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 |
|---|---|---|---|---|
338,480 | <blockquote>
<p>Find the point where equations $x=t^2-t$ and $y= t^3 -3t-1$ cross itself.</p>
</blockquote>
<p>This's the first time I meet this kind of problem, can someone give me some idea? Thank you.</p>
| Yimin | 30,330 | <p>There are some $p,q$ such that</p>
<p>$t^2-t = p$</p>
<p>$t^3-3t-1=q$</p>
<p>has two different solutions $t_1,t_2$,</p>
<p>consider the third solution to the second equation as $t_3$, then $t_1+t_2+t_3 = 0$[why?]. And from the first equation $t_1+t_2 = 1$, thus $t_3 =-1$, plug into the second equation,</p>
<p>$(-1)^3-3*(-1)-1 = q = 1$,</p>
<p>then $t^3-3t-2 = (t+1)(t-2)(t+1)$, then $t_1 =2, t_2=-1$, therefore $p = 2$.</p>
|
91,252 | <p>Let $v_1, \ldots, v_n$ be a set of vectors in a vector space $V$. Show that $v_1, \ldots, v_n$ is a basis of $V$ if and only if for any non-zero linear function $f$ on $V$ there is a vector $v$ in $\operatorname{span}(v_1, \ldots, v_n)$ such that $f(v) \neq 0$. </p>
<p>Suppose that $v_1, \ldots, v_n$ is not a basis of $V$. Then the complement $W$ of $\operatorname{span}(v_1, \ldots, v_n)$ in $V$ is not empty. Let $f$ be a function such that $f(x)=1$ for all $x\in W$ and $f(x)=0$ for all $x$ in $\operatorname{span}(v_1, \ldots, v_n)$. My question is: is $f$ linear in $V$?</p>
| hmakholm left over Monica | 14,366 | <p>Um, unless I'm misreading the question, the thing to be proved is false. Counterexample: Let $V=\mathbb R^1$ (a real vector space) and let $n=2$, $v_1=5$, and $v_2=42$. Then $\mathrm{span}(5,42)$ is $\mathbb R^1$ itself, and it is certainly the case that for any nonzero linear $f$, there is a vector $v$ in $\mathbb R^1$ such that $f(v)\ne 0$ -- that is the <em>definition</em> of $f$ being nonzero.</p>
<p>But the set $(v_1,v_2)$ is not a basis for $\mathbb R^1$ because it is linearly dependent; to wit, $42v_1-5v_2=0$.</p>
|
2,287,312 | <p>So the 12 marbles are distinct with each marble being a different color. Therefore I have to take into account, the different colors and different amount each person have. For example, person 1 can have 2 marbles, person 2 can have 7 marbles, and person 3 can have 3 marbles with each marble being a different color. At first, I thought it was:</p>
<p>$$\binom{12}{1}\binom{11}{1}\binom{10}{1}$$</p>
<p>but I don't think it covers the case of each person having a different amount of marbles for each case. </p>
| Daniel Akech Thiong | 169,316 | <p>Let $x$ be a real number. Then by the density of rational numbers there exists a sequence $\{x_{n} \}$ of rational numbers converging to $x$. But if $f$ is continuous then we have $f(x_{n}) \to f(x)$. But then $f$ would be identically zero, which is a contradiction. </p>
|
2,287,312 | <p>So the 12 marbles are distinct with each marble being a different color. Therefore I have to take into account, the different colors and different amount each person have. For example, person 1 can have 2 marbles, person 2 can have 7 marbles, and person 3 can have 3 marbles with each marble being a different color. At first, I thought it was:</p>
<p>$$\binom{12}{1}\binom{11}{1}\binom{10}{1}$$</p>
<p>but I don't think it covers the case of each person having a different amount of marbles for each case. </p>
| aeyalcinoglu | 87,855 | <p>I'd like to add an answer which uses directly epsilon-delta definition.</p>
<p>Let $x \in \mathbb{R}$ and $x \neq 0$. We want to show that $f$ is not continuous at $x$. So we want to find an $\epsilon > 0$ such that for all $\delta > 0$ the following condition is not satisfied:
$$\forall x' \in \mathbb{R} \text{,} \ (|x-x'| < \delta \implies |f(x)-f(x')| < \epsilon)$$</p>
<p>So, if $x \in \mathbb{Q}$, you can choose $\epsilon = |x/2|$ because it doesn't matter how small you choose $\delta$, there will be always a irrational $x'$ in that interval and then $|f(x)-f(x')| = |x|$ which is always greater then $\epsilon$. Other case is similar.</p>
<p>And for continuity at $0$, you don't actually need the density of $\mathbb Q$, just choose $\delta$ as $\epsilon$.</p>
|
2,287,312 | <p>So the 12 marbles are distinct with each marble being a different color. Therefore I have to take into account, the different colors and different amount each person have. For example, person 1 can have 2 marbles, person 2 can have 7 marbles, and person 3 can have 3 marbles with each marble being a different color. At first, I thought it was:</p>
<p>$$\binom{12}{1}\binom{11}{1}\binom{10}{1}$$</p>
<p>but I don't think it covers the case of each person having a different amount of marbles for each case. </p>
| Ovi | 64,460 | <p>Let $a \in \mathbb{Q}$, $a \not = 0$ and select $\epsilon = \dfrac {|a|}{2}$. In order for the limit to exist, there must exist a $\delta$ such that $|f(x) - f(a)| < \dfrac {|a|}{2}$ whenever $x \in (a- \delta, a+\delta) - \{a\}$. Because of the density of irrationals, for any $\delta$ there exists an irrational $x_0$ such that $x_0 \in (a- \delta, a+\delta) - \{a\}$. But this implies that $|f(x_0) - f(a)| < \dfrac {|a|}{2} \implies |0-a| < \dfrac{|a|}{2}$, which is false.</p>
<p><a href="https://i.stack.imgur.com/BlsAV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BlsAV.png" alt=""></a>
The graph of $f$ is the blue line and the red line (with gaps in the of course). We can see that for any $a$, if we can find $\epsilon$ small enough so that the blue line is outside of $(f(a)-\epsilon, f(a) + \epsilon)$, we could show that there's no $\delta$ which satisfies the required proprety. </p>
<p>If $a=2$, we could pick any $\epsilon$ smaller than $2$, such as $1.5$. But we needed to find an $\epsilon$ that would work no matter what $a$ is, and $\dfrac a2$ does the work. But you need to keep in mind that $a$ could be negative; since $\epsilon>0$, we had to throw in the absolute value. </p>
|
1,536,632 | <p>I have the following system of equations </p>
<p>\begin{align}
\frac{dx}{d \tau} &= x \left(1-x-\frac{y}{x+b} \right) \\
\frac{dy}{d \tau} &= cy \left(-1+a\frac{x}{x+b} \right)
\end{align}</p>
<p>I am asked to show that if $a<1$, the only nonnegative equilibria are $(0,0), (1,0)$.</p>
<p>So first it is obvious that in order to the equations become $0$ is $(x,y)=(0,0)$ </p>
<p>Then I decided $y=0$ and $1-x-\displaystyle \frac{y}{x+b}=0 \Leftrightarrow x=1$ , hence $(x,y)=(1,0)$</p>
<p>In the same way I decided $x=0$ and $-1+a\displaystyle \frac{x}{x+b}=0$, but there is no $y$.</p>
<p>Finally I decided $-1+a\displaystyle \frac{x}{x+b}=0$ and $1-x-\displaystyle \frac{y}{x+b}=0$, and this is difficult.</p>
<p>I can't figure out what to do now. What about the fact that $a<1$?</p>
<p>Can anyone help? </p>
| Bernard | 202,857 | <p>Reducion modulo $61$ is a ring homomorphism, hence
$$\bigl(25^{1202}+3\bigr)^2=\bigl(5^{2404}+3\bigr)^2\equiv\bigl(5^{2404\bmod60}+3\bigr)^2=628^2\equiv 18^2=324\equiv19\mod61.$$</p>
|
2,496,211 | <p>I was wondering if you could help me with this proof I have for metric spaces.
I understand the theorem and the idea I just don't understand how to proceed.
I am attaching the question.</p>
<p>Here is how I think I should proceed: So, Let U = Q(i.e. rational numbers), Y = [0,1], and X = R(i.e. real numbers).</p>
<p>Am I doing it right?</p>
<p><a href="https://i.stack.imgur.com/chfNW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/chfNW.png" alt="enter image description here"></a></p>
| angryavian | 43,949 | <p>Assuming you meant $U = \mathbb{Q} \cap Y$, your example does not work, since $U$ is both not open in $Y$ and not open in $X$.</p>
<p>A counterexample can be made with $U=Y = \{0\}$ in $X=\mathbb{R}$.</p>
|
2,496,211 | <p>I was wondering if you could help me with this proof I have for metric spaces.
I understand the theorem and the idea I just don't understand how to proceed.
I am attaching the question.</p>
<p>Here is how I think I should proceed: So, Let U = Q(i.e. rational numbers), Y = [0,1], and X = R(i.e. real numbers).</p>
<p>Am I doing it right?</p>
<p><a href="https://i.stack.imgur.com/chfNW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/chfNW.png" alt="enter image description here"></a></p>
| bradipolpo | 329,956 | <p>Consider the metric space $X=\mathbb{R}$ and the closed subset $Y=[0,1]$. What are the open sets in $Y$? Are they all open in $X$?</p>
|
236,308 | <p>I solved the following equation the hard way:
$$\sqrt{x+1} +\sqrt{x+33}=\sqrt{x+6} +\sqrt{x+22}$$
The only solution is $x=3$.
I am wondering if there is some easy observation that solves the equation without squaring both sides?</p>
| John Bentin | 875 | <p><strong>Edit:</strong> My answer was wrong. Please refer to Matthew Conroy's correct answer instead. My version of his solution is $$x=\dfrac{(4s_2-s_1^2)^2-64s_4}{64s_3-8s_1(4s_2-s_1^2)},$$where $s_1, s_2, s_3, s_4$ are respectively the cubic, quadratic, linear, and constant coefficients of the polynomial expansion of $(x+a)(x+b)(x+c)(x+d)$.</p>
|
3,012,130 | <p><span class="math-container">$4, 15, 13, 7, 22, -1, 31, -9, 40, -17, 49$</span>.</p>
<p>What comes next? The answer is <span class="math-container">$-25$</span>, but why?</p>
| TurlocTheRed | 397,318 | <p>Break up the sequence into the even ordered terms and odd ordered.</p>
|
86,459 | <p>I have problems with the following logarimthic equation:</p>
<p>$$\log _a \left(\frac{x+\sqrt{x^2+5}}{5}\right) = b$$</p>
<p>How can I compute $ \log _a (x-\sqrt{x^2-5})$ in terms of $b$?</p>
| anon | 11,763 | <p>I assume you have a typo that is preventing others from answering. Let</p>
<p>$$b=\log_a\left(\frac{x+\sqrt{x^2-5}}{5}\right),\qquad \tilde{b}=\log\left(x-\sqrt{x^2-5}\right).$$</p>
<p>Now use the rules </p>
<ul>
<li>$\log(u)+\log(v)=\log(uv)$ </li>
<li>$(z-w)(z+w)=z^2-w^2$</li>
<li>$\log_a(1)=0$</li>
</ul>
<p>in order to add them together:</p>
<p>$$b+\tilde{b}=\log_a\left(\frac{\color{Red}{x}+\color{Blue}{\sqrt{x^2-5}}}{5}\cdot(\color{Red}{x}-\color{Blue}{\sqrt{x^2-5}})\right)$$
$$=\log_a\left(\frac{\color{Red}{x^2}-\color{Blue}{(x^2-5)}}{5}\right)=\log_a(5/5)=0,$$</p>
<p>hence $\tilde{b}=-b$, as you correctly surmised in the comments.</p>
|
129,693 | <p><img src="https://i.stack.imgur.com/tFha8.png" alt="enter image description here"></p>
<p>I'm lost on what's happening here. This is regarding MinML( "an idealized programming language" ) . More pics below: Thank You Very Much</p>
<p><img src="https://i.stack.imgur.com/BYjTF.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/env9T.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/GDo6I.png" alt="enter image description here"></p>
| Harry Peter | 83,346 | <p>$\int\dfrac{\sin x}{1+x^2}dx$</p>
<p>$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$</p>
<p>$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$</p>
<p>$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$</p>
<p>$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$</p>
<p>$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$</p>
<p>$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$</p>
<p>$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$</p>
<p>$=\int\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$</p>
<p>$=\int\left(\dfrac{\sinh1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$</p>
<p>$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$</p>
|
3,963,229 | <p>I tried to find this limit using polar representation <span class="math-container">$x=r \cos\theta $</span> and <span class="math-container">$y=r \sin \theta$</span>
i got something like <span class="math-container">$\ln (\frac{1}{\cos2 \theta})$</span> which makes me feel im in wrong way !</p>
<p>is my approach correct ?? and why alpha wolfram said its not exists if so</p>
| Piotr Piękos | 382,084 | <p>Before calculating you should get some intuition about how the function behaves in the neighbourhood of 0 and whether the limit will depend on the trajectory you choose to converge to zero.
What will be the limit of <span class="math-container">$(x_n, y_n)$</span> where <span class="math-container">$x_n = 0$</span> and <span class="math-container">$y_n = \frac{1}{n}$</span>?
What is the limit if <span class="math-container">$x_n = \frac{1}{n}$</span>, but <span class="math-container">$y_n = \frac{1}{2n}$</span>?
What happens if we try to do <span class="math-container">$x_n = y_n = \frac{1}{n}$</span>?
This should give you a hint of what's going on</p>
|
329,067 | <p>I have a question that is as follows:</p>
<blockquote>
<p>For each integer $n \geq 3$, construct a 3-regular graph on $2n$ vertices such that $G_n$ does not have any 3-cycles.</p>
</blockquote>
<p>Here is what I have:</p>
<p>I have $2n$ vertices numbered $1, 2, \ldots, 2n$, and a vertex $k$ connected to $k-1$, $k+1$, and $k+n$. (All three numbers are interpreted $\mod 2n$, so for instance given $n=5$, we would have edges 1-2, 2-3, 3-4, ..., 9-0, 0-1, and 1-6, 2-7, 3-8, 4-9, 5-0).</p>
<p>Now, to conclude my argument, how can I verify that there exist no 3-cycles in this graph?</p>
| Douglas S. Stones | 139 | <p>What you have described is an example of a <a href="http://en.wikipedia.org/wiki/Circulant_graph" rel="nofollow noreferrer">circulant graph</a>, and your method will pan out (as per Ross Millikan's answer).</p>
<p>I'd also like to add that there's examples that are not only $3$-cycle free, but have no odd length cycles (i.e., they're <a href="http://en.wikipedia.org/wiki/Bipartite_graph" rel="nofollow noreferrer">bipartite graphs</a>).</p>
<p>If we label the vertices $\{u_0,u_1,\ldots,u_{n-1}\} \cup \{v_0,v_1,\ldots,v_{n-1}\}$ and add an edge from each $u_i$ to $v_i$, $v_{i+1}$ and $v_{i+2}$ (with indices modulo $n$), then we obtain a $3$-regular bipartite graph. (The vertex $v_i$ is adjacent to $u_i$, $u_{i-1}$ and $u_{i-2}$, so it is indeed $3$-regular.)</p>
<p>An example when $n=5$ is given below:</p>
<p><img src="https://i.stack.imgur.com/h7l5x.png" alt="Example when $n=5$"></p>
<p>The $2n$-vertex circulant graph examples will have $(n+1)$-cycles, and $n+1$ might be an odd number. For example, when $n=4$ we have a $5$-cycle illustrated below:</p>
<p><img src="https://i.stack.imgur.com/Be8mi.png" alt="$5$-cycle in the circulant graph example"></p>
<p>There's also another "cheats" way to answer the question. Since the question doesn't specify that the graphs be connected, we can find examples $G_3,G_4,G_5$ for $n=3,4,5$, respectively, then we obtain examples in all cases by taking:</p>
<ul>
<li><p>An arbitrary number of disconnected copies of $G_3$,</p></li>
<li><p>The graph $G_4$ together with an arbitrary number of disconnected copies of $G_3$, and</p></li>
<li><p>The graph $G_5$ together with an arbitrary number of disconnected copies of $G_3$.</p></li>
</ul>
<p>Here's a $3$-regular graph on $18$ vertices with no $3$-cycles made from $3$ disconnected copies of $K_{3,3}$:</p>
<p><img src="https://i.stack.imgur.com/Izdbd.png" alt="Disconnected copies of $K_{3,3}$"></p>
|
4,625,921 | <p><a href="https://i.stack.imgur.com/mIKMW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mIKMW.png" alt="Two polar curves." /></a></p>
<p>I am looking for the area outside the circle <span class="math-container">$r = 3\cos\Theta$</span> and inside the limaçon <span class="math-container">$r = 1 + \cos\Theta$</span>:</p>
<p><span class="math-container">$$A = \frac{1}{2}\int_{\frac{\pi }{3}}^{\frac{5\pi }{3}}\left[(1+\cos\Theta )^{2}-(3\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = -2\pi \,\text{units}^{2} $$</span></p>
<p>If the opposite situation is used, I simply reverse the limits and curves. But since the lower limit must be smaller, I use coterminal angles to change one of the limits. Here, after the reversal I changed 5pi/3 to -pi/3</p>
<p><span class="math-container">$$A = \frac{1}{2}\int_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\left[(3\cos\Theta )^{2}-(1+\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = pi \,\text{units}^{2} $$</span></p>
<p>I can't understand why the area I'm getting is negative for the first situation. I get that maybe I have the limits or the curves in reverse but I'm following the correct limits and placement of the curves. What is the general rule for determining the curves and limits so that the result is always positive?</p>
<p>I followed the convention that the outer curve in the graph is the first radius used and the inner curve is the second radius. I've also made sure that the limits start and end appropriately based on the graph. It worked for the opposite situation, as shown in this question.</p>
<p><a href="https://math.stackexchange.com/questions/3519906/find-the-area-inside-r-3-cos-theta-and-outside-r-1-cos-theta">Find the area inside $r=3\cos\Theta $ and outside $r = 1+\cos\Theta $</a></p>
<p>I've seen other problems where that rule works. I'm just wondering why it doesn't work here.</p>
| David Quinn | 187,299 | <p>Hint…consider the region you require to find as two equal parts and work out the upper region.</p>
<p>Draw a straight line from the pole to the point of intersection in the first quadrant.</p>
<p>The upper limits are not the same for both integrals.</p>
<p>The region inside the circle has upper limit <span class="math-container">$\frac{\pi}{2}$</span> but the region inside the cardioid has upper limit <span class="math-container">$\pi$</span>.</p>
<p>So you need to do two separate integrals and subtract them in the correct order - cardioid minus circle.</p>
<p>Then double the result to get the total.</p>
|
33,907 | <p><strong>Note:</strong> Cross-posted at <a href="http://community.wolfram.com/groups/-/m/t/137895?p_p_auth=8QnKtT9I" rel="nofollow">http://community.wolfram.com/groups/-/m/t/137895?p_p_auth=8QnKtT9I</a>.</p>
<p>I am to design a two step gearbox. The first step is to choose the number of teeth in each cog wheel in order to achieve a gear ratio of 17.3. In other words:</p>
<pre><code> (N1 N2)/(n1 n2) == 17.3
</code></pre>
<p>where N1 and N2 are the number of teeth in gear 1 and 2, and n1 and n2 are the number of teeth in pinion 1 and 2. Is it possible to get <em>Mathematica</em> to "guess" the lowest number of teeth possible and still get as close as possible to 17.3? The number of teeth in the pinions must not be lower than 20. In addition, the number of cog teeth in the gears must not be divisible, i.e. only common factor has to be 1.</p>
| Daniel Lichtblau | 51 | <p>I think in general you might have to enforce the gcd (divisibility) constraint after the fact. For this particular example I get a result on the first try that has those numbers relatively prime, so we can accept that solution.</p>
<p>Anyway, here is the code. For objective function I sum the teeth and add a penalty term for that ratio straying (discrepancing itself?) from 17.3.</p>
<pre><code>obj = c1 + c2 + p1 + p2 + (c1*c2 - 17.3*p1*p2)^2;
cs1 = {p1 >= 20, p2 >= 20, c1 >= 2, c2 >= 2};
In[368]:= {min, vals} =
NMinimize[{obj,
Flatten[{cs1, Element[{c1, c2, p1, p2}, Integers]}]}, {c1, c2, p1,
p2}]
Out[368]= {223., {c1 -> 91, c2 -> 76, p1 -> 20, p2 -> 20}}
</code></pre>
<p>As 91 is prime, it is of course relatively prime to 76, so this set of values appears to suit the requirements.</p>
|
3,240,240 | <blockquote>
<p>If a right circular cone has three mutually perpendicular generators
then find its semi-vertical angle.</p>
</blockquote>
<p>We see that if <span class="math-container">$ax^2+by^2+cz^2+2fyx+2gzx+2hxy=0$</span> has three mutually perpendicular generators, then <span class="math-container">$a+b+c=0$</span>. But I don't know what will be the way to find the semi-vertical angle.</p>
<p><strong>Added:</strong> </p>
<p>Let us consider three mutually perpendicular generators with direction cosines <span class="math-container">$l_i,m_i,n_i$</span> for <span class="math-container">$i=1,2,3$</span>. The direction cosines of the axis are <span class="math-container">$\frac{\sum l_1}{3},\frac{\sum m_1}{3},\frac{\sum n_1}{3}=l',m',n'$</span> (say) Since these three generators are mutually perpendicular, we have<br>
<span class="math-container">$l_il_j+m_im_j+n_in_j=0$</span> for <span class="math-container">$i\neq j$</span>. Also we can say that <span class="math-container">$l_i^2+m_i^2+n_i^2=1$</span> for <span class="math-container">$i=1,2,3$</span>. </p>
<p>From the above relation, we have<br>
<span class="math-container">$l_1m_1+l_2m_2+l_3m_3=0$</span> etc.
