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 |
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740,323 | <p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
| Georgy | 139,717 | <p>Well, prime numbers are strongly related to the Riemann zeta function, $\zeta(s)$. This has a product representation which involves the roots of the function. The Riemann Hypothesis now states that all non-trivial roots in the complex plane lie on the "critical line":
$$Re(z)=\frac12$$
which can be thought of as a geometric feature.</p>
|
740,323 | <p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
| Patrick Das Gupta | 344,568 | <p>How about another imagery about prime numbers? A prime number $p = 1 \times p$ and hence, geometrically it is like a one-dimensional segment. On the other hand, a composite number $c= a \times b$, where $a$ and $b$ are its prime factors, is like a rectangle having an area $c$, with side lengths $a$ and $b$ . So, in general, composite numbers $c$ may be imagined as multi-dimensional rectangular parallelepipeds with volumes $c= a \times b \times c \times d \cdots$, having side lengths corresponding to their prime factors. Of course, the question that is pertinent is: does this imagery lead to interesting results or insights? </p>
|
740,323 | <p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
| Alexander Bogomolny | 42,291 | <p>A simple example: number p>2 is prime iff any equiangular p-gon with rational side lengths is regular, see, e.g., <a href="http://www.cut-the-knot.org/Outline/Geometry/EquiangularP-gon.shtml" rel="nofollow noreferrer">http://www.cut-the-knot.org/Outline/Geometry/EquiangularP-gon.shtml</a></p>
|
740,323 | <p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
| Елена Михалькова | 623,221 | <p>If I understand your question correctly, there are several geometrical models for primes (some already mentioned above), that connect algebraic deductions to geometric representations:
1. The Sieve of Eratosthenes, usually depicted on a grid of <span class="math-container">$10x \times 10y$</span>, but can be made cylindrical if columns are set at <span class="math-container">$ \tan^{−1} 1/n $</span> to rows.
2. Ulam Spiral and its variations: Krauber’s triangle, Robert Sack’s number spiral, the hexagonal cloth.
3. Omar E. Pol's prime number diagram.
4. Yuri Matiyasevich and Boris Stechkin's nomogram (a parabola sieve).
5. Gauss-Wantzel theorem (works for Fermat's primes).
6. Models of rectangular shapes in n dimensions. E.g. if n is prime, one cannot build a rectangle with sides a and b so that a and/or b is not 1.</p>
|
740,323 | <p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
| gnuvo | 942,086 | <p>Let there be an infinite triangular lattice. Sequentially enumerate the triangles in a constant direction and a 6 cell hexagon forms. However, if one enumerates the cells such that each time a prime number is encountered the direction of fill switches, the number line exits the 6 cell, but forms a new hexagon (24 cells) - confining all natural numbers to infinity. The reason for confinement is all primes (aside from 2 and 3)are 1Mod6 or 5Mod6 and get stuck in the middle of the 24 cell.
If you are interested in more details, see my woefully outdated website: hexspin.com</p>
|
595,865 | <p>A real number $r$ belonging to $[0,1]$ is said to be computable if there is a simple TM such that for each binary encoding of $n$ ($n$ is a natural number), returns the $n$th bit in the binary expansion of $r$. Give a proof that there are non-computable real numbers.</p>
| Xoff | 36,246 | <p>By a cardinal argument :</p>
<ol>
<li>You have $2^{\aleph_0}$ reals in $[0;1]$</li>
<li>You have $\aleph_0$ Turing machines.</li>
<li>There is no bijection between $\aleph_0$ and $2^{\aleph_0}$</li>
</ol>
|
915,218 | <p>I'd like some help figuring out how to calculate $n$ points of the form $(x,\sin(x))$ for $x\in[0,2\pi)$, such that the Cartesian distance between them (the distance between each pair of points if you draw a straight line between them) is the same.</p>
<p><strong>My background:</strong> I know math up to and through Algebra, have a fairly good grasp of Trig, know the Pythagorean Theorem, but only know the basic principles of Calculus (area under a curve, acceleration of acceleration, etc).</p>
<p>I'd like enough information so that I can either write a computer algorithm to compute the points directly (if possible) or, failing that, write an iterative search function that converges on the proper points. It would also be nice if you could explain how it works, although I could probably figure that out myself if I had the right equation.</p>
| ratchet freak | 17,442 | <p>It is impossible to have more than 3 points that are equidistant.</p>
<p>Having 3 points equidistant on a sine wave is impossible as it requires the points to form an equilateral triangle, and at least one of the sides would need a slope steeper than $60°$ and the steepest slope on a sine wave is $45°$.</p>
<p>The slope of the triangle is made clear when you take one of the sides to be horizontal. The other sides will have a slope of $60°$, when you rotate the triangle one slope will become steeper until it is vertical and the other sides will have a slope of 30°.</p>
|
4,126,124 | <p>I managed to come to the end of a proof regarding a determinant of a linear operator. However, I stuck at the end. I know it is kinda simple but I couldn't see it right away.</p>
<p>Here is my previous work:</p>
<p><span class="math-container">$F$</span> is a field. <span class="math-container">$V = M_{nxn}(F)$</span> is a vector space of <span class="math-container">$nxn$</span> matrices over <span class="math-container">$F$</span>, and <span class="math-container">$B$</span> is some arbitrary element in <span class="math-container">$V$</span>. Now, we are given a linear operator <span class="math-container">$T:V \rightarrow V$</span> defined by <span class="math-container">$T(A) = AB - BA$</span>. We wish to show that <span class="math-container">$det(T)=0$</span>.</p>
<p>Here is my attempt:</p>
<p>If <span class="math-container">$T$</span> has an <span class="math-container">$0$</span> as eigenvalue, then by <span class="math-container">$p(x)=det(T-\lambda I)$</span>, <span class="math-container">$p(0)=det(T)=0$</span>. Thus, showing that <span class="math-container">$\lambda = 0$</span> would suffice.</p>
<p>We split into cases.</p>
<p>Case 1: <span class="math-container">$B$</span> is invertible.</p>
<p>Then choose <span class="math-container">$A=B^{-1}$</span>, we have <span class="math-container">$T(A)=0=0*A$</span>, so <span class="math-container">$\lambda = 0$</span> is an eigenvalue.</p>
<p>Case 2: <span class="math-container">$B$</span> is not invertible.</p>
<p>Let <span class="math-container">$A$</span> be a nonzero matrix, and <span class="math-container">$\lambda \in F$</span>
<span class="math-container">$$T(A)=AB-BA= \lambda A$$</span>
<span class="math-container">$$\lambda A - AB = -BA$$</span>
<span class="math-container">$$A \lambda - AB = -BA$$</span>
<span class="math-container">$$A(\lambda I - B) = -BA$$</span>
Take determinants of each side,
<span class="math-container">$$det(A(\lambda I - B)) = det(-BA)$$</span>
<span class="math-container">$$det(A)det(\lambda I - B) = (-1)^n det(B)det(A)$$</span>
Since <span class="math-container">$B$</span> is not invertible, <span class="math-container">$det(B)=0$</span>, so,
<span class="math-container">$$det(A)det(\lambda I - B)=0$$</span></p>
<p>Here we have:
<span class="math-container">$$det(B-\lambda I) = 0$$</span>
It is clear that <span class="math-container">$\lambda$</span> can be zero. However, how can I show that <span class="math-container">$\lambda$</span> must be zero?</p>
| Henno Brandsma | 4,280 | <p>There is a mapping from the set of metrics on <span class="math-container">$X$</span> to the set of topologies on <span class="math-container">$X$</span>: to a metric assign the topology <span class="math-container">$\mathcal{T}(d)$</span> generated by <span class="math-container">$d$</span>. A topology <span class="math-container">$\mathcal{T}$</span> is metrisable precisely when it is in the image of this mapping (so there is a metric <span class="math-container">$d'$</span> on <span class="math-container">$X$</span> so that <span class="math-container">$\mathcal{T}=\mathcal{T}(d')$</span>.</p>
<p>Typically one shows a topology to be <strong>non</strong>-metrisable by noting that all topologies of the form <span class="math-container">$\mathcal{T}(d)$</span> are special and some topology does not obey one of them:</p>
<ul>
<li>all metrisable topologies are Hausdorff and perfectly normal.</li>
<li>all metrisable topologies are paracompact.</li>
<li>all metrisable topologies that are separable are also second countable, hereditarily Lindelöf etc.</li>
<li>all metrisable topologies are first countable.</li>
</ul>
<p>And the indicrete topology on a space with two or more points is not Hausdorff so cannot be metrisable.</p>
<p>There are theorems that exactly characterise a topology that is metrisable in just finitely many properties to check (and that do not mention metric). Maybe your course will cover them later.</p>
|
819,805 | <p>The polynomial remainder theorem states that when a polynomial $P(x)$ of degree $> 0$ is divided by $x-r$ ($r$ being some constant) the remainder is equal to $P(r)$, that is:<br>
$$\begin{array}l If & \quad P(x) = (x-r)Q(x)+R \\ then & \quad P(r) = R \end{array}$$
The algebra and the graphic representation make sense; the question is <em>why</em>. <em>Why</em> is the functional relationship between $r$ and $R$ the function $P(x)$? What are the <em>mechanics</em>, so to speak, that produce this result? Or is it just a fortunate algebraic "accident"?</p>
| Hakim | 85,969 | <p>Polynomial long division gives a solution of the equation</p>
<p>$$f(x)=q(x)g(x) + r(x)\,,$$</p>
<p>where the degree of $r(x)$, is less than that of $g(x)$.</p>
<p>If we take $g(x) = x-a$ as the divisor, giving the degree of $r(x)$ as $0$, i.e. $r(x) = R$:</p>
<p>$$f(x)=q(x)(x-a) + R\,.$$</p>
<p>Setting $x=a $, we obtain: $f(a)=R.$</p>
|
291,284 | <p>Let $(\Omega, \cal{A}, \mathbb{P})$ be a probability space and $X$ a random variable on $\Omega$. Let, also, $f:\Omega\to\mathbb{R}$ be a Borel function. Then:<br>
$X$ and $f(X)$ are independent $\Longleftrightarrow$ there exists some $t\in\mathbb{R}$ such that $\mathbb{P}[f(X)=t]=1$, that is $f(X)$ is a degenerate r.v. </p>
<p>The only thing that I could make out is that if $X$ and $f(X)$ are independent, then<br>
$\mathbb{P}[f(X)\in B]=0$ or $1$ for every Borel subset of $\mathbb{R}$, since $\sigma(f(X))\subseteq \sigma(X)$ and hence, $f(X)$ is independent of its self. Suppose, now, that $\mathbb{P}[f(X)\leq x]=0$ for all $x\in\mathbb{R}$. Then:<br>
$\mathbb{P}[f(X)\in\mathbb{R}]=\mathbb{P}[\bigcup_{n=0}^{\infty}[f(X)\leq n]]\leq\sum_{n=0}^{\infty}[f(X)\leq n]=0$ which obviously is a contradiction since $\mathbb{P}[f(x)\in\mathbb{R}]=1$. </p>
<p>However, I don't know ow to prove this and my attempt isn't likely to become a complete solution.<br>
Any help would be appreciated.<br>
Thanks in advance! </p>
| Did | 6,179 | <blockquote>
<p>If the random variable $Y$ is independent of itself, then there exists $y$ such that $Y=y$ almost surely.</p>
</blockquote>
<p>To prove this, consider the CDF $F_Y:x\mapsto\mathbb P(Y\leqslant x)$ and note that for every $x$ the event $[Y\leqslant x]$ is independent of itself hence $F_Y(x)$ is $0$ or $1$. Since $F_Y$ has limits $0$ at $-\infty$ and $1$ at $+\infty$, the real number $y=\inf\{x\mid F_Y(x)=1\}$ is well defined and finite.</p>
<p>Since $F_Y$ is nondecreasing, $F(x)=0$ for every $x\lt y$ and $F_Y(x)=1$ for every $x\gt y$, hence $\mathbb P(y-u\lt Y\leqslant y+u)=1$ for every $u\gt0$, in particular, $\mathbb P(Y=y)=1$.</p>
|
3,921,335 | <p>how can I formally proof that for a > 1: <span class="math-container">$$ a> \sqrt a > \sqrt[3]a > \sqrt[4]a ... $$</span>?
Can someone help? ;)</p>
| Mostafa Ayaz | 518,023 | <p>To show <span class="math-container">$\sqrt[n]a>\sqrt[n+1]a$</span>, raise both sides to <span class="math-container">$n(n+1)$</span> to obtain
<span class="math-container">$$
a^{n+1}>a^n
$$</span>
and prove the latter inequality from <span class="math-container">$a>1$</span>.</p>
|
638,922 | <p>If $q \in \mathbb{H}$ satisfies $qi = iq$, prove that $q \in \mathbb{C}$</p>
<p>This seems kinda of intuitive since quaternions extend the complex numbers. I am thinking that $q=i$ because i know that $ij = k , ji = -k$, which is expand to all combinations of $i,j,k,$ which I think means that I have to use $ijk = i^2 = -1$</p>
| Igor Rivin | 109,865 | <p>Any complex number $q$ satisfies $q i = i q,$ so it is not true that $q=i$ However, just writing $q = a + b i + c j + d k,$ and writing down $q i$ and $i q$ ought to do the trick.</p>
|
3,995,913 | <p>We are looking at the following expression:</p>
<p><span class="math-container">$$\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$$</span></p>
<p>The solution is straightforward for this: <span class="math-container">$\frac{d}{dx}\int_{u(x)}^{v(x)}f(t) dt$</span>. Do we evaluate the given expression in like manner? Do we treat the <span class="math-container">$f(x) dx$</span> as if it were <span class="math-container">$f(t) dt$</span>?</p>
| OgvRubin | 468,471 | <p>You should write <span class="math-container">$\int_{u(x)}^{v(x)}f(t)dt$</span>. Then use the fundamental theorem of calculus by first rewriting</p>
<p><span class="math-container">$$\int_{u(x)}^{v(x)}f(t)dt = \int_{c}^{v(x)}f(t)dt+\int_{u(x)}^{c}f(t)dt=\int_{c}^{v(x)}f(t)dt-\int_{c}^{u(x)}f(t)dt$$</span>
for any constant <span class="math-container">$c$</span>. The notation <span class="math-container">$\int_{u(x)}^{v(x)}f(x)dx$</span> does not make sense.</p>
|
3,868,523 | <blockquote>
<p>Given the bases <em>a</em> = {(0,2),(2,1)} and <em>b</em> = {(1,0),(1,1)} compute the change of coordinate matrix from basis <em>a</em> to <em>b</em>.<br />
Then, given the coordinates of <em>z</em> with respect to the basis <em>a</em> as (2,2), use the previous question to compute the coordinates of <em>z</em> with respect to the basis <em>b</em>.</p>
</blockquote>
<p>The way I understood the first part was that I have to multiply the vectors of <em>b</em> by the coordinates of the vectors of <em>a</em> to compute the change of coordinate matrix from <em>a</em> to <em>b</em>. This gives me the following matrix: <span class="math-container">\begin{bmatrix}2&3\\2&1\end{bmatrix}</span></p>
<p>For the second part, I then have to take the inverse of the matrix I got from above and then multiply it by the coordinates of <em>z</em> to get the coordinates of <em>z</em> with respect to the basis <em>b</em>. The inverse of the matrix is: <span class="math-container">\begin{bmatrix}-1/4&3/4\\1/2&-1/2\end{bmatrix}</span> which I then multiply by (2,2) to get the coordinates of <em>z</em> with respect to basis <em>b</em></p>
<p>I am not sure that is correct however.</p>
| Community | -1 | <p>What you can do is use the changes of coordinates between each basis and the standard basis. So, you want: <span class="math-container">$A=\begin{pmatrix}0&2\\2&1\end{pmatrix}$</span> and <span class="math-container">$B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$</span>.</p>
<p>Then what is normally called the change of basis matrix from <span class="math-container">$a$</span> to <span class="math-container">$b$</span> would be the matrix that takes vectors written in terms of <span class="math-container">$b$</span> and returns them written in terms of <span class="math-container">$a$</span>. So we get the matrix <span class="math-container">$C$</span>, where <span class="math-container">$C=A^{-1}B$</span>.</p>
<p>For the second part apply <span class="math-container">$C^{-1}$</span> to <span class="math-container">$z$</span>.</p>
|
3,868,523 | <blockquote>
<p>Given the bases <em>a</em> = {(0,2),(2,1)} and <em>b</em> = {(1,0),(1,1)} compute the change of coordinate matrix from basis <em>a</em> to <em>b</em>.<br />
Then, given the coordinates of <em>z</em> with respect to the basis <em>a</em> as (2,2), use the previous question to compute the coordinates of <em>z</em> with respect to the basis <em>b</em>.</p>
</blockquote>
<p>The way I understood the first part was that I have to multiply the vectors of <em>b</em> by the coordinates of the vectors of <em>a</em> to compute the change of coordinate matrix from <em>a</em> to <em>b</em>. This gives me the following matrix: <span class="math-container">\begin{bmatrix}2&3\\2&1\end{bmatrix}</span></p>
<p>For the second part, I then have to take the inverse of the matrix I got from above and then multiply it by the coordinates of <em>z</em> to get the coordinates of <em>z</em> with respect to the basis <em>b</em>. The inverse of the matrix is: <span class="math-container">\begin{bmatrix}-1/4&3/4\\1/2&-1/2\end{bmatrix}</span> which I then multiply by (2,2) to get the coordinates of <em>z</em> with respect to basis <em>b</em></p>
<p>I am not sure that is correct however.</p>
| tossimmar | 835,571 | <p>Let <span class="math-container">$\mathcal{B} = \{b_1, b_2\}$</span> and <span class="math-container">$\mathcal{B}^{\prime} = \{b_1^{\prime}, b_2^{\prime}\}$</span> be two bases.</p>
<p>If <span class="math-container">$v$</span> is a vector expressed with respect to <span class="math-container">$\mathcal{B}$</span>, then
<span class="math-container">$$v = \begin{bmatrix} b_{1} & b_{2}\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$</span>
for some scalars <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span>. Since <span class="math-container">$\mathcal{B}^{\prime}$</span> is a basis, the elements of <span class="math-container">$\mathcal{B}$</span> can be expressed with respect to <span class="math-container">$\mathcal{B}^{\prime}$</span>. Hence,
<span class="math-container">\begin{align}
v &= \begin{bmatrix} (d_1b_1^{\prime} + d_2b_2^{\prime}) & (d_3b_1^{\prime} + d_4b_2^{\prime}) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \\
&= \begin{bmatrix} b_{1}^{\prime} & b_{2}^{\prime}\end{bmatrix} \begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}
\end{align}</span>
for some scalars <span class="math-container">$d_1, d_2, d_3$</span>, and <span class="math-container">$d_4$</span>.</p>
<p>The vector <span class="math-container">$\begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$</span> is the coordinate vector of <span class="math-container">$v$</span> with respect to <span class="math-container">$\mathcal{B}^{\prime}$</span>.</p>
<p>In your case, <span class="math-container">$\mathcal{B} = a$</span>, <span class="math-container">$\mathcal{B}^{\prime} = b$</span>, and <span class="math-container">$\begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = z$</span>. Can you determine <span class="math-container">$\begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix}$</span>?</p>
|
2,671,384 | <p>I want to solve following difference equation:</p>
<p>$a_i = \frac13a_{i+1} + \frac23a_{i-1}$, where $a_0=0$ and $a_{i+2} = 1$</p>
<p><strong>My approach:</strong>
Substituting $i=1$ in the equation,<br>
$3a_1 = a_2 + 2a_0$ <br>
$a_2 = 3a_1$<br>
Similarly substituting $i = 2, 3 ...$<br>
$a_3 = 7a_1$<br>
$a_4 = 15a_1$<br>
...</p>
<p>Generalizing,<br>
$a_i = (2^i - 1)a_1$<br></p>
<p>I am not sure how to get rid of $a_1$ to get the final answer of
$\frac{(2^i - 1)}{(2^{2+i} - 1)}$</p>
| Ian | 83,396 | <p>If you proceed this way, $a_1$ is an unknown, to be solved for by substituting in the boundary condition at the other endpoint and solving for $a_1$. (This is actually a fairly common trick, for instance it is also used to derive the invariant distribution of a birth-death process by writing all the $\pi_i$'s in terms of $\pi_0$ and then choosing $\pi_0$ based on the normalization requirement.)</p>
|
4,645,440 | <blockquote>
<p>Question: Let <span class="math-container">$X$</span> be a random variable then is the statement <span class="math-container">$\mathbb{P}(B|\{X=i\})$</span> = 0.6(random number), equivalent of saying: <span class="math-container">$X = i \implies \mathbb{P}(B) = 0.6$</span>?</p>
</blockquote>
<blockquote>
<p>Also say in the case A and B are conditionally independent if we condition on the event <span class="math-container">$\{ X = i \}$</span>, which is <span class="math-container">$\mathbb{P}(A \cap B|\{ X = i \}) =\mathbb{P}(A|\{ X = i \}\mathbb{P}(B|\{ X = i\})$</span>. Is it equivalent to:
<span class="math-container">$X = i \implies \mathbb{P}(A\cap B) =\mathbb{P}(A)\mathbb{P}(B)?$</span></p>
</blockquote>
<p>I don't know what conditional probability means from a measure theory perspective, but I was taught in undergraduate probably that <span class="math-container">$\mathbb{P}(A|B)$</span> where A and B are events is defined as <span class="math-container">$\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$</span>. However, treating it as such a definition is somewhat weird when I write proofs with regard to Markov chains, so can I just treat it as the first order logic implication as above?</p>
<p>If I can, I also wonder if it is possible to write it in this first order logic form for some events B, <span class="math-container">$\mathbb{P}(A|B)$</span>, where B does not come from a preimage of what a random variable.</p>
