x^2 - 2x + 1) + 8\\
(y-8) &= -(x-1)^2 \\
(y-8)^2 &= 2*(0.5)*(x-1)^4
\end{align}
$$</p>
<p>But I have read somewhere its wrong, so how do I make it correct?</p>
<p>Or is my solution correct?</p>
| lab bhattacharjee | 33,337 | <p>$$-y=x^2-2x-7=(x-1)^2-8\implies (x-1)^2=-(y-8)$$
which is of the form $(x-\alpha)^2=-4a(y-\beta)$</p>
<p>Now, if we are allowed to make the transformation of axes, we can set $x-1=Y,y-8=X$</p>
<p>So,$Y^2=-X=2\left(-\frac12\right)X$</p>
|
2,225 | <p>If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function and satisfies $f(x)=f(2x+1)$, then its not to hard to show that $f$ is a constant.</p>
<p>My question is suppose $f$ is continuous and it satisfies $f(x)=f(2x+1)$, then can the domain of $f$ be restricted so that $f$ doesn't remain a constant. If yes, then give an example of such a function.</p>
| doraemonpaul | 30,938 | <p>$f(x)=f(2x+1)$</p>
<p>$f(2^x-1)=f(2(2^x-1)+1)$</p>
<p>$f(2^x-1)=f(2^{x+1}-1)$</p>
<p>Note that the general solution is $f(x)=\Theta(\log_2(x+1))$ , where $\Theta(x)$ is any periodic function with unit period</p>
<p>So if the domain of $f$ can be restricted to $f:(-1,\infty)\to\mathbb{R}$ , then $f$ can be non-constant and can be $\Theta(\log_2(x+1))$ , where $\Theta(x)$ is any non-constant continuous periodic function with unit period</p>
|
945,736 | <p>:)</p>
<p>I have this matrix:</p>
<p>B = \begin{bmatrix}
0.626 & 2.56 & 2.15 & \\
0.835 & 6.66 & 5.16 & \\
0 & 0 & -1.65 &
\end{bmatrix}</p>
<p>I was wondering how to find a givens matrix such that I could apply it from the right side of the matrix and eliminate B[2][1] (0.835). </p>
<p>B*g = \begin{bmatrix}
* & * & * & \\
0 & * & * & \\
0 & 0 & * &
\end{bmatrix}
Best regards,
rox</p>
| Serge | 32,384 | <p>If you're familiar with left-applied Givens rotations, you can take the transpose of your desired equation an end up in more familiar territory, where your goal becomes creating a lower-triangular matrix with a left-applied Givens rotation.</p>
<p><span class="math-container">$$(\begin{bmatrix} c & -s & 0 \\ s & c & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0.626 & 0.835 & 0 \\ 2.56 & 6.66 & 0 \\ 2.15 & 5.16 & −1.65\end{bmatrix})^T$$</span></p>
<p>Lets undo this transposition and look at how this affects the desired Givens matrix:</p>
<p><span class="math-container">$$\begin{bmatrix}0.626 & 0.835 & 0 \\ 2.56 & 6.66 & 0 \\ 2.15 & 5.16 & −1.65\end{bmatrix}^T\begin{bmatrix} c & -s & 0 \\ s & c & 0 \\ 0 & 0 & 1\end{bmatrix}^T$$</span></p>
<p><span class="math-container">$$\begin{bmatrix}0.626 & 2.56 & 2.15 \\ 0.835 & 6.66 & 5.16 \\ 0 & 0 & −1.65\end{bmatrix}\begin{bmatrix} c & s & 0 \\ -s & c & 0 \\ 0 & 0 & 1\end{bmatrix}$$</span></p>
<p>So finding the appropriate values for a right-multiplied givens upon some matrix <span class="math-container">$M$</span> is very clearly the transpose of finding the appropriate values for a left-multiplied givens upon <span class="math-container">$M^T$</span>.</p>
<p>But for a more thorough explanation of Givens matrices in general:</p>
<p>Left-applied Givens rotations only affect the rows that <strong>c</strong> and <strong>s</strong> apply to, so lets reduce our visuals to those rows, and the one column we want to zero.</p>
<p><span class="math-container">$$\begin{bmatrix} c & -s \\ s & c \end{bmatrix}\begin{bmatrix}0.835 \\ 6.66 \end{bmatrix} = \begin{bmatrix}0 \\ r\end{bmatrix}$$</span></p>
<p>Note that the sign of s is related to which component gets zeroed out.
Given <strong>a = 0.835</strong> and <strong>b = 6.66</strong>
We want to find <strong>c</strong> & <strong>s</strong> such that:</p>
<p><span class="math-container">$$c*a - s*b = 0$$</span> <span class="math-container">$$s*a + c*b = r$$</span></p>
<p>With <strong>r</strong> being the magnitude of our current vector of focus.
The easy solution would be to say:</p>
<p><span class="math-container">$$c = b/r$$</span> <span class="math-container">$$s = a/r$$</span></p>
<p>The first equation trivially reduces to zero. Our second equation becomes:</p>
<p><span class="math-container">$$\frac{a^2}r + \frac{b^2}r = r$$</span>
<span class="math-container">$$\frac{a^2 + b^2}r = r$$</span>
<span class="math-container">$$r^2/r = r$$</span></p>
<p>So for <span class="math-container">$\begin{bmatrix}0.835 \\ 6.66 \end{bmatrix}$</span>, <span class="math-container">$r = \sqrt{0.835^2 + 6.66^2} = 6.71214$</span></p>
<p>In this instance, <span class="math-container">$c = \frac{6.66}{6.71214}$</span> and <span class="math-container">$s = \frac{0.835}{6.71214}$</span></p>
<p>Once <span class="math-container">$c$</span> and <span class="math-container">$s$</span> are known, constructing the desired Givens matrix becomes trivial.</p>
|
3,746,630 | <p>So I am solving some probability/finance books and I've gone through two similar problems that conflict in their answers.</p>
<h2>Paul Wilmott</h2>
<p>The first book is Paul Wilmott's <a href="https://smile.amazon.com/Frequently-Asked-Questions-Quantitative-Finance/dp/0470748753" rel="nofollow noreferrer">Frequently Asked Questions in Quantitative Finance</a>. This book poses the following question:</p>
<blockquote>
<p>Every day a trader either makes 50% with probability 0.6 or loses 50% with probability 0.4. What is the probability the trader will be ahead at the end of a year, 260 trading days? Over what number of days does the trader have the maximum probability of making money?</p>
</blockquote>
<p><strong>Solution:</strong></p>
<blockquote>
<p>This is a nice one because it is extremely counterintuitive. At first glance it looks like you are going to make money in the long run, but this is not the case.
Let n be the number of days on which you make 50%. After <span class="math-container">$n$</span> days your returns, <span class="math-container">$R_n$</span> will be:
<span class="math-container">$$R_n = 1.5^n 0.5^{260−n}$$</span>
So the question can be recast in terms of finding <span class="math-container">$n$</span> for which this expression is equal to 1.</p>
</blockquote>
<p>He does some math, which you can do as well, that leads to <span class="math-container">$n=164.04$</span>. So a trader needs to win at least 165 days to make a profit. He then says that the average profit <em>per day</em> is:</p>
<blockquote>
<p><span class="math-container">$1−e^{0.6 \ln1.5 + 0.4\ln0.5}$</span> = −3.34%</p>
</blockquote>
<p>Which is mathematically wrong, but assuming he just switched the numbers and it should be:</p>
<blockquote>
<p><span class="math-container">$e^{0.6 \ln1.5 + 0.4\ln0.5} - 1$</span> = −3.34%</p>
</blockquote>
<p>That still doesn't make sense to me. Why are the probabilities in the exponents? I don't get Wilmott's approach here.</p>
<p>*PS: I ignore the second question, just focused on daily average return here.</p>
<hr />
<h2>Mark Joshi</h2>
<p>The second book is Mark Joshi's <a href="https://smile.amazon.com/Quant-Interview-Questions-Answers-Second/dp/0987122827" rel="nofollow noreferrer">Quant Job Interview Question and Answers</a> which poses this question:</p>
<blockquote>
<p>Suppose you have a fair coin. You start off with a dollar, and if you toss an <em>H</em> your position doubles, if you toss a <em>T</em> it halves. What is the expected value of your portfolio if you toss infinitely?</p>
</blockquote>
<p><strong>Solution</strong></p>
<blockquote>
<p>Let <span class="math-container">$X$</span> denote a toss, then:
<span class="math-container">$$E(X) = \frac{1}{2}*2 + \frac{1}{2}\frac{1}{2} = \frac{5}{4}$$</span>
So for <span class="math-container">$n$</span> tosses:
<span class="math-container">$$R_n = (\frac{5}{4})^n$$</span>
Which tends to infinity as <span class="math-container">$n$</span> tends to infinity</p>
</blockquote>
<hr />
<hr />
<p>Uhm, excuse me what? Who is right here and who is wrong? Why do they use different formula's? Using Wilmott's (second, corrected) formula for Joshi's situation I get the average return per day is:</p>
<p><span class="math-container">$$ e^{0.5\ln(2) + 0.5\ln(0.5)} - 1 = 0% $$</span></p>
<p>I ran a Python simulation of this, simulating <span class="math-container">$n$</span> days/tosses/whatever and it seems that the above is not correct. Joshi was right, the portfolio tends to infinity. Wilmott was also right, the portfolio goes to zero when I use his parameters.</p>
<p>Wilmott also explicitly dismisses Joshi's approach saying:</p>
<blockquote>
<p>As well as being counterintuitive, this question does give a nice insight into money management and is clearly related to the Kelly criterion. If you see a question like this it is meant to trick you if the expected profit, here 0.6 × 0.5 + 0.4 × (−0.5) = 0.1, is positive with the expected return, here −3.34%, negative.</p>
</blockquote>
<p>So what is going on?</p>
<p>Here is the code:</p>
<pre><code>import random
def traderToss(n_tries, p_win, win_ratio, loss_ratio):
SIM = 10**5 # Number of times to run the simulation
ret = 0.0
for _ in range(SIM):
curr = 1 # Starting portfolio
for _ in range(n_tries): # number of flips/days/whatever
if random.random() > p_win:
curr *= win_ratio # LINE 9
else:
curr *= loss_ratio # LINE 11
ret += curr # LINE 13: add portfolio value after this simulation
print(ret/SIM) # Print average return value (E[X])
</code></pre>
<p>Use: <code>traderToss(260, 0.6, 1.5, 0.5)</code> to test Wilmott's trader scenario.</p>
<p>Use: <code>traderToss(260, 0.5, 2, 0.5)</code> to test Joshi's coin flip scenario.</p>
<hr />
<hr />
<p>Thanks to the followup comments from Robert Shore and Steve Kass below, I have figured one part of the issue. <strong>Joshi's answer assumes you play once, therefore the returns would be additive and not multiplicative.</strong> His question is vague enough, using the word "your portfolio", suggesting we place our returns back in for each consecutive toss. If this were the case, we need the <a href="https://www.investopedia.com/articles/investing/071113/breaking-down-geometric-mean.asp" rel="nofollow noreferrer">geometric mean</a> not the arithmetic mean, which is the expected value calculation he does.</p>
<p>This is verifiable by changing the python simulation to:</p>
<pre><code>import random
def traderToss():
SIM = 10**5 # Number of times to run the simulation
ret = 0.0
for _ in range(SIM):
if random.random() > 0.5:
curr = 2 # Our portfolio becomes 2
else:
curr = 0.5 # Our portfolio becomes 0.5
ret += curr
print(ret/SIM) # Print single day return
</code></pre>
<p>This yields <span class="math-container">$\approx 1.25$</span> as in the book.</p>
<p>However, if returns are multiplicative, therefore we need a different approach, which I assume is Wilmott's formula. This is where I'm stuck. Because I still don't understand the Wilmott formula. Why is the end of day portfolio on average:</p>
<p><span class="math-container">$$ R_{day} = r_1^{p_1} * r_2^{p_2} * .... * r_n^{p_n} $$</span></p>
<p>Where <span class="math-container">$r_i$</span>, <span class="math-container">$p_i$</span> are the portfolio multiplier, probability for each scenario <span class="math-container">$i$</span>, and there are <span class="math-container">$n$</span> possible scenarios. Where does this (generalized) formula come from in probability theory? This isn't a geometric mean. Then what is it?</p>
| Brian M. Scott | 12,042 | <p>They’re computing two entirely different things. Wilmott is computing the minimum number of days out of <span class="math-container">$260$</span> on which you must make a profit in order to come out ahead; Joshi is computing the expected value of your portfolio. Applying Joshi’s calculation to Wilmott’s setting, we get an expected value after <span class="math-container">$260$</span> days of</p>
<p><span class="math-container">$$(0.6\cdot1.5+0.4\cdot0.5)^{260}=1.1^{260}\approx 57,833,669,934\;.$$</span></p>
<p>Wilmott’s calculation does not take the probabilities of the two outcomes into account: it would yield the same result whether you made a <span class="math-container">$50\%$</span> profit with probability <span class="math-container">$0.99$</span> or with probability <span class="math-container">$0.01$</span>. In the former case, however, you are almost certain to make a net profit, while in the latter you are almost certain to lose virtually everything. No matter what the probabilities are, you need to make a profit on at least <span class="math-container">$165$</span> days in order to come out ahead for the year; your likelihood of actually doing so, however, changes greatly with the probabilities.</p>
<p>In the original problem you might find it odd that the expected number of days on which you make a profit is <span class="math-container">$60\%$</span> of <span class="math-container">$260$</span>, or <span class="math-container">$156$</span> days, and you lose money if you make a profit on exactly <span class="math-container">$156$</span> days, yet your overall expected value is enormous. This is because once you reach the break-even point, your expected final value grows explosively as the number of profitable days (out of <span class="math-container">$260$</span>) increases, and these huge profits more than compensate for the more likely losses.</p>
<p>If you want to know how likely it is that you’ll make a profit, you want Wilmott’s calculation; you can then plug the figure of <span class="math-container">$165$</span> days into a binomial distribution calculator and find that the probability of making a profit on at least <span class="math-container">$165$</span> days is only about <span class="math-container">$0.14$</span>. The fact that the expected profit — expected in the mathematical sense, that is — is considerable would probably not be very comforting, since it results from the fact that relatively unlikely outcomes produce huge profits.</p>
|
85,052 | <p>A housemate of mine and I disagree on the following question: </p>
<p>Let's say that we play a game of yahtzee. Of the five dice you throw, two dice obtain the value 1, two other dice obtain the value 2, and one die shows you six dots on the top side. Given the fact that you haven't thrown a "full house" yet, you start throwing the die which value is 6 again and again, until you throw a one or a two. You get to throw the die of six one or two times. If you throw a one or a two the first time, you stop, because now you have the "full house" already. If you haven't thrown a one or a two with the die of six, you throw it again, hoping for a one or a two this time.</p>
<p>Now, what is the probability that you throw a one or a two with the fifth die after two turns? (Given the way a rational person operates in this situation.)</p>
<p>My take on this question was the following: the probability that you throw a one or a two the first time with the fifth dice is $1/3$, and the probability that you don't throw a dice of which the value is one or two the first time, but you do throw a one or a two the second time is $ 2/3 \cdot 1/3 = 2/9$. Adding these values gives you the probability: $1/3 + 2/9 = 3/9 + 2/9 = 5/9$. </p>
<p>My housemate, however, argues that the chance to throw a one or a two the first time is $1/3$, and believes that throwing the fifth dice again, gives you a probability of throwing a one or a two of $1/3$ again. Adding these values gives the expected probability of throwing a full house of $1/3 + 1/3 = 2/3$. </p>
<p>Who is right, my housemate or me?</p>
<p>I strongly believe I am right, but even if you tell me I'm right, I might not be able to convince my housemate of the truth. He argues that my way of reasoning implies that the probability of throwing a one or a two with the fifth dice the second time is smaller than throwing it the first time. Could you please also provide me with a pedagogically sound way to explain him why the probability is $5/9$? </p>
<p>Thanks in advance</p>
| Ilmari Karonen | 9,602 | <p>We can write the process out as an <a href="http://en.wikipedia.org/wiki/Extensive-form_game" rel="nofollow">extensive-form game tree</a>:</p>
<ul>
<li>First throw:<br><br><ul>
<li>Probability $\frac 13$: rolled 1 or 2, stop.
<li>Probability $\frac 23$: rolled 3, 4, 5 or 6, roll again:<br><br><ul>
<li>Probability $\frac 13$ (total probability $\frac 23 \cdot \frac 13 = \frac 29$): rolled 1 or 2, stop.
<li>Probability $\frac 23$ (total probability $\frac 23 \cdot \frac 23 = \frac 49$): rolled 3, 4, 5 or 6, no more throws left.
</ul></ul></ul>
<p>The overall probability of getting 1 or 2 is then $\frac 13 + \frac 23 \cdot \frac 13 = \frac 59$, while the probability of getting 3, 4, 5 or 6 is simply $\frac 23 \cdot \frac 23 = \frac 49$.</p>
|
92,867 | <p>Suppose we have some random variable $X$ that ranges over some sample space $S$. We also have two probability models $F$ and $G$. Let $f(x)$ and $g(x)$ be the probability density functions for these distributions. Does the following quantity $$ \log \frac{f(x)}{g(x)} = \log \frac{P(F|x)}{P(G|x)}- \log \frac{P(F)}{P(G)}$$ basically tell us how much more likely model $F$ is the true model than model $G$?</p>
| Michael Hardy | 11,667 | <p>I was slightly puzzled by the notation, but I'm assuming that by $P(F)$ you mean the probability that $F$ is the right model, and $P(F\mid x)$ is the conditional probability that $F$ is the right model given the event $X=x$.</p>
<p>The identity you write is then a form of a special case of what is sometimes called Bayes' theorem.</p>
<p>If one assumes the right model must be either $F$ or $G$, but not both, then one can say that
$$
\operatorname{logit} P(F) = \log \frac{P(F)}{1-P(F)} = \log(\operatorname{odds}(F))
$$
increases by $\log(f(x)/g(x))$ when it is observed that $X=x$.</p>
<p>So that quantity tells you by how much the logit of the probability increases when you observe the data.</p>
|
389,675 | <p>I'm trying to use a program to find the largest prime factor of 600851475143.
This is for Project Euler here: <a href="http://projecteuler.net/problem=3">http://projecteuler.net/problem=3</a></p>
<p>I first attempted this with the code that goes through every number up to 600851475143, tests its divisibility, and adds it to an array of prime factors, printing out the largest.</p>
<p>This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory). </p>
<p>Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since. </p>
<p>I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.</p>
| xisk | 73,012 | <p>The thing with Project Euler is that there is usually an obvious brute-force method to do the problem, which will take just about <em>forever</em>. As the questions become more difficult, you will need to implement clever solutions.</p>
<p>One way you can solve this problem is to use a loop that always finds the smallest (positive integer) factor of a number. When the smallest factor of a number is that number, then you've found the greatest prime factor!</p>
<p><em>Detailed Algorithm description:</em></p>
<p>You can do this by keeping three variables:</p>
<ul>
<li>The number you are trying to factor (A)</li>
<li>A current divisor store (B)</li>
<li>A largest divisor store (C)</li>
</ul>
<p>Initially, let (A) be the number you are interested in - in this case, it is 600851475143. Then let (B) be 2. Have a conditional that checks if (A) is divisible by (B). If it is divisible, divide (A) by (B), reset (B) to 2, and go back to checking if (A) is divisible by (B). Else, if (A) is not divisible by (B), increment (B) by +1 and then check if (A) is divisible by (B). Run the loop until (A) is 1. The (3) you return will be the largest prime divisor of 600851475143.</p>
<p>There are numerous ways you could make this more effective - instead of incrementing to the next integer, you could increment to the next necessarily prime integer, and instead of keeping a largest divisor store, you could just return the current number when its only divisor is itself. However, the algorithm I described above will run in seconds regardless.</p>
<p>You said you don't want a straight up answer. Regardless, if you want to see one implementation of it, I've uploaded my code for this problem on Pastebin: <a href="http://pastebin.com/hVzsPSxH">here (C)</a> and <a href="http://pastebin.com/M0qk0Kn8">here (Python)</a>.</p>
<hr>
<p>Example: Let's find the largest prime factor of 105 using the method described above.</p>
<p>Let (A) = 105. (B) = 2 (we always start with 2), and we don't have a value for (C) yet.</p>
<p>Is (A) divisible by (B)? No. Increment (B) by +1: (B) = 3. Is Is (A) divisible by (B)? Yes. (105/3 = 35). The largest divisor found so far is 3. Let (C) = 3. Update <strong>(A) = 35</strong>. Reset (B) = 2.</p>
<p>Now, is (A) divisible by (B)? No. Increment (B) by +1: (B) = 3. Is (A) divisible by (B)? No. Increment (B) by +1: (B) = 4. Is (A) divisible by (B)? No. Increment (B) by +1: (B) = 5. Is (A) divisible by (B)? Yes. (35/5 = 7). The largest divisor we found previously is stored in (C). (C) is currently 3. 5 is larger than 3, so we update (C) = 5. We update (A)=7. We reset (B)=2.</p>
<p>Then we repeat the process for (A), but we will just keep incrementing (B) until (B)=(A), because 7 is prime and has no divisors other than itself and 1. (We could already stop when (B)>((A)/2), as you cannot have integer divisors greater than half of a number - the smallest possible divisor (other than 1) of any number is 2!)</p>
<p>So at that point we return (A) = 7.</p>
<p>Try doing a few of these by hand, and you'll get the hang of the idea. Or just play with the code I've provided. Insert some print statements, see how it iterates through the numbers.</p>
|
389,675 | <p>I'm trying to use a program to find the largest prime factor of 600851475143.