<span class="math-container">$\cos\alpha=\frac{l_1l'+m_1m'+n_1n'}{\sqrt{l'^2+m'^2+n'^2}}=\frac{1}{\sqrt{3}}\implies \alpha=\tan^{-1}(\sqrt{2})$</span>. Is my approach correct?</p>
| Chandrasenachary | 731,992 | <p>Assume 3 mutually perpendicular generators as
coordinate axes
The axis of circular cone makes equal angles with the axes
if <span class="math-container">$(l,m,n )$</span> are DC's of axes then
<span class="math-container">$$l=m=n=\cos\theta=\left(\frac13\right)^\frac12$$</span>
<span class="math-container">$$\tanθ=(2)^\frac12\implies θ=\arctan(2^\frac12)$$</span></p>
|
143,173 | <p>I have a small question that I think is very basic but I am unsure how to tackle since my background in computing inequalities is embarrassingly weak - </p>
<p>I would like to show that, for a real number <span class="math-container">$p \geq 1$</span> and complex numbers <span class="math-container">$\alpha, \beta$</span>, I have
<span class="math-container">\begin{equation}
|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)
\end{equation}</span></p>
<p>I thought it would be best to rewrite this as
<span class="math-container">\begin{equation}
\left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2}
\end{equation}</span></p>
<p>but then I am unsure what to do next - is this a sensible start anyways ? Any help would be great !</p>
<p>(P.S. this is not a homework question - I am currently trying to brush up my knowledge of <span class="math-container">$L^p$</span> spaces, and this inequality came up as a statement. I thought it might be worthwhile to make sure I can fill in the gaps to improve my skills in computing inequalities.)</p>
| Community | -1 | <p>Without loss of generality, we can assume that that $\lvert \alpha \rvert \geq \lvert \beta \rvert$. Now essentially you want to prove that $$\displaystyle \left \lVert \frac{z + 1}{2} \right \Vert^p \leq \displaystyle \frac{\lVert z \rVert^p + 1}{2},$$ where $\displaystyle z = \frac{\alpha}{\beta}$ and $\lVert z \rVert \geq 1$.</p>
<p>Note that $$ \displaystyle \left \lVert \frac{z + 1}{2} \right \Vert \leq \frac{\lVert z \rVert + 1}{2}$$</p>
<p>So if we prove that $\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2}$, where $t = \lVert z \rVert \geq 1$, we are done.</p>
<p>Now this is a one-variable calculus problem. Consider $f(t) = \displaystyle \frac{1+t^p}{2} - \left(\frac{1+t}{2} \right)^p$. We then get that $$\displaystyle f'(t) = \frac{p t^{p-1}}{2} - \frac{p (1+t)^{p-1}}{2^p}.$$ Hence, $$f'(t) = \frac{p}{2} \left( t^{p-1} - \left(\frac{1+t}{2} \right)^{p-1}\right) \geq 0$$ for $t \geq 1$. Hence, $f(t)$ is increasing for all $t \geq 1$. And $f(1) = 0$. Hence, we have that $$f(t) \geq f(1) = 0.$$ Hence, we get that $$\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2},$$ where $t = \lVert z \rVert \geq 1$</p>
|
57,057 | <p>Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$</p>
| user782220 | 21,556 | <p>Replace with reciprocals so the problem becomes</p>
<p>$a + b + c + d + e + {33 \over 2}{1 \over {1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e}} \geq {83 \over 10}$</p>
<p>Sort the numbers so that $a\le b\le c\le d\le e$ and assume that $a,b,c,d,e$ are not all equal to $1$. Since $a<1$ and $e>1$ we have
$(a+b+c+d+e) - (1+b+c+d+ea) = a+e-1-ae = (e-1)(1-a) > 0$</p>
<p>Let $x_1=\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{e}$ and $x_2=\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}$</p>
<p>$x_1 - x_2 = \frac{1}{a}+\frac{1}{e}-\frac{1}{ea} - 1 = \frac{(e-1)(1-a)}{ea} >0$</p>
<p>The geometric harmonic inequality says $(bcdea)^{1/5}\ge \frac{5}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}} $ and thus using $abcde=1$ we conclude $\frac{1}{5}\ge \frac{1}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}}$. Thus we have that $x_2\ge 5$.</p>
<p>The mean value theorem applied to the function $f(x)=1/x$ gives</p>
<p>$\frac{1}{x_1}-\frac{1}{x_2} = f'(\theta)(x_1-x_2)$ where $x_2 \le\theta\le x_1$. This tells us that since $\theta\ge x_2\ge 5$ that $f'(\theta)\ge f'(x_2) \ge f'(5) = -\frac{1}{5^2}$</p>
<p>Since $bcde=1/a$ we cannot have $b,c,d,e$ all less than $1/a^{1/4}$. As $e$ is the largest of them we must have $e\ge 1/a^{1/4}$ and since $0\le a\le 1$ we can conclude $ea\ge a^{3/4}\ge 1$.</p>
<p>Now putting all the computations above together gives
$\left(a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}}\right) - \left(1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\right) =$
$(e-1)(1-a) + \frac{33}{2}\left(\frac{1}{x_1}-\frac{1}{x_2}\right)= (e-1)(1-a) + \frac{33}{2}f'(\theta)(x_1-x_2) =$
$(e-1)(1-a) + \frac{33}{2}f'(\theta)\frac{(e-1)(1-a)}{ea}\ge (e-1)(1-a) - \frac{33}{50}\frac{(e-1)(1-a)}{ea}\ge $</p>
<p>$(e-1)(1-a) - \frac{33}{50}(e-1)(1-a) = \frac{27}{50}(e-1)(1-a)\ge 0$</p>
<p>Thus if we replace $a,b,c,d,e$ with $1,b,c,d,ea$ the left hand side of the inequality decreases while still maintaining $abcde=1$. We sort the numbers $1,b,c,d,ea$ and choose the smallest and largest among them and repeat the process we have just described. Eventually all five numbers will become 1 thus showing</p>
<p>$a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}} \ge 1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\ge \ldots \ge$</p>
<p>$1+1+1+1+1+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{83}{10}$</p>
|
2,848,251 | <p>From: Philip Johnson-Laird <a href="https://dof.princeton.edu/about/clerk-faculty/emeritus/philip-nicholas-johnson-laird" rel="nofollow noreferrer">BA PhD Psychology (UCL)</a>, Stuart Professor of Psychology Emeritus at Princeton. (Author isn't a logician.) <a href="https://www.amazon.ca/How-We-Reason-Philip-Johnson-Laird/dp/0199551332" rel="nofollow noreferrer"><em>How We Reason</em> (1st edn 2008)</a>. p. 44.</p>
<blockquote>
<p> For premises that have several models, a single model consistent with
them—an example—establishes a conclusion about what is possible; but
all the models must be examples to establish a conclusion about what is
necessary. The opposite is the case for refutations: a single counterexample
refutes a conclusion about what is necessary, whereas all models must be
counterexamples to refute a conclusion about what is possible. And so we
should draw a conclusion about what’s possible faster than a conclusion about
what’s necessary, but we should draw a conclusion about what’s not necessary
faster than a conclusion about what’s not possible. This prediction is crucial for
a theory that takes possibilities as fundamental, and several experiments have</p>
</blockquote>
<p>p. 45</p>
<blockquote>
<p>corroborated it. One caveat, learned the hard way, is that it’s no use asking
adults whether something is possible if it's [Author's bungle. This ought be "its".] necessity is obvious. They baulk
at saying, “yes”, in these circumstances; and so inferences have to be hard
enough that when an event is possible its necessity isn’t obvious too. Here is
a typical trial from an experiment. The premises are about a game of one-on-one basketball in which two players take part:</p>
<blockquote>
<p>[1.] If Allan is in [the game] then Betsy is in.<br>
[2.] If Carla is in then David is out. </p>
</blockquote>
<p>The participants were more accurate and faster to infer that Betsy could be in
the game than to infer that she must be in the game. <strong>If you list the possible
games compatible with the premises, you’ll discover that there are three games
and that Betsy is in all of them: Allan versus Betsy, Betsy versus Carla, Betsy
versus David.</strong> </p>
</blockquote>
<p>I refer to the players by only their initials. </p>
<ol start="5">
<li><p>Why are A vs. C and A vs. D impossible? </p></li>
<li><p>How can you efficiently deduce the emboldened sentence, without applying 1 and 2 to each person separately in order (1st: Alan v. B, C, D. 2nd: B v. C, D. 3rd: C v. D)? </p></li>
</ol>
| Eric Wofsey | 86,856 | <p>A vs C and A vs D are impossible because A is in the game and B is not, violating [1].</p>
<p>If B is not in the game, then A cannot be in the game either without violating [1]. But then the players must be C and D, which violates [2]. So B must be in the game. All combinations with B are easily seen to be possible (any combination with B will satisfy [1], and to violate [2] you would have to have both C and D).</p>
|
3,295,973 | <p>Let <span class="math-container">$(\Omega,\mathcal A,\mu)$</span> be a measure space, <span class="math-container">$p,q\ge1$</span> with <span class="math-container">$p^{-1}+q^{-1}=1$</span> and <span class="math-container">$f:\Omega\to\mathbb R$</span> be <span class="math-container">$\mathcal A$</span>-measurable with <span class="math-container">$$\int|fg|\:{\rm d}\mu<\infty\;\;\;\text{for all }g\in L^q(\mu)\tag1.$$</span> By <span class="math-container">$(1)$</span>, <span class="math-container">$$L^q(\mu)\ni g\mapsto fg\tag2$$</span> is a bounded linear fuctional and hence there is a unique <span class="math-container">$\tilde f\in L^p(\mu)$</span> with <span class="math-container">$$(f-\tilde f)g=0\;\;\;\text{for all }g\in L^q(\mu)\tag3.$$</span></p>
<blockquote>
<p>Can we conclude that <span class="math-container">$f=\tilde f$</span>?</p>
</blockquote>
<p><strong>EDIT</strong>: As we can see from <a href="https://math.stackexchange.com/a/3223523/47771">this answer</a>, we need to impose further assumptions; but which do we really need?</p>
| Cornman | 439,383 | <p>Keep in mind, that 'squaring' is not an equivalent term transformation.
It manipulates the domain of your equation and can therefor create 'fake solutions'.</p>
<p>So either you have to be more careful, when squaring, or you have to check every solution if it is fake or not. (Which means that the original equation holds true for this input)</p>
<p>In this case <span class="math-container">$x=-2$</span> is a fake solution and <span class="math-container">$x=1$</span> is a real solution to your equation.</p>
|
3,887,526 | <p>I need to prove <span class="math-container">$\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$</span>. I've tried to use triangle inequality or to explore the form of <span class="math-container">$(a+b+c)^2$</span> but it won't get me anywhere. I would be grateful for some suggestions.</p>
| Calvin Lin | 54,563 | <p>We will prove a stronger statement.</p>
<p>Lemma: <span class="math-container">$f(x) = \frac{x}{1+x}$</span> is an increasing function on <span class="math-container">$ x \geq 0$</span>.<br />
This is obvious by setting <span class="math-container">$ f(x) = 1 - \frac{1}{1+x}$</span>, which is increasing on <span class="math-container">$ x \geq -1$</span>.</p>
<p>Lemma: <span class="math-container">$ |x+y+z| \leq |x| + |y| + |z| $</span>.<br />
This is obvious by the basic properties of absolute values.</p>
<p>Corollary:</p>
<p><span class="math-container">$$ \frac{ |x+y+z| } { 1 + |x+y+z|} \leq \frac{|x|+|y|+|z| } { 1 + |x| + |y| + |z| } \leq \sum \frac{ |x| } { 1 + |y| + |z| }$$</span></p>
|
351,815 | <p>Having trouble understanding this. Is there anyway to prove it?</p>
| Sijo Joseph | 72,833 | <p>Why does the function $exp(x)$ converge?</p>
<p>Since </p>
<p>$$\exp(x)=\sum_{i=0}^{\infty} \frac{x^{n}}{n!}$$ for large $n$, $x^n$ grows slowly compared with $n!$.</p>
|
351,815 | <p>Having trouble understanding this. Is there anyway to prove it?</p>
| ramo | 113,068 | <p>$n!> k^n$ if $n\ge ke$</p>
<p>try it for yourself with various combinations of $n$ and $k$</p>
<p>use $\log(n!)$ and $k \log(n)$ for large $n$</p>
|
2,028,594 | <p>Given a simple undirected connected graph. What is the minimum number of edge-disjoint paths needed to cover all edges of the graph? I have only found information about the vertex-covering one.</p>
<p>My friend suggests that such value equals <em>u/2</em>, with <em>u</em> being number of vertices with odd degree, or 1 if there is no such vertices.</p>
<p>Can anyone confirm if his assumption is correct or not? </p>
| hmakholm left over Monica | 14,366 | <p>Assuming that the "paths" we're talking about are actually trails, that is, walks that may repeat vertices but may <em>not</em> repeat edges (which seems to be the most natural context for requiring them to be edge-disjoint), that sounds right to me.</p>
<p>Clearly <em>at least</em> that many paths will be needed, and it's not too difficult to see that you can construct a covering at set of exactly that many paths greedily:</p>
<ol>
<li><p>Start at an odd-degree vertex, and walk along random unused edges until you there's nowhere you can go. At that time you must have reached another odd degree vertex.</p></li>
<li><p>Continue doing this as long as there are odd-degree vertices left.</p></li>
<li><p>If there are still unused edges in the graph, locate a vertex that is touched by both used and unused edges. Break one of the paths you have already identified through that vertex, and start walking randomly along unused edges from there. Since you've used up all of the odd vertices, the only place you <em>can</em> end is at the vertex where you started, so you can then proceed along the rest of the original path, so the number of paths doesn't increase.</p></li>
<li><p>Repeat until all edges have been covered.</p></li>
</ol>
|
786,271 | <p>Suppose $a, b$ and $n$ are positive integers, all greater than one. If $a^n+b^n$
is prime, what can you relate $n$ with 2?</p>
<p>My approach: for $a^n+b^n$ to be prime $\forall n>1$, $a$ and $b$ has to be coprimes.
But how do I ascertain anything about $n?$</p>
| JEM | 47,652 | <p>Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $\pm\infty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=c\sqrt{x}$ for $c\in\mathbb{R}$ near the origin. Then the differential equation near the origin becomes $y'=\frac{1}{\sqrt{x^2+c^2x}}\approx\frac{1}{c\sqrt{x}}$ and so $y(x)\approx \sqrt{x}/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root. </p>
|
389,269 | <p>Let <span class="math-container">$G$</span> be a nonabelian finite simple group all of whose Sylow subgroups of odd order are cyclic.</p>
<p>If we further assume that its Sylow <span class="math-container">$2$</span>-subgroup is dihedral, then due to Suzuki, we know that <span class="math-container">$G\cong \operatorname{PSL}(2,p)$</span> for a prime <span class="math-container">$p>3$</span>.</p>
<p>Without any further assumption, what is the list of finite nonabelian simple groups all of whose Sylow subgroups of odd order are cyclic?</p>
<p>Secondly, is there any reference for that not appealing to the full classification of finite simple groups?</p>
| Noam D. Elkies | 14,830 | <p>Partial answer:</p>
<p>Any group whose order is a power of <span class="math-container">$2$</span> times a squarefree number
must satisfy this condition. This condition holds for some of
Suzuki's groups <span class="math-container">${\rm Sz}(2^k)$</span> (with <span class="math-container">$k \geq 3$</span> odd),
which have order <span class="math-container">$2^{2k} (2^{2k} + 1) (2^k - 1)$</span>; and it seems from
<a href="https://en.wikipedia.org/wiki/Suzuki_groups#Conjugacy_classes" rel="noreferrer">https://en.wikipedia.org/wiki/Suzuki_groups#Conjugacy_classes</a>
that even the Suzuki groups whose orders do have odd square factors
(such as <span class="math-container">${\rm Sz}(2^5)$</span>, of order <span class="math-container">$2^{10} \, 5^2 \, 31 \!\cdot\! 41$</span>)
still have all their odd-order Sylow groups cyclic.</p>
<p>Another example is the first Janko group <span class="math-container">$J_1$</span>,
of order <span class="math-container">$2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 19$</span>.
The appearance of a sporadic group suggests that
the full answer will require at least some nontrivial part of
the classification theorem.</p>
|
910,020 | <p>Show that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$.</p>
<p><strong>Attempt:</strong> $|G|=168=2^3.3.7$</p>
<p>Then number of sylow $7$ subgroups in $G = n_7 = 1$ or $8$.</p>
<p>Given that $H$ is a normal subgroup of order $4$ in $G$. </p>
<blockquote>
<p>If we prove that $n_7$ cannot be $8$, then $n_7=1$ and as a result, the sylow $7$ subgroup $K$ is normal.Hence, $HK$ will also be a normal subgroup of $G$ and since, $H \bigcap K = \{e\} \implies |HK|=28$.</p>
</blockquote>
<p>Now, suppose $n_7=8$ and hence, $K_1 \cdots K_8$ are the $8$ cyclic subgroups of order $7$.</p>
<p>Each $K_i$ has $\Phi(7)=6$ elements of order $7$ . Hence, total elements of orders $=7$ in the $K_i's$ are $6.8=48$</p>
<p>How do I move forward and bring a contradiction somewhere?</p>
<p>Thank you for your help. </p>
| Morten Krogh | 185,971 | <p>You can show it by using the quotient group as Andrea does, but you can also directly show that $n_7=1$.</p>
<p>A 7-group acts on the normal 4-group by conjugation. This action is an
automorphism. The automorphism group of 4-groups have order 2 or 6 and hence a
7-seven group must act trivially. This means that the 7-group commutes with the 4-group.</p>
<p>You could show the same by looking at the 28 group $P_4P_7$. By Sylow, there is
one normal 7-group. The 4-group was assumed normal in G. Hence the 28-group is the direct product and the 4-group and 7-group commute.</p>
<p>The 4-group and $P_7$ are then both contained in $N_G(P_7)$ which must have index at most $\frac{168}{28}=6$. There can then be at most be 6 conjugates of the 7-group which rules out $n_7=8$.</p>
<p>Also note that the result of a normal 28-group implies one normal 7-group.
There is one 7-group in a 28-group by Sylow and it is hence characteristic.
