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
|
|---|---|---|---|---|
314,416 |
<p>Part of what it means to be a functor between two categories is to have a map of morphisms e.g. $F$ sends $f: A \to B$ to $Ff: FA \to FB$.</p>
<p>Suppose $F$ is a functor from a category to itself, and that category has (some or all) exponential objects. $B^A$ intuitively corresponds to an object of morphisms $A \to B$, so I'd hope to be able to apply the morphism-mapping of the functor $F$ to get $FB^{FA}$. I can do this in $\mathbf{Set}$ in an obvious way, but I can't see how to frame that idea category-theoretically so that I might generalise it. So my questions are:</p>
<ul>
<li>Under what conditions can I "internalise" the morphism map of $F$ to get a morphism $B^A \to FB^{FA}$?</li>
<li>If that doesn't work, what if I replace $B^A$ with the hom-functor of a category enriched over itself, or some other internal notion of function?</li>
</ul>
|
Manos
| 11,921 |
<p>Let $M = U \Sigma V^*$ be an SVD of $M$. Suppose that $V=U$. Then $M$ is symmetric positive-definite. Conversely, if $M$ is symmetric positive semi-definite, then any unitary eigendecomposition $M=U \Lambda U^*$ in which the diagonal entries of $\Lambda$ are ordered from maximal to minimal, is an SVD of $M$.</p>
<p>Hence $U=V$ in the SVD if and only if $M$ is symmetric positive-semidefinite.</p>
<p>Also, the wiki article you reference, does not say that $U=V$ if and only if $M$ is normal. </p>
<p>In your example $M$ is not even symmetric hence it can't possibly be the case that $U=V$.</p>
|
13,922 |
<p>How to sum up this series :</p>
<p>$$2C_o + \frac{2^2}{2}C_1 + \frac{2^3}{3}C_2 + \cdots + \frac{2^{n+1}}{n+1}C_n$$
</p>
<p>Any hint that will lead me to the correct solution will be highly appreciated.</p>
<p>EDIT: Here $C_i = ^nC_i $</p>
|
Timothy Wagner
| 3,431 |
<p>Hint: Use binomial expansion for $(1+x)^n$ and integrate once. Then choose an appropriate value for $x$.</p>
|
602,973 |
<p>I have a problem:</p>
<blockquote>
<p>For $\Omega$ be a domain in $\Bbb R^n$. Show that the function $u(x)=1 \in W^{m,\ 2}(\Omega)$, but not in $W_0^{m,\ 2}(\Omega)$, for all $m \ge 1$.</p>
</blockquote>
<p>=============================</p>
<p>Any help will be appreciated! Thanks!</p>
|
Marek
| 3,223 |
<p>HINT: any smooth function $u_n$ with compact support approximating $1$ well will be close to $1$ on most of $\Omega$ and close to $0$ elsewhere. This implies that there will always be a region where the approximating function has a big gradient. Try to derive a contradiction by showing that the norm of $1 - u_n$ stays bounded away frow $0$ thanks to the contribution from the gradient part of the norm.</p>
|
175,775 |
<p>To clarify the terms in the question above:</p>
<p>The symmetric group Sym($\Omega$) on a set $\Omega$ consists of all bijections from $\Omega$ to $\Omega$ under composition of functions. A generating set $X \subseteq \Omega$ is minimal if no proper subset of $X$ generates Sym($\Omega$).</p>
<p>This might be a difficult question, but perhaps the answer is known already?</p>
|
Jeremy Rickard
| 22,989 |
<p>I think it follows from Theorem 1.1 of "Subgroups of Infinite Symmetric Groups" by Macpherson and Neumann (J. London Math. Soc. (1990) s2-42 (1): 64-84) that there is no minimal generating set of $S(\Omega)$for infinite $\Omega$.</p>
<p>The theorem states that any chain of proper subgroups of $S(\Omega)$ whose union is $S(\Omega)$ must have cardinality strictly greater than $|\Omega|$.</p>
<p>Now suppose $X$ is a minimal generating set. Let $C=\{x_0,x_1,\dots\}$ be a countable subset of $X$. If
$$H_i=\langle X\setminus C,x_0,\dots,x_i\rangle$$
for $i\in\mathbb{N}$, then
$H_0<H_1<\dots$ is a countable chain of proper subgroups whose union is $S(\Omega)$, contradicting the theorem.</p>
<p>(Note: There's a paper of Bigelow pointing out some unstated set-theoretic assumptions in Macpherson and Neumann's paper, but I don't think that affects the theorem I mention.)</p>
|
4,093,406 |
<p>Which of the equations have at least two real roots?
<span class="math-container">\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}</span>
I wasn't able to notice something clever, so I solved each of the equations. The first one has <span class="math-container">$2$</span> real roots, the second one <span class="math-container">$4$</span> real roots and the last one does not have real roots. I am pretty sure that the idea behind the problem wasn't solving each of the equations. What can we note to help us solve it faster?</p>
|
boojum
| 882,145 |
<p>There is a test that can be made for biquadratics without the need for square-roots that is akin to a test for the existence of real zeroes of a quadratic polynomial (there is an evident connection with the Rule of Signs that <strong>Gregory</strong> discusses). There, "completing the square" produces
<span class="math-container">$$ u^2 \ + \ Bu \ + \ C \ \ = \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ + \ \left(C \ - \ \frac{B^2}{4} \right) \ \ = \ \ 0 \ \ \Rightarrow \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ \ = \ \ \frac{B^2}{4} \ - \ C \ \ , $$</span>
which tells us that there are no real zeroes for <span class="math-container">$ \ u \ $</span> if <span class="math-container">$ \ \frac{B^2}{4} - C \ < 0 \ \Rightarrow \ B^2 \ < \ 4C \ \ , $</span> one if <span class="math-container">$ \ B^2 \ = \ 4C \ \ , $</span> and two if <span class="math-container">$ \ B^2 \ > \ 4C \ \ . $</span> This is equivalently to saying that a parabola with its vertex at <span class="math-container">$ \ x \ = \ h \ = \ -\frac{B}{2} \ , \ y \ = \ k \ = \ C \ - \ \frac{B^2}{4} \ $</span> does not intersect the <span class="math-container">$ \ x-$</span>axis in this first case, is just tangent to it in the second, and has two <span class="math-container">$ \ x-$</span>intercepts in the third case.</p>
<p>With the biquadratic <span class="math-container">$ \ x^4 \ + \ Bx^2 \ + \ C \ \ , $</span> which substitutes <span class="math-container">$ \ u = x^2 \ \ , $</span> now requires <span class="math-container">$ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ > \ 0 \ $</span> for <span class="math-container">$ \ B > 0 \ $</span> and the polynomial cannot have a value less than <span class="math-container">$ \ \left( \frac{B}{2} \right)^2 + \frac{B^2}{4} - C \ = \ C \ . $</span> From this, we see that the biquadratic has <strong>no real zeroes</strong> for <span class="math-container">$ \ \mathbf{B > 0 \ , \ C > 0} \ \ $</span> and <strong>two real zeroes</strong> for <span class="math-container">$ \ \mathbf{B > 0 \ , \ C < 0} \ \ . $</span> In terms of factorization, the first of these corresponds to
<span class="math-container">$$ (x^2 + \alpha^2)·(x^2 + \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ , $$</span>
while the second is one case for
<span class="math-container">$$ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 - \beta^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $$</span>
with <span class="math-container">$ \ |\alpha| > |\beta| \ \ , \ \ \alpha \ $</span> being imaginary.</p>
<p>Coming to the cases pertinent to your equations, we have <span class="math-container">$ \ B < 0 \ \ , $</span> which permits <span class="math-container">$ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ \ge \ 0 \ \ . $</span> The significance of this change is that the biquadratic curve has just one "turning point" for <span class="math-container">$ \ B > 0 \ \ , $</span> but <em>three</em> for <span class="math-container">$ \ B < 0 \ \ . $</span> The function <span class="math-container">$ \ f(x) \ = \ x^4 \ + \ Bx^2 \ + \ C \ \ $</span> has even symmetry and can be thought of as "fusing" the portion of the parabola <span class="math-container">$ \ y \ = \ x^2 \ + \ Bx \ + \ C \ $</span> for <span class="math-container">$ \ x \ge 0 \ $</span> with its "reflection" about the <span class="math-container">$ \ y-$</span>axis. For <span class="math-container">$ \ B > 0 \ \ , $</span> the vertex of the parabola is at <span class="math-container">$ \ x \ = \ -\frac{|B|}{2} \ < \ 0 \ \ , $</span> so the biquadratic curve only has a turning point at its <span class="math-container">$ \ y-$</span>intercept <span class="math-container">$ \ (0 \ , \ C ) \ \ . $</span> However, for <span class="math-container">$ \ B < 0 \ \ , $</span> the vertex lies at <span class="math-container">$ \ x \ = \ \frac{|B|}{2} \ > \ 0 \ \ , $</span> leading to the turning point at <span class="math-container">$ \ (0 \ , \ C ) \ \ $</span> and two others at <span class="math-container">$ \ x^2 \ = \ \frac{|B|}{2} \ \Rightarrow \ x \ = \ \pm \sqrt{ \frac{|B|}{2}} \ \ . $</span> The <span class="math-container">$ \ y-$</span>coordinate for these points is then
<span class="math-container">$$ \left(\frac{|B|}{2} \right)^2 \ + \ B·\left(\frac{|B|}{2} \right) \ + \ C \ \ = \ \ \frac{ B^2 }{4} \ - \ |B|·\left(\frac{|B|}{2} \right) \ + \ C $$</span> <span class="math-container">$$ = \ \ \frac{ B^2 }{4} \ - \ \frac{ B^2}{2} \ + \ C \ \ = \ \ C \ - \ \frac{ B^2 }{4} \ \ . $$</span>
[The foregoing can be found more directly by using calculus.]</p>
<p>With this established, we can now sort the cases for <span class="math-container">$ \ B < 0 \ \ : $</span></p>
<p>• with the "high" turning point "below" the <span class="math-container">$ \ x-$</span>axis <span class="math-container">$ \ [ \ \mathbf{C < 0} \ ] \ \ , $</span> the biquadratic curve has <strong>two</strong> <span class="math-container">$ \ x-$</span>intercepts [the factorization is <span class="math-container">$ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\beta^2 - \alpha^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $</span>
with <span class="math-container">$ \ |\alpha| < |\beta| \ \ , \ \ \alpha \ $</span> imaginary]
;</p>
<p>• with the other turning points "above" the <span class="math-container">$ \ x-$</span>axis, for which <span class="math-container">$ \ C \ - \ \frac{ B^2 }{4} \ > \ 0 \ \Rightarrow \ \mathbf{B^2 \ < \ 4C} \ \ , $</span> the curve <em>does not intersect</em> the <span class="math-container">$ \ x-$</span>axis;</p>
<p>• for <span class="math-container">$ \ C \ - \ \frac{ B^2 }{4} \ < \ 0 \ < \ C \ \Rightarrow \ \mathbf{0 \ < \ 4C \ < \ B^2} \ \ , $</span> there are <strong>four</strong> <span class="math-container">$ \ x-$</span>intercepts "surrounding" the three turning points [the corresponding factorization is <span class="math-container">$ \ (x^2 - \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ . \ ] $</span></p>
<p>We see then that for your equation (1), <span class="math-container">$ \ B = -5 \ , \ C = -36 \ $</span> tells us that the equation has two real zeroes; its factorization is <span class="math-container">$ \ (x^2 - 9)·(x^2 + 4) \ \ . $</span> Equation (2) has <span class="math-container">$ \ B = -13 \ , \ C = 36 \ $</span> <span class="math-container">$\Rightarrow \ B^2 = 169 \ > \ 4C = 144 \ > \ 0 \ \ , $</span> hence there are four real zeroes; the expression factors as <span class="math-container">$ \ (x^2 - 9)·(x^2 - 4) \ \ . $</span> Finally, with equation (3), we "factor out" a <span class="math-container">$ \ 4 \ \ $</span> (which has no effect on the solutions) to obtain <span class="math-container">$ \ B = -\frac52 \ , \ C = \frac{25}{4} \ \Rightarrow \ B^2 = \frac{25}{4} \ < \ 4C = 25 \ \ , $</span> indicating that there are no real zeroes. (The factorization here is not very tidy.)</p>
<p>A table of the conditions for the number of real zeroes is presented below.
<a href="https://i.stack.imgur.com/L5mWY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L5mWY.png" alt="enter image description here" /></a></p>
|
2,078,264 |
<p>I've been see the following question on group theory:</p>
<p>Let $p$ be a prime, and let $G = SL2(p)$ be the group of $2 \times 2$ matrices of determinant $1$ with entries in the field $F(p)$ of integers $\mod p$.</p>
<p>(i) Define the action of $G$ on $X = F(p) \cup \{ \infty \}$ by Mobius transformations. [You need not show that it is a group action.] </p>
<p>State the orbit-stabiliser theorem. Determine the orbit of $\infty$ and the stabiliser of $\infty$. Hence compute the order of $SL2(p)$.</p>
<p>I know matrices are isomorphic to Mobius maps, but not how the action of a mobius map can be used to define the action of a matrix (I don't really know what this part means to be honest).
I tried the next part, but wasnn't sure whether the consider the vector $(\infty,\infty)$, $(\infty,a)$ or $(b,\infty)$ $(a,b \in F(p))$.
Any help would be greatly appreciated!!</p>
<p>(Sorry the question title isn't very related to the question, I just didn't know what to put specifically!)</p>
|
AdLibitum
| 210,743 |
<p>Let $F$ be any field. The group ${\rm GL}_2(F)$ (and thus any of each subgroups) has a natural linear action on the $F$-vector space $F^2$.
By linearity, this action descends to an action on the projective line
${\Bbb P}^1(F)$ which is explicitely given by
$$
\begin{pmatrix} a & b\\ c & d\end{pmatrix}\cdot z=\frac{az+b}{cz+d}
$$
Here $z\in F\cup\{\infty\}$ represents a homothety class of a vector in $F^2$ and $z=\infty$ is nothing but the class of the vector $(1,0)$.</p>
|
3,318,735 |
<p>In a solution to a certain Rolle's theorem problem, I have <span class="math-container">$x_k=a+\frac{k}{n}(b-a)$</span> and
<span class="math-container">$$
0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}
$$</span></p>
<p>How exactly does this work? I mean, let's say <span class="math-container">$$\sum_{k=1}^nf(x_k)-f(x_{k-1})$$</span> has <span class="math-container">$$f(x_2)-f(x_{2-1})$$</span> as one of it's terms. How will
<span class="math-container">$x_k=a+\frac{k}{n}(b-a)$</span> be used in this? Also, how we define <span class="math-container">$n$</span> there? Does it mean that <span class="math-container">$k$</span> is divided by <span class="math-container">$n$</span>? Is <span class="math-container">$n$</span> supposed to be some large finite number?</p>
<p>Excuse me if the question sounds silly.</p>
|
John Omielan
| 602,049 |
<p>You're given that</p>
<p><span class="math-container">$$x_k=a+\frac{k}{n}(b-a) \tag{1}\label{eq1}$$</span></p>
<p>Also, it's stated that</p>
<p><span class="math-container">$$0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} \tag{2}\label{eq2}$$</span></p>
<p>It seems you're asking about how & why \eqref{eq2} works. Well, you're dealing with a function <span class="math-container">$f(x)$</span>, with <span class="math-container">$x \in [a,b]$</span> and <span class="math-container">$f(a) = f(b)$</span>. The interval <span class="math-container">$[a,b]$</span> is divided into <span class="math-container">$n$</span> equal parts, for some <span class="math-container">$n \in \mathbb{N}$</span>, with the <span class="math-container">$k$</span>'th point (<span class="math-container">$0 \le k \le n$</span>) being given by \eqref{eq1}. Note that</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
\sum_{k=1}^n \left(f(x_k)-f(x_{k-1})\right) & = (f(x_1) - f(x_0)) + (f(x_2) - f(x_1)) + \ldots + (f(x_n) - f(x_{n-1})) \\
& = f(x_n) - f(x_0) \\
& = f(b) - f(a)
\end{aligned}\end{equation}\tag{3}\label{eq3}$$</span></p>
<p>Note this is a <a href="https://en.wikipedia.org/wiki/Telescoping_series" rel="nofollow noreferrer">Telescoping series</a>, so all of the terms except for <span class="math-container">$f(x_0)$</span> and <span class="math-container">$f(x_n)$</span> cancel each other. Next, from \eqref{eq1}, you have that</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
x_k - x_{k-1} & = \left(a + \frac{k}{n}(b-a)\right) - \left(a + \frac{k - 1}{n}(b-a)\right) \\
& = \frac{b - a}{n}
\end{aligned}\end{equation}\tag{4}\label{eq4}$$</span></p>
<p>Thus, the last part of \eqref{eq2}, i.e.,</p>
<p><span class="math-container">$$\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} \tag{5}\label{eq5}$$</span></p>
<p>is basically multiplying the middle part by the RHS of \eqref{eq4}, then dividing by the LHS of \eqref{eq4}, and moving that last factor into the summation (which you can do as it's just a constant).</p>
|
3,318,735 |
<p>In a solution to a certain Rolle's theorem problem, I have <span class="math-container">$x_k=a+\frac{k}{n}(b-a)$</span> and
<span class="math-container">$$
0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}
$$</span></p>
<p>How exactly does this work? I mean, let's say <span class="math-container">$$\sum_{k=1}^nf(x_k)-f(x_{k-1})$$</span> has <span class="math-container">$$f(x_2)-f(x_{2-1})$$</span> as one of it's terms. How will
<span class="math-container">$x_k=a+\frac{k}{n}(b-a)$</span> be used in this? Also, how we define <span class="math-container">$n$</span> there? Does it mean that <span class="math-container">$k$</span> is divided by <span class="math-container">$n$</span>? Is <span class="math-container">$n$</span> supposed to be some large finite number?</p>
<p>Excuse me if the question sounds silly.</p>
|
GEdgar
| 442 |
<p>For those inexperienced in <span class="math-container">$\Sigma$</span> notation, try writing a particular example. Say <span class="math-container">$n=4$</span>. Then <span class="math-container">$x_0 = a$</span> and <span class="math-container">$x_4 = b$</span>.<br>
So
<span class="math-container">$$
\sum_{k=1}^n \big(f(x_k)-f(x_{k-1}\big) \\=
\big(f(x_1)-f(x_0)\big)+\big(f(x_2)-f(x_1)\big)+\big(f(x_3)-f(x_2)\big)+\big(f(x_4)-f(x_3)\big)
\\ = f(x_4)-f(x_0) = f(b)-f(a).