This is for Project Euler here: <a href="http://projecteuler.net/problem=3">http://projecteuler.net/problem=3</a></p>
<p>I first attempted this with the code that goes through every number up to 600851475143, tests its divisibility, and adds it to an array of prime factors, printing out the largest.</p>
<p>This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory). </p>
<p>Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since. </p>
<p>I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.</p>
| hmakholm left over Monica | 14,366 | <p>The number looks small enough to be brute-forced on a computer. Just try every possible factor, starting with 2, 3, 4, ... and keep dividing them out as long as the division comes out even. Then continue looking for factors of the quotient. You don't even need to explicitly restrict to primes, because any composite number you try simply won't divide the quotient so far (why?).</p>
<p>If you reach a trial factor that's less than the square root of the number you're dividing into, you know that the larger number must be the last prime factor. So there cannot possibly be more than a few million trial divisions to do.</p>
|
3,530,492 | <blockquote>
<p>Evaluate</p>
<p><span class="math-container">$$ \int_0^{e^{\pi}} |\cos\ (\ln x)|dx$$</span></p>
</blockquote>
<p><em>My ideas:</em> I substituted <span class="math-container">$u = \ln x$</span> and tried to evaluate</p>
<p><span class="math-container">$$\int_{-\infty}^\pi |\cos u|\ e^u du$$</span></p>
<p>Integration by parts didn't yield any results however. Does anybody have another idea?</p>
| LHF | 744,207 | <p>Here's a solution based on GEdgar's hint.</p>
<p><span class="math-container">$$\begin{aligned}
\int_0^{e^{\pi}} |\cos (\ln x) |\; dx &= \int_{-\infty}^{\pi} |\cos u |\; e^u \; du\\
&= \int_{\pi/2}^{\pi} |\cos u |\; e^u \; du +\sum_{k = 0}^{\infty} \int_{-k\pi -\pi/2}^{-k\pi +\pi/2} |\cos u |\; e^u \; du \\
&= \left| \int_{\pi/2}^{\pi} \cos u \; e^u \; du \right| +\sum_{k= 0}^{\infty} \left|\int_{-k\pi -\pi/2}^{-k\pi +\pi/2} \cos u \; e^u \; du\ \right| \\
&= \left|\ \left[ \frac12(\sin u+\cos u) \; e^u \right]
_{\pi/2}^{\pi}\
\right|
+
\sum_{k= 0}^{\infty}
\left|\
\left[
\frac12(\sin u+\cos u) \; e^u
\right]
_{-k\pi -\pi/2}^{-k\pi +\pi/2}\
\right|
\\
&=
\frac 12e^\pi
+
2
\sum_{k=0}^{\infty}
\frac 12 e^{-k\pi +\pi/2}
\\
&=
\frac 12e^\pi
+
e^{\pi/2}
\cdot
\frac 1{1-e^{-\pi}}\ .
\end{aligned}$$</span></p>
|
623,819 | <p>I do not understand a remark in Adams' Calculus (page 628 <span class="math-container">$7^{th}$</span> edition). This remark is about the derivative of a determinant whose entries are functions as quoted below.</p>
<blockquote>
<p>Since every term in the expansion of a determinant of any order is a product involving one element from each row, the general product rule implies that the derivative of an <span class="math-container">$n\times n$</span> determinant whose elements are functions will be the sum of <span class="math-container">$n$</span> such <span class="math-container">$n\times n$</span> determinants, each with the elements of one of the rows differentiated. For the <span class="math-container">$3\times 3$</span> case we have
<span class="math-container">$$\frac{d}{dt}\begin{vmatrix}
a_{11}(t) & a_{12}(t) & a_{13}(t) \\
a_{21}(t) & a_{22}(t) & a_{23}(t) \\
a_{31}(t) & a_{32}(t) & a_{33}(t)
\end{vmatrix}=\begin{vmatrix}
a'_{11}(t) & a'_{12}(t) & a'_{13}(t) \\
a_{21}(t) & a_{22}(t) & a_{23}(t) \\
a_{31}(t) & a_{32}(t) & a_{33}(t)
\end{vmatrix}+\begin{vmatrix}
a_{11}(t) & a_{12}(t) & a_{13}(t) \\
a'_{21}(t) & a'_{22}(t) & a'_{23}(t) \\
a_{31}(t) & a_{32}(t) & a_{33}(t)
\end{vmatrix}+\begin{vmatrix}
a_{11}(t) & a_{12}(t) & a_{13}(t) \\
a_{21}(t) & a_{22}(t) & a_{23}(t) \\
a'_{31}(t) & a'_{32}(t) & a'_{33}(t)
\end{vmatrix}.$$</span></p>
</blockquote>
<hr />
<p>It is not difficult to check this equality by simply expanding both sides. However, the remark sounds like using some clever trick to get this result. Can anyone explain it to me, please? Thank you!</p>
| Lutz Lehmann | 115,115 | <p>The determinant is like a generalized product of vectors (in fact, it is related to the outer product). So considering the rows as factors in this generalized product, this formula reflects the product rule of differentiation.</p>
<p>If $D(a,b,c)$ is generally a function of vectors that is linear in each argument, and you apply it to vector functions in one variable, then</p>
<p>\begin{align}
&D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t),c(t))\\[0.5em]
&=\ D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t+h),c(t+h))\\
&\ +D(a(t),b(t+h),c(t+h))-D(a(t),b(t),c(t+h))\\
&\ +D(a(t),b(t),c(t+h))-D(a(t),b(t),c(t))\\[0.5em]
&=\ D([a(t+h)-a(t)],b(t+h),c(t+h))
+D(a(t),[b(t+h)-b(t)],c(t+h))
+D(a(t),b(t),[c(t+h)-c(t)])
\end{align}</p>
<p>and from that the claimed generalized product rule can be obtained.</p>
|
118,873 | <p>I understand that the Mellin transform of a modular form is expected to satisfy RH when it is an eigenform of all Hecke operators, in which case it has an Euler product. Now about when the form is not an eigenform: Is it known a case where the zeros are all in the critical strip?</p>
| Johan Andersson | 10,811 | <p>When it is not a Hecke-Eigen form, the Hecke L-series connected with the modular form does not have an Euler product. However it can still be written as a linear combination of Hecke L-series that have Euler-products. Thus the situation will resemble the case of linear combinations of Dirichlet L-series. In particular we can use joint Voronin universality to obtain $\gg T$ zeroes in any strip $1/2<\sigma_1<\Re(s)<\sigma_2 < 1$, $-T < \Im(s)< T.$ These results follows from the analytic properties of the Rankin-Selberg zeta-function (which gives "independence results" for Hecke eigenvalues of primes attached to different cusp forms). You should be able to find results of this kind in Steuding's SLN "Value Distribution of L-functions".</p>
<p>Also the results of Davenport-Heilbronn for the Hurwitz zeta-function can be proved, i.e. For any $\epsilon>0$ there exists $\gg T$ zeroes with $1<\Re(s)<1+\epsilon$ and $-T < \Im(s) < T $.</p>
<p>Stronger results corresponding to results of Karatsuba, Bombieri, Hejhal and Selberg for Dirichlet L-function that holds close to the critical line likely also holds. I think the russian school (Irina Rezvyakova) has proved results in this direction.</p>
|
2,916,099 | <p>Find a Mobius transformation $T$ from the unit disk to the right half plane with condition $T(0)=3$.</p>
<p>First, the transformation from the unit circle to the upper half plane is $T_1(z)=(1-i)\frac{z-i}{z-1}$.</p>
<p>So from the unit circle to the right half plane, $T_2(z)=-i(1-i)\frac{z-i}{z-1}$</p>
<p>How can I introduce the condition $T(0)=3$ ?</p>
<p>$T(0)=1-i\neq3$</p>
| dxiv | 291,201 | <p>For a purely algebraic derivation, consider the general form of the Möbius transformation <span class="math-container">$\,T(z)=\dfrac{az+b}{cz+d}\,$</span>. Both <span class="math-container">$\,a\,$</span> and <span class="math-container">$\,c\,$</span> cannot be <span class="math-container">$\,0\,$</span>, otherwise it would be a constant transformation. The right half-plane is invariant to the inversion <span class="math-container">$\,T(z) \to \dfrac{1}{T(z)}\,$</span> so it can be assumed WLOG that <span class="math-container">$\,a \ne 0\,$</span>, then after normalizing it can be assumed WLOG that <span class="math-container">$\,a=1\,$</span>. The condition <span class="math-container">$\,T(0)=3\,$</span> translates to <span class="math-container">$\,b = 3d\,$</span>, so in the end <span class="math-container">$\,T(z)=\dfrac{z+3d}{cz+d}\,$</span> for some <span class="math-container">$\,c, d \in \Bbb C\,$</span> with <span class="math-container">$\,d \ne 0\,$</span>.</p>
<p>The unit circle must transform into the imaginary axis, so for <span class="math-container">$\,|z|=1\,$</span>:</p>
<p><span class="math-container">$$
\begin{align}
0 = 2 \operatorname{Re}\left(T(z)\right) &= \dfrac{z+3d}{cz+d} + \dfrac{\bar z+3 \bar d}{\bar c \bar z+ \bar d} \\
&= \frac{(z+3d)(\bar c \bar z + \bar d)+(\bar z + 3 \bar d)(cz + d)}{|cz+d|^2} \\
&= \frac{(c+\bar c)|z|^2+ 6 |d|^2+(\bar d + 3c\bar d) z+(3 \bar cd +d)\bar z}{|cz+d|^2} \\
&= \frac{2 \operatorname{Re}(c)+ 6 |d|^2+(3c+1)\bar d z+(3 \bar c +1)d\bar z}{|cz+d|^2}
\end{align}
$$</span></p>
<p>It follows that <span class="math-container">$\,3c+1=0\,$</span> for the numerator to not depend on <span class="math-container">$\,z\,$</span>, and <span class="math-container">$\,2 \operatorname{Re}(c)+ 6 |d|^2=0\,$</span> for the numerator to be <span class="math-container">$\,0\,$</span>. The first equation gives <span class="math-container">$\,c = -\dfrac{1}{3}\,$</span>, and the second one <span class="math-container">$\,|d|=\dfrac{1}{3}\,$</span>. Therefore, defining <span class="math-container">$\,\omega = 3d\,$</span> the general solution is:</p>
<p><span class="math-container">$$
T(z) \;=\; \frac{z + 3d}{-\dfrac{1}{3}z+d} \;=\; 3\,\dfrac{z + \omega}{-z + \omega} \quad\quad\style{font-family:inherit}{\text{where}}\;\; |\omega|=1
$$</span></p>
<p><hr>
[ <em>EDIT</em> ] For quick verification of the form above:</p>
<p><span class="math-container">$$\small
\frac{1}{3}T(z) = \dfrac{z + \omega}{-z + \omega} \color{red}{\cdot \frac{\bar \omega}{\bar \omega}} = \frac{1+\bar \omega z}{1 - \bar \omega z} \color{red}{\cdot \frac{1 - \omega \bar z}{1 - \omega \bar z}} = \frac{1 - |\omega|^2|z|^2+\bar \omega z - \omega \bar z }{|1 - \bar \omega z|^2} = \frac{1 - |z|^2+ 2i \operatorname{Im}(\bar \omega z)}{|1 - \bar \omega z|^2}
$$</span></p>
<p>Therefore <span class="math-container">$\,\small\operatorname{Re}(T(z)) = 3\,\dfrac{1 - |z|^2}{|1 - \bar \omega z|^2} \ge 0\,$</span> iff <span class="math-container">$\,\small|z| \le 1\,$</span>, and of course <span class="math-container">$\,\small T(0) = 3\,$</span>.</p>
|
4,298,184 | <p><a href="https://i.stack.imgur.com/f5ny0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f5ny0.png" alt="enter image description here" /></a>
as we can see, we are supposed to use stars and bars where n = 10 and r = 3. but what i dont understand is why we use stars and bars when stars and bars is supposed to be used in situations where we try to group 10 objects in 3 distinct groups, and the question is supposed to be asking us the number of ways we can get 3 integers that have a sum of 10. any help?</p>
| Marc van Leeuwen | 18,880 | <p>With new unknowns <span class="math-container">$x_i'=x_i+1\in\Bbb N$</span>, you need to count the number of solutions to <span class="math-container">$x_1'+x_2'+x_3'=10$</span>. Think of writing a solution to that last equation representing in the left hand side each <span class="math-container">$x_i$</span> by that many tally marks (say "<span class="math-container">$1$</span>", so that <span class="math-container">$2+5+3$</span> becomes "<span class="math-container">$11{+}11111{+}111$</span>"). Then you solution is represented by a sequence of <span class="math-container">$12$</span> symbols, <span class="math-container">$10$</span> of which are tally marks and the remaining <span class="math-container">$2$</span> are <span class="math-container">$+$</span> signs; since every arrangement of these symbols is allowed (possibly having no tally marks at all in a row to represent <span class="math-container">$0$</span>), there are <span class="math-container">$\binom{12}{10}=\binom{12}2=66$</span> solutions.</p>
|
3,130,877 | <p>Normal geometry concepts, such as parallel, angle, area, triangle, do they still apply in Mobius band?</p>
<p>If not, in which case will they fail to do so?</p>
<p>For example, what would three lines on a Mobius band form? A triangle if not parallel? or it might be totally something else?</p>
| Travis Willse | 155,629 | <p>A priori a Mobius strip <span class="math-container">$S$</span> is a topological manifold, perhaps with boundary, depending on how we define it---for simplicity I'll assume no boundary.</p>
<p>If one endows <span class="math-container">$S$</span> with a Riemannian metric <span class="math-container">$g$</span>---and there are many ways to do this---then one has available all of the usual trappings of Riemannian geometry. This includes measuring lengths of and angles between vectors, defining geodesics, distances between points, Gaussian curvature and so one, and as usual, these measurements and constructions depend heavily on the choice of <span class="math-container">$g$</span>. Unlike most surfaces we work with, however, <span class="math-container">$S$</span> is nonorientable, so a metric does not determine even up to a fixed choice of sign a global choice of volume form on <span class="math-container">$S$</span>, so we can define (unsigned) area, by integrating a <em>Riemannian density</em> rather than a volume form, but we cannot define a (globally consistent choice of) signed area.</p>
<p>One way to realize the Mobius strip is as the quotient of <span class="math-container">$\tilde S := \Bbb R \times (-1, 1)$</span> by the action <span class="math-container">$\Bbb Z \times \tilde S \to \tilde S$</span> defined by <span class="math-container">$n \cdot (x, y) := (x + n, (-1)^n y)$</span>. In particular, <span class="math-container">$\Bbb Z$</span> acts by isometries of the usual Euclidean metric on <span class="math-container">$\tilde S$</span>, which thus descends to a (locally) flat metric <span class="math-container">$\bar g$</span> on <span class="math-container">$S$</span>.</p>
<p>Locally <span class="math-container">$(S, \bar g)$</span> behaves like any flat manifold, that is, like a patch of Euclidean space. Globally this metric behaves in some peculiar ways, however. For example, one can construct geodesics with an arbitrary number of self-intersections, which does not occur in (global) Euclidean space or on the (round) sphere. Some geodesics close but most do not. If one defines two geodesic to be parallel if they do not intersect, then the parallel postulate fails in general: Given a geodesic <span class="math-container">$\ell$</span> and a point <span class="math-container">$P$</span> not on the line, there may be zero, finitely many, or infinitely many lines through <span class="math-container">$P$</span> parallel to <span class="math-container">$\ell$</span>. Depending on their relative positions, three geodesics may bound zero triangles, one, or many.</p>
|
1,457,838 | <p>I am given position vectors: $\vec{OA} = i - 3j$ and $\vec{OC}=3i-j$.