A characteristic 7-group in a normal 28-group is normal in G.</p>
|
338,383 | <p>Multivariate polynomial indexed by <span class="math-container">${1, \ldots, n}$</span> are acted on by <span class="math-container">$S_n$</span>: for <span class="math-container">$\sigma \in S_n$</span>, define <span class="math-container">$\sigma(x_i) = x_{\sigma(x_i)}$</span>, etc. Symmetric polynomials are those polynomials which are invariant under this action.</p>
<p>If we consider just a subgroup of <span class="math-container">$S_n$</span>, a polynomial which is not symmetric my be invariant. An example: Let <span class="math-container">$S_5$</span> be the symmetric group of degree 5, <span class="math-container">$\tau = (1\ 2\ 3\ 4\ 5)$</span>, and <span class="math-container">$C_5 = \{\tau^n : n = 1, ..., 5\}$</span>. The polynomial <span class="math-container">$$f(x_1, x_2, x_3, x_4, x_5) = x_1x_3 + x_2x_4 + x_3x_5 + x_4x_1+x_5x_2$$</span> is invariant under the elements of <span class="math-container">$C_5$</span>, but not under the elements of <span class="math-container">$S_5$</span>.</p>
<p>I was wondering what sort of theory exists around these sets of polynomials? What are they called?</p>
<p>Also, Every symmetric polynomial can be expressed as some polynomial evaluated on the <a href="https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Fundamental_theorem_of_symmetric_polynomials" rel="noreferrer">elementary symmetric polynomials</a>. Is there some systematic way of generating some set of polynomials which will have this property?</p>
| Sam Hopkins | 25,028 | <p>If you're interested specifically in the cyclic group you may be interested in the "cyclic quasi-symmetric functions" studied recently in this paper of Adin, Gessel, Reiner, and Roichman: <a href="https://arxiv.org/abs/1811.05440" rel="nofollow noreferrer">Cyclic quasi-symmetric functions</a>. These "cyclic quasi-symmetric functions" are not exactly the same as polynomials invariant under the action of the long cycle in <span class="math-container">$S_n$</span> (for instance, they are formal power series rather than polynomials, and also they also have to be "quasi-symmetric functions" in addition to enjoying a cyclic symmetry condition). But the theory here closely parallels the usual theory of symmetric functions (and the theory of quasi-symmetric functions) and does involve a kind of cyclic symmetry.</p>
<p>By the way, the cyclically symmetric polynomials (in your sense) were also discussed in this question of Jim Propp: <a href="https://mathoverflow.net/questions/125454/cyclically-symmetric-functions">Cyclically symmetric functions</a>.</p>
|
2,819,950 | <p>I am a university graduate with a B.S. in mathematics who has been developing software for the past 8 years. I recently discussed a mutual interest in topology with a friend who is just about to complete his degree. Due to time constraints both of us missed our chance to take a topology course during our undergraduate studies, so we have decided to make an independent study of the subject once he finishes school, and are looking for a book to guide us.</p>
<p>Ideally I am looking for a book that...</p>
<ul>
<li>is suitable for our math background</li>
<li>is well laid out and not too difficult to follow (i.e. is suited to self study)</li>
<li>does not assume prior knowledge</li>
<li>is thorough enough to enable future study of topics within the field</li>
<li>contains plenty of exercises</li>
<li>isn't sparse on diagrams where they are appropriate</li>
<li>doesn't waste much time on overly-specific material (I'm the type that likes to prove things about the determinant, <em>not calculate thousands of them</em>)</li>
</ul>
<h2>What topology texts would be appropriate for our self study?</h2>
| Adrian Keister | 30,813 | <p>I'd recommend Crossley's <a href="https://smile.amazon.com/Essential-Topology-Springer-Undergraduate-Mathematics/dp/1852337826/ref=sr_1_1?ie=UTF8&qid=1529007801&sr=8-1&keywords=crossley%20essential%20topology" rel="nofollow noreferrer"><em>Essential Topology</em></a>. It's pitched at a great level, and very easy to read. It covers the basics of topological spaces, homotopy, homotopy groups, simplicial homology, and singular homology.</p>
|
2,066,765 | <p>Are there any examples of subspaces of $\ell^{2}$ and $\ell^{\infty}$ which are not closed?</p>
| Community | -1 | <p>Yes. For instance, the sequences which are eventually $0$.</p>
|
981,181 | <p>I'm stuck with a logic problem like this</p>
<blockquote>
<p>I eat ice cream if I am sad.</p>
<p>I am not sad.</p>
<p>Therefore I am not eating ice cream.</p>
</blockquote>
<p>Is this conclusion logical? The first sentence can be understood both like "ice cream <span class="math-container">$\implies$</span> sad" and vice versa. He stated that he is not sad but does that not mean that he is not eating ice cream ? I'm confused.</p>
| Sebastian Negraszus | 103,176 | <p>In logic, there is a difference between <em><a href="http://en.wikipedia.org/wiki/Logical_consequence" rel="nofollow">implication</a></em> ($a \implies b$) and <em><a href="http://en.wikipedia.org/wiki/Logical_equivalence" rel="nofollow">equivalence</a></em> ($a \iff b$). The word "if" (without "only") usually means the first one:</p>
<p>"I am sad" $\implies$ "I eat ice cream"</p>
<p>If "I am sad" is false, we cannot logically say anything about "I eat ice cream". It may be true or false. This makes sense, because being sad is not the only reason for eating ice cream.</p>
<p>See also: <a href="http://en.wikipedia.org/wiki/Denying_the_antecedent" rel="nofollow">Denying the antecedent</a></p>
|
73,219 | <p>For Riemann surfaces there are at least to possible notions of hyperbolicity. The classical one given by the Uniformization Theorem, or equivalently the type problem, which essentially says that a simply connected Riemann surfaces is conformally equivalent to one of the following:</p>
<ul>
<li>Riemann Sphere $\mathbb{C}\cup\{\infty\}$ (elliptic type).</li>
<li>Complex plane (parabolic type).</li>
<li>Open unit disk (hyperbolic type).</li>
</ul>
<p>On the other hand, given a Riemann surface one can asks if it is hyperbolic in the Gromov's sense. In other words, does there exists $\delta>0$ such that all the geodesic triangles in the surface are $\delta$-thin? </p>
<p>It seems to me that this two notions of hyperbolicity are not equivalent and one can have counterexamples in both directions. For instance, the two dimensional torus $\mathbb{T}^2$ is hyperbolic in Gromov's sense (since it is compact), but it's also a quotient of the Euclidean plane by a free action of a discrete group of isometries and therefore, of parabolic type. </p>
<p>My questions are: what is a sufficient condition for a surface of hyperbolic type to be Gromov's hyperbolic? what is known about the relation of these two notions? </p>
<p><strong>Related Question:</strong> Let $G$ be an infinite planar graph with uniformly bounded degree and assume that the simple random walk is transient. Is the graph necessarily Gromov's hyperbolic? </p>
| Alain Valette | 14,497 | <p>About the related question: it is a result of Babai that a (connected, locally finite) vertex-transitive, planar graph is isomorphic to the 1-skeleton of an Archimedean tiling of the sphere, or the Euclidean plane, or the hyperbolic plane. So assuming transience singles out the hyperbolic plane, and implies Gromov-hyperbolicity for the graph. See
<a href="http://www.cs.uchicago.edu/files/tr_authentic/TR-2001-04.ps" rel="nofollow">http://www.cs.uchicago.edu/files/tr_authentic/TR-2001-04.ps</a></p>
|
180,495 | <blockquote>
<p>Suppose $T$ is an everywhere defined linear map from a Hilbert space $\mathcal{H}$ to itself. Suppose $T$ is also symmetric so that $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y\in\mathcal{H}$. Prove that $T$ is a bounded directly from the uniform boundedness principle and not the closed graph theorem.</p>
</blockquote>
<p>This is problem III.13 in the Reed-Simon volume 1. Hints are welcome.</p>
| user66081 | 66,081 | <p>Both proofs are valid for a linear operator $T : X \to X'$ on a Banach space $X$, that is symmetric in the sense that $\langle T x, y \rangle = \langle T y, x \rangle$, $x, y \in X$, where $\langle \cdot, \cdot \rangle$ is the duality pairing.</p>
|
4,278,202 | <p>Suppose, I am working on <span class="math-container">$Eq(X)$</span> = the set of all equivalence relations on <span class="math-container">$X$</span>. Let <span class="math-container">$\theta_1, \theta_2$</span> denote arbitrary equivalence relations.</p>
<p>Then how does <span class="math-container">$\theta_1 \lor \theta_2$</span> look like? And what about <span class="math-container">$\theta_1 \land \theta_2$</span>?</p>
<p>I know lattices and that these usually correspond to supremum and infimum. But I cannot wrap my head around what does this mean for equivalence relations.</p>
<p>What does <span class="math-container">$\theta_1 \lor \theta_2$</span> mean for arbitrary <span class="math-container">$a, b \in \theta_1 \land \theta_2$</span>?</p>
<p>Thank you.</p>
| Asinomás | 33,907 | <p>Recall that an equivalence relation over <span class="math-container">$X$</span> is a subset of <span class="math-container">$X\times X$</span>. In this case the order is containment.</p>
<p>The meet <span class="math-container">$\land$</span> is going to be the intersection as it usually works out. Note that the intersection of the relations is symmetric, transitive and reflexive.</p>
<p>The join <span class="math-container">$\lor$</span> is going to be be harder to describe, but it's essentially going to be the transitive closure of the union, since transitivity is the thing that can break when you take the union.</p>
|
3,468,537 | <p>I tried to proof this but I'm not sure if it's fine or if I'm missing something. Any help or hints are appreciated.
This is what I got:</p>
<p>By hypothesis we know that <span class="math-container">$p||G|$</span> then <span class="math-container">$|G|=pm$</span>. Supposing that <span class="math-container">$p|m$</span> then we can factorize the prime p out of <span class="math-container">$m$</span> such that <span class="math-container">$|G|=p^{r}n$</span> and <span class="math-container">$p\nmid{n}$</span>. By the Sylow theorem exists <span class="math-container">$H\leq{G}$</span> of order <span class="math-container">$p^{r}$</span>. Suppose there exists another p-subgroup <span class="math-container">$K\leq{G}$</span> which elements have powers of a prime. It's clear that <span class="math-container">$K\leq{H}$</span>. Sylow theorem also tells us that any two p-subgroups are conjugate <span class="math-container">$H=gKg^{-1}$</span>, but remembering that <span class="math-container">$G$</span> is an abelian group so all of its subgroups must be abelian aswell. Then we get <span class="math-container">$H=K$</span> and there is only one Sylow p-group.</p>
<p>In the case <span class="math-container">$p\nmid{m}$</span> we have a Sylow p-group of order <span class="math-container">$p$</span> and we follow the previous steps to get the same result.</p>
| Alex Mathers | 227,652 | <p>You don't have to make this distinction about whether <span class="math-container">$p$</span> divides <span class="math-container">$m$</span> or not; the whole point here is that you can define</p>
<p><span class="math-container">$$H:=\{g\in G\mid\text{ord$(g)$ is a power of $p$}\}$$</span></p>
<p>and under the assumption that <span class="math-container">$G$</span> is abelian, you can prove this is a subgroup of <span class="math-container">$G$</span>. Then you'll find that <span class="math-container">$|H|$</span> is a power of <span class="math-container">$p$</span>, so <span class="math-container">$H$</span> must be contained in some Sylow <span class="math-container">$p$</span>-subgroup <span class="math-container">$P$</span>. But using Lagrange's theorem you can easily prove that <span class="math-container">$P\subseteq H$</span>, so in fact we must have <span class="math-container">$H=P$</span>.</p>
|
766,765 | <blockquote>
<p>Let $f:[0,\infty]\rightarrow \mathbb{R}$ a bounded function in each bounded interval. If $\lim_{x\rightarrow \infty} [f(x+1)-f(x)]=L$ then prove that $\lim_{x\rightarrow \infty}\displaystyle\frac{f(x)}{x}=L$</p>
</blockquote>
<p>I appreciate any hint to solve this problem. </p>
<p>Thanks a lot!</p>
| YTS | 126,222 | <p>This is another proof:</p>
<p>Given $\epsilon>0$ there exist $y>0$ such that if $x>y$ then:</p>
<p>$$L-\epsilon<f(x+1)-f(x)<L+\epsilon$$</p>
<p>we can conclude the following inequalytis valid for every $n$</p>
<p>$$L-\epsilon<f(x+2)-f(x+1)<L+\epsilon$$</p>
<p>$$L-\epsilon<f(x+2)-f(x+1)<L+\epsilon$$</p>
<p>$$\vdots$$</p>
<p>$$L-\epsilon<f(x+n)-f(x+n-1)<L+\epsilon$$</p>
<p>adding each of this inequalitys </p>
<p>$$nL-n\epsilon<f(x+n)-f(x)<nL+n\epsilon $$</p>
<p>taken $n=[x]$ and $h(x)=x-[x]$ then</p>
<p>$$[x]L-[x]\epsilon<f(x)-f(h(x))<[x]L+[x]\epsilon $$</p>
<p>since $0\leq h(x) <1$ and $[0,1)$ is bounded, then there exist $M>0$ such that $|f(h(x))|<M$. This implies that $\lim_{x\rightarrow\infty}\displaystyle\frac{f(h(x))}{x}=0$. Dividing by $x$:</p>
<p>$$\displaystyle\frac{[x]}{x}L-\displaystyle\frac{[x]}{x}\epsilon<\displaystyle\frac{f(x)-f(h(x))}{x}<\displaystyle\frac{[x]}{x}L+\displaystyle\frac{[x]}{x}\epsilon $$</p>
<p>and recalling that $\lim_{x\rightarrow\infty}\displaystyle\frac{[x]}{x}=1$. We take $x\rightarrow\infty$ then</p>
<p>$$L-\epsilon\leq\lim_{x\rightarrow\infty}\displaystyle\frac{f(x)}{x}\leq L+\epsilon$$</p>
<p>since $\epsilon>0$ is aritrary we get the result.</p>
|
184,241 | <p>I am having issues with the code:</p>
<pre><code>CC = p0^2/(p0 - L)
Dbuy = (1 - CC)*(L/p0) + (CC - p0)*Log[CC/p0]
Dwait = (1 - p1)*(L/(p1 - p0))
Dlock = (1 - p1)*(1 - L/p0 - L/(p1 - p0)) + (p1 - CC)*(1 - L/p0) - L*Log[(p1 - p0)/(CC - p0)]
RevLocking1 = Table[NMaximize[{x*Dbuy*p0 + x*Dwait*0.4*p1 + x*Dlock*(p0 + 0.4*L), x*(Dbuy + Dwait + Dlock) <= 1, 0 <= L <= p0 <= p1 <= 1}, {p0, p1, L}, MaxIterations -> 10000], {x, 1, 6, 0.25}]
</code></pre>
<p>The code works, but it ignores the constraint:</p>
<pre><code>0 <= L <= p0 <= p1 <= 1
</code></pre>
<p>Since it reaches values like:</p>
<pre><code>L -> 0.303712, p0 -> -0.0156877, p1 -> 0.900987
</code></pre>
<p>Any ideas?</p>
| Sjoerd Smit | 43,522 | <p>A function that is of great use here, is <code>AssociationTranspose</code> in the <code>GeneralUtilities</code> package. It does the same thing as regular transpose, but works with <code>List</code> and <code>Associations</code> alike (and mixtures of the two):</p>
<pre><code>listData = {
<|"Input" -> 1, "double" -> 2, "squared" -> 1|>,
<|"Input" -> 2, "double" -> 4, "squared" -> 4|>,
<|"Input" -> 3, "double" -> 6, "squared" -> 9|>
};
assocData = <|"Input" -> {1, 2, 3}, "double" -> {2, 4, 6}, "squared" -> {1, 4, 9}|>;
GeneralUtilities`AssociationTranspose[listData] === assocData
GeneralUtilities`AssociationTranspose[assocData] === listData
</code></pre>
<blockquote>
<blockquote>
<p>Out[11]= True</p>
<p>Out[12]= True</p>
</blockquote>
</blockquote>
|
64,491 | <p>I need to change the ticks in the xaxis for representation pH values, but the ticks are not dispersed along the range of xaxis. Another question: why the vertical line defined with Epilog is not visible?</p>
<pre><code>Kb1 := Kw/Ka1
Kb2 := Kw/Ka2
oh := Kw/x
f[x_] := Cs Kb2 oh/(Kb2 oh + oh^2 + Kb1 Kb2)
ff[x_] = Simplify[f[x]]
Ka1 = 6.2 10^-8
Ka2 = 4.8 10^-13
Cs = 0.1
myTicks[xmin_, xmax_] := {#, -N[Log[10, #]]} & /@FindDivisions[{xmin, xmax}, 5]
LogLinearPlot[ff[x], {x, 1. 10^-15, 15 10^-10}, AxesLabel -> {"pH", "[HA]"}, Ticks -> {myTicks, Automatic}, PlotRange -> All, Epilog ->
Line[{{Log[10, 1.725 10^-10], 0}, {Log[10, 1.725 10^-10], 0.1}}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/vACwu.png" alt="enter image description here"></p>
| Junho Lee | 16,245 | <p><strong>Version 9</strong></p>
<p>I made <code>findD</code> for that.</p>
<pre><code>findD[{x1_, x2_}, n_] := FindDivisions[-Log[10, #] & /@ {x1, x2}, n]
myTicks[xmin_, xmax_] := {10^-#, #} & /@ findD[{xmin, xmax}, 10]
</code></pre>
<p>Have try this code.</p>
<pre><code>Kb1 := Kw/Ka1; Kb2 := Kw/Ka2; oh := Kw/x; f[x_] := Cs Kb2 oh/(Kb2 oh + oh^2 + Kb1 Kb2); ff[x_] = Simplify[f[x]]; Ka1 = 6.2 10^-8; Ka2 = 4.8 10^-13; Cs = 0.1;
LogLinearPlot[ff[x], {x, 1. 10^-15, 15 10^-10},
AxesLabel -> {"pH", "[HA]"},
Ticks -> {myTicks, Automatic},
PlotRange -> All
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/5fLeh.png" alt="Blockquote"></p>
</blockquote>
<p>and <code>AbsoluteOptions[%, PlotRange]</code> show that your code
<code>Epilog-> Line[{{Log[10, 1.725 10^-10], 0}, {Log[10, 1.725 10^-10], 0.1}}</code>
is out of range, so this is my trick for your <code>Epilog</code> line.</p>
<pre><code>LogLinearPlot[{ff[x], 1/(x - 1.725 10^-10)}, {x, 1. 10^-15, 15 10^-10},
AxesLabel -> {"pH", "[HA]"},
Ticks -> {myTicks, Automatic},
PlotRange -> {0, 0.1}
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/Hr703.png" alt="Blockquote"></p>
</blockquote>
<p><strong>version 10</strong> </p>
<p><code>LogLinearPlot</code> does not work as function of <code>Ticks</code> in the version 10, and I think that this might be a bug. ( <a href="https://mathematica.stackexchange.com/q/64517/16245">here</a> is related )
So use <code>Ticks -> {myTicks[1. 10^-15, 15 10^-10], Automatic}</code></p>
<pre><code>LogLinearPlot[{ff[x], 1/(x - 1.725 10^-10)}, {x, 1. 10^-15, 15 10^-10},
AxesLabel -> {"pH", "[HA]"},
Ticks -> {myTicks[1. 10^-15, 15 10^-10], Automatic},
PlotRange -> {0, 0.1}
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/Z57lp.png" alt="enter image description here"></p>
</blockquote>
|
3,202,530 | <p>This seems easy. But it isn't. The diameter is given as <span class="math-container">$16$</span> and it asks you to find the coordinates of point <span class="math-container">$9$</span>. It's tempting to say that it is <span class="math-container">$(4, 4\sqrt{3})$</span>, but that isn't the answer. What the heck am I doing wrong? <em>We can also assume that the area of each segment is equal</em></p>
<p><img src="https://i.stack.imgur.com/CpGzi.png" alt="In this triangle, the diameter is 16. Find the coordinates of point (9)"></p>
| Dylan | 135,643 | <p>Let <span class="math-container">$u = \dot y$</span>. Then <span class="math-container">$\ddot y = \dfrac{du}{dt} = \dfrac{du}{dy}\dfrac{dy}{dt} = u \dfrac{du}{dy}$</span></p>
<p>So you have</p>
<p><span class="math-container">$$ u \frac{du}{dy} = \frac{C}{y} $$</span></p>
<p>Integrating both sides gives</p>
<p><span class="math-container">$$ \frac{u^2}{2} = C\ln y + A $$</span></p>
<p>where <span class="math-container">$A$</span> is the integration constant</p>
<p>You're missing a second initial condition. The value of <span class="math-container">$\dot y(0)$</span> would help us find <span class="math-container">$A$</span>.</p>
<p>After that, you can find the inverse function using separation of variables</p>
<p><span class="math-container">$$ t = \pm\int_R^y \frac{ds}{\sqrt{2(C\ln s + A)}} ds $$</span></p>
<p>Again, the sign of <span class="math-container">$\dot y(0)$</span> would determine the sign of the square root</p>
|
3,202,530 | <p>This seems easy. But it isn't. The diameter is given as <span class="math-container">$16$</span> and it asks you to find the coordinates of point <span class="math-container">$9$</span>. It's tempting to say that it is <span class="math-container">$(4, 4\sqrt{3})$</span>, but that isn't the answer. What the heck am I doing wrong? <em>We can also assume that the area of each segment is equal</em></p>
<p><img src="https://i.stack.imgur.com/CpGzi.png" alt="In this triangle, the diameter is 16. Find the coordinates of point (9)"></p>
| Jean Marie | 305,862 | <p>In the case of differential equation <span class="math-container">$y(t)y''(t)=\alpha t$</span> you have asked in your recent question <a href="https://math.stackexchange.com/q/3239474">Anyone know how to solve this differential equation $yy''=\alpha t$</a>, I am afraid that there are no closed form formula.</p>
<p>Nevertheless, I have obtained interesting results by simulation with one of the Runge-Kutta blackboxes available on Matlab. Here are the curves of <span class="math-container">$y=f(t)$</span> for different values of <span class="math-container">$f(0)$</span>, all with <span class="math-container">$f'(0)=0$</span> (Matlab program below).</p>
<p><a href="https://i.stack.imgur.com/HPQVn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HPQVn.jpg" alt="enter image description here"></a></p>
<blockquote>
<pre><code>function diffequ;
ts = 0:0.01:20; % time sampling
sed = @(t,X) SED(t,X); % call to SED
for k=-10:10
X0 = [10*k;0];
[t,X] = ode45(sed, ts, X0); % a version of Runge Kutta
plot(t,X(:,1));hold on;
end;
%
function XP = SED(t,X);%
p=X(1);q=X(2); % p=f, q=f'
pp=q; % q:=p'
qp=10*t./p; % p''=q'=a*t/p;
XP=[pp;qp];
</code></pre>
</blockquote>
|
987,962 | <p>Find an $x$ in $\Bbb R$ for which rank of the matrix $$A=\begin{bmatrix}1 & 1&1&1 \\1 & -1&-1&1\\1 &-3 &-3 &x \end{bmatrix}$$
is as minimal/maximal as possible.</p>
<p>I was thinking of row reducing the matrix and then count the pivot.</p>
| Martingalo | 127,445 | <p>Let us compute $F(x) = \int_{-\infty}^x f(y)dy$ the distribution function:
If $x<0$ then clearly $F(x)=0$.</p>
<p>If $0<x<1$ then
$$F(x) = \int_{-\infty}^x f(y)dy = \int_{0}^x f(y)dy = \int_0^x ydy = \frac{x^2}{2}.$$</p>
<p>If $1<x<2$ then
$$F(x) = \int_{-\infty}^{x} f(y) dy = \int_0^1 f(y)dy+ \int_1^x (2-y)dy = F(1)+ \bigg(2y - \frac{y^2}{2}\bigg|_{y=1}^{y=x} = \frac{1}{2} + 2x-\frac{x^2}{2} - 2+\frac{1}{2} = 2x-\frac{x^2}{2}-1$$
Remember to subtract the last value here, because you want to compute the accumulated area.</p>
<p>Finally, if $x>2$ then $F(x)=1$. Altogether</p>
<p>$$
F(x)=
\begin{cases} 0, \mbox{ if } x<0 \\
\frac{x^2}{2} \mbox{ if } 0<x<1\\
2x-\frac{x^2}{2}-1 \mbox{ if } 1<x<2\\
1 \mbox{ if } x>2
\end{cases}
$$</p>
|
53,596 | <p>It's "well known" that, for any weight $k$ and level $N$, the space $S_k(\Gamma_1(N))$ of cusp forms of that weight and level has a basis in which all the Hecke operators act by matrices with entries in $\mathbb{Z}$; consequently all the Hecke eigenvalues are algebraic numbers (indeed algebraic integers).</p>
<p>I was reflecting on how to prove this while teaching an undergraduate course on modular forms. For $k \ge 2$ it's not hard: there's the Eichler-Shimura machinery which relates it to a question about cohomology, and the cohomology with $\mathbb{Z}$ coefficients does the job. Alternatively, and more or less equivalently, you use the pairing with modular symbols. Both of these methods break down for $k = 1$; the only argument I know that works in this case is to use the fact that $X_1(N)$ has a model as an algebraic variety, and weight $k$ modular forms correspond to sections of the $k$-th power of a line bundle that has a purely algebraic definition. But that's not really something I can stand up and explain to a class of undergraduate students!</p>
<blockquote>
<p>For cusp forms of weight $k = 1$, can the algebraicity of the Hecke eigenvalues be proved without quoting heavy machinery from arithmetic geometry? </p>
</blockquote>
| Marc Palm | 10,400 | <p>Here is another proof: Deligne and Serre have proven that the corresponding L-function equal the Artin L-function of a Galois representation. Deligne had proven similar facts known for weight $k \geq 2$ modular forms before that, and their proof essentially relies on the former results. This implies algebraicity and also is the only approach I know for the same result for Maass Hecke cusp forms of Laplace eigenvalue $1/4$, where the same algebraicity result is unknown.</p>
<p><em>This is probably even harder than what you suggest, or equivalent(?). Quoting that result of Deligne and Serre would be a reasonable choice, and deducing algebraicity from it. The question is pretty old and didn't receive an answer so far, so I guess the class isn't running anymore anyways.</em></p>
<p>There is no trace formula for Hecke eigenvalues of weight one forms available, since the limit of discrete series representations opposed to the discrete series representations are not square integrable, and have no pseudo-matrix coefficients.</p>
|
3,794,148 | <p><span class="math-container">$n$</span> is fixed.