| drhab | 75,923 | <p>If you would have said something like:</p>
<p>"Based on the info that random variable <span class="math-container">$X$</span> has taken value <span class="math-container">$i$</span> we conclude that the probability of occurrence of event <span class="math-container">$B$</span> is now <span class="math-container">$0.6$</span>"</p>
<p>then I would agree, so IMV the intuition on this is okay.</p>
<p>But I disagree with using a logical implication for that. Firstly its premisse is false (because <span class="math-container">$X$</span> is a function and <span class="math-container">$i$</span> is not) so that the implication is always true no matter what is implied.</p>
<p>If you repair this by taking something like <span class="math-container">$X(\omega)=i$</span> then still things are not okay because <span class="math-container">$\mathbb P(B)$</span> is a <em>fixed</em> real number not depending on <span class="math-container">$\omega$</span> or <span class="math-container">$i$</span>.</p>
<p>Let's just do it without that implication. Our intuition does not really need it.</p>
|
4,645,440 | <blockquote>
<p>Question: Let <span class="math-container">$X$</span> be a random variable then is the statement <span class="math-container">$\mathbb{P}(B|\{X=i\})$</span> = 0.6(random number), equivalent of saying: <span class="math-container">$X = i \implies \mathbb{P}(B) = 0.6$</span>?</p>
</blockquote>
<blockquote>
<p>Also say in the case A and B are conditionally independent if we condition on the event <span class="math-container">$\{ X = i \}$</span>, which is <span class="math-container">$\mathbb{P}(A \cap B|\{ X = i \}) =\mathbb{P}(A|\{ X = i \}\mathbb{P}(B|\{ X = i\})$</span>. Is it equivalent to:
<span class="math-container">$X = i \implies \mathbb{P}(A\cap B) =\mathbb{P}(A)\mathbb{P}(B)?$</span></p>
</blockquote>
<p>I don't know what conditional probability means from a measure theory perspective, but I was taught in undergraduate probably that <span class="math-container">$\mathbb{P}(A|B)$</span> where A and B are events is defined as <span class="math-container">$\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$</span>. However, treating it as such a definition is somewhat weird when I write proofs with regard to Markov chains, so can I just treat it as the first order logic implication as above?</p>
<p>If I can, I also wonder if it is possible to write it in this first order logic form for some events B, <span class="math-container">$\mathbb{P}(A|B)$</span>, where B does not come from a preimage of what a random variable.</p>
| ryang | 21,813 | <p><span class="math-container">$$P(B|A)=k\tag1$$</span> <span class="math-container">$$A \text{ happens}\implies P(B)=k\tag2$$</span></p>
<p>It is false that for every pair of events <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and every <span class="math-container">$k\in[0,1],$</span> statements <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> are equivalent.</p>
<p>A counterexample is a toss of a fair coin, with <span class="math-container">$A$</span> and <span class="math-container">$B$</span> being the event of a head and of a tail, respectively, and with <span class="math-container">$k$</span> being <span class="math-container">$0.5;$</span> here, statement <span class="math-container">$(1)$</span> is false while statement <span class="math-container">$(2)$</span> is true.</p>
<hr />
<p><strong>Addendum</strong></p>
<blockquote>
<p>OP: Could you elaborate on the counterexample? I don't really understand what you mean.</p>
</blockquote>
<p>When you plug the counterexample into equation <span class="math-container">$(1),$</span> LHS <span class="math-container">$=0\ne$</span> RHS, so statement <span class="math-container">$(1)$</span> is False; when you plug the counterexample into implication <span class="math-container">$(2),$</span> its consequent(conclusion) is True, so, regardless of its antecedent(premise)'s truth value, statement <span class="math-container">$(2)$</span> is True.</p>
|
3,379,899 | <p>We have two coins. The first one is a fair coin and has <span class="math-container">$50\%$</span> chance of landing on head and <span class="math-container">$50\%$</span> of landing on tails. for the other one it is <span class="math-container">$60\%$</span> for head and <span class="math-container">$40\%$</span> for tails. We choose one of them randomly (with equal chance) and toss it three times.The result is Tails-Head-Head. If we toss it again for the 4th time, what is the probability of the coin landing on tails.</p>
<p>At first, it seems like a bayes theorem question, but after some thought I think because the result of every toss is independent of others, the answer is just:</p>
<p><span class="math-container">$$\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . \frac{4}{10} = \frac{9}{20} $$</span></p>
<p>The first <span class="math-container">$\frac{1}{2}$</span> is because we choose the coins randomly.</p>
<p>I want to know whether my argument is correct for this probability question or not.</p>
| David G. Stork | 210,401 | <p>Call the fair coin <span class="math-container">$A$</span> and the biased coin <span class="math-container">$B$</span>.</p>
<p><span class="math-container">$$P(THT|A) = (.5)^3$$</span></p>
<p><span class="math-container">$$P(THT|B) = (.4)(.6)(.4)$$</span></p>
<p><span class="math-container">$$P(A| THT) = \frac{P(THT|A) P(A)}{P(THT|A) + P(THT|B)}$$</span></p>
<p><span class="math-container">$$P(B| THT) = \frac{P(THT|B) P(B)}{P(THT|A) + P(THT|B)}$$</span></p>
<p>(Note that <span class="math-container">$P(A|THT) + P(B|THT) = 1$</span>, as it must)</p>
<p>Once you have the probability it is <span class="math-container">$A$</span>, you can easily calculate the probability the 4th coin is <span class="math-container">$H$</span> (<span class="math-container">$0.5$</span>). And likewise if the coin is <span class="math-container">$B$</span> (<span class="math-container">$0.6$</span>).</p>
<p>You can plug in the numbers.</p>
|
884,342 | <p>$N$ is a normal subgroup of $G$ if $aNa^{-1}$ is a subset of $N$ for all elements $a $ contained in $G$. Assume, $aNa^{-1} = \{ana^{-1}|n \in N\}$.</p>
<p>Prove that in that case $aNa^{-1}= N.$</p>
<p>If $x$ is in $N$ and $N$ is a normal subgroup of $G$, for any element $g$ in $G$, $gxg^{-1}$ is in $G$. Suppose $x$ is in $N$, and $y=axa^{-1}$ as is defined. Since $N$ is normal, $aNa^{-1}$ is a subset of N.
$x= a^{-1}ya$. Given that $x$ is in $N$, and $x=a^{-1}ya$, $y$ is also in $N$. If $y$ is in N, then $axa^{-1}$ is also in $N$. $X$ is in $aNa^{-1}$.</p>
<p>Does the proof make sense?</p>
| Adriano | 76,987 | <p>I'm a little confused as to how you got to your last step. Here's my version.</p>
<hr>
<p>Suppose that $aNa^{-1} \subseteq N$ for all $a \in G$. Then to show that equality holds, it suffices to show that $bNb^{-1} \supseteq N$ for all $b \in G$. To this end, choose any $b \in G$ and choose any $x \in N$. We want to show that $x \in bNb^{-1}$.</p>
<p>Now recall that $x \in N$ and that $aNa^{-1} \subseteq N$ for all $a \in G$. Thus, by taking $a = b^{-1}$, it follows that:
$$
(b^{-1})x(b^{-1})^{-1} = b^{-1}xb \in b^{-1}Nb \subseteq N
$$
so that $b^{-1}xb \in N$. But then we're done, since:
$$
x = exe = (bb^{-1})x(bb^{-1}) = b\underbrace{(b^{-1}xb)}_{\in ~ N}b^{-1} \in bNb^{-1}
$$
as desired. $~~\blacksquare$</p>
|
1,075,505 | <p>I have the following function:</p>
<p>$$f(x+iy) = x^2+iy^2$$</p>
<p>My textbook says the function is only differentiable along the line $x = y$, can anyone please explain to me why this is so? What rules do we use to check where a function is differentiable?</p>
<p>I know the Cauchy-Riemann equations, and that $u=x^2$ and $v=y^2$ here. </p>
| Community | -1 | <p>Being complex differentiable at a point is <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations#Complex_differentiability" rel="nofollow">equivalent</a> to the combination of</p>
<ol>
<li>Being real differentiable at that point, and</li>
<li>Satisfying the Cauchy-Riemann equations</li>
</ol>
<p>The real and imaginary parts of $f$ are $u=x^2$ and $v= ^2$. They are polynomials, so real-differentiable everywhere. The two Cauchy-Riemann equations take the form $2x=2y$ (from $u_x=v_y$) and $0=0$ (from $ u_y=-v_x$). The second holds everywhere. The first holds when $x=y$ and only then.</p>
|
2,234,737 | <p>$f:\mathbb{R}^m\to\mathbb{R}$ differentiable such that $f(x/2) = f(x)/2$, show $f$ is linear.</p>
<p>I tried to do:</p>
<p>$f(a+b) = f(2(a+b)/2) = f(2(a+b))/2$</p>
<p>but that wouldn't turn into $f(a)+f(b)$</p>
<p>In the same way: $f(a)/2 + f(b)/2 = f((a+b)/2)$ which won't help.</p>
<p>Also, I need to use the differentiability. I know that there must exist $r(v)$ such that</p>
<p>$$f(a+v) = f(a) + grad(f)\cdot v + r(v)$$</p>
<p>where $\lim_{v\to 0}\frac{r(v)}{|v|} = 0$</p>
| Misha Lavrov | 383,078 | <p>We actually only need differentiability at $0$ for this to hold.</p>
<p>First, note that $f(0) = 0$, since the functional equation tells us that $f(0) = f(0)/2$. The existence of a derivative at $0$ tells us that
$$\lim_{v \to 0} \frac{|f(v) - \operatorname{grad}(f) \cdot v|}{\|v\|} = 0.$$
In particular, if we choose any $x \in \mathbb{R}^m$, then this limit converges sequentially for the sequence $v_n = x/2^n$, and we have
$$\lim_{n \to \infty} \frac{|f(x/2^n) - \operatorname{grad}(f) \cdot x/2^n|}{\|x/2^n\|} = 0.$$
The functional equation lets us rewrite $f(x/2^n) = f(x)/2^n$, and everything else in the limit is linear and naturally lets us factor the $2^n$ out of it, so we actually have
$$\lim_{n \to \infty} \frac{|f(x)/2^n - (\operatorname{grad}(f) \cdot x)/2^n|}{\|x\|/2^n} = \lim_{n \to \infty} \frac{|f(x) - \operatorname{grad}(f) \cdot x|}{\|x\|} = 0.$$
But now there is no dependence on $n$ in the limit at all, so the limit can only be $0$ if $f(x) = \operatorname{grad}(f) \cdot x$: in other words, if $f(x)$ is linear with the same constant slope that is its derivative at $0$.</p>
|
1,789,742 | <p>I'm reading the book generatingfunctionology by Herbert Wilf and I came across a partial fraction expansion on page 20 that I cannot understand. The derivation is as follows:</p>
<p>$$
\frac{1}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \frac{\alpha_j}{1-jx}
$$</p>
<p>The book says to fix $r, 1 \leq r \leq k$, and multiply both sides by $1-rx$. Doing so, I get:</p>
<p>$$
\frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)} = \frac{\alpha_1(1-rx)}{1-x} + ... + \frac{\alpha_{r-1}(1-rx)}{1-(r-1)x} + \alpha_r + \frac{\alpha_{r+1}(1-rx)}{1-(r+1)x} + ... + \frac{a_k(1-rx)}{1-kx}
$$</p>
<p>Contrarily, the book has:</p>
<p>$$
\alpha_r = \frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)}
$$</p>
<p>I don't understand how the other other fractions on the right side of my result cancel out to $0$. I tried with a small example where $k=3$ and I couldn't isolate $\alpha_2$ nicely after multiplying both sides by $1-2x$. Any pointers on this would be greatly appreciated.</p>
<p>After this, the book goes on by letting $x=1/r$, resulting in the following:</p>
<p>$$
\begin{aligned}
\alpha_r &= \frac{1}{(1-1/r)(1-2/r)...(1-(r-1)/r)(1-(r+1)/r)...(1-k/r)} \\
&= (-k)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!} && (1 \leq r \leq k)
\end{aligned}
$$</p>
<p>I also can't figure how this is derived (I suspect it's using an identity that I'm not aware of.) Any help would be much appreciated. Thanks a lot!</p>
| Brian M. Scott | 12,042 | <p>My edition of the book doesn’t have</p>
<p>$$\alpha_r = \frac{1}{(1-x)\ldots(1-(r-1)x)(1-(r+1)x)\ldots(1-kx)}\;,$$</p>
<p>which in any case makes no sense, since $\alpha_r$ is a constant and the righthand side is not. However, after letting $x=\frac1r$ it does show</p>
<p>$$\begin{align*}
\alpha_r&=\frac1{(1-1/r)(1-2/r)\cdots(1-(r-1)/r)(1-(r+1)/r)\cdots(1-k/r)}\\
&=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;.
\end{align*}$$</p>
<p>The last step is obtained by first multiplying by $r^{k-1}$ to get</p>
<p>$$\alpha_r=\frac{r^{k-1}}{\underbrace{(r-1)(r-2)\cdot(r-(r-1))}_{(r-1)!}\underbrace{(r-(r+1))\cdots(r-k)}_{(-1)(-2)\cdots(-(k-r))}}$$</p>
<p>and then rewriting this as</p>
<p>$$\alpha_r=\frac{r^{k-1}}{(r-1)!(-1)^{k-r}(k-r)!}=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;.$$</p>
|
128,408 | <p>The following:</p>
<pre><code>mat = {{1., -3, -2}, {-1, 3, -2}, {0., 0., 0.}}
MatrixFunction[Sinc, mat]
</code></pre>
<p>returns:</p>
<pre><code>MatrixFunction::drvnnum: Unable to compute the matrix function because
the subsequent derivative of the function Sinc[#1] & at a numeric value is
not numeric.
</code></pre>
<p>The documentation for this error gives examples where the function is not differentiable. But here, <code>Sinc</code> is, so why do I get this error?</p>
| Michael E2 | 4,999 | <p>Here's a workaround I've used for <code>Sinc</code>. It defines <code>mySinc[k, x]</code> to be the k-th derivative of <code>Sinc[x]</code>, for <code>k</code> a nonnegative integer. (<em>Update notice:</em> Thanks to J.M. for pointing out a simple formula for the derivatives at zero.)</p>
<pre><code>ClearAll[mySinc];
SetAttributes[mySinc, NHoldFirst];
mySinc[x_] := mySinc[0, x];
mySinc[0, x_?NumericQ] := Sinc[x];
mySinc[k_Integer?Positive, 0.] := N[(-1)^k Cos[k π/2]/(k + 1)];
mySinc[k_Integer?Positive, x_] /; x == 0 :=
SetAccuracy[(-1)^k Cos[k π/2]/(k + 1),
Piecewise[{{Log10[2 (k + 3)] + 2 Accuracy[x], EvenQ[k]}}, Log10[k + 2] + Accuracy[x]]];
mySinc[k_Integer?Positive, x_?NumericQ] /; x != 0 :=
Evaluate@D[(# Cos[#] - Sin[#])/#^2, {#, k - 1}] &@x;
Derivative[n_Integer?Positive][mySinc][x_] := mySinc[n, x];
Derivative[0, n_Integer?Positive][mySinc][k_, x_] := mySinc[k + n, x];
</code></pre>
<p>Examples:</p>
<pre><code>MatrixFunction[mySinc, mat] // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/5ppgl.png" alt="Mathematica graphics"></p>
<pre><code>MatrixFunction[mySinc, Round@mat] // Simplify // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/DTeBV.png" alt="Mathematica graphics"></p>
<hr>
<p><em>Caveat:</em>
The derivatives of <code>Sinc</code> are not numerically well-behaved near zero.</p>
<pre><code>Labeled[
TableForm[
Table[SetPrecision[
100 - Precision@Derivative[k][Sinc][SetPrecision[x, 100]],
3], {x, 10^-Range[8]}, {k, 7}],
TableHeadings -> {"10"^Row[{"-", #}] & /@ Range[8], Automatic}],
{"x", "Derivative"}, {Left, Top}]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/JImmy.png" alt="Mathematica graphics"></p>
<p><em>Precision loss of $\hbox{Sinc}^{(k)}(x)$ for $x$ near $0$.</em></p>
</blockquote>
<p>The loss seems fairly predictable, and one could modify the definitions above to use sufficient precision internally to compute a result to a desired precision.</p>
|
3,426,550 | <p>Are there parts in functional analysis that would be easier to learn without real analysis or is this just not possible?</p>
<p>When I say functional analysis I mean</p>
<ol>
<li>Lebesgue measure and integral</li>
<li>spaces of Lebesgue integrable functions</li>
<li>Banach spaces</li>
<li>duality</li>
<li>bounded linear operator and linear operators in general</li>
<li>Hilbert space</li>
<li>reproducing kernel Hilbert space</li>
<li>non-linear analysis in Banach spaces</li>
</ol>
| clm | 591,572 | <p>In my experience, what differentiates "Analysis" from other math regarding functions and the real or complex numbers etc, is the level of depth and rigor. You could probably learn something in each of the topics you listed, but it would be hard to understand the proofs or justification behind a lot of results. Consequently it's difficult to have a deep or truly meaningful understanding of functional analysis without real analysis experience.</p>
<p>Although there would maybe be more specific issues, like not being familiar with specific definitions and concepts stated in Theorems or otherwise. Again, you could learn these definitions but this boils down into what I said above.</p>
|
3,522,681 | <blockquote>
<p>There are 8 people in a room. There are 4 males(M) and 4 females(F). What is the probability that there are no M-F pairs that have the same birthday ? It is OK for males to share a birthday and for females to share a birthday. Assume there are <span class="math-container">$10$</span> total birthdays. </p>
</blockquote>
<p>I give a solution below. Not sure if is correct and is there a more general way to approach it ? I break it into 5 cases-summing these cases gives the total ways M-F do not share. If divide the sum by <span class="math-container">$10^8$</span> would obtain desired probability.</p>
<p>Case 1: all men have different birthdays <span class="math-container">$N_1 = 10 \cdot 9 \cdot 8 \cdot 7 \cdot (10-4)^4$</span></p>
<p>Case 2: one pair men exact + two single men <span class="math-container">$N_2 = {\sideset{_{10}}{_1} C} \cdot {\sideset{_4}{_2} C} \cdot 9 \cdot 8 \cdot (10-3)^4$</span></p>
<ul>
<li>the first term chooses the single BD for the pair of men. </li>
<li>The second term selects the 2 men in the pair. </li>
<li>The <span class="math-container">$9\cdot 8$</span> are the number of ways the two single men can choose their birthdays.</li>
<li>The final term is the number of ways the <span class="math-container">$4$</span> woman can select the remaining <span class="math-container">$10-3 = 7$</span> birthdays which do not equal the men which have used <span class="math-container">$3$</span> birthdays.</li>
</ul>
<p>Case 3: two pair men exact <span class="math-container">$N_3 = {\sideset{_{10}}{_2} C} \cdot {\sideset{_4}{_2} C} \cdot {\sideset{_2}{_2} C} \cdot (10-2)^4$</span></p>
<p>Case 4: one triple and one single man <span class="math-container">$N_4 = {\sideset{_{10}}{_1} C} \cdot {\sideset{_4}{_3} C} \cdot {\sideset{_1}{_1} C} \cdot {\sideset{_9}{_1} C} \cdot (10-2)^4$</span></p>
<p>Case 5: all men have same birthday <span class="math-container">$N_5 = {\sideset{_{10}}{_1} C} \cdot (10-1)^4$</span></p>
<p>The sum of Case <span class="math-container">$1$</span> to <span class="math-container">$5$</span> is the total ways for no M-F pairs. The last term in each case is the number of permutations of the 4 woman with <span class="math-container">$(10-k)^4$</span> choices where <span class="math-container">$k$</span> is the number of unique birthdays used up for the men. I do not believe the order of the people matters: I calculate assuming all the men come first. Please comment on my approach.</p>
<p>I have not found an understandable solution on this website.</p>
| Henry | 6,460 | <p>Your calculation has a minor error in that case 4 should end with <code>*(10-3)^4</code> rather than <code>*(10-2)^4</code>. </p>
<p>If you correct that and add the numbers up then you would get <span class="math-container">$19550250$</span>. Dividing by <span class="math-container">$10^8$</span> would then give the probability of <span class="math-container">$0.1955025$</span></p>
<p>Generalising this is a little messy because your cases 2 and 3 each count possibilities for the mean having two birthdays between them. There is a way round this by using <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling numbers of the second kind</a> and you could say something like</p>
<blockquote>
<p>If there are <span class="math-container">$d$</span> days in a year, and <span class="math-container">$m$</span> men and <span class="math-container">$w$</span> women with their birthdays independently and uniformly distributed across these days, then the probability that there are no cases of a man and a woman sharing a birthday is
<span class="math-container">$$\frac{d! }{d^m}\sum\limits_{n=1}^{\min(m,d)} \frac{S_2(m,n) }{(d-n)!}\left(1-\frac{n}{d}\right)^w $$</span><br>
where <span class="math-container">$S_2(x,y)$</span> is the corresponding Stirling number of the second kind. </p>
</blockquote>
<p>If you applied this to your example with <span class="math-container">$d=10, m=4,n=4$</span>, it would give <span class="math-container">$$362.88\left(\frac{1\times 0.9^{4}}{362880} + \frac{7\times 0.8^{4}}{40320} + \frac{6\times 0.7^{4}}{5040} + \frac{1\times 0.6^{4}}{720}\right)=0.1955025$$</span> </p>
|
3,470,703 | <p><span class="math-container">$\int \left(\frac{1}{2}t+\frac{\sqrt{2}}{4} \right)e^{t^2-\sqrt{2}t}\ dt$</span></p>
| Fareed Abi Farraj | 584,389 | <p><span class="math-container">$\int u'e^u dt =e^u$</span></p>
<p>Try taking <span class="math-container">$u=t^2-\sqrt{2}t$</span> and show your work if you didn't get to your answer if you want us to help you.</p>
|
128,784 | <p>Consider the two lists</p>
<pre><code>list1={1,2,a[1],8,b[4],9};
list2={8,b[4],9,1,2,a[1]};
</code></pre>
<p>it is evident by inspection that <code>list2</code> is just a cyclic rotation of <code>list1</code>. Considering an equivalence class of lists under cyclic rotations, I would like to have a function <code>cycRot[x_List]</code> that takes a list and returns a cyclically rotated representative of that list, which would be independent of the initial cyclic order of the list. Such that</p>
<pre><code>cycRot[list1]==cycRot[list2]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<p>is guaranteed (the exact resulting rotation is irrelevant as long as the function returns the same result for any cyclically equivalent list).