And asked to find a position vector of the point that divides the line $\vec{AC}$ in the ratio $-2:3$. </p>
<p>So I found the vector $\vec{AC}$, and it is $2i+2j$. Then, if the point of interest is $L$, position vector $\vec{OL} = \vec{OA} + \lambda\vec{AC}$. Where $\lambda$ is the appropriate scalar we have to apply. But I am a bit confused with the negative ratio. I tried to apply the logic of positive ratios like $3:2$. In this case, we would split the line into 5 "portions" and we would be applying a ratio of $\frac{3}{5}$. I interpret negative ratio, in the following manner. I first split the $\vec{AC}$ into 5 "portions", then I move two portions outside of the line, to the left; then I move 2 portions back and 1 in. If that makes any sense at all. Therefore the ratio to be applied should be $\frac{1}{5}$. However, I am not getting a required result.</p>
| R.N | 253,742 | <p>Hint: theorem- Every composite number has a proper factor less than or equal to its square root.</p>
<p>Suppose n is composite. I can write $n = ab$ , where $1 < a, b < n$ . If both $a, b > \sqrt{n}$ , then</p>
<p>$$n = \sqrt{n}\cdot \sqrt{n} < a\cdot b = n.$$</p>
<p>This contradiction shows that at least one of a, b must be less than or equal to $\sqrt{n}$</p>
|
2,762,715 | <blockquote>
<p>Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is? </p>
</blockquote>
<p>I don't get how to approach this problem. Any help would be appreciated. </p>
| Ross Millikan | 1,827 | <p>First, replace $a$ in your inequality by $b+10$. Now the left hand side has only one parameter. Take the derivative with respect to $x$, set to zero, and find the $x$ value of the maximum as a function of $b$. Plug that in and find the left hand side maximum as a function of $b$. Take the derivative, set to zero....</p>
|
2,762,715 | <blockquote>
<p>Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is? </p>
</blockquote>
<p>I don't get how to approach this problem. Any help would be appreciated. </p>
| Vishaal Sudarsan | 414,699 | <p>Let $f :\mathbb{R}^+ \to \mathbb{R} \quad $such that$\quad f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx}$</p>
<p>Notice that $f(x)$ is increasing when $a>b$ , $f(x) = 0$ when $a=b$ and $f(x)$ is decreasing when $a<b$.</p>
<p>Now when $a>b$ $\lim_{x \to \infty} f(x) = 5$</p>
<p>Hence $\sqrt{x^2 + ax} - \sqrt{x^2+bx} < 5 \qquad \forall a,b \in \mathbb{R}^+$</p>
|
3,203,282 | <p>Given that <span class="math-container">$C[-\pi,\pi]$</span> is complete:
How can we prove, by using the supremum norm, that the space:</p>
<p><span class="math-container">$$C_p[-\pi,\pi]=\{f\in C[-\pi,\pi]\mid f(-\pi)=f(\pi)\}$$</span></p>
<p>is also complete? thank you!</p>
| carlosayam | 49,844 | <p>The sum initially goes from <span class="math-container">$n=0$</span> to <span class="math-container">$n=\infty$</span>. In the second line, the first term (<span class="math-container">$n=0$</span>), which is <span class="math-container">$1$</span>, is expanded apart, and now the sum starts at <span class="math-container">$n=1$</span>. And don't be confused by the minus "-" in front :)</p>
|
1,765,022 | <p>The problem is:</p>
<blockquote>
<p>$Prove$ $that$ $|\sin^2 (x)-\sin^2 (y)|\le |x-y|$ $ for $ $ all $ $ x,y>0$.</p>
</blockquote>
<p><em>$My$ $work$ :</em> $$\sin^2 (x)\le|\sin x|\le|x|\le|x-y|+|y|$$ and so is $$|\sin^ 2 (x)-\sin^2 (y)|\le |x-y|+|y|$$ But this is not the actual result I want. I think I have done few mistakes. Isn't it? Please help me. Thank you in advance.</p>
| David Heider | 21,026 | <p>The mistake is in my eyes that you allow the polar angle to be in $\phi\in[0,\pi]$ although just integrating over a quadrant integration domain in cartesian coordinates,$(x,y)\in[0,R]^2$. I'd suggest putting $\phi\in[0,\pi/2]$ in order to account for the integration in the first quadrant. This modification of your calculation results in an additional factor of $1/2$ when going to polars $(r,\phi)$.</p>
<p>$I = \sqrt{\frac{\pi}{4i}}\sqrt{1-e^{-iR^2}}$.</p>
<p>The limit $R\to\infty$ works for practical purposes.</p>
<p>Does this help you?</p>
<p>David</p>
|
2,358,838 | <p>I can see the answer to this in my textbook; however, I am not quite sure how to solve this for myself . . . the book has the following:</p>
<blockquote>
<p>To take advantage of the inductive hypothesis, we use these steps:</p>
<p>$ 7^{(k+1)+2} + 8^{2(k+1)+1} = 7^{k+3} + 8^{2k+3} $</p>
<p>$$
= 7\cdot7^{k+2} + 8^{2}\cdot8^{2k+1}\\
= 7\cdot7^{k+2} + 64\cdot8^{2k+1}\\
= 7(7^{k+2}+8^{2k+1})+57\cdot8^{2k+1}\\ $$</p>
</blockquote>
<p>While the answer is apparent to me <em>now</em>; how exactly would I go about figuring out a similar algebraic manipulation if I were to see something like this on a test? Is there an algorithm or a way of thinking about how to break this down that I'm missing? I think I'm most lost regarding the move from the second to last and last equations.</p>
<p><em>Source: Discrete Mathematics and its Applications (7th ed), Kenneth H. Rosen (p.322)</em></p>
| Community | -1 | <p>you would have to think about how to get the form desired. in this case the first part of:</p>
<p>$7(7^{k+2}+8^{2k+1})+57⋅8^{2k+1}$</p>
<p>We have made a factor, of the form desired. Assuming it's divisible by 57, that part of the sum is, and the other part shows it is already. So, the sum of both parts, must divide by 57. Hence, the original number must divide by 57. The original number is of the form suggested but with n replaced by k+1. so it must be true for n=k+1, assuming it works for n=k. then what's left this to show it works for some base value of k. As $7^3+8^3=343+512=855=57\cdot {15}$ it works for n=k=1, and it follows by what we've shown that it then works for all k>1 . <strong>edit:</strong> okay I forgot the 0 case ( see G cab's answer): as 49+8=57, we also have it for the k=0 case, and again all other cases are shown by induction. </p>
|
2,358,838 | <p>I can see the answer to this in my textbook; however, I am not quite sure how to solve this for myself . . . the book has the following:</p>
<blockquote>
<p>To take advantage of the inductive hypothesis, we use these steps:</p>
<p>$ 7^{(k+1)+2} + 8^{2(k+1)+1} = 7^{k+3} + 8^{2k+3} $</p>
<p>$$
= 7\cdot7^{k+2} + 8^{2}\cdot8^{2k+1}\\
= 7\cdot7^{k+2} + 64\cdot8^{2k+1}\\
= 7(7^{k+2}+8^{2k+1})+57\cdot8^{2k+1}\\ $$</p>
</blockquote>
<p>While the answer is apparent to me <em>now</em>; how exactly would I go about figuring out a similar algebraic manipulation if I were to see something like this on a test? Is there an algorithm or a way of thinking about how to break this down that I'm missing? I think I'm most lost regarding the move from the second to last and last equations.</p>
<p><em>Source: Discrete Mathematics and its Applications (7th ed), Kenneth H. Rosen (p.322)</em></p>
| Bram28 | 256,001 | <p>The key is to get <em>very clear</em> on what you <em>have</em>, and what you <em>want</em>.</p>
<p>For the inductive step, you <em>have</em> the inductive assumption:</p>
<p><span class="math-container">$7^{k+2}+8^{2k+1}$</span> is divisible by <span class="math-container">$57$</span></p>
<p>And you <em>want</em> to show that:</p>
<p><span class="math-container">$7^{(k+1)+2}+8^{2(k+1)+1}$</span> is divisible by <span class="math-container">$57$</span></p>
<p>So, you should think to yourself: OK, I want to show that <span class="math-container">$7^{(k+1)+2}+8^{2(k+1)+1}$</span> is divisible by <span class="math-container">$57$</span>, and of course at some point I want to use my inductive hypothesis that <span class="math-container">$7^{k+2}+8^{2k+1}$</span> is divisible by <span class="math-container">$57$</span> to help me with this. So: I probably want to manipulate/rewrite the expression <span class="math-container">$7^{(k+1)+2}+8^{2(k+1)+1}$</span> so that I get some term <span class="math-container">$7^{k+2}+8^{2k+1}$</span> in there. How can I do that? Well, I have <span class="math-container">$1$</span> extra <span class="math-container">$7$</span> in the <span class="math-container">$7^{(k+1)+2}$</span> term, and I have <span class="math-container">$2$</span> extra <span class="math-container">$8$</span>'s in the <span class="math-container">$8^{2k+1}$</span> term, so I should pull those out:</p>
<p><span class="math-container">$$7^{(k+1)+2}+8^{2(k+1)+1} =$$</span></p>
<p><span class="math-container">$$7^{(k+2)+1}+8^{(2k+1)+2} =$$</span></p>
<p><span class="math-container">$$7\cdot 7^{k+2}+8^2\cdot 8^{2k+1}=$$</span></p>
<p><span class="math-container">$$7\cdot 7^{k+2}+64\cdot 8^{2k+1}$$</span></p>
<p>Now, it is not quite clear yet how to isolate a <span class="math-container">$7^{k+2}+8^{2k+1}$</span>, but at this point you might notice something: we got a 7 ... and we got a 64 ... the difference of which is 57 (!). Hmmm.... OK, let's try this:</p>
<p><span class="math-container">$$7\cdot 7^{k+2}+64\cdot 8^{2k+1}=$$</span></p>
<p><span class="math-container">$$7\cdot 7^{k+2}+(57+7)\cdot 8^{2k+1}=$$</span></p>
<p><span class="math-container">$$7\cdot 7^{k+2}+57\cdot 8^{2k+1}+7\cdot 8^{2k+1}=$$</span></p>
<p>(and at this point you'll see it ...)</p>
<p><span class="math-container">$$57\cdot 8^{2k+1}+7\cdot 7^{k+2}+ 7\cdot 8^{2k+1}=$$</span></p>
<p><span class="math-container">$$57\cdot 8^{2k+1}+7\cdot (7^{k+2}+ 8^{2k+1})$$</span></p>
<p>The term <span class="math-container">$57\cdot 8^{2k+1}$</span> is clearly divisible by <span class="math-container">$57$</span>, and we just isolated the <span class="math-container">$7^{k+2}+ 8^{2k+1}$</span> for the second term, so that is divisble by <span class="math-container">$57$</span> as well, meaning that their sum is divisible by <span class="math-container">$57$</span> as well.</p>
|
65,304 | <p>I have a plane curve $C$ described by parametric equations $x(t)$ and $y(t)$ and a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. The line integral of $f$ along $C$ is the area of the "fence" whose path is governed by $C$ and height is governed by $f$.</p>
<p><img src="https://i.stack.imgur.com/4rmZy.png" alt="enter image description here"></p>
<p>How can I generate a picture of the "fence" in Mathematica?</p>
<p>For the sake of a concrete example, let's borrow from Stewart (since I already borrowed his picture). For $0 \leq t \leq \pi$, define
$$
\begin{align*}
x(t) &= \cos t\\
y(t) &= \sin t\\
f(x,y) &= 2 + x^2y
\end{align*}
$$
so that
$$
\begin{align*}
f(x(t),y(t)) &= 2 + \cos^2 t \sin t.
\end{align*}
$$</p>
| Dr. belisarius | 193 | <pre><code>opts = {MeshFunctions -> (#4 &),
MeshShading -> {{Opacity[#2], #1}, {Opacity[#2/2], #1}},
BoxRatios -> {1, 1, 1/2},
BoundaryStyle -> Directive[Thin, Blue]} &;
Show[
ParametricPlot3D[{Cos[t], Sin[t], z (2 + Sin[t] Cos[t]^2)}, {t, 0, π/2}, {z, 0, 1},
Evaluate@opts[Green, .4]],
ParametricPlot3D[{Cos[t], z Sin[t], 0 }, {t, 0, π/2}, {z, 0, 1},
Evaluate@opts[Blue, .2]],
ParametricPlot3D[{z Cos[t], Sin[t], 0 }, {t, 0, π/2}, {z, 0, 1},
Evaluate@opts[Blue, .2]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/BU5cs.png" alt="Mathematica graphics"></p>
|
1,889,957 | <p>I'm a bit rusty on my math notations and I'd like to write that:</p>
<blockquote>
<p>It exists a unique element $z$ such that $z$ belongs to the collection of values returned by $f(x,y)$</p>
</blockquote>
<p>Honestly I'm not just rusty I'm also mostly ignorant of math except from basic functions and basic matrix operations.</p>
<p>I'm in the context of computer programming and I want to write down a specification, and for my own curiosity (and fun) I was wondering how this would be written in a more scientific way.</p>
<p>I'd go with something like:</p>
<blockquote>
<p>$\exists z\in S$ such that...</p>
</blockquote>
<p>And then I'm lost with how to specify that $S$ is the result of $f(x,y)$.</p>
<p>Some usage of $P(z)$ maybe ?</p>
<p>Also $S$ means "set" right? So it doesn't work because $z$ may be present multiple times, but IDK if there's a symbol for such "collection".</p>
<p>I've googled around but it's a bit hard to find the right keywords for searching something like this.</p>
<p>Thank you.</p>
<p><strong>EDIT</strong>:</p>
<p>I knew I'd make a mistake while posting this... I've mistakenly named $x$, $x$, leading to the confusion that it is the same $x$ that is in $f(x,y)$, while actually it is not.</p>
<p>So I have renamed it $z$, sorry about that.</p>
<p><strong>EDIT 2</strong>:</p>
<p>There are multiples solutions that have been provided in the answers and for this I'm thankful, but I can't identify if one matches what I want.</p>
<p>And there are also a lot of questions which I believe are due to me not giving enough details or not expressing myself correctly, and I realize now that I have made a mistake on the way so I will try to add more details and maybe it will help to make the answers converge.</p>
<p>I have a function, say $f$, that given two arguments, say $x\in X$ and $y\in Y$, will return a collection of values, say $S$ whose values are taken from $Z$.</p>
<p>And I want $S$ to contain only $z$ (possibly multiple times).</p>
<p>Given $S1$ and $S2$ the respective results of $f(x1,y1)$ and $f(x2,y2)$, there can not be a given $z$ that would be present in both $S1$ and $S2$.</p>
<p>For the record, $y1$ may be equal to $y2$.</p>
<p>Also $y$ depends on $x$ so I guess we start with the second part of what @celtschk said in his comment and simplify:</p>
<blockquote>
<p>$$S = \bigg\{f(x, g(x)) : x ∈ X \bigg\} ⊂ Z$$</p>
</blockquote>
<p>But the first part should be:</p>
<blockquote>
<p>"$z$ exists at least once and is unique in $S$"</p>
</blockquote>
<p>and I don't know how to write that :)</p>
| S.C.B. | 310,930 | <p>$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac} \ge \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}=\frac{(a+b+c+d)^2}{(a+c)(b+d)+2ac+2bd}$$</p>
<p>From Cauchy. </p>
<p>$$(a+b+c+d)^2=(a+c)^2+(b+d)^2+2(a+c)(b+d) \ge 4ac +4bd+2(a+c)(b+d)$$</p>
<p>Since $(x+y)^2 \ge 4xy$.</p>
<p>So $$\sum_{cyc}\frac{a}{b+c} \ge 2 >\frac{1}{2}$$</p>
|
202,041 | <p>I have been trying to determine the number of metrics of constant curvature on a surface of genus $n$, say $\Sigma$. For low values, the answer is clear, the moduli space is a point for the sphere, and is two dimensional for the torus, but the higher dimensional cases stump me, and I am unable to find the result. Any help or a reference would be appreciated.</p>
| Alexandre Eremenko | 25,510 | <p>First on terminology. "Riemannian surface" is a surface already equipped with a Riemannian metric. So the question "how many metrics of constant curvature exist on a Riemannian surface" makes sense only if you state what is the relation between the metric of
constant curvature and the original metric on the Riemannian surface.</p>
<p>"Riemann surface" is a surface equipped with a conformal structure.</p>
<p>You probably mean "on a RIEMANN surface", this means that the conformal structure is fixed. Then the answer is: one parametric family. Parameter is just the scaling factor.</p>
<p>For the sphere, these metrics are of positive curvature, for the plane, punctured plane and tori of zero curvature, and for the rest of Riemann surfaces negative curvature.</p>
<p>If you change your question and ask about "a topological orientable surface", so that the conformal structure is not fixed then there are usually many conformal structures (and for each there is a metric as above). For example, in the case of compact surfaces of genus $g>1$ the number of conformal structures depends on $6g-6$ parameters. For a torus ($g=1$) there is a 2-parametric family, for the sphere, the structure is unique. For the disk, there are exactly $2$ different structures,
for the annulus, a $1$-parametric family (of non-degenerate annuli) and two other structures (degenerate annuli), etc. (I always mean real parameters when I count, though in many cases there exists also complex analytic structure on the moduli spaces of conformal structures).</p>
<p>All these statements are consequences of the general Uniformization theorem. Some references where a complete proof is given are:</p>
<p>Ahlfors, Conformal invariants</p>
<p>Hubbard, Teichmuller theory</p>
<p>H. P. de Saint-Gervais, Uniformisation des surfaces de Riemann.</p>
<p>(The last book is completely devoted to the discussion of the history, meaning and various proofs of this theorem).</p>
|
1,167,880 | <p>Given $a = [-5, 8, 1]$ and $b = [2, -7, -3]$, find a vector $c$ such that $a \cdot (b × c) = 0$</p>
<p>I don't know how to get it, I've been looking for examples, but I don't know..</p>
| Timbuc | 118,527 | <p>Proof for you to understand and check that $\;f(x)=x^2\;$ is not u.c. on $\;[0,\infty)\;$:</p>
<p>$$x_n:=\sqrt n\implies |x_{n+1}-x_n|\xrightarrow[n\to\infty]{}0\;,\;\;\text{yet nevertheless}\;\;|f(x_{n+1})-f(x_n)|\rlap{\;\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$</p>
|
1,167,880 | <p>Given $a = [-5, 8, 1]$ and $b = [2, -7, -3]$, find a vector $c$ such that $a \cdot (b × c) = 0$</p>
<p>I don't know how to get it, I've been looking for examples, but I don't know..</p>
| Andrew D. Hwang | 86,418 | <p>$\DeclareMathOperator{\sgn}{sgn}\newcommand{\Reals}{\mathbf{R}}\newcommand{\eps}{\varepsilon}$Uniform continuity is a "global" property, depending not only on the formula(s) defining a function but on the domain $X$. To see why, compare the definitions of "$f$ is continuous on $X$" and "$f$ is uniformly continuous on $X$":</p>
<ul>
<li><p>For every $x$ in $X$ and every $\eps > 0$, there exists a $\delta > 0$ such that if $y \in X$ and $|x - y| < \delta$, then $|f(x) - f(y)| < \eps$. (Here, $\delta$ implicitly depends on both $\eps$ and $x$; there may or may not exist a positive lower bound on $\delta$ as $x$ ranges over $X$.)</p></li>
<li><p>For every $\eps > 0$, there exists a $\delta > 0$ such that if $x$, $y$ are in $X$ and $|x - y| < \delta$, then $|f(x) - f(y)| < \eps$. (Here, $\delta$ depends only on $\eps$; a <em>single</em> $\delta$ "works" for every $x$ in $X$. Whether or not this condition is true may depend on the set $X$.)</p></li>
</ul>
<p>Consider the squaring function, $f(x) = x^{2}$, on an unspecified domain $X \subseteq \Reals$. Since
$$
|f(x) - f(y)| = |x^{2} - y^{2}| = |x - y|\, |x + y|,
$$
$f$ is uniformly continuous if $X$ is bounded: If $X \subset [-M, M]$, then $|x + y| \leq |x| + |y| \leq 2M$ on $X$ by the triangle inequality. For every $\eps > 0$ we can choose $\delta = \eps/(2M)$ to see $f$ is uniformly continuous. (Boundedness of $X$ is not necessary; think of what happens for $X = \mathbf{Z}$, the set of integers: You can take $\delta = 1/2$ <em>no matter what function you're considering</em>.)</p>
<p>Another example often surprises students the first time they see it, and may hightlight the global (domain-dependent) nature of uniform continuity: Let $X = \Reals \setminus\{0\}$, and define $\sgn:X \to \Reals$ by
$$
\sgn(x) = \frac{x}{|x|}
= \begin{cases}
1 & \text{if $x > 0$,} \\
-1 & \text{if $x < 0$.}
\end{cases}
$$
The function $\sgn$ is <em>locally constant</em>: For every non-zero $x_{0}$, $\sgn$ is constant on the open interval with endpoints $0$ and $2x_{0}$; that's "as continuous as you can get", pointwise.</p>
<p>But $\sgn$ is <em>not uniformly continuous</em> on $X$: For every $\delta > 0$, the points $x = -\delta/3$ and $y = \delta/3$ satisfy $|x - y| < \delta$, but are of opposite sign, so $|\sgn(x) - \sgn(y)| = 2$.</p>
<p>I leave to you the fun of analyzing the same formula on the set $\Reals\setminus[0, 0.0001]$ obtained by removing an interval of positive length that contains $0$.</p>
|
390,129 | <p>Let <span class="math-container">$O$</span> be a <span class="math-container">$d$</span>-dimensional rotation matrix (i.e., it has real entries and <span class="math-container">$OO^T = O^TO = I$</span>). Let <span class="math-container">$\mathbf{x}$</span> be a uniformly random bitstring of length <span class="math-container">$d$</span>, i.e., <span class="math-container">$\mathbf{x} \sim U(\{0,1\}^d)$</span>. In other words, <span class="math-container">$\mathbf{x}$</span> is a vertex of the Hamming cube, selected uniformly at random. I would like to show that there exists a <span class="math-container">$C > 0$</span> such that
<span class="math-container">$$\mathbb{P}\left[\|O\mathbf{x}\|_1 \leq \frac{d}{4}\right] \leq 2^{-Cd}.$$</span>
I am horribly stuck, any ideas on how to approach this problem would be very much appreciated. Below are some of my own attempts. This question is cross-posted at math stack exchange <a href="https://math.stackexchange.com/questions/4099958/probability-of-ell-1-norms-of-vertices-of-the-rotated-hamming-cube">here</a>.</p>
<hr>
<p><strong>Observation 1:</strong> If <span class="math-container">$O = I$</span>, then the statement holds.</p>
<p>If <span class="math-container">$O = I$</span>, then <span class="math-container">$\|O\mathbf{x}\|_1 = \|\mathbf{x}\|_1$</span> is simply the number of ones in the bitstring. Among the <span class="math-container">$2^d$</span> choices for <span class="math-container">$\mathbf{x}$</span>, the number of choices that satisfies <span class="math-container">$\|\mathbf{x}\|_1 \leq d/4$</span> is</p>
<p><span class="math-container">$$1 + \binom{d}{1} + \binom{d}{2} + \cdots + \binom{d}{\lfloor d/4\rfloor} \leq 2^{dH(\lfloor d/4\rfloor/d)} \leq 2^{dH(1/4)},$$</span>
hence the probability is upper bounded by <span class="math-container">$2^{-d(1-H(1/4))}$</span>. Here, <span class="math-container">$H(\cdot)$</span> is the binary entropy function, i.e., <span class="math-container">$H(p) = -p\log_2(p) - (1-p)\log_2(1-p)$</span>.</p>
<p><strong>Observation 2:</strong> Numerical experiments support this result. Below is a plot of the probability versus the dimension, where <span class="math-container">$O$</span> is selected at random:</p>
<p><a href="https://i.stack.imgur.com/sI9dn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/sI9dn.png" alt="Plot of the probability versus the dimension" /></a></p>
<p>The blue line is the probability. The orange line is the bound derived in the case where <span class="math-container">$O = I$</span>.</p>
<p>For comparison, here is the same numerical experiment, but with <span class="math-container">$O = I$</span>:</p>
<p><a href="https://i.stack.imgur.com/ksyxQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ksyxQ.png" alt="enter image description here" /></a></p>
<p>Thus, it appears that the introduction of <span class="math-container">$O$</span> decreases the probability.</p>
<p>Both plots are obtained by sampling <span class="math-container">$100000$</span> <span class="math-container">$\mathbf{x}$</span>'s at random. The code is here:</p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt
import random
from scipy.stats import ortho_group
H = lambda p : -p * np.log2(p) - (1-p) * np.log2(1-p)
C = 1 - H(1/4)
print(C)
N = 100000
ds,Ps = [],[]
for d in range(2,40):
O = ortho_group.rvs(dim = d)
# O = np.eye(d)
P = 0
for _ in range(N):
x = random.choices(range(2), k = d)
if np.linalg.norm(O @ x, ord = 1) <= d/4: P += 1/N
print(d,P)
ds.append(d)
Ps.append(P)
fig = plt.figure()
ax = fig.gca()
ax.plot(ds, Ps)
ax.plot(ds, [2**(-C*d) for d in ds])
ax.set_yscale('log')
ax.set_xlabel('d')
ax.set_ylabel('P')
plt.show()
</code></pre>
| Guillaume Aubrun | 908 | <p>We prove the weaker bound
<span class="math-container">$$ \mathbf{P} \left[ \|O \mathbb{x}\|_1 \leq \frac{cd}{\sqrt{\log d}} \right] \leq 2^{-Cd} $$</span>
for some constants <span class="math-container">$C, c$</span>.</p>
<p>Define the Gaussian mean width of a compact subset <span class="math-container">$A \subset \mathbf{R}^d$</span> as
<span class="math-container">$$ w(A) = \mathbf{E} \sup_{x \in A} \langle G,x \rangle $$</span>
where <span class="math-container">$G$</span> is a standard Gaussian vector in <span class="math-container">$\mathbf{R}^d$</span>. We use the following properties</p>
<ol>
<li>If <span class="math-container">$\pi$</span> is an orthogonal projection, then <span class="math-container">$w(\pi(A))\leq w(A)$</span>.</li>
<li>If <span class="math-container">$A = \{0,1\}^k \subset \{0,1\}^n$</span>, then <span class="math-container">$w(A)=k/\sqrt{2\pi}$</span>.</li>
<li>If <span class="math-container">$A=B_1^d$</span> (the unit <span class="math-container">$\ell_1$</span> ball), then <span class="math-container">$w(A) \leq \sqrt{2\log d}$</span>.</li>
<li><span class="math-container">$w$</span> is rotation invariant</li>
</ol>
<p>Let <span class="math-container">$A$</span> be the set of <span class="math-container">$x \in \{0,1\}^d$</span> such that <span class="math-container">$\|Ox\|_1 \leq cd/\sqrt{\log d}$</span>. We have <span class="math-container">$w(A) \leq w(cd/\sqrt{\log d} \cdot B_1^d) \leq c\sqrt{2}d$</span>.</p>
<p>If <span class="math-container">$\mathrm{card}(A) \geq 2^{dH(1/4)}$</span>, then the Sauer--Shelah lemma implies that there is a coordinate projection <span class="math-container">$\pi$</span> of rank <span class="math-container">$k=d/4$</span> such that <span class="math-container">$\pi(A)$</span> identifies with <span class="math-container">$\{0,1\}^k$</span>. Therefore, <span class="math-container">$w(A) \geq w(\pi(A))=d/4\sqrt{2\pi}$</span>. If we choose <span class="math-container">$c=1/8\pi$</span>, combining both estimates gives the bound <span class="math-container">$\mathrm{card}(A) < 2^{dH(1/4)}$</span>, as needed.</p>
|
3,451,629 | <p>Let <span class="math-container">$f : \mathbb{R} \to \mathbb{R}$</span> be a continous map then which of the following cannot be the image of </p>
<p><span class="math-container">$[0,1)$</span> under <span class="math-container">$f$</span> ?</p>
<p>(a) <span class="math-container">$0$</span></p>
<p>(b) <span class="math-container">$(0,1)$</span></p>
<p>(c) <span class="math-container">$[0,1)$</span></p>
<p>(d) <span class="math-container">$[0,1]$</span></p>
<p>Now , I know the following theorem if <span class="math-container">$f$</span> is a continuous map iff inverse image of every closed set is a closed set.</p>
<p>So, (a) and (d) must be the correct choices,
But for option (a) I can easily set <span class="math-container">$f(x) = 0$</span>, hence it must be incorrect. </p>
<p>But I don't really understand why isn't option (a) the correct choice, According to the theorem all the hypothesis are satisfied hence <span class="math-container">$0$</span> should be one of the answer.</p>
<p>Can someone please clear my doubt as why is option (a) Not the correct choice ?</p>
| lonza leggiera | 632,373 | <p>If this were a test question, and "<span class="math-container">$0$</span>" were not a typo for "<span class="math-container">$\{0\}$</span>", I would consider it an extremely unfair question, because there are two contradictory answers to the question, both of which seem to me to be reasonable. The problem is that the image of any subset of <span class="math-container">$\ \mathbb{R}\ $</span> must be a <em>subset</em> of <span class="math-container">$\ \mathbb{R}\ $</span>, whereas <span class="math-container">$0$</span> is an <em>element</em> of <span class="math-container">$\ \mathbb{R}\ $</span>.</p>
<p>As <a href="https://math.stackexchange.com/questions/3451629/which-among-the-following-cannot-be-the-image-of-interval-0-1-under-a-contin/3451667#3451667">WoolierThanThou's answer</a> has done, one can reasonably interpret the question as actually asking whether <span class="math-container">$\ \{0\}\ $</span> can be the image of <span class="math-container">$\ [0,1)\ $</span> under a continuous mapping, to which the answer is "yes", as WoolierThanThou shows.</p>
<p>However, in some contexts (in the set-theoretical foundations of Mathematics, for instance), <span class="math-container">$0$</span> is <em>defined</em> to be the <em>empty set</em>. Thus, a bright and well-informed student could very reasonably take the question to be asking whether the <em>number</em> <span class="math-container">$0$</span> itself—considered either as the empty set, or as just a number rather than a set of numbers—can be the image of <span class="math-container">$\ [0,1)\ $</span> under a continuous mapping, to which the correct answer is "no", because the image of a non-empty set must be a <em>non-empty set</em> of numbers.</p>
|
2,723,585 | <p>If $\textbf{A}$ is a square matrix, how can I prove that, by using the power series of matrices that the above equality holds?