<br />
<span class="math-container">$ x \in ]0,1[$</span>
<br />
Coud you help me prove, without theorem of dominated convergence :
<span class="math-container">$$ \int_{1}^{n} \frac{x^t}{t} dt \underset{x \to 1^{-}}{\rightarrow} \int_{1}^{n} \frac{1}{t} dt$$</span></p>
<hr />
<p>My attempt :
<span class="math-container">$ x^t = e^{ t \ln x}$</span> and we could use <span class="math-container">$ e^u \sim 1+u$</span> for <span class="math-container">$u \to 0$</span>. If I do that I obtain a <span class="math-container">$o (t \ln x)$</span> and I cannot conclude.</p>
| RRL | 148,510 | <p>Hint:</p>
<p><span class="math-container">$$\left|\int_1^ n \frac{x^t}{t} \, dt - \int_1^ n \frac{1}{t} \, dt\right| \leqslant \int_1^n\left| \frac{x^t}{t} - \frac{1}{t}\right| \, dt,$$</span></p>
<p>and <span class="math-container">$(x,t) \mapsto x^t/t $</span> is uniformly continuous on <span class="math-container">$[1-\delta,1] \times [1,n]$</span>.</p>
|
3,794,148 | <p><span class="math-container">$n$</span> is fixed.
<br />
<span class="math-container">$ x \in ]0,1[$</span>
<br />
Coud you help me prove, without theorem of dominated convergence :
<span class="math-container">$$ \int_{1}^{n} \frac{x^t}{t} dt \underset{x \to 1^{-}}{\rightarrow} \int_{1}^{n} \frac{1}{t} dt$$</span></p>
<hr />
<p>My attempt :
<span class="math-container">$ x^t = e^{ t \ln x}$</span> and we could use <span class="math-container">$ e^u \sim 1+u$</span> for <span class="math-container">$u \to 0$</span>. If I do that I obtain a <span class="math-container">$o (t \ln x)$</span> and I cannot conclude.</p>
| zhw. | 228,045 | <p>By the MVT, for any fixed <span class="math-container">$t\ge 1$</span></p>
<p><span class="math-container">$$\tag 1 x^t-1 = x^t-1^t = tc_x^{t-1}(x-1)$$</span></p>
<p>for some <span class="math-container">$c_x\in (x,1).$</span> The absolute value of the last expression is therefore bounded above by <span class="math-container">$n\cdot 1 \cdot |x-1|= n|x-1|.$</span></p>
<p>The difference of the two integrals in question is</p>
<p><span class="math-container">$$\tag 2 \int_{1}^{n} \frac{x^t-1}{t} dt.$$</span></p>
<p>Take the absolute value, move the absolute values inside the integral, and then use the bound we found for <span class="math-container">$(1).$</span> This implies the absolute value of <span class="math-container">$(2)$</span> is no more than</p>
<p><span class="math-container">$$n|x-1|\int_1^n \frac{1}{t}\,dt = n|x-1|\ln n.$$</span></p>
<p>This <span class="math-container">$\to 0$</span> as <span class="math-container">$x\to 1^-$</span> as desired.</p>
|
1,341,231 | <p>Let $f$ be continuous on $[0,1]$, and let $\alpha>0$. Find: $\lim\limits_{x\to 0}{x^{\alpha}\int_{x}^{1}{f(t)\over t^{\alpha +1}}dt}$. I tried integration by parts, but I am not sure if $f$ is integrable and to what extent. It also gets really messy. Besides, I am not sure if I am to express the limit using $f$, or to arrive at an actual number. It would be nice if you could take a look.</p>
<p>Using other questions I got: Let us denote $G(x)=\int_{x}^{1}{f(t)\over t^{\alpha +1}}dt$, so I am looking for: $\lim\limits_{x\to 0}{x^{\alpha}G(x)}=\lim\limits_{x\to 0}{G(x)\over {1\over x^{\alpha}}}$. If $g(t)=\int{f(t)\over t^{\alpha +1}}dt$, Then: $G(x)=g(1)-g(x)$ which means: $G'(x)=-g'(x)=-{f(x)\over x^{\alpha +1}}$. Let us use L'Hôpital's rule: $\lim\limits_{x\to 0}{G(x)\over {1\over x^{\alpha}}}=\lim\limits_{x\to 0}{{G'(x)=-{f(x)\over x^{\alpha +1}}}\over {-\alpha\over x^{\alpha+1}}}=\lim\limits_{x\to 0}{f(x)\over \alpha}={f(0)\over \alpha}$.</p>
| vasmous | 249,266 | <p>I will do exactly the same thing. I just finished my degree in mathematics but in our department there is not a single course of Number Theory, and since I will start my graduate courses in October I thought it will be a great idea to study Number Theory on my own. So, I asked one of my professors, who is interested in Algebraic Geometry and Number Theory, what would be a textbook that has everything an undergraduate should know about Number Theory before moving on. He told me that <a href="http://www.bookdepository.com/Classical-Introduction-Modern-Number-Theory-v-84-Kenneth-Ireland/9780387973296" rel="nofollow">A Classical Introduction to Modern Number Theory</a> by <em>Kenneth F. Ireland and Michael Rosen</em> is the perfect choice. He also mentioned that I should definitely study chapters 1-8,10-13 and 17. Another book that he mentioned was <a href="http://rads.stackoverflow.com/amzn/click/0131861379" rel="nofollow">A Friendly Introduction to Number Theory</a> by <em>Joseph H. Silverman</em>. He emphasized though that this book is clearly an introduction whereas the previous one gives you all the tools you need in order to study many things that are connected to Number Theory. I hope that this helped you!</p>
|
4,308,155 | <p>Given an operator <span class="math-container">$T \in B(H)$</span>, denote its spectrum by <span class="math-container">$\sigma(T)$</span>. We write <span class="math-container">$\Omega := \sigma(T)\setminus \{0\}$</span>. The following result is well-known and <em><strong>can be assumed without proof</strong></em>:</p>
<blockquote>
<p>Let <span class="math-container">$T \in B_0(H)$</span> be a self-adjoint compact operator on a Hilbert
space <span class="math-container">$H$</span>. Given <span class="math-container">$\lambda \in \Omega$</span>, let <span class="math-container">$P_\lambda$</span> be the
finite-rank projection on <span class="math-container">$\mathcal{E}_\lambda:= \ker(T-\lambda I)$</span>.
Then <span class="math-container">$$T= \sum_{\lambda \in \Omega} \lambda P_\lambda$$</span> where the
unordered sum converges in the norm topology.</p>
</blockquote>
<p>Given <span class="math-container">$\xi, \eta \in H$</span>, let us define the rank-one operator <span class="math-container">$R_{\xi, \eta}: H \to H$</span> by <span class="math-container">$R_{\xi, \eta}(\zeta):= \langle \zeta, \eta\rangle \xi$</span>. I want to prove the following theorem:</p>
<blockquote>
<p>If <span class="math-container">$T \in B_0(H)$</span> is a self-adjoint operator compact operator, then
there is an orthonormal basis <span class="math-container">$\{e_s\}_{s \in S}$</span> for <span class="math-container">$\ker(T)^\perp$</span>
and a collection <span class="math-container">$\{\mu_s\}_{s \in S} \in c_0(S)$</span> such that <span class="math-container">$$T=
\sum_{s \in S} \mu_s R_{e_s, e_s}$$</span> where the series converges in
norm. In particular, a self-adjoint compact operator admits an
orthonormal basis of eigenvectors.</p>
</blockquote>
<p>The idea of the proof should be more or less the following:</p>
<p>We have an orthogonal direct sum decomposition <span class="math-container">$$\ker(T)^\perp = \bigoplus_{\lambda \in \Omega} \mathcal{E}_\lambda$$</span>
Given <span class="math-container">$\lambda \in \Omega$</span>, let <span class="math-container">$\mathcal{F}_\lambda$</span> be an orthonormal basis for <span class="math-container">$\mathcal{E}_\lambda$</span>. Then we have <span class="math-container">$$P_\lambda = \sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}$$</span> and it follows that
<span class="math-container">$$T = \sum_{\lambda \in \Omega}\lambda \left(\sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}\right) = \sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$</span>
Then <span class="math-container">$\bigcup_{\lambda \in \Omega} \mathcal{E}_\lambda$</span> is an orthonormal basis for <span class="math-container">$\ker(T)^\perp$</span> and this should yield the desired decomposition. However, I feel like the equality
<span class="math-container">$$\sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$</span>
still requires some formal justification. We should at least show that the sum on the right converges in the norm topology! Of course, other approaches to prove my end goal are also welcome!</p>
<p>EDIT: It's a strong requirement to me that the convergence of all series involved is in the norm-topology. Many references only deal with the strong convergence of these series.</p>
| Martin Argerami | 22,857 | <p>That's precisely where you get to use that your <span class="math-container">$T$</span> is compact.</p>
<p>Fix <span class="math-container">$\delta>0$</span>, and consider <span class="math-container">$\Omega_\delta=\{\lambda\in\Omega:\ \lambda>\delta\}$</span>. Suppose <span class="math-container">$\Omega_\delta$</span> is infinite. For each <span class="math-container">$\lambda\in\Omega_\delta$</span>, fix <span class="math-container">$h_\lambda\in P_\lambda H$</span> with <span class="math-container">$\|h_\lambda\|=1$</span>. The set <span class="math-container">$\{h_\lambda\}_{\lambda\in\Omega_\delta}$</span> is orthonormal; we can write <span class="math-container">$Th_\lambda=\lambda h_\lambda$</span> as
<span class="math-container">$$
T(\frac1\lambda\,h_\lambda)=h_\lambda.
$$</span>
As <span class="math-container">$\lambda>\delta$</span>, we have that <span class="math-container">$\tfrac1\lambda\,h_\lambda$</span> sits inside the ball of radius <span class="math-container">$1/\delta$</span>. So <span class="math-container">$\{h_\lambda\}$</span> is in the image through <span class="math-container">$T$</span> of the ball of radius <span class="math-container">$1/\delta$</span>. And, being orthonormal, <span class="math-container">$\{h_\lambda\}$</span> does not have a convergent subsequence, contradicint the compactness of <span class="math-container">$T$</span>. We have thus shown that <span class="math-container">$\Omega_\delta$</span> is finite. Note that this argument also implies that each <span class="math-container">$P_\lambda$</span> is finite-rank.</p>
<p>By consider the finite sets <span class="math-container">$\Omega_{1/n}$</span>, we prove that <span class="math-container">$\Omega$</span> consists of a sequence that converges to <span class="math-container">$0$</span>, as desired.</p>
<p>If <span class="math-container">$\{\xi_n\}$</span> is a finite orthonormal set and <span class="math-container">$P_n$</span> denotes the orthogonal projection onto <span class="math-container">$\mathbb C\xi_n$</span>, then
<span class="math-container">$$\tag1
\Big\|\sum_n\lambda_n\,P_n\Big\|=\sup\{|\lambda_n|:\ n\}.
$$</span>
Indeed, if <span class="math-container">$\xi\in H$</span> with <span class="math-container">$\|\xi\|=1$</span>, then <span class="math-container">$\xi=\eta+\sum_n c_n\xi_n$</span>, where <span class="math-container">$\eta$</span> is orthogonal to all <span class="math-container">$\xi_n$</span>. Then (since <span class="math-container">$P_n\eta=0$</span> for all <span class="math-container">$n$</span>)
<span class="math-container">\begin{align}
\Big\|\sum_n\lambda_nP_n\xi\Big\|^2
&=\Big\|\sum_n\sum_k\lambda_nc_kP_n\xi_k\Big\|^2
=\Big\|\sum_n\lambda_nc_n\xi_n\Big\|^2
=\sum_n|\lambda_n|^2\,|c_n|^2\\[0.3cm]
&\leq\sup\{|\lambda_n|:\ n\}\,\sum_n|c_n|^2\\[0.3cm]
&=\sup\{|\lambda_n|:\ n\}.
