qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,748,719 | <p>So there are a few basic formulas I'd like to start with, $W=\int_0^bFdx$, $F=ma$, and $a=\frac{d^2}{dt^2}x$.</p>
<p>In words, Work $(W)$ is defined as the area under a Force versus Displacement $(F/x)$ graph, Force is defined mass times acceleration $(m\cdot a)$, and acceleration is defined as the second derivative of displacement with respect to time $(t)$.</p>
<p>While this may sound very simple, I am just confused on all of this.</p>
<p>Putting all of this together, I got</p>
<p>$$W=\int_0^bFdx=\int_0^bmadx$$</p>
<p>Assuming mass $(m)$ doesn't change,</p>
<p>$$W=m\int_0^badx$$</p>
<blockquote>
<p>$$W=m\int_0^b\left(\frac{d^2}{dt^2}x\right)dx$$</p>
</blockquote>
<p>At this point, I am slightly confused. How can I take the integral with respect to $x$ when $x$ is part of the definition of acceleration? An example situation and how to tackle it would be nice (assume acceleration is not constant).</p>
<p>I was wondering if it were possible to simplify the integral, given acceleration as a function of time.</p>
| Taiyo Terada | 152,321 | <p>The reason what you have there is tricky, is it usually results in a differential equation in $x$. so you can't in general take that integral. you might be able to find some identities, but that is about it. </p>
<p>Here is an example:
If you want acceleration not constant, a classic example is a spring where you assume it follows <a href="https://en.wikipedia.org/wiki/Hooke%27s_law" rel="nofollow">Hook's law</a>. So that's $F=-kx$ for some constant $k$. Just let k=1 for Simplicity. So you get $ma=-x$, so $\frac{d^2}{dt^2}x=-x/m$, this is a second order differential equation. IN this case it is pretty easy to find a solution, ( try looking it up, or finding one, if you don't already know), once you find it, it is not hard to take the integral.
But the point here is, you can't do much with that integral you wrote in general, because acceleration could be almost any function of t. </p>
|
45,911 | <p>I've been wondering for some time now about the difference between a point and a vector. In high school, it was very important to distinguish them from each other, and we used the notation $(x,y,z)$ for points and $[x,y,z]$ for vectors. We always had to translate the point $P=(a,b,c)$ to the vector $\overrightarrow{OP} =[a,b,c]$ before we started calculating with them. </p>
<p>Now, after I started at the university, people don't seem to care anymore. My professors either say that they're the same, or that they're almost the same, and the books I have seem to share that view. The book I use for my calculus course (Colley's Vector Calculus) says, among other things, the following:</p>
<blockquote>
<p>[...] we adopt the point of view that a vector field assigns to each <em>point</em> $\textbf{x}$ in X a <em>vector</em> $\textbf{F}(\textbf{x})$ in $\mathbb{R}^n$, represented by an arrow whose tail is at the point $\textbf{x}$.</p>
</blockquote>
<p>So it seems like a point is also a vector.</p>
<p>My question is this: Do mathematicians distinguish between points and vectors, and if they do, in what circumstances?</p>
| Benoît Kloeckner | 38,289 | <p>The simplest way to understand the importance of distinguishing points from vector is to consider subspaces. For example, imagine a plane $P$ in $\mathbb{R}^3$ that does not contain the origin. Then, if you add coordinate-wise two points of $P$, the result is not in $P$. The operation point minus point gives as a result a vector in $\vec P$, the direction of $P$, namely the plane parallel to $P$ and that contains the origin. Now the addition of vectors in $\vec P$ stays into $\vec P$, and the addition of a vector (in $\vec P$) and a point (in $P$) gives a point (in $P$).</p>
<p>In a differential geometric setting, vector are all based at a given point. So if you take this point of view, a vector in $\mathbb{R}^n$ should bedefined by $2n$ coordinates. This amount to forget that you can compare (i.e. define equality of) vector based at different points.</p>
|
249,908 | <p>This seems like a very inefficient way of doing what I want. I generate all permutations of (for example) <code>{1, 1, 2, 2}</code>, then for each permutation I generate its rotations, select the first one in <code>Sort</code> order, and use that for <code>GatherBy</code>.</p>
<pre><code>list = {1, 1, 2, 2};
display@GatherBy[Permutations@list, Sort[Table[RotateLeft[#, i], {i, 0, Length@# - 1}]][[1]]& ]
(* {{"1122", "1221", "2112", "2211"}, {"1212", "2121"}} *)
</code></pre>
<p>How can I do this more efficiently?</p>
| Coolwater | 9,754 | <pre><code>With[{p = Permutations[list]}, Extract[p, #] & /@
ConnectedComponents[MapIndexed[UndirectedEdge,
Lookup[PositionIndex[p], RotateLeft[p, {0, 1}]]]]]
{{{1, 1, 2, 2}, {1, 2, 2, 1}, {2, 1, 1, 2}, {2, 2, 1, 1}}, {{1, 2, 1, 2}, {2, 1, 2, 1}}}
</code></pre>
|
1,865,062 | <p>I have seen some examples of inner automorphisms of Lie algebras. Can anyone please give me an example of an automorphism of Lie algebras that is not inner (with proof).
Note - An automorphism is said to be inner if it is of the form $exp(adx)$ for $adx$ nilpotent where $adx(y)$=$[x,y]$.
Thanks for any help.</p>
| Dietrich Burde | 83,966 | <p>The Lie algebra ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$ has an automorphism given by $A\mapsto -A^t$. It is not inner for $n>2$, but inner for $n=2$, in which case it is given by
$$
A\mapsto -A^t=X^{-1}AX, \quad \text{with} \quad X=\begin{pmatrix} 0 & 1 \cr -1 & 0 \end{pmatrix}.
$$
This answers the question in the comment.</p>
|
1,990,804 | <p>I know that we can define the exponential by a function $f: \mathbb{N}^2 \rightarrow \mathbb{N}$ by letting:</p>
<p>$f(m,o) = 1$ and $f(m,n+1) = f_x(m^n,m)$ where $f_x$ is the multiplication function, which we know is recursive.</p>
<p>I would then let $g = s(z)$ and hence $g(n) = 1$ for each $n \in \mathbb{N}$ , and let $h = f_x$ </p>
<p>However I am unsure where to go from here in order to complete a recursive definition for the exponential function. Could somebody please guide me into the approach to finish this proof? </p>
| Bram28 | 256,001 | <p>The $*$ works like a NAND for a binary logic, with $a$ the 'True' element, and $b$ the 'False' element.</p>
|
3,358,592 | <p>In the following quote, what does the notation <span class="math-container">$\{a_n\}$</span> mean?</p>
<blockquote>
<p>Дана последовательность Фибоначчи <span class="math-container">$\{a_n\}$</span>.</p>
</blockquote>
<p><strong>Translation:</strong> "You are given the Fibonacci sequence <span class="math-container">$\{a_n\}$</span>."</p>
| Coffee | 496,684 | <p>I would comment if I only were privileged.</p>
<p>Have in mind that the function you want to construct is only piecewise analytic. Maybe you can obtain this function as the limit of some sequence of functions, which is presumably the exercise here, given the question tags.</p>
<p>I suspect that there won't be any closed series representing the given function.</p>
|
306,848 | <p>The question is in the title. Suppose that $X$ and $Y$ are two pointed connected CW-complexes. I was wondering if there exists a spectral sequence computing the homology of the function space $$H_{\ast}(map_{\ast}(X,Y);k) $$
where $k$ is a fixed field. Could we say something interesting in the case when $H_{\ast}(X;k)$ is trivial. </p>
<p>For simplicity we can assume that $X$ and $Y$ are simply connected. </p>
| John Klein | 8,032 | <p>Let $X$ be a simplicial set and $Y$ a space (or simplicial set) . Then $F_\ast(X,Y)$ is a cosimplicial space and we can consider its homology spectral sequence. Bousfield gave conditions for when this will converge to the homology
of $F_\ast(X,Y)$ with field coefficients. See here:</p>
<p>On the Homology Spectral Sequence of a Cosimplicial Space
A. K. Bousfield
American Journal of Mathematics
Vol. 109, No. 2 (Apr., 1987), pp. 361-394</p>
<p>[This is not my area of expertise, but if I remember correctly, convergence is guaranteed if, say, $X$ is $s$-dimensional and finite, and $Y$ is $(s-1)$-connected.]</p>
<p><strong>Added:</strong></p>
<p>See page 371 of Bousfield's paper, where he attributes the spectral sequence and its convergence to Anderson. Bousfield also claims the $E_2$-term may be identified when $k=\Bbb Z/p$ in terms of the graded vector space $H_\ast(X;k)$ and the algebra $H^\ast(Y;k)$ provided that the latter is "Hopf-like:" i.e., a tensor product of exterior and polynomial algebras.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Stella Biderman | 45,118 | <p>Related to the answer about NP-complete problems, there are a number of theorems that state "either x is true, or P=NP." The most interesting of these in my opinion are hardness of approximation results. For example: "Given two graphs on $n$ vertices, one with max clique size $n^\alpha$ and one with max clique size $n^{1-\alpha}$, there is no polynomial time algorithm that determines which is which, or P=NP."</p>
<p>Most results like this are proven via the <a href="https://en.wikipedia.org/wiki/PCP_theorem" rel="nofollow">PCP Theorem</a>, by showing that if you can approximate a result to a certain extent, you can then convert that into a proof of the statement.</p>
|
3,522,752 | <p>Solve the following equation:
<span class="math-container">$$y=x+a\tan^{-1}p$$</span>
<span class="math-container">$$\text{where p}=\frac{dy}{dx}$$</span>
Differentiating both side w.r.t. x,
<span class="math-container">$$\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}$$</span>
I have tried till this...but what to do next?..please help..</p>
| Ankita Pal | 739,790 | <p><span class="math-container">$$y=x+a\tan^{-1}p\\
\begin{align}
&\implies\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
&\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
&\implies (p-1)(1+p^2)=a\frac{dp}{dx}\\
&\implies\int{\frac{adp}{(p^2+1)(p-1)}}=\int{dx}\\
&\implies\frac a2\int{\frac{1}{p-1}-\frac{p+1}{p^2+1}dp}=\int{dx}\\
&\implies\frac a2\left[\int{\frac{dp}{p-1}}-\int{\frac{pdp}{p^2+1}}-\int{\frac{dp}{p^2+1}}\right]=x+c\\
&\implies\frac a2\left[\log {(p-1)}-\frac 12\log{(p^2+1)}-\tan^{-1}p\right]=x+c\\
\end{align}$$</span></p>
|
125,451 | <p>If $W$ is a subspace of a finite dimensional vector space $V$ and $\{g_{1},g_{2},\cdots, g_{r}\}$ is a basis of the annihilator $W^{\circ}=\{f \in V^{\ast}| f(a)=0, \forall a \in W\}$, then $W=\cap_{i=1}^{r} N_{g_{i}}$, where for $f \in V^{\ast}$, $N_{f}=\{a \in V| f(a)=0\}$ </p>
<p>How shall I prove this?</p>
| mdp | 25,159 | <p>A hint:</p>
<p>To show $W\subset\bigcap^r_{i=1}N_{g_i}$, you just need to unpack definitions.</p>
<p>For the other inclusion, you should use the definition of the linear structure on $V^*$ to show that if $g_i(a)=0$ for all $i$, then in fact $f(a)=0$ for all $f\in W^\circ$. Then you either need to use (if you know it) or prove (if you don't) that for any $v\in V$, if $f(v)=0$ for all $f\in W^\circ$, then $v\in W$.</p>
|
125,451 | <p>If $W$ is a subspace of a finite dimensional vector space $V$ and $\{g_{1},g_{2},\cdots, g_{r}\}$ is a basis of the annihilator $W^{\circ}=\{f \in V^{\ast}| f(a)=0, \forall a \in W\}$, then $W=\cap_{i=1}^{r} N_{g_{i}}$, where for $f \in V^{\ast}$, $N_{f}=\{a \in V| f(a)=0\}$ </p>
<p>How shall I prove this?</p>
| Elchanan Solomon | 647 | <p>We wish to prove that </p>
<p>$$W = \bigcap_{i=1}^{r} N_{g_{i}}$$</p>
<hr>
<p>Step $1$: Proving $W \subset \bigcap_{i=1}^{r} N_{g_{i}}$</p>
<p>Let $w \in W$. We know that the annihilator $W^{o}$ is the set of linear functionals that vanish on $W$. If $g_{i}$ is in the basis for $W^{o}$, it is certainly <strong>in</strong> $W^{o}$. Thus, the $g_{i}$ all vanish on $W$, so that $W \subset N_{g_i}$. We thus see that $w \in N_{g_{i}}$ for all $1 \leq i \leq r$. Hence </p>
<p>$$w \in \bigcap_{i=1}^{r} N_{g_{i}}$$</p>
<p>But $w$ was arbitrary, so </p>
<p>$$W \subset \bigcap_{i=1}^{r} N_{g_{i}}$$</p>
<hr>
<p>Step $2$: Proving $W \supset \bigcap_{i=1}^{r} N_{g_{i}}$</p>
<p>Let $\{\alpha_{1}, \cdots, \alpha_{s}\}$ be a basis for $W$, and extend it to a basis for $V$, $\{ \alpha_{1} ,\cdots, \alpha_{n}\}$. Likewise, extend $\{g_{1}, \cdots, g_{r}\}$ to a basis for $V^{*}$, $\{g_{1}, \cdots, g_{n}\}$, noting that $\mbox{dim}(V) = \mbox{dim} (V^{*})$. It is easily seen that one can choose these two bases to be dual to each other, as the dual basis to $\{\alpha_{1}, \cdots, \alpha_{s}\}$ is not contained in $W^{o}$, and the dual basis for $V \setminus W$ must be contained in $W^{o}$, just by the nature of the dual basis. </p>
<p>(To make it easier to choose the correct basis for $V \setminus W$, try looking at the double dual $V^{**}$, which is naturally isomorphic to $V$, and looking at the dual basis for $W^{o}$ in $V^{**}$)</p>
<p>We get</p>
<p>$$ g_{i}(\alpha_{j}) = \left\{ \begin{array}{cc}
1 & i= j\\
0 & i \neq j
\end{array}
\right.$$</p>
<p>Let $v \in \bigcap_{i=1}^{r} N_{g_{i}}$, so that $g_{i}(v) = 0$ for all $1 \leq i \leq r$, and write</p>
<p>$$v = c_{1}\alpha_{1} + \cdots + \cdots c_{n}\alpha_{n}$$</p>
<p>Taking $g_{i}$ of both sides, where $i$ ranges from $1$ through $r$,</p>
<p>$$g_{i}(v) = c_{1}g_{i}(\alpha_{1}) + \cdots + c_{n}g_{i}(\alpha_{n}) = 0$$</p>
<p>However, these $\{ g_{i}: 1 \leq i \leq r\}$ form a dual basis to $V \setminus W$. Hence, if $v$ had a nonzero component $c_{k}\alpha_{k}$ in $V \setminus W$, it would not vanish on $g_{k}$, $1 \leq k \leq r$. Then $g_{k}(r) \neq 0$, contrary to what we have shown. Thus $v$ must be contained in $W$. </p>
|
2,167,855 | <p>Let $f(t)$ be a differentiable function for $t$ $\in$ $[0,1]$ satisfying the above,</p>
<p>Does $f(t)$ have any fixed points?</p>
<p>I can easily prove there always exists fixed points without the second condition using $MVT$,</p>
<p>does $0$ $\leq$ $\frac{\partial f(t)}{\partial t}$ $\leq$ $\frac 12$ change anything?</p>
<p>I am very curious to know the answer of this problem, and note fixed points are when; $f(x)=x$</p>
| Reginald Dick | 421,431 | <p>Here is a hint. Let g(t)=f(t)-t.
Now use the intermediate value theorem to show that there are fixed points.
For bonus, show that the fixed point is unique!
Good luck!</p>
|
2,223,163 | <p>I don't have any idea on how to prove it, and I need it for one of my questions which is still unanswered: <a href="https://math.stackexchange.com/questions/2192947/what-is-the-largest-number-smaller-than-100-such-that-the-sum-of-its-divisors-is?noredirect=1#comment4521040_2192947">What is the largest number smaller than 100 such that the sum of its divisors is larger than twice the number itself?</a>.</p>
| User8128 | 307,205 | <p>We see $$10^n - 4 = 6 + \sum^{n-1}_{k=1} 9\cdot 10^k$$ and $6$ divides $9\cdot 10^k$ for any $k\ge 1$ since each of those is an even number divisible by $3$. </p>
|
2,374,282 | <p>I am trying to find all connected sets containing $z=i$ on which $f(z)=e^{2z}$ is one to one.
I have no idea how to approach.
Can someone give me some hints?