Note that the $x \in \mathbb{N}$ and $\textbf{A}$ is a square matrix.</p>
| Tsemo Aristide | 280,301 | <p>Hint: $exp(A+B)=exp(A)exp(B)$ if $A$ commutes with $B$</p>
|
148,972 | <p>I am working with solving a linear system that becomes a tridiagonal matrix. In order to speed up the process for large matrices, I want to use sparse matrices. My problem is that the values are not constant along the bands, but change based on their horizontal position in the matrix (x position, if you will). For instance I want to do:</p>
<pre><code>A = SparseArray[{Band[{1,1}]->func1[0,j], Band[{1,2}->func2[0,j], Band[{2,1}]->func3[0,j]},{10,10}];
</code></pre>
<p>where j indexes the x position. Is there a good way to do this?</p>
| george2079 | 2,079 | <p>another way.</p>
<pre><code>Total@MapIndexed[
DiagonalMatrix[SparseArray@#, #2[[1]] - 2] &,
{func3[0, #] & /@ Range[n - 1],
func1[0, #] & /@ Range[n],
func2[0, #] & /@ Range[n - 1]}
]
</code></pre>
<p>This might be faster than using <code>Band</code> (Even for this simple input , computing the vectors dominates the timing though )</p>
|
2,900 | <p>I saved an <code>InterpolationFunction</code> in a ".mx" files using <code>DumpSave</code> on a variable that was scoped by a <code>Module</code>. Here is a stripped-down example:</p>
<pre><code>Module[{interpolation},
interpolation=Interpolation[Range[10]];
DumpSave["interpolation.mx", interpolation];
]
</code></pre>
<p>Is there a way to find out the variable name, presumably of the form <code>interpolation$nnn</code>, of the expression when I <code>Get</code> the interpolation? It is not apparent what the variable is when using</p>
<pre><code><<"interpolation.mx"
</code></pre>
<p>Next time I will not use a <code>Module</code> for scoping the save variable, but meantime I'd like to access the saved data and assign it to a new variable.</p>
| Leonid Shifrin | 81 | <p>You seem to be out of luck (although I will be happy to be proven wrong). This is a rather subtle point, related to the <code>Temporary</code> attribute and garbage-collection. I will just share a few observations. First, note that after loading of .mx file on a fresh kernel, the variable is not found anywhere, it is not among the living:</p>
<pre><code><< "interpolation.mx"
names = Flatten[Names[# <> "*"] & /@ Contexts[]];
Flatten@StringCases[names, ___ ~~ "interpolation" ~~ ___]
{"Manipulate`Dump`interpolationToAnimation",
"Manipulate`Dump`interpolationToManipulate"}
</code></pre>
<p>Here is one way to make it persist:</p>
<pre><code>Module[{interpolation},
ClearAll[interpolation];
interpolation = Interpolation[Range[10]];
DumpSave["interpolation.mx", interpolation];
]
</code></pre>
<p>Now it will be found. By using <code>ClearAll</code>, I removed the <code>Temporary</code> attribute from <code>interpolation</code>. Interestingly, another way to do this is to define a <code>DownValue</code> rather than an <code>OwnValue</code> for it:</p>
<pre><code>Module[{interpolation},
interpolation[1] = Interpolation[Range[10]];
DumpSave["interpolation.mx", interpolation];
]
</code></pre>
<p>In this case, too, it can be found, since garbage-collection of <code>Module</code>-generated variables works differently for <code>DownValues</code>. </p>
<p>I can not fully explain why this affected the code for <code>DumpSave</code>, which is still inside <code>Module</code>. One guess I have is that it gets recorded by <code>DumpSave</code> all right, with all attributes, including <code>Temporary</code>. Then, when you load it, it probably gets garbage-collected immediately after creation, due to this attribute.</p>
<p>Serializing in this manner using <code>Module</code> is also bad (although I am myself guilty of having done this, e.g. <a href="https://mathematica.stackexchange.com/questions/36/file-backed-lists-variables-for-handling-large-data/209#209">here</a>) for another reason: <code>Module</code>-generated variables are only guaranteed to be unique within a single Mathematica session. It is better to use some custom code to generate variables with names guaranteed to be unique for all sessions (which is easy to do. I recommend to read <a href="https://stackoverflow.com/questions/6860391/how-to-generate-new-unique-name-for-context">this discussion</a>). </p>
|
2,416,510 | <p>I have a matrix $A \in R^{n×n}$. I would like to choose two diagonal matrices $D_1,D_2 \in R^{n×n}$ such that $\text{cond}(D_1AD_2)$ should be minimal. How to provide such diagonal matrices? </p>
| Raffaele | 83,382 | <p>Each of the four angles inscribed in the circular segment is half the angle with vertex in the centre of the circle. Each of the $\alpha$ is concave and is $360°$ less the convex part. Adding the $4$ we get $1440°-360°=1080°$ so the sum of the $\beta$ is half that is $540°$</p>
<p>Hope this is useful</p>
<p><a href="https://i.stack.imgur.com/kzbpR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kzbpR.png" alt="enter image description here"></a></p>
|
48,726 | <p>I'm trying to plot a 3d revolution plot from a set of 2d points. These data points form a 2d curve, then we rotate that curve around y axis and get a 3d surface. @<a href="https://mathematica.stackexchange.com/users/50/50">J. M.</a> has a well explained and very helpful post at <a href="https://mathematica.stackexchange.com/a/11738/1364">here</a> which deals exactly the problem I have. However, I tried to use the method, and get a 3d surface that is very rough and not smooth.</p>
<p>Here is the 2d data points:</p>
<pre><code>points=Uncompress["1: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"];
</code></pre>
<p>and this is how it looks like</p>
<pre><code>Graphics[Line[points], Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/nscUw.png" alt="enter image description here"></p>
<p>and then I use J. M.'s method(code copied and modified from <a href="https://mathematica.stackexchange.com/a/11738/1364">here</a>) (it will take about 10 seconds to run)</p>
<pre><code>parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[points];
m = 3;
knots = Join[ConstantArray[0, m + 1], MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i, Length[points]}, {j, Length[points]}];
ctrlpts = LinearSolve[bas, points];
circPoints = {{1, 0}, {1, 1}, {-1, 1}, {-1, 0}, {-1, -1}, {1, -1}, {1,0}};
circKnots = {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1};
circWts = {1, 1/2, 1/2, 1, 1/2, 1/2, 1};
wgpts = Map[Function[pt, Append[#1 pt, #2]], circPoints] & @@@ ctrlpts;
wgwts = ConstantArray[circWts, Length[ctrlpts]];
</code></pre>
<p>and then generate the 3d surface</p>
<pre><code>Graphics3D[{Directive[EdgeForm[]],
BSplineSurface[wgpts, SplineClosed -> {False, True},
SplineDegree -> {3, 2}, SplineKnots -> {knots, circKnots},
SplineWeights -> wgwts]}, Boxed -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/Ds37G.png" alt="enter image description here"></p>
<p>We can see that the surface is not smooth. It looks like the surface is composed by flat rings. So how can we make the surface smooth?</p>
<p>Edit:</p>
<p>I think this unsmoothness may come from my data rather than the rotation process, so I tried to smooth my data using something like</p>
<pre><code>pointsSmooth=ExponentialMovingAverage[points, 1/20];
</code></pre>
<p>then I get the a smoother surface, but <code>ExponentialMovingAverage</code> seems to have removed the end points and there is a hole on the surface, which I don't want.</p>
<p><img src="https://i.stack.imgur.com/oyGlp.png" alt="enter image description here"></p>
<p>Also smoothing using a smooth constant like 1/20 largely modified the original data:</p>
<pre><code>Graphics[{Red, Line[ExponentialMovingAverage[points, 1/20]], Blue,
Line[points]}, Frame -> True, AspectRatio -> 1]
</code></pre>
<p><img src="https://i.stack.imgur.com/kyf6Z.png" alt="enter image description here"></p>
<p>So is it possible to smooth the data while keep the general shape so that it will give a better smooth surface? Or there are other ways to contract a smooth surface from the data?</p>
| m_goldberg | 3,066 | <p>You have a lot of points, so generating two interpolation functions from them, one for the bottom of the surface and the other for the top, should give a smooth surface of revolution. If the default plot isn't smooth enough, you can always increase <code>PlotPoints</code>.</p>
<pre><code>max = Max[First /@ points];
ii = Position[points, {max, _}][[1, 1]];
btm = Interpolation[points[[;; ii]]];
top = Interpolation[points[[ii ;;]]];
RevolutionPlot3D[{{t, btm[t]}, {t, top[t]}}, {t, 0., max}]
</code></pre>
<p><img src="https://i.stack.imgur.com/lXALk.png" alt="plot"></p>
|
48,726 | <p>I'm trying to plot a 3d revolution plot from a set of 2d points. These data points form a 2d curve, then we rotate that curve around y axis and get a 3d surface. @<a href="https://mathematica.stackexchange.com/users/50/50">J. M.</a> has a well explained and very helpful post at <a href="https://mathematica.stackexchange.com/a/11738/1364">here</a> which deals exactly the problem I have. However, I tried to use the method, and get a 3d surface that is very rough and not smooth.</p>
<p>Here is the 2d data points:</p>
<pre><code>points=Uncompress["1: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"];
</code></pre>
<p>and this is how it looks like</p>
<pre><code>Graphics[Line[points], Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/nscUw.png" alt="enter image description here"></p>
<p>and then I use J. M.'s method(code copied and modified from <a href="https://mathematica.stackexchange.com/a/11738/1364">here</a>) (it will take about 10 seconds to run)</p>
<pre><code>parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[points];
m = 3;
knots = Join[ConstantArray[0, m + 1], MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i, Length[points]}, {j, Length[points]}];
ctrlpts = LinearSolve[bas, points];
circPoints = {{1, 0}, {1, 1}, {-1, 1}, {-1, 0}, {-1, -1}, {1, -1}, {1,0}};
circKnots = {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1};
circWts = {1, 1/2, 1/2, 1, 1/2, 1/2, 1};
wgpts = Map[Function[pt, Append[#1 pt, #2]], circPoints] & @@@ ctrlpts;
wgwts = ConstantArray[circWts, Length[ctrlpts]];
</code></pre>
<p>and then generate the 3d surface</p>
<pre><code>Graphics3D[{Directive[EdgeForm[]],
BSplineSurface[wgpts, SplineClosed -> {False, True},
SplineDegree -> {3, 2}, SplineKnots -> {knots, circKnots},
SplineWeights -> wgwts]}, Boxed -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/Ds37G.png" alt="enter image description here"></p>
<p>We can see that the surface is not smooth. It looks like the surface is composed by flat rings. So how can we make the surface smooth?</p>
<p>Edit:</p>
<p>I think this unsmoothness may come from my data rather than the rotation process, so I tried to smooth my data using something like</p>
<pre><code>pointsSmooth=ExponentialMovingAverage[points, 1/20];
</code></pre>
<p>then I get the a smoother surface, but <code>ExponentialMovingAverage</code> seems to have removed the end points and there is a hole on the surface, which I don't want.</p>
<p><img src="https://i.stack.imgur.com/oyGlp.png" alt="enter image description here"></p>
<p>Also smoothing using a smooth constant like 1/20 largely modified the original data:</p>
<pre><code>Graphics[{Red, Line[ExponentialMovingAverage[points, 1/20]], Blue,
Line[points]}, Frame -> True, AspectRatio -> 1]
</code></pre>
<p><img src="https://i.stack.imgur.com/kyf6Z.png" alt="enter image description here"></p>
<p>So is it possible to smooth the data while keep the general shape so that it will give a better smooth surface? Or there are other ways to contract a smooth surface from the data?</p>
| xslittlegrass | 1,364 | <p>The the roughness on the surface is due to the noise in the original data, so that the first derivative is not smooth.</p>
<p>This shows the first derivative of the x and y components using original data, we can see it's very noisy</p>
<pre><code>tvals = parametrizeCurve[points];
m = 3;
knots = Join[ConstantArray[0, m + 1],
MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[
BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i,
Length[points]}, {j, Length[points]}];
ctrlpts = LinearSolve[bas, points];
df1[u_] =
Evaluate@D[
BSplineFunction[ctrlpts, SplineDegree -> 3, SplineKnots -> knots][
u], u];
Plot[{Evaluate[df1[u]], Evaluate[df1[u]][[2]]}, {u, 0, 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/El6rU.png" alt="enter image description here"></p>
<p>We need to remove that noise in order to get a smooth surface. This can be achieved using linear fit. Andy Ross presented very effective a spline model fit at <a href="https://mathematica.stackexchange.com/a/33262/1364">here</a>. I modified his code and create this function to apply to my 2d data.</p>
<pre><code>dataSmoothSplineMethod[ls_?(ArrayQ[#, 1] &), knotNum_Integer] :=
Module[{SplineModel, lth = Length[ls], knots, data},
data = Transpose[{Range[1, lth], ls}];
SplineModel[data_, deg_, knots_] :=
Block[{basis, allKnots, n, kmin, kmax}, n = Length[knots] + 2;
kmin = 0;
kmax = Ceiling[Max[data[[All, 1]]]] + 1;
basis =
Array[\[FormalX]^# &, deg + 1, 0]~Join~
Table[BSplineBasis[{deg, knots}, i, \[FormalX]], {i, 0,
Length[knots] - deg - 2}];
LinearModelFit[data, basis, \[FormalX]]];
knots = {1}~Join~Range[2, lth - 1, Round[lth/(knotNum - 2)]]~
Join~{lth};
SplineModel[data, 3, knots]
]
dataSmoothSplineMethod[ls_?(ArrayQ[#, 2] &), knotNum_Integer] :=
Module[{SplineModel, lth = Length[ls], lsx, lsy},
lsx = ls[[All, 1]];
lsy = ls[[All, 2]];
{dataSmoothSplineMethod[lsx, knotNum],
dataSmoothSplineMethod[lsy, knotNum]}
]
f2 = dataSmoothSplineMethod[points, 20];
</code></pre>
<p>and here shows the comparison of the smoothed data and original data, we can the changes are almost not visible</p>
<pre><code>Legended[Show[
ParametricPlot[Through@f2[x], {x, 1, Length[points]},
PlotPoints -> 100, PlotStyle -> Red],
Graphics[{Blue, Line[points]}]],
Placed[LineLegend[{Red, Blue}, {"spline fit", "original data"}],
ImageScaled@{0.5, 0.5}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/hV5KX.png" alt="enter image description here"></p>
<p>but the first derivative is much more smooth</p>
<pre><code>df2[u_] = {D[f2[[1]][u], u], D[f2[[2]][u], u]};
Plot[Evaluate[df2[u]], {u, 1, Length[points]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/i93f9.png" alt="enter image description here"></p>
<p>Using this smoothed data, we can create a much more smooth 3d surface</p>
<pre><code>pointsSmoothed = Table[Through@f2[x], {x, 1, Length[points]}];
parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[pointsSmoothed];
m = 3;
knots = Join[ConstantArray[0, m + 1],
MovingAverage[ArrayPad[tvals, -1], m], ConstantArray[1, m + 1]];
bas = Table[
BSplineBasis[{m, knots}, j - 1, tvals[[i]]], {i,
Length[pointsSmoothed]}, {j, Length[pointsSmoothed]}];
ctrlpts = LinearSolve[bas, pointsSmoothed];
circpointsSmoothed = {{1, 0}, {1, 1}, {-1, 1}, {-1,
0}, {-1, -1}, {1, -1}, {1, 0}};
circKnots = {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1};
circWts = {1, 1/2, 1/2, 1, 1/2, 1/2, 1};
wgpts = Map[Function[pt, Append[#1 pt, #2]], circpointsSmoothed] & @@@
ctrlpts;
wgwts = ConstantArray[circWts, Length[ctrlpts]];
Graphics3D[{Directive[EdgeForm[]],
BSplineSurface[wgpts, SplineClosed -> {False, True},
SplineDegree -> {3, 2}, SplineKnots -> {knots, circKnots},
SplineWeights -> wgwts]}, Boxed -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/1vN8K.png" alt="enter image description here"></p>
|
4,092,994 | <p>The question is</p>
<blockquote>
<p>Find the solutions to the equation <span class="math-container">$$2\tan(2x)=3\cot(x) , \space 0<x<180$$</span></p>
</blockquote>
<p>I started by applying the tan double angle formula and recipricoal identity for cot</p>
<p><span class="math-container">$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$</span>
<span class="math-container">$$\implies 7\tan^2(x)=3 \therefore x=\tan^{-1}\left(-\sqrt\frac{3}{7} \right)$$</span>
<span class="math-container">$$x=-33.2,33.2$$</span></p>
<p>Then by using the quadrants
<a href="https://i.stack.imgur.com/QFDTs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QFDTs.png" alt="quadrant" /></a></p>
<p>I was lead to the final solution that <span class="math-container">$x=33.2,146.8$</span> however the answer in the book has an additional solution of <span class="math-container">$x=90$</span>, I understand the reasoning that <span class="math-container">$\tan(180)=0$</span> and <span class="math-container">$\cot(x)$</span> tends to zero as x tends to 90 however how was this solution fou<strong>n</strong>d?</p>
<p>Is there a process for consistently finding these "hidden answers"?</p>
| I am a person | 806,777 | <p>In your second equation, the only way you can get from that equation to <span class="math-container">$7\tan^2{x} = 3$</span>, is by assuming that <span class="math-container">$\tan{x}$</span> and <span class="math-container">$1 - \tan^2{x}$</span> is not equal to <span class="math-container">$0$</span> or infinity because you can't divide by <span class="math-container">$0$</span> or infinity. But the hidden case is that <span class="math-container">$1 - \tan^2{x} = 0$</span> or infinity or that <span class="math-container">$\tan{x} = 0$</span> or infinity.</p>
<p>We only see that one case works, and it is when <span class="math-container">$x = \boxed{90}$</span> degrees.</p>
|
4,092,994 | <p>The question is</p>
<blockquote>
<p>Find the solutions to the equation <span class="math-container">$$2\tan(2x)=3\cot(x) , \space 0<x<180$$</span></p>
</blockquote>
<p>I started by applying the tan double angle formula and recipricoal identity for cot</p>
<p><span class="math-container">$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$</span>
<span class="math-container">$$\implies 7\tan^2(x)=3 \therefore x=\tan^{-1}\left(-\sqrt\frac{3}{7} \right)$$</span>
<span class="math-container">$$x=-33.2,33.2$$</span></p>
<p>Then by using the quadrants
<a href="https://i.stack.imgur.com/QFDTs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QFDTs.png" alt="quadrant" /></a></p>
<p>I was lead to the final solution that <span class="math-container">$x=33.2,146.8$</span> however the answer in the book has an additional solution of <span class="math-container">$x=90$</span>, I understand the reasoning that <span class="math-container">$\tan(180)=0$</span> and <span class="math-container">$\cot(x)$</span> tends to zero as x tends to 90 however how was this solution fou<strong>n</strong>d?</p>
<p>Is there a process for consistently finding these "hidden answers"?</p>
| Arthur | 15,500 | <p>The moment you substitute <span class="math-container">$\cot x\mapsto \frac1{\tan x}$</span>, you are implicitly assuming that <span class="math-container">$x\neq 90^\circ$</span>, because that's required for that substitution to make sense. So that's a case you have to manually check in the original equation because it might be a solution that disappears (and in this case it indeed turned out to be a solution that disappears).</p>
|
4,300,993 | <p>In this paper <a href="https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/S0273-0979-2016-01556-4.pdf" rel="nofollow noreferrer">Five stages of accepting constructive mathematics</a> on page 484 (shown in the image below) it contrastingly shows the use of the axiom of choice (<span class="math-container">$\sf AC$</span>) in the first proof and avoidance of its usage in the second proof. However I fail to see how the two proofs are distinct because it seems that the second proof appears exactly like the first except with the absence of the subscript <span class="math-container">$x$</span> in <span class="math-container">$\epsilon_x$</span>. In the image below it says,</p>
<blockquote>
<p>"Such hidden choice happens whenever we introduce a new symbol and
subscript it with another one to indicate a dependence, in our case <span class="math-container">$\epsilon_x$</span>."</p>
</blockquote>
<p>but in the second proof the <span class="math-container">$\epsilon$</span> still depends on the <span class="math-container">$x$</span>, so based on this dependence are we not still implicitly applying <span class="math-container">$\sf AC$</span> in the second proof? It seems to me that the two proofs are "isomorphic", the only difference is in the labelling of <span class="math-container">$\epsilon$</span>.</p>
<p>This leads me to my <strong>question</strong>:</p>
<blockquote>
<ol>
<li>If such a hidden choice happens based on dependence with subscripts appended to a symbol, how is this essentially different to the second proof where <span class="math-container">$\epsilon$</span> still depends on <span class="math-container">$x$</span> (except one didn't write down the subscripts)? How is it then that we are not implicitly applying <span class="math-container">$\sf AC$</span> in the second proof?</li>
</ol>
</blockquote>
<p><a href="https://i.stack.imgur.com/ORbIc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ORbIc.png" alt="enter image description here" /></a></p>
<p><strong>Addendum (November 14, 2021):</strong></p>
<p>I'd like to add what I understand regarding how <span class="math-container">$\sf AC$</span> is used in the first proof. I'm not sure if this is what is really going on so I very much welcome corrections!</p>
<p>Given <span class="math-container">$\sf AC$</span></p>
<blockquote>
<p><span class="math-container">$\forall S \left[ \varnothing \notin S \Rightarrow \exists f:S\to\bigcup S \enspace (\forall A \in S)(f(A)\in A)\right]$</span> <span class="math-container">$\qquad (*)$</span></p>
</blockquote>
<p>I'm under the impression that to show "For every <span class="math-container">$x\in X$</span>, take <span class="math-container">$\epsilon_x>0$</span>" requires more work using <span class="math-container">$\sf AC$</span> as we would have to construct the the following set<sup>1</sup></p>
<blockquote>
<p><span class="math-container">$\displaystyle \mathcal{F}:=\{ Z \in \mathcal{P}\mathcal{P}(X) : (\exists x\in X)(\forall Y\in \mathcal{P}(X)) \left[ Y\in Z \leftrightarrow (\exists \epsilon>0) (Y=B(x,\epsilon) \wedge (\exists A\in\mathcal{A})(B(x,2\epsilon)\subseteq A)\right] \}$</span></p>