\end{align}</span>
Using that <span class="math-container">$\Big\|\sum_n\lambda_nP_n\xi_k\Big\|=|\lambda_k|$</span> it is then easy to check the equality <span class="math-container">$(1)$</span>.</p>
<p>Now allow the set <span class="math-container">$\{\xi_n\}$</span> to be infinite. If <span class="math-container">$\lambda_n\to0$</span>, fix <span class="math-container">$\varepsilon>0$</span>. Choose <span class="math-container">$n_0$</span> such that <span class="math-container">$|\lambda_k|<\varepsilon$</span> for all <span class="math-container">$n\geq n_0$</span>. Then
<span class="math-container">$$
\Big\|\sum_{k=n}^m\lambda_kP_k\Big\|=\sup\{|\lambda_k|:\ k\geq n\}\leq\varepsilon
$$</span>
for all <span class="math-container">$m>n\geq n_0$</span>. This shows that the tails of the series <span class="math-container">$\sum_n\lambda_nP_n$</span> go to zero, and so the series converges.</p>
|
2,682,008 | <p>I’ve been studying general form Ricatti Differential Equations recently, and I’m confused as to why a general form solution is not possible. What would it mean if a general form solution were possible? </p>
| Fred | 380,717 | <p>If a special solution $y_1$ of the Ricatti Differential Equation is known, then the general solution is obtained as $y=y_1+u$, where $u$ satisfies a Bernoulli equation.</p>
<p>If a special solution of the Ricatti Differential Equation is not known, then, in general, the general form of the solutions is not possible.</p>
|
43,690 | <p>I have to apologize because this is not the normal sort of question for this site, but there have been times in the past where MO was remarkably helpful and kind to undergrads with similar types of question and since it is worrying me increasingly as of late I feel that I must ask it.</p>
<p>My question is: what can one (such as myself) contribute to mathematics?</p>
<p>I find that mathematics is made by people like Gauss and Euler - while it may be possible to learn their work and understand it, nothing new is created by doing this. One can rewrite their books in modern language and notation or guide others to learn it too but I never believed this was the significant part of a mathematician work; which would be the creation of original mathematics. It seems entirely plausible that, with all the tremendously clever people working so hard on mathematics, there is nothing left for someone such as myself (who would be the first to admit they do not have any special talent in the field) to do. Perhaps my value would be to act more like cannon fodder? Since just sending in <em>enough</em> men in will surely break through some barrier.</p>
<p>Anyway I don't want to ramble too much but I really would like to find answers to this question - whether they come from experiences or peoples biographies or anywhere.</p>
<p>Thank you.</p>
| ohai | 9,501 | <p>I suspect that most interesting mathematical results raise more questions than they settle, so in this case you would not have anything to worry about.</p>
|
43,690 | <p>I have to apologize because this is not the normal sort of question for this site, but there have been times in the past where MO was remarkably helpful and kind to undergrads with similar types of question and since it is worrying me increasingly as of late I feel that I must ask it.</p>
<p>My question is: what can one (such as myself) contribute to mathematics?</p>
<p>I find that mathematics is made by people like Gauss and Euler - while it may be possible to learn their work and understand it, nothing new is created by doing this. One can rewrite their books in modern language and notation or guide others to learn it too but I never believed this was the significant part of a mathematician work; which would be the creation of original mathematics. It seems entirely plausible that, with all the tremendously clever people working so hard on mathematics, there is nothing left for someone such as myself (who would be the first to admit they do not have any special talent in the field) to do. Perhaps my value would be to act more like cannon fodder? Since just sending in <em>enough</em> men in will surely break through some barrier.</p>
<p>Anyway I don't want to ramble too much but I really would like to find answers to this question - whether they come from experiences or peoples biographies or anywhere.</p>
<p>Thank you.</p>
| Gerry Myerson | 3,684 | <p>If you are an undergraduate, you don't yet know what you can contribute to mathematics, and in particular you don't yet know whether, or at what level, you can create original mathematics. Fortunately, you don't have to work this out entirely for yourself; if you go on to graduate work in mathematics, you will have an advisor, whose job is to help you get the most out of your potential. Even after a student finishes a PhD, she's not alone; much of the best mathematical work today is collaborative, and whatever weaknesses you may think you have can be compensated for by the strengths of your collaborators (while you compensate for their weaknesses with your strengths). </p>
<p>It ain't easy, but it can be done. </p>
|
43,690 | <p>I have to apologize because this is not the normal sort of question for this site, but there have been times in the past where MO was remarkably helpful and kind to undergrads with similar types of question and since it is worrying me increasingly as of late I feel that I must ask it.</p>
<p>My question is: what can one (such as myself) contribute to mathematics?</p>
<p>I find that mathematics is made by people like Gauss and Euler - while it may be possible to learn their work and understand it, nothing new is created by doing this. One can rewrite their books in modern language and notation or guide others to learn it too but I never believed this was the significant part of a mathematician work; which would be the creation of original mathematics. It seems entirely plausible that, with all the tremendously clever people working so hard on mathematics, there is nothing left for someone such as myself (who would be the first to admit they do not have any special talent in the field) to do. Perhaps my value would be to act more like cannon fodder? Since just sending in <em>enough</em> men in will surely break through some barrier.</p>
<p>Anyway I don't want to ramble too much but I really would like to find answers to this question - whether they come from experiences or peoples biographies or anywhere.</p>
<p>Thank you.</p>
| Alok Gandhi | 10,429 | <p>Muad you raised not only a very important question but also a fairly enigmatic one. I completed my Maters in Mathematics 13 years ago and left mathematics then and there, what I realized during these years was that I could not sleep without thinking something new to discover or invent in pure mathematics. The numbers kept on haunting me. I was fascinated by Relativistic Mathematics as well had a great passion for Number Theory.I tried to find a suitable topic to do some original research work but left that to enter a totally different field. I opened a restaurant named e=mc^2. But still couldn't sleep well. So after running a successful talk-of-the-town restaurant for almost 4 years, I decided to turn to Computer Graphics, another of my passions as I was very involved in fine arts and poetry as well. Having graduated from Vancouver Film School, I am now a Computer Graphics Technical Director at a Visual Effects Studio, where I deal with mathematics in a more intimate way on a day to day basis. The point of all this autobiographical ramblings of mine is this : If you have the right motivation, and if you are curious enough, you have something to discover. My nights are still sleepless as I am yet to find something to discover !!!. So I think mathematics is inside you waiting for it to be discovered, just be passionate enough !!</p>
|
43,690 | <p>I have to apologize because this is not the normal sort of question for this site, but there have been times in the past where MO was remarkably helpful and kind to undergrads with similar types of question and since it is worrying me increasingly as of late I feel that I must ask it.</p>
<p>My question is: what can one (such as myself) contribute to mathematics?</p>
<p>I find that mathematics is made by people like Gauss and Euler - while it may be possible to learn their work and understand it, nothing new is created by doing this. One can rewrite their books in modern language and notation or guide others to learn it too but I never believed this was the significant part of a mathematician work; which would be the creation of original mathematics. It seems entirely plausible that, with all the tremendously clever people working so hard on mathematics, there is nothing left for someone such as myself (who would be the first to admit they do not have any special talent in the field) to do. Perhaps my value would be to act more like cannon fodder? Since just sending in <em>enough</em> men in will surely break through some barrier.</p>
<p>Anyway I don't want to ramble too much but I really would like to find answers to this question - whether they come from experiences or peoples biographies or anywhere.</p>
<p>Thank you.</p>
| Bill Thurston | 9,062 | <p>It's not <em>mathematics</em> that you need to contribute to. It's deeper than that: how might you contribute to humanity, and even deeper, to the well-being of the world, by pursuing mathematics? Such a question is not possible to answer in a purely intellectual way, because the effects of our actions go far beyond our understanding. We are deeply social and deeply instinctual animals, so much that our well-being depends on many things we do that are hard to explain in an intellectual way. That is why you do well to follow your heart and your passion. Bare reason is likely to lead you <a href="http://en.wikipedia.org/wiki/Ted_Kaczynski">astray</a>. None of us are smart and wise enough to figure it out intellectually.</p>
<p>The product of mathematics is clarity and understanding. Not theorems, by themselves. Is there, for example any real reason that even such famous results as Fermat's Last Theorem, or the Poincaré conjecture, really matter? Their real importance is not in their specific statements, but their role in challenging our understanding, presenting challenges that led to mathematical developments that increased our understanding.</p>
<p>The world does not suffer from an oversupply of clarity and understanding (to put it mildly). How and whether specific mathematics might lead to improving the world (whatever that means) is usually impossible to tease out, but mathematics collectively is extremely important.</p>
<p>I think of mathematics as having a large component of psychology, because of its strong dependence on human minds. Dehumanized mathematics would be more like computer code, which is very different. Mathematical ideas, even simple ideas, are often hard to transplant from mind to mind. There are many ideas in mathematics that may be hard to get, but are easy once you get them. Because of this, mathematical understanding does not expand in a monotone direction. Our understanding frequently deteriorates as well. There are several obvious mechanisms of decay. The experts in a subject retire and die, or simply move on to other subjects and forget. Mathematics is commonly explained and recorded in symbolic and concrete forms that are easy to communicate, rather than in conceptual forms that are easy to understand once communicated. Translation in the direction conceptual -> concrete and symbolic is much easier than translation in the reverse direction, and symbolic forms often replaces the conceptual forms of understanding. And mathematical conventions and taken-for-granted knowledge change, so older texts may become hard to understand.</p>
<p>In short, mathematics only exists in a living community of mathematicians that spreads understanding and breaths life into ideas both old and new. The real satisfaction from mathematics is in learning from others and sharing with others. All of us have clear understanding of a few things and murky concepts of many more. There is no way to run out of ideas in need of clarification. The question of who is the first person to ever set foot on some square meter of land is really secondary. Revolutionary change does matter, but revolutions are few, and they are not self-sustaining --- they depend very heavily on the community of mathematicians.</p>
|
275,517 | <p>Some mathematical concepts are ended with "-oid", such as Matroid, greedoid, groupoid. What does that mean? Do these concepts share something in common? Thanks!</p>
| Tom Au | 12,506 | <p>If you have a "center" of an object, it means "equidistant" from the edges (spacially).</p>
<p>If you have a <a href="http://en.wikipedia.org/wiki/Centroid" rel="nofollow"><strong>centroid</strong></a>, it means the <strong>weighted</strong> "center" of an object, where the weightS are something like the density of the material.</p>
<p>Here the "oid" is a reference to the "weighted" (e.g. center) property, not the literal one.</p>
|
328,868 | <p>It is easily shown that the function $$\begin{cases} \exp \left(\frac{1}{x^2-1} \right) & |x| < 1 \\ 0 & \text{otherwise} \\ \end{cases}$$
is smooth and has compact support in $\mathbb R$. I tried playing with it to find a function with the following properties:</p>
<p>a. $f(x)=0$ for $x \le 0$</p>
<p>b. $f(x)=1$ for $x \ge 1$</p>
<p>c. $f$ is monotonically increasing.</p>
<p>d. $f$ is smooth.</p>
<p>Is it possible to find an explicit formula for such $f$?</p>
| muzzlator | 60,855 | <p>Let $$h(x) = \left\{\begin{array}{c} 0 & x \leq 0 \\ e^{-1/x} & x > 0 \end{array} \right. $$</p>
<p>Then consider $$ g(x) = \frac{h(x)}{ h(x) + h(1-x)} $$</p>
<p>It is smooth because it is the ratio of smooth functions with the denominator never $0$.</p>
<p>To verify it is increasing, we can calculate the sign of the numerator of $g'(x)$ using quotient rule between $0 < x < 1$:</p>
<p>$$\begin{align} N g'(x) = & h'(x)(h(x) + h(1-x)) - h(x)(h'(x) + h'(1-x)) \\ = & \frac{1}{x^2} h(x) h(1-x) + \frac{1}{(1-x)^2} h(x)h(1-x) \\ > & 0 \end{align}$$</p>
|
3,380,479 | <p>i'm doing a seminar about Galois Theory but I have a problem with the definition of purely inseparable element and the one of purely inseparable extension, I read the definitions but i would like to see an example for each one of them. If somebody can help me please. </p>
| user | 505,767 | <p>We need to specify for what number we need to calculate the "least residue of mod 13".</p>
<p>For example</p>
<p><span class="math-container">$$20 \equiv 7 \mod 13$$</span></p>
<p>then to find the <a href="https://en.wikipedia.org/wiki/Modular_multiplicative_inverse" rel="nofollow noreferrer"><strong>inverse</strong></a> of <span class="math-container">$7 \mod 13$</span> we can use <a href="https://en.wikipedia.org/wiki/Euclidean_algorithm" rel="nofollow noreferrer"><strong>euclidean algorithm</strong></a>, that is</p>
<ul>
<li><span class="math-container">$13=1\cdot 7+6$</span></li>
<li><span class="math-container">$7=1\cdot 6+1$</span></li>
</ul>
<p>then</p>
<p><span class="math-container">$$1=7-6=7-(13-7)=-13+2\cdot 7$$</span></p>
<p>then <span class="math-container">$2$</span> is the inverse of <span class="math-container">$7 \mod 13$</span> (easy to guess in that case).</p>
<p>We can repeat the same procedure for all the other residues <span class="math-container">$\mod 13$</span>.</p>
|
37,977 | <p>I have two lists, say <code>a</code> and <code>b</code>, both of length <code>n</code>. I'd like to compute the following:</p>
<ul>
<li>minimum of $a[i]/b[i]$ where $i=1, 2, ...n$ and $b[i]>0$</li>
</ul>
<p>I'd also like to know the index of the element where the min occurs.</p>
| Gabriel | 744 | <p>I would do this by first combining the lists using Transpose, and then filtering the division pairs using either a pattern with <code>Cases</code> or a predicate using <code>Select</code></p>
<pre><code>Cases[Transpose[{a, b}], {_, _?Positive}]
</code></pre>
<p>or</p>
<pre><code>Select[Transpose[{a, b}], #[[2]] > 0&]
</code></pre>
<p>This will return the filtered list of <code>{numerator, denominator}</code> pairs, to get the divisions I would just expand on the above with Map Apply (<code>@@@</code>)</p>
<pre><code>lst = Divide @@@ Cases[Transpose[{a, b}], {_, _?Positive}]
</code></pre>
<p>or</p>
<pre><code>lst = Divide @@@ Select[Transpose[{a, b}, #[[2]] > 0&]
</code></pre>
<p>This will give back a list with the defined ratios. You can then just use <code>Min</code> and <code>Position</code> as other answers have shown.</p>
|
37,977 | <p>I have two lists, say <code>a</code> and <code>b</code>, both of length <code>n</code>. I'd like to compute the following:</p>
<ul>
<li>minimum of $a[i]/b[i]$ where $i=1, 2, ...n$ and $b[i]>0$</li>
</ul>
<p>I'd also like to know the index of the element where the min occurs.</p>
| Ray Koopman | 8,159 | <p>If the minimum can be assumed to be unique then use <code>f1</code>, which returns the position of only the first occurrence of the minimum value. Otherwise use <code>f2</code>, which is slower but returns the positions of all occurrences.</p>
<pre><code>f1[a_,b_] := With[{i = Ordering[#1,1][[1]]}, {#1[[i]],#2[[i]]}]& [a[[#]]/b[[#]],#]& @
SparseArray[UnitStep[-b], Automatic, 1]["AdjacencyLists"];
f2[a_,b_] := With[{m = Min[#1]}, {m,Pick[#2,#1,m]}]& [a[[#]]/b[[#]],#]& @
SparseArray[UnitStep[-b], Automatic, 1]["AdjacencyLists"];
n = 7;
a = {2, 9, 5, 3, 0, 0, 9};
b = {0,-4, 2, 0, 10, 4, 0};
f1[a,b]
f2[a,b]
</code></pre>
<blockquote>
<p>{0, 5}<br>
{0, {5, 6}}</p>
</blockquote>
|
3,310,599 | <p>Let <span class="math-container">$Y$</span> be a random variable on <span class="math-container">$[0,1]$</span> and let <span class="math-container">$X\sim U[0,1]$</span> be a uniformly distributed random variable that is independent of <span class="math-container">$Y$</span>.</p>
<p>Prove that <span class="math-container">$$P(X<Y\mid Y)=Y.$$</span></p>
<p>Actually that looks kind of trivial, only use the independence of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> and then use the uniform distribution of <span class="math-container">$X$</span>. But I am not able to write down an intermediate step in mathematical formulas, something like
<span class="math-container">$$P(X<Y\mid Y)= \dots=Y.$$</span>
Any help available?</p>
| IljaKlebanov | 692,074 | <p>Start with conditioning on the event <span class="math-container">$Y=y$</span>, which is simpler to think through:
<span class="math-container">$$
P(X<Y|Y=y) = P(X<y|Y=y) \stackrel{\text{indep.}}{=}P(X<y)
= \int_{0}^{y}1\, \mathrm dx
=y.
$$</span>
Conditioning an event <span class="math-container">$A$</span> on a random variable <span class="math-container">$U$</span> yields another random variable defined by <span class="math-container">$P(A|Y)(\omega) := P(A|Y = Y(\omega))$</span>, therefore, by performing the same steps, we obtain:
<span class="math-container">\begin{align*}
P(X<Y|Y)(\omega)
&= P(X<Y|Y=Y(\omega)) = P(X<Y(\omega)|Y=Y(\omega)) \stackrel{\text{indep.}}{=}P(X<Y(\omega))
\\
&=
\int_{0}^{Y(\omega)}1\, \mathrm dx
=Y(\omega).
\end{align*}</span>
Since this identity holds for every <span class="math-container">$\omega\in\Omega$</span>, you get the desired result.</p>
|
2,530,221 | <p>Let $X$ be a topological space, and equip its homeomorphism group $\text{Homeo}(X)$ with the compact-open topology. I probably need to restrict $X$ to be locally compact for it to play off with $\text{Homeo}(X)$ nicely.</p>
<p>The usual definition of $X$ being <b>homogeneous</b> is that given $x,y \in X$ there exists $f \in \text{Homeo(X)}$ with $f(x) = y$. My question: When can this choice of $f$ be made continuous? More precisely, is there a natural class of homogeneous spaces for which there is always a continuous map $$\theta: \{(x,y): x \neq y\} \rightarrow \text{Homeo}(X)$$
such that $\theta(x,y)$ carries $x$ to $y$. Is it true for connected manifolds?</p>
| Moishe Kohan | 84,907 | <p>You are asking the following: Pick $x\in X$ and for $G=Homeo(X)$ consider the orbit map $o_x: G\to X$, $o_x(g)=g(x)$. Assuming that $o_x$ is surjective, when does it have a (continuous) section $s: X\to G$?</p>
<p>A general observation is that if $X$ is not contractible, then no section $s: X\to G$ is homotopic to a constant map. </p>
<p>Here is what happens in the case when $X$ is a closed connected oriented 2-dimensional manifold. </p>
<ol>
<li><p>If $\chi(X)=0$, then $X\cong S^1\times S^1$ and, hence, is homeomorphic to a Lie group and hence, $o_x$ has a section. </p></li>
<li><p>If $\chi(X)=2$, then $G=Homeo(X)$ is homotopy-equivalent to $O(3)$, hence, $\pi_2(G)=0$. However, if there is a section $s: X\to G$, then $\pi_2(G)\ne 0$, proving that a section does not exist. </p></li>
<li><p>If $\chi(X)<0$ then $G$ is homotopy equivalent to ${\mathbb Z}$ (the identity component of $G$ is contractible). Hence, you cannot have a section $s: X\to G$. </p></li>
</ol>
<p>I suspect that when $X$ is the $n$-dimensional sphere then a section $s: X\to G$ exists if and only if $n=1, 3, 7$, i.e. when $S^n$ has structure of a Lie group.</p>
<p>See </p>
<p>M.-E. Hamstorm, <a href="https://projecteuclid.org/download/pdf_1/euclid.ijm/1256054895" rel="nofollow noreferrer">Homotopy groups of the space of homeomorphisms on a 2-manifold</a>, Illinois J. Math. Volume 10, Issue 4 (1966), 563-573</p>
<p>for the details concerning homotopy types of the homeomorphism groups of compact surfaces. (The same conclusion applies when one works with the groups of diffeomorphisms, resp. PL homeomorphisms: Earle and Eells, 1966; resp. Scott, 1970.) </p>
<p>Edit: I just realized that for some reason, you are interested in the existence of a section which is continuous away from one point (I am not sure why though). This changes the answer. Namely, for $X=S^2$ (or sphere of any dimension for this matter), the map $\Theta$ exists, it is a rotation in the plane passing through $x, y$. On the other hand, if $\chi(X)<0$ the map $\Theta$ still does not exist since $X-\{x\}$ is not contractible and the argument in the case 3 still applies. </p>
|
1,123,647 | <p>I am looking into starting an amateur project for online tutoring. I need a math/geometry program that I can use to create shapes, graph functions, and create animation. I know of many types of software (such as geogebra) but they have restrictions for commercial use. In other words, I can't use the material or features (even screenshots) to gain a profit. Does anyone know of any programs/ software that I can use for my project?</p>
<p>Note: I don't mind software or programs that I can purchase for a one time fee.</p>
| Mark Fischler | 150,362 | <p>1) Contact the geogebra support people; there may be some provision to pay a fee for the right to distribute-for-profit screenshots or animations produced. </p>
<p>2) See if Mathematica meets your needs -- but this is not a cheap solution, the commercial cost is about $2500. However, Mathematica is a truly professional, well-made tool and the time it saves you might be worth the cost.</p>
<p>PS- Don't be biased by lack of features in Wolfram Alpha: That is emphatically <em>not</em> like the full-blown Mathematica.</p>
|
3,115,566 | <p>Convergence of the series for <span class="math-container">$a \in \mathbb R$</span> <span class="math-container">$$\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$$</span> </p>
<p>I saw this problem in a calculus book and it gave a hint that says </p>
<p><strong>HINT</strong> First show that <span class="math-container">$$\sin\left(\pi \sqrt{n^2+a^2} \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\sim(-1)^n\frac{\pi a^2}{2n}\qquad (n \to\infty)$$</span> </p>
<p>I was able to show that <span class="math-container">$$\sin\left(\pi \sqrt{n^2+a^2} \right)=\sin\left(\pi \sqrt{n^2+a^2} -\pi n+\pi n\right)=\sin\left(\pi (\sqrt{n^2+a^2}-n)+\pi n \right)=(-1)^n\sin\left(\pi (\sqrt{n^2+a^2}-n) \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}$$</span></p>
<p>But what I don't understand is how did they come up with that equivalence? First I thought that they used the limit comparison test but now I can see that you can't use that test because we're dealing with alternating series. Did they do a mistake or something?</p>
<p>Can somebody help me understand this hint and how to solve this problem?</p>
| robjohn | 13,854 | <p><strong>Hint:</strong>
<span class="math-container">$$
\begin{align}
\sum_{n=1}^\infty\sin\left(\pi\sqrt{n^2+a^2}\right)
&=\sum_{n=1}^\infty(-1)^n\sin\left(\pi\sqrt{n^2+a^2}-\pi n\right)\\
&=\sum_{n=1}^\infty(-1)^n\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)
\end{align}
$$</span>
where <span class="math-container">$\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)$</span> decreases monotonically to <span class="math-container">$0$</span> for <span class="math-container">$n\ge a^2$</span>. That is, for <span class="math-container">$n\ge a^2$</span>, <span class="math-container">$\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\le\frac{\pi a^2}{2n}\le\frac\pi2$</span> and <span class="math-container">$\sin(x)$</span> maps <span class="math-container">$[0,\pi/2]$</span> monotonically onto <span class="math-container">$[0,1]$</span>.</p>
|
3,528,162 | <p>I'm encountering the following interesting math problem:</p>
<p>A device consists of 5 independently working blocks. </p>
<p>Each of them has a damage probability of 1/4. Upon damage of 1, 2 or 3 blocks, the probabilities for shutting the device down are respectively 1/5 , 2/5, 4/5. </p>
<p>If more blocks are damaged the device's probability of shutting down is 1 (100%). What is the expected number of damaged blocks?