$$</span></p>
|
2,230,417 |
<p>I need to formally prove that $f(x)=x^2$ is a contraction on each interval on $[0,a],0<a<0.5$. From intuition, we know that its derivative is in the range $(-1,1)$ implies that the distance between $f(x)$ and $f(y)$ is less then the distance between $x$ and $y$. </p>
<p>But now I need an explicit $\lambda$ such that $0\le \lambda<1$ and $d(f(x),f(y))\le \lambda d(x,y)$, where $d$ is the standard metric on $\Bbb R$.</p>
<p>Thanks a lot!</p>
|
Siminore
| 29,672 |
<p>Remark that
$$
|f(x)-f(y)| = |(x+y)(x-y)| \leq (|x|+|y|)|x-y| < 2a |x-y|
$$
if $x$, $y \in [0,a]$. Now $2a<1$ by assumption.</p>
|
1,920,776 |
<p>As I know there is a theorem, which says the following: </p>
<blockquote>
<p>The prime ideals of $B[y]$, where $B$ is a PID, are $(0), (f)$, for irreducible $f \in B[y]$, and all maximal ideals. Moreover, each maximal ideal is of the form $m = (p,q)$, where $p$ is an irreducible element in $B$ and $q$ is an irreducible element in $(B/(p))[y]$.</p>
</blockquote>
<p>I managed to prove, that any prime non-principal ideal $m \subset B[y]$ is maximal. This gives the first part of the theorem. Now I want to prove, that $m = (p,q)$. I can show, that $m \cap B \neq (0)$. Thus $m$ contains a non-zero element $a$ from B. Since $B$ is PID and $m$ is prime, $m$ must contain an irreducible element from $B$ (we can take an irreducible factor $p$ of $a$). </p>
<p>As I understand, now I need to show, that $m$ contains an element $q$ (see the theorem above). Then it follows that $(p,q) \subseteq m$. And after that I have to prove, that $m \subseteq (p,q)$. How do I make these last steps?</p>
<p>Yes, I have seen similar questions with $B = k[x]$ or $\mathbb{Z}$, but I still cannot prove the last part.</p>
|
Dylan Sommer
| 1,141,434 |
<p>With the help of the other answers on this post (especially Ben Grossmann's), I managed to figure out Two's Complement and why it works for myself, but I wanted to add another complete barebones answer for anyone who still can't understand. This is my first post, so thank you for reading in advance. Also, much of my mathematic notation is likely to be false, so please refer to other answers for more mathematically accurate explanations.</p>
<p>As Ben Grossmann pointed out, understanding that binary addition is modulo is the key to understanding how Two's Complement works. What that means in a binary sense is that the last carry doesn't get used, so:
1111 1111 + 1 = 0000 0000,
not 1 0000 0000.</p>
<p>In decimal, this looks like <span class="math-container">$(255+1)\bmod{2^8}$</span>. A similar example that you may find easier to wrap your head around is <span class="math-container">$(a+b)\bmod{12}$</span>, which should look familiar.</p>
<p>This works for addition, but how about modulo subtraction? Well, continuing with the clock example, if we want to subtract using modulo addition, there is an easy solution: <span class="math-container">$a+(12-b) \pmod{12}$</span>, or in binary: <span class="math-container">$a+(2^8-b)\bmod{2^8}$</span>. The <span class="math-container">$12$</span> and <span class="math-container">$2^8$</span> are canceled out by their corresponding modulo, leaving us with <span class="math-container">$a-b$</span>. The trick is now getting <span class="math-container">$2^8-b$</span>, and that trick lies in Two's Complement.</p>
<p>To derive <span class="math-container">$2^8-b$</span>, or <span class="math-container">$1\ 0000\ 0000 - b$</span> using only 8 bits, we first have to convert that into a familiar format:</p>
<p><span class="math-container">$1\ 0000\ 0000-b=1111\ 1111-b+1$</span></p>
<p>This is the equivalent of <span class="math-container">$11-b+1$</span> using 12 as our modulo. Subtracting a number from <span class="math-container">$1111\ 1111$</span> is the same as inverting it. If this is confusing, then consider the equation</p>
<p><span class="math-container">$1111\ 1111-1101\ 1001$</span></p>
<p>As you may have noticed, no borrows occur, because there are no <span class="math-container">$0$</span>s in <span class="math-container">$2^8$</span>. This effectively means that every bit of b is inverted. And so <span class="math-container">$INV(b)$</span> can be substituted for <span class="math-container">$1111\ 1111 - b$</span>.</p>
<p>Plug that into our previous equation, and we now have <span class="math-container">$a-b=a+INV(b)+1$</span>. There you have Two's Complement.</p>
<p>What this means for negative numbers is that <span class="math-container">$1111/ 1111=-1$</span> (as explained above in more detail by Ben Grossmann) and <span class="math-container">$1000/ 0000=-128$</span>, while <span class="math-container">$0000\ 0000=0$</span> and <span class="math-container">$0111\ 1111=127. The system cycles through first the positives, and once 128 is reached ($</span>1000\ 0000$) it flips the sign and cycles the other way through the negatives. The seventh bit acts like that sign, and no actual sign flip is needed. Without any additional provisions needing to be made, we can now add both negative and positive numbers.</p>
|
2,470,062 |
<blockquote>
<p><span class="math-container">$$\sqrt{k-\sqrt{k+x}}-x = 0$$</span></p>
<p>Solve for <span class="math-container">$k$</span> in terms of <span class="math-container">$x$</span></p>
</blockquote>
<p>I got all the way to
<span class="math-container">$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$</span>
but could not factor afterwards. My teacher mentioned that there was grouping involved</p>
<p>Thanks Guys!</p>
<p>Edit 1 : The exact problem was solve for <span class="math-container">$x$</span> given that <span class="math-container">$$\sqrt{4-\sqrt{4+x}}-x = 0$$</span> with a hint of substitute 4 with k</p>
|
Michael Rozenberg
| 190,319 |
<p>Let $\sqrt{k+x}=y$, where $y\geq0$.</p>
<p>Thus, $\sqrt{k-y}=x$, where $x\geq0$ and we got the following system.
$$k+x=y^2$$ and
$$k-y=x^2,$$
which gives
$$x+y=y^2-x^2$$ or
$$(x+y)(1+x-y)=0.$$
If $x+y=0$, then $x=y=k=0$ otherwise, $y=1+x$ and $k=x^2+x+1$.</p>
<p>Id est, we got the following answer.</p>
<p>If $x<0$ then our equation has no roots;</p>
<p>If $x=0$ then $\{0,1\}$;</p>
<p>If $x>0$ then $\{x^2+x+1\}$.</p>
|
202,043 |
<p>A <a href="http://en.wikipedia.org/wiki/Square-free_word">square-free word</a>
is a string of symbols (a "word") that avoids the pattern $XX$, where $X$ is any
consecutive sequence of symbols in the string.
For alphabets of two symbols, the longest square-free word has length $3$.
But for alphabets of three symbols, there are infinitely long square-free words,
due to Thue.</p>
<p>Define a <em>three-halves-free word</em> as one that avoids the pattern $XYX$, where $|X|=|Y|$.
Another view is that these words avoid $Z(\frac{1}{2}Z)$, i.e., $Z=XY$ with $X$ and $Y$ the
same length. So $Z$ has even length, and we want to avoid immediately
following with the first half of $Z$.</p>
<p>For an alphabet of two symbols, $0$ & $1$, here are the three-halves-free words
of length $5$:
\begin{eqnarray*}
&(0, 1, 1, 0, 0)\\
&(0, 0, 1, 1, 0)\\
&(1, 1, 0, 0, 1)\\
&(1, 0, 0, 1, 1)
\end{eqnarray*}
All the $28$ other length-$5$ words fail.
E.g., $(0,1,0,1,0)$ fails because $(0,1,0)$ matches with $XYX=010$.
But there are no three-halves-free words of length $\ge 6$.
For example, extending $(0, 1, 1, 0, 0)$ with $0$ gives $(0, 1, 1, 0, 0, 0)$, matching
$XYX=000$, and extending with $1$ gives $(0, 1, 1, 0, 0, 1)$, matching with $X=01$.</p>
<blockquote>
<p><strong><em>Q</em></strong>. Are there infinitely long words in three-letter alphabets that are three-halves-free?</p>
</blockquote>
<p>If a word has a square $ZZ$ with $|Z|$ even, then it has a three-halves pattern.
If a word is square-free, it may not be three-halves-free.
For example, both $(1,0,1)$ and $(0,1,2,1,0,1)$ are square-free.
And the square-free infinite word <a href="http://oeis.org/A029883">A029883</a>
$$
(1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, \ldots )
$$
is not three-halves-free: e.g., $(-1,1,-1)$ and $(-1,0,1,0,-1,0)$
are three-halves patterns.</p>
|
JRN
| 12,357 |
<p>This is not an answer, but it may help.</p>
<p>According to <a href="http://www.sciencedirect.com/science/article/pii/S0304397505004019">A generalization of repetition threshold</a> by Lucian Iliea, Pascal Ochemb, and Jeffrey Shallit (<em>Theoretical Computer Science</em>, Volume 345, Issues 2–3, 22 November 2005, Pages 359–369)</p>
<blockquote>
<p>Brandenburg [3] and (implicitly) Dejean [6] considered the problem of determining the
<em>repetition threshold</em>; that is, the least exponent $\alpha=\alpha(k)$ such that there exist infinite words over $\Sigma_k$ that avoid $(\alpha+\epsilon)$-powers for all $\epsilon>0$. Dejean proved that $\alpha(3)=\frac74$.</p>
</blockquote>
<p>(<a href="http://www.sciencedirect.com/science/article/pii/0097316572900118">Dejean's paper</a> is in French, and I can't read French.)</p>
<p>As Bjørn Kjos-Hanssen points out in a comment, the result I cite settles the case of power-free words that avoid all powers greater than or equal to 3/2, but Joseph O'Rourke's question is asking for 3/2-free words. (It is possible for a word to be 3/2-free yet contain powers greater than 3/2, for example, the word $00$ has a square and yet is 3/2-free.)</p>
|
202,043 |
<p>A <a href="http://en.wikipedia.org/wiki/Square-free_word">square-free word</a>
is a string of symbols (a "word") that avoids the pattern $XX$, where $X$ is any
consecutive sequence of symbols in the string.
For alphabets of two symbols, the longest square-free word has length $3$.
But for alphabets of three symbols, there are infinitely long square-free words,
due to Thue.</p>
<p>Define a <em>three-halves-free word</em> as one that avoids the pattern $XYX$, where $|X|=|Y|$.
Another view is that these words avoid $Z(\frac{1}{2}Z)$, i.e., $Z=XY$ with $X$ and $Y$ the
same length. So $Z$ has even length, and we want to avoid immediately
following with the first half of $Z$.</p>
<p>For an alphabet of two symbols, $0$ & $1$, here are the three-halves-free words
of length $5$:
\begin{eqnarray*}
&(0, 1, 1, 0, 0)\\
&(0, 0, 1, 1, 0)\\
&(1, 1, 0, 0, 1)\\
&(1, 0, 0, 1, 1)
\end{eqnarray*}
All the $28$ other length-$5$ words fail.
E.g., $(0,1,0,1,0)$ fails because $(0,1,0)$ matches with $XYX=010$.
But there are no three-halves-free words of length $\ge 6$.
For example, extending $(0, 1, 1, 0, 0)$ with $0$ gives $(0, 1, 1, 0, 0, 0)$, matching
$XYX=000$, and extending with $1$ gives $(0, 1, 1, 0, 0, 1)$, matching with $X=01$.</p>
<blockquote>
<p><strong><em>Q</em></strong>. Are there infinitely long words in three-letter alphabets that are three-halves-free?</p>
</blockquote>
<p>If a word has a square $ZZ$ with $|Z|$ even, then it has a three-halves pattern.
If a word is square-free, it may not be three-halves-free.
For example, both $(1,0,1)$ and $(0,1,2,1,0,1)$ are square-free.
And the square-free infinite word <a href="http://oeis.org/A029883">A029883</a>
$$
(1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, \ldots )
$$
is not three-halves-free: e.g., $(-1,1,-1)$ and $(-1,0,1,0,-1,0)$
are three-halves patterns.</p>
|
Tony Huynh
| 2,233 |
<p>A stronger property is true for four-letter alphabets. That is, in <a href="http://cms.math.ca/cmb/v35/cmb1992v35.0161-0166.pdf">Words without near-repetitions</a>, Currie and Bendor-Samuel prove that there is an infinite word $\omega$ over a four-letter alphabet such that whenever $XYX$ occurs as a subword of $\omega$, $|Y| > |X|$. They also prove that there is no infinite word over a three-letter alphabet with this stronger property. </p>
|
2,263,431 |
<p>Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$</p>
<p>Then find difference between maximum and minimum of $v^2$.</p>
<p>I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?</p>
<p>I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?</p>
|
Maverick
| 171,392 |
<p>Let $v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$</p>
<p>$v^2=a^2+b^2+2\sqrt{\left(a^2\cos^2(x)+b^2\sin^2(x)\right)\left(b^2\cos^2(x)+a^2\sin^2(x)\right)}$</p>
<p>$v^2=a^2+b^2+2\sqrt{a^2b^2\left(\cos^4(x)+\sin^4(x)\right)+\left(a^4+b^4\right)\cos^2(x)\sin^2(x)}$</p>
<p>$v^2=a^2+b^2+\sqrt{4a^2b^2\left(1-2\sin^2(x)\cos^2(x)\right)+\left(a^4+b^4\right)\sin^2(2x)}$</p>
<p>$v^2=a^2+b^2+\sqrt{4a^2b^2-2a^2b^2\sin^2(2x)+\left(a^4+b^4\right)\sin^2(2x)}$</p>
<p>$v^2=|a|^2+|b|^2+\sqrt{4a^2b^2+\left(a^2-b^2\right)^2\sin^2(2x)}$</p>
<p>To find minimumvalue of $v^2$ </p>
<p>put $\sin^2(2x)=0$ </p>
<p>to obtain $v^2=\left(|a|+|b|\right)^2$. So </p>
<p>$v_{min}=|a|+|b|$</p>
<p>And to obtain maximum value of $v^2$ put $\sin^2(2x)=1$</p>
<p>$v_{max}=\sqrt{2(a^2+b^2)}$</p>
|
535,948 |
<p>How do I prove that $(1+\frac{1}{2})^{n} \ge 1 + \frac{n}{2}$ for every $n \ge 1$</p>
<p>My base case is $n=1$ <br/>
Inductive step is $n=k$ <br/>
Assume $n=k+1$<br/></p>
<p>$(\frac{3}{2})^{k} \times \frac{3}{2} \ge (1 + \frac{k+1}{2})$</p>
<p>I'm not sure how to proceed.</p>
|
lab bhattacharjee
| 33,337 |
<p>Let $P(n):(1+x)^n\ge 1+nx$</p>
<p>Clearly $P(n)$ is true for $n=1$</p>
<p>Let $P(n)$ is true for $n=m\implies (1+x)^m\ge 1+mx$</p>
<p>$\implies (1+x)^{m+1}\ge (1+mx)(1+x)=1+(m+1)x+mx^2\ge 1+(m+1)x$ if $m\ge0$ and $x$ is real</p>
|
1,358,574 |
<p>When I searched for the derivative of the Gamma function I got something of the form:</p>
<p>$$\Gamma'(x)=\Gamma(x) \psi(x)$$</p>
<p>But from the definition of the Digamma function to me it's like writing:</p>
<p>$$\Gamma'(x)=\Gamma'(x)$$</p>
<p>And this doesn't seem very useful to me (if I'm wrong feel free to explain me why) so I'm wondering: is there any other form for the derivative(s) of the Gamma function ? This function is defined by an integral so I think that there could be but I'm not sure on how to deal with this.</p>
|
lab bhattacharjee
| 33,337 |
<p>$\dfrac{(n!)^24^n}{(2n)!}=\dfrac{2^n\prod_{r=1}^n(2r)}{2^n\prod_{r=1}^n(2r-1)}=\dfrac{\prod_{r=1}^n(2r)}{\prod_{r=1}^n(2r-1)}$</p>
<p>Now $\dfrac{2r}{2r-1}>1$ for $2r-1>0$</p>
|
1,358,574 |
<p>When I searched for the derivative of the Gamma function I got something of the form:</p>
<p>$$\Gamma'(x)=\Gamma(x) \psi(x)$$</p>
<p>But from the definition of the Digamma function to me it's like writing:</p>
<p>$$\Gamma'(x)=\Gamma'(x)$$</p>
<p>And this doesn't seem very useful to me (if I'm wrong feel free to explain me why) so I'm wondering: is there any other form for the derivative(s) of the Gamma function ? This function is defined by an integral so I think that there could be but I'm not sure on how to deal with this.</p>
|
Mark Viola
| 218,419 |
<p>Recall Stirling's Formula </p>
<p>$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$</p>
<p>Then, we have</p>
<p>$$\begin{align}
\frac{(n!)^2\,4^n}{(2n)!}&=\frac{(2\pi n)\left(\frac{n}{e}\right)^{2n}4^n}{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}\left(1+O\left(\frac1n\right)\right)\\\\
&=\sqrt{\pi n}+O\left(\frac{1}{\sqrt{n}}\right)\\\\
&\to \infty\,\,\text{as}\,\,n\to \infty
\end{align}$$</p>
|
3,812,224 |
<p>I have a circle sector for which I know:</p>
<p><strong>The coordinates of the three points <span class="math-container">$A, B$</span>, and <span class="math-container">$S$</span>, and the radius of circle <span class="math-container">$S$</span>.</strong></p>
<p><img src="https://i.stack.imgur.com/RWzHJ.png" alt="enter image description here" /></p>
<p>I've worked on something similar <a href="https://math.stackexchange.com/questions/3716972/how-to-calculate-the-coordinates-of-the-center-of-the-arc">here</a>, but it was midpoint coordinates.</p>
<p>Now I need <span class="math-container">$4$</span> points to "divide" the arc into five equal parts. How do I calculate it, please?</p>
<p>Thank you.</p>
<p>PS. I calculated the midpoint coordinates using a vector (it's written <a href="https://math.stackexchange.com/a/3717063/799183">here</a>)</p>
|
John Omielan
| 602,049 |
<p>First, determine the angle of the entire arc, i.e., <span class="math-container">$\alpha = \measuredangle ASB$</span>. For this, calculate the perpendicular, i.e., shortest, distance from <span class="math-container">$A$</span> to <span class="math-container">$SB$</span>. This is given by the formula for the distance to a <a href="https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Line_defined_by_two_points" rel="nofollow noreferrer">line defined by two points</a>, where with <span class="math-container">$r = \sqrt{(y_2 - y_3)^2 + (x_2 - x_3)^2}$</span> is the radius of the circle, we get:</p>
<p><span class="math-container">$$p = \frac{\left|(y_2 - y_3)x_1 - (x_2 - x_3)y_1 + x_2y_3 - y_2x_3\right|}{r} \tag{1}\label{eq1A}$$</span></p>
<p>If the point where this perpendicular meets <span class="math-container">$SB$</span> is <span class="math-container">$C$</span>, the <span class="math-container">$\triangle ACS$</span> is right-angled at <span class="math-container">$C$</span>, so we then get</p>
<p><span class="math-container">$$\sin(\alpha) = \frac{p}{r} \implies \alpha = \arcsin\left(\frac{p}{r}\right) \tag{2}\label{eq2A}$$</span></p>
<p>Note, however, the determined value of <span class="math-container">$\alpha$</span> assumes <span class="math-container">$\measuredangle ASB \le \frac{\pi}{2}$</span>. However, with <span class="math-container">$SA$</span> being perpendicular to <span class="math-container">$SB$</span>, there are <span class="math-container">$2$</span> points where <span class="math-container">$A$</span> can be which gives the same value of <span class="math-container">$\alpha$</span> and, otherwise, there are <span class="math-container">$4$</span> possible points, as indicated in the diagram below.</p>
<p><a href="https://i.stack.imgur.com/YaAfR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YaAfR.png" alt="Showing 4 possible places where B can be based to give the same value of alpha" /></a></p>
<p>Note <span class="math-container">$\measuredangle A_1SB = \alpha$</span>, <span class="math-container">$\measuredangle A_2SB = \pi - \alpha$</span>, <span class="math-container">$\measuredangle A_3SB = \pi + \alpha$</span> and <span class="math-container">$\measuredangle A_4SB = 2\pi - \alpha$</span>. If there are bounds or other conditions allowing you to already know which one is correct, e.g., <span class="math-container">$\measuredangle ASB \le \frac{\pi}{2}$</span> so <span class="math-container">$A$</span> is <span class="math-container">$A_1$</span> and the angle is <span class="math-container">$\alpha$</span>, then you can just use that angle. Otherwise, there are several ways to determine which point, and thus angle, is the correct one. Here is a relatively simple method to use.</p>
<p>The vector <span class="math-container">$\mathbf{v_1}$</span> going from <span class="math-container">$S$</span> to <span class="math-container">$B$</span> is</p>
<p><span class="math-container">$$\mathbf{v_1} = (x_2 - x_3, y_2 - y_3) = (x_4, y_4) \tag{3}\label{eq3A}$$</span></p>
<p>Let <span class="math-container">$\theta_i$</span> for <span class="math-container">$1 \le i \le 4$</span> be each of the <span class="math-container">$4$</span> possible values of <span class="math-container">$\measuredangle A_{i}SB$</span> given above, and the vector from <span class="math-container">$S$</span> to <span class="math-container">$A_{i}$</span> be <span class="math-container">$\mathbf{v_{2,i}}$</span>. Then the <a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">rotation matrix</a> formula gives</p>
<p><span class="math-container">$$\mathbf{v_{2,i}} = (x_4\cos(\theta_i) - y_4\sin(\theta_i), x_4\sin(\theta_i) + y_4\cos(\theta_i)) = (x_{5,i}, y_{5,i}) \tag{4}\label{eq4A}$$</span></p>
<p>which means</p>
<p><span class="math-container">$$A_i = (x_3 + x_{5,i}, y_3 + y_{5,i}) = (x_{6,i}, y_{6,i}) \tag{5}\label{eq5A}$$</span></p>
<p>Due to errors, usually quite small, in the determined trigonometric values and rounding errors in the calculations, it's likely none of the <span class="math-container">$A_i$</span> values will match those of <span class="math-container">$A$</span> exactly. You can use something like checking the absolute values of the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> co-ordinates being very close or, alternatively, determine the smallest distance using</p>
<p><span class="math-container">$$d_i = \sqrt{(x_{6,i} - x_1)^2 + (y_{6,i} - y_1)^2} \tag{6}\label{eq6A}$$</span></p>
<p>Once the appropriate <span class="math-container">$\theta_i$</span> angle is determined, the first arc point <span class="math-container">$D$</span> to determine is a rotation of <span class="math-container">$\mathbf{v_1}$</span> by an angle of <span class="math-container">$\beta = \frac{\theta_{i}}{5}$</span>, so let the vector from <span class="math-container">$S$</span> to <span class="math-container">$D$</span> be <span class="math-container">$\mathbf{v_3}$</span>. The rotation formula then gives</p>
<p><span class="math-container">$$\mathbf{v_{3}} = (x_4\cos(\beta) - y_4\sin(\beta), x_4\sin(\beta) + y_4\cos(\beta)) = (x_{7}, y_{7}) \tag{7}\label{eq7A}$$</span></p>
<p>which means</p>
<p><span class="math-container">$$D = (x_3 + x_{7}, y_3 + y_{7}) \tag{8}\label{eq8A}$$</span></p>
<p>You can use a similar procedure for the other <span class="math-container">$3$</span> arc points to determine.</p>
|
3,812,224 |
<p>I have a circle sector for which I know:</p>
<p><strong>The coordinates of the three points <span class="math-container">$A, B$</span>, and <span class="math-container">$S$</span>, and the radius of circle <span class="math-container">$S$</span>.</strong></p>
<p><img src="https://i.stack.imgur.com/RWzHJ.png" alt="enter image description here" /></p>
<p>I've worked on something similar <a href="https://math.stackexchange.com/questions/3716972/how-to-calculate-the-coordinates-of-the-center-of-the-arc">here</a>, but it was midpoint coordinates.</p>
<p>Now I need <span class="math-container">$4$</span> points to "divide" the arc into five equal parts. How do I calculate it, please?</p>
<p>Thank you.</p>
<p>PS. I calculated the midpoint coordinates using a vector (it's written <a href="https://math.stackexchange.com/a/3717063/799183">here</a>)</p>
|
mjw
| 655,367 |
<p>Write <span class="math-container">$s=x_3+i y_3$</span>, compute the radius <span class="math-container">$r=((x_2-x_3)^2+(y_2-y_3)^2)^{1/2}$</span>, compute <span class="math-container">$\theta_0 = \text{atan} \frac{y_2-y_3}{x_2-x_3}$</span>, and <span class="math-container">$\theta_1=\text{atan} \frac{y_1-y_3}{x_1-x_3}.$</span></p>
<p>Now <span class="math-container">$b=s+re^{i\theta_0}$</span> and <span class="math-container">$a=s+re^{i\theta_1}.$</span></p>
<p>The four points are <span class="math-container">$s+re^{i[\theta_0 + \frac{k}{5} (\theta_1-\theta_0)]},\quad k\in\{1,2,3,4\}$</span></p>
|
264,450 |
<p>We know that the class of open intervals $(a,b)$, where $a,b$ are rational numbers is a countable base for $\mathbb R$. </p>
<p>But, $[a,b]$ where $a,b$ are rational numbers does not produce a base for $\mathbb R$.</p>
<p>Can we say that any $(a,b)$ or $[a,b]$ where <strong>$a$ is rational number and $b$ is an irrational number</strong> produce a base for $\mathbb R$?</p>
|
Ittay Weiss
| 30,953 |
<p>I assume that when you say "a base for $\mathbb R$" you mean "a base for the standard topology on $\mathbb R$. With that, the answer to your question is no since $[a,b]$ is never an open set in the standard topology on $\mathbb R$. </p>
<p>If you also meant to ask whether the collection of all $(a,b)$ where $a$ is rational and $b$ is irrational forms a basis for the standard topology on $\mathbb R $ then the answer is yes. </p>
|
1,526,014 |
<p>I have no idea how to compute this infinite sum. It seems to pass the convergence test. It even seems to be equal to $\frac{p}{(1-p)^2}$, but I cannot prove it. Any insightful piece of advice will be appreciated.</p>
|
Surb
| 154,545 |
<p><strong>Hint</strong></p>
<p>$$\sum_{k=1}^\infty kp^k=p\sum_{k=1}^\infty kp^{k-1}.$$</p>
|
1,526,014 |
<p>I have no idea how to compute this infinite sum. It seems to pass the convergence test. It even seems to be equal to $\frac{p}{(1-p)^2}$, but I cannot prove it. Any insightful piece of advice will be appreciated.</p>
|
Jack D'Aurizio
| 44,121 |
<p>$$\begin{eqnarray*}(1-p)^2\sum_{k\geq 1}kp^k &=& \sum_{k\geq 1}kp^k -2\sum_{k\geq 1}kp^{k+1} +\sum_{k\geq 1} kp^{k+2}\\&=&(p+2p^2)-(2p^2)+\sum_{k\geq 3}\left(k-2(k-1)+(k-2)\right)p^{k}\\&=&\color{red}{p}.\end{eqnarray*}$$</p>
|
2,590,537 |
<p>Let $A$ be a proper subset of of $X$, $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X\times Y)-(A\times B)$ is connected.</p>
<p>I have already seen the solution of this problem here on math.stackexchange, but my confusion is this.