</blockquote>
<p>and then we apply <span class="math-container">$\sf AC$</span> to the set <span class="math-container">$\mathcal{F}$</span> by substituting <span class="math-container">$\mathcal{F}$</span> into <span class="math-container">$S$</span> in <span class="math-container">$(*)$</span> to deduce the consequent of <span class="math-container">$(*)$</span>, then apply existential instantiation to the consequent so that we have <span class="math-container">$f:\mathcal{F}\to \bigcup\mathcal{F}$</span> such that <span class="math-container">$(\forall Z\in\mathcal{F})(f(Z)\in Z)$</span>. Then define<sup>2</sup> <span class="math-container">$g:X\to\mathcal{F}$</span> where</p>
<p><span class="math-container">$$g(x):=\{B(x,\epsilon): \epsilon>0 \wedge (\exists A\in\mathcal{A})(B(x,2\epsilon)\subseteq A)\} $$</span></p>
<p>It is then evident that <span class="math-container">$f\circ g :X\to \bigcup X$</span> such that <span class="math-container">$f\circ g:x\mapsto \epsilon_x$</span> is the function in the paper that includes a choice function <span class="math-container">$f$</span> (via <span class="math-container">$\sf AC$</span>). Now we can obtain the open cover <span class="math-container">$\{B(x,\epsilon_x)\}_{x\in X}$</span> via <span class="math-container">$f\circ g$</span>. Therefore by existential instantiation we have our open cover <span class="math-container">$\{B(x,\epsilon_x)\}_{x\in X}$</span>.</p>
<p>Thus I'm led to further <strong>additional questions</strong>:</p>
<blockquote>
<ol start="2">
<li>So based on this it seems like invoking <span class="math-container">$\sf AC$</span> required more work to actually spell it out? Is this what is really going on in the phrase "For every <span class="math-container">$x\in X$</span>, take <span class="math-container">$\epsilon_x>0$</span>"?</li>
</ol>
</blockquote>
<blockquote>
<ol start="3">
<li>Admittedly the way I initially (before this addendum) interpreted the first proof was really in accordance with what the second proof was saying. So when we do "introduce a new symbol and subscript it with another one to indicate dependence", does this really mean that I implicitly used <span class="math-container">$\sf AC$</span>?</li>
</ol>
</blockquote>
<blockquote>
<ol start="4">
<li>Constructing the function <span class="math-container">$g$</span> and the set <span class="math-container">$\mathcal{F}$</span> did not require <span class="math-container">$\sf AC$</span> (at least as far as I'm aware), but asserting the existence of <span class="math-container">$f$</span> necessarily requires (based on my intuitive understanding of heuristic arguments for usage of <span class="math-container">$\sf AC$</span>) <span class="math-container">$\sf AC$</span>. So how do we know a priori (if this is even possible to "know"?) that there is a plausible argument without <span class="math-container">$AC$</span> before we find one? Conversely, what types of formulations indicate to us that a priori (again, is this even possible to "know"?) we know <span class="math-container">$\sf{AC}$</span> will be necessary so that trying to find an argument without <span class="math-container">$AC$</span> is futile? Can you provide examples?</li>
</ol>
</blockquote>
<hr />
<p><em>Footnotes.</em></p>
<ol>
<li><p>Apologies for the readability as I eschewed using <span class="math-container">$\sf AC$</span> by ensuring that the set formed did not use it. The set <span class="math-container">$\{\{B(x,\epsilon) : (\exists A\in\mathcal{A})(B(x,2\epsilon)\subseteq A)\}_{x} \in \mathcal{P}\mathcal{P}(X) : x\in X \}$</span> is exactly the set <span class="math-container">$\mathcal{F}$</span> above, but I'm not sure if I needed <span class="math-container">$\sf AC$</span> to form this (In a sense it doesn't really matter if this implicitly uses <span class="math-container">$\sf AC$</span> or not since we will end up using <span class="math-container">$\sf AC$</span> to obtain the function <span class="math-container">$f:\mathcal{F}\to \bigcup\mathcal{F}$</span> anyway, but I just wanted to push and see how far I could go without using <span class="math-container">$\sf AC$</span>).</p>
</li>
<li><p>I'm aware that this is subscripting <span class="math-container">$x$</span> to the family of balls so it looks like I may have used <span class="math-container">$\sf AC$</span> but it seems to me that I haven't because I'm not making an arbitrary choice. In fact we can show that <span class="math-container">$(\forall x\in X)(\exists ! Z\in \mathcal{F})(\forall Y\in Z)\left[x\in Y\right]$</span> so in fact I did not make an arbitrary choice and <span class="math-container">$g(x)$</span> is that unique <span class="math-container">$Z$</span> that satisfies the given formula. I'm not so sure whether it is necessary to show that this formula holds in order to completely assert that choice is not used, but sufficiently so implies we did not. So we really only used choice to obtain the function <span class="math-container">$f$</span> and not <span class="math-container">$g$</span>.</p>
</li>
</ol>
| HallaSurvivor | 655,547 | <p>Here is a clear way (imo) to see that the first proof uses choice:</p>
<blockquote>
<p>Let <span class="math-container">$\mathcal{A}$</span> be an open cover of <span class="math-container">$X$</span>. Then each <span class="math-container">$x$</span> lies in some <span class="math-container">$A \in \mathcal{A}$</span>, and so each <span class="math-container">$x$</span> has an open ball <span class="math-container">$B(x, 2 \epsilon) \subseteq A$</span>. Of course, there are many choices of <span class="math-container">$\epsilon$</span> which work for a given <span class="math-container">$x$</span>. Now <strong>choose</strong> an <span class="math-container">$\epsilon_x$</span> for every <span class="math-container">$x \in X$</span>. Then <span class="math-container">$\{ B(x, 2 \epsilon_x) \mid x \in X\}$</span> is an open cover, thus has a finite subcover, and taking the minimum of the resulting <span class="math-container">$\epsilon_x$</span>s finishes the job.</p>
</blockquote>
<p>Notice what we've done here: We first argue that every <span class="math-container">$x$</span> has an associated family
<span class="math-container">$E_x \triangleq \{ \epsilon \mid B(x,2 \epsilon) \subseteq A \text{ for some } A \in \mathcal{A} \}$</span>. Since <span class="math-container">$\mathcal{A}$</span> is an open cover, we know that each <span class="math-container">$E_x$</span> is nonempty, so by <span class="math-container">$\mathsf{AC}$</span> we can choose an <span class="math-container">$\epsilon_x$</span> from each <span class="math-container">$E_x$</span>. This is what the author means by "hidden choice": We have a family of objects (in this case the <span class="math-container">$E_x$</span>s) associated to each <span class="math-container">$x$</span>, and then we want to choose one <span class="math-container">$\epsilon_x$</span> from each family. Anytime we have to make an arbitrary choice for (infinitely many) points, we need <span class="math-container">$\mathsf{AC}$</span> to actually do it. Again, in this example we know that for every <span class="math-container">$x$</span> there exists some <span class="math-container">$\epsilon$</span> which does the trick, but if you actually want to <em>choose</em> a specific <span class="math-container">$\epsilon_x$</span> for each <span class="math-container">$x$</span>, we need <span class="math-container">$\mathsf{AC}$</span> to do so.</p>
<p>Now let's take a look at the second proof.</p>
<blockquote>
<p>Let <span class="math-container">$\mathcal{A}$</span> be an open cover of <span class="math-container">$X$</span>. Note <span class="math-container">$\{ B(x, 2 \epsilon) \mid x \in X, \epsilon \in (0,\infty), B(x,2 \epsilon) \subseteq A \text{ for some } A \in \mathcal{A} \}$</span> is an open cover too. By compactness, it admits a finite subcover, and taking the minimum of the resulting <span class="math-container">$\epsilon$</span>s does what we want.</p>
</blockquote>
<p>So, what's different here? The idea is that we don't make any choices! Instead of choosing an <span class="math-container">$\epsilon$</span> for each <span class="math-container">$x$</span>, we just (inefficiently!) take <em>every possible <span class="math-container">$\epsilon$</span> at once</em>. This is a very common strategy for avoiding arbitrary choices: Instead of choosing one thing, find a way to use everything, even if it contains irrelevant information. This is the same reason that the <a href="https://en.wikipedia.org/wiki/Fundamental_group" rel="nofollow noreferrer">fundamental group</a> is not canonical (it depends on a choice of base point) while the <a href="https://en.wikipedia.org/wiki/Fundamental_groupoid" rel="nofollow noreferrer">fundamental groupoid</a> <em>is</em> canonical (we can choose <em>every</em> point to be a base point).</p>
<hr />
<p>As for your follow-up questions:</p>
<ol start="2">
<li><p>It's not infrequent that proofs requiring <span class="math-container">$\mathsf{AC}$</span> are <em>formally</em> less efficient than their <span class="math-container">$\mathsf{AC}$</span>-free counterparts (in exactly the way you've noticed). Philosophically one could say that this is because a proof without <span class="math-container">$\mathsf{AC}$</span> has computational content (which makes it easier to formalize), but I don't know of a way to make this precise. That said, oftentimes the <span class="math-container">$\mathsf{AC}$</span>-free proofs <em>feel</em> inefficient at a higher level. After all, why would we consider the whole huge family <span class="math-container">$\{ B(x, \epsilon) \mid x \in X, \epsilon \in (0,\infty) \}$</span> when the (much smaller looking) family <span class="math-container">$\{ B(x, \epsilon_x) \mid x \in X \}$</span> suffices? We as humans really like efficiency, and so it can be hard to see the option of taking everything all at once when the option of taking only what's required (even if it requires making <em>choices</em>) is available.</p>
</li>
<li><p>Introducing a symbol with a subscript implicitly assumes <span class="math-container">$\mathsf{AC}$</span> exactly when you have to arbitrarily <em>choose</em> what that symbol should mean from a set of possible options. For instance, in the above example any element of <span class="math-container">$E_x$</span> will do, but if you want to fix a specific <span class="math-container">$\epsilon_x$</span> then we have to <em>choose</em> an <span class="math-container">$\epsilon_x \in E_x$</span> for every <span class="math-container">$x$</span>! However, there are two situations where this is ok. First, if you're only making <em>finitely many</em> choices. There's nothing wrong with choosing a finite number of things -- we only need <span class="math-container">$\mathsf{AC}$</span> to simultaneously make an <em>infinite</em> number of choices! Second, if your "choices" are actually <em>canonical</em> in some sense. In the above example, no <span class="math-container">$\epsilon_x$</span> was better than any other. But suppose instead we let
<span class="math-container">$\epsilon_x = \max \{ \frac{1}{k} \mid k \in \mathbb{N}, B(x, \frac{2}{k}) \subseteq A \text{ for some } A \in \mathcal{A} \}$</span>. Provided <span class="math-container">$\mathcal{A}$</span> is definable in some sense, then our function <span class="math-container">$x \mapsto \epsilon_x$</span> is definable too! Here instead of needing to <em>choose</em> an <span class="math-container">$\epsilon_x$</span> arbitrarily, we come up with a rule for determining <span class="math-container">$\epsilon_x$</span> for us! The study of cleverly coming up with these rules to sidestep <span class="math-container">$\mathsf{AC}$</span> is part of <a href="https://en.wikipedia.org/wiki/Descriptive_set_theory" rel="nofollow noreferrer">descriptive set theory</a>, and there is a <em>wealth</em> of machinery for solving problems like this.</p>
</li>
<li><p>Knowing whether <span class="math-container">$\mathsf{AC}$</span> is necessary is a fairly subtle matter. There are a <em>ton</em> of statements that are equivalent to <span class="math-container">$\mathsf{AC}$</span> (in fact, there's an <a href="https://en.wikipedia.org/wiki/Equivalents_of_the_Axiom_of_Choice" rel="nofollow noreferrer">entire book</a> dedicated to them!) and some are quite surprising. Moreover, I don't have a ton of intuition for knowing which results necessarily depend on choice, so I can't provide any words of wisdom here. Hopefully someone more knowledgeable is able to chime in!</p>
</li>
</ol>
<hr />
<p>I hope this helps ^_^</p>
|
3,258,249 | <p><span class="math-container">$\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$</span></p>
<p>It's clear that we can simplify the limit a little bit, after which we get:</p>
<p><span class="math-container">$\lim\limits_{n\to\infty}{(\frac{1}{4})^{n}\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{4}{5})^{k}}$</span></p>
<p>I could further simplify the expression, but I feel like there's a more elegant solution. </p>
<p>Give me a hint, please</p>
| user600016 | 545,151 | <p>Using identify <span class="math-container">${n\choose r} = \frac{n}{r}\cdot {{n-1} \choose {r-1}}$</span></p>
<p><span class="math-container">$$\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$$</span></p>
<p>= <span class="math-container">$$ \lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n} \frac{n}{k}{k \choose n}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}} $$</span>.</p>
<p>Now, Divide both sides in the expansion of <span class="math-container">$(1+x)^n=nC0+nC1 x+...+nCn x^n$</span> by x and integrate both sides from 0 to 1.
Substitute the identity in the limit expression and evaluate it.</p>
|
77,504 | <p>I'm in the embarrassing situation that I want to ask a question that
was <a href="https://mathoverflow.net/questions/14175/how-to-learn-about-shimura-varieties">already asked</a>, but (for complicated reasons) never answered. I'd
like to try with a blank slate.</p>
<p>Shimura varieties show connections to a lot of interesting
mathematical subjects. They're a topic of active research and have
been of importance in number theory and the Langlands program.</p>
<p>However, the theory has a bit of a reputation: for heavy
prerequisites; for a large and difficult-to-penetrate body of
literature; for seminar talks that spend a minimum of half an hour
getting past the definitions. ("Aren't you assuming that the
polarizations are principal here?" "I don't see why that has
cocompact center.")</p>
<p>Let's suppose that a poor graduate student doesn't have the best
access to the experts, but has gone to lengths to make themselves
familiar with "the basics" on modular curves and Shimura curves.
There is still a bewildering abundance of new material and new
ideas to absorb:</p>
<ul>
<li><p>Abelian schemes.</p></li>
<li><p>Reductive algebraic groups and the switch to the adelic perspective.</p></li>
<li><p>Representation theory and the switch in perspective on
modular/automorphic forms.</p></li>
<li><p>$p$-divisible groups and their various equivalent formulations.</p></li>
<li><p>Moduli problems and geometric invariant theory.</p></li>
<li><p>Deformation theory.</p></li>
<li><p>Polarizations. (Yes, I think this deserves its own bullet point.)</p></li>
<li><p>(This is a placeholder for any and all major topics that I forgot.)</p></li>
</ul>
<p>Obviously there is a lot to learn, and there's no magic way to obtain
enlightenment.</p>
<p>But for an outsider, it's not clear where to start, what a good place
to read is, what really constitutes the "core" of the subject, or even
if one might cobble together a basic education while learning things
that will prove useful outside of this specialty.</p>
<p>Is a route from modular curves to Shimura varieties that will help
both with understanding the basics of the subject, and with getting an
idea of where to learn more?</p>
<p>Thank you (and sorry for the side commentary).</p>
<p>(Often these kinds of questions ask for "<a href="https://mathoverflow.net/questions/11219/what-is-a-good-roadmap-for-learning-shimura-curves">roadmaps</a>"; but "roadmap"
seems like it presupposes the existence of roads.)</p>
| jvo | 6,121 | <p>I think the general wisdom is that Deligne's <em>Travaux de Shimura</em> and Milne's <em>Introduction to Shimura Varieties</em> are the most comprehensive references, with the latter being somewhat lighter on prerequisites (but heavier on examples). </p>
<p>I've heard it suggested by people who work in the area that the best way to learn the theory is via special cases and examples, motivated by focused research problems. I suppose this is true of many things, though. </p>
<p>It might also help to learn about quaternionic Shimura curves first, assuming that you know a bit about modular curves. </p>
|
4,459,221 | <p>I'm looking to count ways to make <span class="math-container">$n$</span> pairs if there are two groups of <span class="math-container">$n$</span> people and each pair must consist of one person from each group.</p>
<p>My initial thought is <span class="math-container">$^nP_n = n!$</span>, as we can line the groups up, fix the positions of one line, and only permute the other line to make all possible pairings.</p>
<p>I'd also be interested in hearing the intuition behind why this is different from picking <span class="math-container">$n$</span> pairs from a single group of <span class="math-container">$2n$</span> people, which seems to have <span class="math-container">$$\frac{(2n)!}{n!2^n}$$</span> possible sets of pairs.</p>
| Stephen Donovan | 869,084 | <p>I like your reasoning for the case of two separate groups, and I believe that it is correct: we can simply look at each of the <span class="math-container">$n$</span> elements of one group in order, and for the <span class="math-container">$i^{th}$</span> element we will have <span class="math-container">$n - i + 1$</span> options when we select from the other group without replacement, giving us <span class="math-container">$\prod_{i = 1}^n n - i + 1 = \prod_{k = 1}^n k = n!$</span> total groupings. (this isn't really necessary, I just thought that expanding the argument out a bit would help us check why it's true)</p>
<hr>
<p>Regarding why the case for a single group of <span class="math-container">$2n$</span> is different, it's because the restriction of the two groups has been removed, or equivalently, we have the choice to select the groups ourselves. If we want to use a similar argument, we can consider splitting any two permutations of our <span class="math-container">$2n$</span> elements into two groups of <span class="math-container">$n$</span> and matching from the two groups as we did before.</p>
<p>This gives us <span class="math-container">$(2n)!$</span> pairs of ordered groups, but once we have our two groups we need to note that if we rearrange the two groups of <span class="math-container">$n$</span> in the same way then we get the same set of pairs, so we're overcounting by a factor of the number of orders of one of our lines of <span class="math-container">$n,$</span> which is <span class="math-container">$n!.$</span></p>
<p>This is the only way to reorder within groups without changing the pairs, but if we have that the pairs are unordered then we can also swap any two paired people between their groups without changing the pairs. For any set of pairs there are <span class="math-container">$2^n$</span> such ways we can choose to swap the <span class="math-container">$n$</span> pairs, so we also need to divide by that. Because these are the only two ways we can change the order of our <span class="math-container">$2n$</span> elements without changing the pairs, we get that the total number of ways to pair <span class="math-container">$2n$</span> people is</p>
<p><span class="math-container">$$\frac{(2n)!}{n! \cdot 2^n} = (2n-1)!!$$</span></p>
<p>where the double-factorial representation matches an intuitive method of putting everyone in a line, matching the first person with any of the other <span class="math-container">$2n - 1$</span> people, then matching the next person with any of the remaining <span class="math-container">$2n - 3,$</span> and so on, matching the <span class="math-container">$i^{th}$</span> person considered with any of the remaining <span class="math-container">$2n - 2i + 1$</span> people, giving us</p>
<p><span class="math-container">$$\prod_{i = 1}^n 2n - 2i + 1 = \prod_{k = 1}^n 2k - 1 = (2n - 1)!!$$</span></p>
|
4,459,221 | <p>I'm looking to count ways to make <span class="math-container">$n$</span> pairs if there are two groups of <span class="math-container">$n$</span> people and each pair must consist of one person from each group.</p>
<p>My initial thought is <span class="math-container">$^nP_n = n!$</span>, as we can line the groups up, fix the positions of one line, and only permute the other line to make all possible pairings.</p>
<p>I'd also be interested in hearing the intuition behind why this is different from picking <span class="math-container">$n$</span> pairs from a single group of <span class="math-container">$2n$</span> people, which seems to have <span class="math-container">$$\frac{(2n)!}{n!2^n}$$</span> possible sets of pairs.</p>
| true blue anil | 22,388 | <p>For the first part, you have already given a nice intuitive explanation.</p>
<p>For an intuitive explanation of the second formula, in the <span class="math-container">$(2n)!$</span> permutations of the items which we can pair serially, neither the order of the pairs nor the order <em>within</em> the pairs matter, thus the division by <span class="math-container">$n!2^n$</span>,</p>
<p>I much prefer, though, to think of the ways for sequential choices being <span class="math-container">$[(2n-1)(2n-3)(2n-5)...1 = (2n-1)!!$</span>,</p>
|
2,807,611 | <p>I know the answer is $n=6$, but can't figure out how to solve.