If the device has shut down, what is the probability exactly 2 blocks to be damaged?</p>
<p>What I tried:
I tried combining the probabilities for the damaged probability blocks, and then dividing them, but I don't seem to do it properly.</p>
<p>I also thought of:
P(A or B) = P(A) + P(B) - P(A and B), but still... to no avail.</p>
<p>Any idea how to proceed will be of use!</p>
| jvdhooft | 437,988 | <p>Since there are <span class="math-container">$n = 5$</span> blocks, and each block has a probability <span class="math-container">$p =\frac{1}{4}$</span> of being damaged, the expected number of damaged blocks <span class="math-container">$d$</span> equals:</p>
<p><span class="math-container">$$d = n \cdot p = 5 \cdot \frac{1}{4} = \frac{5}{4}$$</span></p>
<p>Call <span class="math-container">$X$</span> the number of damaged blocks. Using the binomial distribution, we find:</p>
<p><span class="math-container">$$P(X = 1) = {5 \choose 1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^4 = 5 \cdot \frac{3^4}{4^5} \approx 0.3955$$</span></p>
<p>Call <span class="math-container">$A$</span> the event in which the device is being shut down. We find:</p>
<p><span class="math-container">$$P(A, X = 1) = P(X = 1) P(A | X = 1) = 5 \cdot \frac{3^4}{4^5} \cdot \frac{1}{5} \approx 0.0791$$</span></p>
<p>Likewise, we can determine the probability of the device being shut down because exactly two, three, four or five blocks were damaged:</p>
<p><span class="math-container">$$P(A, X = 2) = {5 \choose 2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3 \cdot \frac{2}{5} \approx 0.1055$$</span></p>
<p><span class="math-container">$$P(A, X = 3) = {5 \choose 3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^2 \cdot \frac{4}{5} \approx 0.0703$$</span></p>
<p><span class="math-container">$$P(A, X = 4) = {5 \choose 4} \left(\frac{1}{4}\right)^4 \left(\frac{3}{4}\right)^1 \cdot 1 \approx 0.0146$$</span></p>
<p><span class="math-container">$$P(A, X = 5) = {5 \choose 5} \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^0 \cdot 1 \approx 0.0010$$</span></p>
<p>Given that the device has shut down, the probability of two blocks being damaged thus equals:</p>
<p><span class="math-container">$$P(X = 2 | A) = \frac{P(A, X = 2)}{P(A)} = \frac{P(A, X = 2)}{P(A, X = 1) + P(A, X = 2) + P(A, X = 3) + P(A, X = 4) + P(A, X = 5)} \approx \frac{0.1055}{0.0791 + 0.1055 + 0.0703 + 0.0146 + 0.0010} \approx 0.3899$$</span></p>
|
3,250,593 | <p>Revising the modal logic principles, I have encountered an negative introspection axiom:
<span class="math-container">$$ \vDash \neg \square \alpha \longrightarrow \square \neg \square \alpha $$</span>
with additional information, that it is a direct effect from Euclidean property of accessibility relation <span class="math-container">$\mathbb{R}$</span>. Square operator is defined as follows (for Kripke structure <span class="math-container">$\kappa = (\mathbb{W}, \mathbb{R}, m)$</span>):
<span class="math-container">$$ \kappa, w \vDash \square \alpha \equiv \forall w' \in \mathbb{W} \quad w\mathbb{R}w' \implies \kappa, w' \vDash \alpha $$</span></p>
<p>I got stuck in the point where I have to apply the Euclidean property. We basically know that for any Kripke structure <span class="math-container">$\kappa$</span> and state <span class="math-container">$w \in \mathbb{W}$</span>:
<span class="math-container">$$ \exists w' \in \mathbb{W} \quad w\mathbb{R}w' \wedge \kappa, w' \nvDash \alpha$$</span>
and from the fact that <span class="math-container">$\mathbb{R}$</span> is Euclidean:
<span class="math-container">$$ \forall w_{1}, w_{2}, w_{3} \in \mathbb{W} \quad w_{1}\mathbb{R}w_{2} \wedge w_{1}\mathbb{R}w_{3} \implies w_{2}\mathbb{R}w_{3} $$</span></p>
<p>I've tried to make it similarily to <a href="https://math.stackexchange.com/questions/1835246/the-positive-introspection-axiom">this</a>, but there we have all universal quantifiers, so it was moreorless sensible to me. I would be really grateful, if someone would provide a solution or any hint to that. </p>
| mattrym | 679,574 | <p>By the time, I have found a possible solution, I would appreciate some comments:</p>
<p>Let's get any possible world <span class="math-container">$w \in \mathbb{W}$</span>, such as <span class="math-container">$\kappa, w \vDash \neg \square \alpha $</span>. Then,from a definition of <span class="math-container">$\square$</span>:
<span class="math-container">$$ \exists w' \in \mathbb{B} w\mathbb{R}w' \wedge \kappa, w' \vDash \neg \alpha $$</span>
Let's consider any state <span class="math-container">$w'' \in \mathbb{W}$</span> such as <span class="math-container">$w\mathbb{R}w''$</span>. Then, from Euclideanity of <span class="math-container">$\mathbb{R}$</span> we have <span class="math-container">$w\mathbb{R}w'' \wedge w\mathbb{R}w' \implies w''\mathbb{R}w'$</span>. That concludes to the following:
<span class="math-container">$$\forall w'' \in \mathbb{W} \, \exists w' \in \mathbb{W} \, w\mathbb{R}w'' \implies w''\mathbb{R}w' \wedge \kappa, w' \vDash \neg \alpha $$</span>
which is equivalent to:
<span class="math-container">$$ \forall w'' \in \mathbb{W} \,, w\mathbb{R}w'' \implies \kappa, w'' \vDash \neg \square \alpha$$</span>
and is in fact a definition of <span class="math-container">$\square$</span> operator: <span class="math-container">$\kappa, w \vDash \square \neg \square \alpha$</span>. To sum up, for any Kripke structure <span class="math-container">$\kappa$</span> and state <span class="math-container">$w \in \mathbb{W}$</span> we have <span class="math-container">$\kappa, w \vDash \neg \square \alpha \implies \kappa, w \vDash \square \neg \square \alpha$</span>, thus thesis: <span class="math-container">$\vDash \neg \square \alpha \longrightarrow\square \neg \square \alpha$</span>. <span class="math-container">$\blacksquare$</span></p>
|
2,326,072 | <p>We define 3 sequences $(a_n),(b_n),(c_n)$ with positive terms so that
$$ a_{n+1}\leq\frac{b_n+c_n}{3}\ ,\ b_{n+1}\leq\dfrac{a_n+c_n}{3}\ ,\ c_{n+1}\leq\dfrac{a_n+b_n}{3} $$
Check if any of $(a_n),(b_n),(c_n)$ converge, and if they do find their limit.</p>
<p>PROOF</p>
<p>My part of the proof is this: By adding the above inequalities we get
$$ a_{n+1}+b_{n+1}+c_{n+1}\leq\frac{2}{3}(a_n+b_n+c_n) $$
We define the sequence $ (x_n) $ with $ x_n=a_n+b_n+c_n $ so we have
$$ x_{n+1}\leq\frac{2}{3}x_n\Rightarrow \frac{x_{n+1}}{x_n}\leq\frac{2}{3}<1 $$
which implies that $ (x_n) $ is decreasing. We also have $ x_n>0, \forall n $ thus it's bounded, and so it converges to $ 0 $. Is it correct to say that since $ a_n<x_n $ then $ a_n\to0 $?</p>
<p>EDIT</p>
<p>I forgot to mention that we also prove that $x_n\to0$.</p>
| Andre | 303,581 | <p>Yes because you have the "sandwiched" inequality $0 < a_{n} < x_{n} \rightarrow 0$, so $a_{n} \rightarrow 0$.</p>
<p>EDIT: Of course, you have to make sure that $x_n \rightarrow 0$ in the first place, but you have that it is bounded by a geometric sequence which converges to zero which implies the stated convergence.</p>
|
1,012,652 | <p>I have a problem with the following question.</p>
<p>For which $n$ does the following equation have solutions in complex numbers</p>
<p>$$|z-(1+i)^n|=z $$</p>
<p>Progress so far.</p>
<ol>
<li><p>Let $z=a+bi$.</p></li>
<li><p>Since modulus represents a distance, the imaginary part of RHS has to be 0. This immediately makes $b=0$.</p></li>
<li><p>If solutions are in the complex domain $|a-(1+i)^n|=a $ by 2., and $a$ is Real. </p></li>
<li><p>?</p></li>
</ol>
<p>I don't know where to go from here. </p>
| Matt Samuel | 187,867 | <p>Note that $1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$. We have </p>
<p>$|z-(1+i)^n|=|z-(\sqrt{2})^n(\cos(n\pi/4)+i\sin(n\pi/4))|=z$</p>
<p>$\sqrt{(z-(\sqrt{2})^n\cos(n\pi/4))^2+(\sqrt{2})^{2n}\sin^2(n\pi/4)}=z$</p>
<p>so</p>
<p>$(z-2^{n/2}\cos(n\pi/4))^2+2^n\sin^2(n\pi/4)=z^2$,</p>
<p>$-2^{n/2+1}z\cos(n\pi/4)+2^n\cos^2(n\pi/4)+2^n\sin^2(n\pi/4)=0$,</p>
<p>$2^{n/2+1}z\cos(n\pi/4)=2^n$,</p>
<p>$$z=\frac{2^{n/2-1}}{\cos(n\pi/4)}$$</p>
<p>whenever $n$ is odd; for $n$ even there is no solution.</p>
|
1,012,652 | <p>I have a problem with the following question.</p>
<p>For which $n$ does the following equation have solutions in complex numbers</p>
<p>$$|z-(1+i)^n|=z $$</p>
<p>Progress so far.</p>
<ol>
<li><p>Let $z=a+bi$.</p></li>
<li><p>Since modulus represents a distance, the imaginary part of RHS has to be 0. This immediately makes $b=0$.</p></li>
<li><p>If solutions are in the complex domain $|a-(1+i)^n|=a $ by 2., and $a$ is Real. </p></li>
<li><p>?</p></li>
</ol>
<p>I don't know where to go from here. </p>
| Lucian | 93,448 | <ul>
<li>Obviously, <em>z</em> is a point on the positive semi-axis, since absolute values are always positive reals.</li>
<li>$|z_1-z_2|$ represents the distance between two points on the complex plane.</li>
<li>Therefore, <em>z</em> is equally distanced from the origin <em>O</em>, and from $A_n=(1+i)^n$. This implies that it lies on the perpendicular bisector of the segment $(OA_n)$.</li>
<li>Since <em>n</em> is a natural number, and since the angle between $A_1=1+i$ and the positive semiaxis is $45^\circ$, it follows that, with each successive multiplication, the angle between the segment $OA_k$ and the positive semiaxis grows by $45^\circ$, until it resets on the eighth turn, since $360^\circ/45^\circ=8$.</li>
<li>But on which of these eight cyclically repeating situations does the perpendicular bisector of the segment $(OA_n)$ actually meet or intersect the positive semiaxis ? Obviously, it can only happen for $n\equiv0,\pm1\mod8$.</li>
</ul>
|
3,340,374 | <p>I'm writing up several proofs for myself, all of which have a particular sticking point.</p>
<p>Essentially, I want to prove that for a function <span class="math-container">$f$</span> of two real variables, we have</p>
<p><span class="math-container">$\lim_{h \to 0} \frac{f(x \;+\; h, \; y \;+\; h) \; - \; f(x, \; y \;+ \;h)}{h} = \frac{\partial f}{\partial x}$</span>.</p>
<p>When I've seen this crop up in textbooks and lecture notes (say, in the context of proving the multivariable chain rule), it's generally just been taken for granted, rather than proven.</p>
<p>Intuitively, how can we be sure that the function is sufficiently well behaved that having <span class="math-container">$y + h$</span> instead of <span class="math-container">$y$</span> in the second argument doesn't 'mess things up' and not allow this limit to be the partial derivative? And moreover, how could I give a rigorous proof of this?</p>
<p>Thanks all!</p>
| Allawonder | 145,126 | <p>There's no way anything could be messed up since <em>the second coordinate is always fixed.</em> It's only the first that's varying. That is, we're considering the change from the point <span class="math-container">$(x,y+h)$</span> to the point <span class="math-container">$(x+h,y+h).$</span> Throughout this change, the second coordinate is clearly fixed. Thus, it's not concerned with the limiting process at all. This is why we can be sure that the limit, if it exists, is indeed the derivative with respect to <span class="math-container">$x,$</span> <em>fixing <span class="math-container">$y.$</span></em> Thus, there's nothing more to justify than in the usual derivative. Hence, what you noticed about the books.</p>
|
2,508,103 | <p>Could someone, please, define a cycle in a $3-$Uniform Hypergraph? I looked at different resources, but they have different definitions and I am so confused!!</p>
| Donald Splutterwit | 404,247 | <p>An example of a ("pentagon") $5-$cycle in a $3-$ uniform hypergraph.</p>
<p><a href="https://i.stack.imgur.com/lCAAF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lCAAF.png" alt="enter image description here"></a></p>
<p>The definition of a cycle is very similar to that for a graph it is a sequence of verticies and hyperedges such that
\begin{eqnarray*}
a_1 E_{1,2} a_2 E_{2,3} a_3 \cdots a_k E_{k,1} a_1
\end{eqnarray*}
such that $a_i, a_{i+1} \in E_{i,i+1}$.</p>
|
2,508,103 | <p>Could someone, please, define a cycle in a $3-$Uniform Hypergraph? I looked at different resources, but they have different definitions and I am so confused!!</p>
| Misha Lavrov | 383,078 | <p>The problem is that there <em>are</em> different notions of what cycles are in hypergraphs. You always have to specify which you mean.</p>
<p>In a $3$-uniform hypergraph:</p>
<ul>
<li>A <em>loose</em> cycle consists of edges $\{v_1,v_2,v_3\}, \{v_3,v_4,v_5\}, \dots, \{v_{2k-1}, v_{2k}, v_1\}$. From each edge of the cycle, we pick a "first vertex" and a "last vertex" (this information is <em>not</em> built into the hypergraph). The last vertex of each edge is the first vertex of the next, the last vertex of the last edge is the first vertex of the first edge, and no other vertices are shared.</li>
<li>A <em>tight</em> cycle consists of edges $\{v_1,v_2,v_3\}, \{v_2,v_3,v_4\}, \dots, \{v_{k-1},v_{k},v_1\}, \{v_k, v_1,v_2\}$. There is a sequence of vertices $v_1, v_2, \dots, v_k$, and every three consecutive vertices form an edge, as do $\{v_{k-1}, v_k, v_1\}$ and $\{v_k, v_1, v_2\}$.</li>
<li>A <em>Berge</em> cycle is a sequence of distinct vertices and edges $v_1, e_{12}, v_2, e_{23}, v_3, \dots, v_k, e_{k1}, v_1$ such that the edge $e_{i,i+1}$ contains $v_i$ and $v_{i+1}$ (and $e_{k1}$ contains $v_k$ and $v_1$). This is different from a loose cycle because we don't ask for other vertices not to be shared between the edges.</li>
</ul>
<p>For $r$-uniform hypergraphs when $r>3$, all these notions also exist, but we can interpolate between "loose" and "tight". We could ask, for instance, that adjacent edges share two out of $r$ vertices.</p>
|
3,402,800 | <p>Let,<span class="math-container">$E$</span> be a zero dimensional sheaf on an algebraic surface <span class="math-container">$S$</span> over the field of complex numbers,i.e <span class="math-container">$dim(Supp(E)) =0$</span></p>
<p>Then my question is the following : Is it true that <span class="math-container">$H^{1}(E) =0$</span> or in general <span class="math-container">$H^{i}(E) =0$</span> for all <span class="math-container">$i \geq 1$</span>?</p>
<p>Any help from anyone is welcome</p>
| Slup | 348,401 | <p>This can be also viewed as a consequence of a general fact due to Grothendieck (which can be proved by noetherian induction).</p>
<p><em>Let <span class="math-container">$X$</span> be a noetherian topological space of dimension <span class="math-container">$n$</span>. Then for all <span class="math-container">$i>n$</span> and all sheaves <span class="math-container">$\mathcal{F}$</span> of abelian groups on <span class="math-container">$X$</span> we have <span class="math-container">$\mathrm{H}^i(X,\mathcal{F}) = 0$</span>.</em></p>
<p>For this fact c.f. Hartshorne's <strong>Algebraic Geometry, chapter III, Theorem 2.7</strong>. The proof uses the fact (as a basis of induction) that on zero dimensional neotherian spaces (on finite discrete spaces) all higher cohomologies of sheaves vanish. So the best approach to your question is showing that sheaves on zero dimensional noetherian spaces are flabby. </p>
|
893,259 | <p>I have tried this problem multiple times, I have the solution but not the steps. I keep getting the wrong answer. I believe it may be in the algebra after I have taken the Laplace on both sides.</p>
<p>$y''+y = \sin(t) ;\:\: y(0) = 1, \:\: y'(0) = -1 $</p>
| Dmoreno | 121,008 | <p>Follow the next steps I describe for you below:</p>
<ul>
<li>Take Laplace transform on both sides of the ODE to have:</li>
</ul>
<p>$$ s^2 Y(s) - s y_0 - y'_0 + Y(s) = \frac{1}{s^2+1}, \quad Y(s) = \mathcal{L}_s [y(t)]$$</p>
<ul>
<li>Substitute data and solve for $Y(s)$:</li>
</ul>
<p>$$Y(s) = \frac{1}{s^2+1} \left( s - 1 + \frac{1}{s^2 +1} \right)$$</p>
<ul>
<li>Inverse-Laplace-transform both sides to find:</li>
</ul>
<p>\begin{align}
\color{blue}{y(t)} & = \mathcal{L}^{-1}_t[Y(s)] = \mathcal{L}^{-1}_t \left[ \underbrace{\color{green}{\frac{s-1}{s^2+1}}}_{\text{tabulated}} + \color{red}{\frac{1}{(s^2+1)^2}} \right] = \\
&=\color{green}{\cos{t} - \sin{t} }+ \color{red}{\frac{1}{2} (\sin{t} - t\cos{t})} = \color{blue}{ \cos{t} - \frac{1}{2} \sin{t} - \frac{t}{2}\cos{t} } .