Suppose $X$=(a,b) & $Y$=(c,d) and let $A$={m} & $B$={n} be singleton sets such that $A\subset X$ & $B\subset Y$ then $(X\times Y)-(A\times B)$= $[(a,m)\cup(m,b)]\times[(c,n)\cup(n,d)]$ = $[(a,m)\times(c,n)]\cup[(m,b)\times(n,d)]$
which implies that $(X\times Y)-(A\times B)$ is not connected. </p>
<p>Can someone explain me what I have done wrong & what is this question asking. </p>
|
Sri-Amirthan Theivendran
| 302,692 |
<p>A binary $n+1$ vector $(x_1,x_2,\dotsc, x_n, y)$ can be written as $(x,y)$ where $x$ is a binary n-vector. There are $2^n$ choices for $x$ by the inductive hypothesis and $2$ choices for $y$. Hence there are
$$
2^n\times 2
$$
binary $n+1$ vectors.</p>
|
2,309,900 |
<p>EDIT: How many ways are there to distribute $0$ and $1$ in a vector of length $N$ such that the number of zeros and 1 is the same (Suppose $N$ is even). I'm guessing it's $\binom{N}{2}$, but I'm not sure.</p>
|
Zubin Mukerjee
| 111,946 |
<p>There are $N/2$ zeroes and $N/2$ ones. It suffices to choose the position of all of the zeroes, so we are choosing $N/2$ slots out of $N$ possibilities:</p>
<p>$$\binom{N}{N/2} = \frac{N!}{\left((N/2)!\right)^2}$$</p>
<p>This is a <a href="https://en.wikipedia.org/wiki/Central_binomial_coefficient" rel="nofollow noreferrer">central binomial coefficient</a>. </p>
|
2,811,908 |
<p>I have this monster as the part of a longer calculation. My goal would be to somehow make it... nicer.</p>
<p>Intuitively, I would try to somehow utilize the derivate and the integral against each other, but I have no idea, exactly how.</p>
<p>I suspect, this might be a relative common problem, i.e. if we want to derivate a parametrized definite integral against one of its parameters. Maybe it has even some trick... or methodology to handle it.</p>
<p>Of course it is not a problem if the integral remains. The primary goal would be in this stage, to eliminate or significantly simplify the derivative operator.</p>
|
Disintegrating By Parts
| 112,478 |
<p>I looked for a way to simplify such things, and found out that trading differentiation for integration was an efficient way to simplify the task. In your case,
\begin{align}
\int_{0}^{1}e^{bx}f(x)dx &= \int_{0}^{1}\left(\int_{0}^{b}xe^{cx}dc+1\right) f(x)dx\\
& = \int_{0}^{b}\int_{0}^{1}e^{cx}xf(x)dx dc+\int_{0}^{1}f(x)dx.
\end{align}
The function $c\mapsto \int_{0}^{1}e^{cx}xf(x)dx$ is a continuous function of $c$. So the Fundamental Theorem of Calculus now implies that the left side is differentiable in $b$ because the right side is, and
$$
\frac{d}{db}\int_{0}^{1}e^{bx}f(x)dx = \int_{0}^{1}e^{bx}xf(x)dx.
$$
It turns out that trading differentiation for integration in this way is a handy way to justify such operations. Continuity in $c$ of the function mentioned above also follows from integral theorems.</p>
|
4,480,276 |
<p>[This question is rather a very easy one which I found to be a little bit tough for me to grasp. If there is any other question that has been asked earlier which addresses the same topic then kindly link this here as I am unable to find such questions by far.]</p>
<p>let <span class="math-container">$f(x)=x.$</span> If we are to find limit of it at <span class="math-container">$a$</span> by the definition <span class="math-container">$|f(x)-a|<\epsilon \implies |x-a|<\epsilon$</span>. Again <span class="math-container">$|x-a|<\delta$</span>. Then how does it prove that <span class="math-container">$\epsilon=\delta$</span>. Can't <span class="math-container">$\delta$</span> be larger or smaller than <span class="math-container">$\epsilon$</span>?</p>
<p>N.B: I am self learning limits from Calculus Early Transcendentals by James Stewart. I could reach upto the lesson "Precise Definition of Limits" which involves such a problem and didn't explain much about the explained problem</p>
|
Connor Gordon
| 1,049,415 |
<p>The textbook is not <em>proving</em> that <span class="math-container">$\delta=\epsilon$</span>. What it means for a limit to exist is that for every such <span class="math-container">$\epsilon$</span> you can choose <em>a</em> <span class="math-container">$\delta$</span> that works. There is no requirement for uniqueness (in fact uniqueness is never the case, given that you can divide by <span class="math-container">$2$</span> and get something that still works). The textbook is merely claiming that <span class="math-container">$\delta=\epsilon$</span> works, because it does.</p>
|
1,509,502 |
<p>I am trying to prove the following statement:</p>
<blockquote>
<p>Suppose <span class="math-container">$\mu$</span> and <span class="math-container">$\nu$</span> are finite measures on the measurable space <span class="math-container">$(X,\mathcal A)$</span> which have the same null sets. Show that there exists a measurable function <span class="math-container">$f$</span> such that <span class="math-container">$0 < f < \infty$</span> <span class="math-container">$\mu$</span>-a.e. and <span class="math-container">$\nu$</span>-a.e. and for all <span class="math-container">$A \in \mathcal A$</span> one has</p>
<p><span class="math-container">$$ \nu(A) = \int_A f d\mu \quad \text{and} \quad \mu(A) = \int_A \frac{1}{f} d\nu. $$</span></p>
</blockquote>
<p>My approach is to define <span class="math-container">$f = \frac{d\nu}{d\mu}$</span> (applying the Radon-Nikodym theorem) and show that the second equality also holds. Showing the almost-everywhere-requirements is trivial.</p>
<p>I am able to prove this when <span class="math-container">$f$</span> is a simple function, but I run into trouble when I try to extend this argument. I tried to do this by writing <span class="math-container">$f$</span> as the limit of a sequence of increasing simple positive functions <span class="math-container">$f_k$</span>, defining <span class="math-container">$\nu_k(A) = \int_A f_k d\mu$</span>, and then showing that <span class="math-container">$\lim\limits_{k\to\infty} \int_A \frac{1}{f_k} d\nu_k = \mu(A)$</span>. I also showed that <span class="math-container">$\nu(A) = \lim\limits_{k\to\infty}\nu_k(A)$</span>. However, I was unable to relate <span class="math-container">$\int_A \frac{1}{f} d\nu$</span> to these limits.</p>
<p>This is a practice question for an exam, so I'm not looking for complete answers, but I would really appreciate it if someone pointed me in the right direction. Is this the correct approach, or am I overlooking a much simpler argument?</p>
|
Umberto P.
| 67,536 |
<p>You know from Radon-Nikodym that there exists a nonnegative measurable function $g$ with the property that $$\mu(A) = \int_A g \, d\nu$$ for every measurable set $A$. Consequently $$\int \phi \, d\mu = \int \phi\, g \, d\nu$$ for every nonnegative simple function $\phi$, and by taking monotone limits you have $$\nu(A) = \int_A f \, d\mu = \int_A fg \,d\nu$$ for any measurable set $A$. Can you show that $fg = 1$ $\nu$-almost everywhere from here?</p>
|
679,712 |
<p>Find the volume bounded by the cylinder $x^2 + y^2=1$ and the planes $y=z , x=0 ,z=0$ in the first octant.</p>
<p>How do I go about doing this?</p>
|
G Tony Jacobs
| 92,129 |
<p>This might be easier in cylindrical coordinates. You just want the region in which $r$ runs from $0$ to $1$, $\theta$ runs from $0$ to $\frac{\pi}2$, and $z$ runs from $0$ to $y=r\sin\theta$. That gives:</p>
<p>$$\int_0^{\frac\pi2}\int_0^1\int_0^{r\sin\theta}r\,dz\,dr\,d\theta$$,</p>
<p>Which I believe comes out to $\frac13$.</p>
<p>The hard part of these things is visualizing the shape and setting up the integral. Do you see where the limits on each of those variables comes from?</p>
|
2,253,676 |
<p>$f(x)$ continuous in $[0,\infty)$. Both $\int^\infty_0f^4(x)dx, \int^\infty_0|f(x)|dx$ converge. </p>
<p>I need to prove that $\int^\infty_0f^2(x)dx$ converges.</p>
<p>Knowing that $\int^\infty_0|f(x)|dx$ converges, I can tell that $\int^\infty_0f(x)dx$ converges.</p>
<p>Other than that, I don't know anything. I tried to maybe integrate by parts but I got to nowhere</p>
<p>$\int^\infty_0f^4(x)dx = \int^\infty_0f^2(x)|f(x)|\cdot |f(x)| dx$.. </p>
|
user296113
| 296,113 |
<p><strong>Hint</strong> Notice that if $a\ge1$ then $a^4\ge a^2$ and if $ \vert a\vert\le1$ then $a^2\le \vert a\vert$ so in all cases
$$a^2\le \max(\vert a\vert,a^4)$$</p>
|
3,497,919 |
<p><span class="math-container">$$\sum_{k=0}^{\Big\lfloor \frac{(n-1)}{2} \Big\rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} = \frac{n(n+1)}{2} $$</span></p>
<p>So I feel like <span class="math-container">$(-1)^k$</span> is almost designed for the inclusion-exclusion principle. And the left-hand side looks like some sort of pairing, so I am interested in some combinatorics proof like below-linked question. But using a generating function is always helpful. </p>
<p>[EDIT] now I am probably equally, if not more interested in a generating function solution now that I see below answer that completely makes sense to me, but with some issues in signs..</p>
<p><a href="https://math.stackexchange.com/questions/897948/evaluation-of-a-sum-of-1k-n-choose-k-2n-2k-choose-n1">Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$</a></p>
|
Marko Riedel
| 44,883 |
<p>We seek to show that</p>
<p><span class="math-container">$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n}
= \frac{1}{2} n (n+1).$$</span></p>
<p>The LHS is</p>
<p><span class="math-container">$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n-1-2k}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k}
z^{2k} (1+z)^{-2k}.$$</span></p>
<p>Now the coefficient extractor <span class="math-container">$[z^{n-1}]$</span> combined with the <span class="math-container">$z^{2k}$</span>
term enforces the range, making for a zero contribution when <span class="math-container">$2k\gt
n-1$</span> and we may continue with</p>
<p><span class="math-container">$$[z^{n-1}] (1+z)^{2n-1}
\sum_{k\ge 0} (-1)^k {n+1\choose k}
z^{2k} (1+z)^{-2k}
\\ = [z^{n-1}] (1+z)^{2n-1}
\left(1-\frac{z^2}{(1+z)^2}\right)^{n+1}
\\ = [z^{n-1}] \frac{1}{(1+z)^3} (1+2z)^{n+1}.$$</span></p>
<p>This is</p>
<p><span class="math-container">$$\sum_{q=0}^{n-1} (-1)^q {q+2\choose q}
{n+1\choose n-1-q} 2^{n-1-q}.$$</span></p>
<p>Observe that</p>
<p><span class="math-container">$${q+2\choose q} {n+1\choose n-1-q}
= \frac{(n+1)!}{q!\times 2! \times (n-1-q)!}
= {n+1\choose 2} {n-1\choose q}. $$</span></p>
<p>This yields for the sum</p>
<p><span class="math-container">$${n+1\choose 2} \sum_{q=0}^{n-1} (-1)^q {n-1\choose q}
2^{n-1-q}
\\ = {n+1\choose 2} (-1+2)^{n-1}
= {n+1\choose 2}.$$</span></p>
<p>We have the claim.</p>
|
1,703,332 |
<p>Let $f \in L^2 \backslash L^\infty$ on some bounded domain. Let $h$ be a function such that $h(p) \to \frac 32$ as $p \to \infty$.</p>
<p>Consider $$I_p = \left(\int |f|^{h(p)}\right)^{\frac{p}{h(p)}}.$$</p>
<p>Is it true that $\lim_{p \to \infty} I_p \leq C$ exist for some finite constant $C$?</p>
<p>I don't think so, since there is a $p$ in the exponent. </p>
|
MikeP
| 323,373 |
<p>I think that you can do it in n deletes. Delete the first column (below the useful file). Note that the useful files in 2nd - nth rows now are one element closer to the left. Delete the first column below the (now two!) useful files. Continue this a total of n-1 times, at which point all useful files are along the left-most column and simply delete all other columns. The pictures shows the files (red are useful and green are deleted in each of 4 steps).</p>
<p><a href="https://i.stack.imgur.com/PY3y6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PY3y6.jpg" alt="enter image description here"></a></p>
<p>if the shift is column first vs row first, then only 2 steps</p>
<p><a href="https://i.stack.imgur.com/rPOuF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rPOuF.jpg" alt="enter image description here"></a></p>
|
805,596 |
<p>Evaluate $$\lim_{x\to\infty}\dfrac{5^{x+1}+7^{x+2}}{5^x + 7^{x+1}}$$</p>
<p>I'm getting a different result but not the exact one.