I tried dividing by $n!$, but didn't work because there isn't one in RHS to simplify... also tried using Gamma function properties, but didn't work either... </p>
<p>Any help would be appreciated.</p>
<p>Thanks.</p>
| N8tron | 32,820 | <p>I don't see a way to solve for n but I do see a way to see the only solution is 6</p>
<p>Multiply both sides by 5! And factor out $n!$ to get the equivalent equation</p>
<p>$$
(n^2+3n+1)n!=55(6!)
$$</p>
<p>The left hand side is a strictly increasing function for $n>0$. So for any value of the right hand side, if there is a solution it is unique.</p>
<p>A quick check off $n=6$</p>
<p>$$
36+18+1=55
$$</p>
<p>So $n=6$ is the only solution</p>
|
21,238 | <p>Would someone be able to point me to a good resource explaining step by step the process for solving inhomogenous recurrence relations? (ie something of the form $ a_n = \sum{{b_i}{a_{n-i}}} + f(n)$ )</p>
| Csar Lozano Huerta | 1,547 | <p>For the case of singular plane curves though $C\subset \mathbb{P}^2$, the result is well known. The blowup of the projective plane itself is non singular yeah, but what about the image of the curve $C$ in the blowup? That is, the proper transform of $C$? Is this proper transform $\tilde{C}$ smooth?. Such an information lies in the intersection of $\tilde{C}\cap E$, where $E$ is the exceptional divisor of the blowup. Basically, if this intersection is still singular, you will need to blowup again up to the point of getting smooth point within such intersections. This is a finite process due to the fact that every single time you blowup the curve $C$, the arithmetic genus drops by one. Since there is no negative arithmetic genus, that means the process needs to finish eventually. Notice though, that the (potential) smooth curve $\tilde{C}$ no longer live in the projective plane. You can send it back to $\mathbb{P}^2$ though, but the price to pay is that you can get singular points again. These singular points though, sometimes are milder than the original ones. But then we have got something after all!. This can be understood in terms of Cremona transformations of the plane. Here is an example of this phenomena:</p>
<p>$$X^2Y^2+X^2+Y^2=0$$ is singular at $[0:0:1]$. Apply the cremona transformation (which is a rational map from $\mathbb{P}^2$ to itself) $$\phi:[x:y:z]\mapsto [\tfrac{1}{x}:\tfrac{1}{y}:\tfrac{1}{z}]$$. Under such a map the curve $C$ becomes the
conic $Z^2+X^2+Y^2=0$ which is indeed smooth. According to another answer above, roughly speaking, this rational map corresponds to blowup $\mathbb{P}^2$ at $[1:0:0],[0:1:0],[0:0:1]$. That's why under $\phi$ the singular curve $C$ maps to a smooth one.</p>
|
21,238 | <p>Would someone be able to point me to a good resource explaining step by step the process for solving inhomogenous recurrence relations? (ie something of the form $ a_n = \sum{{b_i}{a_{n-i}}} + f(n)$ )</p>
| Blup | 5,998 | <p>For monomial ideals there is a combinatorial smoothness criterion, see "Blowups in tame monomial ideals" <a href="https://arxiv.org/abs/0905.4511" rel="nofollow noreferrer">https://arxiv.org/abs/0905.4511</a></p>
|
2,604,844 | <p>I have the folowing induction :</p>
<p>"Every graph with n vertices and zero edges is connected"</p>
<ul>
<li><p>Base:</p>
<p>For $n=1$ graph with one vertice is connected, hence a graph with $1$ vertex and zero edge.</p></li>
<li><p>Assumpution:</p>
<p>Every graph with $n-1$ vertices and zero edges is connected.</p></li>
</ul>
<p>For every graph with $n$ vertices and zero edges lets remove one vertice hence we get a graph with $n-1$ vertices and zero edges, by the assumpution the graph is connected, therefore the original graph is connected.</p>
<p>I feel that when we remove one vertice from the graph with $n$ vertice it not must be a connected graph, Yet I'm not sure if thats the flaw, or how to proof it.</p>
<p>I read about the "all horses has the same color" yet I don't find how to use the method of it here.</p>
<p>Any ideas?</p>
<p>Thanks!</p>
| Gregory Fenn | 389,331 | <p>The induction step (from $n-1$ to $n$, or from all '$k < n$' to $n$) is missing here. All you've done is proven that </p>
<pre><code>if all graphs of n-1 nodes with no edges are connected, then all graphs with n nodes and zero edges have a connected subgraph of n-1 nodes and no edges.
</code></pre>
<p>That's really not very interesting! From your basis $n=1$, you have proved successfully that the graph with $n=2$ vertices and no edges has a subgraph with one vertex that is connected. But that's ALL you've proved. [You would need to replace the basis step with $n=2$ to prove $n=3$, and the basis step obviously fails for $n=2$.]</p>
|
872,017 | <p>$$\int_0^1 xe^{\sqrt{x}} dx = ? $$</p>
<p>All I can think of is the integration by parts rule, where
$ u = x $ and $ dv= e^{\sqrt(x)} $ $ \Rightarrow du = 1$ and $ v= e^{\sqrt(x)} $ </p>
<p>The answer I get is $e^{\sqrt(x)}(x-1)$ , which is wrong.</p>
<p>Can anyone please explain in detail?</p>
| 2'5 9'2 | 11,123 | <p>You can do integration by parts like this, without substituting. Of course, substituting is fine and all, but you'll have to use Integration by Parts three times either way.</p>
<p>$$\begin{align}
\int_0^1xe^{\sqrt{x}}\,dx
&=\int_0^1\frac{2x\sqrt{x}}{2\sqrt{x}}e^{\sqrt{x}}\,dx\\
&=\int_0^12x\sqrt{x}\left(\frac{1}{2\sqrt{x}}e^{\sqrt{x}}\,dx\right)\\
&=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx\\
&=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx
\end{align}$$</p>
<p>And we've reduced the intebrand from $cx^1e^{\sqrt{x}}$ to $cx^{1/2}e^{\sqrt{x}}$. Repeat the technique two more times to bring it to $cx^{-1/2}e^{\sqrt{x}}$, and then you can just directly antidifferentiate.</p>
|
3,732,648 | <p>So I was reading Dugundji's topology and I found myself in trouble trying to prove the following.</p>
<blockquote>
<p>Prove that the following statemets are equivalent:</p>
<ul>
<li><span class="math-container">$f:X\rightarrow{}Y$</span> is continuous</li>
<li><span class="math-container">$f(A')\subseteq \overline{f(A)}$</span> for each <span class="math-container">$A\subseteq X$</span></li>
<li><span class="math-container">$\operatorname{Fr}(f^{-1}(B))\subseteq f^{-1}(\operatorname{Fr}(B))$</span></li>
</ul>
</blockquote>
<p>Where<span class="math-container">$A'$</span> is the set of all limit points of <span class="math-container">$A$</span>, <span class="math-container">$\overline{A}$</span> is the closure and <span class="math-container">$Fr(A)$</span> is the boundary</p>
<p>I was trying to prove it using the other image and pre-image properties like
<span class="math-container">$$f(\overline{A})\subseteq\overline{f(A)}\text{ and }f^{-1}(\operatorname{int}(A))\subseteq\operatorname{int}(f^{-1}(A))$$</span>
But so far I haven't been able to get anywhere, I would be glad if someone could help me.</p>
| zkutch | 775,801 | <p>You have <span class="math-container">$y=x$</span> as tangent line in <span class="math-container">$x=1$</span> and function <span class="math-container">$x^x$</span> is convex.</p>
|
3,732,648 | <p>So I was reading Dugundji's topology and I found myself in trouble trying to prove the following.</p>
<blockquote>
<p>Prove that the following statemets are equivalent:</p>
<ul>
<li><span class="math-container">$f:X\rightarrow{}Y$</span> is continuous</li>
<li><span class="math-container">$f(A')\subseteq \overline{f(A)}$</span> for each <span class="math-container">$A\subseteq X$</span></li>
<li><span class="math-container">$\operatorname{Fr}(f^{-1}(B))\subseteq f^{-1}(\operatorname{Fr}(B))$</span></li>
</ul>
</blockquote>
<p>Where<span class="math-container">$A'$</span> is the set of all limit points of <span class="math-container">$A$</span>, <span class="math-container">$\overline{A}$</span> is the closure and <span class="math-container">$Fr(A)$</span> is the boundary</p>
<p>I was trying to prove it using the other image and pre-image properties like
<span class="math-container">$$f(\overline{A})\subseteq\overline{f(A)}\text{ and }f^{-1}(\operatorname{int}(A))\subseteq\operatorname{int}(f^{-1}(A))$$</span>
But so far I haven't been able to get anywhere, I would be glad if someone could help me.</p>
| Martin R | 42,969 | <p>This can be solved without taking derivatives if you know that the logarithm is a strictly increasing function. We have (for all <span class="math-container">$x > 0$</span>)
<span class="math-container">$$
x^x \ge x \iff x \ln x \ge \ln x \iff (x-1) \ln x \ge 0
$$</span>
and the last inequality is always true because the factors on the left are simultaneously negative, zero, or positive.</p>
<p>Equality holds exactly for <span class="math-container">$x=1$</span>.</p>
|
43,282 | <p>This question is somewhat related to Tilmans notorious problem in <a href="https://mathoverflow.net/questions/17532/does-linearization-of-categories-reflect-isomorphism">this post</a>. Let $(M,\cdot)$ be a monoid with unit $1$ and set
$$(M,\cdot)^{\times} := \lbrace x \in M \mid \exists y \in M : xy=yx=1 \rbrace.$$
Let $k$ be a field (say $\mathbb Z/ 2 \mathbb Z$) and let $k[M]$ be the monoid ring of $M$ with coefficients in $k$. Consider now $$GL(k[M]) := (k[M],\cdot)^{\times}.$$</p>
<blockquote>
<p><strong>Question:</strong>(answered by Torsten Ekedahl) Can it happen that $(M,\cdot)^{\times} = \lbrace 1 \rbrace$ but $\lbrace 1 \rbrace \subsetneq GL(k[M])$?</p>
</blockquote>
<p><strong>EDIT:</strong> Torsten Ekedahl has nicely answered the above question. However, since I was really missing a condition, I will take the opportunity to change it slightly.</p>
<blockquote>
<p><strong>Question:</strong> If $(M,\cdot)^{\times} = \lbrace 1 \rbrace$, can $k[M]$ contain an invertible element $z \in GL(k[M])$, such that the coefficient of $z$ at $1$ is zero?</p>
</blockquote>
| Torsten Ekedahl | 4,008 | <p>Let $R$ be a finite dimensional algebra over $\mathbb Z/2$. Then $\{1\}\neq
R^\times$ unless $R=(\mathbb Z/2)^n$. Indeed, if $N$ is the radical of $R$, then
$1+N\subseteq R^\times$ so we may assume $R$ is semi-simple. Then $R=\prod_iR_i$
where the $R_i$ are simple algebras and $R^\times=\prod_iR_i^\times$ so we may
assume that $R$ is simple and hence a matrix algebra over some extension field
of $\mathbb Z/2$. The only such algebra with only the trivial unit is $\mathbb
Z/2$.</p>
<p>Now, pick any finite monoid $M$ with $M^\times=\{1\}$ and apply the above to
$R=\mathbb Z/2[M]$. This gives that $R^\times=\{1\}$ precisely when $M$ is a
commutative monoid where every element is idempotent. As an explicit example
where this is not the case we may let $M$ be the identity matrix plus all
non-invertible matrices of fixed size $>1$ over some finite field.</p>
|
791,719 | <p>I have this inequation:
$$5-3|x-6|\leq 3x -7$$</p>
<p>i solved this this way: </p>
<p>i said, for $x\geq6$ is the modulus positive, so I made 2 cases in which the modulus gives + or - : </p>
<p>1) for $x\geq6$ (positive): </p>
<p>$5-3x+6\leq 3x -7\\
6x\geq30\\
x\geq5$</p>
<p>2) for $x<6$ (negative): </p>
<p>$5-3(-x+6)\le3x-7\\
-13\leq-7$</p>
<p>But i dont understand what those $x\geq6$ and $x\geq5$ say to me about $x$. </p>
| John Joy | 140,156 | <p>Sometimes drawing a diagram is helpful.</p>
<p><img src="https://i.stack.imgur.com/h5mbF.png" alt="enter image description here"></p>
<p>From the diagram it is clear that
$$\tan A = x$$
$$A = \arctan(x)$$
and that also
$$\tan(\pi/2 - A) = 1/x$$
$$\arctan(\tan(\pi/2 - A)) = \arctan(1/x)$$
$$\pi/2 - A = \arctan(1/x)$$
$$\pi/2 - \arctan(x) = \arctan(1/x)$$
or alternatively,
$$\pi/2 - \arctan(1/x) = \arctan(x)$$</p>
|
3,766,042 | <p>I was doing the problem</p>
<blockquote>
<p>Find all real solutions for <span class="math-container">$x$</span> in:</p>
<p><span class="math-container">$$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$$</span></p>
</blockquote>
<p>There was a hint, to prove that <span class="math-container">$2^{x} - 1$</span> has the same sign as <span class="math-container">$x$</span>, although with basic math, if <span class="math-container">$2^0 - 1$</span> = 0, and x in this case = 0, The two expressions will have the same sign, so I am just puzzled on where to go next with this problem, any help would be welcome!</p>
<p>Messing around with basic values, I got <span class="math-container">$\pm 1, 0$</span> as the answers, although I have not checked for extra solutions, and do not know how to prove these, as it was just basic guessing and checking with the three most basic numbers for solving equations.</p>
| Michael Rozenberg | 190,319 | <p>Now, for <span class="math-container">$x\neq1$</span>, <span class="math-container">$x\neq0$</span> and <span class="math-container">$x\neq-1$</span> rewrite our equation in the following form:
<span class="math-container">$$\frac{2^x-1}{x}+\frac{2^{x^2-1}-1}{x^2-1}=0.$$</span>
Can you end it now?</p>
|
2,594,837 | <p>Let $R$ be a Noetherian domain with quotient field $K$,<br>
$I \subseteq R$ a finitely generated ideal, $I \neq (0)$ and<br>
$x \in K$ such that $x \cdot I \subseteq I$.<br>
I want to show that $x$ is <a href="https://en.wikipedia.org/wiki/Integral_element" rel="nofollow noreferrer">integral</a> over $R$.</p>
<p>Supposedly, this is also true when dropping the requirement of $R$ being Noetherian.</p>
<p>My ideas so far: </p>
<ul>
<li>Let $x = \frac a b$ with $a,b \in R$. If we could show that $\frac 1 b$ is integral over $R$, then also $\frac a b$ is integral over $R$.</li>
<li>Using something similar as <a href="https://math.stackexchange.com/a/371657/514331">https://math.stackexchange.com/a/371657/514331</a> :<br>
We have $\frac 1 b \cdot I \subseteq \frac 1 {b^2} \cdot I \subseteq \dots$<br>
If we could show that $\frac 1 b \cdot I \subseteq I$, then by $R$ being Noetherian, we would get $\frac 1 {b^n} \cdot I = \frac 1 {b^{n+1}} \cdot I$ for some $n$. But even with this, I fail to see that $\frac 1 b$ is integral over $R$.</li>
</ul>
| Jesko Hüttenhain | 11,653 | <p>Your statement is not quite correct, you need to assume that $I$ is also nonzero. In that case however, there is a very elementary way to see this.</p>
<p>So you have $I=(f_1,\ldots,f_n)$ and $x\cdot I\subseteq I$. In particular,
$$
x\cdot f_i = \sum_{j=1}^n x_{ij} f_j
$$
for a matrix $X:=(x_{ij})\in R^{n\times n}$. Denote by $f=(f_1,\ldots,f_n)\in R^n$ the vector containing the $f_i$, then we can write this as $X\cdot f = x\cdot f$. Let $\chi=\sum_{k=0}^n a_k t^k \in R[t]$ be the characteristic polynomial of $X$, then we have
$$
\chi(x)\cdot f= \sum_{k=0}^n a_k x^k\cdot f = \sum_{k=0}^n a_k X^k \cdot f = \chi(X)\cdot f = 0
$$
Since $I$ is not the zero ideal, we may assume that $f_1\ne 0$ and the above implies $\chi(x)\cdot f_1 = 0$, which as an equation inside the field $K$ implies $\chi(x)=0$. Now $\chi$ is the integral relation for $x$ which you seek.</p>
<p>PS: Do you have any interesting examples of such a situation?</p>
|
2,594,837 | <p>Let $R$ be a Noetherian domain with quotient field $K$,<br>
$I \subseteq R$ a finitely generated ideal, $I \neq (0)$ and<br>
$x \in K$ such that $x \cdot I \subseteq I$.<br>
I want to show that $x$ is <a href="https://en.wikipedia.org/wiki/Integral_element" rel="nofollow noreferrer">integral</a> over $R$.</p>
<p>Supposedly, this is also true when dropping the requirement of $R$ being Noetherian.</p>
<p>My ideas so far: </p>
<ul>
<li>Let $x = \frac a b$ with $a,b \in R$. If we could show that $\frac 1 b$ is integral over $R$, then also $\frac a b$ is integral over $R$.</li>
<li>Using something similar as <a href="https://math.stackexchange.com/a/371657/514331">https://math.stackexchange.com/a/371657/514331</a> :<br>
We have $\frac 1 b \cdot I \subseteq \frac 1 {b^2} \cdot I \subseteq \dots$<br>
If we could show that $\frac 1 b \cdot I \subseteq I$, then by $R$ being Noetherian, we would get $\frac 1 {b^n} \cdot I = \frac 1 {b^{n+1}} \cdot I$ for some $n$. But even with this, I fail to see that $\frac 1 b$ is integral over $R$.</li>
</ul>