Thank you</p>
| Claude Leibovici | 82,404 | <p>There is a very simple solution (assuming that you can use programs) : convert dates to Julian days and subract the two numbers.</p>
<p>For example, in the book <a href="http://numerical.recipes/" rel="nofollow noreferrer">"Numerical Recipes"</a> you could find a subroutine named <strong>julday</strong>.</p>
<p>You also could find a source code <a href="http://ecosystems.mbl.edu/MEP/DataModels/julday.for" rel="nofollow noreferrer">here</a>.</p>
<p>For example, today $(07/28/2017)$ is $2457963$ and your birthday $(08/17/1996)$ is $2450313$ which makes a difference of $7650$.</p>
<p>You can use the method for any other situation.</p>
|
604,836 | <p>Prove if <span class="math-container">$a \equiv c \pmod{n}$</span> and <span class="math-container">$b \equiv d \pmod n$</span> then <span class="math-container">$ab \equiv cd \pmod{n}$</span>.</p>
<p>I tried to use <span class="math-container">$(a-c)(b-d) = ab-ad-cb+cd$</span>, but it seem doesn't work.</p>
| Brian M. Scott | 12,042 | <p>You need to subtract and add the same thing:</p>
<p>$$ab-cd=ab-ad+ad-cd=\ldots$$</p>
|
4,002,925 | <p>I'm struggling a bit to understand some aspects of compactness in infinite sets.</p>
<p>We say that <span class="math-container">$[0,1]$</span> is compact by Heine-Borel, but that means that for all open covers of <span class="math-container">$[0,1], \exists$</span> a finite subcover of <span class="math-container">$S$</span>. Does that mean that no infinite collection of open sets between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> inclusive exists? Furthermore, if that's true does that mean there's a minimum non-empty size/interval for an open set? How is that possible?</p>
<p>What about the case of <span class="math-container">$$\mathscr{F} = \{ (\frac{1}{(n+1)}, \frac{1}{n}) \mid n \in \mathbb{N} \} \cup \{ (\frac{1}{n} - \frac{1}{n^3 + 1}, \frac{1}{n} + \frac{1}{n^3 + 1}) \mid n \in \mathbb{N} \}$$</span>?</p>
<p>It seems to me that here that the only subcover that would exist for <span class="math-container">$[0,1]$</span> would be infinite.</p>
| Son Gohan | 865,323 | <p>No, it does not mean that "no infinite collection of open sets between 0 and 1 inclusive exists". It means that for every (also infinite) cover (of open sets) that contains <span class="math-container">$[0,1]$</span>, we can find a finite number of them (open sets) that contains <span class="math-container">$[0,1]$</span>.</p>
<p>Important remark: the condition for compactness is that <em><strong>every</strong></em> open cover has a finite subcover, not just that a single given open cover. A proof that [0,1] is compact with respect to the euclidean topology can be found <a href="http://www-groups.mcs.st-andrews.ac.uk/%7Ejohn/MT4522/Lectures/L21.html" rel="nofollow noreferrer">here.</a></p>
<p>I suggest that you try to understand the main idea of the proof: it contains all the ingredients to understand what you have to do when dealing with compactness.</p>
|
4,002,925 | <p>I'm struggling a bit to understand some aspects of compactness in infinite sets.</p>
<p>We say that <span class="math-container">$[0,1]$</span> is compact by Heine-Borel, but that means that for all open covers of <span class="math-container">$[0,1], \exists$</span> a finite subcover of <span class="math-container">$S$</span>. Does that mean that no infinite collection of open sets between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> inclusive exists? Furthermore, if that's true does that mean there's a minimum non-empty size/interval for an open set? How is that possible?</p>
<p>What about the case of <span class="math-container">$$\mathscr{F} = \{ (\frac{1}{(n+1)}, \frac{1}{n}) \mid n \in \mathbb{N} \} \cup \{ (\frac{1}{n} - \frac{1}{n^3 + 1}, \frac{1}{n} + \frac{1}{n^3 + 1}) \mid n \in \mathbb{N} \}$$</span>?</p>
<p>It seems to me that here that the only subcover that would exist for <span class="math-container">$[0,1]$</span> would be infinite.</p>
| Mark Bennet | 2,906 | <p>To take one of your questions.</p>
<p>Suppose we take as a cover of the closed unit interval the set containing every every open sub-interval.</p>
<p>Note that the endpoints have to be included in a cover so there will be sets of the form <span class="math-container">$[0, a)$</span> and <span class="math-container">$(b, 1]$</span> included as well as the whole set itself <span class="math-container">$[0,1]$</span>. These are open in the topology we are considering.</p>
<p>This has a finite sub-cover by Heine Borel.</p>
<p>Note that any such sub-cover can be refined to a cover with smaller intervals. eg Take <span class="math-container">$(a, b)$</span> with <span class="math-container">$a \lt b$</span> (where it may involve <span class="math-container">$[0, .$</span> or <span class="math-container">$.,1]$</span> to the two intervals <span class="math-container">$(a, \frac {2a+3b}5)$</span> and <span class="math-container">$(\frac {3a+2b}5, b)$</span>. The cover remains finite, but the length of the largest interval is reduced. We can make the cover as fine as we choose while it remains finite.</p>
<p>The ability to refine covers in this way becomes significant in the theory of measure/integration.</p>
|
723,707 | <p>I'm trying to understand what the relation is between the direct product and the quotient group. </p>
<p>If we let $H$ be a normal subgroup of a group $G$, then it is not too difficult to show that the set of all cosets of $H$ in $G$ forms a quotient group $G/H$:
\begin{equation}
G/H = \{ g H \mid g \in G \}
\end{equation}</p>
<p>On the other hand, the Cartesian product of two groups $G$ and $H$ is defined as:
\begin{equation}
G \times H = \{ (g,h) \mid g \in G \text{ and } h \in H \}
\end{equation}
where $(g,h)$ denotes the set of ordered pairs. The direct product operation on this set is defined as:
\begin{equation}
(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2) \in G \times H
\end{equation}
and it is easy to see that the direct product forms a group.</p>
<p>Is the following statement true:
\begin{equation}
K = G \times H \implies G \simeq K / (\{e_G \} \times H)
\end{equation}
If so, under what conditions is it true? And how can we see it is true (or false)?</p>
<p><strong>Edit 26/03</strong>:</p>
<p>Up to this point, I believe I have found a method (see below) of showing the isomorphism relation. I would be really grateful if someone could tell me whether this proof is correct or not.</p>
<p>Let us identity the elements of $h \in H$ with element of $K$ by setting $h \equiv (e_G,h)$. The elements of $K/H$ are as usual defined by:
\begin{equation}
K/H = \{ (g,h) H \mid g \in G \text{ and } h \in H \}
\end{equation}
Since $h_1H=H$ for some $h_1 \in H$, we have:
\begin{equation}
(g,h)H = (g',h')H \iff g=g' \text{ and } h' = h h_1 \tag{1}
\end{equation}
and so without loss of generality we can write every element of $K/H$ in the form $(g,e_H)H$. Now, let the map:
\begin{equation}
f : G \to K/H
\end{equation}
be defined by:
\begin{equation}
f(g) = (g,e_H) H \tag{2}
\end{equation}
The map is one-to-one. This can be seen by equation $(1)$, because if:
\begin{equation}
f(g) = f(g')
\end{equation}
then:
\begin{equation}
(g,e_H) H = (g',e_H) H \implies g=g'
\end{equation}
Furthermore, the map is trivially onto:
\begin{equation}
\forall (g,h) H \in K/H \; \exists \; g \in G \; , \; f(g)=(g,h) H
\end{equation}
and thus the map is bijective. Finally, the map is also a homomorphism, because:
\begin{equation}
f(gg') = (gg',e_H) H = (g,e_H)(g',e_H) H = (g,e_H)(g',e_H) HH = (g,e_H) H (g',e_H) H = f(g) f(g')
\end{equation}
and so $f$ is a isomorphism. Thus, by definition of equation $(2)$, we have shown that $G \simeq K/H$.</p>
<p>Any input is much appreciated.</p>
| mesel | 106,102 | <p>Your solution is not wrong but it has unnecassary steps. You can simply use following arguments.</p>
<p>Let $\pi:G\times H\to G$ be projection map .i.e. $\pi(g,h)=g$. It is clear that map is onto.</p>
<p><strong>Claim$1:$</strong> $\pi$ is an homomorphism;</p>
<p>$$\pi((g_1,h_1)(g_2,h_2))=\pi((g_1g_2,h_1h_2))=g_1g_2=\pi((g_1,h_1))\pi((g_2,h_2))$$
and since it is onto map it is an epimorphism.</p>
<p><strong>Claim$2$:</strong> $Ker(\pi)=e_G\times H$</p>
<p>Let $\pi(x,y)=e_g$ then you can say that $x=e_g$ and $y$ is any element in $H$ so result follows. By the way it also show that $e_G\times H$ is normal in $G\times H$.</p>
<p><strong>Result:</strong> By first isomorphim theorem;
$$(G\times H)/ker(\pi)=(G\times H)/(e_G\times H)\cong G$$</p>
<p><strong>Notes:</strong> By using the other projection you can show smiliar argument for $G\times e_H$ and $H$. I hope you are familiar with isomorphism theorems.</p>
|
815,661 | <p>Let $m$ be the product of first n primes (n > 1) , in the following expression :</p>
<p>$$m=2⋅3…p_n$$</p>
<p>I want to prove that $(m-1)$ is not a complete square.</p>
<p>I found two ways that might prove this . My problem is with the SECOND way . </p>
<p><strong>First solution (seems to be working) :</strong> </p>
<p>The first way that I used is this : </p>
<p>Proof by negation : assume that $m-1$ is a complete square , i.e. $m-1 = x^2$ , then </p>
<p>$m=x^2+1=x^2-(-1)=(x-(-1))(x+(-1))=(x+1)(x-1)$</p>
<p>So we have either : </p>
<ol>
<li><p>$(x+1)$ is even and $(x-1)$ is even </p></li>
<li><p>$(x+1)$ is even and $(x-1)$ is odd</p></li>
<li><p>$(x-1)$ is even and $(x+1)$ is odd</p></li>
</ol>
<p>First case : $(x+1)$ is even and $(x-1)$ is even , then $m$ looks like this : </p>
<p>$m=2⋅otherNumbersA⋅2⋅otherNumbersB$ </p>
<p>If we disregard $2$ then $m$ is a multiplication of $n-1$ prime numbers , then </p>
<p>$m$ is a multiplication of : $2 \cdot bigPrimeNumber$ . Contradiction . </p>
<p>The other two cases are just the same .</p>
<p><strong>Second solution (my problem) :</strong></p>
<p>What I'm interested in is the following solution (that I'm stuck in) :</p>
<p>Proof by negation : assume that : $m-1 = x^2$ and $m=2⋅3…p_n$ , means that $m$ divides by 3 , so we can write : $m-1≡2(mod 3)$ , which means that : </p>
<p>$m-1≡2(mod 3) ===> (m-1)-2=3q , q\in N ===> m-3=3q=m=3(1+q)$</p>
<p>Meaning : </p>
<p>$m-1=x^2$</p>
<p>$m-1≡2(mod 3)$</p>
<p>$x^2≡2(mod 3)$</p>
<p>How do I continue from here ? how can I use : $x^2≡2(mod 3)$ to reach a contradiction ?</p>
<p>Thanks</p>
| Mr.Fry | 68,477 | <p>All squares modulo $3$ are: $\{0,1\} $.</p>
|
815,661 | <p>Let $m$ be the product of first n primes (n > 1) , in the following expression :</p>
<p>$$m=2⋅3…p_n$$</p>
<p>I want to prove that $(m-1)$ is not a complete square.</p>
<p>I found two ways that might prove this . My problem is with the SECOND way . </p>
<p><strong>First solution (seems to be working) :</strong> </p>
<p>The first way that I used is this : </p>
<p>Proof by negation : assume that $m-1$ is a complete square , i.e. $m-1 = x^2$ , then </p>
<p>$m=x^2+1=x^2-(-1)=(x-(-1))(x+(-1))=(x+1)(x-1)$</p>
<p>So we have either : </p>
<ol>
<li><p>$(x+1)$ is even and $(x-1)$ is even </p></li>
<li><p>$(x+1)$ is even and $(x-1)$ is odd</p></li>
<li><p>$(x-1)$ is even and $(x+1)$ is odd</p></li>
</ol>
<p>First case : $(x+1)$ is even and $(x-1)$ is even , then $m$ looks like this : </p>
<p>$m=2⋅otherNumbersA⋅2⋅otherNumbersB$ </p>
<p>If we disregard $2$ then $m$ is a multiplication of $n-1$ prime numbers , then </p>
<p>$m$ is a multiplication of : $2 \cdot bigPrimeNumber$ . Contradiction . </p>
<p>The other two cases are just the same .</p>
<p><strong>Second solution (my problem) :</strong></p>
<p>What I'm interested in is the following solution (that I'm stuck in) :</p>
<p>Proof by negation : assume that : $m-1 = x^2$ and $m=2⋅3…p_n$ , means that $m$ divides by 3 , so we can write : $m-1≡2(mod 3)$ , which means that : </p>
<p>$m-1≡2(mod 3) ===> (m-1)-2=3q , q\in N ===> m-3=3q=m=3(1+q)$</p>
<p>Meaning : </p>
<p>$m-1=x^2$</p>
<p>$m-1≡2(mod 3)$</p>
<p>$x^2≡2(mod 3)$</p>
<p>How do I continue from here ? how can I use : $x^2≡2(mod 3)$ to reach a contradiction ?</p>
<p>Thanks</p>
| Krishnaar | 155,879 | <p>IT IS CLEAR THAT ONE CAN WRITE ANY INTEGER IN THE FORM OF - 3K OR 3K+1 OR 3K+2 AS 0,1,2 ARE THE RESIDUES SET FOR 3. THUS WHEN WE SQUARE IT AND DIVIDE THE EXPRESSION BY 3- WE GET ONLY 0 AND 1 AS THE REMAINDER- IE;- FOR 3K IT IS OBVIOUSLY 0, FOR THE OTHER TWO IT IS 1. THUS WE CAN NEVER WRITE IT AS X^2 CONGRUENT TO 2 ( MOD 3)</p>
|
3,281,503 | <blockquote>
<p>For natural numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, what is the greatest value of <span class="math-container">$b$</span> so that <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits?</p>
</blockquote>
<p>I knew that the greatest value of <span class="math-container">$b$</span> is <span class="math-container">$21$</span>, where <span class="math-container">$9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$</span>, but I am not sure how to prove that there is no greater value of <span class="math-container">$b$</span>.</p>
| Taha Direk | 676,723 | <p>You are trying to find the maximum <span class="math-container">$b$</span> that satisfies <span class="math-container">$10^{b-1}<9^b<10^b$</span> (it is easy to show <span class="math-container">$a=9$</span> is best possible) and as you said <span class="math-container">$b=21$</span> satisfy that. Also if a "<span class="math-container">$b$</span>" doesn't satisfy the inequality, then <span class="math-container">$b+1$</span> also doesn't. (easy to show that) And <span class="math-container">$b=22$</span> doesn't satisfy the inequality. Thus <span class="math-container">$b=21$</span> is the maximum number.</p>
|
3,281,503 | <blockquote>
<p>For natural numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, what is the greatest value of <span class="math-container">$b$</span> so that <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits?</p>
</blockquote>
<p>I knew that the greatest value of <span class="math-container">$b$</span> is <span class="math-container">$21$</span>, where <span class="math-container">$9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$</span>, but I am not sure how to prove that there is no greater value of <span class="math-container">$b$</span>.</p>
| fleablood | 280,126 | <p>If <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits then </p>
<p><span class="math-container">$10^{b-1} \le a^b < 10^b$</span>.</p>
<p><span class="math-container">$\log 10^{b-1} \le \log a^b < \log 10^b$</span> (Notation: <span class="math-container">$\log$</span> means <span class="math-container">$\log_{10}$</span>)</p>
<p><span class="math-container">$b-1 \le b\log a < b$</span></p>
<p><span class="math-container">$1 - \frac 1b \le \log a <1$</span>. so <span class="math-container">$a < 10$</span></p>
<p>The largest we can allow <span class="math-container">$b$</span> to be, the closer to <span class="math-container">$1$</span> that <span class="math-container">$\log a$</span> will be. </p>
<p>Smaller values of <span class="math-container">$a$</span> will result is smaller options for <span class="math-container">$b$</span> and large options of <span class="math-container">$a$</span> will allow for larger options for <span class="math-container">$b$</span>. And if <span class="math-container">$a <a'$</span> and <span class="math-container">$b < b'$</span> we have <span class="math-container">$a^b < a'^{b'}$</span> we clearly will get larger values of <span class="math-container">$a^b$</span> if <span class="math-container">$a$</span> is as large as possible. So we can make <span class="math-container">$b$</span> largest (and therefore <span class="math-container">$a^b$</span> largest) when <span class="math-container">$a = 9$</span>.</p>
<p>so <span class="math-container">$1-\frac 1b \le \log 9< 1$</span></p>
<p>which means <span class="math-container">$b \le \frac 1{1-\log 9}$</span></p>
<p>Welp.... calculator time... <span class="math-container">$b \le 21.85$</span> so <span class="math-container">$9^{21}$</span> is indeed the largest such number.</p>
|
2,545,226 | <p>Suppose $a_n$ is a positive sequence but not necessarily monotonic. </p>
<p>For the series $\sum_{n=1}^\infty \frac{1}{a_n}$ and $\sum_{n=1}^\infty \frac{a_n}{n^2}$ I can find examples where both diverge: $a_n = n$, and where one converges and the other diverges: $a_n = n^2$.</p>
<p>Can we find example where both converges?</p>
| Eric Thoma | 35,667 | <p>By Cauchy-Schwarz, we have
$$
\sum_{n=1}^N \frac{1}{n}= \sum_{n=1}^N \frac{1}{\sqrt{a_n}} \cdot \sqrt{\frac{a_n}{n^2}} \leq \left(\sum_{n=1}^N \frac{1}{{a_n}} \right)^{1/2}\left(\sum_{n=1}^N \frac{a_n}{n^2}\right)^{1/2}
$$
and the left hand side goes to $\infty$ as $N \to \infty$. So the right must as well, meaning one of the two series diverges.</p>
|
912,002 | <p>I'm solving some programming puzzle and it has come down to this:</p>
<p>I've a fraction, say 12/13, and I need to multiply it with a smallest possible natural number (say x) to get a whole number. How do I solve for x?</p>
<p>I intuitively feel I need to use LCM to solve this but haven't been able to pin down on a method.</p>
| GuiguiDt | 171,756 | <p>If the fraction is allready reduce ($\implies gcd(num;den)=1$) the smallest $x$ is denominator.</p>
<p>Or to reduce the fraction you have to divide both num and den by $gcd(num;den)$ so the new denominator is $den/gcd(num;den)$.</p>
<p>So the smallest $x$ is $den/gcd(num;den)$.</p>
|
2,704,102 | <p>Let $X,Y,Z$ be topological spaces. Is the following statement true?
$X \times Z \cong Y \times Z \implies X \cong Y$?
how would you prove it? </p>
<p>and I know that if $A \cong B$, and $a \in A$ that there is a $b \in B$, such that $A\setminus{\{a\}} \cong B\setminus{\{b\}}.$ How would you prove the same for removing lines from product topology, instead of point of normal topological spaces?</p>
| Henno Brandsma | 4,280 | <p>Not true. Some examples, let C be the Cantor set :</p>
<p>$C \times C \simeq C \times \{0\}$ but $C \not\simeq \{0\}$.</p>
<p>$\mathbb{Q} \times \mathbb{Q} \simeq \mathbb{Q} \times \mathbb{Z}$ but $\mathbb{Q} \not\simeq \mathbb{Z}$ </p>
<p>$[0,1] \times [0,1]^{\mathbb{N}} \simeq [0,1]^2 \times [0,1]^{\mathbb{N}}$
but $[0,1] \not\simeq [0,1]^2$.</p>
<p>$[0,1) \times [0,1) \simeq [0,1) \times [0,1]$ etc.</p>
|
4,398,873 | <p>Given f: (0,1) <span class="math-container">$\rightarrow$</span> <span class="math-container">$\mathbb R$</span>. f is a continuous function and improper integrable.</p>
<p>If <span class="math-container">$\int_{0}^{x}f(t)dt = 0$</span> <span class="math-container">$\forall x \in [0,1] $</span>,</p>
<p>Does <span class="math-container">$f(x) = 0 $</span> <span class="math-container">$\forall x \in [0,1] $</span> ?</p>
| Peter Szilas | 408,605 | <p>Assume there is a <span class="math-container">$z \in (0,1)$</span> with</p>
<p><span class="math-container">$f(z) >0$</span> (or <span class="math-container">$f(z)<0$</span>).</p>
<p>Since f is continuos there is
a</p>
<p><span class="math-container">$e>0$</span> s.t. for <span class="math-container">$z-e <x<z+e$</span></p>
<p>we have <span class="math-container">$f(x)>0$</span>.</p>
<p>Then</p>
<p><span class="math-container">$I:=\int_{z-e}^{z+e}f(x)dx \not =0;$</span></p>
<p>But</p>
<p><span class="math-container">$I=$</span></p>
<p><span class="math-container">$\int_{0}^{z+e}f(x)dx - \int_{0}^{z-e}f(x)dx=0$</span>,</p>
<p>a contradiction.</p>
|
2,354,609 | <p>I have to approximate $\sqrt2$ using Taylor expansion with an error $<10^{-2}$.</p>
<p>I noticed that I can do MacLaurin expansion of $\sqrt{x+1}$ then put $x=1$</p>
<p>So: $$\sqrt{x+1}=1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \dfrac{x^3}{16} + {{\frac1{2}}\choose{n+1}}x^{n+1}(1+\xi)^{-\frac1{2}-n}$$</p>
<p>I have to find the order of the polynomial at which </p>
<p>$\sqrt2 -($<strong>the value I found with the polynomial</strong>$)<10^{-2}$</p>
<p>I can check the value of the polynomial in $x=1$ order by order until I find that </p>
<p>$\sqrt2 -($<strong>the value I found with Taylor</strong>$)<10^{-2}$</p>
<p>Or is there a faster way to find the desired order?</p>
| Community | -1 | <p>The series$\sum_{k=0}^\infty\binom{1/2}{k}$ is alternating (which is the reason why it converges, by the way). For such series this estimate holds $$\left\lvert\sum_{k=0}^\infty a_k-\sum_{h=0}^n a_h\right\rvert\le \rvert a_{n+1}\rvert$$</p>
<p>See, for instance, the proof of <a href="https://en.wikipedia.org/wiki/Alternating_series_test" rel="nofollow noreferrer">Leibniz criterion</a>.</p>
|
587,878 | <p>Let $X$ be a reflexive Banach space of infinite dimension. </p>
<p>a) Prove that there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.</p>
<p>b) Let $x_n$ be a sequence such that $\forall f \in X' \quad \exists \lim\limits_{n\to\infty} f(x_n)<\infty$ .Prove that $x_n$ converges weakly.</p>
<p>c) Find a Banach space non riflexive where b is false.</p>
<p>My idea:
the weak closure of $S^1$ is the closed unit ball, thus $0\in$ the weak closure of $S^1$. But to prove that there exists a sequence in $S^1$ that converges weakly to 0 we must prove that the weak topology is metrizzable.Is it always true in riflexive spaces?</p>
<p>I would appreciate any idea on the solution.Thank you in advance.</p>
| Norbert | 19,538 | <p>a) Let $\{z_n\}_{n\in\mathbb{N}}\subset S_X$ be such a sequence that $\Vert z_n-z_m\Vert\geq 1/2$ provided $n\neq m$. This sequence exist thanks to <a href="https://math.stackexchange.com/questions/163500/an-application-of-riesz-lemma">Riesz lemma</a>. Since $X$ is reflexive its <a href="https://math.stackexchange.com/questions/199478/equivalence-of-reflexive-and-weakly-compact">unit ball is weakly compact</a>, so we can extract weakly convergent subsequence $\{z_{n(k)}\}_{k\in\mathbb{N}}$. Define $y_k=z_{n(2k+1)}-z_{n(2k)}$ for each $k\in\mathbb{N}$. Clearly $\{y_k\}_{k\in\mathbb{N}}$ weakly converges to $0$. Define $x_k=\Vert y_k\Vert^{-1} y_k$. By construction $1/2\leq\Vert y_k\Vert\leq 2$ for all $k\in\mathbb{N}$, then $\{x_k\}_{k\in\mathbb{N}}$ also weakly converges to $0$ and what is more $\Vert x_k\Vert=1$ for all $k\in\mathbb{N}$</p>
<p>b) Let $i:X\to X^{**}$ be the natural embedding into the second dual. From assumption the family of operators $\{i(x_n)\}_{n\in\mathbb{N}}\subset\mathcal{B}(X^*,\mathbb{C})$ is pointwise bounded family, so by Banach-Steinhaus theorem it is uniformly norm bounded by some constant $C>0$. By assumption we have well defined function $\varphi(f):=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}i(x_n)(f)$.