I got $$\dfrac{5\cdot\dfrac{5^n}{7^n} +49}{\dfrac{5^n}{7^n} + 7}.$$
I know the result is $7$ but I cannot figure out the steps.</p>
|
Michael Hardy
| 11,667 |
<p>The powers of $5$ are negligible compared to the powers of $7$ when $x$ is big, so you have in effect $7^{x+2}/7^{x+1}$, so the limit is $7$.</p>
<p>One can write
$$
\lim_{x\to\infty} \frac{5^{x+1}+7^{x+2}}{5^x + 7^{x+1}} = \lim_{x\to\infty}\frac{\left(\frac 5 7\right)^{x+1} + 7}{\frac 1 7\left(\frac 5 7 \right)^x + 1} = \frac{0+7}{0+1}
$$
(since the powers of $5/7$ go to $0$ as $x\to\infty$).</p>
|
1,363,074 |
<p>Q- If roots of quad. Equation $x^2-2ax+a^2+a-3=0$ are real and less than $3$ then,</p>
<p>a) $a<2$ </p>
<p>b)$2<a<3$ </p>
<p>c)$a>4$</p>
<p>In this ques., i used $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and then if $a$ will be $1,2$ only then the root will be defined but if we use $3$ then there will be only one root but in ques. Roots are mentioned. Is the right.</p>
|
Bhaskar Vashishth
| 101,661 |
<p>Here roots are $a \pm\sqrt{3-a}$. </p>
<p>So roots to be real, $3-a>0 \implies a<3$.</p>
<p>And to satisfy second condition that roots be less than $3$, we see that </p>
<p>$a -\sqrt{3-a}<3$ iff $a<3$ and $a +\sqrt{3-a}<3$ iff $a<2$ , </p>
<p>hence combining them all we get $a<2$.</p>
|
91,766 |
<p>Hello,
I am looking for a proof for the Chern-Gauss-Bonnet theorem. All I have found so far that I find satisfactory is a proof that the euler class defined via Chern-Weil theory is equal to the pullback of the Thom class by the zero section, but I would like a proof of the fact that this class gives the Euler characteristic when coupled to the fundamental class. Thanks in advance. </p>
|
drbobmeister
| 8,472 |
<p>Chern's original paper may be found at: Chern, Shiing-Shen (1945), "On the curvatura integra in Riemannian manifold", Annals of Mathematics 46 (4): 674–684; this citing quoted from the Wikipedia entry, though I have a copy of the
original paper somewhere in the piles in my office.</p>
|
31,790 |
<p>See following three snippet code:</p>
<pre><code>(*can compile*)
range = Range[-2, 2, 0.005];
Compile[{},
With[{r = range}, Table[ArcTan[x, y], {x, r}, {y, r}]]][];
(*can compile*)
Compile[{},
With[{r = Range[-2, 2, 0.005 - 10^-8]}, Table[ArcTan[x, y], {x, r}, {y, r}]]][];
(*can't compile*)
Compile[{},
With[{r = Range[-2, 2, 0.005]}, Table[ArcTan[x, y], {x, r}, {y, r}]]][];
</code></pre>
<p>Why the last code can't be compiled? I used Mathematica 9.0.1.</p>
|
Mr.Wizard
| 121 |
<p>The warning messages tell you:</p>
<blockquote>
<p>CompiledFunction::cfn: Numerical error encountered at instruction 33; proceeding with uncompiled evaluation. >></p>
<p>ArcTan::indet: Indeterminate expression ArcTan[0.,0.] encountered. >></p>
</blockquote>
<p>By using an offset of <code>10^-8</code> you avoid having exactly zero in the list, and avoid the error.</p>
<p>The first version does not compile in version 7 so I can't tell you about that one.</p>
|
2,411,081 |
<p>How can I show (with one of the tests for convergence , <strong>not by solving</strong>) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?</p>
|
Fred
| 380,717 |
<p>Show that $ \lim_{x \to \infty} \frac{\ln^5(x)}{\sqrt{x}}=0$. Hence there is $a>1$ sucht that </p>
<p>$0 < \frac{\ln^5(x)}{\sqrt{x}} \le 1$ for $x>a$. It follows:</p>
<p>$0 < \frac{\ln^5(x)}{x^2} \le \frac{1}{x^{3/2}}$ for $x>a$. </p>
<p>Can you proceed ?</p>
|
2,078,943 |
<p>My book says this but doesn't explain it:</p>
<blockquote>
<p>Row operations do not change the dependency relationships among columns.</p>
</blockquote>
<p>Can someone explain this to me? Also what is a dependency relationship? Are they referring to linear dependence? </p>
|
hardmath
| 3,111 |
<p>Consider an $m\times n$ matrix with columns:</p>
<p>$$\begin{pmatrix} v_1 & v_2 & \ldots & v_n \end{pmatrix}$$</p>
<p>Now a dependence relation on the columns would express zero as a nontrivial linear combination of them:</p>
<p>$$ \sum_{i=1}^n c_i v_i = 0 $$</p>
<p>where $c_1,c_2,\ldots,c_n$ are scalars.</p>
<p>We can consider the effect of various <a href="https://en.wikipedia.org/wiki/Elementary_matrix#Operations" rel="noreferrer">elementary row operations</a> on the matrix composed of the columns above, one definition of which would be:</p>
<ul>
<li>switching places of two rows</li>
<li>multiplication of a row by a nonzero scalar</li>
<li>adding a multiple of one row to another</li>
</ul>
<p>[NB: The first of these row operations can be effected by a combination of the other two, so this is not a minimal list.]</p>
<p>A concise way to show that such operations preserve any linear dependence relation that may exist among the columns is by appealing to a matrix multiplication form. That is, the linear dependence relation above is:</p>
<p>$$\begin{pmatrix} v_1 & v_2 & \ldots & v_n \end{pmatrix}
\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}
$$</p>
<p>If $E$ is an elementary matrix that effects one of the elementary row operations, the multiplying by $E$ on the left in this last equation:</p>
<p>$$ E \begin{pmatrix} v_1 & v_2 & \ldots v_n \end{pmatrix}
\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}
$$</p>
<p>leaves the linear dependence relation unchanged. Thus the associative property of matrix multiplication gives an easy way to prove the required claim once we understand that elementary row operations correspond to multiplication by elementary matrices.</p>
|
1,397,646 |
<blockquote>
<p>Given $f_n:= e^{-n(nx-1)^2} $, any suggested approaches for showing that $\displaystyle \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$? </p>
</blockquote>
<p>I have already shown that $f_n \to 0$ pointwise (and not uniformly) on $[0,1]$ and have been trying to show that $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$$ I know that $ \forall n \in \mathbb {N}, $ $\exists x_n = \frac {1}{n} \in [0,1]$ such that $f_n(x_n) = 1$, which has me thinking of doing something where I split the integral in such a way that isolates the interval to which the point $x_n$ belongs. Something along the lines of $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = \lim_{n \to \infty}\Big(\int^1_{x_n}f_n(x)dx + \int_0^{x_n} f_n(x)dx\Big)$$ However, this particular split doesn't help much because I can't really do anything with the first integral like I could with the second, bounding it with the sup norm and then observing that $x_n \to 0 $ as $ n \to \infty$ since $\sup_{[0,1]}|f_n| = 1$ (or so I think I can't). Any hints to other approaches I could use or to any modifications to my initial approach?</p>
<p>Thanks in advance.</p>
|
Cameron Williams
| 22,551 |
<p>Here is a somewhat.. cheap way of doing the problem that you might feel a bit unsatisfied with but consider the following instead:</p>
<p>$$\int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx.$$</p>
<p>Clearly</p>
<p>$$\int_0^1 e^{-n(nx-1)^2}\,dx \le \int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx$$</p>
<p>since the integrand is positive. Making a change of variable of $y = nx-1$, we see that $dy = n\,dx$ and we get</p>
<p>$$\int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx \Longrightarrow \frac{1}{n} \int_{-\infty}^{\infty} e^{-ny^2}\,dy.$$</p>
<p>Making another change of variables of $z= \sqrt{n}y$, we get $dz = \sqrt{n}\,dy$ and so</p>
<p>$$\frac{1}{n} \int_{-\infty}^{\infty} e^{-ny^2}\,dy \Longrightarrow \frac{1}{n^{\frac{3}{2}}} \int_{-\infty}^{\infty} e^{-z^2}\,dz.$$</p>
<p>What happens to this integral as you let $n$ tend to infinity? Do you see how this answers the question at hand?</p>
|
1,634 |
<p>I am not too certain what these two properties mean geometrically. It sounds very vaguely to me that finite type corresponds to some sort of "finite dimensionality", while finite corresponds to "ramified cover". Is there any way to make this precise? Or can anyone elaborate on the geometric meaning of it?</p>
|
Peter Arndt
| 733 |
<p>If you have a morphism X-->Y of schemes, finite type means that the fibers are finite dimensional and finite, that the fibers are zero-dimensional. </p>
<p>Take for a finite type example K[x]-->K[x,y]. This corresponds to the projection A^2-->A^1, the fibers are 1-dimensional, which is reflected by K[x,y] being a K[x]-algebra of rank one (or K(x,y) having transcendence degree 1 over K(1)).</p>
<p>For a finite type example take K[x]-->K[x,y]/(y^2-x). For every prime ideal P in K[x] you find two prime ideals in K[x,y]/(y^2-x) whose preimage is P, so the fibers of the corresponding scheme map have cardinality 2 (and dimension zero).</p>
|
20,683 |
<p>The most elementary construction I know of quantum groups associated to a finite dimensional simple Hopf algebra is to construct an algebra with generators $E_i$ and $F_i$ corresponding to the simple positive roots, and invertible $K_j$'s generating a copy of the weight lattice. Then one has a flurry of relations between them, and a coproduct defined on the generators by explicit formulas. These are not mortally complicated, but are still rather involved. Then come explicit checks of coassociativity, and compatibility between multiplication and comultiplication. Finally, one has the $R$-matrix which is an infinite sum with rather non-obvious normalizations. Enter more computations to verify $R$-matrix axioms.</p>
<p>I recall learning about a nice way to construct the quantum group, which in addition to requiring less formulas has the advantage of making it clear conceptually why it's braided.</p>
<blockquote>
<p>I'm hoping someone can either point me to a reference for the complete picture, or perhaps fill in some of the details, since I only remember the rough outline. That, precisely, is my question.</p>
</blockquote>
<p>I include the remarks below in hopes it will jog someone's memory.</p>
<p>You start with the tensor category $Vect_\Lambda$ of $\Lambda$-graded vector spaces, where $\Lambda$ is the weight lattice. We have a pairing $\langle,\rangle:\Lambda\times\Lambda\to \mathbb{Z}$, and we define a braiding $\sigma_{\mu,\nu}:\mu \otimes \nu \to \nu\otimes\mu$ to be $q^{\langle \mu,\nu \rangle}$. Here $q$ is either a complex number or a formal variable. We may need to pick some roots of $q$ if we regard it as a number; I don't remember (and am not too worried about that detail). Also, here we denoted by $\mu$ and $\nu$ the one dimensional vector space supported at $\mu$ and $\nu$ respectively, and we used the fact that both $\mu\otimes\nu$ and $\nu\otimes\mu$ are as objects just $\mu+\nu$.</p>
<p>Okay, so now we're supposed to build an algebra in this category, generated by the $E_i$'s, which generators we regard as living in their respective gradings, corresponding to the simple roots. Here's where things start to get fuzzy. Do we take only the simples as I said, or do we take all the $E_\alpha$'s, for all roots $\alpha$? Also, what algebra do we build with the $E_i$'s? Of course it should be the positive nilpotent part of the quantum group, but since we build it as an algebra in this category, there may be a nicer interpretation of the relations? Anyways, let's call the algebra we are supposed to build here $U_q(\mathfrak{n}^+)$. I definitely remember that it's now a bi-algebra in $Vect_\Lambda$, and the coproduct is just $\Delta(E_i)=E_i\otimes 1 + 1\otimes E_i$ (the pesky $K$ that appears there usually has been tucked into the braiding data). Now we take $U_q(\mathfrak{n}^-)$ to be generated by $F_i$'s in negative degree, and we construct a pairing between $U_q(\mathfrak{n}^+)$ and $U_q(\mathfrak{n}^-)$. The pairing is degenerate, and along the lines of Lusztig's textbook, one finds that the kernel of the pairing is the q-Serre relations in each set of variables $E_i$ and $F_i$.</p>
<p>Finally, once we quotient out the kernel, we take a relative version of Drinfeld's double construction (the details here I also can't remember, but would very much like to), and we get a quasi-triangular Hopf algebra in $Vect_\Lambda$. As an object in $Vect_\Lambda$ it's just an algebra generated by the $E_i$'s and $F_i$'s, so no torus. But since we're working in this relative version, we can forget down to vector spaces, and along the way, we get back the torus action, because that was tucked into the data of $Vect_\Lambda$ all along.</p>
<p>So, the construction (a) gives neater formulas for the products, coproducts, and relations (including the $q$-Serre relations), and (b) makes it clear why there's a braiding on $U_q(\mathfrak{g})$ by building it as the double.</p>
<p>The only problem is that I learned it at a seminar where to my knowledge complete notes were never produced, and while I remember the gist, I don't remember complete details. Any help?</p>
|
Simon Lentner
| 22,709 |
<p><strong>Yes!</strong> It is indeed, to get the <strong>quantum group</strong> $\mathcal{U}_q(\mathfrak{g})$, you have to...</p>
<ul>
<li><p>start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the <strong>Cartan algebra</strong> $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the <strong>rank</strong> of $\mathfrak{g}$ </p></li>
<li><p>consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with</p>
<ul>
<li>an <strong>action</strong> of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$</li>
<li>a <strong>graduation/coaction</strong> to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$</li>
<li>therefore a <strong>braiding</strong> $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$</li>
</ul></li>
<li>form the <strong>tensor algebra</strong> $\mathcal{T}M$ modulo the <strong>Serre relations</strong> for the Cartan matrix a_{ij}
$$ad_{E_i}^{1-a_{ij}}E_j=0$$
with the <strong>braided adjoint action</strong> resp. <strong>commutator</strong>
$$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$
This is the <strong>Borel part</strong> $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$,
in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$</li>
<li>glue Borel part and Cartan algebra to a <strong>Radford biproduct</strong>
$$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$
Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations
$$K_iE_j=q_{ij}E_jK_j$$
$$\Delta(E_i)=K_i\otimes E_i\otimes 1$$</li>
<li>Form the generalized <strong>Drinfel'd double</strong>, which means...
<ul>
<li>Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$
to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct
$$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$ </li>
<li>Quotient out an identification of both Cartan algebras
$$U_q(\mathfrak{g})=(U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g}))^{\sigma}/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$
the last step is called <strong>linking</strong> can nowadays (by A. Masuoka) be done via a <strong>Doi twist</strong> with a 2-cocycle, which causes the additional relations
$$E_i F_i-F_i E_i=\frac{K_i-K_i^{-1}}{q_{ii}-q_{ii}^{-1}}$$</li>
</ul></li>
</ul>
<p><strong>LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes)</strong>, 2008 (<a href="http://www.mi.uni-koeln.de/~iheckenb/na.pdf" rel="noreferrer">http://www.mi.uni-koeln.de/~iheckenb/na.pdf</a>) page 35ff. </p>
<p><em>PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.</em> </p>
|
2,409,377 |
<p>I'm trying to prove that if $F \simeq h_C(X)$ or "$X$ represents the functor $F$", then $X$ is unique up to unique isomorphism. I already know that if $h_C(X) \simeq F \simeq h_C(Y)$ that $s: X \simeq Y$ since Yoneda says that $h_C(X)$ is fully faithful, so reflects isomorphisms (in either direction). If $h_C(X) \simeq h_C(Y)$ is unique then I'm done as then $\psi(s) = \psi(s')$ ans so $s = s'$, where $\psi : h_C(X,Y) \to \text{Hom}_{C^{\wedge}}(h_C(X), h_C(Y))$ is the Yoneda bijection. </p>
<p>But how do I know that $h_C(X) \simeq h_C(Y)$ is unique?</p>
|
Derek Elkins left SE
| 305,738 |
<p>There are as many isomorphisms between $h_C(X)$ and $h_C(Y)$ as there are isomorphisms between $X$ and $Y$.</p>
<p>If $\varphi: h_C(X)\cong F$ and $\psi : h_C(Y)\cong F$, then there is a unique isomorphism $\alpha : h_C(X) \cong h_C(Y)$ <em>such that</em> $\psi \circ \alpha = \varphi$ (just post-compose both sides by $\psi^{-1}$), but there are many ways of exhibiting a representation (i.e. many possible natural isomorphisms even for a given representing object $X$) so simply knowing that there is <em>some</em> natural isomorphism that exhibits $X$ as a representation of $F$ is not enough to uniquely pick out an isomorphism between $X$ and $Y$ where $Y$ is another object that represents $F$ via some unspecified natural isomorphism.</p>
<p>Consider coproducts, i.e. representations of $Z\mapsto\mathsf{Hom}(A,Z)\times\mathsf{Hom}(B,Z)$. What does a representation of this functor look like? What is the universal property of coproducts? How does that universal property relate to representability?</p>
|
308,426 |
<p>Assume that a sequence $(x_n)_{n\in\omega}$ of points of a locally convex topological vector space converges to zero. Is it always possible to find increasing number sequences $(n_k)_{k\in\omega}$ and $(m_k)_{k\in\omega}$ such that the sequence $(m_kx_{n_k})_{k\in\omega}$ still converges to zero? </p>
<p><strong>Added in Edit.</strong> So we already know that this property (called <em>the Mackey convergence condition</em>) does not hold in any locally convex space. But we can ask another </p>
<p><strong>Problem.</strong> Assume that a locally convex space $X$ admits an indexed family $(B_\alpha)_{\alpha\in\omega^\omega}$ of bounded sets such that (i) $B_\alpha\subset B_\beta$ for all $\alpha\le \beta$ in $\omega^\omega$ and (ii) each bounded subset $B\subset X$ is contained in some $B_\alpha$, $\alpha\in\omega^\omega$.</p>
<p>Does $X$ satisfy the Mackey convergence condition?</p>
|
Taras Banakh
| 61,536 |
<p>Oh, sorry! I posed this question too quickly. </p>
<p>A simple example is the dual space $\ell_1=c_0^*$ to the Banach space $c_0$, endpowed with the weak$^*$ topology. The sequence $(e^*_n)_{n\in\omega}$ of coordinate functionals tends to zero, but for any increasing number sequences $(m_k)_{n\in\omega}$ and $(n_k)_{k\in\omega}$ the sequence $(m_ke^*_{n_k})_{k\in\omega}$ does not converge to zero on any sequence $(x_n)_{n\in\omega}\in c_0$ such that $x_{n_k}=\frac1{\sqrt{m_k}}$ for all $k\in\omega$.</p>
|
4,070,810 |
<p>I'm trying to find the number of subgroups of <span class="math-container">$\mathbb{Z}^n$</span> such that the quotient is a <span class="math-container">$\bf{cyclic}$</span> group of order <span class="math-container">$p^k$</span> (<span class="math-container">$p$</span> a prime).</p>
<p>I guess I don't know enough about the finitely generated abelian groups, though I consulted some results about that. So I hope for a hint, thank you very much.</p>
|
Futurologist
| 357,211 |
<p><span class="math-container">$AB = BD$</span> implies that their corresponding circular arcs are equal too, so <span class="math-container">$EB$</span> is the angle bisector of <span class="math-container">$\angle \, AED$</span> and thus <span class="math-container">$\angle \, AEB = \angle \, DEB$</span>. Let <span class="math-container">$R$</span> be the intersection point of <span class="math-container">$BE$</span> and <span class="math-container">$AD$</span>. Then, after a short angle chasing, triangles <span class="math-container">$AEP$</span> and <span class="math-container">$BEQ$</span> are similar, so
<span class="math-container">$$\frac{EP}{EQ} = \frac{EA}{EB}$$</span>
and triangles <span class="math-container">$AER$</span> and <span class="math-container">$BED$</span> are similar, so
<span class="math-container">$$\frac{ER}{ED} = \frac{EA}{EB}$$</span><br />
Consequently,
<span class="math-container">$$\frac{EP}{EQ} = \frac{EA}{EB} = \frac{ER}{ED}$$</span>
which by the Thales' intercept theorem yields
<span class="math-container">$$RD \, || \, PQ$$</span> and since <span class="math-container">$R$</span> lies on <span class="math-container">$AD$</span> you get
<span class="math-container">$$AD \, || \, PQ$$</span></p>
|
2,662,792 |
<p>I'm looking for a Galois extension $F$ of $\mathbb{Q}$ whose associated Galois group $\mbox{Gal}(F, \mathbb{Q})$ is isomorphic to $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. I'm wondering if I should just consider some $u \notin Q$ whose minimum polynomial in $Q[x]$ has degree 9. In that case we'd have $[\mathbb{Q}(u) : \mathbb{Q}] = 9$ and thus $|\mbox{Gal}(\mathbb{Q}(u), \mathbb{Q})| = 9$ (assuming the extension is Galois). But since there are two distinct groups of order 9 (up to isomorphism), I'm not sure that this will yield the desired result. </p>
|
hunter
| 108,129 |
<p>It turns out that if you pick a polynomial at random, it's unlikely that the extension will be Galois (even if you could guarantee what it's Galois group would be). This can be quantified in various ways.</p>
<p>if $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are linearly disjoint, Galois extensions, then the Galois group of the compositum $LK$ is equal to the product of the Galois groups. So we're reduced to finding linearly disjoint Galois extensions of degree $3$. (Conversely, all extensions with your Galois group arise this way, by Galois theory). </p>
<p>Since these cubics are Galois and $3$ is prime, "linearly disjoint" is the same as "intersection equals $\mathbb{Q}$. A cubic polynomial generates a cyclic cubic field if and only if its discriminant is a square, so you can write down any two irreducible cubic polynomials with coprime square discriminant, for example. For more on cubic extensions, this is nice:</p>
<p><a href="http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf" rel="nofollow noreferrer">http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf</a></p>
|
2,439,278 |
<p>I wanted to solve the following problem.</p>
<blockquote>
<p>In $\triangle ABC$ we have $$\sin^2 A + \sin^2 B = \sin^2 C + \sin A \sin B \sin C.$$ Compute $\sin C$.</p>
</blockquote>
<p>Since it's an equation for a triangle, I assumed that $\pi = A + B + C$ would be important to consider.</p>
<p>I've tried solving for $\sin C$ as a quadratic, rewriting $\sin C = \sin (\pi - A - B)$, but nothing seemed to work.</p>
<p>How does one approach this problem? Any help would be appreciated.</p>
<p>The answer is</p>
<blockquote class="spoiler">
<p>$$\frac{2\sqrt{5}}{5}$$</p>
</blockquote>
|
lab bhattacharjee
| 33,337 |
<p>Using <a href="https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b">Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $</a>,</p>
<p>$$\sin^2A+\sin(B-C)\sin(B+C)=\sin A\sin B\sin C$$</p>
<p>As $\sin(B+C)=\sin(\pi-A)=\sin A$ and $\sin A>0$ cancelling it in both sides</p>
<p>$$\sin A+\sin(B-C)=\sin B\sin C$$</p>
<p>$$\sin(B+C)+\sin(B-C)=\sin B\sin C$$</p>
<p>$$\iff2\sin B\cos C=\sin B\sin C$$</p>
<p>As $\sin B>0,$ $$\dfrac{\sin C}2=\dfrac{\cos C}1=\pm\sqrt{\dfrac{\sin^2C+\cos^2C}{2^2+1^2}}=?$$</p>
<p>But again $\sin C>0$</p>
|
2,229,244 |
<p>combinatorial proof of $2^{n+1}\nmid (n+1)(n+2)\dots (2n)$.</p>
<p>I don't have any ideas about proving $sth \nmid sthx$ using combinatorics for showing $sth \mid sthx$ we show that there is a problem that gives $\frac{sthx}{sth*k}$ but what about proving sth doesn't divide sth using combinatorics?</p>
|
the_firehawk
| 416,286 |
<p>by definition $\phi$ is surjective if $\forall z\in \mathbb{C} \;\exists \mathcal{P}\in \mathbb{R}[X]$ such that $\phi(\mathcal{P}) = z = a+ib$
and here the $\mathcal{P}$ is simply $a+bx$</p>
|
3,162,294 |
<p>I've been trying to prove this statement by opening up things on the left hand side using the chain rule but am really getting nowhere. Any tips/hints would be very helpful and appreciated!</p>
|
btilly
| 6,708 |
<p>This is the same as the answer I gave to the now closed <a href="https://stats.stackexchange.com/questions/399396/probability-that-rolling-n-dice-that-the-two-larger-dice-sum-to-x/399409#399409">https://stats.stackexchange.com/questions/399396/probability-that-rolling-n-dice-that-the-two-larger-dice-sum-to-x/399409#399409</a>.</p>