| Mohan | 245,104 | <p>Here is a proof which uses Noetherianness (not that it is much simpler). You have inclusions of $R$-algebras, $R\subset R[x]\subset \mathrm{End}_R(I)$. Noetherian property implies the last is a finite type $R$-module and hence again by Noetherian property, so is $R[x]$. This immediately implies $x$ is integral over $R$. </p>
|
3,338,388 | <p>I tried to calculate the expression:
<span class="math-container">$$\lim_{n\to\infty}\prod_{k=1}^\infty \left(1-\frac{n}{\left(\frac{n+\sqrt{n^2+4}}{2}\right)^k+\frac{n+\sqrt{n^2+4}}{2}}\right)$$</span>
in Wolframalpha, but it does not interpret it correctly. </p>
<p>Could someone help me type it in and get the answer? Is it <span class="math-container">$1/2$</span>?</p>
<hr>
<p><strong>Edit:</strong> This was the <a href="https://www.mat.uniroma2.it/~tauraso/AMM/AMM12110.pdf" rel="nofollow noreferrer">AMM problem 12110</a>, whose deadline passed on 31 August 2019.</p>
<p>As an alternative numerical method, I could calculate the value in MS Excel.</p>
| David G. Stork | 210,401 | <pre><code>Limit[
Product[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2),
{k, 1, \[Infinity]}],
n -> \[Infinity]]
</code></pre>
|
3,066,967 | <p>Prove that <span class="math-container">$k^2+k+1$</span> is not divisible by <span class="math-container">$101$</span> for any natural <span class="math-container">$k.$</span></p>
| Seewoo Lee | 350,772 | <p>Hint: If <span class="math-container">$101|k^{2}+k+1$</span>, then <span class="math-container">$101|4(k^{2} + k + 1) = (2k+1)^{2} + 3$</span>, which implies that <span class="math-container">$\left(\frac{-3}{101}\right) = 1$</span>. However, you may show that <span class="math-container">$\left (\frac{-3}{101}\right) = -1$</span> by using the quadratic reciprocity. </p>
|
187,459 | <p>What are all 4-regular graphs such that every edge in the graph lies in a unique-4 cycle?</p>
<p>Among all such graphs, if we impose a further restriction that any two 4-cycles in the graph have at most one vertex in common, then can we characterize them in some way?</p>
<p>When is it possible to draw such a graph on a plane such that every 4-cycle is of the form: (a,c)-(b,c)-(b,d)-(a,d)-(a,c) for some a,b,c,d ?</p>
| Aaron Meyerowitz | 8,008 | <p>There should be lots of these, even with the second condition. So many that I can't imagine a classification. I'll call a $4$-cycle a <em>square.</em></p>
<p>One construction is as follows: Start with an appropriate connected graph such that </p>
<ul>
<li>Each edge is on a unique square </li>
<li>All vertices have degree $2$ or $4$. </li>
<li>Two $4$-cycles have at most one vertex in common. </li>
</ul>
<p>Then identify various degree $2$ vertices in pairs without making any new squares. So <em>appropriate</em> means you can do this. For example take an $N$ cycle for $N$ not too small, and attach a square to each edge. You could even take several of these and glue them together at vertices (in a sort of tree structure.) Here is a $12$-cycle decorated with squares and an extra square glued on just to show that we could grow this out in a wild variety of ways. I think in my identification of degree two vertices I did not create any new squares.<img src="https://i.stack.imgur.com/MSQkc.png" alt="enter image description here"></p>
<hr>
<p>The examples below are included because they look nice and are already in the comments. The graph at the top right meets my three conditions if the two red edges are deleted. I am confident that the degree $2$ vertices could be identified in pairs without creating any new $4$-cycles. </p>
<p>With the red edges one has four vertices of degree $3$ along with two edges not in a square. One could take a mirror image of that graph and add $4$ more edges to connect them, create two more squares and bring the degrees up (there are also easier remedies such as using two more edges to make a square with the red ones.)</p>
<p>I previously had a claim about the graph with the octagons which was too optimistic.It has $42$ edges missing and I won't specify how to draw them. In each half (ignoring the two curved edges) there are $22$ <em>deficient</em> edges which do not belong to a square and connect vertices of degree $3.$ I thought that one could connect corresponding degree $3$ vertices in each half. I now see that that would put the edge $AD$ on the left into two squares. I've indicated a different match for edge $CD$ and imagine that it is not hard, with a little care, to match up deficient edges on the left with those on the right to create just enough squares. I don't think one needs to preserve orientation. The bottom (fragment of a) graph could probably be treated similarly. One could even try to match up deficient edges without adding any new vertices. <img src="https://i.stack.imgur.com/JXVDS.png" alt="enter image description here"></p>
|
2,205,042 | <p>I want to show that there exists some $M$ such that for any $n$ and any $x \in [\epsilon, 2\pi - \epsilon]$ we have $\left |\sum_{m = 1}^n e^{imx} \right|\leq M$. Geometrically, it is like starting at the origin facing east, then turning left by $x$ degrees and moving forward by 1 unit of distance, and repeating this $n$ times. We want to show that the spot you end up is always less than some distance $M$ from the origin. From this geometric interpretation, it is clear that this is the case since $x$ is at least $\epsilon$ and at most $2\pi - \epsilon$. However, how can I show this algebraically? </p>
| J.-E. Pin | 89,374 | <p>Just use regular expressions instead of automata. Indeed,
<span class="math-container">$$S(L_1,L_2) = (L_1L_2)^* \cup (L_1L_2)^*L_1$$</span></p>
|
510,633 | <p>A independent variable is "the input" and the dependent variable is the "output", atleast thats how it was explained to us.</p>
<p>But if you have some random function can't both variables be seen as "affecting" the other variable?</p>
<p>For example, in $ y = 1/x$, "x" could be seen as an input and "y" the output, but isn't the opposite also true. Why can't "y" be the input and "x" be the output? </p>
<p>You could say that the dependent variable is the variable that has a unique independent variable associated with it, but thats not necessary true with the independent variable. (Think $y = x^2$. $x$ is the independent variable because for every $x$ value there is only one $y$ value, but there isn't a a unique $x$ value for every $y$ value)</p>
<p>But how can this definition be extended to functions that has unique domain values for all range values and unique range values for all domain values? Think $y= 1/x$.</p>
| Felix Marin | 85,343 | <p>\begin{align}
\left(1 + {1 \over n}\right)^{n}
&=
\sum_{\ell = 0}^{n}{n \choose \ell}{1 \over n^{\ell}}
=
\sum_{\ell = 0}^{n}
{n\left(n - 1\right)\ldots\left(n - \ell + 1\right) \over \ell!}\,
{1 \over n^{\ell}}
\\[3mm]&<
\sum_{\ell = 0}^{n}
{\left(n + 1\right)n\ldots\left(n - \ell + 2\right) \over \ell!}\,
{1 \over \left(n + 1\right)^{\ell}}
=
\sum_{\ell = 0}^{n}{n + 1 \choose \ell}\,{1 \over \left(n + 1\right)^{\ell}}
\tag1
\\[3mm]&<
\sum_{\ell = 0}^{n + 1}{n + 1 \choose \ell}\,{1 \over \left(n + 1\right)^{\ell}}
=
\left(1 + {1 \over n + 1}\right)^{n + 1}
\end{align}</p>
<p>$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\left(1 + {1 \over n}\right)^{n}
\color{#000000}{\large\ <\ }
\left(1 + {1 \over n + 1}\right)^{n + 1}
\quad}
\\ \\ \hline
\end{array}
$$</p>
<p>In $\left(1\right)$, I used your textbook information.</p>
|
3,756,970 | <p>I know which step is wrong in the following argument, but would like to have contributors' explanations of <em>why</em> it is wrong.</p>
<p>We assume below that weather forecasts always predict whether or not it is going to rain, so <em>not forecast to rain</em> means the same as <em>forecast not to rain</em>. We shall also assume that forecasts are not always right.</p>
<p>It is not generally true that the probability of rain when forecast is equal to that of its having been forecast to rain when it does rain. Indeed let us assume that
<span class="math-container">$$P(\text{R}|\text{F}_{\text R}) \neq P(\text{F}_{\text R}|\text{R}).$$</span>
But, having been forecast to rain, it will either rain or not rain (<span class="math-container">$\bar{\text{R}}$</span>), so
<span class="math-container">$$P(\text{R}|\text{F}_{\text R})+P(\overline {\text{R}}|\text{F}_{\text R})=1\ \ \ \ \ \ \mathbf{eq. 1}$$</span>
Likewise, if it rains, it will either have been forecast to rain or (we are assuming) not forecast to rain (<span class="math-container">$\overline{\text{F}_{\text R}}$</span>), so
<span class="math-container">$$P(\text{F}_{\text R}|\text{R})+P(\overline{\text{F}_{\text R}}|\text{R})=1 \ \ \ \ \ \ \mathbf{eq. 2}$$</span>
But we know that "If rain then not forecast to rain" is logically equivalent to "If forecast to rain then no rain". So the corresponding conditional probabilities must be equal, that is
<span class="math-container">$$P(\overline{\text{F}_{\text R}}|\text{R})=P(\overline {\text{R}}|\text{F}_{\text R})\ \ \ \ \ \ \ \ \ \ \ \ \mathbf{eq. 3}$$</span>
It follows immediately from <span class="math-container">$\mathbf {eqs 1,\ 2\ and\ 3}$</span> that
<span class="math-container">$$P(\text{R}|\text{F}_{\text R}) = P(\text{F}_{\text R}|\text{R}).$$</span>
which is contrary to our hypothesis.</p>
| José Carlos Santos | 446,262 | <p>No, there is no such set. You can always define the discrete metric on any set.</p>
|
2,847,419 | <p>I know that <br/>
$\sigma , \delta$ be 2 function then <br/>
$1)$ $\sigma \circ \delta$ is onto or one-one if both $\sigma $ and $\delta$ is onto or one one.<br/>
I can prove this fact .
I wanted to find the counterexample for both cases if the converse is not true.
<br/> Any Help will be appreciated </p>
| Chris2018 | 578,559 | <p>At critical points $f'(x)=0$ so</p>
<p>$0=k(x+e^x)^{k-1} \times (1+e^x)$<br>
k and $(1+e^x)$ are positive so $0=x+e^x$<br>
$x_n=-e^{x_{n+1}}$ so $x \approx -0.56714329$</p>
|
1,822,160 | <p>Why is the "column space" on the vertical in a matrix? In my mind the column space is that space that the vectors in the matrix have created.
I mean, for example take the equations:</p>
<pre><code>3x + 4y = 5
2x + 8y = 6
</code></pre>
<p>Then the matrix will be:</p>
<p>\begin{pmatrix}
3 & 4 \\
2 & 8
\end{pmatrix}</p>
<p>But why is the space defined by this matrix on the vertical?</p>
<p>Aren't the two vectors:</p>
<pre><code>3i + 4j
</code></pre>
<p>and</p>
<pre><code>2i + 8j
</code></pre>
<p>defining the space we're working in?</p>
| Aloizio Macedo | 59,234 | <p>I think what you are meaning to ask is:</p>
<blockquote>
<p>Why is the collumn space the range of the matrix, in the
appropriate bases?</p>
</blockquote>
<p>Otherwise, the collumn space is just a definition per se.</p>
<p>Therefore, we must see who is the image of the basis. That is, compute</p>
<p>$\begin{pmatrix}a_{1,1} & a_{1,2} & a_{1,3} &\cdots & a_{1,m} \\
a_{2,1} & a_{2,2} & a_{2,3} &\cdots & a_{2,m} \\
\cdots & \cdots & \cdots &\cdots & \cdots \\
\cdots & \cdots & \cdots &\cdots & \cdots \\
a_{n,1} & a_{n,2} & a_{n,3} &\cdots & a_{n,m}
\end{pmatrix} \cdot \begin{pmatrix}0 \\
\cdots \\
1 \\
\cdots \\
0
\end{pmatrix},$</p>
<p>where the $1$ appears on the $i-$th coordinate. But note that this is precisely</p>
<p>$\begin{pmatrix}a_{1,i} \\
a_{2,i} \\
\cdots \\
a_{(n-1),i} \\
a_{n,i}
\end{pmatrix}.$</p>
|
770,538 | <p>Let $\alpha$ be cylindrical helix with unit vector $u$, angle $\theta$, and arc length $s$ (measured from $0$). The only curve $\gamma$ such that $$\alpha(t)=\gamma(t)+s(t)\cos(\theta)u$$ is called the cross-section curve of the cylinder where $\alpha$ lies. </p>
<p><strong>$\bf (a)$</strong> How to show $\gamma(t)$ lies on the plane through $\alpha(0)$ orthogonal to $u$? I know I must show $$\langle \gamma(t)-\alpha(0), u\rangle=0$$ for every $t$, however I wasn't able to do that yet, maybe I'm missing some property of this kind of curve. </p>
<p><strong>Obs</strong> A cylindrical helix is a curve $\alpha$ such that $\langle \alpha^{'}(t), u\rangle=\cos(\theta)$ for every $t$, where $u$ is fixed unit vector.</p>
<p>$\bf (b)$ How to show the curvature of $\gamma$ is $\kappa/\sin^2(\theta)$ where $\kappa$ is the curvature of $\alpha$?</p>
| Mick | 42,351 | <p>Referring to the diagram below:-
<img src="https://i.stack.imgur.com/GR8wR.png" alt="enter image description here"></p>
<p>It is not that difficult to see that:-</p>
<p>(1) The two tangents are equal in length (and = 2)</p>
<p>(2) By properties of tangents, $\angle KHO = \angle PHO = x$)</p>
<p>(3) $H$, $K$, $O$, $P$ are con-cyclic. </p>
<p>(4) By ext. angle cyclic quadrilateral, $\angle POA = 2x$</p>
<p>(5) $HO = \sqrt{5}$</p>
<p>(6) $\cos x = \frac{2} {\sqrt{5}}$ and $\sin x = \frac{1}{\sqrt{5}}$</p>
<p>$a = \frac{a}{1} = \cos 2x$</p>
<p>$= cos^2x – sin^2 x$</p>
<p>$= \left(\frac {2} {\sqrt 5}\right)^2 – \left(\frac {1}{\sqrt 5}\right)^2$</p>
<p>$= \frac {3}{5}$</p>
|
94,134 | <p>I have a feeling that the following inequality should be very easy to prove:</p>
<p>$$
x^n \geq \prod_{i=1}^n{(x+k_i)},\quad\text{where } \sum_{i=1}^{n}{k_i}=0,\quad \text{and } x+k_i>0\text{ for all } i
$$</p>
<p>(and the equality only holds when all the $k_i=0$).</p>
<p>It seems intuitively obvious (when $n=2$, a square has a greater area than a rectangle with the same perimeter, when $n=3$, a cube has greater volume than a rectangular prism with the same surface area, etc.) but I can't find an appropriately easy proof.</p>
<p>I think I can show it analytically by finding the local maximum for $f(x_1,\ldots,x_n)=\prod_{i=1}^n{x_i}$ within the box $\max{x_i}=r$ in the upper-right quadrant, but I feel like there should be a neat algebraic/geometric argument, since it's such an intuitive statement.</p>
| Davide Giraudo | 9,849 | <p>We can show it by induction: for $n=2$, we have for $a+b=0$: $x^2-(x+a)(x+b)=-(a+b)x-ab=-ab\geq 0$, since $ab\leq 0$. We assume that the result is true for $n$. Let $(k_1,\ldots,k_{n+1})$ such that $\sum_{j=1}^{n+1}k_j=0$. We can assume that $k_nk_{n+1}\leq 0$. We put $k_j'=k_j$ if $j\leq n-1$, and $k_n'=k_n+k_{n+1}$. We get by induction hypothesis
$$x^n\geq \prod_{j=1}^n(x+k_j')=(x+k_n+k_{n+1})\prod_{j=1}^{n-1}(x+k_j),$$
so $$x^{n+1}\geq x(x+k_n+k_{n+1})\prod_{j=1}^{n-1}(x+k_j),$$
and we have to show that $x(x+k_n+k_{n+1})\geq (x+k_n)(x+k_{n+1})$, since $\prod_{j=1}^{n-1}(x+k_j)\geq 0$. But
\begin{align*}
x(x+k_n+k_{n+1})-(x+k_n)(x+k_{n+1})&=x(k_n+k_{n+1})-xk_{n+1}-xk_n-k_nk_{n+1}\\
&=-k_nk_{n+1}\geq 0.
\end{align*}</p>
|
1,221,639 | <p>Consider two random variables <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>. If X and Y are independent random
variables, then it can be shown that:
<span class="math-container">$$E(XY) = E(X)E(Y).$$</span></p>
<p>Let <span class="math-container">$X$</span> be the random variable that takes each of the values <span class="math-container">$-1\!\!\!$</span>, <span class="math-container">$0$</span>, and <span class="math-container">$1$</span> with probability <span class="math-container">$1/3$</span>. Let <span class="math-container">$Y$</span> be the random variable with value <span class="math-container">$Y = X^2$</span>.</p>
<blockquote>
<p>Prove that <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are not independent.</p>
<p>Prove that <span class="math-container">$E(XY) = E(X)E(Y)$</span>.</p>
</blockquote>
<hr />
<p>I understand that <span class="math-container">$E(XY) = E(X^3)$</span> since <span class="math-container">$Y = X^2$</span> so that makes each side of the equation equal to zero.</p>
<p>But I am not sure how to go about proving that <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are not independent.</p>
| megas | 191,170 | <p><strong>Hint</strong>: By the definition of independence,
two discrete random variables $X$ and $Y$ are independent if the joint probability mass function $P(X = x \text{ and } Y = y )$ satisfies
$$
P(X = x \text{ and } Y = y ) = P(X = x) \cdot P(Y = y)
$$
for all $x$ and $y$.</p>
<p>What are the possible values $x$ for the random variable $X$ in your example?
What are the possible values $y$ for $Y$?