Clearly $\varphi$ is linear, but it is also bounded because $|\varphi(f)|
=\lim_{n\to\infty}|i(x_n)(f)|\leq\limsup_{n\to\infty}\Vert i(x_n)\Vert\Vert f\Vert\leq C\Vert f\Vert$. Thus $\varphi\in X^{**}$. Since $X$ is reflexive, we have $x\in X$ such that $\varphi=i(x)$. Now for all $f\in X^*$ we have $f(x)=i(x)(f)=\varphi(f)=\lim_{n\to\infty}f(x_n)$ i.e. $\{x_n\}_{n\in\mathbb{N}}$ weakly converges to $x$.</p>
<p>c) Let $X=c_0$ and $\{e_n:n\in\mathbb{N}\}$ be its natural basis. Define $x=\sum_{k=1}^n e_k$, then for all $f\in c_0^*\cong_1 \ell_1$ we have $\lim_{n\to\infty} f(x_n)=\sum_{n=1}^\infty f_n\in\mathbb{C}$, though there is no $x\in c_0$ such that $f(x)=\sum_{n=1}^\infty f_n$ for all $f\in c_0^*$. Indeed the last equality would imply $x(k)=1$ for all $k\in\mathbb{N}$ which is impossible since $x\in c_0$.</p>
|
4,065,797 | <p>Just to give a simple numerical example but in general the variables <span class="math-container">$x,y,z,u,v$</span> are not equal.</p>
<p><span class="math-container">$113= 2*4^2 + 2*4^2 +2*4^2 + 4^2 +1^2$</span></p>
<p>I am looking for a general method to solve this type of equation or a piece of software to do the same. I already looked in this site for methods that could help but could not find anything dealing with this ind of case.</p>
<p><strong>Question 2</strong> It is also useful to know if there is a test that can tell if the equation does not have a solution.</p>
| Giulio R | 807,789 | <p>We have
<span class="math-container">$$
I_n=2\int_0^{\pi/2}\sin^n(x)\log(\sin(x)) dx=
\int_0^{\pi/2}\sin^n(x)\log(1-\cos^2(x)) dx
$$</span>
<span class="math-container">$$
=-\sum_{k\geq 1}\frac 1k\int_0^{\pi/2}\sin^n(x)\cos^{2k}(x) dx.
$$</span></p>
<p>With the help of Euler's Beta function
<span class="math-container">$$
\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}=B(u,v)=2\int_0^{\pi/2}\sin^{2u-1}(x)\cos^{2v-1}(x) dx
$$</span>
we obtain
<span class="math-container">$$
I_n=-\Gamma(\frac {n+1}2)\sum_{k\geq 1}\frac 1{2k}\frac{\Gamma(k+\frac 12)}{\Gamma(\frac n2+k+1)}.
$$</span>
This can be further evaluated in closed terms by means of Harmonic Numbers of integer / half integer arguments, as in Igor Rivin's answer, but it may be already good for estimates.</p>
|
114,664 | <p>How would one evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$?</p>
<p>I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.</p>
| Ragib Zaman | 14,657 | <p>$$ \int^1_0 \frac{ \log (1+x) }{x} dx = \int^1_0 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n} dx$$ </p>
<p>$$ =\sum_{n=1}^{\infty} (-1)^{n-1} \int^1_0 \frac{x^{n-1} }{n} dx = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2}. $$</p>
<p>Denote $\displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$ Then $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} = S - 2\left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right) = S - \frac{S}{2} = \frac{\pi^2}{12}.$$</p>
<p>Thus $$\int^1_0 \frac{ \log (1+x) }{x} dx = \frac{\pi^2}{12}.$$</p>
|
139,105 | <p>Can a (finite) collection of disjoint circle arcs in $\mathbb{R}^3$ be interlocked in the sense in that they cannot be separated, i.e. each moved arbitrarily far from one another while remaining disjoint (or at least never crossing) throughout?
(Imagine the arcs are made of rigid steel; but infinitely thin.)
The arcs may have different radii; each spans strictly less than $2 \pi$ in angle, so each has a positive "gap" through which arcs may pass:
<br /> <img src="https://i.stack.imgur.com/hd2l0.jpg" alt="Arcs4"><br />
Of course, if one could prove that in any such collection, one arc can be removed to infinity, the result would follow by induction.
But an impediment to that approach is that sometimes there is no arc than can be removed while all the others remain fixed.</p>
<p>Another approach would be to reduce the <em>piercing number</em> of the configuration:
the number of intersections of an arc with the disks on whose boundary the arcs lie. If the piercing number could always be reduced in any configuration, then it would "only" remain to prove that if there are no disk-arc piercings at all, the configuration can be separated.</p>
<p>Intuitively it seems that no such collection can interlock, but I am not seeing a proof.
I'd appreciate any proof ideas—or interlocked configurations!</p>
| Cristi Stoica | 10,095 | <p>There is already a beautiful correct <a href="https://mathoverflow.net/a/139114/10095">answer</a>.</p>
<p>Mine is just an illustrated comment, for the case $\mathbb R^2$, although the question was about $\mathbb R^3$.</p>
<p>Here is how three arcs can be locked, in $\mathbb R^2$:</p>
<p><img src="https://i.stack.imgur.com/UOegM.gif" alt="enter image description here"></p>
<p>Here is why two can't be locked, in $\mathbb R^2$:</p>
<p><img src="https://i.stack.imgur.com/IX0lX.gif" alt="enter image description here"></p>
<hr>
<p><strong>Update</strong></p>
<p>Ooops! Actually, both configurations with 3 circles can be unlocked. </p>
<p><img src="https://i.stack.imgur.com/arGw5.gif" alt="enter image description here"></p>
<p>So the problem remains open for the $\mathbb R^2$ case too. I made it into <a href="https://mathoverflow.net/questions/140571/can-a-tangle-of-arcs-interlock-in-plane">a question</a>.</p>
<hr>
<p>The question <a href="https://mathoverflow.net/a/140575/10095">got answered</a>.</p>
|
1,687,868 | <p>This might be a really obvious question so I apologize in advance, but I'm having trouble seeing when matrices are commutative for general nxn matrices. For example, when proving tr(AB)=tr(BA), I can easily prove this in a 2x2 matrix but I'm getting confused for proving it in a nxn matrix.</p>
<p>I've searched online and the recurring solution that comes up is:</p>
<p><span class="math-container">$$ \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{nk}B_{kn} = \sum_{k=1}^{n} \sum_{i=1}^{n} B_{kn}A_{nk} = \sum_{i=1}^{n} (BA)_{ii} $$</span></p>
<p>How come we're able to switch the sums in line 3? I originally tried solving the question by trying to do TR(AB) and TR(BA) separately:</p>
<p><span class="math-container">$$ \mbox{tr}(AB) = \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{nk}B_{kn} $$</span></p>
<p><span class="math-container">$$ \mbox{tr}(BA) = \sum_{i=1}^{n} (BA)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} B_{nk}A_{kn} $$</span></p>
<p>I would greatly appreciate it if someone could perhaps point out where my reasoning went wrong.</p>
| Fryie | 282,149 | <p>Basically, you have <span class="math-container">$$(A_{11}B_{11}+A_{12}B_{21}+\ldots)+(A_{21}B_{12}+A_{22}B_{22}+\ldots)+\ldots$$</span> and by simply regrouping the terms you get <span class="math-container">$$(A_{11}B_{11}+A_{21}B_{12}+\ldots)+(A_{12}B_{21}+A_{22}B_{22}+\ldots)+\ldots.$$</span></p>
<p>The products in the sum are products of scalars, thus commutative, and therefore the two double sums are equal.</p>
<p>If you really want to be rigorous about this, you can probably show it by induction.</p>
|
3,114,208 | <p>Say I have a biased coin that shows heads with probability <span class="math-container">$p \in ]1/3,1/2[$</span> and I initially have capital of <span class="math-container">$100 $</span>EUR. Every time heads is shown, my capital is doubled, in the other case I pay half of my capital. Let <span class="math-container">$X_{n}$</span> denote my capital after the <span class="math-container">$n$</span>th flip. </p>
<p><span class="math-container">$1.$</span> Show that <span class="math-container">$\lim_{n \to \infty}\mathbb E[X_{n}]=\infty$</span></p>
<p><span class="math-container">$2.$</span> Show that <span class="math-container">$X_{n}\to 0$</span> a.s.</p>
<p>My idea on <span class="math-container">$1.$</span>:</p>
<p>Let <span class="math-container">$R_{n}$</span> denote whether heads <span class="math-container">$(1)$</span> or tails <span class="math-container">$(2)$</span> is flipped on the <span class="math-container">$nth$</span> attempt. It follows that </p>
<p><span class="math-container">$R_{n}$</span>~<span class="math-container">$\operatorname{Ber}(p)$</span>.</p>
<p>Note that <span class="math-container">$X_{0}=100$</span>, and <span class="math-container">$X_{n+1}=100\prod_{i=1}^{n+1}(\frac{1}{2}+\frac{3}{2}R_{i})$</span></p>
<p><span class="math-container">$\mathbb E[X_{n+1}]=100\mathbb E[\prod_{i=1}^{n+1}(\frac{1}{2}+\frac{3}{2}R_{i})]=100\prod_{i=1}^{n+1}\mathbb E[(\frac{1}{2}+\frac{3}{2}R_{i})]$</span></p>
<p>and then by law of expectation of discrete distributions:</p>
<p><span class="math-container">$100(\mathbb E[(\frac{1}{2}+\frac{3}{2}R_{1})^{n}])=100[P(R_{1}=1)2^{n}+(\frac{1}{2})^{n}P(R_{1}=0)]=100[p2^{n}+(1-p)(\frac{1}{2})^{n}]\xrightarrow{n\to \infty}\infty$</span> </p>
<p>since <span class="math-container">$p \neq 0$</span></p>
<p>Any tips on <span class="math-container">$2.$</span>?</p>
<p>Another question, it seems very counterintuitive that if <span class="math-container">$P(X_{n} \to 0)=1$</span> then theres is still a chance that <span class="math-container">$\lim_{n \to \infty}\mathbb E[X_{n}]=\infty$</span>, likewise as <span class="math-container">$\lim_{n \to \infty}\mathbb E[X_{n}]=\infty$</span> that then <span class="math-container">$P(X_{n} \to 0)=1$</span> is still a possibility, is there any intuitive explanation for this behaviour?</p>
| Mike Earnest | 177,399 | <p>Let <span class="math-container">$X_n$</span> be your wealth after <span class="math-container">$n$</span> flips, and let <span class="math-container">$Q_{n}=X_n/X_{n-1}$</span>. Then
<span class="math-container">$$
X_n=X_0\times Q_1\times Q_2\times\dots \times Q_n
$$</span>
Let <span class="math-container">$Y_n=\log X_n$</span>. Then
<span class="math-container">$$
Y_n=Y_0+\log Q_1+\log Q_2+\dots +\log Q_n
$$</span>
Note that <span class="math-container">$\log Q_i$</span> is an iid sequence of random variables. Find its mean, and then apply the strong law of large numbers, to conclude that <span class="math-container">$\sum_{i=1}^n Q_i\to-\infty$</span> almost surely. This shows <span class="math-container">$Y_n\to-\infty$</span> a.s, so that <span class="math-container">$X_n=\exp(Y_n)\to 0$</span> a.s.</p>
<hr>
<p>I agree this is counter-intuitive. What is happening here is that the distribution of <span class="math-container">$X_n$</span> is getting progressively more right skewed; it has a positive tail which has a low probability, but is far enough out to have a high expectation. <span class="math-container">$X_n$</span> for large <span class="math-container">$n$</span> is like a lottery; most of the time you lose most of your <span class="math-container">$\$100$</span>, but with a small probability you win huge. Unlike a real-life lottery, the balance of probability is such that <span class="math-container">$E[X_n]$</span> is large. </p>
|
939,747 | <p>Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.</p>
<p>The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. In addition, we can put a bound on $a_n$, namely $a_n \le 2+3+5+7+9+11+...+(2n+1)=n^2+2n+5$.</p>
| Robert Lewis | 67,071 | <p>Not in general, since the rank of $XX^T$ is must be either $0$ or $1$, whereas the rank of $A$ is unconstrained, i.e. we have <em>a priori</em> $1 \le \text{rank}(A) \le n$. To see that $0 \le \text{rank}(XX^T) \le 1$, note that with</p>
<p>$X = (x_1, x_2, \ldots, x_n)^T, \tag{1}$</p>
<p>a column vector, we have</p>
<p>$XX^T = \begin{bmatrix} x_1 X^T \\ x_2X^T \\ \vdots \\ x_nX^T \end{bmatrix}; \tag{2}$</p>
<p>that is, the $i$-th row of $XX^T$ is $x_iX^T$, where $x_iX^T$ denotes ordinary scalar-vector multiplication. If $X = 0$, then $XX^T = 0$ so $\text{rank}(XX^T) = 0$; if $X \ne 0$, then $x_j \ne 0$ for some $j$, so $x_jX^T \ne 0$ and $\text{rank} (XX^T) \ge 1$. But then all the non-zero rows lie in the $1$-dimensional subspace $\text{span} (X^T)$ and the dimansion of the row-space must be at most $1$; thus $\text{rank}(XX^T) = 1$. If $n \ge 2$, $XX^T$ is singular, but obviously, <em>a priori</em>, $A$ need not be. So we can't in general find $X$ with $A = XX^T$.</p>
<p>In light of the above, one might ask if any $A$ with $\text{rank}(A) = 1$ is given by $XX^T$ for some column vector $X$. I don't believe this is the case, since $\text{rank}(A) = 1$ implies that $A$ may be written</p>
<p>$A = \begin{bmatrix} a_1V \\ a_2V \\ \vdots \\ a_nV \end{bmatrix}, \tag{3}$</p>
<p>for some row vector $V \ne 0$ and set of scalars $\{a_1, a_2, \ldots, a_n \}$; but for non-zero $V$ we may always choose $\{a_1, a_2, \ldots, a_n\}$ so that $A$ is not symmetric, as any matrix of the form $XX^T$ must be.</p>
<p>I think these results hold over any field $\Bbb F$, not only over $\Bbb R$.</p>
<p>Just to clarify: as I read the question, we are given any $n \times n$ real matrix $A$ and asked if $A = XX^T$ for some column vector $X$.</p>
<p>Hope this helps. Cheers,</p>
<p>and as always, </p>
<p><strong><em>Fiat Lux!!!</em></strong></p>
|
3,225,553 | <p>Show that <span class="math-container">$4x^2+6x+3$</span> is a unit in <span class="math-container">$\mathbb{Z}_8[x]$</span>.</p>
<p>Once you have found the inverse like <a href="https://math.stackexchange.com/questions/3172556/show-that-4x26x3-is-a-unit-in-mathbbz-8x">here</a>, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = <span class="math-container">$1$</span> imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?</p>
| Ehsaan | 78,996 | <p>If <span class="math-container">$R$</span> is a commutative ring: the units in <span class="math-container">$R[x]$</span> are the polynomials whose constant term is a unit, and whose higher order coefficients are nilpotent. You can apply this directly to your example.</p>
|
2,568,157 | <p>Consider the following:</p>
<p>$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$</p>
<p>$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$</p>
<p>$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$</p>
<p>In General is it true for further increase i.e.,</p>
<p>Is</p>
<p>$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$</p>
| achille hui | 59,379 | <p>Notice $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. For the identity at hand,</p>
<p>$$\sum_{k=1}^n k^5 + \sum_{k=1}^n k^7 \stackrel{?}{=} 2 \left(\sum_{k=1}^n k\right)^4$$
If one compute the difference of successive terms in RHS, we find</p>
<p>$$\begin{align}{\rm RHS}_n - {\rm RHS}_{n-1}
&= 2\left(\frac{n(n+1)}{2}\right)^4-2\left(\frac{n(n-1)}{2}\right)^4\\
&= \frac{n^4}{8}\left((n+1)^4 - (n-1)^4\right)
= \frac{n^4}{8}\left(8n^3 + 8n\right) = n^7 + n^5\end{align}$$
This clearly equals to ${\rm LHS}_n - {\rm LHS}_{n-1}$.
As a result,
$${\rm LHS}_n - {\rm RHS}_{n} = {\rm LHS}_{n-1} - {\rm RHS}_{n-1}$$
and the expression ${\rm LHS}_n - {\rm RHS}_{n}$ is independent of $n$. Since this difference vanishes at $n = 0$, we can conclude
${\rm LHS}_n = {\rm RHS}_n$ for all $n$ and hence establishes the identity at hand.</p>
|
73,111 | <blockquote>
<p>A trapezoid was inscribed into a semicirle of radius R. The side of
the trapezoid is slanting alpha against the base which is the diameter
of the semicirlce. Compute the area of the trapezoid.</p>
</blockquote>
<p>So the base is 2R. The bad thing is: that's all I know. How should I move on? Don't know what to do with that angle.</p>
| Ross Millikan | 1,827 | <p>Hint: both base angles are $\alpha$ to make the top parallel to the base. Draw a radius to the upper corners and you have three isosceles triangles.</p>
|
2,303,163 | <blockquote>
<p>Let $T$ be a linear operator on the finite-dimensional space $V.$ Suppose there is a linear operator $U$ on $V$ such that $TU=I.$ Prove that $T$ is invertible and $U=T^{-1}.$</p>
</blockquote>
<p>Attempt: Let $\dim V=n$ and $\{\alpha_i\}_{i=1}^n$ a basis for $V$. We claim that $\{U(\alpha_i)_{i=1}^n\}$ is a basis for $V.$ If not then there exists scalars $c_i$'s $\in F$ not all zero such that $\sum c_iU(\alpha_i)=0.$ Applying $T$ on both sides we get $\sum c_iTU(\alpha_i)=\sum c_i\alpha_i=0,$ a contradiction. Thus $\{U(\alpha_i)_{i=1}^n\}$ is a basis for $V.$</p>
<p>Now we observe that $T[U(\alpha_i)]=\alpha_i$ since $TU=I.$ We infer that $T$ and $U$ are invertible since they map basis vectors to basis vectors. It remains to show that $U=T^{-1}.$</p>
<p><strong>Can I say that $U=T^{-1}$ using $TU=I$?</strong>
I am not sure if I can use this since I don't know if inverses are unique.</p>
| Sahiba Arora | 266,110 | <p>As $TU =I$, therefore $TU$ is onto. This implies $T$ is onto.