Is there such a function in Mathematica? Or maybe one can implement it efficiently? Thanks for any suggestion!</p>
| Kagaratsch | 5,517 | <p>My own attempt at a solution is this</p>
<pre><code>cycRot[x_List] := Block[{p},
p = Position[x, Sort[x][[1]], 1][[1, 1]];
{x[[p ;;]], x[[1 ;; p - 1]]} // Flatten
]
</code></pre>
<p>However, I am not sure if this is going to be slow for larger lists, since the cyclic property is not being utilized to improve performance when performing a complete sorting. Maybe there are better more efficient solutions?</p>
|
332,001 | <p><a href="https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(geometry)#Generalizations_and_related_results" rel="nofollow noreferrer">Wikipedia states</a> that A. D. Alexandrov generalized <em>Cauchy's rigidity theorem for polyhedra</em> to higher dimensions. </p>
<p>The relevant statement in the article is not linked to any source. The sources at the end of the Wikipedia page seem to be only about <span class="math-container">$3$</span>-dimensional polyhedra as well, in particular Alexandrov's book "Convex polyhedra".</p>
<blockquote>
<p>Where can I find a reference for that statement?</p>
</blockquote>
| Ivan Izmestiev | 98,590 | <p>Wikipedia is correct. This is discussed in Alexandrov's book "Convex polyhedra" in Section 3.6.5.</p>
|
2,204,217 | <p>I am attempting to find a formula for $$(\mu * \mu)(n)$$ where * represents the Dirichlet Convolution operator. I know this can be expressed as $$\sum_{d|n} \mu(d)\mu(\frac{n}{d})$$ but I'd like the formula to not include any sums over divisors. I know it will be necessary to include information about the factorization of n, but I'm not sure how. For reference, $$\mu(n)= \begin{cases}0,&\text{if $n$ has one or more repeated prime factors}\\1,&\text{if $n$=1}\\(-1)^k,&\text{if $n$ is a product of $k$ distinct primes}\end{cases}$$
Some initial thoughts: the Dirichlet Convolution of two multiplicative functions is multiplicative, and since $$\mu(n)$$ is multiplicative, then so is $$(\mu * \mu)(n)$$
Any information to point me in the right direction on this will be greatly appreciated.</p>
| Especially Lime | 341,019 | <p>As you say, this is multiplicative. This means that once you know how to calculate it for numbers of the form $p^a$ you can calculate it for arbitrary $n$ by writing $n$ as a product of powers of different primes $p^aq^b\cdots$, and then multiplying the corresponding values $(\mu*\mu)(n)=((\mu*\mu)(p^a))((\mu*\mu)(q^b))\cdots$.</p>
<p>If $a=1$, $\sum_{d\mid p}\mu(d)\mu(p/d)=-2$ (the two factors each contribute $-1$). If $a=2$, $\sum_{d\mid p^2}\mu(d)\mu(p^2/d)=1$, since the only way for $\mu(d)\mu(p^2/d)$ to be non-zero is if $d\leq p$ and $p^2/d\leq p$, which requires $d=p$. If $a>2$ then $\sum_{d\mid p^a}\mu(d)\mu(p^2/d)=0$, since for each term in the sum either $d$ or $p^a/d$ is divisible by $p^2$.</p>
<p>So your function is $0$ if $n$ is divisible by the cube of any prime. Otherwise it is $(-2)^k$, where $k$ is the number of primes that divide $n$ exactly once (i.e. their squares do not divide $n$).</p>
|
2,204,217 | <p>I am attempting to find a formula for $$(\mu * \mu)(n)$$ where * represents the Dirichlet Convolution operator. I know this can be expressed as $$\sum_{d|n} \mu(d)\mu(\frac{n}{d})$$ but I'd like the formula to not include any sums over divisors. I know it will be necessary to include information about the factorization of n, but I'm not sure how. For reference, $$\mu(n)= \begin{cases}0,&\text{if $n$ has one or more repeated prime factors}\\1,&\text{if $n$=1}\\(-1)^k,&\text{if $n$ is a product of $k$ distinct primes}\end{cases}$$
Some initial thoughts: the Dirichlet Convolution of two multiplicative functions is multiplicative, and since $$\mu(n)$$ is multiplicative, then so is $$(\mu * \mu)(n)$$
Any information to point me in the right direction on this will be greatly appreciated.</p>
| mds | 445,155 | <p><strong>Note on the Previous Solution</strong>: Let $\operatorname{rad}(n)$ denote the <em>radix</em>, or squarefree part, of $n$, i.e., so that if $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the factorization of $n$ into powers of distinct primes then $\operatorname{rad}(n) = p_1 p_2 \cdots p_k$. Then we see that $\mu(\operatorname{rad}(n))$ is always non-zero and that $\mu^2(n / \operatorname{rad}(n)$ is the characteristic function of the cube-free integers. Also, if $\omega(n)$ counts the number of distinct prime factors of $n$ (the <em>prime omega function</em>) then the number of primes that divide $n$ exactly once is given by
$$\#\{p \text{ prime } : p|n \text{ and } p^2 \nmid n\} = \omega(n) - \omega(n / \operatorname{rad}(n)).$$
This implies that the solution given by <strong>Essentially Lime</strong> in the previous response has the following exact closed-form expression for any $n \geq 1$:
$$(\mu \ast \mu)(n) = (-2)^{\omega(n) - \omega(n / \operatorname{rad}(n))} \cdot \mu^2(\omega(n) - \omega(n / \operatorname{rad}(n)).$$
Curiously enough, parts of this formula for the solution can be re-expressed in terms of 1) the identity that $\sum_{d|n} \mu^2(d) = 2^{\omega(n)}$, or in other words the convolution $\mu^2 \ast 1 = 2^{\omega}$; and 2) the fact that $(-1)^{\omega(n)} = \mu(n)$ whenever $n$ is squarefree. </p>
<p><strong>Alternate Solution Using Dirichlet Inverses:</strong>
Another, perhaps shorter and more direct way to find this solution is to consider the <a href="https://en.wikipedia.org/wiki/Dirichlet_convolution#Dirichlet_inverse" rel="noreferrer">Dirichlet inverse</a> of the <em>divisor function</em> $d(n) := \sum_{d|n} 1 = (1 \ast 1)(n)$. Recall that the <em>Dirichlet inverse</em> $f^{-1}$ of any arithmetic function $f$ is the unique function (if one exists) satisfying $(f \ast f^{-1})(n) = \varepsilon = \delta_{n,1}$ where $\varepsilon$ is the multiplicative identity of Dirichlet convolutions. This inverse exists precisely when $f(1) \neq 0$. </p>
<p>Now consider the following phrasing of the problem: write $f := mu \ast \mu$. Then since $\mu \ast 1 = \varepsilon$ (an elementary fact which is seen <a href="https://terrytao.wordpress.com/2009/08/30/an-elementary-inequality-involving-the-mobius-function/" rel="noreferrer">here</a>, for example, or proved by Dirichlet generating functions), we can invert the right-hand-side to obtain that $\varepsilon = f \ast 1 \ast 1 = f \ast d$. We can easily verify that $d(1) = 1 \neq 0$, so that the divisor function has a Dirichlet inverse. This implies that $f = d^{-1}$. The values of this inverse function are defined by $d(1) = 1$ and for $n > 1$ by the recurrence relation:
$$d^{-1}(n) = -\sum_{\substack{d|n \\ d>1}} d(n) d^{-1}(n/d).$$
By computation, the first few values of this sum correspond to the sequence $\{d^{-1}(n)\}_{n \geq 1} = \{1,-2,-2,1,-2,4,-2,0,1,4,\ldots\}$ (<a href="https://oeis.org/A007427" rel="noreferrer">A007427</a>). </p>
|
2,279,392 | <p>I could use some help again.
Let $f,g$ be functions from $\mathbb{N}\to \mathbb{N}$.
Also known is that $f(n) = g(2n)$ for every $n\in\mathbb{N}$.
Assuming that $f$ is a surjective function, how do you prove that $g$ is not a one-to-one function?</p>
<p>Cheers</p>
| Siong Thye Goh | 306,553 | <p>Since $f$ is a surjection, $\exists k \in \mathbb{Z}$ such that $f(k)=g(1)$.</p>
<p>$$g(2k)=g(1)$$.</p>
<p>Hence $g$ is not injective.</p>
|
1,115,389 | <p>Given: </p>
<p>$$\sum_{n = 0}^{\infty} a_nx^n = f(x)$$</p>
<p>where:</p>
<p>$$a_{n+2} = a_{n+1} - \frac{1}{4}a_n$$</p>
<p>is the recurrence relationship for $a_2$ and above ($a_0$ and $a_1$ are also given).</p>
<p>Is there a nice closed form to this pretty recurrence relationship?</p>
| drhab | 75,923 | <p>First of all: your approach is okay and your answer is <em>almost</em> okay. There is a factor $72$ in your denominator that must be left out and that's all. As you will understand it concerns $9$ factors (and not $10$). An alternative way:</p>
<p>For selecting $9$ spots in such a way that the rooks placed on them cannot capture other rooks there are $9!$ possibilities. For selecting $9$ spots of $81$ without any condition there are $\binom{81}{9}$ ways. This gives a probability of $$9!\times\binom{81}{9}^{-1}$$ to place the rooks at random in such a way that they cannot capture other rooks.</p>
<p>Working this out you come to $$\frac{9^2\times8^2\times\cdots\times1^2}{81\times80\times\cdots\times 73}$$</p>
<p>matching your answer (if factor $72$ has been taken away there).</p>
|
4,412,700 | <p>I've been working on a problem but have been stuck for several hours finishing it.</p>
<p>The problem is to show that
<span class="math-container">$$
\frac{x}{(e+x) \ln(e+x)} \leq \ln(\ln(e+x)) \leq \frac{x}{e}
$$</span>
for all <span class="math-container">$x > 0$</span>.</p>
<p>I proceeded by first integrating each component as the problem suggested, which changed it to proving that
<span class="math-container">$$
\int_0^x \frac{1}{(e+x) \ln(e+x)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{(e+t) \ln(e+t)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{e} \,\mathrm{d}t \,.
$$</span></p>
<p>From here I was able to show quite easily that when <span class="math-container">$x>0$</span>, the second inequality holds. However, I'm unsure of how to now prove the first inequality: is there a result which could help with this or something that I'm missing?</p>
<p>Thanks very much for any help, and sorry for the poor formatting, it's my first time using stack exchange.</p>
<p>Mark</p>
| Calvin Lin | 54,563 | <p>Let <span class="math-container">$ f(x) = \ln ( \ln (e+x) ) $</span>.</p>
<ol>
<li>Show that <span class="math-container">$ f'(x) = \frac{1}{(e+x)( \log e+x)} $</span>.</li>
<li>Show that <span class="math-container">$f(0) = 0 $</span>.</li>
<li>Show that for <span class="math-container">$X> x$</span>, <span class="math-container">$$ \frac{1}{ ( e+X) \log (e + X) } \leq f'(x) \leq \frac{1}{e}.$$</span></li>
<li>Consider <span class="math-container">$ \int_0^X f'(x) \, dx$</span> and derive the desired inequality
<span class="math-container">$$ \frac{X}{ ( e+X) \log (e + X) } \leq \ln ( \ln (e+X)) \leq \frac{X}{e}.$$</span></li>
</ol>
|
243,510 | <p>So this question is seriously flooring me.</p>
<p>Let $G$ be drawn in the plane so that it satisfies: </p>
<ol>
<li>The boundary of the infinite region is a cycle $C$</li>
<li>Every other region has boundary cycle of length $3$</li>
<li>Every vertex of $G$ not in $C$ has even degree</li>
</ol>
<p>Show that $\chi(G) \le 3$, where $\chi(G)$ is the chromatic number.</p>
<p>I know I have to use induction and consider two cases for the first step: Whether some two non-consecutive vertices of $C$ are adjacent and in the second case I would delete an edge of $C$ and apply the induction hypothesis.</p>
<p>Can anyone help?</p>
| William Macrae | 49,817 | <p>This came up on CS Theory a while back: <a href="https://cstheory.stackexchange.com/questions/4027/coloring-planar-graphs">https://cstheory.stackexchange.com/questions/4027/coloring-planar-graphs</a></p>
<p>Unfortunately some of the links in that post are dead / not free.</p>
<p><a href="http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBMQFjAA&url=http://www.math.uiuc.edu/~west/pubs/eultri.ps&rct=j&q=coloring%20near%20triangulations&ei=sVsdTZz3HISClAe2k9HhDA&usg=AFQjCNHFOybQ5YRN-gZA8184kA1VfYxJ1w&cad=rja" rel="nofollow noreferrer">This paper</a> proves a special case of what you mention, that if all vertices are of even degree in a near-triangulation, then the graph is 3-colorable. The key is that such a graph is eulerian, and therefore has an eulerian trail that doesn't cross over itself. Coloring vertices 1, 2, 3, 1, 2, 3, etc. along this trail is a proper 3-coloring.</p>
<p>For the case when $G$ has odd vertices on its boundary, you can construct a planar Eulerian near-triangulation $H$ such that $G \subset H$ and $H$ is Eulerian. Let $C = v_1v_2\ldots v_x$ and suppose $v_j$ and $v_k$ on $C$ are of odd degree. WLOG $k > j$. Add vertices $u_ju_{j+1}\ldots u_{k-1}$ in the exterior region in the obvious order. Then add the edges $v_iu_i$ and $v_{i+1}u_i$ for $j \le i \le k-1$. Now $u_j$ and $u_{k-1}$ are both of odd degree unless $j = k-1$, but we can repeat this process until these two odd vertices are resolved (since the odd vertices continue to get closer). We now have fewer odd vertices, so repeating this process can make all vertices even.</p>
|
801,574 | <p>Let $A$ be a complex $n\times m$ matrix ($n<m$), which is row full rank. For any $m\times m$ matrix $M$, such that $AM$ is still row full rank, can we find an invertible matrix $X$, such that $XA=AM$?</p>
| fuglede | 10,816 | <p><strong>Hint:</strong> $A = \begin{pmatrix} 1 & 0 \end{pmatrix}$, and $M = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$.</p>
|
2,149,006 | <p>While learning the power rule, one thing popped up in my mind which is confusing me. We know what the power rule states :</p>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}$$ where $n$ is a real number.</p>
<blockquote>
<p>But instead of $n$, if we have a trig function like $\sin(x)$, <strong>will the power rule still apply?</strong></p>
</blockquote>
<p>Eg. We have a function $y = x^{\sin(x)}$, and thus by the power rule;</p>
<p>$$\frac{dy}{dx} = sin(x)x^{sin(x)-1}$$. </p>
<p>Is this possible? Please tell me if even the function I wrote above really does exist or not.</p>
<p>I know this may seem a stupid question to many, but please help because I cannot find any explanation to this. </p>
| R.W | 253,359 | <p>No. It will not maintain the rule as you pointed out. The reason for that is that the derivative act on a function like an operator much more generalized as the power rule: It is a limit of some quantity. Because of that if you change the power to some function you will have to understand how this new functions changes as the limit is been taken. </p>
<p>The function $\sin(x)x^{\sin(x)-1}$ exists and is the function plotted below
<a href="https://i.stack.imgur.com/It5J8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/It5J8.png" alt="enter image description here"></a></p>
<p>But if you make the calculations of the derivative of $x^{\sin(x)}$ the function that you'll get is $x^{\sin(x)-1}(\sin(x)+x\ln(x)\cos(x))$ plotted below</p>
<p><a href="https://i.stack.imgur.com/6jN3r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6jN3r.png" alt="enter image description here"></a></p>
<p>So note how you have to use the full tolls to get the correct answer, these two functions are different because that you have to think that, because the power is now a function, other tools pointed out in other answers will have to be used.</p>
|
654,263 | <p>My intention is neither to learn basic probability concepts, nor to learn applications of the theory. My background is at the graduate level of having completed all engineering courses in probability/statistics -- mostly oriented toward the applications without much emphasis on mathematical rigor.</p>
<p>Now I am very interested in learning the core logic and mathematical framework of probability theory, as a math branch. More specifically, I would like to learn answers to the following questions:</p>
<blockquote>
<p>(1) What are the necessary axioms from which we can build probability theory?</p>
<p>(2) What are the core theorems and results in the mathematical theory of probability?</p>
<p>(3) What are the derived rules for reasoning/inference, based on the theorems/results in probability theory?</p>
</blockquote>
<p>So I am seeking a book that covers the "heart" of mathematical probability theory -- not needing much on applications, or discussion on extended topics.</p>
<p>I would like to appreciate your patience for reading my post and any informative responses.</p>
<p>Regards,
user36125</p>
| arsmath | 4,880 | <p>I personally liked Davidson's <i>Stochastic Limit Theory</i>. It rigorously introduces measure-theoretic probability, and then proves the key convergence results. Towards the end it rigorously develops continuous-time stochastic processes.</p>
<p>You mention inference, so if you are particularly interested in the statistics rather than just probability, I would also recommend Schervich's <i>Theory of Statistics</i>.</p>
<p>If you'd like to see a mathematical development, but you find you don't care for measure theory, then Lehman's <i>Elements of Large-Sample Theory</i> is pretty good.</p>
|
2,372,035 | <p>$ABCD$ is a square with side-length $1$ and equilateral triangles $\Delta AYB$ and $\Delta CXD$ are inside the square. What is the length of $XY$?</p>
| lioness99a | 401,264 | <p>First draw a diagram of the problem:</p>
<p><a href="https://i.stack.imgur.com/LEyyy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LEyyy.png" alt="Diagram"></a></p>
<p>Remember that angles in an equilateral triangle are all $60^\circ$ and each point is directly above the midpoint of the opposite side.</p>
<p>We can calculate the distance between $X$ and the midpoint of $AB$, let's call it $E$, using Pythagoras' Theorem:</p>
<p>\begin{align}(AE)^2+(EX)^2&=(AX)^2\\
\left(\frac 12\right)^2+(EX)^2&=1^2\\
\frac 14+(EX)^2&=1\\
EX^2&=\frac 34\\
EX&=\sqrt{\frac 34}\\
&=\frac{\sqrt3}2\end{align}</p>
<p>If we say $F$ is the midpoint of $CD$ then we can see that \begin{align}XF&=1-EX\\
&=1-\frac{\sqrt3}2\end{align}</p>
<p>Now, because of symmetry, we can say that $YE=1-\frac{\sqrt3}2$ as well</p>
<p>Therefore \begin{align}XY&=1-YE-XF\\
&=1-\left(1-\frac{\sqrt3}2\right)-\left(1-\frac{\sqrt3}2\right)\\
&=\sqrt 3-1\\
&\approx 0.73\end{align}</p>
|
1,849,809 | <p>Let $n$ be a positive integer. Define $$\textbf{A}_n(x):= \left[\frac{1}{x+i+j-1}\right]_{i,j\in\{1,2,\ldots,n\}}$$ as a matrix over the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in variable $x$. </p>
<p>(a) Prove that the <em>Hilbert matrix</em> $\textbf{A}_n(0)$ is an invertible matrix over $\mathbb{Q}$ and all entries of the inverse of $\textbf{A}_n(0)$ are integers.</p>
<p>(b) Determine the greatest common divisor (over $\mathbb{Z}$) of all the entries of $\big(\textbf{A}_n(0)\big)^{-1}$.</p>
<p>(c) Show that $\textbf{A}_n(x)$ is an invertible matrix over $\mathbb{Q}(x)$ and every entry of the inverse of $\textbf{A}_n(x)$ is a polynomial in $x$.</p>
<p>(d) Prove that $x+n$ is the greatest common divisor (over $\mathbb{Q}[x]$) of all the entries of $\big(\textbf{A}_n(x)\big)^{-1}$.</p>
<p><strike>Parts (a) and (c) are known.</strike> <strike>Parts (b) and (d) are open.</strike> <strike>Now, Part (d) is known (see i707107's solution below), but Part (b) remains open, although it seems like the answer is $n$.</strike></p>
<hr>
<p>Recall that
$$\binom{t}{r}=\frac{t(t-1)(t-2)\cdots(t-r+1)}{r!}$$
for all $t\in\mathbb{Q}(x)$ and $r=0,1,2,\ldots$. According to i707107, the $(i,j)$-entry of $\big(\textbf{A}_n(x)\big)^{-1}$ is given by
$$\alpha_{i,j}(x)=(-1)^{i+j}\,(x+n)\,\binom{x+n+i-1}{i-1}\,\binom{x+n-1}{n-j}\,\binom{x+n+j-1}{n-i}\,\binom{x+i+j-2}{j-1}\,.\tag{*}$$
This means that, for all integers $k$ such that $k\notin\{-1,-2,\ldots,-2n+1\}$, the entries of $\big(\textbf{A}_n(k)\big)^{-1}$ are integers. <strong><em>I now have a new conjecture, which is the primary target for the bounty award.</em></strong></p>
<blockquote>
<p><strong>Conjecture:</strong> The greatest common divisor $\gamma_n(k)$ over $\mathbb{Z}$ of the entries of $\big(\textbf{A}_n(k)\big)^{-1}$, where $k$ is an integer not belonging in the set $\{-1,-2,\ldots,-2n+1\}$, is given by $$\gamma_n(k)=\mathrm{lcm}(n,n+k)\,.$$</p>
</blockquote>
<p>It is clear from (*) that $n+k$ must divide $\gamma_n(k)$. However, it is not yet clear to me why $n$ should divide $\gamma_n(k)$. I would like to have a proof of this conjecture, or at least a proof that $n \mid \gamma_n(k)$.</p>
<hr>
<p>Let $M_n$ denote the (unitary) cyclic $\mathbb{Z}[x]$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)$. Then, the (unitary) $\mathbb{Z}[x]$-module $N_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}[x]$-submodule of $M_n$. </p>
<p>We also denote by $\tilde{M}_n$ for the (unitary) $\mathbb{Z}$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)\,x^l$ for $l=0,1,2,\ldots,2n-2$. Then, the (unitary) $\mathbb{Z}$-module $\tilde{N}_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}$-submodule of $\tilde{M}_n$.</p>
<p>For example, $M_2/N_2$ is isomorphic to the (unitary) $\mathbb{Z}[x]$-module $\mathbb{Z}/2\mathbb{Z}$ (in which $x$ acts trivially), and $\tilde{M}_2/\tilde{N}_2$ is isomorphic to the (unitary) $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$. Hence, $\left|M_2/N_2\right|=2=\left|\tilde{M}_2/\tilde{N}_2\right|$. For $n=3$, Mathematica yields
$$\tilde{M}_3/\tilde{N}_3\cong (\mathbb{Z}/2\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/4\mathbb{Z})^{\oplus 3}\,,$$
as abelian groups. That is, $\left|\tilde{M}_3/\tilde{N}_3\right|=1152$. On the other hand,
$$M_3/N_3\cong \mathbb{Z}[x] \big/\left(12,2x^2+6x+4,x^4-x^2\right)$$
as $\mathbb{Z}[x]$-modules, which gives $\left|M_3/N_3\right|=576$. </p>
<blockquote>
<p><strong>Question:</strong> Describe the factor $\mathbb{Z}[x]$-module $M_n/N_n$ and the factor $\mathbb{Z}$-module $\tilde{M}_n/\tilde{N}_n$. It is easily seen that $\left|M_n/N_n\right|\leq\left|\tilde{M}_n/\tilde{N}_n\right|$. What are $\left|M_n/N_n\right|$ and $\left|\tilde{M}_n/\tilde{N}_n\right|$? It can be shown also that the ratio $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ is an integer, provided that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite. Compute $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ for all integers $n>0$ such that $\left|\tilde{M}_n/\tilde{N}_n\right|<\infty$. Is it always the case that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite?</p>
</blockquote>
<p><strong><em>Apart from the conjecture above, this question is also eligible for the bounty award.</em></strong> I have not yet fully tried to deal with any case involving $n>3$. However, for $n=4$, the module $\tilde{M}_4/\tilde{N}_4$ is huge:
$$ \tilde{M}_4/\tilde{N}_4\cong (\mathbb{Z}/2\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 3}\oplus(\mathbb{Z}/8\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/9\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/16\mathbb{Z})\oplus(\mathbb{Z}/27\mathbb{Z})$$
as abelian groups.</p>
| Batominovski | 72,152 | <p>For Part (b), according to i707107's answer, the $(i,j)$-entry of $\textbf{H}_n:=\big(\textbf{A}_n(0)\big)^{-1}$ is given by $$h_{i,j}:=(-1)^{i+j}\,n\,\binom{n+i-1}{i-1}\,\binom{n-1}{j-1}\,\binom{n+j-1}{i+j-1}\,\binom{i+j-2}{i-1}\,.$$
Hence, $n$ is a divisor of the greatest common divisor $g_n$ over $\mathbb{Z}$ of the entries of $\textbf{H}_n$.