\end{align}</p>
<ul>
<li>You can hereby identify the solution of the homogeneous equation as $y_h(t) = \cos{t} - 1/2 \, \sin{t}$ and the particular solution as $y_p(t) = - t/2 \, \cos{t} $.</li>
</ul>
<p>Hope this helps!</p>
<p>Cheers.</p>
<p><strong>Edit:</strong> you can check that the inverse Laplace transform of $1/(s^2+1)^2$ can be computed as (convolution property):</p>
<p>$$ \mathcal{L}^{-1}_t [G(s)F(s)] = \int^t_0 g(t-\tau) f(\tau) \, \mathrm{d}\tau = \int^t_0 \sin(t-\tau) \sin{\tau} \, \mathrm{d} \tau = \frac{1}{2}(\sin{t} - \cos{t}).$$</p>
|
3,990,717 | <p>I'm currently a self-learner on Abstract Algebra and Group theory. The book that I'm currently reading is <em>A First Course In Abstract Algebra</em> by <em>John B. Fraleigh</em>. So I know this question might truly be a basic question which every student would be taught in class and there is confusion about it. However, maybe it is me who is not careful in reading the book that raises in my mind this question. So given a binary operation <span class="math-container">$\ast $</span> on a set <span class="math-container">$S$</span>, how can we compute:<br />
<span class="math-container">$$a \ast b \ast c$$</span>
Actually, on page 23 of the book, the author does mention this but I still can not get it. I don't know if we should compute this from left to right or reversely or we will just leave it there if <span class="math-container">$\ast$</span> is not associative. Moreover, if this binary operation is associated, is it true to define that:
<span class="math-container">$$a \ast b \ast c = \left(a \ast b \right)\ast c$$</span>
Then, after learning about <strong>Cyclic group</strong> there is a problem of which can be stated as below:</p>
<blockquote>
<p>Let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be elements of a group <span class="math-container">$G$</span>. Show that if <span class="math-container">$ab$</span> has finite order <span class="math-container">$n$</span>, then <span class="math-container">$ba$</span> also has order <span class="math-container">$n$</span>.</p>
</blockquote>
<p>I know that we should show that <span class="math-container">$(ba)^n = e$</span> which can be driven from <span class="math-container">$(ab)^n = e$</span>. Unfortunately, since <span class="math-container">$G$</span> might not be an Abelian group so what I had gone so far before looking for a solution is just manipulating <span class="math-container">$(ab)^n = e$</span> by using the fact that <span class="math-container">$(ab)^{-1} = b^{-1} a^{-1}$</span> since I don't know how to move further after splitting <span class="math-container">$(ab)^n = (ab)(ab)^{n-1}$</span>. I also get confused whether <span class="math-container">$(ab)^n = (ab)(ab)^{n-1}$</span> or <span class="math-container">$(ab)^n = (ab)^{n-1} (ab)$</span>. So I checked up for the solution and this is the way they manipulate it that I found: <a href="https://math.stackexchange.com/questions/622305/show-that-if-ab-has-finite-order-n-then-ba-also-has-order-n-fraleigh">Show that if ab has finite order n, then ba also has order n. - Fraleigh p. 47 6.46.</a>
So I completely have no idea about this can be infered:</p>
<p><span class="math-container">\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies
\color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e
\end{align}</span>
I also tried to check another link in order to understand the solution: <a href="https://math.stackexchange.com/questions/3659250/let-g-be-a-group-and-a-b%E2%88%88g-suppose-order-of-ab-in-g-is-n-show-that?noredirect=1&lq=1">same question</a> And the asker states that:
<span class="math-container">$$(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$$</span>
<span class="math-container">$$=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$$</span>
<span class="math-container">$$=a(ba)^{n-1} b$$</span>
It is here that the number of terms which we're dealing with is <span class="math-container">$n$</span> now. I don't have any idea how could he manipulates from the first line to the second line so I think that I need some helps on this. In summary, how can we compute binary operation of more than 3 elements? Pleas explain the problems that I state above and give some other examples if possible. Thank you so much.</p>
| Community | -1 | <p>Conjugation by any element of the group is an automorphism, and thus preserves order. All that is left is to note that <span class="math-container">$ab$</span> and <span class="math-container">$ba$</span> are conjugates: <span class="math-container">$b(ab)b^{-1}=ba$</span>.</p>
|
137,253 | <p>We have the following: $A,B,C$ are sets. </p>
<p>$C = \{ab: a \in A, b \in B\}$. </p>
<p>What is the relationship between $\sup(C),\sup(A)$, and $\sup(B)$?. </p>
<p>Is it: $$\sup(C) \le \sup(A) \sup(B)\;,$$ and why?.</p>
| Yuval Filmus | 1,277 | <p>Suppose for simplicity that $A$ and $B$ are finite. For all $a \in A$, we have $a \leq \max A$. For all $b \in B$, we have $b \leq \max B$. What can you say about $ab$ (assuming $A,B\subseteq \mathbb{R}_{\geq 0}$)? What can you deduce about $\max C$?</p>
<p>You can also think about the extreme case $A = \{a\}$, $B = \{b\}$, $C = \{ab\}$. What does it imply about the inequality?</p>
|
546,083 | <p>Thanks to internet, I found and understand how to solve diophantine $x^2 - Dy^2 = 1$. Now I would like to solve the following diophantine equation :
$$x^2 - 2y^2 = x - 2y$$
but I don't know how to do it, even if I could read articles explaining how to solve the diophantine $x^2 - Dy^2 = c$.</p>
| Shaun | 104,041 | <p>That's not the only answer in fact.</p>
<p>For any $\alpha\in\mathbb{C}$, we define, for any complex $z=a e^{i\theta }\neq 0$, the multifunction
$$[z^{\alpha }]:=\{e^{\alpha (\log |z|+i\theta )} : \theta\in\{\arg (z)+2k\pi i: k\in\mathbb{Z}\}\},$$
taking $z^{\alpha }$ as $w=be^{i\phi }\in [z^{\alpha }]$ such that $\phi\in (-\pi , \pi ]$; we take the "principle value" in the set defined above. [Recall that $e^{2\pi }=1.$.]</p>
<p>Take $z=\alpha =i$.</p>
<p>[This definition is from "Introduction to Complex Analysis: second edition," by H. A. Priestley.]</p>
|
2,115,451 | <p>$$\lim_{n\to \infty}\frac{\frac{1}{2\ln2}+\frac{1}{3\ln3}+\ldots+\frac{1}{n\,\ln\,n}}{\ln(\ln\,n)}$$</p>
<p>The result is $1$ (according to the book, though it does not show the steps, which I'm interested in).</p>
<p>I've applied the theorem and it led me to an equally unhelpful limit.</p>
| Rene Schipperus | 149,912 | <p>If you apply Stolz you get</p>
<p>$$\frac{1}{n\ln n(\ln \ln n-\ln\ln(n-1))}$$ Now take $e$ to the power of the denominator you get
$$\left(\frac{\ln n}{\ln(n-1)}\right)^{n\ln n}$$
Write this as
$$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}$$</p>
<p>Where $$a_n=n\ln n \frac{\ln (1+\frac{1}{n-1})}{\ln (n-1)}$$ its easy to see that $a_n\to 1$.
And thus $$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}\to e$$</p>
<p>Thus the denominator limits to $\ln e=1$ so the original expression above limits to $1$.</p>
|
134,424 | <p>I would like to create a function where I can define which case I want to use to create a path.</p>
<pre><code>p1 = {40, 48}; p2 = {50, 116}; p3 = {63, 160};
listPurple = Symbol["p" <> ToString[#]] & /@ Range[3];
disksPurple = {Purple, Disk[#, 2] & /@ listPurple};
Graphics[{disksPurple}, ImageSize -> 200]
</code></pre>
<p><a href="https://i.stack.imgur.com/X8otFm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X8otFm.png" alt="Imagem"></a></p>
<p>I do not want two functions, as I created:</p>
<pre><code>lVertical[p1_, p2_] := {p1, {p1[[1]], p2[[2]]}};
lHorizontal[p1_, p2_] := {p1, {p2[[1]], p1[[2]]}};
</code></pre>
<p>With them I define whether it will be a horizontal or vertical line:</p>
<pre><code>l1 = lVertical[p1, p2];
l2 = lHorizontal[p2, p1];
l3 = lHorizontal[p2, p3];
l4 = lVertical[p3, p2];
lines = Sort@Symbol["l" <> ToString[#]] & /@ Range[4];
l = {Red, Dashed, Line[#] & /@ lines};
Graphics[{l, disksPurple}, ImageSize -> 200]
</code></pre>
<p><a href="https://i.stack.imgur.com/cl4HEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cl4HEm.png" alt="Imagem"></a></p>
<p>I would like it in a format similar to this:</p>
<pre><code>f[p1_, p2_, lVertical or lHorizontal]
</code></pre>
| glS | 27,539 | <p>Here is an attempt at automatically finding the best paths to use to join the points.
This avoids having to explicitly give the horizontal and vertical specification for every point.</p>
<pre><code>ClearAll[generateDirections, hvPath]
generateDirections[pts_, initialDirection_] :=
Module[{d = initialDirection /. None :> RandomChoice[{1, 2}]},
Join[{d},
(Ordering[#[[All, d = 3 - d]]][[2]] == 2 || (d = 3 - d); d) & /@
Partition[pts, 3, 1]]
]
hvPath[{pt1 : {a_, b_}, pt2 : {i_, j_}}, dir_] :=
{pt1, Switch[dir, 1, {i, b}, 2, {a, j}], pt2}
hvPath[pts : {{_, _} ..}, initialDirection_ : None] :=
Join @@ MapThread[
hvPath, {Partition[pts, 2, 1], generateDirections[pts, initialDirection]}]
</code></pre>
<p>The idea is to use <code>generateDirections</code> to generate the best horizontal-vertical pathways, using the rule of thumb of initiating every new path using alternativily horizontal and vertical lines, except when this results in going over the same line used to arrive to the given point.
Once the optimal directions are found the rest is easily handled by <code>hvPath</code> which generates the list with all the middle points.</p>
<p>Here is a couple of usage example:</p>
<pre><code>With[{pts = {{0, 0}, {1, 1}, {2, 2}, {3, 3}}},
Graphics[{
PointSize@0.02, Point@pts,
Dashed, Line@hvPath@pts
}]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/Jirb3m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jirb3m.png" alt="enter image description here"></a></p>
<p>and a more complex one in which we join 20 random points:</p>
<p><a href="https://i.stack.imgur.com/AtztS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AtztS.png" alt="enter image description here"></a></p>
|
134,424 | <p>I would like to create a function where I can define which case I want to use to create a path.</p>
<pre><code>p1 = {40, 48}; p2 = {50, 116}; p3 = {63, 160};
listPurple = Symbol["p" <> ToString[#]] & /@ Range[3];
disksPurple = {Purple, Disk[#, 2] & /@ listPurple};
Graphics[{disksPurple}, ImageSize -> 200]
</code></pre>
<p><a href="https://i.stack.imgur.com/X8otFm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X8otFm.png" alt="Imagem"></a></p>
<p>I do not want two functions, as I created:</p>
<pre><code>lVertical[p1_, p2_] := {p1, {p1[[1]], p2[[2]]}};
lHorizontal[p1_, p2_] := {p1, {p2[[1]], p1[[2]]}};
</code></pre>
<p>With them I define whether it will be a horizontal or vertical line:</p>
<pre><code>l1 = lVertical[p1, p2];
l2 = lHorizontal[p2, p1];
l3 = lHorizontal[p2, p3];
l4 = lVertical[p3, p2];
lines = Sort@Symbol["l" <> ToString[#]] & /@ Range[4];
l = {Red, Dashed, Line[#] & /@ lines};
Graphics[{l, disksPurple}, ImageSize -> 200]
</code></pre>
<p><a href="https://i.stack.imgur.com/cl4HEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cl4HEm.png" alt="Imagem"></a></p>
<p>I would like it in a format similar to this:</p>
<pre><code>f[p1_, p2_, lVertical or lHorizontal]
</code></pre>
| Simon Woods | 862 | <p>You can simply provide two definitions for <code>f</code></p>
<pre><code>f[p1_, p2_, lVertical] := {p1, {p1[[1]], p2[[2]]}}
f[p1_, p2_, lHorizontal] := {p1, {p2[[1]], p1[[2]]}}
</code></pre>
|
14,858 | <p>Let $V$ be a complex vector space of dimension 6 and let $G\subset {\mathbb P}^{14}\simeq {\mathbb P}(\Lambda^2V)$ be the image of the Plucker embedding of the Grassmannian $Gr(2, V)$.</p>
<ol>
<li>Why the degree of $G$ is 14? or in general, how to calculate the degree of a Plucker embedding?</li>
</ol>
<p>Let ${\mathbb P}^8\simeq L\subset {\mathbb P}(\Lambda^2V)$ be a generic 8-plane and $S$ be the intersection of $L$ with $G$.</p>
<ol>
<li>How to prove that $S$ is a K3 surface?</li>
</ol>
<p>Another question: the paper said that this construction depends on 19 parameters. I know that this is the dimension of the deformation family of the polarized K3 we get here. But I think that in this statement 19 is coming from varying the generic 8-plane. How can we obtain this number?</p>
| Andrea Ferretti | 828 | <p>Guess you are reading Beauville-Donagi :-)</p>
<p>To compute the degree of the Grassmannian you can note that the hyperplane divisor under the Plucker embedding has class $\sigma_1$, so the degree of the Grassmannian is just $\sigma_1^8$, which you can compute by Schubert calculus. See the paragraph about Grassmannians in Griffiths-Harris.</p>
<p>Namely $\sigma_1^2 = \sigma_{1,1} + \sigma_2$ hence $\sigma_1^3 = 2\sigma_{2,1} + \sigma_3$ by Pieri's formula and finally $\sigma_1^4 = \sigma_4 + 3 \sigma_{3,1} + 2 \sigma_{2,2}$ again by Pieri.</p>
<p>Since $\sigma_4$, $\sigma_{3,1}$ and $\sigma_{2,2}$ are Poincaré dual to themselves, you find $\sigma_1^8 = 1 + 3^2 + 2^2 = 14$</p>
<p>The other question has already been answered by Dmitri.</p>
|
807,795 | <p>Task is to find an eigenvector of the following linear operator:<br>
$f \to \int^{x}_{-x} f(t)dt$ in the linear span $\langle cos(x), sin(x), ...,cos(mx),sin(mx)\rangle$.</p>
<p>I know how to find eigenvectors and have the general idea about what linear span and linear operators are, but still I feel like I'm a bit out of my depth with this problem. </p>
<p>Since we see integral in function, mapping transforms original vector to $\mathbb{R}$. But I keep wondering what is an original vector space, what's $x$ and $t$ here.</p>
<p>Please note, I probably would fail to understand the complete solution and it's not what I want. I would appreciate just some hints on how to approach this problem. Thanks!</p>
| Just_a_fool | 114,428 | <p>Note: </p>
<p>$\sum_{k=1}^n\sqrt{k^2+1}$ > $a_k=\sum_{k=1}^n\sqrt{k^2}$</p>
<p>for any positive n</p>
<p>As the definition of the RHS series (lets call it $c_n$) suggests, it is equal to the sum of the positive integers up to n.
This sum is given by the formula:</p>
<p>$$c_n=(n/2)(n+1)$$</p>
<p>Plugging this into your original definition for $b_n$ yields: $n + 1 - n$ or simply $1$.</p>
<p>Also note that the sum for $a_n$ approaches $c_n$ as $n$ approaches infinity. Now, with each increase in n, the difference between the 2 series steadily decreases. </p>
<p>Proving that the difference steadily decreases (from one term to the next) proves that $b_n$ steadily decreases (and as it seems, approaches 1). It's up to you to prove this critical bit.</p>
<p>Good day! :)</p>
|
807,795 | <p>Task is to find an eigenvector of the following linear operator:<br>
$f \to \int^{x}_{-x} f(t)dt$ in the linear span $\langle cos(x), sin(x), ...,cos(mx),sin(mx)\rangle$.</p>
<p>I know how to find eigenvectors and have the general idea about what linear span and linear operators are, but still I feel like I'm a bit out of my depth with this problem. </p>
<p>Since we see integral in function, mapping transforms original vector to $\mathbb{R}$. But I keep wondering what is an original vector space, what's $x$ and $t$ here.</p>
<p>Please note, I probably would fail to understand the complete solution and it's not what I want. I would appreciate just some hints on how to approach this problem. Thanks!</p>
| Mick A | 153,109 | <p>Try to show $b_{n+1} - b_n < 0$:</p>
<p>$b_{n+1} - b_n = \left[\dfrac{2a_{n+1}}{n+1} - (n+1)\right] - \left[\dfrac{2a_n}n - n\right]$</p>
<p>$ = \dfrac1{n(n+1)}\left[2n \sum_{k=1}^{n+1}\sqrt{k^2+1} - 2(n+1) \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]$</p>
<p>$ = \dfrac1{n(n+1)} \left[2n \sqrt{(n+1)^2 + 1} - 2 \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]\qquad\qquad\mbox{(1)}$</p>
<p>$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2n}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2k}\right) - n(n+1)\right]$</p>
<p>$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2n}\right) - n(n+1) - \ln(n+1) - n(n+1)\right]$ ... [Ref: <a href="http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Integral_test" rel="nofollow">http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Integral_test</a> to show $\sum_{k=1}^n\dfrac1k > \ln(n+1)$]</p>
<p>$ = \dfrac1{n(n+1)} \left[1 - \ln(n+1)\right]$</p>
<p>$ < 0$ for all $n \geq 2$.</p>
<p>For $n = 1$ case: calculate: $b_2 - b_1 = \sqrt5 - \sqrt2 - 1 < 0$.</p>
<p>Revised part to fix mistake GL5 found, following from (1):</p>
<p>$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2(k+1)}\right) - n(n+1)\right]$</p>
<p>$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - n(n+1) - \sum_{k=2}^{n+1} \dfrac1{k} - n(n+1)\right]$</p>
<p>$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2(n+1)}\right) - n(n+1) - \ln(n+1) + 1 - \dfrac1{n+1} - n(n+1)\right]$</p>
<p>$ = \dfrac1{n(n+1)} \left[2 - \dfrac2{n+1} - \ln(n+1)\right]$</p>
<p>$ < 0$ for all $n \geq 4$.</p>
<p>Initial cases for $1 \leq n \leq 3$ to be done by individual calculation. Having 3 initial cases is inelegant. There might be a way to tighten up the inequalities to reduce it.</p>
<p>Mick</p>
|
3,213,253 | <p>We're given dynamical system:</p>
<p><span class="math-container">$$
\dot x = -x + y + x (x^2 + y^2)\\
\dot y = -y -2x + y (x^2 + y^2)
$$</span></p>
<p>Question is what's the largest constant <span class="math-container">$r_0$</span> s.t. circle <span class="math-container">$x^2+y^2 < r_0^2$</span> lies in the origins basin of attraction.</p>
<p>So far with relatively easy algebra I've got:</p>
<p><span class="math-container">$$ \dot r = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
\dot \phi = -(1+\cos^2(\phi))
$$</span></p>
<p>Which immediately shows <span class="math-container">$r_0 \geq \sqrt{1/2}$</span>. How to show that there is no better bound?</p>
| AVK | 362,247 | <p>(This is only a "numerical" answer to the question)</p>
<p>A basin of attraction can be found numerically using reverse-time (or backward) integration. If we choose some initial point and integrate backwards for a sufficient time, we can obtain the set of the system states that it had before it reached the selected initial point. So, if we choose sufficiently many initial points that are contained in the basin of attraction (i.e. sufficiently close to the attracting steady state) and integrate backwards, we can get some idea of how the basin of attraction looks like.</p>
<p>The following picture demonstrates the aprroximation of the basin of attraction of the system:
<a href="https://i.stack.imgur.com/keHGG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/keHGG.png" alt="enter image description here"></a>
The blue curves fill the basin of attraction. The red circle is the largest circle that fits in the basin of attraction.
I have choosen 36 initial points on the circle of radius 0.05. Here is the Matlab code:</p>
<pre><code>axes
hold on
r= 0.05; % radius of the circle of initial points
rpm= @(t,x)[-x(1)+x(2)+x(1)*(x(1)^2+x(2)^2);...
-x(2)-2*x(1)+x(2)*(x(1)^2+x(2)^2)]; % the right part of the system
for fi= 0:pi/36:2*pi
% notice the backward in time direction of integration
[t,z]= ode45(rpm,15:-0.1:0,r*[cos(fi) sin(fi)]);
plot(z(:,1),z(:,2),'b');
end
grid on
h= ezplot('x^2+y^2=0.847^2',[-2 2],[-1.5 1.5]); % draw the circle
h.LineWidth= 1.7;
h.Color= 'red';
axis equal
</code></pre>
|
2,615,427 | <p>I saw the following question in my linear algbra book, and found it rather strange:</p>
<p>"Let $V$ be a real vector space of dimension $n$. Let $L,K \colon V \rightarrow \mathbb{R}$ be linear transformations, so that $\ker(L) \subset \ker(K)$.