As $V$ is finite dimensional therefore this also shows that $T$ is one-one. Thus, $T$ is invertible and inverse has to be $U$.</p>
|
117,933 | <p>I couldn't find similar question being asked here. The closest one I can find is <a href="https://mathoverflow.net/questions/11366/when-to-split-merge-papers">When to split/merge papers?</a>. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is: Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part? </p>
| Andreas Blass | 6,794 | <p>Since the two papers together prove one main theorem (if I correctly understand the first few lines of the question), it seems reasonable to submit them to the same journal. I can imagine a referee or editor being unhappy about being asked to publish part 1, which builds up to a big theorem that will appear in a different journal.</p>
<p>It's true that it can be hard to find room in a journal for very long papers, but I assume the journal doesn't mind publishing papers of 50+ pages at least occasionally (otherwise you wouldn't have sent part 1 there), and the difficulty of finding space is no worse for two of your papers than for one of yours and one of someone else's. The main effect of space shortage in such situations is that the two parts of your paper might appear in different issues of the journal. </p>
<p>I've published multi-part papers with all parts in the same journal. The splitting into parts was forced by the journal's upper bound on the length of papers, but I don't recall any complaints from the editors about my submitting several parts, each close to the upper bound. In at least one such case, the parts ended up appearing consecutively in the same issue of the journal. (The only time I've submitted "parts" to different journals was when they were really separate papers, though on the same topic. Two parts were really logic and appeared in the Notre Dame Journal of Formal Logic, and one part was, at least in my opinion, of broader interest and appeared in the AMS Transactions.)</p>
|
117,933 | <p>I couldn't find similar question being asked here. The closest one I can find is <a href="https://mathoverflow.net/questions/11366/when-to-split-merge-papers">When to split/merge papers?</a>. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is: Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part? </p>
| Kevin R. Vixie | 30,269 | <p><strong>What I would do</strong>: Put them both on arXiv.org and then submit the second one to the same journal. </p>
<p>Readers will thank you (for the arXiv versions <strong>and</strong> for only having to go to one journal to read the whole result).</p>
|
1,684,124 | <p>Here is my attempt:</p>
<p>$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$</p>
<p>Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.</p>
| Mathematics | 290,994 | <p>Your answer is almost correct, but you should put modulus sign in your log's argument as it must be positive i.e your answer should be;
$$2\ln|x+1|+\frac{2}{x+1}+C$$</p>
<p>You can try one more method too.</p>
<p>$$\int \frac{2x}{x^2+2x+1}dx$$
Try to create derivative of denominator in numerator
$$\int \frac{2x+2-2}{x^2+2x+1}dx$$
$$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{x^2+2x+1}dx$$</p>
<p>$$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{(x+1)^2}dx$$
For first integral simply take $x^2+2x+1=t$, then $(2x+2)dx=dt$, hence you will get:
$$\ln(x^2+2x+1)+\frac{2}{x+1}+C$$
$$\ln(x+1)^2+\frac{2}{x+1}+C$$
$$2\ln|x+1|+\frac{2}{x+1}+C$$</p>
|
1,684,124 | <p>Here is my attempt:</p>
<p>$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$</p>
<p>Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.</p>
| Itakura | 229,346 | <p>$$\int\frac{2x}{x^2+2x+1}\text{d}x$$</p>
<p>let $u=x^2+2x+1$ and $\frac{du}{dx}=2x+2$ and now $dx=\frac{du}{2(x+1)}$</p>
<p>$$\int\frac{2x}{u}\cdot \frac{du}{2(x+1)}$$</p>
<p>Now since $x^2+2x+1=(x+1)^2$, therefore $u^\frac{1}{2}=x+1$ then $u^\frac{1}{2}-1=x$</p>
<p>$$\int\frac{u^\frac{1}{2}-1}{u^\frac{3}{2}}du$$</p>
<p>$$\int\frac{u^\frac{1}{2}}{u^\frac{3}{2}}du-\int\frac{1}{u^\frac{3}{2}}du$$</p>
<p>$$\int\frac{1}{u}du--2u^{-\frac{1}{2}}$$</p>
<p>$$2\cdot ln|x+1|+\frac{2}{x+1}+C$$</p>
|
3,896,709 | <ul>
<li>Any idea to evaluate the sum
<span class="math-container">$$
\sum_{j=m}^{k}\frac{\binom{m}{2m - j\,\,}}{\binom{k}{j}}
\quad\mbox{with}\quad m \leq k < 2m - 1.
$$</span></li>
<li>I have found the sum for <span class="math-container">$k=2m-1$</span>. In fact, it is verified that
<span class="math-container">$$
\sum_{j = m}^{2m - 1}\ \frac{\binom{m}{2m-j\,\,}}
{\binom{2m - 1\,\,}{j}} = 2m\left(H_{2m} - H_{m}\right),
$$</span>
where <span class="math-container">$H_{j}$</span> is the <span class="math-container">$j$</span>-th harmonic number.</li>
</ul>
| vonbrand | 43,946 | <p>Here <a href="https://maxima.sourceforge.net" rel="nofollow noreferrer">maxima</a>'s Zeilberger package (take a peek at <a href="https://www.math.upenn.edu/%7Ewilf/AeqB.html" rel="nofollow noreferrer">Petkovsek, Wilf, Zeilberger "A = B"</a> for the gory details) says your sum isn't Gosper summable. That means there is no simple closed form for it.</p>
|
2,930,003 | <p>Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.</p>
<p>If <span class="math-container">$A = \{ a,b,c \}$</span>. Let <span class="math-container">$R$</span> be a relation which is reflexive, symmetric, transitive, and antisymmetric.</p>
<p><span class="math-container">$R = \{ (a,a), (b,b), (c,c) \}$</span></p>
<p>Is this correct? If I'm wrong, can you help me understand it?</p>
<p>Since if <span class="math-container">$(a, b)$</span> and <span class="math-container">$(b, c)$</span> are elements of <span class="math-container">$R$</span> by transitive there would be <span class="math-container">$(a, c)$</span>, but then there should be <span class="math-container">$(b, a)$</span>, <span class="math-container">$(c, b)$</span> and <span class="math-container">$(c, a)$</span> by symmetry, but then it would not be antisymmetric. If I'm not mistaken.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Your answer is correct and you can easily generalize it to a set with more elements </p>
<p>Apparently the only solution to your question is the diagonal relation, <span class="math-container">$$R=\{(x,x)|x\in A \}$$</span> for any set A.</p>
|
3,884,891 | <p>I am reading an article about "Longest Paths in Digraphs". In the proof there is a step that they considered as a trivial step, it seems easy but I am not being able to write or concieve an exact proof for it.</p>
<p>We have a strong digraph <span class="math-container">$D$</span>, that has minimum in degree <span class="math-container">$h$</span> and minimum out degree <span class="math-container">$k$</span>. Let <span class="math-container">$C$</span> be the longest circuit in <span class="math-container">$D$</span> of length <span class="math-container">$c$</span>; then <span class="math-container">$c\ge \max(h,k)+1$</span>.</p>
<p>Why is that true?</p>
<p>I think it is solved by contradiction and we can find a larger circuit.</p>
<p>Suppose to contradiction that <span class="math-container">$c< \max(h,k)+1$</span> and suppose without loss of generality that <span class="math-container">$\max(h,k)=k$</span>, then <span class="math-container">$c \le k$</span> which means that every vertex on the circuit has at least <span class="math-container">$1$</span> out neighbor outside <span class="math-container">$C$</span>.</p>
<p>But how can we continue to find a larger circuit?</p>
| Fareed Abi Farraj | 584,389 | <p>I will use @Mike idea in his answer, but write a much shorter version of the answer.</p>
<p>Let <span class="math-container">$P=v_p\cdots v_1$</span> be the longest path in <span class="math-container">$D$</span>, (starting from <span class="math-container">$v_p$</span> to <span class="math-container">$v_1$</span>). All the out neighbors of <span class="math-container">$v_1$</span> are on <span class="math-container">$P$</span>, since otherwise we get a longer path. <span class="math-container">$d^+(v_1)\ge k$</span>, let <span class="math-container">$c$</span> be the maximum integer such that <span class="math-container">$(v_1,v_c)\in E(D)$</span> then we have a circuit <span class="math-container">$C=v_cv_{c-1}\cdots v_1$</span> of length greater than or equal to <span class="math-container">$k+1$</span>.</p>
|
88,511 | <p>In version <a href="http://reference.wolfram.com/language/guide/SummaryOfNewFeaturesIn102.html">10.2</a> there is a new experimental function: <a href="http://reference.wolfram.com/language/ref/FindFormula.html"><code>FindFormula[]</code></a>.</p>
<p>I suspect that a <a href="https://en.wikipedia.org/wiki/Symbolic_regression">genetic programming algorithm (symbolic regression)</a> is behind this new feature, but I can't find any references.</p>
<p><strong>Question</strong></p>
<ul>
<li>What is behind this new function?</li>
</ul>
| Giorgia | 30,942 | <p>The Experimental function <code>FindFormula[]</code> at the moment is using a combination of different methods: it combines non linear regression with Markov chain Monte Carlo methods (e.g. Metropolis–Hastings algorithm). In the future (possibly in V$10.3$) there will be an option allowing the user to choose which method to use. </p>
|
1,991,950 | <p>Let $m$ and $n$ be relatively prime integers, with $n \ne 0,1$, so that $a=m/n$ is a non-integer rational fraction. Let $p$ be an odd prime.</p>
<blockquote>
<p><strong>QUESTION 1</strong>: Can Fermat's Little Theorem be applied, <em>i.e.</em>, can one say $a^p \equiv a\!\pmod{p}$, without any further consideration/explanation?</p>
</blockquote>
<p>To that end,</p>
<blockquote>
<p><strong>QUESTION 2</strong>: Are there any proofs (preferably well-known results) that do this?</p>
</blockquote>
| Robert Israel | 8,508 | <p>If $p$ divides $n$, $m/n \mod p$ makes no sense. If $p$ divides $m$, $a^p \equiv a \equiv 0 \mod p$. If $p$ divides neither, $a^{p-1} \equiv m^{p-1}/n^{p-1} \equiv 1 \mod p$.</p>
|
197,877 | <p>According to answer of Denis Serre to <a href="https://mathoverflow.net/questions/197773/a-geometric-property-of-singular-matrices">this question</a>, the manifold of singular matrices in $M_{n}(\mathbb{R})$ is defined as follows:
$$M=\{A\in M_{n}(\mathbb{R})\mid \text{rank}(A)=n-1\}$$</p>
<p>So we define a (line bundle) over this manifold:
$$\{(A,v)\in M\times\mathbb{R}^{n}\mid Av=0\}$$.</p>
<blockquote>
<p>Is it a trivial line bundle?</p>
</blockquote>
| Daniel Valenzuela | 52,936 | <p>Consider the natural map $M \to\mathbb RP^n$ assigning $x\mapsto ker x$. Then obviously your line bundle arises as the pullback of the tautological bundle. But now you can write down a homotopically non-trivial map $S^1 \to M \to \mathbb RP^n$ (try to restrict your attention to hit $\mathbb RP^1\subset \mathbb RP^n$). No it is a matter of taste to say that the Stiefel Whitney class will be non trivial or considering the non-trivial pullback to $S^1$ etc.</p>
|
90,112 | <p>When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following result which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.</p>
<p>The result (slightly simplified) is</p>
<p>If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.</p>
<p>Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.</p>
<p>I'm sure this should be easy and I'm just missing an obvious trick, but oh well.</p>
| Stephan Fackler | 21,704 | <p>In addition to S. Ivanov's proof, I give you a reference to the literature. A proof can be found in Classical Banach Spaces II by J. Lindenstrauss & L. Tzafriri in Section 1.e directly after the definition of uniform convexity on p. 60.</p>
|
3,979,686 | <ul>
<li><a href="https://www.britannica.com/science/derivative-mathematics" rel="nofollow noreferrer">Derivative</a></li>
</ul>
<p>This article says the following:</p>
<blockquote>
<p>To find the slope at the desired point, the choice of the second point needed to calculate the ratio represents a difficulty because, in general, the ratio will represent only an average slope between the points, rather than the actual slope at either point (see figure).</p>
</blockquote>
<p>I have simplified this as follows:</p>
<blockquote>
<p>To find the slope at the desired point we need a second point to calculate the ratio. The choice of the second point represents difficulty. Because, in general, the ratio will not represent the actual slope at either point. Rather, it will represent an average slope between the points.</p>
</blockquote>
<p>What is the "average slope"? What is the "actual slope"? What is the difference between these two?</p>
| Leonard Neon | 818,617 | <p><span class="math-container">$$
\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C
\label{eq1} \tag{eq1}\\
$$</span></p>
<p><span class="math-container">$$
\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C
\label{eq2} \tag{eq2}\\
$$</span></p>
<p><span class="math-container">$$
I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C
\label{eq3} \tag{eq3}\\
$$</span></p>
<p><strong>First, \eqref{eq1} and \eqref{eq3} are the same.</strong>
<span class="math-container">$$\text{arcsinh} (x/a) = \ln(x + \sqrt{x^2 + a^2})-\ln a.$$</span></p>
<p><strong>Second, \eqref{eq2} and \eqref{eq3} are different.</strong></p>
<p>\eqref{eq2} holds, when <span class="math-container">$a<0$</span> and <span class="math-container">$b^2-4ac> 0$</span>.</p>
<p>\eqref{eq2} holds, when <span class="math-container">$a>0$</span> and <span class="math-container">$b^2-4ac< 0$</span>.</p>
|
1,999,194 | <p>Determine the following system of equations has 'a unique solution', 'many solutions' or 'no solution':
$$\begin{cases}
& x + 2y + z &= 1\\
&2x + 2y - 2z &= 4\\
&-x + 2y - 3z &= 5
\end{cases}
$$</p>
<p>Answer = A unique solution</p>
<p>How is it a unique solution? Could anyone explain clearly? </p>
<p>Thanks</p>
| Casper Thalen | 385,675 | <p>The easiest way is to show that the corresponding coefficient matrix is non-singular. </p>
<p>Define A = $\begin{bmatrix}1&2&1\\2&2&-2\\-1&2&-3\end{bmatrix}$</p>
<p>It is a property that if this matrix is non-singular, that the system of linear equations corresponding to this matrix has exactly one solution for any combination of outcomes.</p>
<p>The logic behind that it would have one solution is that you have three variables, but also three different ways these variables are described. When they are linear you can combine them to eventually find one solution for every variable.</p>
<p>But if you were too add some fourth equation that is different from the other three, for example 2x + y - 2z = 1, then you couldn't find any solutions since you cannot find a combination of three variables too satisfy these four lines simultaneously.</p>
<p>On the other hand, if you ignored one of the equations, lets say the third, then you could find infinity many solutions. Since for every x you choose, you can always find some y and z such that only the top two equations hold.</p>
<p>Bottem line: If you have equal variables and equations you only have one unique solution. If you have fewer equations than variables you have infinite solutions. If you have more equation than variables you have no solutions.</p>
<p>Note: It is important that we are talking about different lines, for example:</p>
<p>$x+y+z = 3$</p>
<p>$2x+2y+2z=6 $</p>
<p>are the same constraint, and do not yield additional information. Therefore they are counted as one line.</p>
|
400,838 | <p>I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$</p>
<p>I tried and tried... friends of mine tried as well and we don't know how to get out of:</p>
<p>$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$</p>
<p>(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)</p>
<p>How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.</p>
| Alex | 38,873 | <p>Multiply both numerator and denominator by $2+\sqrt{3+x}$, simplify, cancel out. I get $-\frac{1}{4}$</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| Siong Thye Goh | 306,553 | <p>Hint:$$\lim_{x\to 1} \frac{\sin(1-\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+1)}$$</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| Rebellos | 335,894 | <p>Let's try to work around our <em>limit</em> :</p>
<p>$$\lim_{x\to 1} \frac{\sin(1-\sqrt{x})}{x-1} = \lim_{x\to 1} \frac{\sin(1-\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+1)}$$</p>
<p>since $x-1 = (\sqrt{x}-1)(\sqrt{x}+1). $</p>
<p>A known standar limit tells us that : </p>
<p>$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$</p>
<p>so :</p>
<p>$$ \lim_{x\to 1} \frac{\sin(1-\sqrt{x})}{\sqrt{x}-1} \cdot \lim_{x \to 1}\frac{1}{(\sqrt{x}+1)} = 1 \cdot \frac{1}{2} = \frac{1}{2}$$</p>
|
3,936,102 | <p>Can you have a function <span class="math-container">$f \notin L^1$</span> but its Fourier transform <span class="math-container">$\hat{f} \in L^1$</span>? Ive been playing around with examples and I cant find one, but I also cant prove one doesn't exist.</p>
| GEdgar | 442 | <p>More on the example of @md2pepe.
Line 18 on this page below from a textbook says that the Fourier transform of
<span class="math-container">$\frac{W}{\pi}\operatorname{sinc}(Wt)$</span> is <span class="math-container">$\operatorname{rect}(\frac{\omega}{2W})$</span></p>
<p><a href="https://i.stack.imgur.com/WHOul.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WHOul.png" alt="table" /></a></p>
<p>I just Googled "table of fourier transforms" and took the first image.<br />
This table by itself does not show what definition of Fourier transform is used. That is, "where is the <span class="math-container">$2\pi$</span>?" Possibilities are
<span class="math-container">$$
\widehat{f}(\omega) = \int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \int_{-\infty}^\infty e^{-j 2\pi \omega t}f(t)\;dt
$$</span>
and maybe more?</p>
|
3,657,026 | <p>I need to prove expression using mathematical induction <span class="math-container">$P(1)$</span> and <span class="math-container">$P(k+1)$</span>, that:</p>
<p><span class="math-container">$$
1^2 + 2^2 + \dots + n^2 = \frac{1}{6}n(n + 1)(2n + 1)
$$</span></p>
<p>Proving <span class="math-container">$P(1)$</span> gave no difficulties, however I was stuck with <span class="math-container">$P(k+1)$</span>, I've reached this point:</p>
<p><span class="math-container">$$
1^2 + \dots + (k+1)^2 = 1^2 + \dots + k^2 + (k+1)^2 = \\ \frac{1}{6}k(k+1)(2k+1) + (k+1)^2
$$</span></p>
<p>I've checked answer from the exercise book, the next step would be:</p>
<p><span class="math-container">$$
= \frac{1}{6}(k+1)(k(2k+1)+6(k+1))
$$</span></p>
<p>How it was converted like that? Could you provide some explanation?</p>
<p>Thank you in advance</p>
| JCH | 467,371 | <p>True for <span class="math-container">$n=1.$</span></p>
<p>Let <span class="math-container">$n \geq 1$</span> and suppose true for <span class="math-container">$ n. $</span></p>
<p>Then </p>
<p><span class="math-container">$1^2 + 2^2 + ...+ n^2 + (n+1)^2 = \frac{1}{6}n(n+1)(2n+1)+ (n+1)^2 = \frac{1}{6}(n+1)(2n^2 + n + 6n +6)= \frac{1}{6}(n+1)[(n+1)+1][2(n+1)+1]$</span></p>
<p><span class="math-container">$ \Box $</span></p>
|
571,955 | <p>I've tried solving this problem every way I know how and I just can't get it. I've looked at similar problems of this type, and I still cannot get an answer that seems right.</p>
<p>Parametric Equations:</p>
<p><strong>a) Write the distance between the line and the point as a function of s</strong></p>
<p><strong>b) Find the value of s such that the above distance is minimum</strong></p>
<blockquote>
<p><span class="math-container">$$x = -t+3$$</span></p>
<p><span class="math-container">$$y = \frac{t}{2} +1$$</span></p>
<p><span class="math-container">$$z = 2t - 1$$</span></p>
</blockquote>
<p>Point:</p>
<blockquote>
<p>(4,3,s)</p>
</blockquote>
<p>The most obvious thing seems to be to use the distance formula, but this gives a function of s and t. I can take partials of this equation to find s and t, but for some reason this seems wrong to me.</p>
<p><span class="math-container">$$D = \sqrt{(4-(-t + 3))^2 + (3 - (1/2t + 1))^2 + (s - (2t- 1))^2}$$</span></p>
<p><span class="math-container">$$D^2 = (1 + t)^2 + (2 - 1/2t)^2 + (s - 2t -1)^2$$</span></p>
<p><span class="math-container">$$D^2 = 19/4t^2 + 4t + s^2 - 4st - 2s + 6$$</span></p>
<p>partial w/ respect to s = 2s - 4t -2</p>
<p>partial w/ respect to t = 19/2t + 4 - 4s</p>
<p>t = 0, s = 1 ???</p>
<p>I've tried all sorts of other methods, including taking the cross product of the two direction vectors to find a normal vector perpendicular to both, as this is the minimum distance from a vector to a point. At this point my head is spinning and I just don't know the right approach. Any hints? Am I on the right track?</p>
| Cameron Buie | 28,900 | <p>In the first part, we are taking $s$ to be some <em>fixed</em> (but unspecified) value. The distance from <em>a point</em> on the parametric line to the given point is indeed $$\sqrt{(1+t)^2+\left(2-\frac12t\right)^2+(s-2t-1)^2}.\tag{$\star$}$$ To find the distance from <em>the parametric line, itself</em> to the given point, we must minimize this. It suffices to minimize its square. To do this, we will take the partial derivative of its square with respect to $t,$ set it equal to $0,$ solve for $t$, and plug it back into $(\star),$ to give us a function in terms of $s.$</p>
<p>Then, to find $s$ such that the distance is at a minimum, we will square the resulting function and minimize, by taking the derivative with respect to $s,$ setting equal to $0,$ and solving for $s$.</p>
|
129,287 | <p>Suppose $p(x_1, x_2, \cdots, x_n)$ is a symmetric polynomial. Given any univariate polynomial $u$, we can define a new polynomial $q(x_1, x_2, \cdots, x_{n+1})$ as</p>
<p>$q(x_1, x_2, \cdots, x_{n+1}) = u(x_1)p(x_2, x_3, \cdots, x_{n+1}) + u(x_2)p(x_1, x_3, \cdots, x_{n+1}) + \cdots \\ \phantom{q(x_1, x_2, \cdots, x_{n+1}) = } \qquad + u(x_{n+1})p(x_1, x_2, \cdots, x_n).$</p>
<p>It is easy to verify that $q$ is a symmetric polynomial. My question is: Is there a name already defined for such a mapping from $(p, u)$ to $q$? Thanks.</p>
| David Wehlau | 16,684 | <p>Your sum is (up to a scalar) the transfer or trace of $u(x_{n+1})p(x_1,x_2,\dots,x_n)$. Given any
polynomial $h(x_1,x_2,\dots,x_n,x_{n+1})$ its transfer or trace (with respect to the symmetric group, $S_{n+1}$ is the symmetric polynomial $\sum_{\sigma\in S_{n+1}} \sigma \cdot h(x_1,x_2,\dots,x_n,x_{n+1})$. This construction works for any finite group.</p>
<p>More precisely, you have a relative transfer $\sum_{\sigma\in S_{n+1}/S_n} \sigma \cdot h(x_1,x_2,\dots,x_n,x_{n+1})$ where $h=u(x_{n+1})p(x_1,x_2,\dots,x_n)$ is $S_n$-invariant and
the sum is over a (any) set of (left) coset representatives of $S_n$ in $S_{n+1}$. </p>
|
3,718,347 | <p>If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are real numbers and <span class="math-container">$$x^2+y^2-4x-6y-1=0$$</span>what is the minimum value of <span class="math-container">$x + y$</span>?</p>
<p>I tried to factor it but I couldn't. Then I tried to make squares so all the squares could equal to 0. But I couldn't find a way. So please kindly give me the solution for this question.</p>
| J.G. | 56,861 | <p>The circle <span class="math-container">$(x-2)^2+(y-3)^2=14$</span> minimizes <span class="math-container">$x+y$</span> when meeting its leftmost tangent of the form <span class="math-container">$x+y=c$</span>, hence of gradient <span class="math-container">$-1$</span>, with a radius of gradient <span class="math-container">$1$</span>. This radius joins the centre <span class="math-container">$(2,\,3)$</span> to <span class="math-container">$(2-\sqrt{7},\,3-\sqrt{7})$</span>, where it meets the tangent. So the minimum <span class="math-container">$x+y$</span> is <span class="math-container">$5-2\sqrt{7}$</span>.</p>
|
1,743,542 | <p>I'm looking for a continuous random variable with the following properties</p>
<ul>
<li>It is not bounded towards $+\infty$.</li>
<li>The expected value of the <em>maximum</em> of x-many draws out of that random variable has a closed-form solution.</li>
</ul>
<p>The more standard and well-known it is, the better. I have no idea how to perform this search. The first property is immediately obvious. I've tried looking at Normal and Pareto, and compute</p>
<p>$$\int x F(j)^{x-1} F'(j) j dj$$</p>
<p>where $F(j)$ is the CDF of the RV, and $x$ denotes the number of draws. In both cases, this integral became quite messy (See <a href="https://math.stackexchange.com/questions/1743535/expected-maximum-of-pareto">here</a> for a question regarding Pareto).</p>
<p>What is a recommended way of finding such a RV? Is there perhaps an obvious candidate?</p>
| FooBar | 112,643 | <p>Something else I was hinted at is to use the Frechet distribution, as it preserves maxima. </p>
<p>More precisely, if $X_i \sim Frechet(\alpha, s, m)$, then $\max \{X_1, \cdot X_n\} = Frechet(\alpha, n^\frac{1}{\alpha}s, m)$.</p>
|
4,028,534 | <p>I have three points with coordinates: <span class="math-container">$A (5,-1,0),B(2,4,10)$</span>, and <span class="math-container">$C(6,-1,4)$</span>.</p>
<p>I have the following vectors <span class="math-container">$\overrightarrow {CA} = (-1, 0, -4)$</span> and <span class="math-container">$\overrightarrow{CB} = (-4, 5, 6)$</span>.</p>
<p>To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula <span class="math-container">$A=0.5ab\sin{C}$</span> to find the area of the triangle which gave me <span class="math-container">$15.07(2dp)$</span>.</p>
<p>However in the given solutions the answer is given as <span class="math-container">$(3*\sqrt(102))/2$</span></p>
<p>I think they have used the trig identity <span class="math-container">$\cos^2(\theta) + \sin^2(\theta) = 1$</span> to find the value of <span class="math-container">$\sin(\theta)$</span> rather than <span class="math-container">$\arccos(\theta)$</span> to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using <span class="math-container">$\arccos$</span> and the other using the trig identity.</p>
| Andrei | 331,661 | <p>If you calculate the angle, you introduce numerical errors. The most elegant way to calculate the area is to use the cross product. You will not need any trigonometric functions:
<span class="math-container">$$A=\frac12|\vec{CA}\times\vec{CB}|$$</span>
So all you need will be some multiplications, additions(subtractions), and a square root. Let me know if you can follow this path.