Note that
$$\left|h_{1,j}\right|=n\,\binom{n-1}{j-1}\,\binom{n+j-1}{j}=n\,\binom{n+j-1}{n-1}\,\binom{n-1}{j-1}=n\,\binom{n+j-1}{j-1}\,\binom{n}{j}\,;$$
in particular,
$$h_{1,1}=n^2\,.$$
Ergo, $$n\mid g_n\mid n^2\,.$$
If $p$ is a prime divisor of $n$ such that $p^k$ is the largest power of $p$ that divides $n$, then using <a href="https://en.wikipedia.org/wiki/Lucas%27_theorem" rel="nofollow">Lucas's Theorem</a>, we know that
$$\binom{n+p^k-1}{p^k-1}\equiv 1\pmod{p}$$
and
$$\binom{n}{p^k}\equiv \frac{n}{p^k}\pmod{p}\,.$$
Therefore, $p$ does not divide $\dfrac{h_{1,p^k}}{n}$, whence $p\nmid \dfrac{g_n}{n}$. Hence, the greatest common divisor of the entries of $\textbf{H}_n=\big(\textbf{A}_n(0)\big)^{-1}$ must be
$$g_n=n\,.$$</p>
|
3,382,561 | <p>I just encountered the random variable <span class="math-container">$Y = |X|$</span>, where <span class="math-container">$X \sim \text{N}(\mu, \sigma^2)$</span>. Now, based on what we know about the absolute value function, this random variable is still continuous; however, the absolute value function means that there exists a cusp at <span class="math-container">$X = 0$</span>, and so the derivative is undefined at this point. </p>
<p>This makes me wonder: How does this affect the PDF and CDF? How would we go about calculating such things in this case? </p>
<p>I would greatly appreciate it if people could please take the time to clarify this situation.</p>
| Community | -1 | <p><span class="math-container">$$P(\lvert X\rvert\le\alpha)=\begin{cases}P(-\alpha\le X\le\alpha)&\text{if }\alpha>0\\ 0&\text{if }\alpha<0\end{cases}$$</span>
Therefore the cdf is <span class="math-container">$$F_{\lvert X\rvert}(\alpha)=\begin{cases}0&\text{if }\alpha<0\\ F_X(\alpha)-\sup_{\beta<-\alpha} F_X(\beta)&\text{if }\alpha\ge 0\end{cases}$$</span></p>
<p>Since <span class="math-container">$F_X$</span> is continuous, <span class="math-container">$F_{\lvert X\rvert}(\alpha)=F_X(\alpha)-F_X(-\alpha)$</span> for all <span class="math-container">$\alpha\ge 0$</span>.</p>
<p><span class="math-container">$F_{\lvert X\rvert}$</span> will therefore be differentiable on <span class="math-container">$\Bbb R\setminus\{0\}$</span>, because <span class="math-container">$F_X$</span> is. Since <span class="math-container">$F_{\lvert X\rvert}$</span> also turns out to be continuous on the whole <span class="math-container">$\Bbb R$</span>, any function in this form <span class="math-container">$$\begin{cases}F'_{\lvert X\rvert}(\alpha)&\text{if }\alpha\ne0\\ c&\text{if }\alpha=0\end{cases}$$</span> will be a pdf of <span class="math-container">$\lvert X\rvert$</span>.</p>
|
1,170,708 | <p>What functions satisfy $f(x)+f(x+1)=x$?</p>
<p>I tried but I do not know if my answer is correct.
$f(x)=y$</p>
<p>$y+f(x+1)=x$</p>
<p>$f(x+1)=x-y$</p>
<p>$f(x)=x-1-y$</p>
<p>$2y=x-1$</p>
<p>$f(x)=(x-1)/2$</p>
| N. S. | 9,176 | <p>Replacing $x$ by $x+1$ you get</p>
<p>$$f(x+1)+f(x+2)=x+1=f(x)+f(x+1)+1$$</p>
<p>This shows that
$$f(x+2)=f(x)+1$$</p>
<p>Let $g(x)=f(x)-\frac{x}{2}$. Then $g(x)$ is periodic with period $2$, and </p>
<p>$$g(x)+g(x+1)=-\frac{1}{2}$$</p>
<p><strong>This gives your answer:</strong></p>
<p>Define $g(x)$ to be any function on $[0,1)$. Define $g:[1,2) \to \mathbb R$ by
$$g(x):=-\frac{1}{2}-g(x-1)$$</p>
<p>This defines $g:[0,2) \to \mathbb R$. Extend it to $\mathbb R$ by asking that $g$ is 2-periodic.</p>
<p>Now, let $f(x)=g(x)+\frac{x}{2}$.</p>
<p>It is easy to check that $f$ satisfies the given condition, and the above shows that any function which satisfies the given condition has this form.</p>
<p>Now, the answer can change if one adds extra conditions like continuity, differentiability, but even so I think there is enough freedom to generate uncountably many examples.</p>
|
1,170,708 | <p>What functions satisfy $f(x)+f(x+1)=x$?</p>
<p>I tried but I do not know if my answer is correct.
$f(x)=y$</p>
<p>$y+f(x+1)=x$</p>
<p>$f(x+1)=x-y$</p>
<p>$f(x)=x-1-y$</p>
<p>$2y=x-1$</p>
<p>$f(x)=(x-1)/2$</p>
| Akiva Weinberger | 166,353 | <p>The simplest answer is:
$$f(x)=\frac x2-\frac14$$
If you have any function $g$ of period 2, such that $g(x)=-g(x+1)$ for all $g$ ($\sin(\pi x)$ is a good example), then $f(x)=g(x)+\frac x2-\frac14$ also works. For example, the function $\sin(\pi x)+\frac x2-\frac14$ also satisfies it.</p>
|
1,734,819 | <p>I think I'm on the right track with constructing this proof. Please let me know.</p>
<p>Claim: Prove that there exists a unique real number $x$ between $0$ and $1$ such that
$x^{3}+x^{2} -1=0$</p>
<p>Using the intermediate value theorem we get
$$r^{3}+r^{2}-1=c^{3}+c^{2}-1$$
......
$$r^{3}+r^{2}-c^{3}-c^{2}=0$$</p>
<p>Factoring gives us</p>
<p>$$(r-c)[(r^{2}+rc+c^{2})+(r+c)]=0$$
I'm lost now. How do I prove that there exists a number between $0$ and $1$.</p>
| User8128 | 307,205 | <p>You just need the intermediate value theorem and monotonicity. Put $p(x) = x^3 + x^2 - 1$. Since $p(0) = -1$ and $p(1)= 1$, the intermediate value theorem says that there is at least one value $r \in (0,1)$ such that $p(r) = 0$. </p>
<p>Consider $p'(x) = 3x^2 + 2x > 0$ for $x \in (0,1]$. Hence $p$ is increasing. Thus for any $s < r < t$ we have $p(s) < 0 < p(t)$. This shows that there is no other point in $(0,1]$ where $p$ is zero. </p>
<p>EDIT: Alternatively, your approach works: if $p(r) = p(c) = 0$, as you stated, we have $$(r-c) (r^2+rc+c^2+r+c) = 0.$$ Thus $r - c = 0$ or $r^2+rc+c^2+r+c = 0$. In the latter case, we would have $r= 0, c = 0$ which doesn't work, since $p(0) \neq 0$, so instead we must have $r- c = 0$ which gives $r=c$. Thus the root is unique. </p>
|
1,966,128 | <p>I know that by De Morgan's law that it is false. But how to disprove it?</p>
| Siong Thye Goh | 306,553 | <p>Hint:</p>
<p>Let the universal set be $\left\{ 0,1 \right\}$</p>
<p>Let $A=\left\{ 0\right\}$ and $B=\left\{ 1\right\}$</p>
<p>Verify that it works.</p>
|
1,966,128 | <p>I know that by De Morgan's law that it is false. But how to disprove it?</p>
| Alexis Olson | 11,246 | <p>All you need is a single counterexample.</p>
<p>Let $A = \{1,2,3\}$ and $B = \{3,4,5\}$ and the universal set $U = \{1,2,3,4,5,6\}$.</p>
<p>Then $A^c = \{4,5,6\}$ and $B^c = \{1,2,6\}$ and</p>
<p>$$A^c \cup B^c = \{1,2,4,5,6\} = (A \cap B)^c$$</p>
<p>but</p>
<p>$$(A \cup B)^c = \{6\} = A^c \cap B^c$$</p>
<p>and clearly $\{1,2,4,5,6\} \not\subseteq \{6\}$.</p>
|
3,788,283 | <p>I am searching for the solution to</p>
<p><span class="math-container">$$\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{2\pi\sigma_{X}\sigma_{Y}\sqrt{1-\rho^{2}}}\exp\left(-\frac{1}{2(1-\rho^{2})}\left[\frac{(\bar{x}-\mu_{X})^{2}}{\sigma_{X}^{2}}+\frac{(\bar{y}-\mu_{Y})^{2}}{\sigma_{Y}^{2}}-\frac{2\rho(\bar{x}-\mu_{X})(\bar{y}-\mu_{Y})}{\sigma_{X}\sigma_{Y}}\right]\right)\bar{x}~\mathrm{d}\bar{x}~\mathrm{d}\bar{y}$$</span></p>
<p>Does anyone know how to derive it? I hit only unsolvable integrals!</p>
| Benedict W. J. Irwin | 330,257 | <p>I've thought about this for a few days now, I didn't originally intend to answer my own question but it seems best to write this as an answer rather than add to the question. I think there is nice interpretation in the following:
<span class="math-container">$$
f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h}
$$</span>
also consider the Abel shift operator
<span class="math-container">$$
e^{h D_x}f(x) = f(x+h)
$$</span>
from the limit form of the derivative we have (in the sense of an operator)
<span class="math-container">$$
D_x = \lim_{h \to 0} \frac{e^{h D_x}-e^{0 D_x}}{h} = \lim_{h \to 0} \frac{e^{h D_x}-1}{h}
$$</span>
now we can also manipulate the first equation to get
<span class="math-container">$$
\log f(x) = \lim_{h \to 0} \frac{f^h(x)-1}{h}
$$</span>
so by (a very fuzzy) extrapolation, we might have
<span class="math-container">$$
\log(D_x) = \lim_{h \to 0} \frac{D_x^h-1}{h}
$$</span>
and applying that to a <em>function</em> we now get
<span class="math-container">$$
\log(D_x) f(x) = \lim_{h \to 0} \frac{D_x^h f(x)-f(x)}{h}
$$</span>
which is the <span class="math-container">$\alpha = 0$</span> case of the 'super-derivative'. So <strong>one interpretation</strong> of this case is the logarithm of the derivative?
If we apply the log-derivative to a fractional derivative then we have
<span class="math-container">$$
\log(D_x) D^\alpha_x f(x) = \lim_{h \to 0} \frac{D_x^h D^\alpha_x f(x)-D^\alpha_x f(x)}{h}
$$</span>
there might be a question of the validity of <span class="math-container">$D_x^h D^\alpha_x = D_x^{\alpha+h}$</span> which I believe isn't always true for fractional derivatives.</p>
<p>This interpretation would explain the <span class="math-container">$\log(x)$</span> type terms arising in the series above. I'd be interested to see if anyone has any comments on this? I'd love to see other similar interpretations or developments on this. What are the eigenfunctions for the <span class="math-container">$\log D_x$</span> operator for example? Can we form meaningful differential equations?</p>
<p><strong>Edit:</strong> For some functions I have tried we do have the expected property
<span class="math-container">$$
n \log(D_x) f(x) = \log(D_x^n) f(x)
$$</span>
with
<span class="math-container">$$
\log(D_x^n) f(x) = \lim_{h \to 0} \frac{D_x^{n h} f(x)-f(x)}{h}
$$</span></p>
|
3,788,283 | <p>I am searching for the solution to</p>
<p><span class="math-container">$$\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{2\pi\sigma_{X}\sigma_{Y}\sqrt{1-\rho^{2}}}\exp\left(-\frac{1}{2(1-\rho^{2})}\left[\frac{(\bar{x}-\mu_{X})^{2}}{\sigma_{X}^{2}}+\frac{(\bar{y}-\mu_{Y})^{2}}{\sigma_{Y}^{2}}-\frac{2\rho(\bar{x}-\mu_{X})(\bar{y}-\mu_{Y})}{\sigma_{X}\sigma_{Y}}\right]\right)\bar{x}~\mathrm{d}\bar{x}~\mathrm{d}\bar{y}$$</span></p>
<p>Does anyone know how to derive it? I hit only unsolvable integrals!</p>
| Tom Copeland | 27,786 | <p>Seems like you have happened upon some relations similar to ones I've written about over several years. Try for starters the <a href="https://math.stackexchange.com/questions/125343/lie-group-heuristics-for-a-raising-operator-for-1n-fracdnd-betan-fra">MSE-Q&A</a> "Lie group heuristics for a raising operator for <span class="math-container">$(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}$</span>." There are several posts on my blog (see my user page) on this topic, logarithm of the derivative operator (see also <a href="https://oeis.org/A238363" rel="nofollow noreferrer">A238363</a> and links therein, a new one will be added soon, my latest blog post), and fractional differ-integral calculus.</p>
|
551,252 | <p>I'm sure I've made a trivial error but I cannot spot it.</p>
<p>Fix R>0
Consider the cube $C_R$ as the cube from (0,0,0) to (R,R,R) (save me from listing the 8 vertices) </p>
<p>Consider $S_R$ as the surface of $C_R$</p>
<p>Consider the vector field $v:\mathbb{R}^3\rightarrow\mathbb{R}^3$
given by $v(x,y,z) = (3x+z^2,2y,R-z)$</p>
<p><strong>Part 1</strong></p>
<p>Calculate $\nabla.v$</p>
<p>$\nabla.v=\frac{\partial{v_x}}{\partial{x}}+\frac{\partial{v_y}}{\partial{y}}+\frac{\partial{v_z}}{\partial{z}}=3+2-1=4$</p>
<p><strong>Part 2</strong></p>
<p>Calculate $\iiint_{C_R}\nabla.vdV$</p>
<p>$=\int_0^R\int_0^R\int_0^R4dxdydz$</p>
<p>This is trivial it is $=4R^3$</p>
<p><strong>Part 3</strong></p>
<p>Calculate the flux $\iint_{S_R}v.ndA$ where n is the unit normal to $S_R$ at the point.</p>
<p>Every keystroke is lagging now.</p>
<p>I did this by doing it over all 6 sides of the cube. The normals are trivial and the sides look like $R(1,s,t)$ for $s,t\in[0,1]$ (this is the right side) or $R(0,s,t)$ which is the left side.</p>
<p>I'm not even sure how you'd get an $R^3$ in there, I can provide more working if needed but it really isn't hard.</p>
<p>What have I done? Is this result right and perhaps I have misunderstood something?</p>
<p><strong>Addendum</strong></p>
<p>I think my error may come from my parameter ranges, I should be going from 0 to R not 0 to 1. This is essentially a substitution where I didn't multiply by the rate of that substitution with respect to the thing it replaces. </p>
| AlexR | 86,940 | <p><strong>Hint</strong><br>
You can do
$$\iint_{S_R} v\cdot n\ dA = \int_0^R \int_0^R v(0,x',y') \cdot -e_1 + v(R, x', y') \cdot e_1 + v(x', 0, y')\cdot-e_s + \ldots dx' dy'$$
The double integral will yield a $R^2$-like term and the integrand should give another (By Gauss, the results of (1) and (2) should be equal).</p>
<hr>
<p>$$\begin{align*}
v(0,s,t) \cdot -e_1 & = -t^2 \\
v(R,s,t) \cdot +e_1 & = 3R + t^2 \\
v(s,0,t) \cdot -e_2 & = 0 \\
v(s,R,t) \cdot +e_2 & = 2R \\
v(s,t,0) \cdot -e_3 & = -R \\
v(s,t,R) \cdot +e_3 & = 0 \\
\Rightarrow \iint_{S_R} v\cdot n\ dA & = \iint_{(0,0)}^{(R,R)} 4R\ ds\ dt = 4R^3
\end{align*}$$</p>
|
4,566,926 | <p><span class="math-container">$a_n$</span> converges to <span class="math-container">$a \in \mathbb{R}$</span>, find <span class="math-container">$\lim\limits_{n \rightarrow \infty}\frac{1}{\sqrt{n}}(\frac{a_1}{\sqrt{1}} + \frac{a_2}{\sqrt{2}} +\frac{a_3}{\sqrt{3}} + \cdots + \frac{a_{n-1}}{\sqrt{n-1}} + \frac{a_n}{\sqrt{n}})$</span></p>
<p>I've tried to use Squeeze theorem. Nothing good. Maybe I've tried it in a wrong way. According to my assumptions, this sequence diverges (I mean "goes" to infinity).</p>
<p>Can we solve this using Stolz–Cesàro theorem? Is there easier/better/more correct approach? If there is, please, share your thoughts</p>
| Oliver Díaz | 121,671 | <p>Given <span class="math-container">$\varepsilon>0$</span>, there is <span class="math-container">$N$</span> such that <span class="math-container">$\sup_{n\geq N}|a_n-a|<\varepsilon$</span>
<span class="math-container">$$\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}=\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k-a}{\sqrt{k}} +\frac{a}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}$$</span>
The second term is the Riemann sum of the convergent integral <span class="math-container">$a\int^1_0\frac{1}{\sqrt{x}}$</span> and thus, as <span class="math-container">$x\mapsto\frac{1}{\sqrt{x}}$</span> is monotone nondecreasing,
<span class="math-container">$$\frac{a}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}\xrightarrow{n\rightarrow\infty}2a$$</span>
The first term can be estimated similarly.