<p>Here is a program to solve the question:</p>
<pre><code>import fractions
# The usual recursive factorial implementation.
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
# The standard formula for n choose m
def choose(n, m):
return factorial(n) / factorial(m) / factorial(n-m)
# The number of ways to get a roll given a number of dice
# of a given size.
def num_dice_outcomes(num_of_dice, dice_size):
return dice_size**num_of_dice
# Recursive way to calculate rolling the dice and coming to a
# specific answer. Note, the cache uses memoization to make it
# faster.
cached_dice_outcomes = {}
def num_dice_outcomes_of_result(num_of_dice, dice_size, result):
# First the trivial answers.
if result < num_of_dice:
return 0
elif num_of_dice * dice_size < result:
return 0
elif num_of_dice < 0:
return 0
elif num_of_dice < 2:
# Either 0 from 0 dice, or something in the range
# 1..size_of_dice from 1 die. Either way there is 1 way.
return 1
else:
# The answer depends only on t.
t = (num_of_dice, dice_size, result)
# If we don't have a cached answer
if t not in cached_dice_outcomes:
answer = 0
# For each possible roll of the first die
for i in range(1, dice_size + 1):
# We add the number of ways that the rest adds up.
answer = answer + num_dice_outcomes_of_result(num_of_dice - 1, dice_size, result - i)
# And now cache it so that we don't repeat this calculation.
cached_dice_outcomes[t] = answer
return cached_dice_outcomes[t]
# This is the function that computes our answer. We will calculate
# the number of ways to get the right answer over the number of
# possible dice outcomes.
def prob_of_sum(result, num_of_dice, top_n_dice=2, dice_size=6):
# Get the number of possible dice outcomes.
total_count = num_dice_outcomes(num_of_dice, dice_size)
# Now count the number of outcomes that get the result.
result_count = 0
# cutoff is the smallest roll we will include in our top n dice.
# It is in the range 1..dice_size.
for cutoff in range(1, dice_size + 1):
# How more do we need to get from the dice that are above
# our cutoff?
result_above = result - cutoff * top_n_dice
# We can have 0..(top_n_dice - 1) dice above the cutoff.
for dice_above in range(top_n_dice):
# How many ways do the dice above get to the needed
# result above?
ways_above = num_dice_outcomes_of_result(dice_above, dice_size - cutoff, result_above)
# How many combinations of dice can be part of our dice above?
ways_dice_above = choose(num_of_dice, dice_above)
# How many dice are at our cutoff? This includes all of
# the top n that are not above, plus any number of the rest.
# That range works out to be:
# (top_n_dice - dice_above)..(num_of_dice - dice_above)
for dice_at_cutoff in range(top_n_dice - dice_above, num_of_dice - dice_above + 1):
# How many ways can we choose which dice are at the cutoff?
ways_dice_cutoff = choose(num_of_dice - dice_above, dice_at_cutoff)
# How many ways can the dice below the cutoff be rolled?
ways_dice_below = num_dice_outcomes(num_of_dice - dice_above - dice_at_cutoff, cutoff-1)
# We now know the number of ways to get this result from
# this cutoff, this many dice_above cutoff, and this many
# dice_at_cutoff.
this_ways = ways_above * ways_dice_above * ways_dice_cutoff * ways_dice_below
# Add that to our running total.
result_count = result_count + this_ways
# We return a fraction to get exact arithmetic, even if the numbers
# involved are very large.
return fractions.Fraction(result_count, total_count)
# We print our answer as a float for convenience.
print(float(prob_of_sum(12, 14, 2, 6))) # 0.704031049874
</code></pre>
|
1,063,774 |
<blockquote>
<p>Prove that the <span class="math-container">$\lim_{n\to \infty} r^n = 0$</span> for <span class="math-container">$|r|\lt 1$</span>.</p>
</blockquote>
<p>I can't think of a sequence to compare this to that'll work. L'Hopital's rule doesn't apply. I know there's some simple way of doing this, but it just isn't coming to me. :(</p>
|
Richard
| 173,949 |
<p>The case where $r=0$ is trivial. WLOG, suppose that $0<r<1$. </p>
<p>We
let $M= \frac{1}{r}-1$. </p>
<p>Now take any $\epsilon >0$. There exists $N \in \mathbb{N}$ such that $N > \frac{1}{\epsilon M}$.</p>
<p>By Binomial expansion, for $n \geq N$,</p>
<p>\begin{eqnarray}
r^n &= & \frac{1}{(1+M)^n} \\
& \leq & \frac{1}{1+nM} \\
& \leq & \frac{1}{nM} \\
& \leq & \frac{1}{NM} \\
& < & \epsilon.
\end{eqnarray}</p>
|
1,659,224 |
<p>Find a sequence $a_n$ such that $\lim_{n \to \infty}|a_{n+1} - a_n | = 0$ while the sequence does not converge.</p>
<p>I am thrown for a loop. For a sequence not to converge it means that for all $\epsilon \geq 0$ and for all $N$, when $n \geq N$, $|a_n - L| \geq \epsilon$ as $n \longrightarrow \infty$</p>
<p>or I could change some terms and state it in terms of a Cauchy sequence, but either way how can the terms have a distance of 0 but the sequence not be consider converging? hints??</p>
|
Daniel R. Collins
| 266,243 |
<p>How about $a_n = \sum_1^n \frac{1}{m}$? The sequential differences are of course equal to $\frac{1}{n}$, but the limit of the sequence is the sum of all such fractions. </p>
<p>Of course, that's just one concrete example of what @Jake said.</p>
|
4,075,667 |
<p>[this is question. I want to know about iv.] I want to know that without being defined everywhere ,can a mapping be onto and one-to-one?
In iv - D has four elements and B has three elements, while in question only three elements are used. So it cannot be function as per definition. Then how it will onto or one to one without function? Is it possible?
<a href="https://i.stack.imgur.com/QfArD.png" rel="nofollow noreferrer">1</a></p>
|
DanielWainfleet
| 254,665 |
<p>This is a matter of terminology and notation. In Set Theory a function <span class="math-container">$is$</span> its graph, and when we say "<span class="math-container">$f$</span> is a function from <span class="math-container">$A$</span> to <span class="math-container">$B\,$</span>", or write <span class="math-container">$f: A\to B$</span> we mean that <span class="math-container">$f\subseteq A\times B$</span> and <span class="math-container">$\forall a\in A\,\,\exists! \,b\in B\,(\,(a,b)\in f)$</span>... (where "<span class="math-container">$\exists!$</span>" means "there exists exactly one"). And <span class="math-container">$A$</span> is called the domain of <span class="math-container">$f$</span>.</p>
<p>A source of confusion is that strict adherence to this definition is sometimes tedious or inconvenient. E.g. suppose <span class="math-container">$f:\Bbb R\setminus \{0\}$</span> where <span class="math-container">$f(x)= e^{-1/|x|},\,$</span> and suppose you say that we extend <span class="math-container">$f$</span> to all of <span class="math-container">$\Bbb R$</span> by letting <span class="math-container">$f(0)=0$</span>. No one will find fault with this. But technically there is no such thing as <span class="math-container">$f(0)$</span>. A pedantic adherence to the def'n would require you to say "Let <span class="math-container">$g=f\cup \{(0,0)\}.\,$</span>" Just remember that there are no exceptions to the def'n, but that sometimes it is easier to act as if there were.</p>
|
1,108,787 |
<p><img src="https://i.stack.imgur.com/YfmKh.png" alt="enter image description here"></p>
<p>I know we have to prove these 10 properties to prove a set is a vector space. However, I don't understand how to prove numbers 4 and 5 on the list.</p>
|
DeepSea
| 101,504 |
<p>Its better to give an example of a set with a given binary operation on it and we can find the zero vector, and the additive inverse of any given vector. I would take the set $V = \{M: M \in R^{2\times 2}\}$, and the operation is addition of matrices $\text{+}$ , then you can find the zero vector and the additive inverse of a matrix fairly easily. This helps you with understanding $\text{#4, 5}$ above.</p>
|
868,384 |
<p>Let's assume I have a set like $S = \{2,1,3,4,8,10\}$. </p>
<p>What's the math notation for the smallest number in the set? </p>
|
Mr.Fry
| 68,477 |
<p>In general for a given set $S$ which is nonempty and a subset of an ordered field we define the smallest element in the set to be the element $x \in S$ such that $x\leq y, \ \forall y \in S$. Since you said in a set, I will not introduce the notion of inf. I hope this helps.</p>
|
1,167,438 |
<p>I am currently solving a differential equation but I am having a little trouble figuring out my integrative factor. </p>
<p>I have </p>
<p>$$\exp\bigg({∫\frac{3}{t}dt}\bigg)$$</p>
<p>so to integrate I made it $\exp(\ln(t^3))$ what is the final solution? $t^3$?</p>
|
Tad
| 85,024 |
<p>$$\int_0^1 da \int_0^1 db \int_0^1 dc \int_0^{{\rm min}(1,a+b+c)} dd \int_0^{{\rm min}(1,a+b+c-d)} de = \frac{31}{40}=0.775$$</p>
|
1,535,914 |
<p>When I plot the following function, the graph behaves strangely:</p>
<p><span class="math-container">$$f(x) = \left(1+\frac{1}{x^{16}}\right)^{x^{16}}$$</span></p>
<p>While <span class="math-container">$\lim_{x\to +\infty} f(x) = e$</span> the graph starts to fade at <span class="math-container">$x \approx 6$</span>. What's going on here? (plotted on my trusty old 32 bit PC.)
<a href="https://i.stack.imgur.com/YL7WQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YL7WQ.png" alt="Approximating e" /></a></p>
<p>I guess it's because of computer approximation and loss of significant digits. So I started calculating the binary representation to see if this is the case. However in my calculations values of <span class="math-container">$x=8$</span> should still behave nicely.</p>
<p>If computer approximation would be the problem, then plotting this function on a 64 bit pc should evade the problem (a bit). I tried the Wolfram Alpha servers:</p>
<p><a href="https://i.stack.imgur.com/joNc2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/joNc2.png" alt="Wolfram alpha" /></a></p>
<p>The problem remains for the same values of <span class="math-container">$x$</span>.</p>
<h3>Questions</h3>
<ol>
<li>Could someone pinpoint the problem? What about the 32 vs 64 bit plot?</li>
<li>Is there a way to predict at which <span class="math-container">$x$</span>-value the graph of the function below would start to fail?
<span class="math-container">$$f_n(x) = \left(1+\frac{1}{x^n}\right)^{x^n}$$</span></li>
</ol>
|
Hagen von Eitzen
| 39,174 |
<p>32bit vs. 64bit affects which integer types are used by default, which is of no interest here. Rather, the floting point computations are made (by default) with IEEE <code>double</code> type.
With this <code>double</code> precision (53 bit mantissa), the relative error of $(1+\frac1{x^{16}})$ is approximately $2^{-53}$. Raising to the $x^{16}$th power roughly multiplies the relative error, so it's $\frac{8^{16}}{2^{53}}=2^{-5}$ (so absolutely $\approx 0.1$) for $x=8$, which does match your plot. At $x=9$ we can expect a relative error $\frac{9^{16}}{2^{53}}\approx0.2$, also consistent with your plot.
For $x>10$ we have $x^{16}>2^{53}$, so $(1+\frac1{x^{16}})\dot=1$, whereas righ tbefore this cutoff to a terminally too small $(1+\frac1{x^{16}})$, we have a longer phase of too big in the last and only bit, i.e., we essentially compute $(1+\frac2{x^{16}})^{x^{16}}\approx e^2\approx 7.4$.</p>
|
40,463 |
<p>If a rectangle is formed from rigid bars for edges and joints
at vertices, then it is flexible in the plane: it can flex
to a parallelogram.
On any smooth surface with a metric, one can define a linkage
(e.g., a rectangle) whose edges are geodesics of fixed length,
and whose vertices are joints, and again ask if it is rigid
or flexible on the surface.
This leads to my first, specific question:</p>
<blockquote>
<p><b>Q1</b>.
Is a rhombus, or a rectangle,
always flexible on a sphere?</p>
</blockquote>
<p><br /><img src="https://i.stack.imgur.com/FRW0R.jpg" alt="alt text" /><br /></p>
<p>It seems the answer should be <em>Yes</em> but I am a bit uncertain
if there must be a restriction on the edge lengths.
(In the above figure, the four arcs are each <span class="math-container">$49^\circ$</span> in length, comfortably short.)</p>
<blockquote>
<p><b>Q2</b>.
The same question for other surfaces: Arbitrary convex surfaces? A torus?</p>
</blockquote>
<p>I am especially interested to learn if there are situations where a linkage that is flexible
in the plane is rendered rigid when embedded on some surface.
It seems this should be possible...?</p>
<blockquote>
<p><b>Q3</b>.
More generally,
<a href="https://mathworld.wolfram.com/LamansTheorem.html" rel="nofollow noreferrer">Laman's theorem</a>
provides a combinatorial characterization of the rigid linkages in the plane.
The <span class="math-container">$n{=}4$</span> rectangle is not rigid because it has fewer than <span class="math-container">$2n-3 = 5$</span> bars:
it needs a 5th diagonal bar to rigidify.
Has Laman's theorem been extended to arbitary (closed, smooth) surfaces
embedded in <span class="math-container">$\mathbb{R}^3$</span>?
Perhaps at least to spheres, or to all convex surfaces?</p>
</blockquote>
<p>Thanks for any ideas or pointers to relevant literature!</p>
<p><b>Addendum</b>.
I found one paper related to my question:
"Rigidity of Frameworks Supported on Surfaces"
by A. Nixon, J.C. Owen, S.C. Power.
<a href="https://arxiv.org/abs/1009.3772v1" rel="nofollow noreferrer" title="Journal version at doi:10.1137/110848852. zbMATH review at https://zbmath.org/?q=an:1266.52018">arXiv:1009.3772v1 math.CO</a>
In it they prove an analog of Laman's theorem for
the circular cylinder in <span class="math-container">$\mathbb{R}^3$</span>.
If one phrases Laman's theorem as requiring for
rigidity that the number of edges <span class="math-container">$E \ge 2 V - 3$</span>
in both the graph and in all its subgraphs,
then their result (Thm. 5.3) is that, on the cylinder, rigidity requires
<span class="math-container">$E \ge 2 V -2$</span> in the graph and in all its subgraphs.
This is not the precise statement of their theorem.
They must also insist that the graph be <em>regular</em>
in a sense that depends on the rigidity matrix achieving maximal rank
(Def. 3.3).
They give as examples of <em>irregular</em> linkages on a sphere
one that
contains an edge with antipodal endpoints, or one that includes
a triangle all three of whose vertices lie on a great circle.
But modulo excluding irregular graphs and other minor technical
details, they essentially replace the constant 3 in Laman's
theorem for the plane with 2 for the cylinder.</p>
<p>Theirs is a very recent paper but contains few citations to related
work on surfaces, suggesting that perhaps the area
of linkages embedded on surfaces is
not yet well explored.