By the way, the answer should be <span class="math-container">$\frac32\sqrt{101}$</span>, if I did the math right, which is closer to your calculations</p>
|
2,333,847 | <p>A function $f(x) = k$ and the domain is $\{-2,-1,\dotsc,3\}$. Would I say
$$x = \{-2,-1,\dotsc,3\}\quad\text{or}\quad x \in \{-2,-1,\dotsc,3\} \ ?$$
Thanks. </p>
| MPW | 113,214 | <p>Your second alternative is the correct one.</p>
<p>The value of $x$ is not a <em>set</em> of numbers, rather it is <em>in a set</em> of numbers.</p>
|
756,662 | <p>I am trying to figure out what random variables are measurable with respect to sigma algebra given by $[1-4^{-n}, 1]$ where $n= 0, 1, 2, ....$ if $[0,1]$ is the sample space. I believe I can do with with indicator functions but I'm not sure how to write this.
Thanks!</p>
| Davide Giraudo | 9,849 | <p>The collection of intervals $[1-4^{-n}]$, $n\geqslant 0$ does not form a $\sigma$-algebra, but if we define $I_n:=[1-4^{-n},1-4^{-n-1})$, then the collection of sets of the form $\bigcup_{n\in J}I_n$, $J\subset\mathbb N$, plus singleton $\{1\}$, is a $\sigma$-algebra. </p>
<p>If a $\sigma$-algebra is generated by a countable partition $(A_n)_{n\in\mathbb N}$, we can find an explicit characterization of measurable functions, namely, functions of the form $x\mapsto \sum_{n=0}^{+\infty}c_n\chi_{A_n}$, where $c_n$'s are real numbers. </p>
|
2,709,832 | <p><a href="https://en.wikipedia.org/wiki/Cantor_set#Construction_and_formula_of_the_ternary_set" rel="noreferrer">Cantor set</a>
See the link, I am referring to cantor set on the real line. I wish to show that it is compact. I am doing this by pointing following arguments. I am not sure if this is enough.</p>
<ol>
<li>Cantor set is bounded by definition in the region <span class="math-container">$[0,1]$</span></li>
<li>Cantor set is the union of closed intervals, and hence it is a closed set.</li>
<li>Since the Cantor set is both bounded and closed it is compact by Heine-Borel Theorem.</li>
</ol>
| Xander Henderson | 468,350 | <p>This question is tagged <a href="/questions/tagged/solution-verification" class="post-tag" title="show questions tagged 'solution-verification'" rel="tag">solution-verification</a>, which indicates that the asker is seeking a critique of their argument. As neither of the previous answers have addressed this, and as there is an important (and common!) error in the argument, I am answering in order to fill that gap.</p>
<p>The argument presented in the question is incorrect, though the spirit of the argument is good. The error is in the second line, where the asker writes:</p>
<blockquote>
<ol start="2">
<li>Cantor set is the union of closed intervals, and hence it is a closed set.</li>
</ol>
</blockquote>
<p>There are at least two errors in this statement:</p>
<h3>The union of closed sets needn't be closed</h3>
<p>It is not true that the union of closed sets is closed. For example, it is possible to write the open unit interval as the union of closed sets:
<span class="math-container">$$ (0,1) = \bigcup_{n=1}^{\infty} \left[ \frac{1}{n}, 1-\frac{1}{n} \right]. $$</span>
For all integers <span class="math-container">$n$</span>, the inequalities
<span class="math-container">$$ 0 < \frac{1}{n} \qquad\text{and}\qquad 1-\frac{1}{n} < 1, $$</span>
which implies that neither <span class="math-container">$0$</span> nor <span class="math-container">$1$</span> is in any of the intervals in the union (therefore neither <span class="math-container">$0$</span> nor <span class="math-container">$1$</span> is in the union itself). However, for any real number <span class="math-container">$x \in (0,1)$</span>, there exists some <span class="math-container">$n$</span> so large that
<span class="math-container">$$ \frac{1}{n} \le x \le 1 - \frac{1}{n}; $$</span>
if <span class="math-container">$x < 1/2$</span>, take <span class="math-container">$n$</span> to be larger than <span class="math-container">$1/x$</span>. Otherwise, take <span class="math-container">$n$</span> to be larger than <span class="math-container">$1/((1/2)-x)$</span>.</p>
<p>In short, an arbitrary union of closed sets needn't be closed. Indeed, the <em>correct</em> statements are:</p>
<ol>
<li>arbitrary unions and finite intersections of open sets are open</li>
<li>finite unions and arbitrary intersections of closed sets are closed</li>
</ol>
<p>The first statement will be useful, below.</p>
<h3>The Cantor set is not a union of closed intervals</h3>
<p>The Cantor set is not a union of intervals. In fact, the Cantor set doesn't contain any intervals. For contradiction, suppose that there is some interval <span class="math-container">$[a,b]$</span> contained in the Cantor set. While there are more elegant arguments, it might be simplest to observe that if <span class="math-container">$m,n\in\mathbb{Z}$</span> (with <span class="math-container">$0 < n < 3^m$</span>, then the Cantor set cannot contain both of the points
<span class="math-container">$$\frac{n}{3^m}-\frac{1}{2\cdot 3^m}, \qquad\text{and}\qquad
\frac{n}{3^m}+\frac{1}{2\cdot 3^m}. \tag{1}$$</span>
This is because the point <span class="math-container">$n/3^m$</span> is the endpoint of an interval that is removed in the <span class="math-container">$m$</span>-th step of the construction. The interval removed is either
<span class="math-container">$$ \left( \frac{n-1}{3^m}, \frac{n}{3^m} \right)
\qquad\text{or}\qquad
\left( \frac{n}{3^m}, \frac{n+1}{3^m} \right). $$</span>
As one of these intervals is removed in the construction of the Cantor set, at least one of the two points listed in (1) must be removed.</p>
<p>To show that the Cantor set contains no intervals, choose <span class="math-container">$m$</span> and <span class="math-container">$n$</span> such that both of the points given in (1) are in the interval <span class="math-container">$(a,b)$</span> (note that this shows not only that <span class="math-container">$[a,b]$</span> is not contained in the Cantor set, but also that <span class="math-container">$(a,b)$</span> is not contained in the Cantor set). This is possible, as the points are dense in <span class="math-container">$[0,1]$</span>.</p>
<p>In short, the Cantor set contains no intervals, and cannot be written as the union of intervals.</p>
<h3>Fixing the argument presented</h3>
<p>Cyriac Antony gave <a href="https://math.stackexchange.com/users/120721/cyriac-antony">a good argument</a>, using the fact that arbitrary intersections of closed sets are closed. Summarizing that argument, take <span class="math-container">$C_0 = [0,1]$</span>, and construct <span class="math-container">$C_{n+1}$</span> by removing the open middle third from each interval contained in <span class="math-container">$C_n$</span>. Then each <span class="math-container">$C_n$</span> is closed and the Cantor set is given by
<span class="math-container">$$ \bigcap_{n=0}^{\infty} C_n, $$</span>
which is a countable intersection of closed sets, therefore closed.</p>
<p>An alternative is to look at the sets which are removed: the core idea is that the Cantor set can be thought of as the complement of the sets which are removed at each stage of the construction.</p>
<p>For each <span class="math-container">$n$</span>, define
<span class="math-container">$$ U_n = \bigcup_{j=1}^{\frac{3^n-1}{2}} \left(\frac{2j-1}{3^n}, \frac{2j}{3^n} \right). $$</span>
Observe that <span class="math-container">$U_n$</span> is the set of all intervals of length <span class="math-container">$1/3^n$</span> where the left endpoint is an odd multiple of <span class="math-container">$1/3^n$</span>. All of these intervals are removed in the construction of the Cantor set (there is redundancy here, but who cares?). Then define
<span class="math-container">$$ U = \bigcup_{n=1}^{\infty} U_n, $$</span>
which is a countable union of open sets, and therefore open itself.</p>
<p>Observe that the Cantor set is precisely <span class="math-container">$[0,1] \setminus U$</span>, and the complement of an open set is closed by definition (or, as this might make the point a little more clearly, the Cantor set is given by
<span class="math-container">$$ \mathbb{R} \setminus ( (-\infty,0) \cup U \cup (1, \infty), $$</span>
which gives the Cantor set as the complement of an open set in <span class="math-container">$\mathbb{R}$</span>, so that one doesn't need to muck about with subspace topologies at all).</p>
<p>In any event, the Cantor set is the complement of an open set, therefore closed.</p>
|
208,883 | <p>Let $\bar{\rho}: G_K\to PGL_n(\mathbb{C})$ be projective representation of the absolute Galois group of a number field $K$ and $\varphi\in Aut(G_K)$.</p>
<p>A theorem of Tate tells us that we can always lift $\bar{\rho}$ to some $\rho: G_K \to GL_n(\mathbb{C})$. I am wondering if there is a lift $\rho$ whose kernel is preserved by $\varphi$, i.e. $\varphi(\ker\rho)=\ker\rho$.</p>
<p><strong>Edit. A better question would be</strong>: Do you have any idea about how to determine necessary and sufficient conditions for the existence of a lift with kernel stable under the automorphism $\varphi$?</p>
| Lasse Rempe | 3,651 | <p><strong>EDIT.</strong> I am returning to this many years after the fact. The essence of the answer remains unchanged, but I am writing it now as a more detailed proof.</p>
<p>Suppose that we are given a surface <span class="math-container">$S$</span> with a countable "<span class="math-container">$K$</span>-qc atlas", where the change-of-charts maps are <span class="math-container">$K$</span>-qc. (As already mentioned in the comments, it is important that the <span class="math-container">$K$</span> is uniform across <span class="math-container">$S$</span>. I am assuming countable topology - this is true for Riemann surfaces by Rado's theorem; it should be true for your "qc" surfaces also, but I will just make it part of the definition.)</p>
<p>More precisely, the surface <span class="math-container">$S$</span> is covered by open sets <span class="math-container">$(U_i)_{i=1}^{\infty}$</span>, and there are homeomorphisms <span class="math-container">$\phi_j\colon U_i\to D_i\subset \newcommand{\C}{\mathbb{C}}\C$</span> such that the transition maps <span class="math-container">$\phi_j\circ \phi_k^{-1}$</span> (when and where defined) are all <span class="math-container">$K$</span>-qc.</p>
<p>We may assume for simplicity that the closures of the U_i are closed topological discs, every D_i is an actual Euclidean disc, and that any <span class="math-container">$U_i$</span> intersects at most finitely many other <span class="math-container">$U_k$</span>. We won't actually use these, but they may simplify the mental picture.</p>
<p><strong>CLAIM.</strong> There exist open sets <span class="math-container">$W_j\subset\C$</span> and <span class="math-container">$K$</span>-quasiconformal homeomorphisms <span class="math-container">$\theta_i\colon D_i\to W_i$</span> such that the charts
<span class="math-container">$$ \psi_i\colon D_i\to W_j; \quad \psi_i := \theta_i\circ \phi_i $$</span>
define a Riemann surface structure. (I.e., the transition maps <span class="math-container">$\psi_i\circ \psi_k^{-1}$</span> are all holomorphic.)</p>
<p><strong>PROOF OF THE CLAIM.</strong> We construct the maps <span class="math-container">$\theta_i$</span> inductively, with the additional property that the dilatation of <span class="math-container">$\theta_i$</span> is trivial on
<span class="math-container">$$ \phi_i\left(U_i\setminus \bigcup_{k=1}^{i-1} U_k\right).$$</span>
For <span class="math-container">$i=1$</span>, this means that <span class="math-container">$\theta_1$</span> is conformal; we set <span class="math-container">$\theta_1 = \operatorname{id}$</span> and <span class="math-container">$W_1=D_1$</span>. Suppose <span class="math-container">$i>1$</span> and the maps are defined up to <span class="math-container">$i-1$</span> with the desired properties. We define a Beltrami differential <span class="math-container">$\mu_i$</span> on <span class="math-container">$D_i$</span> as follows: for points in <span class="math-container">$\phi_i(U_k)$</span> with <span class="math-container">$k<i$</span>, it is the pullback of the standard complex structure under <span class="math-container">$\phi_i\circ \psi_k^{-1}$</span>. At other points, it is the standard complex structure. By the induction hypothesis, <span class="math-container">$\psi_{k'}^{-1}\circ \psi_k$</span> is holomorphic for <span class="math-container">$k'\leq k<i$</span>, which means that the differential is well-defined.</p>
<p>Moreover, the dilatation of the differential is bounded by <span class="math-container">$K$</span>. Indeed, for any point <span class="math-container">$z\in D_i$</span>, let <span class="math-container">$k$</span> be minimal such that <span class="math-container">$z\in \phi(U_k)$</span>. If <span class="math-container">$k=i$</span>, then the dilatation is <span class="math-container">$1$</span> at <span class="math-container">$z$</span> by definition. Otherwise, it is given by the pullback of the standard dilatation under <span class="math-container">$(\phi_i\circ \phi_k^{-1})\circ \theta_k^{-1}$</span>. Now <span class="math-container">$\theta_k$</span> is holomorphic at <span class="math-container">$\phi_k(z)$</span> by our additional inductive hypothesis, and the change of charts map is <span class="math-container">$K$</span>-qc by assumption. So the dilatation is indeed bounded by <span class="math-container">$K$</span>.</p>
<p>Now we can solve the Beltrami equation and find <span class="math-container">$W_i$</span> and a quasiconformal map <span class="math-container">$\theta_i\colon D_i \to W_i$</span> such that <span class="math-container">$\mu_i$</span> is the pullback of the standard complex structure on <span class="math-container">$\theta_i$</span>. (If <span class="math-container">$U_i$</span> is connected and simply-connected, we can let <span class="math-container">$W_i$</span> be a disc.) This means that <span class="math-container">$\psi_i\circ \psi_k^{-1}$</span> is holomorphic for <span class="math-container">$k<i$</span>, as desired. This completes the inductive construction and the proof of the claim.</p>
<p>Observe that the map from our original "qc surface" to this new Riemann surface is, in charts, given by the <span class="math-container">$\theta_i$</span>. So it is <span class="math-container">$K$</span>-quasiconformal, and we have found the desired Riemann surface structure.</p>
<p>The argument works for any surface; I see no issue with extending it to surfaces with boundaries.</p>
<p>Note that, if we choose a different enumeration of our charts, then we will get a different Riemann surface structure. So this structure is not canonical, as mentioned in the original version of my answer.</p>
|
2,747,509 | <p>How would you show that if <span class="math-container">$d\mid n$</span> then <span class="math-container">$x^d-1\mid x^n-1$</span> ?</p>
<p>My attempt :</p>
<blockquote>
<p><span class="math-container">$dq=n$</span> for some <span class="math-container">$q$</span>. <span class="math-container">$$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$</span> in fact, <span class="math-container">$$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$</span></p>
<p>By multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$(x-1)$</span> we get that <span class="math-container">$1-x^d\mid 1-x^n$</span> which is the final result</p>
</blockquote>
<p>Is this an ok proof?</p>
| Mark Bennet | 2,906 | <p>You can always do:</p>
<p>$f(y)=1+y+ \dots +y^{r-1}$ so that $yf(y)=y+y^2+\dots +y^r$ and $yf(y)-f(y)=(y-1)f(y)=y^r-1$</p>
<p>Then put $y=x^d$ with $dr=n$ and obtain $(x^d-1)f(x^d)=x^n-1$ and by construction $f(x^d)$ is a polynomial.</p>