<span class="math-container">$$\Big|\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k-a}{\sqrt{k}}\Big|\leq\frac{1}{\sqrt{n}}\sum^N_{k=1}\frac{|a_k-a|}{\sqrt{k}}+\varepsilon\frac{1}{\sqrt{n}}\sum^n_{k=N+1}\frac{1}{\sqrt{k}}$$</span>
The term <span class="math-container">$\frac{1}{\sqrt{n}}\sum^N_{k=1}\frac{|a_k-a|}{\sqrt{k}} \xrightarrow{n\rightarrow\infty}0$</span> (sińce <span class="math-container">$N$</span> is fixed), the second one is bounded by <span class="math-container">$\frac{\varepsilon}{n}\sum^n_{k=1}\frac{\sqrt{n}}{\sqrt{k}}\xrightarrow{n\rightarrow\infty}\varepsilon\int^1_0\frac{1}{\sqrt{x}}=2\varepsilon$</span>.</p>
<p>Putting things together,
<span class="math-container">$$\limsup_n\Big|\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k}{\sqrt{k}} -2a\Big|\leq2\varepsilon$$</span>
Therefore limit of the expression in the OP exists and is <span class="math-container">$2a$</span>.</p>
|
90,130 | <p>So now that my term's over, I've been brushing up on my quantum field theory, and I came across the following line in my textbook without any justification:</p>
<p>$$\frac{1}{4\pi^2}\int_m^{\infty}\sqrt{E^2-m^2}e^{-iEt}dE \sim e^{-imt}\text{ as }t\to\infty$$</p>
<p>Well, I can see intuitively that <em>if</em> most of the integral cancels out, the main contribution will be from the region $E\approx m$, since (under a coordinate transformation) that's the region of stationary phase. But I'm a mathematician, dangit, not a physicist, and I want this to be rigourous.</p>
<p>The Riemann-Lebesgue lemma, if I'm not mistaken, doesn't apply since $\sqrt{E^2-m^2}$ is unbounded as $E\to\infty$, and it certainly isn't $L^1$. And I guess I could shift the path of the integral off the real axis in the complex plane, but I don't see why that would be the <em>right</em> way to take the integral. The whole thing is giving me the heebie-jeebies, and I was hoping one of you folks could assuage my fears.</p>
| Sasha | 11,069 | <p>Clearly the integral as stated diverges, so one needs to regularize it. To that end, consider $t$ complex with small negative imaginary part, to ensure that it converges at $E\to\infty$.</p>
<p>The integral directly matches the following <a href="http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/01/01/0001/" rel="nofollow">integral representation</a> of the Bessel function of the second kind:
$$
K_{\nu }(z)=\frac{\sqrt{\pi } z^{\nu } }{2^{\nu } \, \Gamma \left(\nu
+\frac{1}{2}\right)} \, \int_1^{\infty }
\left(t^2-1\right)^{\nu -\frac{1}{2}} e^{-z t} \, \mathrm{d}t
$$
valid for $\mathfrak{Re}(\nu) > -\frac{1}{2}$ and $\mathfrak{Re}(z) > 0$.</p>
<p>Thus:
$$
\frac{1}{4 \pi^2} \int_m^\infty \mathrm{e}^{-i t \mathcal{E}} \sqrt{ \mathcal{E}^2-m^2} \mathrm{d} \mathcal{E} = -i \frac{m}{4 \pi^2 t} K_1\left( i m t \right)
$$
By means of regularization we proclaim that integral equal to the rhs even for real $t$. Expanding:
$$
-i \frac{m}{4 \pi^2 t} K_1\left( i m t \right) = -i \frac{m}{8 \pi t} H_1(m t) + i \frac{m}{4 \pi t} \operatorname{sign}(t) J_1(m t)
$$
This allows to conclude that for $t \to +\infty$, the expression is proportional to $\mathrm{e}^{-i m t} \frac{\mathrm{e}^{-i 3 \pi/4}}{8} \sqrt{m} (\pi t)^{-3/2}$</p>
|
90,130 | <p>So now that my term's over, I've been brushing up on my quantum field theory, and I came across the following line in my textbook without any justification:</p>
<p>$$\frac{1}{4\pi^2}\int_m^{\infty}\sqrt{E^2-m^2}e^{-iEt}dE \sim e^{-imt}\text{ as }t\to\infty$$</p>
<p>Well, I can see intuitively that <em>if</em> most of the integral cancels out, the main contribution will be from the region $E\approx m$, since (under a coordinate transformation) that's the region of stationary phase. But I'm a mathematician, dangit, not a physicist, and I want this to be rigourous.</p>
<p>The Riemann-Lebesgue lemma, if I'm not mistaken, doesn't apply since $\sqrt{E^2-m^2}$ is unbounded as $E\to\infty$, and it certainly isn't $L^1$. And I guess I could shift the path of the integral off the real axis in the complex plane, but I don't see why that would be the <em>right</em> way to take the integral. The whole thing is giving me the heebie-jeebies, and I was hoping one of you folks could assuage my fears.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>A '<em>qualitative approach</em>':</p>
</blockquote>
<p>\begin{align}
&{1 \over 4\pi^{2}}\int_{m}^{\infty}\root{E^{2} - m^{2}}\expo{-\ic Et}\,\dd E
\,\,\,\stackrel{E\ =\ m + \varepsilon^{2}}{=}\,\,\,
{\expo{-\ic mt} \over 2\pi^{2}}\int_{0}^{\infty}\root{\varepsilon^{2} + 2m}\expo{-\ic \varepsilon^{2} t}\varepsilon^{2}\,\dd\varepsilon
\\[5mm] \stackrel{\varepsilon/\root{2m}\ \mapsto\ \varepsilon}{=}\,\,\, &
\expo{-\ic mt}\,{2m^{2} \over \pi^{2}}
\int_{0}^{\infty}\root{\varepsilon^{2} + 1}
\exp\pars{-2mt\varepsilon^{2}\,\ic}\varepsilon^{2}\,\dd\varepsilon
\end{align}</p>
<blockquote>
<p>As $\ds{t \to \infty}$, the '<em>main contribution</em>' arises ftom values of $\ds{\varepsilon \lesssim 1/\pars{2mt}^{1/2}}$ such that</p>
</blockquote>
<p>\begin{align}
&{1 \over 4\pi^{2}}\int_{m}^{\infty}\root{E^{2} - m^{2}}\expo{-\ic Et}\,\dd E \sim
\expo{-\ic mt}\,{2m^{2} \over \pi^{2}}
\int_{0}^{1/\pars{2mt}^{1/2}}\varepsilon^{2}\,\dd\varepsilon =
\expo{-\ic mt}\,{2m^{2} \over \pi^{2}}\bracks{\pars{2mt}^{-3/2} \over 3}
\\[5mm] = &\
\color{#f00}{{\root{2} \over 6\pi^{2}}}\,\root{m}t^{-3/2}\expo{-\ic mt}
\end{align}</p>
<blockquote>
<p>The $\color{#f00}{prefactors}$ can only be determined from the 'exact evaluation' ( see the <a href="https://math.stackexchange.com/a/1619372/85343">@Brightsun fine answer</a> ) but the qualitative evaluation yields the correct asymptotic behaviours $\bbx{\ds{\root{m}t^{-3/2}\expo{-\ic mt}}}$</p>
</blockquote>
|
1,679,044 | <blockquote>
<p>Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle. Then find the area.</p>
<p>$$y = 2x^2, y = 8x^2, 3x + y = 5, x ≥ 0$$ </p>
</blockquote>
<p>Here is my drawing: <a href="http://www.webassign.net/waplots/d/a/986f170c68ee080b539049f410cbba.gif" rel="nofollow">http://www.webassign.net/waplots/d/a/986f170c68ee080b539049f410cbba.gif</a> </p>
<p>If I'm integrating with in terms of x, I know I'm going to need more than one integral, but I don't know how to set it up. Any help is appreciated</p>
| DylanSp | 308,461 | <p>You'll use one integral from $x = 0$ to $x = \frac{5}{8}$, where $y = 8x^2$ intersects $3x + y = 5$, and another integral from $x = \frac{5}{8}$ to $x = 1$. Put together, this gives the expression
$$
\left[\int_{0}^{5/8} 8x^2 - 2x^2\,dx\right] + \left[\int_{5/8}^{1} (-3x + 5) - 2x^2\,dx \right]
$$</p>
|
27,951 | <p>Something I notice is when there's an advanced/specialized question, it often receives very few upvotes. Even if it is seemingly well written. I try to upvote advanced questions <strong>that I might not even understand</strong>, if they appear well written. </p>
<p>Is this good behaviour? Should we encourage upvoting seemingly well-written questions <strong>even if you don't understand it</strong>? </p>
| Stella Biderman | 123,230 | <p>I think that the general guidelines for "asking a good question" on this site are nearly content agnostic. I might write a horribly thought out and poorly reasoned question about category theory which belies a fundamental misunderstanding of what a morphism is, but that isn't grounds to call it a bad question. The only real reason a question might be a bad question and require you to know a lot about the subject is for the question to be so fundamentally confused that it's unclear what is being asked. However, I think that this is generally a rare phenomenon and so I would be comfortable with people upvoting questions based solely on their non-mathematical content, such as their context, attempts at work, formatting, and length.</p>
|
1,753,263 | <p>I've been struggling with understanding how surface integrals come to be. Can someone please explain why they have such a format with a normal vector to the surface?
thanks!</p>
| Chris Rackauckas | 244,186 | <p>Part of it is because of the fact that these methods were first used to understand physics, and there are clear physical interpretations. The vector coming out of the surface is the "flow", and so a lot of physical laws like Guass' law and Fick's ask about balancing flows (i.e. for Gauss' law, if a ball contains electrical charge, then the total flow of "electric force" out of the ball is proportional to the total charge, since any electrical charge outside the ball has a force which flows inward and one that flows outward and together those cancel).</p>
<p>Geometrically, it's because you have to scale the quantity by the "infinitesimal area", and the normal vector comes from the cross product of vectors on the surface and is thus a measure of area. If you need to dot the vector with something to get a scalar, then this is the "natural" quantity.</p>
<p>More mathematically, it's because the gradient (or differential form) is a covector, and thus is a function that takes in a vector and spits out a number (this is something that you go into depth in differential geometry). This is because of how it has to handle coordinate transformations. </p>
|
1,753,263 | <p>I've been struggling with understanding how surface integrals come to be. Can someone please explain why they have such a format with a normal vector to the surface?
thanks!</p>
| spinoza | 310,540 | <p>Let $F$ be a vector field, $n$ a normal vector to the surface, with $u=\frac{n}{|n|}$ the unit normal vector. </p>
<p>The integral of the flux of a vector field through a surface requires an orientation (direction) that defines an "inside" and an "outside" relative to the surface that is the boundary between the "inside" and the "outside" . </p>
<p>We want to ask, for example, how much of the vector field $F$ is "escaping" from the the inside to the outside. To do this, we decompose the vector field into two orthogonal components: (1) the part normal to the surface; (2) the part tangent to the surface. At any given point on the surface, only the part normal to the surface escapes. We find this part by taking the dot product with the unit normal vector, which gives the orthogonal projection of the vector field onto the normal line: $F\cdot u=|F|\cos\theta,$ where $\theta$ is the angle between the two vectors.</p>
<p>The local surface area about a point on the surface is given by the magnitude of the cross product of two vectors tangent to the surface at that point. This cross product depends on orientation, $u\times v=-v\times u,$ but both of these are normal vectors to the surface (they point in opposite directions). Depending on our choice of orientation, we choose one of these to be $n.$ Then
$n=u|n|$, where $u$ is our unit normal vector, and $n=dS$ is the infinitesimal area. </p>
<p>At each point on the surface, we see that the total of the vector field escaping through the surface is the normal component $F\cdot u$ multiplied by the local (infinitesimal) surface area $|n|=dS.$ Summing over all points on the surface we get
$$
\int_S F\cdot u\ dS.
$$</p>
|
3,605,368 | <p>Imagine a <span class="math-container">$9 \times 9$</span> square array of pigeonholes, with one pigeon in each pigeonhole. Suppose that all at once, all the pigeons move up, down, left, or right by one hole. (The pigeons on the edges are not allowed to move out of the array.) Show that some pigeonhole winds up with two pigeons in it.</p>
<p>Let each side of the square be n. There are <span class="math-container">$n^2$</span> pigeons and pigeonholes. If the pigeons are shifted in any direction, then there will be n empty pigeonholes on the side opposite to the direction. Furthermore, now <span class="math-container">$n^2$</span> pigeons are trying to fit into <span class="math-container">$n^2 - n$</span> pigeonholes. We can invoke the pigeon hole principle as follows: Let the entire set of pigeons be <span class="math-container">$X$</span> and the set of pigeonholes to be populated after the shift be <span class="math-container">$Y$</span>. For <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> and for some integer <span class="math-container">$k$</span>, if <span class="math-container">$X > k Y$</span>, and <span class="math-container">$f X: \to Y$</span>, then <span class="math-container">$f(x) = \ldots = f(x {\rm till\ index}\ k+1)$</span>.</p>
<p>So, <span class="math-container">$81 > 72 k$</span> which means <span class="math-container">$k > 1.125$</span> which means <span class="math-container">$k = 2$</span>. This means that there are at least <span class="math-container">$3$</span> instances with <span class="math-container">$2$</span> pigeons in it.</p>
<p>Now intuitively I know there ought to be <span class="math-container">$9$</span> instances. Where did I go wrong? Forgive me if I have butchered the whole thing. I am new to this type of math. </p>
| Christian Blatter | 1,303 | <p>A hint: Color the <span class="math-container">$9^2$</span> squares checkerboard like!</p>
|
38,915 | <p>It has to be the silliest question, but it’s not clear to me how to calculate eigenvectors quickly. I am just talking about a very simple 2-by-2 matrix.</p>
<p>When I have already calculated the eigenvalues from a characteristic polynomial, I can start to solve the equations with $A\mathbf{v}_1 = e_1\mathbf{v}_1$ and A\mathbf{v}_2 = e_2\mathbf{v}_2$, but in this case it always requires writing lines of equations and solving them.</p>
<p>On the other hand I figured out that just by looking at the matrix you can come up with the eigenvectors very quickly. But I'm a bit confused in this part.</p>
<p>When you have the matrix with subtracted $e_1$ values like this:
$$\left(\begin{array}{cc}
A&B\\
C&D
\end{array}\right).$$</p>
<p>Then for me, it always worked to use the eigenvector.
$$\left(\begin{array}{r}-B\\A\end{array}\right)$$</p>
<p>But in some guides I find that they are using A C as an eigenvector.
$$\left(\begin{array}{c}A\\C
\end{array}\right).$$</p>
<p>And when I check it, they are indeed multiples of each other. But this other method is not clear to me, how could $A$ and $C$ mean anything about the eigenvector, when both of them are connected to $x$, without having to do anything with $y$. But it’s still working. Was it just a coincidence?</p>
<p>So is the recommended method for calculating them is just to subtract the eigenvalues from the matrix and look at
$$\left(\begin{array}{r}
-B\\A\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{r}
-D\\A
\end{array}\right).$$</p>
| Jiangwei Xue | 9,430 | <p>Please goes through this list one by one. </p>
<ol>
<li>Your method will work as long as either $A$ or $B$ is not zero. </li>
<li>When doesn't it work? Try calculate the eigenvector of $\left(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix} \right )$.</li>
<li>Does the method of taking $\left(\begin{smallmatrix} A\\C \end{smallmatrix} \right )$ work? Try calculate the eigenvector of $\left(\begin{smallmatrix} 3 & 3 \\ 4 & 7 \end{smallmatrix} \right )$ and see it for yourself.</li>
<li>What is an obvious solution to the equation $Ax+By=0$? Hint: it has something to do with your method. </li>
<li>Say when either $A$ or $B$ is nonzero, what will happen to the second row when you do row reduction? Why? Remember, the determinant of $\left(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right )$ is zero. </li>
<li>Can you see why your method works by now?</li>
</ol>
|
70,500 | <p>I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.</p>
| Michael Hardy | 11,667 | <p>$$
e^y+e^{-y}=2
$$
Letting $u = e^y$, this becomes
$$
u + \frac 1u = 2
$$
Multiplying both sides by $u$:
$$
u^2 + 1 = 2u
$$
That's just a quadratic equation.</p>
|
1,976,733 | <p>What does the following statement mean?</p>
<blockquote>
<p>Suppose that $X$ is a random variable for the experiment, taking values in a set $T$.
The function $B↦P(X∈B)$ defines a probability measure on $T$.</p>
</blockquote>
<p>Let, </p>
<p>$Experiment$ = a dice rolls.</p>
<p>$\Omega = \{1,2,3,4,5,6\}$</p>
<ul>
<li>What are the element of $T$?</li>
<li>Is $B$ a function of a set?</li>
<li>What would be the elements of set $B$?</li>
<li><p>What does it mean by $X∈B$?</p></li>
<li><p>What does the symbol $↦$ represent?</p></li>
</ul>
| Anthony P | 377,521 | <p>Consider the Real interval $[0,\epsilon]$. How many points are inside that interval? Divide the interval so that the new interval is $[0,\epsilon/2] $. How many points are inside this interval? Still infinitely many. We say that $\mathbb{Q}$ is dense in $\mathbb{R}$ in the language of Real Analysis. This means you can always find rational numbers in a neighbourhood of a real number (loosely speaking). In fact, you can always find infinitely many rational numbers inside an arbitrarily small (non zero) real interval </p>
|
1,976,733 | <p>What does the following statement mean?</p>
<blockquote>
<p>Suppose that $X$ is a random variable for the experiment, taking values in a set $T$.