In light of this apparent paucity of information, it seems appropriate that I 'accept'
one of the excellent answers received. Thanks!</p>
<p><b>Addendum [31Jan11].</b>
I just learned of a 2010 paper by Justin Malestein and Louis Theran,
"Generic combinatorial rigidity of periodic frameworks"
<a href="https://arxiv.org/abs/1008.1837v2" rel="nofollow noreferrer" title="Journal version at doi:10.1016/j.aim.2012.10.007. zbMATH review at https://zbmath.org/?q=an:1268.52021">arXiv:1008.1837v2 (math.CO)</a>,
which pretty much completely solves the problem of linkages on a flat 2-torus,
generalizing to flat orbifolds. They obtain a combinatorial characterization for generic minimal
rigidity for "planar periodic frameworks," which encompass these surfaces.</p>
|
maproom
| 9,547 |
<p>There is an obvious metric on the sphere, which you are assuming applies. There is more than one way to put a nice metric on a torus - but I think you mean a torus embedded in three-space, which does not admit such a nice metric. You need to specify what metric you are assuming on the surfaces you are interested in.</p>
|
2,565,763 |
<p>I was playing a standard solitaire game on my mobile app and I came across a round where I couldn't perform a single move thus resulting in a loss. I was then thinking as to what the probability of this event to happen in a single game. Would anybody know this probability and show the calculations given a standard deck of 52 cards? </p>
<p>Here is a link to the rules: <a href="https://www.wikihow.com/Play-Solitaire" rel="nofollow noreferrer">https://www.wikihow.com/Play-Solitaire</a></p>
<p>Note: There are three cards turned at a time but you can only play the second card after you played the top card. Also, the game is klondike.</p>
|
symplectomorphic
| 23,611 |
<p>I think neither of the other two answers succinctly answers your question "why $Z$?"</p>
<p>The point is that $Z$ only appears in this example <em>because of the distributional assumption made in the problem statement</em>. If you make different distributional assumtions, the likelihood ratio inequality may not yield the $Z$ statistic.</p>
<p>Likelihood tests are a <em>general framework</em> for hypothesis testing. In <em>certain cases</em> they are equivalent to doing the $Z$ tests any good high school student learns. But in other cases they aren't. The point of the example you link to is to show that using the likelihood test framework recovers a standard sort of hypothesis test you are assumed to already be familiar with.</p>
|
4,036,975 |
<p><span class="math-container">$\require{begingroup} \begingroup$</span>
<span class="math-container">$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$</span></p>
<p><a href="https://math.stackexchange.com/q/3590905/122782">Related question</a></p>
<p>Is there a known closed form solution to</p>
<p><span class="math-container">\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}</span></p>
<p>It checks out numerically, for <span class="math-container">$n=1,\dots,7$</span> that</p>
<p><span class="math-container">\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}</span></p>
<p>so \eqref{0} seems to follow the pattern</p>
<p><span class="math-container">\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}</span></p>
<p>for some function <span class="math-container">$f$</span>.</p>
<p>Items \eqref{5} and \eqref{7} look promising
as they agree to <span class="math-container">$f(n)=2n\cot^2(\tfrac\pi{n})$</span>,
but the other fail on that.</p>
<hr />
<p><strong>Edit:</strong></p>
<p>Also, it looks like
<span class="math-container">\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}</span></p>
<p>and</p>
<p><span class="math-container">\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}</span></p>
<p>with the same <span class="math-container">$f$</span>.</p>
<hr />
<p><strong>Edit</strong></p>
<p>Thanks to <a href="https://math.stackexchange.com/a/4037168/122782">the great answer by @Quanto</a>,
the function <span class="math-container">$f$</span> can be defined as</p>
<p><span class="math-container">\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}</span></p>
<p><span class="math-container">$\endgroup$</span></p>
|
Svyatoslav
| 869,237 |
<p>Just to put in more closed form what @Varu Vejalla has evaluated</p>
<p><span class="math-container">$I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx$</span></p>
<p><span class="math-container">$\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$</span>
<span class="math-container">$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx [t=\frac{1}{x}]=\int_0^1\frac{\ln(1+t^{2n})}{1+t^2} dt-\int_0^1\frac{\ln(t^{2n})}{1+t^2} dx=I+2nG$</span></p>
<p><span class="math-container">$$I=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx-nG$$</span></p>
<p>Branch points of <span class="math-container">$\log$</span> (roots of <span class="math-container">$1+x^{2n}$</span>) are <span class="math-container">$e^{\frac{i\pi}{2n}},e^{-\frac{i\pi}{2n}}, e^{\frac{3i\pi}{2n}}, e^{\frac{-3i\pi}{2n}},... e^{\frac{(2n-1)i\pi}{2n}}, e^{-\frac{(2n-1)i\pi}{2n}} $</span></p>
<p>Next we integrate in the complex plane, closing the contour in the upper half-plane for the roots <span class="math-container">$e^{-\frac{(2k-1)i\pi}{2n}}$</span> and in the lower half-plane for the roots <span class="math-container">$e^{+\frac{(2k-1)i\pi}{2n}}$</span> - to integrate the single-valued function.</p>
<p>Finally we get
<span class="math-container">$$I=\frac{2\pi{i}}{4*2i}\log\Bigl((-i-e^{\frac{i\pi}{2n}})(i-e^{-\frac{i\pi}{2n}})...(-i-e^{\frac{(2n-1)i\pi}{2n}})((-i-e^{-\frac{(2n-1)i\pi}{2n}})\Bigr)-nG=$$</span>
<span class="math-container">$$=\frac{\pi}{4}\log\Bigl((2+2\sin\frac{\pi}{2n})...(2+2\sin\frac{(2n-1)\pi}{2n})\Bigr)-nG$$</span></p>
<p><span class="math-container">$$I=\frac{\pi}{4}\log\Bigl((1+\sin\frac{\pi}{2n})...(1+\sin\frac{(2n-1)\pi}{2n})\Bigr)+\frac{\pi{n}}{4}\log2-nG$$</span></p>
|
603,291 |
<p>Suppose $f:(a,b) \to \mathbb{R} $ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some $\alpha >1$
and all $x,y \in (a,b) $. Prove that $f$ is constant on $(a,b)$. </p>
<p>I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?</p>
|
Did
| 6,179 |
<blockquote>
<p>Derivatives are not needed here (which is a good thing since proofs based on them can lead to erroneous arguments, see the accepted answer for an example)...</p>
</blockquote>
<p>Choose $y\gt x$ in the interval $(a,b)$. For every $n\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\leqslant M(x_{i+1}-x_i)^\alpha=M(y-x)^\alpha n^{-\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\leqslant \sum\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\leqslant M(y-x)^\alpha n^{1-\alpha}.$$ If $\alpha\gt1$, the RHS goes to zero when $n\to\infty$ because $n^{1-\alpha}\to0$, hence $f(x)=f(y)$, QED.</p>
|
2,463,052 |
<p>I admit $\frac{1}{1+b}<1$ is trivial for $b>0.$ However, the above claim boils down to suggesting that, any $y\in\{x\in\mathbb{R}:x>1\}$ can be uniquely represented by $1+b$ for some $b>0.$ Can you give me some hints how to prove it?</p>
|
Piquito
| 219,998 |
<p>HINT.-Because if $f(x)=\dfrac{1}{1+x}$ then $f([0,+\infty[)=]0,1]$</p>
|
597,986 |
<p>It's a classical theorem of Lie group theory that any compact connected abelian Lie group must be a torus. So it's natural to ask what if we delete the connectedness, i.e. the problem of classification of the compact abelian Lie groups.</p>
|
Cronus
| 229,396 |
<p>Let <span class="math-container">$G$</span> be an abelian Lie group. The connected component <span class="math-container">$G_0$</span> is a connected abelian Lie group, so it's isomorphic to <span class="math-container">$\Bbb{R}^n\times\Bbb{T}^m$</span> for some <span class="math-container">$n,m\in\omega$</span> (this is not hard to see: mostly one just needs to observe <span class="math-container">$\exp$</span> is a surjective open homomorphism with a discrete kernel). In particular, it is divisible. </p>
<p><strong>Lemma</strong>. Suppose <span class="math-container">$D$</span> is a divisible subgroup of an abelian group <span class="math-container">$G$</span> and <span class="math-container">$h:D\to H$</span> is a homomorphism into a divisible abelian group <span class="math-container">$H$</span>. Then <span class="math-container">$f$</span> extends to a homomorphism <span class="math-container">$\tilde f:G\to H$</span>.</p>
<p>For a proof, see <a href="https://planetmath.org/ExampleOfInjectiveModule" rel="noreferrer">here</a>.</p>
<p>Therefore the identity <span class="math-container">$\mathrm{id}:G_0\to G_0$</span> extends to a homomorphism <span class="math-container">$f:G\to G_0$</span>. Denote <span class="math-container">$K=\ker(f)$</span>; clearly <span class="math-container">$K\cdot G_0=G$</span> and <span class="math-container">$K\cap G_0={0}$</span> so (since <span class="math-container">$G$</span> is abelian) <span class="math-container">$G=K\times G_0$</span>. Since <span class="math-container">$G_0$</span> is open, we moreover know that <span class="math-container">$K$</span> must be discrete (and therefore closed). This shows <span class="math-container">$G=K\times G_0$</span> as a topological group and not just an abstract group.</p>
<p>Thus, we have seen abelian Lie groups are exactly the direct products of an abelian discrete group and a connected abelian Lie group.</p>
<p>If we moreover require <span class="math-container">$G$</span> to be compact, then <span class="math-container">$K$</span> must be finite (since it is compact, as a closed subgroup of <span class="math-container">$G$</span>, and discrete) and <span class="math-container">$G_0$</span> a torus (since all connected compact abelian Lie groups are tori). So a group <span class="math-container">$G$</span> is a compact abelian Lie group if and only if it is isomorphic to <span class="math-container">$F\times \Bbb{T}^m$</span> for a finite abelian group <span class="math-container">$F$</span> and <span class="math-container">$m\in\omega$</span>.</p>
|
169,919 |
<blockquote>
<p>If $p$ is a prime, show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$.</p>
</blockquote>
<p>I know that if $a^k$ is a primitive root of $p$ if gcd$(k,p-1)=1$.And sum of all those $k's$ is $\frac{1}{2}p\phi(p-1)$,but then I don't know how use these $2$ facts to show the desired result.<br>
Please help.</p>
|
lab bhattacharjee
| 33,337 |
<p>We know $ ord_m(a^k) =\frac{d}{(d, k)} $ where d=$ord_ma$ => $ord_pa=ord_p(a^{-1})$</p>
<p>There are $\phi(p-1)$ primitive roots for prime $p$.</p>
<p>As $\phi(n)$ is even for $n>2$ ,i.e. for $p-1>2$.</p>
<p>So,for $p>3$, we can always find $a^{-1}$ for each primitive root $a$ of $p$.</p>
<p>Now if $a≡a^{-1}(mod\ p)$ =>$a^2≡1(mod\ p)$ =>$ord_pa|2$.</p>
<p>But $\phi(p-1)$>2 for n>6.</p>
<p>So, there will be $\frac{\phi(p-1)}{2}$ pairs each having product $≡1(mod\ p)$ if $p>3$.
The product is ≡$(-1)^{\phi(p-1)}\pmod p$ , for $p>3$ .</p>
<p>For $p=3$, the only primitive root is $=2 ≡-1(mod 3)=(-1)^{(\phi(3)-1)}\,$ , as $\,\phi(3)=2$</p>
|
268,482 |
<p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p>
<p>The work was reported on a few seminars, and several experts found the conjectures interesting. </p>
<p>One would like to publish this paper in a refereed journal. The paper was rejected from a certain journal just two days after its submission because "this genre of article does not fit the journal".</p>
<blockquote>
<p><strong>QUESTION.</strong> Are there examples of publications of this genre in refereed journals?</p>
</blockquote>
<p><strong>ADD:</strong> The mentioned paper explains the background, states the conjectures, discusses various special cases and consequences, and lists known cases. It is 20 pages long. </p>
|
T. Amdeberhan
| 66,131 |
<p>This is interesting to me too. In my case, I decided for a simple approach: by posting on the arXiv, here is <a href="https://arxiv.org/abs/1207.4045" rel="nofollow noreferrer">one such paper</a>. I also created a separate route on my own website. Since then the conjectures have received several attention as you can see <a href="http://dauns01.math.tulane.edu/%7Etamdeberhan/conjectures.html" rel="nofollow noreferrer">at these coordinates</a> where I keep updating new papers proving the claims or certain partial progress. That is one possible choice though, you may still opt to find a home for your work in some journals, such as in Experimental Mathematics as Carlo Beenakker pointed out. Perhaps otheres can name the journals they like suggesting.</p>
|
268,482 |
<p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p>
<p>The work was reported on a few seminars, and several experts found the conjectures interesting. </p>
<p>One would like to publish this paper in a refereed journal. The paper was rejected from a certain journal just two days after its submission because "this genre of article does not fit the journal".</p>
<blockquote>
<p><strong>QUESTION.</strong> Are there examples of publications of this genre in refereed journals?</p>
</blockquote>
<p><strong>ADD:</strong> The mentioned paper explains the background, states the conjectures, discusses various special cases and consequences, and lists known cases. It is 20 pages long. </p>
|
Brian Hopkins
| 14,807 |
<p>For open problems/conjectures in combinatorial and discrete geometry, there's the journal <a href="http://geombina.uccs.edu/" rel="nofollow noreferrer"><em>Geombinatorics</em></a> focusing on "live mathematics," i.e., research in progress. </p>
<p>One big title is Aubrey de Grey's 2018 "The chromatic number of the plane is at least 5."</p>
<p>The explanation of the journal's topical parameters includes the advice, "We encourage mathematicians in other areas to start similar lively publications." It seems there are some further examples now.</p>
|
1,265,987 |
<p>I am working on some Automata practice problems. I am working a 2 part question. Here it is:</p>
<blockquote>
<p>Let $\Sigma = \{a,b\}$ be an alphabet.
Let $L = \left\{w \in \Sigma^* \mid n_a(w) \le 4\right\}$</p>
<p>a) Create a transition graph for $L$</p>
<p>b) Find a regular expression for $L$</p>
</blockquote>
<hr>
<p>Here is my solution. I am no too sure about part b. If anyone can help me out, I'd really appreciate it. Thank you.</p>
<p><img src="https://i.stack.imgur.com/uktBo.jpg" alt="enter image description here"></p>
|
Brian M. Scott
| 12,042 |
<p>Your work for (a) is fine, but your regular expression is wrong: its last term matches $aaaaa$, for instance, which is not in $L$, and none of it matches $bab$, which is in $L$. It looks as if you tried to generate it from the transition graph and made quite a few mistakes; in particular, it looks as if you forgot to keep track of which states in the original DFA are acceptor states. In this case the language is simple enough that it’s probably easier just to write it down from scratch.</p>
<p>First, $b^*$ covers every word in $L$ that has no $a$s. The words with exactly one $a$ can have any number of $b$s before or after that $a$; they’re covered by $b^*ab^*$. The words with exactly two $a$s can have any number of $b$s before the first $a$, between the two $a$s, or after the second $a$; they’re covered by $b^*ab^*ab^*$. The cases of three and four $a$s are handled similarly, and you end up with</p>
<p>$$b^*+b^*ab^*+b^*ab^*ab^*+b^*ab^*ab^*ab^*+b^*ab^*ab^*ab^*ab^*\;.$$</p>
|
4,097,262 |
<p><a href="https://i.stack.imgur.com/E1UoR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E1UoR.png" alt="Cubic equation with a unit radius circle" /></a>
A cubic equation and circle (unit radius) has intersection at A,B,C,D. ABCD is a square. <em>Find the angle <span class="math-container">$\theta$</span>.</em></p>
<p>I tried:</p>
<ol>
<li><p><span class="math-container">$(0,0)$</span> is a solution so constant term is <span class="math-container">$0$</span></p>
</li>
<li><p>Substituting A(x,y) and C(-x,-y) and adding them gives coefficient of <span class="math-container">$x^2$</span> is 0.</p>
</li>
</ol>
<p>Then the cubic becomes f(x) = <span class="math-container">$ax^3+bx$</span>.</p>
<p>3.Substituting A and B and added the two equations.</p>
<p>I found it interesting-<a href="https://math.stackexchange.com/questions/3680693/prove-that-from-n1-given-points-a-unique-polynomial-of-n-degree-will-pass">for n points given we can find a unique n+1 degree polynomial</a></p>
<p>Also - Can complex number be used here?</p>
<p><strong>Please note</strong>: I am not sure whether we can find the angle(integer) without knowing the coefficients of the cubic.</p>
<p><strong>EDIT</strong>: From the answers<br />
1.putting A <span class="math-container">$(cos\theta,sin\theta)$</span> in f(x) :
<span class="math-container">$acos^3\theta + b cos\theta = sin\theta$</span></p>
<p>2.putting B <span class="math-container">$(-sin\theta,cos\theta)$</span> in f'(x) :
<span class="math-container">$asin^2\theta + b = tan\theta$</span> [ as circle has <span class="math-container">$tan\theta$</span> slope at B]</p>
<p><span class="math-container">$1,2 $</span> eqn gives <span class="math-container">$3asin^2\theta = acos^2\theta$</span></p>
<p>So, <span class="math-container">$sin^2\theta = \frac{1}{4}$</span></p>
<p><em>But I getting the value of <span class="math-container">$\theta$</span> but a answer shows plot of many cubics</em>
-> because in my case <span class="math-container">$ABCD$</span> is a square.</p>
|
RicardoCruz
| 36,340 |
<p>Let
<span class="math-container">$$f(x)=ax^3+bx. \quad \quad (1)$$</span>
Let's find out the values of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> for a specific value of <span class="math-container">$R$</span>.</p>
<p>Let the points <span class="math-container">$ A=(R \cos \theta, R \sin \theta)$</span> and <span class="math-container">$B=(-R \sin \theta, R \cos \theta)$</span> two intersection points of the circle <span class="math-container">$\Lambda (O, R)$</span> with <span class="math-container">$f(x)$</span>, so that B is a tangent point.</p>
<p>Substituting the coordinates of <span class="math-container">$A$</span> in <span class="math-container">$f(x)$</span>, we get:
<span class="math-container">$$\tan \theta =aR^2\cos^2\theta+b. \quad \quad (2)$$</span></p>
<p>Substituting the coordinates of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> in <span class="math-container">$f(x)$</span>, we get with some algebra:
<span class="math-container">$$a =\frac{4}{R^2 \sin4\theta}. \quad \quad (3)$$</span></p>
<p>Substituting the <span class="math-container">$x$</span> of <span class="math-container">$B$</span> in <span class="math-container">$f'(x)$</span>, which is equal to <span class="math-container">$\tan \theta$</span>, we get:
<span class="math-container">$$\tan \theta =3aR^2 \sin^2\theta +b. \quad \quad (4)$$</span></p>
<p>Choosing <span class="math-container">$R =2$</span> and substituting in <span class="math-container">$(2)$</span> and <span class="math-container">$(4)$</span>, we get:
<span class="math-container">$$\theta = 30°.$$</span></p>
<p>The values of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are:</p>
<p><span class="math-container">$$a =\frac{2}{\sqrt 3}$$</span></p>
<p>and</p>
<p><span class="math-container">$$b =-\frac{5\sqrt 3}{3}.$$</span></p>
<p>Plot:
<a href="https://i.stack.imgur.com/LGrBc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LGrBc.png" alt="CircleAndPoly" /></a></p>
|
545,728 |
<p>So I have $X \sim \text{Geom}(p)$ and the probability mass function is:</p>
<p>$$p(1-p)^{x-1}$$</p>
<p>From the definition that:</p>
<p>$$\sum_{n=1}^\infty ns^{n-1} = \frac {1}{(1-s)^2}$$</p>
<p>How would I show that the $E(X)=\frac 1p$</p>
|
thepiercingarrow
| 274,737 |
<p>A simpler way would be to plug in $q=1-p$ and solve it that way using formula for geometric sequences:</p>
<p>\begin{align*}
E(X) &= \sum\limits_{k=1}^\infty kpq^{k-1}\\
&= \frac{p}{q} \sum\limits_{k=1}^\infty kq^{k}\\
&= \frac{p}{q} \frac{q}{(1-q)^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{1}{p}.
\end{align*}</p>
|
357,138 |
<blockquote>
<p>If $a_1,a_2,\dotsc,a_n $ are positive real numbers, then prove that</p>
</blockquote>
<p>$$\lim_{x \to \infty} \left[\frac {a_1^{1/x}+a_2^{1/x}+.....+a_n^{1/x}}{n}\right]^{nx}=a_1 a_2 \dotsb a_n.$$ </p>
<p>My Attempt: </p>
<p>Let $P=\lim_{x \to \infty} \left[\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} \implies \ln P=\lim_{x \to \infty} \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} =\lim_{x \to \infty} nx \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]= \lim_{x \to \infty} n \left[\frac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\right]$ </p>
<p>and this is $0/0$ form and so I have to apply L'Hospital's rule. Now things get a bit complicated during derivative. </p>
<p>Can someone point me in the right direction? Thanks in advance for your time.</p>
|
Xiaolang
| 71,857 |
<p>I will just prove this limit
<span class="math-container">$$
\lim_{x\to\infty} \left(\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}\right)^{x}=(a_1.a_2.....a_n)^{\frac{1}{n}}
$$</span>
We have
<span class="math-container">$$\begin{align}
\lim_{x\to\infty} \left(\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}\right)^{x}&=\lim_{x\to\infty} \left(1+\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}-1\right)^{x} \\&=e^{\displaystyle\lim_{x\to\infty}x\left(\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right)}
\end{align}$$</span>
we know <span class="math-container">$a^x-1$</span> ~ <span class="math-container">$x\ln a$</span></p>
<p>so
<span class="math-container">$$
e^{\displaystyle\lim_{x\to\infty}\left(x\cdot\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right)}=e^{\dfrac{(\ln a_1+\ln a_2+...+\ln a_n)}{n}}=(a_1a_2\cdots a_n)^{\frac{1}{n}}
$$</span></p>
|
3,752,948 |
<p>Find <span class="math-container">$\displaystyle \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $</span><br />
I tried to do like this :
Let <span class="math-container">$A=\displaystyle \lim_{x \to\frac{\pi}{2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $</span></p>
<p>Then <span class="math-container">$\ln A=\lim_{x \to\frac{\pi} {2}} \cos^2 x \ln\{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}$</span><br />
which implies
<span class="math-container">$$\ln A = \lim_{x \to\frac {\pi}{2}} \frac{1^{\sec^2 x} + 2^{\sec^2 x}+ \cdots + n^{\sec^2 x}}{\sec^2 x}$$</span>
I'm stuck here. If I am on the right way please guide me to reach conclusion. Otherwise please describe the actual way. Thanks in advance.</p>
|
Mark Viola
| 218,419 |
<p>Let <span class="math-container">$f_n(x)$</span> be the sequence given by</p>
<p><span class="math-container">$$f_n(x)=\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}$$</span></p>
<p>Then, we have</p>
<p><span class="math-container">$$\begin{align}
f_n(x)&=\left(n^{\sec^2(x)}\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
&=n\left(\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
\end{align}$$</span></p>
<p>For any fixed <span class="math-container">$n\ge1$</span></p>
<p><span class="math-container">$$\lim_{x\to\pi/2}(k/n)^{\sec^2(x)}=\begin{cases}0&1\le k<n\\\\1&,k=n\end{cases}$$</span></p>
<p>Therefore, we have</p>
<p><span class="math-container">$$\lim_{x\to \pi/2}f_n(x)=n$$</span></p>
|
478,523 |
<p>I'm trying to reason through whether $\int_{x=2}^\infty \frac{1}{xe^x} dx$ converges.</p>
<p>Intuitively, It would seem that since $\int_{x=2}^\infty \frac{1}{x} dx$ diverges, then multiplying the denominator by something to make it smaller would make it converge.</p>
<p>If I apply a Taylor expansion to the $\frac{1}{e^x}$, then I get</p>
<p>$\displaystyle \int_{x=2}^\infty \frac{1}{x} e^{-x} dx = \int_{x=2}^\infty \frac{1}{x}(1 + (-x) + \frac{(-x)^2}{2} + \cdots) dx$</p>
<p>My initial thought was to multiply everything out and just look at the leading order behavior, $ x^{n-1}$, but this would seem to mean the integral diverges.</p>
<p>What is wrong with this approach?</p>
|
André Nicolas
| 6,312 |
<p>The Taylor expansion works. Note from $e^x=1+x+\frac{x^2}{2!}+\cdots$, we have $e^x\gt x$ for positive $x$.</p>
<p>It follows that
$$\frac{1}{xe^x}\lt \frac{1}{x^2}.$$
But we know that $\displaystyle\int_2^\infty \frac{1}{x^2}\,dx$ converges, and therefore by Comparison so does our integral.</p>
<p>As has been pointed out, there are better ways to handle the problem without using the Taylor series. </p>
|
1,197,875 |
<p>I'm just starting partials and don't understand this at all. I'm told to hold $y$ "constant", so I treat $y$ like just some number and take the derivative of $\frac{1}{x}$, which I hope I'm correct in saying is $-\frac{1}{x^2}$, then multiply by $y$, getting $-\frac{y}{x^2}$.</p>
<p>But apparently the correct answer is $\frac{1}{x}$. What am I missing?</p>
|
ASB
| 111,607 |
<p>By first principles,</p>
<p>$ \dfrac{\partial}{\partial y}\left( \dfrac{y}{x}\right)=\lim\limits_{\delta y\to 0}\left( \dfrac{\dfrac{y+\delta y}{x}-\dfrac{y}{x}}{\delta y}\right)=\lim\limits_{\delta y\to 0}\dfrac{\delta y}{x\delta y}=\lim\limits_{\delta y\to 0}\dfrac{1}{x}=\dfrac{1}{x} $</p>
|
84,982 |
<p>I am a new professor in Mathematics and I am running an independent study on Diophantine equations with a student of mine. Online I have found a wealth of very helpful expository notes written by other professors, and I would like to use them for guided reading. <strong>I am wondering whether it is customary to ask permission of the author before using his or her online notes for my own reading course.</strong> </p>
<p>Also, if anyone has suggestions for good sources Diophantine Equations please feel free to enlighten me.</p>
|
anon
| 18,030 |
<p>Rule of thumb: it's OK to print out one copy of online notes for your own private use without requesting permission (that's why they are posted). So if you and your students will each be printing your own copies, there should be no need to request permission. If you are going to print a stack of copies to hand out, or place them in the library, you should probably request permission.