|
65,192 | <p>Multivariate parameters appear to present a jagged appearance of integrands (using default Runge-Kutta ODE integration intervals?) in ParametricPlot3D plotting on a single argument. </p>
<p>Higher Mesh.. ing to 200 improves sectors' jagging (large step secants appearing instead of tangent) somewhat, but still color from PlotStyle for lines does not come through, e.g., like in case of the straight computed helical lines shown. How to get PlotStyle line colors with smoother lines? </p>
<pre><code>thmax=6Pi ;
EQU={SI'[th]==Sin[PH[th]],SI[0]==.123,
PH'[th]==Cos[th],PH[0]==-1.234,
Z'[th]==2 Cos[PH[th]]Cos[SI[th]],
R'[th]==3 Sin[PH[th]] Cos[SI[th]],
R[0]==1.321,Z[0]==0.};
NDSolve[EQU,{SI,PH,R,Z},
{th,0,thmax}];
{ph[u_],si[u_],
r[u_],z[u_]}={PH[u],
SI[u],R[u],Z[u]}/.First[%];
ParametricPlot[{z[th],r[th]},{th,0,thmax},
AspectRatio->Automatic,PlotLabel->MERIDIAN_,PlotStyle->{Thick,Magenta}]
ParametricPlot3D[{r[th] Cos[th+t],z[th],r[th] Sin[th+t]},{t,0,2 Pi},
{th,0,thmax},PlotLabel->SURFACE_]
SpC1=ParametricPlot3D[{r[th] Cos[th],z[th],r[th] Sin[th]},{t,0,2 Pi},
{th,0,thmax},PlotStyle->{Thick,Magenta},PlotLabel->SPACE_CURVE,Mesh->20];
SpC2=ParametricPlot3D[{2 Sin[th],th,1.8 Cos[th]},{th,0,thmax/3},
PlotLabel->HELIX_,PlotStyle->{Thick,Magenta}];
Show[SpC1,SpC2]
</code></pre>
| kglr | 125 | <p>Remove <code>{t, 0, 2 Pi}</code> from your <code>SpC1</code> to get a 3D parametrized line </p>
<pre><code>SpC1b = ParametricPlot3D[{r[th] Cos[th], z[th], r[th] Sin[th]}, {th, 0, thmax},
PlotStyle -> {Thick, Magenta}, PlotLabel -> "SPACE_CURVE"];
Show[SpC1b, SpC2]
</code></pre>
<p><img src="https://i.stack.imgur.com/NS7Pt.png" alt="enter image description here"></p>
<p>Or, to get fancier,</p>
<pre><code>Show[SpC1b /. {Magenta -> Orange, Line -> (Tube[#, .1] &)},
SpC2 /. Line -> (Tube[#, .1] &), Background -> Black, Boxed -> False, Lighting -> "Neutral"]
</code></pre>
<p><img src="https://i.stack.imgur.com/vE6wD.png" alt="enter image description here"></p>
<p>and, increasing <code>thmax</code> to <code>36 Pi</code>,</p>
<p><img src="https://i.stack.imgur.com/VHh57.png" alt="enter image description here"></p>
<p><strong>Note:</strong> As is, your code for <code>SpC1</code>produces a 3D <em>surface</em> parametrized by <code>t</code> and <code>th</code>. That is, it produces polygons. So, you can change the setting for <code>PlotStyle</code> in your code to <code>PlotStyle -> EdgeForm[{Thick, Magenta}]</code> to make the polygons "look like" a <code>Magenta</code>-colored line. However, this approach is both unnecessary and slow because of the number of polygons produced: In your original version with <code>Mesh->20</code> it produces</p>
<pre><code>Total@Cases[SpC1[[1]], Polygon[x_] :> Length[x], {0, Infinity}]
(* 10 471 *)
</code></pre>
<p>of them. Changing to <code>Mesh->200</code> produces <code>90 643</code> polygons; and using the options settings <code>Mesh -> 100, PlotPoints -> 200, MaxRecursion -> 1</code> produces <code>485 208</code> tiny polygons. Since each polygon is small, together they appear as a <code>Line</code>. </p>
|
1,253,778 | <p>When we say that an inequality is sharp, does it mean that it is "the best" inequality we can get between the two quantities involved?</p>
<p>For example, I read that we would say that the inequality
$$
\frac{a^2+b^2}{2}\geq ab
$$
is sharp, but wouldn't $\frac{a^2+b^2}{2}$ on the RHS be sharper than $ab$?</p>
<p>Do we just mean that we can't <em>multiply</em> the RHS of $\cdot\geq\cdot$ by a constant $>1$ (or equivalently that we can't multiply the LHS by a constant in $[0,1)$)? So that would be <em>a</em> "best" inequality <em>in this sense</em>?</p>
| Nate 8 | 226,768 | <p>As was said, many times bounds are not as strong as they theoretically could be. In math, this comes because a specific quality is hard to compute. A basic example is computing world population-- you could say "The population of the world is greater than 3 billion." but this inequality might not be very good.</p>
<p>So by saying an inequality is sharp, it means that a theoretical maximum has been reached and there's no room for improvement for that particular inequality. However, there could of course be other inequalities with the same LHS or RHS.</p>
|
733,908 | <p>How do i start off with integrating the below function? i tried applying trig substitution and U substitution. how do i go about solving this function? should i split them up further into 2 separate functions ? need some help in this as i can't seem to figure out how to continue on with it </p>
<p>$$\int\frac{x^{3}}{({x^{2}-1})^{0.5}}dx$$</p>
| lab bhattacharjee | 33,337 | <p>$$I=\int\frac{x^3}{\sqrt{x^2-1}}dx=\int\frac{x^2}{\sqrt{x^2-1}}x\ dx$$</p>
<p>Setting $\displaystyle\sqrt{x^2-1}=y,\frac{2x}{2\sqrt{x^2-1}}dx=dy$ and $\displaystyle x^2=y^2+1$</p>
<p>$$I=\int(y^2+1)dy=\cdots $$</p>
<hr>
<p>Alternatively,</p>
<p>Using Trigonometric substitution (<a href="http://www.sosmath.com/calculus/integration/trigsub/trigsub.html" rel="nofollow">1</a>,<a href="http://en.wikipedia.org/wiki/Trigonometric_substitution" rel="nofollow">2</a>), $\displaystyle x=\sec y\implies dx=\sec y\tan y\ dy$</p>
<p>$$\int\frac{x^3}{\sqrt{x^2-1}}dx=\int\frac{\sec^3y}{\tan y}\sec y\tan y\ dy=\int\sec^4y\ dy$$</p>
<p>$$=\int(\tan^2y+1)\sec^2y\ dy$$</p>
<p>Now set $\displaystyle\tan y =u$</p>
|
657,047 | <p>So I have $a^n = b$. When I know $a$ and $b$, how can I find $n$?</p>
<p>Thanks in advance!</p>
| Community | -1 | <p>Exponent problems like finding $n$ if we know the value of both $a$ and $b$ in the equation $a^n = b$ can be solved using <a href="http://en.wikipedia.org/wiki/Logarithm" rel="nofollow">logarithms</a>:</p>
<p>$$
\begin{align}
a^n=b&\Rightarrow \log \left(a^n\right)=\log (b) \\
&\Rightarrow n\log(a)=\log(b) \\
&\Rightarrow n=\dfrac{\log(b)}{\log(a)}=\log_a (b)
\end{align}$$</p>
|
271,259 | <p>My goal is to have NumericQ[h[j]]=True for any j regardless of whether j may be symbolic with no defined value.</p>
<p>Setting NumericQ[h[j_]]=True does not work and as I understand the NumericFunction attribute only works when the input is also numeric.</p>
<p>One solution might be to unprotect NumericQ and then set NumericQ[h[j_]]=True but I have not tried this as I am afraid of using Unprotect.
Another solution might be to define my own NumericQ function, that I can call numericq for example, and set numericq[h[j_]]=True but I would have to replace all previous NumericQ with numericq in my code and test that everything works.</p>
<ul>
<li>edit *</li>
</ul>
<p>I want to do this because I want my code to ignore certain symbols. My solution was to treat these symbols as numeric but this does not work for symbols like h[j] when j is a generic symbol. Any other function that works like NumericQ could work although I would have to change my code a bit.</p>
| user293787 | 85,954 | <p>Instead of unprotecting <code>NumericQ</code>, one can define an <a href="https://reference.wolfram.com/language/ref/UpValues.html" rel="noreferrer">upvalue</a> for <code>h</code>:</p>
<pre><code>h/:HoldPattern[NumericQ[h[_]]]=True;
</code></pre>
<p>This means that the definition is stored with <code>h</code> rather than with <code>NumericQ</code>:</p>
<pre><code>UpValues[h]
(* {HoldPattern[HoldPattern[NumericQ[h[_]]]]:>True} *)
</code></pre>
<p>When an expression such as <code>NumericQ[h[4]]</code> is evaluated, then upvalue definitions for <code>h</code> are tried before downvalue definitions for <code>NumericQ</code>, hence we get <code>True</code>. See <a href="https://reference.wolfram.com/language/tutorial/EvaluationOfExpressions" rel="noreferrer">this tutorial</a>.</p>
<p><strong>Warning.</strong> Though I have used upvalues before with <code>NumericQ</code>, I do not know if it can break something.</p>
|
482,003 | <p>I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$</p>
<p>Thanks.</p>
| Mikasa | 8,581 | <p>Think about $$\int_0^1 f(x)dx,~~~f(x)=\frac{1}{\sqrt{x}}
$$</p>
<p>Indeed: $$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=1}^nf\left(a+\frac{b-a}{n}i\right)\left(\frac{b-a}{n}\right)$$</p>
|
482,003 | <p>I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$</p>
<p>Thanks.</p>
| mrf | 19,440 | <p>Your sum can be interpreted as a Riemann sum:</p>
<p>$$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}}.
$$</p>
<p>Let $f(x) = 1/\sqrt{x}$ and let $x_k = k/n$. Then
$$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n \sum_{k=1}^n \sqrt{\frac{n}{k}} = \frac1n \sum_{k=1}^n f(x_k) \to \int_0^1 f(x)\,dx
$$
as $n \to \infty$.</p>
<p>(Since the integral is improper, a little care is needed to justify the last step.)</p>
|
2,419,529 | <p>I am asked to state whether the following is true or if false to give a counterexample:</p>
<blockquote>
<p>If $A_1 \supseteq A_2 \supseteq A_3 \supseteq \ldots $ are all sets containing an infinite number of elements, then the intersection $$\bigcap_{k=1}^\infty A_k$$ is infinite as well.</p>
</blockquote>
<p>I believe this statement to be false but I am not sure if the counterexample I have thought up makes sense. I said: </p>
<p>Let $A_n = \{m \in \mathbb{Z} | m> n\}$ for $n \in \mathbb{N}$. Would this be okay?</p>
| gt6989b | 16,192 | <p>Not quite, because then you have a finite intersection, i.e. each $A_n$ has a finite amount of elements. How about trying $A_n = [n,\infty) \cap \mathbb{N}$? What is the intersection then?</p>
|
2,816,965 | <p>Hey so I was wondering how to differentiate $(e^2)^x$ without using the chain rule.</p>
<p>I tried but I always end up using the chain rule in this case.</p>
<p>Would appreciate some help! (No hints please).</p>
| Community | -1 | <p>If you know the rule</p>
<p>$$(a^x)'=\ln a\,a^x$$ then</p>
<p>$$((e^2)^x)'=\ln e^2\,(e^2)^x=2(e^2)^x.$$</p>
<hr>
<p>Alternatively,</p>
<p>$$(e^{2x})'=\lim_{h\to0}\frac{e^{2(x+h)}-e^{2x}}h=e^{2x}\lim_{h\to0}\frac{e^{2h}-1}h=2e^x\lim_{2h\to0}\frac{e^{2h}-1}{2h}=2e^{2x}.$$</p>
|
868,943 | <p>Can you please tell me the sum of the seires</p>
<p>$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $ </p>
<p>where the numerator is the series of triangular numbers?</p>
<p>Is there a simple way to find the sum?</p>
<p>Thank you.</p>
| André Nicolas | 6,312 | <p>Your expression is equal to $g(1/10)$, where
$$g(x)=\frac{x}{2}\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\cdots\right)$$</p>
<p>Take the power series $1+x+x^2+x^3+\cdots$ for $\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\cdots$ if we do it term by term, and $\frac{2!}{(1-x)^3}$ if we do it the usual way. Thus
$$g(x)=\frac{x}{2}\cdot\frac{2!}{(1-x)^3}$$
(when $|x|\lt 1$, and in particular at $x=1/10$). </p>
<p><strong>Remark:</strong> The idea generalizes. The $n$-th triangular nunber is $\binom{n}{2}$. The same idea can be used to calculate $\sum \binom{n}{k}x^n$ for $|x|\lt 1$ and fixed positive integer $k$.</p>
|
2,908,361 | <p>I tried to solve this inequality by taking the square outside the floor function $[y]$ (greatest integer less than $y$)but it was wrong since if $x=2.5$ then $[x]= 2$ and $x^2=4$ while $[x^2]=[6.25]=6$.</p>
| Iti Shree | 433,761 | <p>We can you Calvin's suggested method here (see his comment) by solving the equation as we would with the quadratic equation.
You solve this equation as following, let us take $[x]$ another variable say $n$, and $[x^2]$ as $n^2 + c$. Now your equation becomes :</p>
<p>$$n^2 + c + 5n + 6 = 2$$
$$n (n + 5) = -4 - c$$</p>
<p>Now, we know from above equation that,
n varies from , $$ n = -4 -c$$
to $$ n = -9 - c$$</p>
<p>here, for our required equation to be true, the equation should be strictly greater than $2$. Hence, the least possible value for $x$ in greatest integer function should be greater than $-4$.</p>
<p>I hope this clears your doubt.</p>
|
1,774,084 | <p>I think it is convergent to $1$ because as $n$ tends to $\infty$ , $1/\sqrt(n)$ tends to $0$. Is it true?</p>
<p>Thanks!</p>
| Bernard | 202,857 | <p>$$n^{\frac1{\sqrt n}}\stackrel{\text{def}}{=}\mathrm e^{\frac{\log n}{\sqrt n}}.$$
Now a basic limit is $\;\lim_{n\to\infty}\dfrac{\log n}{n}=0$, from which we deduce, for any $\alpha>0$: $$\;\lim_{n\to\infty}\dfrac{\log(n^\alpha)}{n^\alpha}=\alpha\lim_{n\to\infty}\dfrac{\log n}{n^\alpha}=0\;$$
whence $\;\lim_{n\to\infty}\dfrac{\log n}{n^\alpha}=0$. Thus, the exponent of $\mathrm e$ tends to $0$, and $n^{\frac1{\sqrt n}}$ tends to $1$.</p>
|
1,774,084 | <p>I think it is convergent to $1$ because as $n$ tends to $\infty$ , $1/\sqrt(n)$ tends to $0$. Is it true?</p>
<p>Thanks!</p>
| David Holden | 79,543 | <p>for $x \in \mathbb{R^+} $let
$$
A_x=x^{1/\sqrt{x}}
$$
giving
$$
\log A_x = \frac2{\sqrt{x}}\log \sqrt{x}
$$
differentiating $\log A_x$ wrt $x$
$$
\frac{A'_x}{A_x} = x^{-\frac32}(1-\log \sqrt{x})
$$
so $A_x$ is a decreasing function of $x$ for $x \gt e^2$ and is bounded below by 1. This implies that a limit exists as $x \to \infty$, and
$$
\lim_{n \to \infty} \log A_n = \lim_{n^2 \to \infty} \log A_{n^2}
$$
set $N=n^2$,
$$
\log A_{N} = \frac{2\log n}{n}
$$
since $\lim_{n^2 \to \infty} \frac{\log n}n =0 $ we have $\log A_n \to 0$ and $A_n \to 1$</p>
|
106,396 | <p>An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)</p>
<p>$$\sin x \approx \frac{{16x\left( {\pi - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi - x} \right)}}$$</p>
<p>for $(0,\pi)$</p>
<p>Here's an image. Cyan for the sine and blue for the approximation.
<img src="https://i.stack.imgur.com/mgmoA.jpg" alt="enter image description here"></p>
<p>¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations? </p>
| Steven Stadnicki | 785 | <p>This is very close to a Padé approximant, and in this case the formula is simple enough that it's easy to derive. Firstly, we know that $\sin(x)$ is $0$ at $x=0, x=\pi$; this suggests recasting in terms of the variable $y=x(\pi-x)$. What we're after is a first-order rational approximation $\sin(x) = f(y) = \frac{ay+b}{cy+d}$; since we know that $f(y) = 0$ at $y=0$ (i.e., as $x$ approaches $0$ or $\pi$) then the constant term in the numerator is $0$, and after dividing out the approximation takes the form $\frac{y}{a+by}$.</p>
<p>Now, we certainly want our approximation to give $\sin(\pi/2) = f(\pi^2/4) = 1$; this means $\displaystyle{\frac{\pi^2/4}{a+b\pi^2/4}} = 1$, or $4a+b\pi^2 = \pi^2$, or $a=\frac{1-b}{4}\pi^2$. The other relation between $a$ and $b$ presumably comes from trying to match the derivative at $0$, $\left.d(\sin(x))/dx\right|_{x=0} = 1$; the condition for this this can easily be written out in terms of $df/dy$ at $y=0$. I'll spare the arithmetic (unless someone's really curious), but the result works out to be that $a=\pi$; this would give $b=(\pi-4)/\pi$ and the overall approximant $f(y) = \frac{\pi y}{\pi^2+(\pi-4)y}$, but instead the formula uses a second approximation by setting $a=5\pi^2/16$, which (thanks to the first relation) gives a rational value of $b$ (and in fact, the 'nice' value $1/4$). This approximation is equivalent to saying that $5\pi^2/16\approx\pi$, or in other words that $\pi\approx 16/5 = 3.2$; it means a slight error in the slope of the approximation at $x=0$, but that's a fair tradeoff for the ease of calculation gained.</p>
|
106,396 | <p>An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)</p>
<p>$$\sin x \approx \frac{{16x\left( {\pi - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi - x} \right)}}$$</p>
<p>for $(0,\pi)$</p>
<p>Here's an image. Cyan for the sine and blue for the approximation.
<img src="https://i.stack.imgur.com/mgmoA.jpg" alt="enter image description here"></p>
<p>¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations? </p>
| Tatar Elemér | 82,923 | <p>That's a pretty neat visualization of how can you intiutively build, proove and understand the approximation.