The function $B↦P(X∈B)$ defines a probability measure on $T$.</p>
</blockquote>
<p>Let, </p>
<p>$Experiment$ = a dice rolls.</p>
<p>$\Omega = \{1,2,3,4,5,6\}$</p>
<ul>
<li>What are the element of $T$?</li>
<li>Is $B$ a function of a set?</li>
<li>What would be the elements of set $B$?</li>
<li><p>What does it mean by $X∈B$?</p></li>
<li><p>What does the symbol $↦$ represent?</p></li>
</ul>
| Mauro ALLEGRANZA | 108,274 | <p>See the full discusssion in <a href="https://books.google.it/books?id=jF1j0y2-hM0C&pg=PA65" rel="nofollow noreferrer"><strong>CH:VII RATIONAL, REAL, AND COMPLEX NUMBERS</strong></a>; he is defining <em>fractions</em> :</p>
<blockquote>
<p>We shall define the <em>fraction</em> <span class="math-container">$\dfrac m n$</span> as being that relation which holds between two inductive numbers [see page 27 : We shall use the phrase “inductive numbers” to mean thesame set as we have hitherto spoken of as the “<em>natural numbers</em>.”] <span class="math-container">$x, y$</span> when <span class="math-container">$xn = ym$</span>. This definition enables us to prove that <span class="math-container">$\dfrac m n$</span> is a one-one relation, provided neither <span class="math-container">$m$</span> nor <span class="math-container">$n$</span> is zero.</p>
<p>It will be seen that <span class="math-container">$\dfrac 0 n$</span> is always the same relation, whatever inductive number <span class="math-container">$n$</span> may be; it is, in short, the relation of <span class="math-container">$0$</span> to any other inductive cardinal. [Thus :] Zero is <em>one-many</em>.</p>
<p>Conversely, the relation <span class="math-container">$\dfrac m 0$</span> is always the same, whatever inductive number <span class="math-container">$m$</span> may be. There is not any inductive cardinal to correspond to <span class="math-container">$\dfrac m 0$</span>. We may call it “the infinity of rationals.” It is an instance of the
sort of infinite that is traditional in mathematics, and that is represented by “<span class="math-container">$\infty$</span>.” [Thus:] infinity is <em>many-one</em>.</p>
</blockquote>
<p>In other words, if we agree to "name" with the symbol “<span class="math-container">$\infty$</span>” the relation <span class="math-container">$\dfrac m 0$</span>, this holds for any <span class="math-container">$m$</span>.</p>
|
389,425 | <blockquote>
<p>What kind of math topics exist?</p>
</blockquote>
<p>The question says everything I want to know, but for more details: I enjoy studying mathematics but the problem is that I can't find any information with a summary of all math topics, collected together. I also googled this and took a look at other websites and searched this website but without success.</p>
<p>So if someone knows most of the topics, then please let me know them. The topics I am looking for are the ones from basics, to the university, and beyond university limit. Any comprehensive information you can provide would be useful.</p>
<p>Further thanks and have a nice day.</p>
| amWhy | 9,003 | <p>You might want to start at the <a href="http://www.math-atlas.org/">Mathematical Atlas</a>. It gives a visual map of domains of mathematics, and how they are inter-related: you can click on any "bubble" to learn more about each area. There is also a menu to the right-hand side: for example, see <a href="http://www.math-atlas.org/">A Gentle Introduction to the Mathematics Subject Classification Scheme</a>, also referred to as MSC.</p>
<p>The Mathematics Subject Classification was developed to help categorize <em>publications</em> and research in mathematics. </p>
<hr>
<p>See also the entry in Wikipedia: <a href="http://en.wikipedia.org/wiki/Lists_of_mathematics_topics">Lists of Mathematical Topics</a>. This list includes topics which may sound more familiar to you, and groups them together in a sort of classification, each topic with its own link, so you can further explore any/all of them, at your leisure. </p>
<blockquote>
<p>The purpose of this list is not similar to that of the Mathematics Subject Classification formulated by the American Mathematical Society....This list has some items that would not fit in such a classification, such as list of exponential topics and list of factorial and binomial topics, which may surprise the reader with the diversity of their coverage.</p>
</blockquote>
|
3,645,263 | <p>I have recently started studying Set Theory in a self-thaught way, for that purpose I have been following Kunen's book: Set Theory: An Introduction to Independence Proofs. I'm in Chapter I section 7 and it has been defined the ordinals addition but I don't quite understand that definition. I have seen that in other books authors defines the addition using transfinite induction and it seems easier but now I want to understand Kunen's one.</p>
<p><span class="math-container">$$\alpha + \beta=type(\alpha \times \{ 0 \} \cup \beta \times \{1\}, R) \:\text{where } $$</span>
<span class="math-container">$$R=\{ \langle \langle \xi,0 \rangle, \langle \eta , 0\rangle \rangle : \xi<\eta<\alpha\} \; \cup \{\langle \langle \xi,1 \rangle, \langle \eta , 1\rangle \rangle : \xi<\eta<\beta\} \; \cup [(\alpha\times\{0\})\times(\beta\times\{1\})]. $$</span></p>
<p>With <span class="math-container">$type(A,R)$</span> is the unique ordinal <span class="math-container">$C$</span> such that <span class="math-container">$\langle A, R\rangle \cong C$</span> when <span class="math-container">$\langle A,R\rangle$</span> is a well-ordering set.</p>
<p>What I think I understood so far is that this definition tries to order two non-disjoint sets having that <span class="math-container">$\alpha<\beta$</span> and keeping the order inside <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>. What I can't understand is how I can get that cardinal (in for example a finite case), maybe I'm being stubborn and I should "ignore" this definition and trying to understand the more simplified one given in the later results.</p>
<p>Thank you for your time. </p>
| Ross Millikan | 1,827 | <p>As Kunen explains it you are putting a copy of <span class="math-container">$\beta$</span> after <span class="math-container">$\alpha$</span> and looking at the resulting order type. Whether <span class="math-container">$\alpha \lt \beta$</span> is not important. If they are both finite it is just regular addition. If you add <span class="math-container">$\omega+2$</span> and <span class="math-container">$\omega+1$</span> the order type is <span class="math-container">$\omega, 2, \omega, 1$</span>. The <span class="math-container">$2$</span> gets absorbed into the start of the following <span class="math-container">$\omega$</span> and the result is <span class="math-container">$\omega + \omega + 1$</span></p>
|
3,645,263 | <p>I have recently started studying Set Theory in a self-thaught way, for that purpose I have been following Kunen's book: Set Theory: An Introduction to Independence Proofs. I'm in Chapter I section 7 and it has been defined the ordinals addition but I don't quite understand that definition. I have seen that in other books authors defines the addition using transfinite induction and it seems easier but now I want to understand Kunen's one.</p>
<p><span class="math-container">$$\alpha + \beta=type(\alpha \times \{ 0 \} \cup \beta \times \{1\}, R) \:\text{where } $$</span>
<span class="math-container">$$R=\{ \langle \langle \xi,0 \rangle, \langle \eta , 0\rangle \rangle : \xi<\eta<\alpha\} \; \cup \{\langle \langle \xi,1 \rangle, \langle \eta , 1\rangle \rangle : \xi<\eta<\beta\} \; \cup [(\alpha\times\{0\})\times(\beta\times\{1\})]. $$</span></p>
<p>With <span class="math-container">$type(A,R)$</span> is the unique ordinal <span class="math-container">$C$</span> such that <span class="math-container">$\langle A, R\rangle \cong C$</span> when <span class="math-container">$\langle A,R\rangle$</span> is a well-ordering set.</p>
<p>What I think I understood so far is that this definition tries to order two non-disjoint sets having that <span class="math-container">$\alpha<\beta$</span> and keeping the order inside <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>. What I can't understand is how I can get that cardinal (in for example a finite case), maybe I'm being stubborn and I should "ignore" this definition and trying to understand the more simplified one given in the later results.</p>
<p>Thank you for your time. </p>
| Brian M. Scott | 12,042 | <p>The definition that Ken is using amounts to placing a copy of the ordinal <span class="math-container">$\beta$</span> after the ordinal <span class="math-container">$\alpha$</span>. Since the sets <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> are not actually disjoint (unless one of them is <span class="math-container">$0$</span>), we first use a small trick to make disjoint copies of them, replacing <span class="math-container">$\alpha$</span> by <span class="math-container">$\alpha\times\{0\}$</span> and <span class="math-container">$\beta$</span> by <span class="math-container">$\beta\times\{1\}$</span>. We give these the obvious orders, which I’ll call <span class="math-container">$\le_\alpha$</span> and <span class="math-container">$\le_\beta$</span>: for <span class="math-container">$\xi,\eta\in\alpha$</span> we set <span class="math-container">$\langle\xi,0\rangle\le_\alpha\langle\eta,0\rangle$</span> iff <span class="math-container">$\xi\le\eta$</span>, and we define <span class="math-container">$\le_\beta$</span> similarly. As sets these relations are</p>
<p><span class="math-container">$$\le_\alpha=\{\langle\langle\xi,0\rangle,\langle\eta,0\rangle\rangle:\xi\le\eta<\alpha\}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\le_\beta=\{\langle\langle\xi,1\rangle,\langle\eta,1\rangle\rangle:\xi\le\eta<\beta\}\;.$$</span></p>
<p>Now we have disjoint copies of <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> — copies in the sense that they are order-isomorphic to <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>, respectively — and we define an order that places the copy of <span class="math-container">$\beta$</span> after the copy of <span class="math-container">$\alpha$</span>. We do this by imposing the reverse lexicographic order on <span class="math-container">$(\alpha\times\{0\})\cup(\beta\times\{1\})$</span>. That is, we order first on the <strong>second</strong> coordinate and then on the first: we define</p>
<p><span class="math-container">$$\langle\xi,i\rangle\,R\,\langle\eta,j\rangle\text{ iff }i<j,\text{ or }i=j\text{ and }\xi\le\eta\;.$$</span></p>
<p>If you check the various possibilities, you’ll see that this makes </p>
<p><span class="math-container">$$\langle\xi,0\rangle\,R\,\langle\eta,0\rangle\text{ iff }\langle\xi,0\rangle\le_\alpha\langle\eta,0\rangle\text{ iff }\xi\le\eta$$</span></p>
<p>for <span class="math-container">$\xi,\eta\in\alpha$</span>,</p>
<p><span class="math-container">$$\langle\xi,1\rangle\,R\,\langle\eta,1\rangle\text{ iff }\langle\xi,1\rangle\le_\beta\langle\eta,1\rangle\text{ iff }\xi\le\eta$$</span></p>
<p>for <span class="math-container">$\xi,\eta\in\beta$</span>, and <span class="math-container">$\langle\xi,0\rangle\,R\,\langle\eta,1\rangle$</span> whenever <span class="math-container">$\xi\in\alpha$</span> and <span class="math-container">$\eta\in\beta$</span>. In short, <span class="math-container">$R$</span> orders <span class="math-container">$\alpha\times\{0\}$</span> just like <span class="math-container">$\le_\alpha$</span> and <span class="math-container">$\beta\times\{1\}$</span> just like <span class="math-container">$\le_\beta$</span>, and it places all of <span class="math-container">$\alpha\times\{0\}$</span> before all of <span class="math-container">$\beta\times\{1\}$</span>.</p>
|
1,090,658 | <p>I'm doing some previous exams sets whilst preparing for an exam in Algebra.</p>
<p>I'm stuck with doing the below question in a trial-and-error manner:</p>
<p>Find all $ x \in \mathbb{Z}$ where $ 0 \le x \lt 11$ that satisfy $2x^2 \equiv 7 \pmod{11}$</p>
<p>Since 11 is prime (and therefore not composite), the Chinese Remainder Theorem is of no use? I also thought about quadratic residues, but they don't seem to solve the question in this case.</p>
<p>Thanks in advance</p>
| kingW3 | 130,953 | <p>$$2x^2\equiv 7\pmod{11}\\2x^2\equiv -4\pmod{11}\\x^2\equiv -2\pmod{11}\\x^2\equiv 9\pmod{11}\\x^2\equiv\pm3\pmod{11}$$
As HSN mentioned since $11$ is prime than there are at most 2 solutions,</p>
|
46,446 | <p>I'm an highschool graduate who is currently waiting for college.</p>
<p>Meanwhile, I'm trying to do a little project by myself. (Computer stuff) And yesterday, I found that I needed to deal with something called "De Casteljau algorithm"</p>
<p>I know calculus (single-variable, but I'm thinking about learning multi-)
and I don't want an empty 'memorize-without-understanding-or-proving' approach.</p>
<p>Which path will take me there?
(I'm hoping for answers like:
"Calculus -> Differential Eq -> ...")</p>
<p>p.s: I would also appreciate book/video lecture recommendations :)</p>
<p>Thank you!</p>
| Edison | 11,857 | <p>What exactly do you need the De Casteljau algorithm for? You might be able to use other methods and get good results. Anyway, it believe you have the required mathematical tools to work and understand De Casteljau's algorithm. </p>
<p>Here is a similar question
<a href="https://stackoverflow.com/questions/6271817/casteljaus-algorithm-practical-example">https://stackoverflow.com/questions/6271817/casteljaus-algorithm-practical-example</a></p>
<p>search on computer graphics websites and you will get more information with examples. here is one such example
<a href="http://www.cc.gatech.edu/classes/AY2007/cs3451_fall/bezier.pdf" rel="nofollow noreferrer">http://www.cc.gatech.edu/classes/AY2007/cs3451_fall/bezier.pdf</a></p>
|
197,393 | <p>Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?</p>
| Robert Israel | 8,508 | <p>Note that
$$ \tan \left(\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) \right)=z
$$
so that $$\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) = \arctan(z) + n \pi $$
for the appropriate integer $n$.
For integers $z$ we get interesting arctan identities from this. </p>
<p>$$\eqalign{ \arctan(1) + \arctan\left(2\right)+ \arctan\left(3\right) &= \pi \cr
\arctan(2) + \arctan(4) + \arctan(13) &= \arctan(1) + \pi \cr
\arctan(3) + \arctan(8) + \arctan(57) &= \arctan(2) + \pi \cr
\arctan(4) + \arctan(14) + \arctan(183) &= \arctan(3) + \pi \cr}$$
etc.</p>
|
1,653,806 | <p>This is a but of a more mathematically juvenile question but I'm trying to get all my intuition in order. When taking a limit we can cancel things that might be zero because in taking a limit, we allow ourselves to avoid trouble spots. If the limit exists, it is very much a number/function which could be zero. So, if I have a function $f$ differentiable at a point $a$ then I have </p>
<p>$$\lim_{z \to a}\frac{f(z)-f(a)}{z-a}=f'(z)$$</p>
<p>Now, this implies continuity since</p>
<p>$$\lim_{z \to a}f(z)-f(a)=\lim_{z \to a}\frac{f(z)-f(a)}{z-a}\cdot f'(z)\lim_{z \to a}z-a$$</p>
<p>Now, this idea makes perfect sense to me but we know that the limit $z-a$ approaches is $0$. How is it we are comfortable cancelling them? Is it within the $\epsilon, \delta$ language? If my question is unclear let me know. Thanks for your help!</p>
| Markus Heinrich | 127,899 | <p>Since you said you have a mathematical background and know about the unification in terms of the exterior derivative, I thought I should add some information about the connection to the exterior derivative here.</p>
<p>As it is often the case, the generalisation of concepts can be performed in various ways, because the aspects they rely on fall together in special cases. For the divergence of vector fields $V$ we have two coinciding definitions, namely as the trace of $\nabla V$, which is locally $\nabla_a V^a$ and motivated by the divergence on $\mathbb{R}^n$, and via the canoncial identification with 1-forms, namely $\operatorname{div} V=*d*V^\flat$. Here $V^\flat=g(V,\cdot)$ and $*$ is the Hodge dual.</p>
<p>Differential forms are only a subset of all tensor fields, so it is natural to use the first definition to define the divergence of a $(p,q)$-tensor $T$ by taking a trace of $\nabla T$. How this is precisely done depends on the author. In Riemannian geometry you sometimes encounter the definition
$$
\operatorname{div}T(Y_1,\dots,Y_{q-1}) = \operatorname{tr}(X\mapsto (\nabla_XT(\cdot,Y_1,\dots,Y_{q-1}))^\sharp)=\sum_i\nabla_{E_i}T(E_i,Y_1,\dots,Y_{q-1}).
$$
Here the dependance on forms is implicit. The last equality holds locally after choosing a local frame $\{E_i\}$. With this you can show that $\operatorname{div}(fg)=df$ for any function $f$ and that $2\operatorname{div} \operatorname{Ric}=dS$, so that
$$
\operatorname{div}\left(\operatorname{Ric}-\frac{1}{2}Sg\right)=0,
$$
which is the mentioned relation for the Einstein tensor (here $\operatorname{Ric}$ is the Ricci tensor and $S$ the Ricci scalar).</p>
<p>This also sometimes coincides with the GR convention, noting that the Einstein, energy-momentum and field strength tensors are naturally covariant tensors and you first have to raise indices to compute the divergence.</p>
<p>For $p$-forms, this tensor definition again coincides with the definition using the exterior derivative, so that we find, e.g. for the field strength tensor $F$:
$$
\operatorname{div} F = *d*F,
$$
which plays a role in the equations of motion of Einstein-Maxwell theory.</p>
|
503,827 | <p>I need help to prove the following:</p>
<p>Let $a$, $b$, and $c$ be any integers. If $a \mid b$, then $a \mid bc$</p>
<p>My brain is in overload and just not working.</p>
| user96646 | 96,646 | <p>If $a\mid b$ then $b=an$ for $n\in$ $\mathbb{Z}$. Then, $bc = (an)c$ and so $a\mid bc$. </p>
|
3,754,225 | <p>Let S be the circle with centre <span class="math-container">$(0,0)$</span>, radius <span class="math-container">$r$</span> units. The chord <span class="math-container">$C$</span> of the circle S subtends an angle of 2π/3 at its center. If R represents the region consisting of all points inside S which are closer to C than to circumference of S, then</p>
<p>(1) What is area of region <span class="math-container">$R$</span>, and,</p>
<p>(2) In what ratio does <span class="math-container">$C$</span> divide the region <span class="math-container">$R$</span> ?</p>
<p>I found the equation of a circle intersecting circle radius <span class="math-container">$r$</span> for the chord subtending angle <span class="math-container">$2\pi/3$</span> at center we should have</p>
<p><span class="math-container">$$ x=\dfrac{r}{2} $$</span></p>
<p>but don't know what to do next. Please help.</p>
| Vishu | 751,311 | <p>The circle of interest is <span class="math-container">$x^2+y^2=r^2$</span>. Since there are infinitely many chords which subtend an angle <span class="math-container">$\frac{2\pi}{3}$</span> at the center, we might as well assume the chord to be horizontal (slope <span class="math-container">$0$</span>). By some simple trigonometry , you’ll find that that equation of the chord is <span class="math-container">$$y=\frac r2$$</span> Now, let <span class="math-container">$(a,b)$</span> with <span class="math-container">$a^2+b^2\lt 1$</span> be any point which is closer to <span class="math-container">$C$</span> than to the circumference. First, you need to determine the distance of <span class="math-container">$(a,b)$</span> from the circumference, which I’ll call <span class="math-container">$d_1$</span>.</p>
<ol>
<li><p>Find the equation of the line joining the center of the circle and <span class="math-container">$(a,b)$</span>.</p>
</li>
<li><p>Solve for the point of intersection of this line with the circle (you’ll get two points, of which you must choose the closer one to <span class="math-container">$(a,b)$</span>).</p>
</li>
<li><p>Then, apply the formula for the distance between two points to get <span class="math-container">$d_1$</span>.</p>
</li>
</ol>
<p>Now, <span class="math-container">$d_2$</span> (distance of <span class="math-container">$(a,b)$</span> from <span class="math-container">$C$</span>) is just <span class="math-container">$$|b-\frac r2|$$</span> Then to get the locus of all such points, we need <span class="math-container">$$d_2\lt d_1 \implies d_2^2\lt d_1^2$$</span> You’ll get a quadratic inequality in <span class="math-container">$b$</span> from here, which you have to solve by finding its roots. After that, you’ll get that the value of <span class="math-container">$b$</span> must be between two functions, something like <span class="math-container">$$f(a)\lt b\lt g(a)$$</span> and this will be the required region <span class="math-container">$R$</span>. Finding the area is simple from here by integration.</p>
<p>This is a lot of work, but it will work. Hope I was able to help.</p>
|
1,435,189 | <p>A map $T:X \rightarrow Y$ is $A, B$- measurable if the pre-image of every set in $B$ is a set in $A$, where $A,B$ are $\sigma$-algebras in $X,Y$. </p>
<p>But I was thinking that isn't any map $T: X \rightarrow Y$ measurable then? Because since pre-images behave so nicely, the pre-image of all sets from $B$ make up a $\sigma$-algebra in $X$, and thus the definition is satisfied?</p>
| parsiad | 64,601 | <p><strong>Hint</strong>: The more sets you have in the $\sigma$-algebra of $X$ and the less sets you have in the $\sigma$-algebra of $Y$, the more "likely" your function is to be measurable.</p>
<p>I'm sure you can think of some trivial nonmeasurable examples now.</p>
|
1,160,076 | <p>Consider $f:\left[a,b\right]\rightarrow \mathbb{R}$ continuous at $\left[a,b\right]$ and differentiable at $\left(a,b\right)$.<br>
$\forall x\:\ne \frac{a+b}{2}$, $f'\left(x\right)\:>\:0$, but at $x\:=\frac{a+b}{2}$, $f'\left(x\right)\:=\:0$.<br>
Show that $f$ is <strong>strictly increasing</strong> at $\left[a,b\right]$. </p>
<p>So far i don't find any approach how to handle with $x\:=\:\frac{a+b}{2}$.<br>
I know that $x\:=\:\frac{a+b}{2}$ can't be minimum or maximum.
I thought about assuming there is point $c\:\ne \frac{a+b}{2}$ such that $f\left(c\right)=f\left(\frac{a+b}{2}\right)$ and get a contradict somehow but i can't figure out how..