If your $X$ and $Y$ are not independent, then you should be able to find an $x$ and $y$ pair that violates the above condition.</p>
|
1,790,612 | <p>Let $G$ be a compact connected semisimple Lie group and $\frak g$ its Lie algebra. It is known that the Killing form of $\frak g$ is negative definite. What about the Killing form $B$ of the complex semisimple Lie algebra ${\frak g}_{\Bbb C}={\frak g}\otimes\Bbb C$?</p>
<p>In particular:</p>
<blockquote>
<p>If $X\in{\frak g}_{\Bbb C}$ lies in the Cartan subalgebra ${\frak h}_{\Bbb C}\subseteq{\frak g}_{\Bbb C}$ and $B(X,X)=0$, does that imply that $X=0$?</p>
</blockquote>
| Dietrich Burde | 83,966 | <p>Take the compact simple real Lie algebra $\mathfrak{su}(2)$. The Killing form is negative definite. Its complexification is isomorphic to the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$, where the Killing form is non-degenerate, but does not satisfy $B(x,x)=0$ implies $x=0$. If $x,y,h$ is the standard basis, then $B(x,x)=0$.</p>
|
587,275 | <p>I was trying to understand why $e^{x}$ is special by finding the derivatives of other exponential functions and comparing the results. So I tried ${\rm f}\left(x\right) = 2^{x}$, but now I'm stuck.</p>
<p>Here's my final step:
<strong>$\displaystyle{{\rm f}'\left(x\right)
= \lim_{h \to 0}{2^{x}\left(2^{h} - 1\right) \over h}}$.</strong> </p>
| Community | -1 | <p>Write $2^h$ as $(e^{\log(2)})^h$. Hence, $$\dfrac{2^h-1}h = \log(2) \cdot \dfrac{e^{\log(2)h}-1}{h\log 2}$$ Now finish off using the fact that
$$\lim_{x \to 0} \dfrac{e^x-1}{x} = 1$$</p>
|
3,753,819 | <p><span class="math-container">$\textbf{Question:}$</span> Let <span class="math-container">$(X,\mathcal{F},\mu)$</span> be an arbitrary measure space. Let <span class="math-container">$\varphi: \mathbb{R} \rightarrow \mathbb{R}$</span> be continuous and satisfy for some <span class="math-container">$K>0$</span>:</p>
<p><span class="math-container">$$ \vert \varphi(t) \vert \leq K \vert t \vert, \forall t\in \mathbb{R} (*)$$</span></p>
<p>If <span class="math-container">$f \in L^p$</span>, then <span class="math-container">$\varphi \circ f$</span> belongs to <span class="math-container">$L^p$</span>. Conversely, if <span class="math-container">$\varphi$</span> does not satisfy (*), there exists a measure space <span class="math-container">$(X,\mathcal{F},\mu)$</span> and a function <span class="math-container">$f \in L^p$</span> such that <span class="math-container">$\varphi \circ f$</span> does not belong to <span class="math-container">$L^p$</span>.</p>
<p><span class="math-container">$\textbf{My attempt:}$</span> If <span class="math-container">$\varphi$</span> satisfy <span class="math-container">$(*)$</span> we have for each <span class="math-container">$(X,\mathcal{F},\mu)$</span> and <span class="math-container">$x\in X$</span></p>
<p><span class="math-container">$$ \vert (\varphi \circ f)(x) \vert = \vert \varphi(f(x)) \vert \leq K \vert f(x) \vert $$</span></p>
<p><span class="math-container">$$ \implies \vert \varphi \circ f \vert^p \leq K^p \vert f \vert^p $$</span></p>
<p>So <span class="math-container">$\varphi \circ f \in L^p$</span>. I can't solve the other statement, help please.</p>
| Oliver Díaz | 121,671 | <p>Observe that he Converse part is equivalent to the statement</p>
<p><strong>Theorem:</strong> If <span class="math-container">$\phi\in C(\mathbb{R})$</span> and for any measure space <span class="math-container">$(X,\mathscr{F},\mu)$</span>
<span class="math-container">$$\phi\circ f\in L_p(\mu)$$</span>
whenever <span class="math-container">$f\in L_p(\mu)$</span>, then <span class="math-container">$\phi$</span> satisfies (*).</p>
<p>We now prove this Theorem. We argue by contradiction. Suppose <span class="math-container">$\phi$</span> satisfies <span class="math-container">$\phi\circ f\in L_p(\mu)$</span> whenever <span class="math-container">$f\in L_p(\mu)$</span> for any measure space <span class="math-container">$(X,\mathscr{F},\mu)$</span>, but that that for any <span class="math-container">$n\in\mathbb{N}$</span> there is <span class="math-container">$t_n$</span> such that <span class="math-container">$|\phi(t_n)|> n|t_n|$</span>.</p>
<p>In particular, consider the Lebesgue space <span class="math-container">$(\mathbb{R},\mathscr{B},\lambda_1)$</span>. Since <span class="math-container">$f=\mathbb{1}_{[-1/2,1/2]}\in L_p$</span> and <span class="math-container">$\phi(f)=\phi(0)\mathbb{1}){[-1/2,1/2]^c}+\phi(1)\mathbb{1}_{[-1/2,1/2]}$</span>, it follows that
<span class="math-container">$$\|\phi\circ f\|^p_p=|\phi(1)|^p+|\phi(0)|^p\cdot\infty<\infty$$</span>
and so, <span class="math-container">$\phi(0)=0$</span>. This shows that each <span class="math-container">$|t_n|>0$</span>.</p>
<p>Fix <span class="math-container">$0<\varepsilon<p$</span> and define <span class="math-container">$a_n=\frac{1}{|t_n|^p n^{1+\varepsilon}}$</span>, and let <span class="math-container">$\{A_n:n\in\mathbb{N}\}$</span> be a sequence of pairwise disjoint measurable sets in the real line such that <span class="math-container">$\lambda_1(A_n)=a_n$</span>. Define
<span class="math-container">$$ f=\sum_nt_n\mathbb{1}_{A_n}$$</span>
Clearly <span class="math-container">$\|f\|^p_p=\sum_n|t_n|^p\frac{1}{|t_n|^p n^{1+\varepsilon}}<\infty$</span>; however
<span class="math-container">$$
\|\phi(f)\|_p=\sum_n|\phi(t_n)|^p\lambda_1(A_n)>\sum_nn^p|t_n|^p\frac{1}{|t_n|^pn^{1+\varepsilon}}=\infty$$</span>
contraditing the assumption that <span class="math-container">$\phi\circ f\in L_p$</span> whenever <span class="math-container">$f\in L_p$</span>. Therefore, there exist <span class="math-container">$k>0$</span> such that <span class="math-container">$|\phi(t)|\leq k|t|$</span> for all <span class="math-container">$t$</span>.</p>
<hr />
<p>Some remarks:</p>
<ul>
<li>The result hold for <span class="math-container">$0<p<\infty$</span>.</li>
<li>If <span class="math-container">$\phi\in C(\mathbb{R})$</span> and <span class="math-container">$(\Omega,\mathscr{F},\mu)$</span> is an arbitrary but fixed nonatomic finite measure space, then the following slightly weaker result holds:
<span class="math-container">$\phi\circ f\in L_p(\Omega,\mathscr{F},\mu)$</span> whenever <span class="math-container">$f\in L_p(\Omega,\mathscr{F},\mu)$</span> iff there is <span class="math-container">$k>0$</span> such that
<span class="math-container">$$|\phi(t)|\leq k|t|\quad\text{for all}\quad|t|>k$$</span>
<a href="https://math.stackexchange.com/a/3755045/121671">https://math.stackexchange.com/a/3755045/121671</a></li>
</ul>
|
1,812,914 | <p>What is the correct approach to solving a log equation with more than one non log value? Please demonstrate using the following equation:
$$\log(2x-1)=-x+3$$</p>
| Claude Leibovici | 82,404 | <p>As alex jordan answered, there is an analytical solution involving Lambert function. If you cannot or do not want to use it, numerical methods should be required.</p>
<p>Consider the function $$f(x)=\log(2x-1)+x-3$$ and its first derivative $$f'(x)=\frac 2{2x-1}+1$$ It is always positive since, because of the logarithn, $x>\frac12$; so $f(x)$ is increasing and there is a single root.</p>
<p>By inspection $f(1)=-2$, $f(2)=\log(3)-1>0$. So, the root is beween $1$ and $2$. </p>
<p>Let us be very lazy and start Newton iterations at $x_0=\frac 32$ and apply the iterative scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ This will generate as successive iterates $$\left(
\begin{array}{cc}
n & x_{n} \\
1 & 1.903426410 \\
2 & 1.941095404 \\
3 & 1.941304324 \\
4 & 1.941304331
\end{array}
\right)$$</p>
|
831,618 | <p>Please help me to prove that this integral converges.</p>
<p>$$\int_{0}^1 \frac{1}{\sqrt[3]{1-x^3}}\ dx $$</p>
<p>No ideas. Tried to find function which is bigger and converges, equivalent fun-s, but no result still.</p>
| Ron Gordon | 53,268 | <p>Using the factorization $a^3-b^3 = (a-b)(a^2+ab +b^2) $, rewrite the integral as </p>
<p>$$\int_0^1 dx \, \frac{(1-x)^{-1/3}}{(1+x+x^2)^{1/3}} $$</p>
<p>Sub $y=1-x$ and observe that</p>
<p>$$\int dy \, y^{-1/3} = \frac{3}{2} y^{2/3} + C$$</p>
|
147,363 | <blockquote>
<p>If $\alpha$ is an algebraic element of $\mathbb{C}$, then there is a unique non-zero polynomial $f \in \mathbb{Q}[x]$ with leading coefficient $1$ such that $f(\alpha) = 0$, and $f$ is irreducible. </p>
</blockquote>
<p>The first part of this proof would be proving that $f$ is not a unit, but what does the concept of a unit mean in the set of polynomials? I can't see how a polynomial would have a $2$ sided inverse under multiplication? </p>
| Gerry Myerson | 8,269 | <p>It would, if it were of degree zero. </p>
|
3,129,248 | <p>I am solving ordinary differential equation in <span class="math-container">$S'$</span> (dual to Schwartz space) given as:</p>
<p><span class="math-container">$y' + ay = \delta$</span>, where <span class="math-container">$\delta$</span> is a Dirac delta function.</p>
<p>The general solution of homogenous equation is <span class="math-container">$Ce^{-ax}$</span>, where <span class="math-container">$C$</span> is a constant.</p>
<p>I actually started solving it via Fourier transform, but it is not probably efficient and I got for <span class="math-container">$x \lt 0$</span> a zero solution. But according to my textbook the solution is:</p>
<p><span class="math-container">$y(x) =
\begin{cases}
(C+1)e^{-ax}, & x \gt 0 \\[2ex]
Ce^{-ax}, & x \lt 0
\end{cases}$</span></p>
<p>And no matter how long I am staring at it, I don't understand. My textbook solves it via fundamental solution of the equation given as this in general: <span class="math-container">$Lu =f$</span>, where <span class="math-container">$L$</span> is an ordinary differential operator. And then I suppose is used the gluing of the solution (which I don't know how to proceed, nor I found any good example on the internet).</p>
<p>Can anyone help me to understand this?</p>
| xpaul | 66,420 | <p>Let <span class="math-container">$Y(s)=L\{y(t)\}$</span> be the Laplace transform of <span class="math-container">$y$</span>. Using
<span class="math-container">$$ L\{y'(t)\}=sY(s)-y(0), L\{\delta\}=1, $$</span>
one has
<span class="math-container">$$ sY(s)-y(0)+aY(s)=1 $$</span>
from which,
<span class="math-container">$$ Y(s)=\frac{y(0)+1}{s+a}. $$</span>
Hence
<span class="math-container">$$ y(t)=L^{-1}\{Y(s)\}=(y(0)+1)L^{-1}\{\frac{1}{s+a}\}=(y(0)+1)e^{-at}u(t) $$</span>
where <span class="math-container">$u(t)$</span> is the unit step function.</p>
|
2,526,695 | <p>I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$</p>
<p>where $ a_{0}=a_{1}=0$</p>
<p>I know what to do when I deal with sequence in form like this:</p>
<p>$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this type of problems?
What's the general algorithm behind solving those?</p>
| adfriedman | 153,126 | <p>One approach that works to get to the final form is to take the formal power series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
and try and rewrite it in terms of itself. Applying the initial conditions where necessary:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_n x^n\\
&= a_0 + a_1 x +\sum_{n=2}^{\infty} \left(2a_{n-1}-a_{n-2} + 2^n + 4\right) x^n\\
%
&= 2\sum_{n=2}^{\infty}a_{n-1}x^n - \sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=2}^{\infty}2^nx^n + 4\sum_{n=2}^{\infty}x^n\\
%
&= 2\sum_{n=1}^{\infty}a_nx^{n+1} - \sum_{n=0}^{\infty}a_n x^{n+2} + \sum_{n=0}^{\infty}(2x)^{n+2} + 4\sum_{n=0}^{\infty}x^{n+2}\\
%
&= 2x\left(\sum_{n=0}^{\infty}a_nx^n - a_0\right) - x^2\sum_{n=0}^{\infty}a_n x^n + 4x^2\sum_{n=0}^{\infty}(2x)^n + 4x^2\sum_{n=0}^{\infty}x^n\\
%
&= 2xf(x) - x^2f(x) + 4x^2 \frac{1}{1-2x} + 4x^2 \frac{1}{1-x}\\
%
&= (2x-x^2)f(x) + 4x^2\frac{(1-x)+(1-2x)}{(1-2x)(1-x)}\\
%
&= (2x-x^2)f(x) + \frac{4x^2(2-3x)}{(1-2x)(1-x)}
\end{align}</p>
<p>Solving for $f(x)$:
$$f(x) = \frac{4x^2(2-3x)}{(1-2x)(1-x)(1-2x+x^2)} = \frac{4x^2(2-3x)}{(1-2x)(1-x)^3}$$</p>
<p>Applying partial fraction expansion and using the well-known result that
$$\sum_{n=0}^{\infty} (n+1)(n+2)\dotsb(n+m)x^n
= \frac{d^m}{dx^m}\left(\frac{1}{1-x}\right)
= \frac{m!}{(1-x)^{m+1}}$$
we get</p>
<p>\begin{align}
\sum_{n=0}^{\infty} a_n x^n &=
\frac{4x^2(2-3x)}{(1-2x)(1-x)^3}\\
&= \frac{4}{1-2x} + \frac{4}{1-x} - \frac{12}{(1-x)^2} + \frac{4}{(1-x)^3}\\
&= 4\frac{1}{1-2x} + 4\frac{1}{1-x} - 12\frac{1!}{(1-x)^{1+1}} + 2 \frac{2!}{(1-x)^{2+1}}\\
&= 4 \sum_{n=0}^{\infty} (2x)^n + 4 \sum_{n=0}^{\infty} x^n - 12 \sum_{n=0}^{\infty} (n+1) x^n + 2 \sum_{n=0}^{\infty} (n+1)(n+2) x^n\\
&= \sum_{n=0}^{\infty} \left(4\cdot 2^n + 4 - 12(n+1) + 2(n+1)(n+2)\right)x^n\\
&= \sum_{n=0}^{\infty} \left(2^{n+2} + 2n^2 - 6n -4\right)x^n
\end{align}</p>
<p>Equating coefficients, we find</p>
<p>$$a_n = 2^{n+2} + 2n^2 - 6n -4$$</p>
|
104,195 | <p>Background: This year I'll do another Group Theory course ( Open University M336 ). In the past I have used Mathematica's AbstractAlgebra package but (although visually appealing ) this is no longer sufficient (i.e. listing subgroups of <span class="math-container">$S_4$</span> takes ages). So, I want to learn more about GAP. I worked through beginner tutorials that I found via the <a href="https://www.gap-system.org" rel="nofollow noreferrer">GAP website</a>. Currently, I am not making much progress with GAP. The <a href="https://www.gap-system.org/Manuals/doc/ref/chap0.html" rel="nofollow noreferrer">reference manual</a> does not help me much at this stage.</p>
<p>Question: <strong>Which resources are available to self-study GAP?</strong> How does one become proficient in GAP? What ( books, tutorials ) should you study?</p>
| Olexandr Konovalov | 70,316 | <p>In addition to some answers given in comments to this question (cf. <a href="http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments">http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments</a>?), let me add that the following.</p>
<p>The <a href="http://www.gap-system.org/Doc/Learning/learning.html" rel="nofollow noreferrer">Learning GAP</a> section of the GAP website contains "a variety of material intended to help people to learn on their own the GAP language and the use of the GAP system". </p>
<p>Various tutorials, including the <a href="http://www.gap-system.org/Manuals/doc/tut/chap0.html" rel="nofollow noreferrer">GAP Tutorial</a>, are a good point to start, indeed. As for the reference manual, it is not assumed that one should read all its chapters sequentially. To start with, it may be worth to look at chapter titles to have a better idea of capabilities of the core GAP system, and look in more details on chapters which are most relevant to your current mathematical interests. Note that a lot of the functionality is contained in <a href="http://www.gap-system.org/Packages/packages.html" rel="nofollow noreferrer">GAP packages</a> which are developed independently and come with their own documentation.</p>
<p>It is also recommended to subscribe to the <a href="http://www.gap-system.org/Contacts/Forum/forum.html" rel="nofollow noreferrer">GAP Forum</a> where you may find not only news about the GAP
system, but also discussions and questions from other users. Reading these may provide further insight into the system. Finally, if there are any questions, please do not hesitate to send them to the GAP Forum or GAP Support.</p>
<hr>
<p><strong>Update 1</strong>: I’ve recently developed the <a href="https://carpentries-incubator.github.io/gap-lesson/" rel="nofollow noreferrer">Software Carpentry lesson "Programming with GAP"</a>. This lesson is intended for GAP beginners. It is suitable for self-study, and has been taught at various events, such as at training schools in discrete computational mathematics, organised by the <a href="http://www.codima.ac.uk/" rel="nofollow noreferrer">CoDiMa project</a> in 2015 and 2016 (see more slides from these events <a href="http://www.codima.ac.uk/schools/" rel="nofollow noreferrer">here</a>), <a href="https://www.codima.ac.uk/gsta2017/" rel="nofollow noreferrer">Groups St Andrews 2017 in Birmingham</a>, <a href="http://www-groups.mcs.st-and.ac.uk/~pgtc2018/#gap" rel="nofollow noreferrer">PGTC 2018</a> and <a href="https://lbfm-rwth.github.io/gap-in-algebraic-research-2018/" rel="nofollow noreferrer">"GAP in Algebraic Research" (2018)</a>.</p>
<hr>
<p><strong>Update 2</strong>: This book has been mentioned in the comment above, but should be made more visible: <a href="http://www.math.colostate.edu/~hulpke/CGT/howtogap.pdf" rel="nofollow noreferrer">Abstract Algebra in GAP</a> by Alexander Hulpke. From its preface: "<em>This book aims to give an introduction to using GAP with material appropriate for an undergraduate
abstract algebra course. It does not even attempt to give an introduction to abstract algebra —there
are many excellent books which do this.
Instead it is aimed at the instructor of an introductory algebra course, who wants to incorporate
the use of GAP into this course as an calculatory aid to exploration</em>".</p>
|
1,448,427 | <p>Consider a lot consisting of 3 blue balls and 1 red ball.</p>
<p>Suppose I pick 2 balls one after another without replacement, now the probability of 2 balls being blue:</p>
<p>$$\frac{3\choose 2}{4 \choose 2}$$</p>
<p>Now taking different approach using conditional probability, the solution is also:</p>
<p>$$\frac34\times\frac23$$</p>
<p>Now it turns out that both of them evaluate to $1/2$.</p>
<p>Why this happens and does it happen always?</p>
<p>Is solving the problem using combinations an efficient approach always?</p>
<p>Or am I missing something basic out here!</p>
| got it--thanks | 160,720 | <p>Here's how to draw $$\vec {BC} = -\frac 25\vec{AB}$$</p>
<p><a href="https://i.stack.imgur.com/yxJCc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yxJCc.jpg" alt="enter image description here"></a></p>
<p>I drew $\vec {AB}$ a little to the side so that it wasn't overtop $\vec{BC}$.</p>
|
1,505,336 | <p>Please can someone explain if this identity is correct:</p>
<p>|a| = $\sqrt{a^2} \ $ <p>
I thought it should be:<p> |a| = $(\sqrt{a})^2\ $</p>
<p>being that the former would produce an answer that is either positive or negative.</p>
<p>Thank you for your help.</p>
<p>PS: The full question was comparing say:</p>
<p>|2x|/|3y| with $(\sqrt{(-2x/3y)} )^2 \ $</p>
| Thomas Russell | 32,374 | <p>If we examine your suggestion, we would have:</p>
<p>$$|(-3)|=\left(\sqrt{-3}\right)^{2}=\left(\sqrt{3}i\right)^{2} = 3i^{2} = -3 \neq 3$$</p>
<p>Moreover, the $\sqrt{\cdot}$ operator is defined to be non-negative for all non-negative real arguments, so $\sqrt{x^{2}}$ does give us a well defined value $\forall x \in \mathbb{R}$.</p>
<p>In response to your comment that $x \geq 0$ is not stated, we must remember that $x^{2}:\mathbb{R}\to \mathbb{R}^{*}$, where $\mathbb{R}^{*}$ is the set of non-negative real numbers. And so the square-root operator in $\sqrt{x^{2}}$ only ever has a non-negative argument.</p>
|
3,415,845 | <p>Using the technique proof by cases, show that</p>
<p><strong>"For integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, if <span class="math-container">$xy$</span> is odd, then <span class="math-container">$x$</span> is odd and <span class="math-container">$y$</span> is odd."</strong></p>
<p>There are solutions online that go through every combination of odd and even values. (Case 1: <span class="math-container">$x$</span> is even and <span class="math-container">$y$</span> is even, Case 2: <span class="math-container">$x$</span> is even and <span class="math-container">$y$</span> is odd, Case 3: <span class="math-container">$x$</span> is odd and <span class="math-container">$y$</span> is even ... etc).</p>
<p>I don't think this is the proper way of doing a proof by cases because the cases should be in terms of what is given in the hypothesis (<span class="math-container">$xy$</span> is odd) instead of the conclusion we are trying to prove.</p>
<p>However, the only thing I can think of is Case 1: <span class="math-container">$xy$</span> is odd, but there is no other case which makes it more of a direct proof. What is the proper way to prove this?</p>
| ironX | 534,898 | <p>If <span class="math-container">$xy$</span> is odd, then:</p>
<p>case 1: <span class="math-container">$x$</span> is even, <span class="math-container">$y$</span> is odd. But <span class="math-container">$xy$</span> would be even since any integer (<span class="math-container">$y$</span>) multiple of an even number (<span class="math-container">$x$</span>) is an even number. Hence, contradiction. </p>
<p>case 2: <span class="math-container">$x$</span> is odd, <span class="math-container">$y$</span> is even. Again, contradiction. </p>
<p>case 3: both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are even. Then, again <span class="math-container">$xy$</span> would be even, since <span class="math-container">$xy$</span> is an integer (<span class="math-container">$y$</span>) multiple of an even number <span class="math-container">$x$</span>. Contradiction. </p>
<p>Therefore, if <span class="math-container">$xy$</span> is odd, then neither <span class="math-container">$x$</span> nor <span class="math-container">$y$</span> can be even. </p>
|
3,415,845 | <p>Using the technique proof by cases, show that</p>
<p><strong>"For integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, if <span class="math-container">$xy$</span> is odd, then <span class="math-container">$x$</span> is odd and <span class="math-container">$y$</span> is odd."</strong></p>
<p>There are solutions online that go through every combination of odd and even values. (Case 1: <span class="math-container">$x$</span> is even and <span class="math-container">$y$</span> is even, Case 2: <span class="math-container">$x$</span> is even and <span class="math-container">$y$</span> is odd, Case 3: <span class="math-container">$x$</span> is odd and <span class="math-container">$y$</span> is even ... etc).</p>
<p>I don't think this is the proper way of doing a proof by cases because the cases should be in terms of what is given in the hypothesis (<span class="math-container">$xy$</span> is odd) instead of the conclusion we are trying to prove.</p>
<p>However, the only thing I can think of is Case 1: <span class="math-container">$xy$</span> is odd, but there is no other case which makes it more of a direct proof. What is the proper way to prove this?</p>
| YiFan | 496,634 | <p>The proof is perfectly fine. I suspect your confusion comes from the idea that to show <span class="math-container">$p\rightarrow q$</span> by breaking into cases, we have to break <span class="math-container">$p$</span> itself up into cases and discuss. That is not really true. In particular, the statement <span class="math-container">$p\rightarrow q$</span> is equivalent to its contrapositive <span class="math-container">$\neg q\rightarrow\neg p$</span>, so we can just as well break into cases regarding <span class="math-container">$\neg q$</span> (the negation of <span class="math-container">$q$</span>) instead.</p>
<p>In particular, we want to show if <span class="math-container">$xy$</span> is odd then <span class="math-container">$x,y$</span> are odd. We may just as well show that if one of <span class="math-container">$x,y$</span> is even (i.e. they are not both odd), then <span class="math-container">$xy$</span> is even (the contrapositive statement), which you can do for example by breaking into three cases: when <span class="math-container">$x$</span> is even and <span class="math-container">$y$</span> is odd, then <span class="math-container">$x$</span> is odd and <span class="math-container">$y$</span> is even, or when <span class="math-container">$x,y$</span> are both even. Of course, there's no necessity to do that here, since it is easy to observe that if one of the factors of <span class="math-container">$xy$</span> is even then it automatically is too.</p>
|
1,692,346 | <p>I have heard of a statement like this:</p>
<blockquote>
<p>A car can technically never run out of gas (when still moving) if the driver uses half of the gas left each time.</p>
</blockquote>
<p>Is this possible (mathematics wise)?</p>
| Count Iblis | 155,436 | <p>"A car can technically never run out of gas (when still moving) if the driver uses half of the gas left each time."</p>
<p>A more practical restatement: If you reduce the speed of the car as a function of time like $v(t) = v(0)\exp(-a t)$, then the car will always keep on moving, yet the fuel consumption will be bounded. </p>
|
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