Some notes say what can be done with them.</p>
<p>Added: to answer your specific question, no, it is not customary. I have several online notes which googling shows are often used in courses. I never get requests for permission, except sometimes when the instructor is planning to print copies and hand them out (or sell them at cost) to the class, or when the notes are to be placed in a library.</p>
|
815,418 |
<p>Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the </p>
<pre><code>nth x = √(1st x √ 2nd x ... √nth x);
</code></pre>
<p>Then</p>
<p>$$\text{the "infinith" } x = x$$</p>
<p>Example: </p>
<p>$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$<br>
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).</p>
<p>I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.</p>
<p>However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.</p>
|
AnalysisStudent0414
| 97,327 |
<p>You're basically doing this:</p>
<p>$x_0 = \sqrt x$</p>
<p>$x_1 = \sqrt{x x_0}$</p>
<p>$\displaystyle x_n = \sqrt{x x_{n-1}} = x^{\sum_{k=1}^n \frac{1}{2^k}} $</p>
<p>So it's pretty obvious that converges to $x$</p>
|
815,418 |
<p>Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the </p>
<pre><code>nth x = √(1st x √ 2nd x ... √nth x);
</code></pre>
<p>Then</p>
<p>$$\text{the "infinith" } x = x$$</p>
<p>Example: </p>
<p>$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$<br>
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).</p>
<p>I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.</p>
<p>However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.</p>
|
robjohn
| 13,854 |
<p>It is important to show that the limit exists. Let define the sequence
$$
a_k=\sqrt{\vphantom{A}na_{k-1}}
$$
Since $\dfrac{a_k}{a_{k-1}}=\sqrt{\dfrac{n}{a_{k-1}}}$ and $\dfrac{a_k}{n}=\sqrt{\dfrac{a_{k-1}}{n}}$, we have</p>
<ol>
<li><p>if $a_{k-1}\le n$, then $a_{k-1}\le a_k\le n$; that is, $a_k$ is increasing and bounded above by $n$.</p></li>
<li><p>if $a_{k-1}\ge n$, then $a_{k-1}\ge a_k\ge n$; that is, $a_k$ is decreasing and bounded below by $n$.</p></li>
</ol>
<p>In either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get
$$
\lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{\vphantom{A}na_{k-1}}=\sqrt{n\lim_{k\to\infty}a_k}
$$
Squaring and dividing by $\lim_{k\to\infty}a_k$, we get that
$$
\lim_{k\to\infty}a_k=n
$$</p>
<hr>
<p><strong>Another Approach</strong></p>
<p>Not as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get
$$
\begin{align}
\log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right)
&=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\
&=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\
&=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt]
&=\log(n)
\end{align}
$$</p>
|
815,418 |
<p>Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the </p>
<pre><code>nth x = √(1st x √ 2nd x ... √nth x);
</code></pre>
<p>Then</p>
<p>$$\text{the "infinith" } x = x$$</p>
<p>Example: </p>
<p>$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$<br>
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).</p>
<p>I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.</p>
<p>However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.</p>
|
Community
| -1 |
<p>Assume that you iterated infinitely many times (can take a while), and observed a convergence to $n$.</p>
<p>One more iteration yields $\sqrt{n.n}=n$.</p>
|
3,967,118 |
<p>How is it possible to define <span class="math-container">$\lim_{x \rightarrow a} f(x)$</span> if <span class="math-container">$f(x)$</span> is undefined at <span class="math-container">$a$</span>? There is an infinite amount of <span class="math-container">$x$</span>-values on either side of <span class="math-container">$x=a$</span>, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?</p>
<p>I am using this definition of a limit [1].</p>
<blockquote>
<p>A function <span class="math-container">$f(x)$</span> approaches a limit <span class="math-container">$A$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$a$</span> if, and
only if, for each positive number <span class="math-container">$\epsilon$</span> there is another,
<span class="math-container">$\delta$</span>, such
that whenever <span class="math-container">$0 < |x-a| < \delta $</span> we have <span class="math-container">$|f(x) - A|< \epsilon$</span>.
That is, when <span class="math-container">$x$</span> is near <span class="math-container">$a$</span> (within a distance <span class="math-container">$\delta$</span>
from it), <span class="math-container">$f(x)$</span> is near <span class="math-container">$A$</span> (within a distance <span class="math-container">$\epsilon$</span> from it).
In symbols we write <span class="math-container">$\lim_{x \rightarrow a} f(x) = A$</span>.</p>
</blockquote>
<p>[1] David V. Widder. Advanced Calculus. Dover 1989.</p>
|
Community
| -1 |
<p>A limit uses the values of <span class="math-container">$f(x)$</span> everywhere <em>except</em> at <span class="math-container">$x=a$</span> ! Whether <span class="math-container">$f(a)$</span> is defined or not and whether <span class="math-container">$f(a)$</span> coincides with the limit or not is irrelevant. This is expressed in the definition by <span class="math-container">$0<|x-a|$</span>.</p>
<p>The goal of a limit is precisely to "guess" what <span class="math-container">$f(a)$</span> is <em>should be</em>.</p>
|
3,967,118 |
<p>How is it possible to define <span class="math-container">$\lim_{x \rightarrow a} f(x)$</span> if <span class="math-container">$f(x)$</span> is undefined at <span class="math-container">$a$</span>? There is an infinite amount of <span class="math-container">$x$</span>-values on either side of <span class="math-container">$x=a$</span>, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?</p>
<p>I am using this definition of a limit [1].</p>
<blockquote>
<p>A function <span class="math-container">$f(x)$</span> approaches a limit <span class="math-container">$A$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$a$</span> if, and
only if, for each positive number <span class="math-container">$\epsilon$</span> there is another,
<span class="math-container">$\delta$</span>, such
that whenever <span class="math-container">$0 < |x-a| < \delta $</span> we have <span class="math-container">$|f(x) - A|< \epsilon$</span>.
That is, when <span class="math-container">$x$</span> is near <span class="math-container">$a$</span> (within a distance <span class="math-container">$\delta$</span>
from it), <span class="math-container">$f(x)$</span> is near <span class="math-container">$A$</span> (within a distance <span class="math-container">$\epsilon$</span> from it).
In symbols we write <span class="math-container">$\lim_{x \rightarrow a} f(x) = A$</span>.</p>
</blockquote>
<p>[1] David V. Widder. Advanced Calculus. Dover 1989.</p>
|
DonAntonio
| 31,254 |
<p>I present you with the grap of the function <span class="math-container">$\;f(x)=\cfrac{\sin x}x\;$</span> :</p>
<p><a href="https://i.stack.imgur.com/gFnXl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gFnXl.png" alt="enter image description here" /></a></p>
<p>Clearly the functions isn't defined at <span class="math-container">$\;x=0\;$</span> , yet the limit if the function when <span class="math-container">$\;x\to0\;$</span> is <span class="math-container">$\;1\;$</span> .</p>
<p>In words: the limit <span class="math-container">$\;\lim\limits_{x\to x_0}f(x)\;$</span> doesn't depend <em>at all</em> on whether <span class="math-container">$\;f(x_0)\;$</span> is defined or what its possible value could be, but only depends on the behavior of the function on an open <strong>punctured</strong> neighborhood of <span class="math-container">$\;x_0\;$</span> , meaning: without the point <span class="math-container">$\;x_0\;$</span> itself.</p>
|
261,156 |
<p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</p>
<p><span class="math-container">$$123456789101112131415$$</span></p>
<p>And when <span class="math-container">$n=4$</span> we get:</p>
<p><span class="math-container">$$1234$$</span></p>
|
Syed
| 81,355 |
<p>Identical to the solution by @murray but written as a composition:</p>
<pre><code>f = FromDigits@*
Flatten@*
IntegerDigits@*
Range@
# &
f /@ Range[8, 15]
</code></pre>
<hr />
<p>Using Reap/Sow:</p>
<pre><code>g = FromDigits@
Flatten@Last@Reap@Scan[Sow[IntegerDigits[#]] & ]@Range[#] &
g /@ Range[8, 15]
</code></pre>
<hr />
<p>Result:</p>
<blockquote>
<pre><code>{12345678, 123456789, 12345678910, 1234567891011, 123456789101112, \
12345678910111213, 1234567891011121314, 123456789101112131415}
</code></pre>
</blockquote>
|
2,801,294 |
<p>I cam across an interesting claim that:</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>In trying to understand it I consulted Bromiley:</p>
<p><a href="http://www.tina-vision.net/docs/memos/2003-003.pdf" rel="nofollow noreferrer">http://www.tina-vision.net/docs/memos/2003-003.pdf</a></p>
<p>Bromiley concludes that:</p>
<p>if </p>
<p>$f(x) = \frac{1}{\sqrt{2\pi\sigma_f^2}} e^{-\frac{(x-\mu_f)^2}{2 \sigma_f^2}}$
and
$g(x) = \frac{1}{\sqrt{2\pi\sigma_g^2}} e^{-\frac{(x-\mu_g)^2}{2 \sigma_g^2}}$</p>
<p>then:</p>
<p>$f(x)g(x) = D_{fg} \frac{1}{\sqrt{2\pi\sigma_{fg}^2}}
e^{-
\frac
{ (x - \mu_{fg})^2 }
{2 \sigma_{fg}^2 }
}$</p>
<p>where:</p>
<p>$\mu_{fg} = \frac { \sigma_g^2\mu_f + \sigma_f^2 \mu_g } {\sigma_f^2 + \sigma_g^2}$
and
$\sigma_{fg}^2 = \frac {\sigma_f^2 \sigma_g^2} {\sigma_f^2 + \sigma_g^2}$</p>
<p>$S_{fg} = \frac {1} {\sqrt{2\pi(\sigma_f^2+\sigma_g^2)}}
e^{
-\frac{(\mu_f-\mu_g)^2}{2(\sigma_f^2+\sigma_g^2)}
}$</p>
<p>Note that if $\mu_f$, $\mu_g$ , $\sigma_f$ and $\sigma_f$ are known constants then the $S_{fg}$ is a known constant too.</p>
<p>To wit, if I cast Bromiley's result in the format of the claim I'm exploring:</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = S_{fg} \; \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>In short there is a constant scaling factor $S_{fg}$. In fact Bromiley describes the product as a scaled Gaussian. </p>
<p>Given $f(x)$ and $g(x)$ are both functions of $x$ the original claim, which reads (as a reminder):</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>implies that:</p>
<p>$\int_{-\infty}^{\infty} f(x) g(x) \;dx = 1$</p>
<p>But Bromiley's result suggests this implication is false. I presume it inetgrates to S_{fg}, or:</p>
<p>$\int_{-\infty}^{\infty} f(x) g(x) \;dx = S_{fg}$</p>
<p>My tentative conclusion is that the claim I am exploring is false, and my questions would be:</p>
<ol>
<li>Is my tentative conclusion true? (is the explored claim false?)</li>
<li>Am I right in concluding the integral would be $S_{fg}$?</li>
</ol>
<p>Those are the areas I'm a little shakey on at present and seek some review on I guess.</p>
|
Nadiels
| 394,719 |
<p>My (subjective) opinion:</p>
<h3>Do write</h3>
<p>$$
\mathcal{N}(x \mid \mu_1,\sigma_1^2)\cdot\mathcal{N}(x \mid \mu_2,\sigma_2^2)
$$</p>
<h3>Don't write</h3>
<p>$$
\mathcal{N}(\mu_1,\sigma_1^2)\mathcal{N}(\mu_2, \sigma^2)
$$</p>
<hr>
<p>The notation $\mathcal{N}(x|\mu_1,\sigma_1^2)$ makes it clear that this is a <em>function</em>, this particular function can have a particular significance in a statistical/probability setting question but that does not <em>need</em> to be the case, it is simply a function and the product of functions is a well defined and familiar operation with little ambiguity.</p>
<p>On the other hand the notation $\mathcal{N}(\mu, \sigma^2)$ is to be used as short hand to make $X \sim \mathcal{N}(\mu, \sigma^2)$ equivalent to "the random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$". </p>
<p>In this setting $\mathcal{N}(\mu, \sigma^2)$ <em>is not a function</em>, it isn't even a distribution function it is a symbolic representation of a particular statement, a sign post that allows us to retrieve certain information should we wish, but it is not a mathematical object for which we have a well defined product.</p>
|
1,110,652 |
<p>Would anyone happen to know any introductory video lectures / courses on partial differential equations? I have tried to find it without success (I found, however, on ODEs). </p>
<p>It does not have to be free material, but something not to expensive would be nice.</p>
|
Community
| -1 |
<p>Khan Academy is very good with teaching Differential Equations. You should check it out:</p>
<p><a href="https://www.khanacademy.org/math/differential-equations" rel="nofollow">Khan Academy</a></p>
|
1,110,652 |
<p>Would anyone happen to know any introductory video lectures / courses on partial differential equations? I have tried to find it without success (I found, however, on ODEs). </p>
<p>It does not have to be free material, but something not to expensive would be nice.</p>
|
DilipSahu
| 95,033 |
<p>May Be these Lectures will be useful to you </p>
<p><a href="http://www.math.lamar.edu/faculty/maesumi/PDE1.html" rel="nofollow">http://www.math.lamar.edu/faculty/maesumi/PDE1.html</a></p>
|
3,657,751 |
<p>Consider the series <span class="math-container">$$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$</span> This is clearly not absolutely convergent. On the other hand, obvious choice, alternating series does not work here. Seems like the partial sum sequence is bounded but it is not monotone.</p>
<p>How can we prove that this series converges? and, where does it converge to? </p>
|
user780985
| 780,985 |
<p>The triangular numbers alternate odd, odd, even, even, odd, odd, even, even, etc. The reason is that to move from <span class="math-container">$T_n$</span> to <span class="math-container">$T_{n+2}$</span>, we add <span class="math-container">$n + (n + 1) = 2n + 1$</span>, an odd number, so we obtain an alternating pattern of parities for <span class="math-container">$T_n$</span>, over the odd and even integers <span class="math-container">$n$</span>.</p>
<p>So, consider grouping the terms in pairs:
<span class="math-container">$$\left(1+\frac12\right)-\left(\frac13+\frac14\right)+\left(\frac15+\frac16\right)-\ldots = \sum_{n=1}^\infty (-1)^n\left(\frac{1}{2n-1} + \frac{1}{2n}\right).$$</span>
This series is convergent, using the alternating series test.</p>
<p>Because the terms of the original series converge to <span class="math-container">$0$</span>, this also implies that the original series is convergent too.</p>
|
3,657,751 |
<p>Consider the series <span class="math-container">$$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+1}}{n}=1+\dfrac12-\dfrac13-\dfrac14+\dfrac15+\dfrac16-\cdots.$$</span> This is clearly not absolutely convergent. On the other hand, obvious choice, alternating series does not work here. Seems like the partial sum sequence is bounded but it is not monotone.</p>
<p>How can we prove that this series converges? and, where does it converge to? </p>
|
wjmccann
| 426,335 |
<p>The triangle numbers themselves follow the pattern of </p>
<p><span class="math-container">$$odd, odd, even, even, odd, odd, even, even, \ldots$$</span></p>
<p>This can be shown as this pattern alternates whether or not the even number in <span class="math-container">$\frac{n(n+1)}{2}$</span> is <span class="math-container">$0,2 \mod 4$</span>. So as you can see in your sum, the terms will alternate in a pattern of twos. </p>
<p>Now if we group these terms of the series as follows </p>
<p><span class="math-container">$$
(1+1/2)-(1/3+1/4) +(1/5+1/6) - \ldots = \sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)
$$</span></p>
<p>which converges via the alternating series test, and according to Wolfram alpha, it converges to the value of
<span class="math-container">$$
\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right) = \dfrac{1}{4}(\pi+2\log(2))
$$</span></p>
<p><strong>However</strong> you can only associate the series if it converges, so that associative step we made assumes that your initial series converges. I'd expect your initial series to converge, however I am much to tired and need to sleep!</p>
|
1,341,896 |
<p>given three lines $\ell_1,\ell_2, \ell_3 $ which intersect in one point $P$. How can one construct a triangle such that the given lines become its angle bisectors?</p>
<p>So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays $\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2$ - any line $\ell_j$ determines the two rays $\omega_j^1,\omega_j^2$- emanating from the common point $P$.</p>
<p>Any of the three vertices $V_1,V_2,V_3$ of the desired triangle has to lie on exactly one line, i.e. $V_j\in \ell_j=\omega_j^1\cup \omega_j^2$. Hence we have either $V_j\in \omega_j^1$ or $V_j\in \omega_j^2$.</p>
<p>Therefore the triangle should exist if and only if it is possible to choose three rays $\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace $, $\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace$, $\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace$ such that any three of the rays $\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}$ form an angle less than $\pi$.</p>
<p>But I do not know how to construct the triangle?! </p>
<p>I started to draw a circle around $P$ with any radius. Now I can choose a Point $V_1$ on the ray $\tilde{\omega_1}$ outside the circle. And now I can draw the tangents through $V_1$ to the circle. I think those tangents will meet the other rays at some points $V_2,V_3$ (Why?). Now we can draw the line $V_2,V_3$. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way?</p>
<p>Best regards </p>
|
Narasimham
| 95,860 |
<p>I was attempting to address the general question you posed earlier <a href="http://%20[in%20Mathematica][1]" rel="nofollow noreferrer">(by Mathematica)</a> with an assumption of existence of common angular bisectors of given the three lines for variable triangles schematically hand sketched here:</p>
<p><a href="https://i.stack.imgur.com/49xFU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/49xFU.png" alt="enter image description here"></a></p>
<p>which would perhaps be also interesting. Whether they would be a set of parallel similar triangles of parallel sides needs yet to be confirmed.</p>
|
4,228,535 |
<p>First the geometric inversion map <span class="math-container">$f:\mathbb{R}^n\setminus \{0\}\rightarrow \mathbb{R}^n$</span> is defined by
<span class="math-container">$$f(x)=\frac{x}{|x|^2}=:X.$$</span>
The one of its properties is following:</p>
<p>For any <span class="math-container">$c\in\mathbb{R}^n, c\not=0$</span>, the sphere <span class="math-container">$|x-c|=|c|$</span> is mapped to the hyperplane <span class="math-container">$2X\cdot c=1$</span>.</p>
<p>I can't understand what it means intuitively. As I know, <span class="math-container">$2X\cdot c=1$</span> means the set of <span class="math-container">$X$</span> with length of <span class="math-container">$1/2$</span> when projection maps <span class="math-container">$X$</span> to <span class="math-container">$c$</span>. However, I failed to find some connection with sphere <span class="math-container">$|x-c|=|c|$</span>. Can I get any help?</p>
|
paul garrett
| 12,291 |
<p>Your argument is correct. Indeed, characteristic functions <span class="math-container">$\chi_\varepsilon$</span> of smaller and smaller intervals <span class="math-container">$[-\varepsilon,\varepsilon]$</span> do not converge to <span class="math-container">$\delta$</span> (in a distributional sense), but, rather, to <span class="math-container">$0$</span>. Still, the renormalizations to have "total mass" <span class="math-container">$1$</span>, namely, <span class="math-container">${1\over 2\varepsilon}\chi_{\varepsilon}$</span>, <em>do</em> converge to <span class="math-container">$\delta$</span>, by a similar computation.</p>
|
3,954,066 |
<p>How do you show that for some <span class="math-container">$A\subset \mathbb R$</span>, <span class="math-container">$A\not \in \mathbb R$</span>.</p>
<p>Intuitively it makes sense, but how do you actually prove it?</p>
|
saulspatz
| 235,128 |
<p>It depends upon how the real numbers are defined. For example, if a real number is an equivalence class of Cauchy sequences of rational numbers, then a set of real numbers is not a real number, because its elements are not equivalence classes of Cauchy sequences, but sets of such equivalence classes.</p>
|
4,557,576 |
<p>Working on a 3U CubeSat as part of a project for a Space Engineering club. To calculate the maximum solar disturbance force, we are trying to calculate the largest shadow a 0.1 * 0.1 * 0.3 rectangular prism can cast.</p>
<p>If the satellite was oriented with the largest side facing the sun directly, the shadow cast would be 0.03 m^2. It is our thought that there is a certain orientation in which the shadow is larger than this, so the solar disturbance force will be maximized. Is there a function we could use to maximize this value?</p>
<p>Any help would be greatly appreciated!</p>
|
Andrew Draganov
| 432,295 |
<p>I can work out the answer more thoroughly if you'd like but here's a write-up of the steps you want to take to find a solution.</p>
<p>First, the size of the shadow is a continuous function of (I think) seven variables. We have already defined the length, width and height, though, so I'll just treat those as constants for now. You probably want to set them to 1, though, to make the calculations easier. You can make them the real numbers later.</p>
<p>The first three <em>unknown</em> variables that affect the shadow of the box, then, are the <a href="https://en.wikipedia.org/wiki/Aircraft_principal_axes" rel="nofollow noreferrer">yaw, pitch and roll</a> of the box. Using those, we can fully represent the rotation of the box by the <a href="https://en.wikipedia.org/wiki/Rotation_matrix#General_rotations" rel="nofollow noreferrer">3D rotation matrix</a>. These are super clean to work with, so if you haven't learned about them yet then don't worry -- things will cancel out.</p>
<p>Rather than working with the box itself, though, I'd recommend working with the vectors representing the box. It is really just three vectors, after all, where each is parallel to one of the <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, or <span class="math-container">$Z$</span> axes. So what does the rotation matrix do to those vectors?</p>
<p>Next, I believe a fourth necessary variable would be the angle of the sun. Let's just say that the sun is directly above the box, though. That's the easiest place to start and then you can consider what changes when you remove that assumption.</p>
<p>Okay, we know we are representing our rotation in terms of the operations <span class="math-container">$\{\text{rotate in x}, \text{rotate in y}, \text{rotate in z}\}$</span>. We now want a function that relates this rotation to the area of the shadow.</p>
<p>The size of the shadow can be represented as flattening the box onto some plane that is flat in <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> (assuming the sun is above us). If we assume the box is centered at the origin (which we probably want to), this means that we take all of the <span class="math-container">$Z$</span> values for every point in the box and set them to 0. <a href="https://en.wikipedia.org/wiki/Projection_matrix" rel="nofollow noreferrer">This operation is called a projection and can also be described by a matrix</a>. Now what is the area of the following 2D shape? I trust you to work that part out. It will come out being some sum of trigonometric functions of the yaw, pitch, and roll and should, likely, have a lot of things cancel.</p>
<p>Our second to last step is to take the derivative of this function and set it to 0 to find the maximum. I am sure there are both mins and maxes (most likely an even number of both), so you only need to solve for the cases where the derivative is 0 at a local maximum of the shape.</p>
<p>Then my last step would be to double-check the work. Does the resulting answer make sense? Is it a function of all three variables as we hoped? Does it maintain the symmetries that we'd expect of such an operation? Does our area for the shadow work for the easy 90 degree rotation cases? etc. etc.</p>
<p>Have fun :)</p>
|
3,331,865 |
<p>Find the Orthonormal basis of vector space <span class="math-container">$V$</span> of the linear polynomials of the form <span class="math-container">$ax+b$</span> such that <span class="math-container">$\:$</span>, <span class="math-container">$p:[0,1] \to \mathbb{R}$</span>. with inner product</p>
<p><span class="math-container">$$\langle p,q \rangle= \int_0^1 p(x)q(x) dx$$</span> for <span class="math-container">$q, p \in V$</span></p>
<p>The polynomial in <span class="math-container">$V$</span> are linear so the basis of <span class="math-container">$V$</span> are <span class="math-container">$\left \{1,x \right \}$</span> further i am not getting how to proceedes further </p>
<p>please help</p>
<p>Thankyou.</p>
|
projectilemotion
| 323,432 |
<p>A standard approach for these types of problems is to use the Gram-Schmidt procedure:</p>
<hr>
<p>You already found a basis of <span class="math-container">$V$</span>, so let <span class="math-container">$\mathbf{v}_1=1$</span> and <span class="math-container">$\mathbf{v}_2=x$</span>. Then an orthogonal basis <span class="math-container">$\{\mathbf{u}_1,\mathbf{u}_2\}$</span> is given by:
<span class="math-container">$$\mathbf{u}_1=\mathbf{v}_1$$</span>
<span class="math-container">$$\mathbf{u}_2=\mathbf{v}_2-\frac{\langle \mathbf{v}_2,\mathbf{u}_1\rangle}{\langle \mathbf{u}_1,\mathbf{u}_1\rangle}\mathbf{u}_1$$</span>
So in your case, all you have to do is compute two simple integrals.