Any ideas? thanks in advance!</p>
| Timbuc | 118,527 | <p>Suppose there are $\;x,y\in [a,b]\;,\;\;x,y\;$ with $\;f(x)=f(y)\;$ . Since the function is <strong>at least</strong> monotonic increasing "weakly", as $\;f'(x)\ge 0\;$ , the above means </p>
<p>$$f(w)=0\;\;\;\forall\,w\in (x,y)\implies f'(w)=0\;\;\;\forall\,w\in (x,y)\implies\;\text{contradiction}$$</p>
|
1,682,961 | <p>I was making a few exercises on set proofs but I met an exercise on which I don't know how to start:</p>
<blockquote>
<p>If $A \cap C = B \cap C $ and $ A-C=B-C $ then $A = B$</p>
</blockquote>
<p>Where should I start? Should I start from $ A \subseteq B $ or should I start from this $ ((A\cap C = B\cap C) \land (A-C = B-C)) \Rightarrow (A = B)$ ? </p>
| user 1 | 133,030 | <p>$$A=(A \cap C)\cup(A-C)=(B \cap C)\cup(B-C)= B.$$
1st and last equality proved <a href="https://math.stackexchange.com/questions/1360411/how-to-prove-a-a-setminus-b-cup-a-cap-b">here</a>, and the middle one is your assumptions.</p>
|
785,844 | <p>I've got the following limit to solve:</p>
<p>$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$</p>
<p>I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?</p>
| Mark Bennet | 2,906 | <p>Try putting $t=\sqrt s$ to get $$\frac {t-t^4}{1-t}=\frac {t(1-t^3)}{1-t}=t(1+t+t^2)$$You can notice this without the substitution, of course, but sometimes a substitution like this helps to clarify what is going on.</p>
|
2,835,290 | <p>I am given two variables $Y_1$ and $Y_2$ obeying an exponential distribution with mean $\beta= 1$</p>
<p>We are asked what the distribution of their average is and the solution must be found using moment generating functions.</p>
<p>The solution to this exercise says:</p>
<p><a href="https://i.stack.imgur.com/eOilN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eOilN.png" alt="enter image description here"></a></p>
<p>Well first, the solution already has a typo since it should be $E[e^{t((1/2)Y_1+(1/2)Y_2)}]$ but moreover, </p>
<p>I don't agree that $E[e^{t((1/2)Y_1+(1/2)Y_2)}]=M_{Y_1}(t)M_{Y_2}(t)$</p>
<p>Since $M_{Y_1}(t)M_{Y_2}(t)=E[e^{t(Y_1+Y_2)}]$, from my understanding.</p>
<p>So what is the actual solution to this problem?</p>
| Jack D'Aurizio | 44,121 | <p>Assuming $a>1$, the original integral equals
$$ \frac{1}{2}\int_{0}^{1}\log\left(\frac{a-x}{a+x}\right)\frac{dx}{x\sqrt{x(1-x)}}\,dx=-\int_{0}^{1}\frac{\text{arctanh}(x/a)}{x\sqrt{x(1-x)}}\,dx \tag{1}$$
or
$$ -\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n-1/2}}{(2n+1)a^{2n+1}\sqrt{1-x}}\,dx=-\sum_{n\geq 0}\frac{\pi\binom{4n}{2n}}{(2n+1)16^n a^{2n+1}}\tag{2}$$
hence it is enough to recall the generating function for <a href="https://en.wikipedia.org/wiki/Catalan_number" rel="nofollow noreferrer">Catalan numbers</a> to recover the wanted identity.</p>
<hr>
<p>It is worth noticing that <a href="https://math.stackexchange.com/questions/878477/a-closed-form-of-sum-k-0-infty-frac-1k1k-gamma2-left-frack2/878499#878499">the Maclaurin series of the squared arcsine</a> allows to compute the similar integral
$$ \int_{0}^{1}\log\left(\frac{a^2+x^2}{a^2-x^2}\right)\frac{dx}{\color{red}{x}\sqrt{1-x^2}} $$
or the hypergeometric function $\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{4},\tfrac{5}{4},\tfrac{3}{2};z\right)$ in terms of $\arcsin^2$ and $\text{arcsinh}^2$, in a very similar and efficient way.</p>
|
3,367,903 | <p>Let <span class="math-container">$(\Omega,\leq)$</span> be a dense linear order, with
<span class="math-container">$\Omega \subset \mathbb{Q}$</span>, and where "<span class="math-container">$\leq$</span>" is induced by the usual order relation in <span class="math-container">$\mathbb{Q}$</span>. </p>
<p>Is there an elegant general description for <span class="math-container">$\Omega$</span> (as a subset of <span class="math-container">$\mathbb{Q}$</span>) ? </p>
| Mark Kamsma | 661,457 | <p>Yes, after possibly deleting its endpoints, it is isomorphic to <span class="math-container">$\mathbb{Q}$</span>. That is, <span class="math-container">$\Omega$</span> will be isomorphic to one of the following:</p>
<ul>
<li><span class="math-container">$\mathbb{Q}$</span>,</li>
<li><span class="math-container">$\mathbb{Q} \cup \{\infty\}$</span>,</li>
<li><span class="math-container">$\{-\infty\} \cup \mathbb{Q}$</span>,</li>
<li><span class="math-container">$\{-\infty\} \cup \mathbb{Q} \cup \{\infty\}$</span>.</li>
</ul>
<p>I think the interpretation of <span class="math-container">$-\infty$</span> and <span class="math-container">$\infty$</span> should be clear.</p>
<p>The reason for this is that the theory of dense linear orders without endpoints is <span class="math-container">$\omega$</span>-categorical. This is a well-known fact, and a proof of it appears in pretty much every book on model theory. It is not hard to see from that point that the other variants of the theory of dense linear orders (i.e. left endpoint, right endpoint or both) are also all <span class="math-container">$\omega$</span>-categorical.</p>
<p>In particular, any dense linear order is going to be infinite, and since <span class="math-container">$\Omega \subseteq \mathbb{Q}$</span> it must be countable. Thus by <span class="math-container">$\omega$</span>-categoricity it must be isomorphic to one of the models mentioned above.</p>
<p><strong>Edit:</strong> you also ask "(as a subset of <span class="math-container">$\mathbb{Q}$</span>)", unfortunately this is generally not possible I think. Although it might depend on what you call "an elegant description". For example, any union of open intervals in <span class="math-container">$\mathbb{Q}$</span> will be an example of such an <span class="math-container">$\Omega$</span>. But also weirder sets like the set of all <span class="math-container">$\frac{a}{b}$</span> where this fraction is in reduced form and <span class="math-container">$b$</span> is odd.</p>
|
3,367,903 | <p>Let <span class="math-container">$(\Omega,\leq)$</span> be a dense linear order, with
<span class="math-container">$\Omega \subset \mathbb{Q}$</span>, and where "<span class="math-container">$\leq$</span>" is induced by the usual order relation in <span class="math-container">$\mathbb{Q}$</span>. </p>
<p>Is there an elegant general description for <span class="math-container">$\Omega$</span> (as a subset of <span class="math-container">$\mathbb{Q}$</span>) ? </p>
| Alex Kruckman | 7,062 | <p>Following up on Mark Kamsma's answer, there are <span class="math-container">$2^{\aleph_0}$</span>-many suborders of <span class="math-container">$\mathbb{Q}$</span> which are isomorphic to <span class="math-container">$\mathbb{Q}$</span>, even up to automorphisms of <span class="math-container">$\mathbb{Q}$</span>. This suggests that there are too many to be "nicely classifiable". </p>
<p>To see this, let <span class="math-container">$\eta\in 2^\omega$</span> be any infinite binary sequence, and define for all <span class="math-container">$n\in \omega$</span>: <span class="math-container">$$I_n = \begin{cases} (n,n+1) & \text{if }\eta(n) = 0 \\ (n,n+\frac{1}{2}) & \text{if } \eta(n) = 1\end{cases}$$</span>
Let <span class="math-container">$\Omega_\eta = \bigcup_{n\in \omega} I_n$</span>. Then <span class="math-container">$\Omega_\eta$</span> is a dense linear order without endpoints. Namely, it is the interval <span class="math-container">$(0,\infty)$</span> in <span class="math-container">$\mathbb{Q}$</span>, with a sequence of gaps: <span class="math-container">$\{n+1\}$</span> is removed when <span class="math-container">$\eta(n) = 0$</span>, and the entire interval <span class="math-container">$[n+\frac{1}{2},n]$</span> is removed when <span class="math-container">$\eta(n) = 1$</span>. Any automorphism of <span class="math-container">$\mathbb{Q}$</span> which maps <span class="math-container">$\Omega_\eta$</span> to <span class="math-container">$\Omega_{\eta'}$</span> must map <span class="math-container">$(0,\infty)$</span> to <span class="math-container">$(0,\infty)$</span>, and must map the <span class="math-container">$n^\text{th}$</span> gap in <span class="math-container">$\Omega_\eta$</span> to the <span class="math-container">$n^{\text{th}}$</span> gap in <span class="math-container">$\Omega_{\eta'}$</span>. So these gaps must have the same type, and we conclude that <span class="math-container">$\eta = \eta'$</span>. </p>
<hr>
<p>Now consider the space of suborders of <span class="math-container">$\mathbb{Q}$</span> (by which I mean the Cantor space <span class="math-container">$2^\mathbb{Q}$</span>, where an element encodes whether each <span class="math-container">$q\in \mathbb{Q}$</span> is in or out of the suborder). This is a Polish space, and the subspace of suborders which are themselves dense linear orders (without endpoint) is a <span class="math-container">$G_\delta$</span> subspace, hence also Polish. The Polish group <span class="math-container">$\text{Aut}(\mathbb{Q})$</span> acts on both spaces by automorphisms of <span class="math-container">$\mathbb{Q}$</span>. So above I showed that this group action has <span class="math-container">$2^{\aleph_0}$</span>-many orbits. </p>
<p>But there is one orbit which is much larger than all the others, in the sense that it is comeager in <span class="math-container">$2^{\mathbb{Q}}$</span>. This is the orbit of any suborder of <span class="math-container">$\mathbb{Q}$</span> which is dense and codense in <span class="math-container">$\mathbb{Q}$</span>. For example, the suborder of all dyadic rationals in <span class="math-container">$\mathbb{Q}$</span> is in this "generic" orbit. </p>
|
3,291,303 | <p>this very same problem appeared in a different thread but the questions was slightly different. In my case, I'm looking precisely for the answer.</p>
<p>This is how I solved it, I only need confirmation of whether this actually is correct:</p>
<p>So I assumed that 1111... (100 ones) is going to be exactly divided by 1111111 (7 ones) 14 times (100/7 = 14). Hence, if 14 times 7 equals 98 '1's, then the remainder is 2 '1's.</p>
<p>Thanks.</p>
| Ross Millikan | 1,827 | <p>Yes, it is correct. If you envision setting up the long division, the first subtraction strips off the first seven <span class="math-container">$1$</span>s, the next strips the next seven, and so on.</p>
|
1,282,420 | <p>In his book of Differential Topology, Hirsch starts a little detour in his theory in order to present a way to see things in a general perspective. The precise fragment I'm referring to is the following:</p>
<p><img src="https://i.stack.imgur.com/rfzFr.png" alt="enter image description here"></p>
<p>I have never encountered a inverse limit before, and I don't have any intuition about what it is. I've searched the internet and it didn't help me either (for instance, the Wikipedia Page on <a href="http://en.wikipedia.org/wiki/Inverse_limit" rel="nofollow noreferrer">Inverse Limit</a>). He then proceeds to give an example:</p>
<p><img src="https://i.stack.imgur.com/5V3tB.png" alt="enter image description here"></p>
<p>It is not clear at all to me why it is continuous. My question is three-fold:</p>
<ol>
<li>Is there a (less general) definition of inverse limit adequate to this setting, or is the general definition adequate and acessible enough?</li>
<li>Is there an intuition?</li>
<li>Why is the structure functor above obviously continuous?</li>
</ol>
| Thomas Andrews | 7,933 | <p>The first cases of inverse limits I saw were of the form:</p>
<p>$$X_1\mathop{\leftarrow}_{f_1} X_2\mathop{\leftarrow}_{f_2} X_3\mathop{\leftarrow}_{f_3}\cdots$$</p>
<p>Then the inverse limit of this sequence of sets is the set of tuples:</p>
<p>$$\left\{(x_1,x_2,\cdots,x_n,\cdots)\in X_1\times X_2\cdots\mid \forall i( f_i(x_{i+1})=x_i)\right\}$$</p>
<p>An example of this, from the categories of rings, is the sequence:</p>
<p>$$\mathbb Z_p\leftarrow \mathbb Z_{p^2}\leftarrow \mathbb Z_{p^3}\cdots$$</p>
<p>Where $$f_i: \mathbb Z_{p^{i+1}}\rightarrow \mathbb Z_{p^i}$$ is the natural map.</p>
<p>The inverse limit here is the $p$-adic integers.</p>
<p>An even more basic example is if $X_{i+1}\subseteq X_i$ with $f_i$ the usual inclusion. Then the inverse limit is the same as $\bigcap X_i$. </p>
|
1,374,660 | <p>The IQ of actuarial science majors is assumed to be normally distributed with mean 112 and standard deviation of 14. In a class of 19 students, find the probability that the mean IQ of all 19 students is greater than 120. </p>
<p>I know this question is based of the CLT, but I don't know what to change when everyone in the group has to be above a certain mean. Shouldn't the mean just be a group average? </p>
| gogurt | 29,568 | <p>If $X_1, \ldots, X_{19}$ are the 19 students you sample from the class, then you can assume they're iid with distribution $N(112, 14)$ as stated in the problem. Then write $\bar{X} = (X_1 + \ldots + X_{19})/19$ for the sample mean you're dealing with.</p>
<p>Just from rules for means and variances of sums of independent random variables, you know that
$$ \mathbb{E} \bar{X} = \mathbb{E} X_1 = 112 \quad \textrm{and} \quad SD(\bar{X}) = \frac{SD(X_1)}{\sqrt{19}} = \frac{14}{\sqrt{19}}$$</p>
<p>As Augustin mentioned, note that nothing in the above relies on the Central Limit Theorem. The last fact you need is that sums of Normal random variables are still Normal (still, not relying on the CLT), so all in all you know that</p>
<p>$$ \bar{X} \sim N(112, 14/\sqrt{19})$$</p>
<p>That should help you find the answer.</p>
|
3,657,106 | <blockquote>
<p>Sam was adding the integers from <span class="math-container">$1$</span> to <span class="math-container">$20$</span>. In his rush, he skipped one of the numbers and forgot to add it. His final sum was a multiple of <span class="math-container">$20$</span>. What number did he forget to add?</p>
</blockquote>
<p>My idea was to use Gauss's trick to find this relatively simply so I proceeded as follows.</p>
<p>We have <span class="math-container">$S=1+2+3+ \dots+ 18+19+20$</span>. Using Gauss's trick we get <span class="math-container">$\frac{n(n+1)}{2} = \frac{20(21)}{2} = 210$</span>. Since we want this to equal some multiple of <span class="math-container">$20$</span> we have that <span class="math-container">$210 = 20n$</span>, but solving for <span class="math-container">$n$</span> results in <span class="math-container">$\frac{21}{2} = 10.5$</span>.</p>
<p>The correct answer for this was <span class="math-container">$10$</span>, but it seems that I'm missing something?</p>
| fleablood | 280,126 | <blockquote>
<p>but it seems that I'm missing something?</p>
</blockquote>
<p>Oh, I can't resist. </p>
<p>You aren't missing anything. That's the problem.</p>
<p>(....<em>laughs in the corner to himself for hours</em>....)</p>
<p>You were supposed to skip a number but you included them all.</p>
<p>The numbers add to <span class="math-container">$1+2+3 + 4+.... + 20 = 210$</span> and you did that correctly. But Sam did <em>not</em> do it correctly. Sam left out a number. So Sam did <em>not</em> get <span class="math-container">$210$</span>.</p>
<p>So what did Sam get. If the number he skipped was <span class="math-container">$k$</span> then same got <span class="math-container">$1 + 2 + 3 + .... + (k-1) + (k+1) + .... +20 = 210 - k$</span>.</p>
<p>So we have <span class="math-container">$210 - k = 20n$</span>. Now <span class="math-container">$k$</span> can be as small as <span class="math-container">$1$</span> and is <span class="math-container">$210-k = 20n$</span> can be as big as <span class="math-container">$209$</span>. And <span class="math-container">$k$</span> can be as big as <span class="math-container">$20$</span> so <span class="math-container">$210-k = 20n$</span> can be as small as <span class="math-container">$190$</span>. So <span class="math-container">$190 \le 210-k = 20n \le 209$</span>. Then only number in that range that is divisible by <span class="math-container">$20$</span> is ... <span class="math-container">$200$</span>. So <span class="math-container">$20n = 210 -k = 200$</span>. And that means <span class="math-container">$k$</span> is .... <span class="math-container">$10$</span>.</p>
<p>And indeed, if <span class="math-container">$1 + 2 + 3 + 4 + ..... + 9 + 11 + 12 + 13 + .... + 20 = $</span></p>
<p><span class="math-container">$(1 + 2+ .... + 9) + (11+12 + .... + 20) = $</span></p>
<p><span class="math-container">$(\underbrace{1 + \underbrace {2 + .... +8}+9}) + (\underbrace{11 + \underbrace {12 + .... +19}+20})=$</span></p>
<p><span class="math-container">$(4*10+5) + (5*31) =$</span></p>
<p><span class="math-container">$45 + 155 = 200$</span>.</p>
<p>Then <span class="math-container">$9+8+7+4 +..... + 1 + 20 + 12 + 13 + .... + 20 = N$</span> and </p>
<p><span class="math-container">$10+10+10+ .... + 10 + 31 + 31+.... + 31 = N+N$</span></p>
<p><span class="math-container">$90 + 310 = 2N$</span></p>
<p><span class="math-container">$400 = 2N$</span></p>
<p><span class="math-container">$N = 200$</span>.</p>
<p>....</p>
<p>More elegant, if you know modular arithmetic is:</p>
<p><span class="math-container">$20$</span> divides <span class="math-container">$(1+2+3+ ..... + 20) -k$</span> so</p>
<p><span class="math-container">$(1+2+3 + ..... + 20) -k \equiv 0 \pmod {20}$</span></p>
<p>So <span class="math-container">$210 -k \equiv 0 \pmod {20}$</span></p>
<p><span class="math-container">$k \equiv 210 \equiv 10 \pmod {20}$</span>.</p>
<p>And as <span class="math-container">$1\le n \le 20; k \equiv 10 \pmod{20}$</span> then <span class="math-container">$k =10$</span>.</p>
|
3,163,355 | <p>I found that the Wieferich prime <span class="math-container">$1093$</span> divides <span class="math-container">$3^{1036}-1$</span>.
Does <span class="math-container">$1093$</span> divides infinitely many <span class="math-container">$3^{k}-1$</span>, with k a positive integer? And what features must <span class="math-container">$3^{k}-1$</span> have to be divisible by <span class="math-container">$1093$</span>?</p>
| Mostafa Ayaz | 518,023 | <p>Note that if <span class="math-container">$$r_k\triangleq 3^k-1\mod 1093$$</span>then <span class="math-container">$$r_{k+1}{=3^{k+1}-1\mod 1093\\=3^{k+1}-3+2\mod 1093\\=\Big(3\cdot(3^{k}-1)\mod 1093\Big)+2\mod 1093\\=3r_k+2\mod 1093}$$</span>Based on this and with <span class="math-container">$r_1=2\mod1093=2 $</span> we obtain<span class="math-container">$$r_1=2\\r_2=8\\r_3=26\\r_4=80\\r_5=242\\r_6=728\\r_7=0\\r_8=2\\.\\.\\.$$</span>by a simple induction we conclude that:<span class="math-container">$$r_k=0\iff k=7u\quad\text{for some }u\in\Bbb Z$$</span>therefore<span class="math-container">$$3^n\equiv 1\mod 1093\iff n=7u$$</span>and all the answers are as follows:<span class="math-container">$$3^{7n}\equiv 1\mod 1093\quad,\quad n\in \Bbb N$$</span></p>
|
581,958 | <p>I am working on a project about Mathematics and Origami. I am working on a section about how origami can be used to solve cubic equations. This is the source I am looking at:<br>
<a href="http://origami.ousaan.com/library/conste.html">http://origami.ousaan.com/library/conste.html</a><br>
I understand the proof they go on to explain about how the slope of the crease line is the solution to the general cubic equation. I am also interested in understanding how this is used to double the cube and trisect the angle which they go on to mention at the end. However, I am having difficulty filling in the missing gaps and understanding how to use this origami method to solve those equations given and how this tells us we can trisect the angle, and double the cube. Any help is much appreciated</p>
| amorvincomni | 95,697 | <p>For the trisection of an angle note that $\cos3\alpha=4\cos^3\alpha-3\cos\alpha$ with $3\alpha$ constructible (this is equivalent to say that $\cos 3 \alpha$ is constructible). It is an equation of degree 3. In order to obtain the desired $\alpha$ angle, take a squared paper and do the crease $L_1$ corresponding to $3\alpha$ (the angle between $L_1$ and the $x-$axe). Make now the crease $L_2$ parallel to the $x-$axe and that divide in two stripes of equal heigh the entire paper. Call $P_1$ the intersection point of $L_2$ and the $y-$axe. Do the same with the first stripe, constructing a new crease $L_3$. Now using one of the Huzita-Hatori axioms you can do the crease that sends $P_1$ over $L_1$ and the origin over $L_3$. Call $P_1'$ and $O'$ the points obtained by this reflection. It turns that the axis (call it $L_4$) of the segment $\overline{P_1'O'}$ passes through the origin. The reflection of the line $L_1$ (call it $L_5$) with respect $L_4$ also passes through $O$, and in particular it trisect the angle $3\alpha$.</p>
|
2,309,613 | <p>Find a general solution to $x^2-2y^2=1$</p>
<p>I found that (3,2) is a solution.
Now what should I do?