<a href="https://i.stack.imgur.com/ACoat.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ACoat.jpg" alt="enter image description here" /></a></p>
<p>The full <a href="https://www.youtube.com/watch?v=ltTaTT6aUls&ab_channel=MathematicsVisuallyExplained" rel="nofollow noreferrer">proof with video</a> and explanation.</p>
|
3,048,702 | <p>Let <span class="math-container">$S$</span> be a orientable closed surface with genus <span class="math-container">$g \geq 1$</span> and let <span class="math-container">$\gamma \subset S$</span> be an immersed curve. Does there exist a finite cover of <span class="math-container">$S$</span> where <span class="math-container">$\gamma$</span> lifts to a curve that is homotopic to an embedded curve? </p>
<p>If so, and <span class="math-container">$\gamma$</span> has <span class="math-container">$k$</span> double points, is there an explicit bound on the degree of the finite cover that we need to take in order to lift <span class="math-container">$\gamma$</span> to a curve homotopic to an embedded curve?</p>
| Moishe Kohan | 84,907 | <p>The links in Lee's answer give you a part of the story but not the whole story. There is a missing step for going from residual finiteness to the "lifting property". To explain the missing part I will need two definitions. </p>
<p><strong>Definition 1.</strong> A subgroup <span class="math-container">$H$</span> of a group <span class="math-container">$G$</span> is called "separable" (or "closed in profinite topology") if for every finite subset <span class="math-container">$F\subset G-H$</span> there exists a finite index subgroup <span class="math-container">$G_0< G$</span> which contains <span class="math-container">$H$</span> and is disjoint from <span class="math-container">$F$</span>. </p>
<p>Thus, residual finiteness of <span class="math-container">$G$</span> amounts to separability of the trivial subgroup of <span class="math-container">$G$</span>. In general, residual finiteness of <span class="math-container">$G$</span> does not imply separability of cyclic subgroups. </p>
<p><strong>Example.</strong> Let <span class="math-container">$G=BS(p,1)=<a,b| bab^{-1}=a^p>$</span> for some <span class="math-container">$p>1$</span>. This group is residually finite, even linear: It embeds in the subgroup of upper triangular matrices in <span class="math-container">$SL(2,{\mathbb Q})$</span>. However, the cyclic subgroup <span class="math-container">$H=<a^p>$</span> is not separable. Indeed, <span class="math-container">$a\notin H$</span>. On the other hand, for every homomorphism <span class="math-container">$\phi: G\to F$</span> to a finite group, the subgroups <span class="math-container">$<\phi(a)>$</span> and <span class="math-container">$<\phi(a^p)>$</span> are conjugate in <span class="math-container">$F$</span>, hence, have the same order. Since <span class="math-container">$<\phi(a^p)>\subset <\phi(a)>$</span> it follows that these cyclic groups are equal. Hence, <span class="math-container">$\phi(a)\in \phi(H)$</span> for every homomorphism <span class="math-container">$\phi$</span> from <span class="math-container">$G$</span> to a finite group and, thus, <span class="math-container">$H$</span> is not a separable subgroup of <span class="math-container">$G$</span>. </p>
<p><strong>Definition 2.</strong> An abelian subgroup <span class="math-container">$H< G$</span> is "self-centralizing" if the centralizer of <span class="math-container">$H$</span> in <span class="math-container">$G$</span> equals <span class="math-container">$H$</span>. </p>
<p>This condition clearly fails for the cyclic subgroup <span class="math-container">$H=<a^p>$</span> in the example. </p>
<p><strong>Lemma 1.</strong> If <span class="math-container">$H$</span> is finitely generated, abelian, self-centralizing and <span class="math-container">$G$</span> is residually finite, then <span class="math-container">$H$</span> is separable. </p>
<p><strong>Proof.</strong> Let <span class="math-container">$F\subset G-H$</span> be a finite subset. Let <span class="math-container">$h_1,...,h_n$</span> be a finite generating set of <span class="math-container">$H$</span>. Define the finite subset <span class="math-container">$C\subset G$</span> to be the set of commutators <span class="math-container">$[h_i,f], f\in F$</span>. Since <span class="math-container">$G$</span> is residually finite, there exists a finite group <span class="math-container">$K$</span> and a homomorphism <span class="math-container">$\phi: G\to K$</span> such that <span class="math-container">$\phi(C)$</span> does not contain the neutral element <span class="math-container">$e\in K$</span>. Since <span class="math-container">$H$</span> is self-centralizing, for each <span class="math-container">$f\in F$</span>, <span class="math-container">$\phi(f)\notin \phi(H)$</span> (otherwise, <span class="math-container">$\phi(f)$</span> commutes with all the generators of <span class="math-container">$\phi(H)$</span>). Then let
<span class="math-container">$$
G_0:= \phi^{-1}(\phi(H)).$$</span>
This is clearly a finite index subgroup containing <span class="math-container">$H$</span>, but disjoint from <span class="math-container">$F$</span> (as noted above). <strong>qed</strong></p>
<p>As a corollary we obtain that maximal cyclic subgroups of surface groups are separable. In fact, all finitely generated subgroups of surface groups are separable, but this is a much harder result (due to Peter Scott). </p>
<p><strong>Lemma 2.</strong> Let <span class="math-container">$G=\pi_1(S)$</span> be a hyperbolic surface group, <span class="math-container">$A< G$</span> is a maximal cyclic subgroup. Then there exists a finite covering <span class="math-container">$S_0\to S$</span> such that <span class="math-container">$A$</span> is realized by the fundamental group of a simple loop in <span class="math-container">$S_0$</span>. </p>
<p><strong>Proof.</strong> Let <span class="math-container">${\mathbb H}^2\to S$</span> be the universal covering of <span class="math-container">$S$</span>. Let <span class="math-container">$L\subset {\mathbb H}^2$</span> be the unique geodesic invariant under the action of <span class="math-container">$A$</span>. Let <span class="math-container">$D=pq\subset L$</span> be a subinterval which is a fundamental domain for the action of <span class="math-container">$A$</span> on <span class="math-container">$L$</span>, i.e. if <span class="math-container">$a$</span> is a generator of <span class="math-container">$A$</span>, then <span class="math-container">$a(p)=q$</span>. Since <span class="math-container">$D$</span> is compact and the action of <span class="math-container">$G$</span> on <span class="math-container">${\mathbb H}^2$</span> is properly discontinuous, there is a finite subset <span class="math-container">$F\subset G$</span> such that for all <span class="math-container">$g\in G-F$</span>, either <span class="math-container">$g(D)\cap D=\emptyset$</span> or <span class="math-container">$g=e$</span>, the neutral element. By Lemma 1, there exists a finite index subgroup <span class="math-container">$G_0< G$</span> disjoint from <span class="math-container">$F$</span> and containing <span class="math-container">$A$</span>. Then, the projection of <span class="math-container">$L$</span> (which is the same as the projection of <span class="math-container">$D$</span>) to <span class="math-container">$S_0;{\mathbb H}^2/G_0$</span> is a simple loop. <strong>qed</strong></p>
<p>From this proof you can see that residual finiteness of <span class="math-container">$G$</span> is not enough here, one needs separability of <span class="math-container">$A$</span> in <span class="math-container">$G$</span>. You also get an answer to the question asked in a comment to Lee's answer. The subgroup separability (even for surface subgroups) of hyperbolic 3-manifold groups was unknown before the work of Agol and Wise. You cannot use the "commutator trick" from Lemma 1 since
surface subgroups are non-abelian. A version of this trick works for maximal surface subgroups represented by immersed totally geodesic subsurfaces, it is due to D.Long, see <a href="http://web.math.ucsb.edu/~long/pubpdf/imm&emb.pdf" rel="nofollow noreferrer">here</a>. But this case is too special. </p>
<p>We are not done yet, as we still have to deal with non-maximal cyclic subgroups. </p>
<p><strong>Lemma 3.</strong> Given a maximal cyclic subgroup <span class="math-container">$A< G$</span> as above, there is a finite covering <span class="math-container">$S_1\to S$</span> such that <span class="math-container">$A$</span> is represented by a simple <strong>nonseparating</strong> loop in <span class="math-container">$S_1$</span>. </p>
<p>I will skip the proof, for instance, you again can use the separability of <span class="math-container">$A$</span>. </p>
<p><strong>Lemma 4.</strong> The conclusion of Lemma 2 holds for all infinite cyclic subgroups of <span class="math-container">$G$</span>. </p>
<p><strong>Proof.</strong> Let <span class="math-container">$C< G$</span> be an infinite cyclic subgroup and <span class="math-container">$C< A$</span> is a maximal cyclic subgroup of <span class="math-container">$G$</span> containing <span class="math-container">$C$</span>. Let <span class="math-container">$i$</span> denote the index of <span class="math-container">$C$</span> in <span class="math-container">$A$</span>. By lemma 3, we can assume that <span class="math-container">$A$</span> is realized by a simple nonseparating loop <span class="math-container">$\alpha$</span> on our hyperbolic surface <span class="math-container">$S$</span>. Let <span class="math-container">$\sigma\in H^1(S; {\mathbb Z})$</span> be the cohomology class Poincare dual to the homology class represented by <span class="math-container">$\alpha$</span>. Then <span class="math-container">$\sigma$</span> lifts to a homomorphism <span class="math-container">$\psi: G\to H_1(S; {\mathbb Z})\to {\mathbb Z}$</span>. Let <span class="math-container">$G_2:= \psi^{-1}(i{\mathbb Z})$</span>; it is an index <span class="math-container">$i$</span> subgroup of <span class="math-container">$G$</span> containing <span class="math-container">$C$</span>. The loop <span class="math-container">$\alpha^i$</span> then lifts to a simple loop on <span class="math-container">$S_2:= {\mathbb H}^2/G_2$</span>. <strong>qed</strong></p>
<p>This proof actually works in much greater generality and provides a topological interpretation of subgroup separability due to Peter Scott:</p>
<p><strong>Lemma 5.</strong> A subgroup <span class="math-container">$H< G$</span> of a countable group <span class="math-container">$G$</span> is separable if and only if the following holds: For every locally compact Hausdorff topological space <span class="math-container">$X$</span> (say, a manifold), a properly discontinuous action <span class="math-container">$G\times X\to X$</span> and a compact subset <span class="math-container">$K\subset X/H$</span>, there exists a finite index subgroup <span class="math-container">$G_0<G$</span> containing <span class="math-container">$H$</span> such that the projection
of the covering map <span class="math-container">$p: X/H\to X/G_0$</span> restricts to a one-to-one map on
<span class="math-container">$K$</span>. </p>
<p>In Lemma 4, the space <span class="math-container">$X$</span> is the universal covering of <span class="math-container">$S$</span>, the action of <span class="math-container">$G$</span> on <span class="math-container">$X$</span> is the covering action and the compact subset <span class="math-container">$K$</span> is a simple loop which carries the fundamental group of the cylinder <span class="math-container">$X/H$</span>. </p>
<p>The proof of Lemma 4 of course, will not give you an estimate on the degree of a finite covering of <span class="math-container">$S$</span> to which the given loop <span class="math-container">$\gamma$</span> in <span class="math-container">$S$</span> lifts to a loop homotopic to a simple loop. Getting such an estimate is an interesting project worth pursuing. If you are interested: The first step is to get such an estimate in the case when <span class="math-container">$S$</span> is replaced with a finite graph <span class="math-container">${\mathcal G}$</span>. The construction of a finite covering <span class="math-container">${\mathcal G}_0\to {\mathcal G}$</span> to which given "tight" loop <span class="math-container">$\gamma$</span> in <span class="math-container">${\mathcal G}$</span> lifts to a simple loop is pretty explicit, you can find it for instance in the paper </p>
<p>P.Scott, T.Wall "<a href="https://www.maths.ed.ac.uk/~v1ranick/papers/wallhomgroup.pdf" rel="nofollow noreferrer">Topological methods in group theory</a>". </p>
<p>From this proof you can extract an estimate on the degree of the covering. After that, if you are still interested in the problem, you can read the proof of subgroup separability of surface groups which is due to Wise, his argument is combinatorial, inspired by the one by Scott and Wall, and you can get an explicit degree estimate. </p>
|
3,632,431 | <blockquote>
<p>Consider the function <span class="math-container">$f: \mathbb{N} \to \mathbb{N}$</span> defined by <span class="math-container">$f(x)=\frac{x(x+1)}{2}$</span>. Show that <span class="math-container">$f$</span> is injective but not surjective.</p>
</blockquote>
<p>So I started by assuming that <span class="math-container">$f(a)=f(b)$</span> for some <span class="math-container">$a,b \in \mathbb{N}$</span>.
I want to show that <span class="math-container">$a=b$</span>.</p>
<p><span class="math-container">$$\Rightarrow \frac{a(a+1)}{2} = \frac{b(b+1)}{2}\\
\Rightarrow a(a+1)=b(b+1) \\
\Rightarrow a^2+a=b^2+b$$</span></p>
<p>I don't know where to go from here.</p>
| Tuvasbien | 702,179 | <p>First notice that <span class="math-container">$\sum_{n=1}^x n=\frac{x(x+1)}{2}=f(x)$</span> for all <span class="math-container">$x\in\mathbb{N}$</span>, thus if <span class="math-container">$f(a)=f(b)$</span> with <span class="math-container">$a\neq b$</span>, we can suppose without loss of generality that <span class="math-container">$a<b$</span>, thus
<span class="math-container">$$ \sum_{k=a+1}^b k=\sum_{k=1}^b k-\sum_{k=1}^a k=f(b)-f(a)=0 $$</span>
which is not because <span class="math-container">$\sum_{k=a+1}^b k\geqslant a+1>0$</span> so <span class="math-container">$a=b$</span>.
You can also say that if <span class="math-container">$f(a)=f(b)$</span>, then <span class="math-container">$a^2+a=b^2+b$</span> as you wrote, so that <span class="math-container">$$b-a=a^2-b^2=(a+b)(a-b)$$</span> if <span class="math-container">$a\neq b$</span>, then <span class="math-container">$a-b\neq 0$</span> and thus <span class="math-container">$1=-(a+b)<0$</span> which is not so that <span class="math-container">$a=b$</span>.</p>
|
3,392,171 | <p>We have a partial fraction equation:
<span class="math-container">$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$</span></p>
<p>I multiplied the equation by the common denominator <span class="math-container">$(x+5)(x-5)$</span> and got <span class="math-container">$0=1$</span>. Is this correct?</p>
| Quanto | 686,284 | <p>Rearrange the equation as </p>
<p><span class="math-container">$$\frac{2x}{x^2-25}-\frac{1}{x-5}-\frac{1}{x+5}
=\frac{2x+1}{x^2-25}-\frac{2x}{x^2-25}= \frac{1}{x^2-25}=0$$</span></p>
<p>which leads to the solutions </p>
<p><span class="math-container">$$x=\pm \infty$$</span></p>
|
223,955 | <p>How can we convert a list to an integer correctly? </p>
<p><strong>{5, 22, 4, 5} -> 52245?</strong></p>
<p>When I use the command <code>FromDigits</code> in Mathematica </p>
<pre><code>FromDigits[{5, 22, 4, 5}]
</code></pre>
<p>The result is incorrect, namely <strong>7245</strong></p>
| flinty | 72,682 | <p>Here are three ways to do it. The first is the best I think:</p>
<pre><code>FromDigits@Flatten[IntegerDigits /@ {5, 22, 4, 5}]
ToExpression@StringJoin[ToString /@ {5, 22, 4, 5}]
ToExpression@StringJoin@StringCases[Characters@ToString@{5, 22, 4, 5}, DigitCharacter]
</code></pre>
|
223,955 | <p>How can we convert a list to an integer correctly? </p>
<p><strong>{5, 22, 4, 5} -> 52245?</strong></p>
<p>When I use the command <code>FromDigits</code> in Mathematica </p>
<pre><code>FromDigits[{5, 22, 4, 5}]
</code></pre>
<p>The result is incorrect, namely <strong>7245</strong></p>
| user1066 | 106 | <pre><code>ToExpression@StringJoin@IntegerString[{5, 22, 4, 5}]
</code></pre>
<blockquote>
<p>52245</p>
</blockquote>
<pre><code>ToExpression@StringJoin@IntegerString[{5, 22, 4, 5},#]&/@{2,10,16}
</code></pre>
<blockquote>
<p>{10110110100101, 52245, 51645}</p>
</blockquote>
|
223,955 | <p>How can we convert a list to an integer correctly? </p>
<p><strong>{5, 22, 4, 5} -> 52245?</strong></p>
<p>When I use the command <code>FromDigits</code> in Mathematica </p>
<pre><code>FromDigits[{5, 22, 4, 5}]
</code></pre>
<p>The result is incorrect, namely <strong>7245</strong></p>
| kcr | 49,048 | <pre><code>list = {5, 22, 4, 5};
FromDigits[StringJoin[IntegerString[list]]]
</code></pre>
<p>which gives</p>
<pre><code>52245
</code></pre>
|
1,215,273 | <blockquote>
<p>Describe the cosets of the subgroup $\langle 3\rangle$ of $\mathbb{Z}$</p>
</blockquote>
<p>The problem I have is $\mathbb{Z}$ is infinite.</p>
<p>So we know that $\langle 3\rangle=\{0,3,6,9,12,\ldots\}$ and I know the definition of cosets (in this case right cosets) is the set of all products of ha, as a remains fixed and $h$ ranges over $H$.</p>
<p>So $H$ is the subgroup $\langle 3\rangle$ which does not remain fixed and the elements of $\mathbb{Z}$ remains fixed.</p>
<p>I started writing some numbers out to see how I can possibly describe it but I didn't see any help.</p>
<p>I defined: $\mathbb{Z}=\{\ldots,-4,-3,-2,-1,0,1,2,3,4,\ldots\}$ and I already defined $\langle 3\rangle=\{0,3,6,9,12,\ldots\}$</p>
<p>Therefore here are some numbers:</p>
<p>Lets pick $-4$ as our fixed $a$, so then we obtain:</p>
<p>$0 \times -4=0$</p>
<p>$3 \times -4=-12$</p>
<p>$6 \times -4=-24$</p>
<p>and so on</p>
<p>If I pick $2$ as our fixed $a$, then we have:</p>
<p>$0 \times 2=0$</p>
<p>$3 \times 2=6 $</p>
<p>$6 \times 2=12 $</p>
<p>I probably can't see it but then how would I describe the cosets?</p>
| pjs36 | 120,540 | <p>Since $\Bbb Z$ is an <em>additive</em>, your cosets are $a + \Bbb \langle 3 \rangle$, not $a\langle 3 \rangle$, like you're calculating.</p>
<p>So, to given an example, one would be the coset $1 + \langle 3 \rangle = \{1 + n: n \in \langle 3 \rangle\} = \{\ldots, -5, -2, 1, 4, \ldots\}$.</p>
<p>So that was your only problem, not using the right group operation; everything else looked good!</p>
|
994,620 | <p>I need to solve $f(2x)=(e^x+1)f(x)$. I am thinking about Frobenius type method:
$$\sum_{k=0}^{\infty}2^ka_kx^k=\left(1+\sum_{m=0}^{\infty}\frac{x^m}{m!}\right)\sum_{n=0}^{\infty}a_nx^n\\
\sum_{k=0}^{\infty}(2^k-1)a_kx^k=\left(\sum_{m=0}^{\infty}\frac{x^m}{m!}\right)\left(\sum_{n=0}^{\infty}a_nx^n\right)=\sum_{m=0}^{\infty}\left(\frac{x^m}{m!}\sum_{n=0}^{\infty}a_nx^n\right)=\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}\frac{a_nx^{m+n}}{m!}\right)$$
So I think
$$(2^k-1)a_k=\sum_{m=0}^k\frac{a_{k-m}}{m!}.$$
Now I think about a recurrence relation:
$$(2^k-1)r^k=\sum_{m=0}^k\frac{r^{k-m}}{m!}.$$
Can somebody help me? Firstly, is my Frobenius technique right? Useful? Is there another easier way? Recurrence relation?</p>
| doraemonpaul | 30,938 | <p>$\because$ the trivial solution is $f(x)=e^x-1$</p>
<p>$\therefore$ the general solution is $f(x)=\Theta(\log_2x)(e^x-1)$ , where $\Theta(x)$ is an arbitrary periodic function with unit period</p>
|
326,094 | <p>Suppose <span class="math-container">$(f_n)_n$</span> is a countable family of entire, surjective functions, each <span class="math-container">$f_n:\mathbb{C}\to\mathbb{C}$</span>. Can one always find complex scalars <span class="math-container">$(a_n)_n$</span>, not all zero, such that <span class="math-container">$\sum_{n=1}^{\infty} a_n f_n$</span> is entire but not-surjective? In fact, I am interested in this question under the additional assumption that <span class="math-container">$(f_n)_n$</span> are not polynomials. </p>
| Alexandre Eremenko | 25,510 | <p>The answer is no. If something does not hold for polynomials, don't expect that it will hold for entire functions:-)</p>
<p>For example, all non-constant functions of order less than <span class="math-container">$1/2$</span> are surjective.