An orthonormal basis <span class="math-container">$\{\mathbf{w}_1,\mathbf{w}_2\}$</span> can be found by normalizing the two vectors <span class="math-container">$\mathbf{u}_1$</span> and <span class="math-container">$\mathbf{u}_2$</span>. Namely:
<span class="math-container">$$\mathbf{w}_1=\frac{\mathbf{u}_1}{\|\mathbf{u}_1\|}=\frac{\mathbf{u}_1}{\sqrt{\langle \mathbf{u}_1,\mathbf{u}_1\rangle}}$$</span>
<span class="math-container">$$\mathbf{w}_2=\frac{\mathbf{u}_2}{\|\mathbf{u}_2\|}=\frac{\mathbf{u}_2}{\sqrt{\langle \mathbf{u}_2,\mathbf{u}_2\rangle}}$$</span>
This requires the computation of one additional (easy) integral.</p>
|
14,541 |
<p>I know this is a quite general question , but I've always heard about creating mathematical models in economics, social sciences, engineering,... And I would want to know what are the starting points and roadmap to understand what a mathematical model is and how to create them. Thanks</p>
|
Joseph Malkevitch
| 1,369 |
<p>COMAP, the Consortium for Mathematics and Its Applications has developed lots of materials at different levels that address your question:</p>
<p><a href="http://www.comap.com/" rel="nofollow">http://www.comap.com/</a></p>
|
14,541 |
<p>I know this is a quite general question , but I've always heard about creating mathematical models in economics, social sciences, engineering,... And I would want to know what are the starting points and roadmap to understand what a mathematical model is and how to create them. Thanks</p>
|
Mike Spivey
| 2,370 |
<p>The last time I taught mathematical modeling we used <em><a href="http://books.google.com/books?id=XPxrItVsXpQC&printsec=frontcover&dq=a+first+course+in+mathematical+modeling&source=bl&ots=4EcHhIII_M&sig=K1KZ1lq9OmZWj-aMKMdJiv5wg64&hl=en&ei=pVQKTeTAI4nUtQPioODDCg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CEMQ6AEwBQ#v=onepage&q&f=false" rel="nofollow noreferrer">Introduction to Mathematical Modeling</a>,</em> by Giordano, Fox, Horton, and Weir. The first part of Chapter 2 contains a nice introduction to the modeling process.</p>
<p>A <a href="http://www.google.com/search?tbs=bks%3A1&tbo=1&q=mathematical+modeling&btnG=Search+Books" rel="nofollow noreferrer">search on Google books for mathematical modeling texts</a> should yield some more good results.</p>
<p>My <a href="https://math.stackexchange.com/questions/11231/probability-in-single-lane-traffic-flow-what-are-the-odds-of-choke-points-bein/11316#11316">answer to a recent math.SE question</a> also contains some discussion of the modeling process.</p>
|
2,214,231 |
<blockquote>
<p>Find the derivative of $\tan^3[\sin(2x^2-17)]$. </p>
</blockquote>
<p>Sorry if my question is a little too specific but I am confused on this trig equation. After completing the derivative I was wondering why does the $3tan^2$ not distribute to $sec^2$? Is there a rule for this? How come the exponents and the power of $2$ don't get placed onto $sec^2$? Am I missing out on some of the properties of the chain rule? </p>
|
PiE
| 110,208 |
<p>What you have to do is to solve the following equation:</p>
<p>$$F(x) = \ln(x) - 1 = 0.$$</p>
<p>When you do, you'll find out that $x = e$.</p>
<p>This is where the function crosses the $x$-axis. Thus, the point is $(e,0)$.</p>
|
20,802 |
<p>Look at the following example:</p>
<p>Which picture has four apples?</p>
<p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p>
<hr />
<p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p>
<hr />
<p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p>
<hr />
<p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p>
<p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p>
<blockquote>
<p>"Do you have three dollars?"
"Yes."</p>
</blockquote>
<p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p>
<p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as "which picture has <strong>exactly</strong> four apples"?</p>
|
Thierry
| 12,671 |
<p>I don't think there's anything wrong with the wording; it's clear what is being asked. Your example with the three dollars is also not always the way we speak in everyday language. If you ask someone with three children if they have two children, they're unlikely to say "yes" and leave it at that.</p>
<p>Getting more silly, a bicycle isn't a unicycle despite the fact that bicycles have at least one wheel. The root words for unicycle are "one" and "wheel," but a unicycle is defined to have exactly one wheel even though no root word for "exactly" appears. Exactly one wheel is just the more natural interpretation, just as exactly four apples is the more natural interpretation.</p>
|
20,802 |
<p>Look at the following example:</p>
<p>Which picture has four apples?</p>
<p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p>
<hr />
<p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p>
<hr />
<p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p>
<hr />
<p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p>
<p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p>
<blockquote>
<p>"Do you have three dollars?"
"Yes."</p>
</blockquote>
<p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p>
<p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as "which picture has <strong>exactly</strong> four apples"?</p>
|
Jason E
| 15,949 |
<p>Nearly every test like this includes instructions to choose the "best answer" to cover exactly this scenario. This looks like it's part of a test of basic counting skills, and in that context, the best answer is B. While one could make an argument for either C or D, I can't imagine an argument for either of those being the <em>best</em> answer when B is present.</p>
|
2,530,298 |
<p>I tried putting y alone and got y=(-6x-5)/5. Which I then put into the distance formula sqrt((x-1)^2+(y+5) and substitute the number above in for y but my answer never comes out correct.. Wondering if I could get some help.</p>
|
lab bhattacharjee
| 33,337 |
<p><strong>Method</strong>$\#1:$</p>
<p>As the perpendicular distance is the shortest find the equation of the perpendicular of $$6x+5y+5=0$$ passing through $(1,-5)$</p>
<p>Find the intersection of the two lines.</p>
<p><strong>Method</strong>$\#2:$</p>
<p>$$6x+5y+5=0\iff\dfrac x5=\dfrac{y+1}{-6}$$ $=k$(say)</p>
<p>$\implies x=\cdots, y=\cdots$</p>
<p>We need to minimize $$\sqrt{(x-1)^2+(y+5)^2}$$</p>
<p>$\iff $ to minimize $$(x-1)^2+(y+5)^2$$ which will be quadratic in $k$</p>
<p>Do you know how to find the possible extreme values of a quadratic in real?</p>
|
2,530,298 |
<p>I tried putting y alone and got y=(-6x-5)/5. Which I then put into the distance formula sqrt((x-1)^2+(y+5) and substitute the number above in for y but my answer never comes out correct.. Wondering if I could get some help.</p>
|
Michael Rozenberg
| 190,319 |
<p>By C-S
$$\sqrt{(6^2+5^2)\left((x-1)^2+(y+5)^2\right)}\geq6(x-1)+5(y+5)=14.$$
Thus, $$\sqrt{(x-1)^2+(y+5)^2}\geq\frac{14}{\sqrt{61}}.$$
The equality occurs for $(x-1,y+5)||(6,5),$ which gives the needed point.</p>
<p>I got $\left(\frac{145}{61},-\frac{235}{61}\right).$</p>
|
28,389 |
<p>I've been studying the axiomatic definition of the real numbers, and there's one thing I'm not entirely sure about.</p>
<p>I think I've understood that the Archimedean axiom is added in order to discard ordered complete fields containing infinitesimals like the hyperreal numbers. Additionally, this property clearly cannot be derived solely from the axioms of ordered field and completeness, since $^*\mathbb{R}$ and $\mathbb{R}$ are two complete ordered fields, two models of the axioms, one of them Archimedean and the other non-Archimedean. Are these ideas correct?</p>
<p>Thanks.</p>
|
Pete L. Clark
| 299 |
<p>The answer to your question depends critically on what you mean by a "complete ordered field" <span class="math-container">$(F,<)$</span>. Here are two rival definitions:</p>
<ol>
<li><p>[<b>added</b>: <em>sequentially</em>] Cauchy complete: every Cauchy sequence in <span class="math-container">$F$</span> converges.</p>
</li>
<li><p>Dedekind complete: every nonempty subset <span class="math-container">$S \subset F$</span> which is bounded above has a least upper bound.</p>
</li>
</ol>
<p>(There are in fact many other axioms equivalent to 2): that every bounded monotone sequence converges, that <span class="math-container">$F$</span> is connected in the order topology, the <a href="http://alpha.math.uga.edu/%7Epete/instructors_guide_shorter.pdf" rel="nofollow noreferrer">principle of ordered induction holds</a>, and so forth.)</p>
<p>It turns out that there is a unique Dedekind complete ordered field up to (unique!) isomorphism, namely the real numbers <span class="math-container">$\mathbb{R}$</span>. Famously <span class="math-container">$\mathbb{R}$</span> is also Cauchy complete -- or, if you like, Dedekind complete ordered fields satisfy the Bolzano-Weierstass theorem, which is enough to make Cauchy sequences converge -- so that Dedekind completeness implies Cauchy completeness.</p>
<p>The converse is true <strong>with an additional hypothesis</strong>: an <em>Archimedean</em> Cauchy-complete field is Dedekind complete. I show this in <span class="math-container">$\S 12.7$</span> of <a href="http://alpha.math.uga.edu/%7Epete/FieldTheory.pdf" rel="nofollow noreferrer">these notes</a> using somewhat more sophisticated methods (namely <em>Cauchy nets</em>). For a more elementary proof, see e.g. Theorem 3.11 of <a href="http://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article=1004&context=mathsp" rel="nofollow noreferrer">this nice undergraduate thesis</a>.</p>
<p>On the other hand, just as one can take the "Cauchy" completion of any metric space (or normed field) and get a complete metric space (or complete normed field), one can take the Cauchy completion of a non-Archimedean ordered field and get an ordered field which is Cauchy complete but not Dedekind complete. The easiest example of such a field is probably the
rational function field <span class="math-container">$\mathbb{R}(t)$</span> with the unique ordering that makes <span class="math-container">$t$</span> positive and infinitely large.</p>
<p>For some reason these subtleties seem to be hard to find in standard analysis texts. I myself didn't learn about them until rather recently (so, several years after my PhD). I actually wrote up some of this material as supplemental notes for a sophomore-junior level course I am currently teaching on sequences and series...but I have not as yet been able to make myself inflict these notes on my students. I talked about ordered fields in several lectures and it seemed to be one level of abstraction beyond what they could even meaningfully grapple with (so it started to seem a bit pointless).</p>
|
326,600 |
<p>The intervals are: </p>
<p>$I_1=\{ x\in\mathbb{R}:a\leq x\leq b\}$ with $-1<a<b<1$.</p>
<p>$I_2=\{ x\in\mathbb{R}:-1< x< 1\}$.</p>
<p>I've shown $f_n$ is pointwise convergent to $0$ on $I_1$ and $I_2$. But what is the difference for showing uniform convergence on these two intervals? I have a few other questions that have very similar intervals and I'm not sure what I should be doing different. </p>
<p>Basically, what is different in showing uniform convergence on $I_1$ and $I_2$?</p>
|
Cameron Buie
| 28,900 |
<p>To show uniform convergence to a function $f$ on an interval $I,$ you must show that $$\lim_{n\to\infty}\left[\sup_{x\in I}|f_n(x)-f(x)|\right]=0.$$ In this particular case, you must show that $$\lim_{n\to\infty}\left[\sup_{x\in I}|f_n(x)|\right]=0.$$</p>
<p>Since $|f_n(x)|\leq |x^n|$ for all $x$, then to see uniform convergence on $I_1$, it suffices to show that $$\lim_{n\to\infty}\left[\sup_{x\in I_1}|x^n|\right]=0,$$ which is fairly straightforward.</p>
<p>To see that there <em>isn't</em> uniform convergence on $I_2$, consider the sequence $x_n=1-\frac1n$, and note that $|f_n(x_n)|$ fails to converge <em>at all</em>, which would not be the case if we had uniform convergence to $0$. (Why?)</p>
|
79,782 |
<p>Calculate $17^{14} \pmod{71}$</p>
<p>By Fermat's little theorem:<br>
$17^{70} \equiv 1 \pmod{71}$<br>
$17^{14} \equiv 17^{(70\cdot\frac{14}{70})}\pmod{71}$</p>
<p>And then I don't really know what to do from this point on. In another example, the terms were small enough that I could just simplify down to an answer, but in this example, I have no idea what to do with that $17^{(70\cdot\frac{14}{70})}$</p>
<p>What do I do from here?</p>
|
Arthur
| 15,500 |
<p>We have that</p>
<p>$$
17^{14}\equiv 17^{70} \cdot 17^{14} \equiv 17^{84}
$$</p>
<p>or</p>
<p>$$
17^{14}\equiv 17^{-70} \cdot 17^{14} \equiv 17^{-56}
$$</p>
<p>I don't see any of these as especially easy to calculate, as $17^{14} = 289^{7} \equiv 5^{7}$. If you don't have to use Fermat's, I'd suggest going with that.</p>
|
2,439,863 |
<p>I was working on the series </p>
<p>$\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}z^{n(n + 1)}}$ and I was to consider when $z = i$. I have that $$\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}i^{n(n + 1)}} = \sum_{n=1}^{\infty}{\frac{(-1)^{\frac{3}{2}n+\frac{1}{2}n^2}}{n}} = 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + ... \cong 0.43882457311697565541$$</p>
<p>I believe it converges. Does anyone have any suggestions to find an exact value for the infinite series?</p>
|
Community
| -1 |
<p>Your sum is
\begin{align}
s&=\sum^\infty_{n=0}\left(\frac1{4n+1}-\frac1{4n+2}-\frac1{4n+3}+\frac1{4n+4}\right)\\&=\frac14\sum^\infty_{n=0}\left(\frac1{n+1/4}-\frac1{n+1/2}-\frac1{n+3/4}+\frac1{n+1}\right)
\\&=\frac14\left(-\psi(1/4)+\psi(1/2)+\psi(3/4)-\psi(1)\right)
\end{align}
using the series representation of the digamma function $\psi=\psi_0$ (cf. <a href="https://en.wikipedia.org/wiki/Digamma_function#cite_ref-AbramowitzStegun_1-1" rel="nofollow noreferrer">here</a>). Now we have the well-known special values
\begin{align}
\psi_0(1/4)&=\frac12\left(-2\gamma-\pi-6\ln2\right)\\
\psi_0(1/2)&=-\gamma-2\ln2\\
\psi_0(3/4)&=\frac12\left(-2\gamma+\pi-6\ln2\right)\\
\psi_0(1)&=-\gamma
\end{align}
(given <a href="http://mathworld.wolfram.com/GausssDigammaTheorem.html" rel="nofollow noreferrer">here</a>),
so $$s=\frac{\pi}4-\frac{\ln 2}2.$$</p>
|
2,438,236 |
<p>Suppose a finite group $G$ has order smaller than the dimension of a vector space $V$ over any field, how to prove that the representation of $G$ on $V$ is reducible？</p>
|
Long
| 41,744 |
<p>The following works over the complex numbers -- I'm not sure if there is an analogous result for arbitrary fields:</p>
<p>Let $Irr(G)$ be the set of all irreducible representations of $G$, then
$$|G| = \sum_{\rho \in Irr(G)} \deg(\rho)^2$$</p>
<p>(See for instance Corollary 4.4.5 of Benjamin Steinberg's <a href="http://users.metu.edu.tr/sozkap/513-2013/Steinberg.pdf" rel="nofollow noreferrer">Representation Theory of Finite Groups</a>)</p>
|
3,327,094 |
<p>Give an example of a non abelian group of order <span class="math-container">$55$</span>.</p>
<p>To find non abelian group the simplest way is to find one non abelian group whose order divides the order of given group and then we take the group which is the external direct product of the non abelian group and some other abelian group.</p>
<p>For example to find a non abelian group of order <span class="math-container">$36$</span> we take the permutation group <span class="math-container">$S_3$</span> and take the group <span class="math-container">$S_3\otimes \Bbb Z_6$</span>. But using this way we can not have a group of order <span class="math-container">$55$</span> since any group of order <span class="math-container">$5$</span> or <span class="math-container">$11$</span> will be abelian.</p>
<p>So how do we proceed?</p>
|
Community
| -1 |
<p>Notince that <span class="math-container">$\operatorname{aut}\Bbb Z_{11}\cong \Bbb Z_{11}^*\cong \Bbb Z_{10}$</span>. Therefore you can consider an element <span class="math-container">$g\in\operatorname{aut}\Bbb Z_{11}$</span> of order <span class="math-container">$5$</span> and the homomorphism <span class="math-container">$\phi:\Bbb Z_5\to \operatorname{aut}\Bbb Z_{11}$</span>, <span class="math-container">$\phi(m)=g^m$</span>. Then you have a non-abelian semidirect product <span class="math-container">$\Bbb Z_{11}\rtimes_\phi \Bbb Z_5$</span> with operation <span class="math-container">$(a,b)*(c,d)=(a+\phi(b)(c),b+d)$</span>.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.