I can not catch what the question really want.</p>
<p>It is about pell's equation. Would you give me a form of general solution? </p>
| Leviathan | 448,426 | <p>The function is holomorphic and can therefore be decomposed in a power series;
\begin{align*}
f(z) = \sum_{n=0}^{\infty} a_n z^{n}
\end{align*}
Furthermore, it is known that $f(i\mathbb{R}) \in i \mathbb{R}$ and $f(\mathbb{R}) \in \mathbb{R}$ . That is why
\begin{align}
f(r) &= \sum_{n= 0}^{\infty} a_n r^{n}, \quad r\in \mathbb{R} \\
f(ir) &= \sum_{n=0}^{\infty} a_n (ir)^{n} , \quad r\in \mathbb{R} \label{ir}\, .
\end{align}
This equation can be written as
\begin{align*}
f(ir) = \sum_{n=0}^{\infty} i^{n} a_n r^{n} = \sum_{n=0}^{\infty} (-1)^{n} a_{2n} r^{2n} + i \sum_{n=0}^{\infty} (-1)^{n} a_{2n+1} r^{2n+1}
\end{align*}
By assumption, $f(i\mathbb{R}) \in i\mathbb{R}$, that is, $a_{2n} = 0$ consequently we have an odd function because only the odd powers are present:
\begin{align*}
f(z) = \sum_{n=0}^{\infty}a_{2n+1} z^{2n+1}
\end{align*}
Which is equivalent for saying $z\in G \cap (-G)$ because $a_n(-z)^n=-a_nz^n$. </p>
|
89,669 | <p>If we fix a locally profinite group $G$ , we note $R(G)$ the category of smooth representations of $G$, $\mathcal{E}$ the set of equivalence classes of $R(G)$, and finaly $Irr(G)$ the set of irreducible equivalence calasses. I recall that the theorem of Cantor-Bernstein says : If $E$ and $F$ two sets. If there is an injection from $E$ to $F$ and an injection from $F$ to $E$, then there is a bijection between $E$ and $F$. This enables us to define an ordering relation in the set of equipotence classes of sets : If $E$ and $F$ two sets, we note $E\leq F$ if $E$ injects in $F$. </p>
<p>Similarly, we define a relation $\leq$ in $\mathcal{E}$, but in general is not an ordering relation, I think that is an ordering relation if $R(G)$ is semisimple (for example, for compact locally profinite group). </p>
<p>If $L$ a non empty subset of $Irr(G)$, we define a $L$-minimal representation as a smooth representation $\pi$ of $G$ such that :</p>
<p>1) For every $\sigma\in L$, $\sigma \leq \pi$. </p>
<p>2) For every $\tau \in R(G)$, if $\sigma\leq\tau$ for every $\sigma\in L$, then $\pi\leq\tau$.</p>
<p>I ask the following questions: </p>
<p>Q1) An $L$-minimal representation exits ?</p>
<p>Q2) unicity ? </p>
<p>Q3) If $\pi$ an $L$-minimal (if there exist) representation, $dim\mathbf{Hom}_{G}(\sigma,\pi)$, where $\sigma\in L$, is minimal ?</p>
<p>I'm interested of this question for the set $L_{k}$ of equivalence classes of irreducible supercuspidal representation of $PGL(n,F)$ with conductor=$k$.</p>
| Rami | 4,690 | <p>Your group has compact center (in fact trivial center). So supercuspidal representation indeed form a semi-simple category, as pm said (see <a href="http://www.math.uchicago.edu/~mitya/langlands/Bernstein/Bernstein93new.dvi" rel="nofollow">http://www.math.uchicago.edu/~mitya/langlands/Bernstein/Bernstein93new.dvi</a> pg. 22-25, 36). </p>
<p>Your Set $L_k$ is finite set of irreducible representations. So the representation that you are looking for is the direct sum of all the representations in $L_k$</p>
|
1,602,392 | <p>I think I have a proof using Pythagoras for $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.</p>
<p><em>I'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and <strong>also</strong> in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).</em></p>
<p><strong><em>Lemma</em></strong>:</p>
<p>Let positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle. </p>
<p>Then $\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.</p>
<p>And $\sqrt{a_1^2} + \sqrt{a_2^2}$ is the sum of the length of the two legs.</p>
<p>By the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.</p>
<p>Therefore, for any real numbers $a_1, a_2$, $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.</p>
<p><em>I'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to "extract" pairs of elements from under $\sqrt{a_1^2 + a_2^2 +...+a_n^2}$.
I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.</em></p>
| Kibble | 277,229 | <p>Here is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \ldots + |a_n|$. </p>
<p>$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$<a href="https://i.stack.imgur.com/DzcrF.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DzcrF.png" alt="enter image description here"></a></p>
<p>Comparing the area of the whole square to the area of the small squares contained within it we see that</p>
<p>$$(|a_1| + |a_2| + \ldots + |a_n|)^2 \geq a_1^2 + a_2^2 + \ldots + a_n^2$$</p>
<p>and by taking the square root we get the desired inequality. Equality can only happen when one of the small squares covers the whole square which can only happen when atleast $n-1$ of the $a_i$'s are zero.</p>
|
3,714,418 | <p>I need to calculate the angle between two 3D vectors. There are plenty of examples available of how to do that but the result is always in the range <span class="math-container">$0-\pi$</span>. I need a result in the range <span class="math-container">$\pi-2\pi$</span>.</p>
<p>Let's say that <span class="math-container">$\vec x$</span> is a vector in the positive x-direction and <span class="math-container">$\vec y$</span> is a vector in the positive y-direction and <span class="math-container">$\vec z$</span> is a reference vector in the positive z-direction. <span class="math-container">$\vec z$</span> is perpendicular to both <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span>. Would it then be possible to calculate the angle between <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span> and get a result in the range <span class="math-container">$\pi-2\pi$</span>? </p>
<p>The angle value should be measured counter clockwise. I have not been able to figure out how to do that.
I am no math guru but I have basic understanding of vectors at least.
Thank you very much for the help! </p>
| Saaqib Mahmood | 59,734 | <p>Let <span class="math-container">$x$</span> be an arbitrary element. In the table below, let <span class="math-container">$0$</span> denote that <span class="math-container">$x$</span> is not in the set while <span class="math-container">$1$</span> denote that <span class="math-container">$x$</span> is in the set.</p>
<p><span class="math-container">$$
\begin{matrix}
A \ & B \ & A-B \ & B-A \ & A-(B-A) \\
0 \ & 0 \ & 0 \ & 0 \ & 0 \\
0 \ & 1 \ & 0 \ & 1 \ & 0 \\
1 \ & 0 \ & 1 \ & 0 \ & 1 \\
1 \ & 1 \ & 0 \ & 0 \ & 1
\end{matrix}
$$</span></p>
<p>Note that corresponding to every <span class="math-container">$1$</span> (in fact the only one <span class="math-container">$1$</span>) in the column headed <span class="math-container">$A-B$</span>, we have a <span class="math-container">$1$</span> in the column headed <span class="math-container">$A-(B-A)$</span>, and there is a <span class="math-container">$1$</span> in the column headed <span class="math-container">$A-(B-A)$</span> for which the corresponding entry in the column headed <span class="math-container">$A-B$</span> is a <span class="math-container">$0$</span>.</p>
<p>This shows that, for any sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, we always have
<span class="math-container">$$
A - B \subset A-(B-A),
$$</span>
but there are examples when
<span class="math-container">$$
A - (B-A) \not\subset A-B.
$$</span>
For example, let us consider the sets
<span class="math-container">$$
A = \{ p \} = B.
$$</span>
Then <span class="math-container">$A-B = \emptyset = B-A$</span>, but
<span class="math-container">$$ A-(B-A) = A - \emptyset = A \neq \emptyset. $$</span></p>
<p>In the table above, the discrepency between the columns headed <span class="math-container">$A-B$</span> and <span class="math-container">$A-(B-A)$</span> occurs in the very last line where our arbitrary element is in <em>both</em> the sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. And, this precisely is the hint we have used in constructing our counter-example above.</p>
<p>Hope this helps.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Anton Geraschenko | 1 | <p>You and infinitely many other people are wearing hats. Each hat is either red or blue. Every person can see every other person's hat color, but cannot see his/her own hat color; aside from that, you cannot share any information (but you are allowed to agree on a strategy before any of the hats appear on your heads). Everybody simultaneously guesses the color of his/her hat. You win if all but finitely many of you are right. Find a strategy so that you always win.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| rgrig | 840 | <p>When they came to diner some shook hands. Ask them to prove that that two of them shook hands the same number of times.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Kiochi | 4,241 | <p>Here's an easy but fun one (probably suitable for a class or for mathematicians who have had some drinks). You have ten bags of coins, one of which contains fake coins. We may of course assume that each bag contains infinitely many coins. The real coins weigh 1 gram each, while the fake coins weigh .9 grams each. You have a scale, which is capable of only one accurate reading before breaking. Determine which bag contains the fake coins.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p>Oldie but a goodie (Monty-hall problem): </p>
<p>You are on a game show with three doors, behind one of which is a car and behind the other two are goats. You pick door #1. Monty, who knows what’s behind all three doors, reveals that behind door #2 is a goat. Before showing you what you won, Monty asks if you want to switch doors. Should you switch?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| JBL | 4,658 | <p>Take a convex polyhedron and any point inside of it. To every face, drop a normal line from the point. Note that it is both possible to land inside the face or outside. Construct a polyhedron where every such normal line drops outside of the corresponding face, or prove such a polyhedron cannot exist.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Daniel Krenn | 8,153 | <p>There is a plane with 100 seats and we have 100 passengers entering the plane one after the other. The first one cannot find his ticket, so chooses a random (uniformly) seat. All the other passengers do the following when entering the plane (they have their tickets). If the seat written on the ticket is free, one sits on this seat, if not he chooses a other (free) seat at random (uniformly). What is the probability the last passenger entering the plane gets the correct seat?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Tony Huynh | 2,233 | <p>Here's one that I like that I just heard a few days ago. Alice and Bob play the following game. Alice is randomly dealt 5 cards from an ordinary deck of cards. She is allowed to show Bob 4 of the 5 cards (in order). Bob must then guess what the 5th card is. </p>
<p>Prove that Alice and Bob have a strategy where Bob can always guess correctly. </p>
<p><strong>Edit.</strong> Actually, there is a strategy that works for 124 cards, but it is probably not human implementable (unlike the 52 card problem). </p>
|
2,623,621 | <p>I'm struggling to find the derivative of this function :</p>
<p>$$y=\bigg\lfloor{\arccos\left(\frac{1}{\tan\left(\sqrt{\arcsin x}\right)}\right)}\bigg\rfloor$$</p>
<p>I've been told it should be 0 but how can I find that ?!</p>
| Community | -1 | <p>For the "trivial" inclusion: note that for every $i\in I$ you have $S_i\subseteq\bigcup_{i\in I}S_i$ so $\overline{S_i}\subseteq\overline{\bigcup_{i\in I}S_i}$. Taking the union on the left side, we have $\bigcup_{i\in I}\overline{S_i}\subseteq\overline{\bigcup_{i\in I}S_i}$.</p>
<p>For the converse, it is enough to prove that $\bigcup_{i\in I}\overline{S_i}$ is closed. (This is because it trivially contains $\bigcup_{i\in I}S_i$, so it must then contain its closure (i.e. $\overline{\bigcup_{i\in I}S_i}$).)</p>
<p>For that purpose, pick $x\not\in \bigcup_{i\in I}\overline{S_i}$. Due to local finiteness, there is an neigbourhood / open set $U, x\in U$ such that $U$ intersects only finitely many $S_i$'s: let's say $S_{i_1}, S_{i_2},\ldots,S_{i_n}$. Now create a new neighbourhood $U'=U\setminus(\overline{S_{i_1}}\cup\overline{S_{i_2}}\cup\cdots\cup\overline{S_{i_n}})$. It is an open set, it still contains $x$, and it does not intersect any of $S_i$'s, which implies (as it is open) that it also does not intersect any of $\overline{S_i}$'s.</p>
<p>Thus, the complement of $\bigcup_{i\in I}\overline{S_i}$ is open, i.e. $\bigcup_{i\in I}\overline{S_i}$ is closed.</p>
|
3,648,462 | <p>I need to find a solution to this equation, being <span class="math-container">$z\in\mathbb{C}$</span>,
<span class="math-container">$$2e^3\cos (6z)-e^6=1.$$</span>
I don't know what method can I follow to solve it.</p>
| J. W. Tanner | 615,567 | <p><strong>Hint:</strong></p>
<p>The sum of coefficients of a polynomial <span class="math-container">$f(x)$</span> is <span class="math-container">$f(1)$</span>.</p>
|
2,004,895 | <p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
| Chill2Macht | 327,486 | <p>This is to just to elaborate on why someone would use the notation $$f: x \mapsto 2x-3$$ When we treat a function $f$ as an object, i.e. do more with it than just evaluate it at points, then we need to be able to distinguish between the <em>object</em>, the function $f$, and its rule of assignment determining its values at points $x$, i.e. after evaluating at points $x$.</p>
<p>Since $f(x)$ denotes both the function $f$, as an object in its own right, as well as the value of that function evaluated at a point $x$, it is too ambiguous for such purposes, because it does not allow us to distinguish between the function and its rule of assignment.</p>
<p>Often we can identify the function $f$, as an object in its own right, with its rule of assignment taking $x$ to $f(x)$, since we are in such instances only considering the latter, so no ambiguity arises.</p>
<p>However, the notation $$f:x\mapsto 2x-3$$ is nice because it both presents the function $f$ as a distinct object while also specifying its rule of assignment. The colon : is meant to be read "such that", which implies that the expression $f:x\mapsto 2x-3$ reads "the function $f$ such that $x$ is mapped to $2x-3$".</p>
<p>The benefit of this notation is that it allows us to distinguish between the function $f$ and its rule of assignment when we need to distinguish between the two, while taking exactly as long to write as $f(x)=2x-3$, when we don't need to distinguish between the function and its rule of assignment, and thus economy of notation becomes a priority. </p>
<p>Thus, the notation $f:x\mapsto2x-3$ is <em>both</em> exactly as efficient as the classical notation $f(x)=2x-3$ <em>and</em> less ambiguous.</p>
|
2,932,799 | <p><strong>Notations.</strong></p>
<p>Let <span class="math-container">$\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb( R\setminus\mathbb Q)^4$</span> such that <span class="math-container">$\xi_1\xi_4-\xi_2\xi_3\ne 0$</span> and the <span class="math-container">$\xi_i$</span> are linearly independent over <span class="math-container">$\mathbb Q$</span>.</p>
<p>I have the following linear form:</p>
<p><span class="math-container">$$\begin{matrix}L\colon & \mathbb R^6 & \to & \mathbb R \\ &(\eta_1,\ldots,\eta_6) & \mapsto & \eta_1(\xi_1\xi_4-\xi_2\xi_3)-\eta_2\xi_4+\eta_3\xi_3-\eta_4\xi_2+\eta_5\xi_1-\eta_6.\end{matrix}$$</span></p>
<p>We consider the norm <span class="math-container">$\Vert\cdot\Vert$</span> to be the euclidean norm on <span class="math-container">$\mathbb R^6$</span>.</p>
<p><strong>The problem.</strong></p>
<p>I am interesting in finding a constant <span class="math-container">$\gamma>0$</span>, such that if we <em>choose</em> <span class="math-container">$\xi$</span> properly, then the resulting linear form <span class="math-container">$L_\xi$</span> will verify:</p>
<p><span class="math-container">$$\forall \eta\in\mathbb Z^6\setminus\{0\},\quad L_\xi(\eta)\geqslant \frac c{\Vert\eta\Vert^\gamma}\gcd(\eta_1,\ldots,\eta_6),$$</span></p>
<p>where <span class="math-container">$c=c_\xi$</span> is a constant which depends only on <span class="math-container">$\xi$</span>.</p>
<p><strong>The conjecture.</strong></p>
<p>There are hopes for this to be true, since if we choose <span class="math-container">$\xi$</span> properly (for instance badly approximated by rationals), then for <span class="math-container">$\eta\in\mathbb Z^6\setminus\{0\}$</span>, <span class="math-container">$L_\xi(\eta)$</span> will have troubles being too small.</p>
<p>I believe that the constant <span class="math-container">$\gamma=2$</span> would work for a fine choice of <span class="math-container">$\xi$</span>.</p>
<p><strong>Additional remarks.</strong></p>
<p>This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated. </p>
<p>I do believe that <span class="math-container">$\gamma=2$</span> would work (and it would be the best), but any proof that would work for a <span class="math-container">$\gamma<4$</span> would be great.</p>
| Exodd | 161,426 | <p>Take for example
<span class="math-container">$$
A =
\begin{pmatrix}
1&1&1\\
1&1&1\\
1&3&2\\
\end{pmatrix}
\quad
E =
\begin{pmatrix}
0&1&0\\
0&0&0\\
0&0&0\\
\end{pmatrix}
$$</span>
and <span class="math-container">$B_\delta = A+\delta E$</span>. </p>
<p>You get (after lot of computations) that <span class="math-container">$r(A)=4$</span> and <span class="math-container">$r(B_\delta) = 4 + O(\sqrt{\delta})$</span> for <span class="math-container">$\delta$</span> converging to zero, so
<span class="math-container">$$|r(A)-r(B_\delta)|= O(\sqrt{\delta}) > C\delta = C\|B_\delta-A\|$$</span>
for every constant <span class="math-container">$C$</span> for <span class="math-container">$\delta$</span> small enough.</p>
|
150,295 | <p>I will appreciate any enlightenment on the following which must be an exercise in a certain textbook. (I don't recognize where it comes from.)
I understand that the going down property does not hold since $R$ is not integrally closed (in fact, it is not a UFD), but I have no idea how to show that $q$ is such a counterexample.</p>
<blockquote>
<p>Let $k$ be a field, $A = k[X, Y]$ be a polynomial ring, $R = \lbrace f \in A \colon f(0, 0) = f (1, 1) \rbrace \subset A$ be a subring.
Define $q = (X)\cap R$, $p = (X - 1, Y - 1) \cap R$, $P = (X - 1, Y - 1)$.
Show that there is no $Q \in \operatorname{Spec} A$, $Q\subset P$ that goes down to $q$. </p>
</blockquote>
| wxu | 4,396 | <p>We show that there is not a prime $Q\subset P$ such that $Q\cap R=q$.</p>
<p>Let us compute $p,q$. The prime $q=(X)\cap R=\{Xh\mid h(1,1)=0,h(X,Y)\in k[X,Y]\}$, the prime $p=P\cap R=\{g(X,Y)\mid g(0,0)=g(1,1)=0, g\in k[X,Y]\}$.
Now it is clear $q\subset p$ and not equal $p$, since $X-Y\in p\setminus q$. </p>
<p>We are ready to show our statement. Suppose there is a prime ideal $Q\subset P$ such that $Q\cap R=q$, then $Q\neq P,0$. Thus $Q$ must be a principal ideal $(f)$ with an irreducible polynomial $f$ such that $f(1,1)=0$. But in this case, $(f)\cap R=\{fh\mid f(0,0)h(0,0)=0,h\in k[X,Y]\}$. We may find an irreducible polynomial $g\in k[X,Y]$ such that $g(1,1)=0$ but $(g)\neq (f)$, then $Xg(X,Y)\in q$ but $Xg(X,Y)\notin (f)$. We are done.</p>
|
1,535,376 | <p>I have a hard time understanding that when and under what conditions we can use Gauss elimination with complete pivoting, and when with partial pivoting, and when with no pivoting? (I mean what is the exact feature of a matrix that will tell us which one to choose?)</p>
| Community | -1 | <blockquote>
<p>I have a hard time understanding that when and under what conditions
we can use Gauss elimination with complete pivoting, and when with
partial pivoting, and when with no pivoting? (I mean what is the exact
feature of a matrix that will tell us which one to choose?)</p>
</blockquote>
<p>My professor explained this to the class with an example which I can no longer specifically recall. He noted that the when you use the solver it will pretty much by default use partial pivoting as the error is too bad without it, however, the cost for using complete pivoting is not worth the improvement you get in precision typically compared to other decompositions.</p>
<p>That is if you need complete pivoting you should probably just use the QR decomposition. You should be checking for the error you get when you solve the system of equations I believe but I've never seen how someone would implement everything. </p>
<p>You won't be able to tell immediately but if you solve</p>
<p><span class="math-container">$$ Ax=b \tag{1} $$</span></p>
<p><span class="math-container">$$ LUx =b \tag{2}$$</span></p>
<p>For partial pivoting you'd have</p>
<p><span class="math-container">$$ PA = LU \implies P^{'}LUx b \tag{3}$$</span></p>
<p>then you'd back sub and front sub. Instead with pivoting you get a pivot matrix <span class="math-container">$P$</span>. You'd like to get the norm </p>
<p><span class="math-container">$$ \| \hat{x} - x \| \tag{4}$$</span> </p>
<p>where <span class="math-container">$\hat{x}$</span> is the solution vector you get and <span class="math-container">$x$</span> is the real solution . Alternatively the relative error. </p>
<p><span class="math-container">$$ \frac{\| \hat{x} - x \|}{\|x\|} \tag{5}$$</span></p>
|
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