This follows from an old theorem of Wiman that for such function <span class="math-container">$f$</span> there exists
a sequence <span class="math-container">$r_k\to\infty$</span> such that <span class="math-container">$\min_{|z|=r_k}|f(z)|\to\infty$</span> as <span class="math-container">$k\to \infty.$</span>
And of course linear combinations of functions of order less than <span class="math-container">$1/2$</span> are of order less
than <span class="math-container">$1/2$</span>.</p>
<p>Edit. To construct a counterexample with infinite sums, one can use lacunary series. Let <span class="math-container">$\Lambda$</span> be a sequence of integers <span class="math-container">$n_k$</span> which grows sufficiently fast,
for example, such that <span class="math-container">$n_k/k\to\infty$</span>,
and consider the class of entire functions of the form
<span class="math-container">$$f(z)=\sum_{n\in\Lambda}c_nz^n.$$</span>
It is known that all such functions are surjective. And of course any linear combination of such functions, finite or infinite, belongs to the class.</p>
<p>Reference: L. Sons, An analogue of a theorem of W.H.J. Fuchs on gap series,
Proc. LMS, 1970, 21 525-539.</p>
|
3,532,033 | <p>Let <span class="math-container">$(M_1,+,\times)$</span> be an algebraic structure, lets say, for example, a ring. If we have another structure <span class="math-container">$(M_2,+,\times)$</span> isomorphic to the first one does that mean that <span class="math-container">$(M_2,+,\times)$</span> is also a ring ?</p>
<p>Edit : To give more context, in 2 different exercises I had to prove that the following are rings :</p>
<p>1) <span class="math-container">$(\mathbb{C},+,•)$</span> where "<span class="math-container">$+$</span>" is simple addition and <span class="math-container">$z_1•z_2 = z_1z_2 + \operatorname{Im}(z_1) \operatorname{Im}(z_2)$</span>, where <span class="math-container">$\operatorname{Im}(z) = $</span> the imaginary component of z.</p>
<p>2) <span class="math-container">$(\mathbb{R} \times \mathbb{R},+,•)$</span> where <span class="math-container">$(a,b)+(x,y) = (a+x,b+y)$</span> and <span class="math-container">$(a,b)•(x,y) = (ax,ay+bx)$</span> </p>
<p>In fact, these are the same with different notations: <span class="math-container">$x+yi = (x,y)$</span></p>
| Shaun | 104,041 | <p>Sometimes.</p>
<p>It depends on what aspects of the two algebras are isomorphic; for example, <span class="math-container">$(M_1, +_1)$</span> could be isomorphic to <span class="math-container">$(M_2, +_2)$</span> <em>as groups</em>, but not necessarily as, say, <em>rings</em> <span class="math-container">$(M_1, +_1, \times_1)$</span> and <span class="math-container">$(M_2, +_2, \times_2)$</span>. If they're isomorphic <em>as the same structure, including the ring axioms</em>, then, yes, if one is a ring, so is the other.</p>
|
3,532,033 | <p>Let <span class="math-container">$(M_1,+,\times)$</span> be an algebraic structure, lets say, for example, a ring. If we have another structure <span class="math-container">$(M_2,+,\times)$</span> isomorphic to the first one does that mean that <span class="math-container">$(M_2,+,\times)$</span> is also a ring ?</p>
<p>Edit : To give more context, in 2 different exercises I had to prove that the following are rings :</p>
<p>1) <span class="math-container">$(\mathbb{C},+,•)$</span> where "<span class="math-container">$+$</span>" is simple addition and <span class="math-container">$z_1•z_2 = z_1z_2 + \operatorname{Im}(z_1) \operatorname{Im}(z_2)$</span>, where <span class="math-container">$\operatorname{Im}(z) = $</span> the imaginary component of z.</p>
<p>2) <span class="math-container">$(\mathbb{R} \times \mathbb{R},+,•)$</span> where <span class="math-container">$(a,b)+(x,y) = (a+x,b+y)$</span> and <span class="math-container">$(a,b)•(x,y) = (ax,ay+bx)$</span> </p>
<p>In fact, these are the same with different notations: <span class="math-container">$x+yi = (x,y)$</span></p>
| mrtaurho | 537,079 | <p>I think what you are looking for is the idea of <a href="https://en.wikipedia.org/wiki/Transport_of_structure" rel="nofollow noreferrer"><em>transport of structure</em></a>. Lets stick with your example for now. Suppose, we know that <span class="math-container">$(M_1,+,\times)$</span> is a ring and we have a set <span class="math-container">$M_2$</span> equipped with two operations <span class="math-container">$\dot+$</span> and <span class="math-container">$\dot\times$</span>. Without further investigation we do not know anything else about <span class="math-container">$(M_2,\dot+,\dot\times)$</span> rather than being a set equipped with two operations.</p>
<p>Now, suppose we can define a function <span class="math-container">$f:M_1\to M_2$</span> which satisfies the following properties</p>
<ol>
<li><span class="math-container">$f$</span> is a <em>bijection</em> of sets, saying that <span class="math-container">$M_1$</span> and <span class="math-container">$M_2$</span> are basically the same set up to renaming</li>
<li><span class="math-container">$f$</span> satisfies the <em>axioms of a ring homomorphism</em>, i.e. we have <span class="math-container">$\forall x,y\in M_1$</span> that
<span class="math-container">\begin{align*}
f(x+y)&=f(x)\dot+f(y)\\
f(x\times y)&=f(x)\dot\times f(y)
\end{align*}</span></li>
</ol>
<p>Then we can <em>transport</em> the given ring structure from <span class="math-container">$M_1$</span> to <span class="math-container">$M_2$</span> via the function <span class="math-container">$f:M_1\to M_2$</span>. For example, take the (left) distributive law in <span class="math-container">$M_2$</span>. Via the surjectivity of <span class="math-container">$f$</span> and the ring structure of <span class="math-container">$M_1$</span> we may conclude
<span class="math-container">\begin{align*}
\dot x\dot\times(\dot y\dot+\dot z)&=f(x)\dot\times(f(y)\dot+f(z))\\
&=f(x)\dot\times f(y+z)\\
&=f(x\times (y+z))\\
&=f(x\times y+x\times z)\\
&=f(x\times y)\dot+f(y\times z)\\
&=f(x)\dot\times f(y)\dot+f(y)\dot\times f(z)\\
&=\dot x\dot\times\dot y\dot+\dot x\dot\times\dot z
\end{align*}</span>
In a similiar manner you can in fact prove that all the ring axioms hold in <span class="math-container">$M_2$</span> simply by using the properties of the given map <span class="math-container">$f$</span> over and over again. I just took the (left) distributive law as an illustration. The <a href="https://www.reddit.com/r/learnmath/comments/em1i74/abstract_algebra_isomorphismlike_map_inducing/" rel="nofollow noreferrer">same idea applies when <span class="math-container">$(M_1,+,\times)$</span> is a field</a>, or a different algebraic structure defined as a set equipped with operations.</p>
<blockquote>
<p>But <span class="math-container">$f$</span> is <em>not</em> an isomorphism in the first place! The concept of an isomorphism does not apply in this setting as an isomorphism is commonly defined between objects of the <em>same</em> category (as <em>Maryam Ajorlou</em> pointed out). In this setting we <em>use</em> that we have a bijective, <em>structure-preserving</em> map to <em>transport</em> (or define) structure from <span class="math-container">$M_1$</span> to <span class="math-container">$M_2$</span>.</p>
</blockquote>
|
160,165 | <p>For an elliptic curve $E$ over $\Bbb{Q}$, we know from the proof of the Mordell-Weil theorem that the weak Mordell-Weil group of $E$ is $E(\Bbb{Q})/2E(\Bbb{Q})$. It is well known that
$$
0 \rightarrow E(\Bbb{Q})/2E(\Bbb{Q}) \rightarrow S^{(2)}(E/\Bbb{Q}) \rightarrow Ш(E/\Bbb{Q})[2] \rightarrow 0
$$
is an exact sequence which gives us a procedure to compute the generators for $E(\Bbb{Q})/2E(\Bbb{Q})$. </p>
<p>(Relatively) recently I found out that there is another way to compute the rank of $E$ using $3$-descent. I was wondering, since the natural structure of the weak Mordell-Weil group is $E(\Bbb{Q})/2E(\Bbb{Q})$, what is the motivation behind using $3$-descent? Also does $3$-descent similarly produce the generators of $E(\Bbb{Q})/2E(\Bbb{Q})$ or does it simply tell us the structure of $E(\Bbb{Q})$ via the Mordell-Weil theorem by giving us only the rank of $E$? Finally does it help us get around the issue of $Ш(E/\Bbb{Q})$ containing an element that is infinitely $2$-divisible?</p>
| Matt E | 221 | <p>An $n$-descent will compute the $n$-Selmer group, which sits in a s.e.s.
$$0 \to E(\mathbb Q)/n E(\mathbb Q) \to S^{(n)}(E/\mathbb Q) \to Ш(E/\mathbb Q)[n] \to 0.$$</p>
<p>If you do a $2$-descent, it will give an upper bound on the size of $E(\mathbb Q)/2E(\mathbb Q).$ If you do a $3$-descent, it will give you an upper bound on the size of $E(\mathbb Q)/3 E(\mathbb Q).$</p>
<p>The advantage of one over the other will depend on the structure as a Galois module of $E(\mathbb Q)[n]$, and the structure of the $n$-torsion in Sha.</p>
<p>If $E$ contains a rational $2$-torsion point, then the $2$-Selmer group may be easier to compute than the $n$-Selmer groups for other $n$. </p>
<p>As an example of a descent at a different choice of $n$, note that
the elliptic curve $X_0(11)$ contains a rational $5$-torsion point, and Mazur does a $5$-descent to prove that it has no other rational points besides the five points generated by this $5$-torsion point. (See <a href="https://math.stackexchange.com/a/35341/221">this answer</a> for more details on this.)
The elliptic curve $X_0(17)$ has a rational $3$-torsion point, and for this Mazur does a $3$-descent.</p>
|
105,071 | <p>As one may know, a <b>dynamical system</b> can be defined with a monoid or a group action on a set, usually a manifold or similar kind of space with extra structure, which is called the <i>phase space</i> or <i>state space</i> of the dynamical system. The monoid or group doing the acting is what I call the <i>time space</i> of the dynamical system, and is usually the naturals, integers, or reals. Often, one may require the evolution map to be continuous, differentiable, etc.</p>
<p>But has anyone studied a generalization in which we allow the <i>time space</i> to be something more general & exotic, a multidimensional space like $\mathbb{R}^n$, $\mathbb{C}$ (viewed as "2-dimensional" by considering it to be like $\mathbb{R}^2$), etc.? I'm especially curious about the case where the time space is $\mathbb{C}$ and the phase space is $\mathbb{C}^n$ or another complex manifold and the map is required to be holomorphic in both its arguments, as that holomorphism provides a natural linkage between the two dimensions that lets us think of the complex time as a single 2-dimensional time as opposed to two real times (any dynamical system with a timespace of $\mathbb{R}^n$ can be decomposed into a bunch of mutually-commutative evolution maps with timespaces of $\mathbb{R}$). My questions are:</p>
<ol>
<li><p>Is it true that the only dynamical system with phase-space $\mathbb{C}$ and time-space $\mathbb{C}$ where the evolution map is required to be holomorphic in both its arguments (the time and point to evolve) is the linear one given by
$$\phi^t(z) = e^{ut} z + K \frac{1 - e^{ut}}{1 - e^u}, K, u \in \mathbb{C}$$
? I suspect so, because an injective entire function is linear (in the sense of a “linear equation” not <em>necessarily</em> a “linear map”), and $\phi^{-t}$ must be the inverse of $\phi^t$. Thus, $\phi^t$ must have the form $a(t) z + b(t)$ with $a: \mathbb{C} \rightarrow \mathbb{C}$, $b: \mathbb{C} \rightarrow \mathbb{C}$. Am I right?</p></li>
<li><p>Are there any interesting (e.g. with complicated, even chaotic behavior) dynamical systems of this kind on $\mathbb{C}^n$? On more sophisticated complex manifolds? (for the phase space, that is) If so, can you provide an example? Or does the holomorphism requirement essentially rule this out? EDIT: I provide one below.</p></li>
<li><p>There is something else here, an interesting observation I made. Consider the above complex-time, holomorphic dynamical system. We can investigate the two prime behaviors represented by the real and imaginary times. We'll just set u = 1, K = 0 for here.</p></li>
</ol>
<p>In “real time”, the dynamics looks like an “explosion” in the plane: all points “blast” away from z = 0 at exponentially increasing velocity.</p>
<p>In “imaginary time”, the result is cyclic motion, swirling around $z = 0$ with constant angular velocity that depends only on the distance of the point from $z = 0$.</p>
<p>But if we trace the contours of these two evolutions, formed from different points on the plane, and then superimpose them, we have contours intersecting in what looks like contours from a contour graph of the image of the complex plane under the function $\exp(z)$! Conversely, we could say it looks like a countour plot of $\log(z)$ with a cut along a ray from $0$. So, somehow “naturally” related to the dynamical system $\phi^t(z) = \exp(t) z$ is the function $\exp(z)$ (or, perhaps, $\log(z)$).</p>
<p><img src="https://i.stack.imgur.com/kPq32.png" alt="plot of evoluton countours for first CTHDS"></p>
<p>Note that my plotting facilities are unfortunately pretty limited, so I can't give really nice graphs with lots of contours, just a few selected ones taken from evolutions of various points in both real and imag times.</p>
<p>But we have another case. To see this, we must turn our attention away from a phase space given by the complex plane to one given by the Riemann sphere, $\hat{\mathbb{C}}$. In this case, we still have the dynamical system as above, but we have an additional class of dynamical systems given by the “Moebius transformations”, which include the above linear-function dynamical systems as a special case. One example is
$$\phi^t(z) = \frac{(1 + e^{i\pi t}) z + (1 – e^{i\pi t})}{(1 – e^{i\pi t})z + (1 + e^{i\pi t})}$$.
It is easy to check that this is indeed a Moebius transformation of the Riemann sphere. This map is holomorphic everywhere on the Riemann sphere. Note that for integer step t, the unit-step map is the reciprocal map.</p>
<p>Now we consider the contour lines of the real and imaginary evolution, as before. They look like this:</p>
<p><img src="https://i.stack.imgur.com/MECR7.png" alt="plot of evolution contours for second CTHDS"></p>
<p>(Physics buffs may notice that the real evolution (concentric circles) reminds one of the lines of a <i>magnetic</i> field of a <i>magnetic</i> dipole (like a bar magnet), while the imaginary evolution (arcs joining at points) looks like the lines of an <i>electric</i> field of an <i>electric</i> dipole.)</p>
<p>Again, notice how the lines meet at right angles. It looks again like the image of the plane (or the Riemann sphere, perhaps?) under some function which may be holomorphic, though I'm not sure what that function is in this case. Is this the case? Is there such a function, with a special relation to this CTHDS in the same way as $\exp$ (or $\log$) is to the other?</p>
<p>But in any case, it appears that for a complex-time holomorphic dynamical system, or CTHDS, there exists an associated natural function. What is the significance of this function/map? How does it relate to the CTHDS? If you give me a CTHDS that I don't have a closed form for, can I find its natural map?</p>
| Pablo Shmerkin | 11,009 | <p>This is more a comment than an answer but it's too long for a comment.</p>
<p>In ergodic theory (as opposed to dynamical systems), systems in which the acting group is not the integers or the reals have been widely studied, both for their own intrinsic interest and because of deep and striking applications to number theory and other areas. For example, the proof by Einsiedler, Katok, Lindenstrauss that the set of exceptions to <a href="http://en.wikipedia.org/wiki/Littlewood_conjecture" rel="nofollow">Littlewood's conjecture</a> has zero Hausdorff dimension uses an ergodic-theoretic result on higher rank group actions.</p>
<p>As soon as the acting group has higher rank (even in the simplest case, i.e. $\mathbb{Z}^2$), the study of their ergodic theoretic properties becomes dramatically more complicated, even if the phase space is as simple as possible (the circle). A famous example is Furstenberg's $\times 2\times 3$ problem: what are the measures which are simultaneously invariant under $x\to 2x \bmod 1$ and $x\to 3x\bmod 1$? (this corresponds to the action of $\mathbb{N}^2$ on $[0,1]$ given by $(a,b)\cdot x=2^a 3^b x\bmod 1$). It is suspected that there are very few invariant measures, which illustrates a general (conjectured or proved) phenomenon: higher rank dynamical systems tend to have few invariant measures, all of them with some algebraic structure. Indeed, Einsiedler, Katok and Lindenstrauss use such a rigidity result.</p>
<p>The book by Einsiedler and Ward "Ergodic thory (with a view towards number theory)" is an excellent reference for this general topic.</p>
|
271,343 | <p>I need help finding the integral of $\sin(\sqrt{x})dx$. I have the answer here but would like to know how to get there. </p>
| Mhenni Benghorbal | 35,472 | <p><strong>Hint:</strong> Use the substitution $\sqrt{x}=u$ or $x=u^2$ and then integration by parts.</p>
<p>$$\int \sin(\sqrt{x})dx = 2\int u\sin(u)du $$</p>
|
2,466,947 | <p>In class we are beginning discrete probability and we are being introduced to counting. He said in an example today that the word "anagram" (which has $7$ letters) can be rearranged to have $7!/3!$ possible words when we don't count the order of the a's. To me this seemed like a combination but this result reduces to $7 \cdot 6 \cdot 5 \cdot 4$ which means we are choosing $7$ letters on $4$ spots. So a bit confused if anyone could help?</p>
| N. F. Taussig | 173,070 | <p>Your instructor is counting distinguishable arrangements of the word <em>anagram</em>.</p>
<p>The word <em>anagram</em> has seven letters, so we have seven positions to fill with $3$ <em>a</em>s, $1$ <em>g</em>, $1$ <em>m</em>, and $1$ <em>r</em>. We can fill three of these seven positions with <em>a</em>s in $\binom{7}{3}$ ways. The remaining four letters are distinct, so they can be arranged in the remaining four positions in $4!$ ways. Hence, the number of distinguishable arrangements of the word <em>anagram</em> is
$$\binom{7}{3}4! = \frac{7!}{3!4!} \cdot 4! = \frac{7!}{3!}$$
The factor of $3!$ in the denominator represents the number of ways the three <em>a</em>s can be permuted among themselves within a given arrangement of the letters of the word <em>anagram</em> without producing an arrangement distinguishable from the given arrangement.</p>
|
2,466,947 | <p>In class we are beginning discrete probability and we are being introduced to counting. He said in an example today that the word "anagram" (which has $7$ letters) can be rearranged to have $7!/3!$ possible words when we don't count the order of the a's. To me this seemed like a combination but this result reduces to $7 \cdot 6 \cdot 5 \cdot 4$ which means we are choosing $7$ letters on $4$ spots. So a bit confused if anyone could help?</p>
| Allawonder | 145,126 | <p>I think what you notice is that the process reminds you of the number of permutations (not selections, as you wrote) of <span class="math-container">$4$</span> out of <span class="math-container">$7$</span> objects. This is indeed the case.</p>
<p>Now, first note that to count the number of arrangements of <span class="math-container">$n$</span> <em>distinct</em> objects, you may proceed as follows:</p>
<p>Partition the set in two, one containing <span class="math-container">$r$</span> of these <span class="math-container">$n$</span> objects. Then the number of arrangements is (by the multiplication principle) <span class="math-container">$$nPr\times (n-r)P(n-r)=nPr\times (n-r)!=n!.$$</span> This means that for each arrangement of the <span class="math-container">$r$</span> objects in the first part, we have <span class="math-container">$(n-r)!$</span> arrangements of the <span class="math-container">$n$</span> objects. <em>Now, if these <span class="math-container">$n-r$</span> objects are indistinguishable, then the number of permutations of the <span class="math-container">$n$</span> (previously distinct) objects should be divided by <span class="math-container">$(n-r)!.$</span></em> This is what you noticed; and, as I have explained, hopefully you can see why it is true that <strong>the number of arrangements of <span class="math-container">$n$</span> objects, <span class="math-container">$n-r$</span> of which are indistinguishable, is the same as the number of arrangements of <span class="math-container">$r$</span> out of <span class="math-container">$n$</span> <em>distinct</em> objects.</strong></p>
|
9,168 | <p>I'm having a doubt about how should we users encourage the participation of new members. So far I have only presented MSE to three of my fellow colleagues in grad school. In an overall way I feel like if MSE becomes too open and wide known, some of the high-rank researchers and top-class grad and undergrads users will be frustrated. Maybe I get this impression for seeing how some really complicated questions are well-received.</p>
<p>But today I saw someone asking some really simple algebra questions like "solve this equation for x" and started thinking what is the general feel of MSE when this kind of question arises. I'm not judging anything and no one. I just would like to get some opinions on this.</p>
| Eric Naslund | 6,075 | <p>To answer your main concern, questions of any level are welcome on Math Stack Exchange. </p>
<p>Having said that, try to actively look to see if a question has been answered before, as this is far more likely with lower level questions. Questions about solving basic limits, or testing the convergence of certain series, have been answered many times in the past. The reputation system encourages posting questions without looking for duplicates first, and it encourage answering questions without looking for duplicates - so let me try to discourage doing so. The <a href="http://meta.math.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions">List of Generalizations of Common Questions</a> was created to help out with exactly this. If you see a duplicate or similar questions, try to find the original, or something from the list of generalized common questions. If that is not possible, depending on the situation, flag for moderator attention.</p>
|
1,434,420 | <p>Why is $\bigcap\limits_{n=1}^{\infty} \left( \bigcup\limits_{i=1}^{n} G_i \right)^c = \left( \bigcup\limits_{n=1}^{\infty} \left( \bigcup\limits_{i=1}^{n} G_i \right) \right)^c$? What set properties are being applied here? (The $^c$ is set complement)</p>
| Brian M. Scott | 12,042 | <p>For each $n\in\Bbb Z^+$ let $A_n=\bigcup_{i=1}^nG_i$, and let $C_n=A_n^c$; then $A_1\subseteq A_2\subseteq A_3\subseteq\ldots$, and therefore $C_1\supseteq C_2\supseteq C_3\supseteq\ldots\;$, a fact that may help you visualize the situation. You’re being asked to show that </p>
<p>$$\bigcap_{n\ge 1}C_n=\left(\bigcup_{n\ge 1}A_n\right)^c\;.$$</p>
<p>You can do this directly by element-chasing: assume that $x\in\bigcap_{n\ge 1}C_n$ and show that $x\notin\bigcup_{n\ge 1}A_n$, and then assume that $x\notin\bigcup_{n\ge 1}A_n$ and show that $x\in\bigcap_{n\ge 1}C_n$.</p>
|
1,295,453 | <p>In my assignment I have to calculate to following limit. I wanted to know if my solution is correct. Your help is appreciated:</p>
<p>$$\lim_{n \to \infty}n\cos\frac{\pi n} {n+1} $$</p>
<p>Here's my solution:</p>
<p>$$=\lim_{n \to \infty}n\cos \pi \frac{n} {n+1} $$</p>
<p>Since $\frac {n} {n+1}\to 1 $ and $\cos \pi \to (-1)$ we can use the "infinity times a number" rule, since $n \to \infty$. </p>
<p>Therefore, the limit will be $(- \infty) $</p>
<p>Did I Get it right? </p>
<p>Thanks, </p>
<p>Alan </p>
| Paul | 17,980 | <p>You are right. However, we usually say its limit does not exist.</p>
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