Prove that $K=\lambda L$ for a $\lambda \in \mathbb{R}$, using the following steps:</p>
<ol>
<li>Prove this for $K=0$.</li>
<li>Assume $K \neq 0$. Prove that $\dim(\ker(L)) = \dim(\ker(K))$.</li>
<li>Now prove that $K = \lambda L$ for $K \neq 0$."</li>
</ol>
<p>What is meant by "$K=0$"? How is a linear transformation ever equal to a number? I have a feeling that they might mean something else than just "$K$", but I'm not sure.</p>
<p>I tried something for "K maps every vector to $0$", but that doesn't really seem to be what is meant, because then I obviously get that $K(v) = \lambda L(v)$ when $\lambda = 0$, which would mean that this question allows me to just choose whatever $\lambda$ to make the statement true and I'm not sure that is the case here. Also, how would I continue for $K \neq 0$?</p>
<p>Can anyone give me a hint?</p>
<p>Thanks!</p>
| The Phenotype | 514,183 | <p>Assume $0\leq a<b$.</p>
<p>Suppose $a^2\geq b^2$, then $a^2\geq b^2=b\cdot b>ab>a\cdot a=a^2$, which is a contradiction in itself ($a^2<a^2$ is not possible), so $a^2<b^2$.</p>
<p>Suppose $\sqrt{a}\geq \sqrt{b}$, then $a=\sqrt{a}\cdot \sqrt{a}\geq \sqrt{b}\cdot\sqrt{a}\geq\sqrt{b}\cdot \sqrt{b}=b$, which is in contradiction with $a<b$, so $\sqrt{a}<\sqrt{b}$.</p>
|
301,905 | <p>I'm having troubles to solve this integration: $\int_0^1 \frac{x^{2012}}{1+e^x}dx$</p>
<p>I've tried a lot using so many techniques without success. I found $\int_{-1}^1 \frac{x^{2012}}{1+e^x}dx=1/2013$, but I couldn't solve from 0 to 1.</p>
<p>Thanks a lot.</p>
| Community | -1 | <p>You have $$\int_{-1}^{1} \frac{x^{2012}}{1+e^{x}} \ dx =\underbrace{\int_{-1}^{0}\frac{x^{2012}}{1+e^{x}}}_{I_{1}} \ dx + \int_{0}^{1}\frac{x^{2012}}{1+e^{x}} \ dx \qquad \cdots (1)$$</p>
<p>In $I_{1}$ put $x=-t$, then you have $dx = -dt$, and so the limits range from $t=0$ to $t=1$. So you have $$I_{1}= -\int_{1}^{0} \frac{e^{t}\cdot t^{2012}}{1+e^{t}} \ dt = \int_{0}^{1} \frac{e^{x}\cdot x^{2012}}{1+e^{x}} \ dx$$</p>
<p>Put this in equation $(1)$ to get the value.</p>
|
46,505 | <p>What do I do if I have to solve the usual quadratic equation $X^2+bX+c=0$ where $b,c$ are in a field of characteristic 2? As pointed in the comments, it can be reduced to $X^2+X+c=0$ with $c\neq 0$.</p>
<p>Usual completion of square breaks. For a finite field there is <a href="http://www.ams.org/mathscinet-getitem?mr=680147">Chen Formula</a> that roughly looks like $X=\sum_{m} c^{4^m}$. I am more interested in the local field $F((z))$ or actually an arbitrary field of characteristic 2. </p>
| Tim Dokchitser | 3,132 | <p>I think this solves $X^2+X+c=0$ over $F((t))$:</p>
<p>I want to assume that $c\in F[[t]]$. If not, say $c=at^{-m}+...$, then the quadratic has no solutions when $m$ is odd or $a$ is not a square, and otherwise the substitution $X\mapsto X+\sqrt{a}t^{-m/2}$ gives a new equation with smaller $m$. So, after finitely many steps $c=c_0+c_1t+...$ is integral.</p>
<p>Because $X^2+X+c$ has derivative $1$, by Hensel's lemma the equation has a solution if and only the constant term $c_0$ is of the form $d^2+d$ for some $d$ in $F$. And if it is, Hensel's approximations are obtained by starting with an approximate solution $x_0=d$ and recursively computing $x_{m+1}=x_m-f(x_m)/f'(x_m)=x_m^2+c$. This gives
$$
x = d + \sum_{n=0}^\infty (c-c_0)^{2^n}
$$
as the solution (the partial sums are the $x_m$). Actually, the approach seems to work over any complete field, reducing the problem to the residue field. Hope this helps. </p>
|
1,075,983 | <p>Consider the sum $$S = \sum_{k=1}^n k$$</p>
<p>As I was computing the first triangle number with over 500 divisors (Project Euler), I came across the hypothesis that most triangle numbers have an even number of divisors (if $n=8$, then $S = 36$ has $9$ divisors).</p>
<p>Example:</p>
<pre><code>The number 3 has 2 divisors for triangle number 2
The number 6 has 4 divisors for triangle number 3
The number 10 has 4 divisors for triangle number 4
The number 15 has 4 divisors for triangle number 5
The number 21 has 4 divisors for triangle number 6
The number 28 has 6 divisors for triangle number 7
The number 36 has 9 divisors for triangle number 8 <--- only exception
The number 45 has 6 divisors for triangle number 9
The number 55 has 4 divisors for triangle number 10
The number 66 has 8 divisors for triangle number 11
The number 78 has 8 divisors for triangle number 12
The number 91 has 4 divisors for triangle number 13
The number 105 has 8 divisors for triangle number 14
The number 120 has 16 divisors for triangle number 15
The number 136 has 8 divisors for triangle number 16
The number 153 has 6 divisors for triangle number 17
The number 171 has 6 divisors for triangle number 18
The number 190 has 8 divisors for triangle number 19
The number 210 has 16 divisors for triangle number 20
The number 231 has 8 divisors for triangle number 21
The number 253 has 4 divisors for triangle number 22
The number 276 has 12 divisors for triangle number 23
The number 300 has 18 divisors for triangle number 24
The number 325 has 6 divisors for triangle number 25
The number 351 has 8 divisors for triangle number 26
The number 378 has 16 divisors for triangle number 27
The number 406 has 8 divisors for triangle number 28
The number 435 has 8 divisors for triangle number 29
The number 465 has 8 divisors for triangle number 30
The number 496 has 10 divisors for triangle number 31
The number 528 has 20 divisors for triangle number 32
The number 561 has 8 divisors for triangle number 33
The number 595 has 8 divisors for triangle number 34
The number 630 has 24 divisors for triangle number 35
The number 666 has 12 divisors for triangle number 36
The number 703 has 4 divisors for triangle number 37
The number 741 has 8 divisors for triangle number 38
The number 780 has 24 divisors for triangle number 39
The number 820 has 12 divisors for triangle number 40
The number 861 has 8 divisors for triangle number 41
The number 903 has 8 divisors for triangle number 42
The number 946 has 8 divisors for triangle number 43
The number 990 has 24 divisors for triangle number 44
The number 1035 has 12 divisors for triangle number 45
</code></pre>
<p>Which triangle numbers $S_n$ will have an odd number of divisors?</p>
| Mark Bennet | 2,906 | <p>Here is a way of making substantial progress.</p>
<p>Divisors come in pairs except when a number is square, so you are looking for triangle numbers which are also squares i.e. $$\frac {n(n+1)}2=m^2$$</p>
<p>Next multiply both sides by $8$ and add $1$ to obtain $$4n^2+4n+1=8m^2+1$$</p>
<p>If you then put $p=2n+1, q=2m$ this gives $$p^2-2q^2=1$$ where $p$ is odd and $q$ is even.</p>
<p>This is a well known equation (Pell's) and any solution (there are infinitely many) can be worked back to a solution to the original problem.</p>
<hr>
<p>To expand on the comments made by others, if we start with $(p,q)$ then $(3p+4q,2p+3q)$ is another solution, and we have $(p,q)=(3,2)$ to start us off - or we could start with $(1,0)$.</p>
<hr>
<p>Again if $(p,q)$ is a solution so is $(3p-4q, 3q-2p)$ and you can use this to show that there is only one sequence of solutions beginning with $(1,0)$</p>
<p>This all links with the fact that $p^2-2q^2=(p+q\sqrt 2)(p-q\sqrt 2)$ which is why the $\sqrt 2$ terms come in the comments.</p>
|
1,075,983 | <p>Consider the sum $$S = \sum_{k=1}^n k$$</p>
<p>As I was computing the first triangle number with over 500 divisors (Project Euler), I came across the hypothesis that most triangle numbers have an even number of divisors (if $n=8$, then $S = 36$ has $9$ divisors).</p>
<p>Example:</p>
<pre><code>The number 3 has 2 divisors for triangle number 2
The number 6 has 4 divisors for triangle number 3
The number 10 has 4 divisors for triangle number 4
The number 15 has 4 divisors for triangle number 5
The number 21 has 4 divisors for triangle number 6
The number 28 has 6 divisors for triangle number 7
The number 36 has 9 divisors for triangle number 8 <--- only exception
The number 45 has 6 divisors for triangle number 9
The number 55 has 4 divisors for triangle number 10
The number 66 has 8 divisors for triangle number 11
The number 78 has 8 divisors for triangle number 12
The number 91 has 4 divisors for triangle number 13
The number 105 has 8 divisors for triangle number 14
The number 120 has 16 divisors for triangle number 15
The number 136 has 8 divisors for triangle number 16
The number 153 has 6 divisors for triangle number 17
The number 171 has 6 divisors for triangle number 18
The number 190 has 8 divisors for triangle number 19
The number 210 has 16 divisors for triangle number 20
The number 231 has 8 divisors for triangle number 21
The number 253 has 4 divisors for triangle number 22
The number 276 has 12 divisors for triangle number 23
The number 300 has 18 divisors for triangle number 24
The number 325 has 6 divisors for triangle number 25
The number 351 has 8 divisors for triangle number 26
The number 378 has 16 divisors for triangle number 27
The number 406 has 8 divisors for triangle number 28
The number 435 has 8 divisors for triangle number 29
The number 465 has 8 divisors for triangle number 30
The number 496 has 10 divisors for triangle number 31
The number 528 has 20 divisors for triangle number 32
The number 561 has 8 divisors for triangle number 33
The number 595 has 8 divisors for triangle number 34
The number 630 has 24 divisors for triangle number 35
The number 666 has 12 divisors for triangle number 36
The number 703 has 4 divisors for triangle number 37
The number 741 has 8 divisors for triangle number 38
The number 780 has 24 divisors for triangle number 39
The number 820 has 12 divisors for triangle number 40
The number 861 has 8 divisors for triangle number 41
The number 903 has 8 divisors for triangle number 42
The number 946 has 8 divisors for triangle number 43
The number 990 has 24 divisors for triangle number 44
The number 1035 has 12 divisors for triangle number 45
</code></pre>
<p>Which triangle numbers $S_n$ will have an odd number of divisors?</p>
| Travis Willse | 155,629 | <p>First, note that for any positive integer $S$, its factors come in pairs $(n, \frac{S}{n})$; if $n$ and $\frac{S}{n}$ are the same, then $S = n^2$, that is, $S$ is a square number and $n$ is its square root. So, a positive integer has an odd number of factors iff it is square, and hence we may rephrase the question as one about numbers that are both triangular and square, and there are infinitely many of these---</p>
<p>$$0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, \ldots$$</p>
<p>---(this is <a href="https://oeis.org/A001108" rel="nofollow">OEIS A001108</a>) as one can show by translating the problem into one about the <a href="http://en.wikipedia.org/wiki/Pell%27s_equation" rel="nofollow">classical Pell equation</a>, which I see now Mark has done in his excellent solution.</p>
|
2,520,669 | <p>How are these equal? </p>
<blockquote>
<p>$$|\sqrt{x} - \sqrt{c}| = \frac{|x-c|}{|\sqrt{x} + \sqrt{c}|},$$ </p>
</blockquote>
| symplectomorphic | 23,611 | <p>Hint:</p>
<p>You're supposed to spot derivatives here. Write your quotient as </p>
<p>$$\frac{\cos(x+h)-\cos x}{h}\cdot\frac{h}{(x+h)^{1/2}-x^{1/2}}$$</p>
|
2,520,669 | <p>How are these equal? </p>
<blockquote>
<p>$$|\sqrt{x} - \sqrt{c}| = \frac{|x-c|}{|\sqrt{x} + \sqrt{c}|},$$ </p>
</blockquote>
| Michael Hardy | 11,667 | <p>\begin{align}
& \lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}} = \frac{\lim_{h\to0} \Big(\cos(x+h) - \cos x\Big) / h}{\lim_{h\to0} \Big( (x+h)^{1/2} - x^{1/2} \Big) / h} = \frac{\dfrac d {dx} \cos x }{\dfrac d {dx} x^{1/2}} = \cdots\cdots
\end{align}</p>
|
1,750,998 | <p>It is given that a and b belong to non negative real numbers and $a^3+b^3=2$. Then prove that $3(a^4+b^4)+2a^4b^4\leq8$</p>
| Macavity | 58,320 | <p>By some quick AM-GM we have the following:</p>
<ol>
<li>$3=a^3+b^3+1 \ge 3ab \implies ab \le 1$. </li>
<li>Also $a^3+1+1 \ge 3a \implies 3a \le 4-b^3$ and similarly $3b \le 4-a^3$. </li>
</ol>
<p>Now we have all the ingredients to do:
$$LHS = 3a (a^3)+3b (b^3)+2ab(ab)^3 \le (4-b^3)a^3+(4-a^3)b^3+ 2\cdot 1(ab)^3 = 4(a^3+b^3)=8$$</p>
|
1,750,998 | <p>It is given that a and b belong to non negative real numbers and $a^3+b^3=2$. Then prove that $3(a^4+b^4)+2a^4b^4\leq8$</p>
| Jean Marie | 305,862 | <p>(This is not a new solution but a comment with a graphical display)</p>
<p>It might be of interest to have an idea of the tightness of the condition. This can be put in evidence by displaying the graphical representation of </p>
<p>$$y=3((2 - b^3)^{4/3} + b^4) + 2(2 - b^3)^{4/3} b^4 - 8$$</p>
<p>obtained by plugging relationship $a=(2-b^3)^{1/3}$ into the difference between LHS and RHS of the constraint.</p>
<p>It appears that </p>
<ul>
<li><p>there is a case of equality $a=b=1$, but</p></li>
<li><p>this constraint is far from being "hard" on the rest of the values in $[0,2^{1/3}]$.</p></li>
</ul>
<p><a href="https://i.stack.imgur.com/al8Eh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/al8Eh.png" alt="enter image description here"></a></p>
|
2,939,808 | <p><img src="https://i.stack.imgur.com/5DeV4.jpg" alt="enter image description here"></p>
<p>q 23</p>
<p>my solution<img src="https://i.stack.imgur.com/urLMH.jpg" alt="enter image description here"></p>
<p>I attempted like above.Then to check i plotted graph
on desmos. so it showed me three roots.One before zero also. So i tried using first dervivative to get monotonocity of the function, yet am not able to reach a proper solution.</p>
| Avinash N | 253,506 | <p>Number of roots=3.
<a href="https://i.stack.imgur.com/gNP8F.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gNP8F.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/zRB3c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zRB3c.jpg" alt="enter image description here"></a></p>
<p>I am using<strong>Geogebra</strong> for drawing this curve. </p>
|
427,596 | <p>Just a quick question. I'm trying to understand the answer to <a href="https://math.stackexchange.com/questions/424063/group-action-on-a-manifold-with-finitely-many-orbits">one of my previous questions</a>. The precise problem I want to show is as follows. </p>
<blockquote>
<p>Let $G$ be a group acting faithfully on a manifold $X$. If $G$ is such that $\dim G$ makes sense (for example, $G$ is also a manifold), then $\dim G \leq \dim X$. </p>
</blockquote>
<p>Suppose not, i.e., $\dim G > \dim X$. Then we wish to show that the kernel of the homomorphism $G \to \operatorname{Aut}(X)$ is nontrivial. This should follow if $\dim \operatorname{Aut}(X) = \dim X$, but in general $\operatorname{Aut}(X)$ can be much larger than $X$. (Also, does $\dim \operatorname{Aut}(X)$ make sense?)</p>
<p>Any suggestions?</p>
<p>EDIT: It turns out that this proposition is false as the examples below show, and the answer I linked to has been retracted (and is being rewritten?). Thanks to everyone for the help. </p>
| Jyrki Lahtonen | 11,619 | <p>This is not true. $G=GL_2(\mathbb{R})$ acts faithfully on $\mathbb{R}^2$, no?</p>
|
3,206,838 | <blockquote>
<p>The number of 5-digit numbers of the form abcde where a,b,c,d,e belong to <span class="math-container">${0,1,2,...9}$</span> and <span class="math-container">$b = a + c$</span>, <span class="math-container">$d = c + e $</span> are?</p>
</blockquote>
<p>I tried to reason out that out of the 5 digits we need to choose only <span class="math-container">$3$</span>, that is <span class="math-container">$a,c$</span>, and <span class="math-container">$e$</span>, while <span class="math-container">$b$</span> and <span class="math-container">$d$</span> will become fixed on the basis of those. Now, we also need to satisfy <span class="math-container">$a + c ≤ 9$</span> and <span class="math-container">$c + e ≤ 9$</span>. Solving the former equation gives us some pairs of <span class="math-container">$a$</span> and <span class="math-container">$c$</span>. But, this fixes c, I cannot do the same thing with the latter equation. It just seems like a intertwined puzzle I can't get hold of from any end.</p>
<p>I may also add the solution given to this problem:
<a href="https://i.stack.imgur.com/8ocrS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8ocrS.png" alt="Solution"></a></p>
<p>I don't understand what they are trying to do here exactly.</p>
| Community | -1 | <p>For each choice of <span class="math-container">$c$</span>, there are <span class="math-container">$10-c$</span> choices for <span class="math-container">$d$</span> (any of <span class="math-container">$c,\dots,9$</span>), and <span class="math-container">$9-c$</span> for <span class="math-container">$a$</span> (<span class="math-container">$b$</span> and <span class="math-container">$e$</span> are determined). So <span class="math-container">$(10-c)(9-c)$</span> choices. Hence the sum.</p>
|
2,878,073 | <p><strong>Problem:</strong> $\{a_n\}_{n\in \mathbb N}, \quad a_{n+1}=\sqrt{2+a_n}, \quad \forall n\geq 1, \quad a_1=\sqrt{2}$</p>
<p><strong>Solution:</strong></p>
<p>We assume $a_n$ converges. Then is $\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}$</p>
<p>So we get</p>
<p>$$$a_n=\sqrt{2+a_n} \Rightarrow a_n^2-a_n-2=0$$</p>
<p>which is solved by $a_n=-1$ and $a_n=2$ but since $a_n\geq 0 \forall n$ we only find 2 to be a limit point.</p>
<p><strong>Question:</strong> Is that argument okay? I'm not sure, the main part I don't like about is, is the assumption of convergence. Does it work? </p>
<p>I know there are differet solutions, but how good is the one above? Is it valid?</p>
| Fred | 380,717 | <p>If $(a_n)$ is convergent, then denote the limit of the sequence by $a$. Then we get</p>
<p>$$a=\sqrt{2+a}.$$</p>
<p>Hence $a=-1$ or $a=2$ (and not $a_n=-1$ or $a_n=2$ as you wrote). Since all $a_n \ge 0$, we see that $a_n \to 2$.</p>
<p>All this considerations are made under the assumption that $(a_n)$ is convergent !</p>
<p>Thus it remains to show that $(a_n)$ is convergent .</p>
<p>To this end show that $(a_n)$ is increasing and bounded.</p>
|
3,243,906 | <p>I am doing a problem where I am differentiating from first principles, but I can't simplify the final expression:</p>
<p><span class="math-container">$\frac{-2xh - h^2}{x^4h + 2x^3h^2+x^2h^3}$</span></p>
<p>Could someone explain it in steps?</p>
| learningstudent | 677,133 | <p>Factorising h from both N&D to get<span class="math-container">$$=-\frac{h+2x}{x^2(x^2+2xh+h^2)}$$</span>
<span class="math-container">$$=-\frac{h+2x}{x^2(x+h)^2}$$</span></p>
|
749,731 | <p>if $G' <H < G$ then $H$ is normal in $G$. ($G'$ is the commutator subgroup of $G$.)</p>
<p>This is what I do:</p>
<p>because $G' < H$ we have $\frac{H}{G'} \triangleleft \frac{G}{G'}$. because $\frac{G}{G'}$ is abelian then $\frac{\frac{G}{G'}}{\frac{H}{G'}} \approx \frac{G}{H} $ is abelian and it means $\frac{G}{H}$ is abelian. </p>
<p>now we have $Hg_1Hg_2=Hg_2Hg_1 \Rightarrow Hg_1g_2=Hg_2g_1 \Rightarrow (g_2g_1)^{-1}(g_1g_2)=h $ for some $h \in H$.</p>
<p>now I stuck here.I need some help to finish this, I feel that I am in right path. </p>
<p>Thank you very much.</p>
| tchappy ha | 384,082 | <p>I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.<br />
This problem is the same problem as Problem 5(d) on p.65 in Herstein's book.<br />
I could not solve this problem, so I referred to "An Introduction to Algebraic Systems" (in Japanese) by Kazuo Matsuzaka.</p>
<blockquote>
<p>Let <span class="math-container">$a\in G$</span> and <span class="math-container">$x\in H$</span>.<br />
Since <span class="math-container">$H$</span> includes all the commutators of <span class="math-container">$G$</span>, <span class="math-container">$axa^{-1}x^{-1}\in H$</span>.<br />
So, <span class="math-container">$axa^{-1}=hx$</span> for some <span class="math-container">$h\in H$</span>.<br />
So, <span class="math-container">$axa^{-1}=hx\in H$</span>.<br />
So, <span class="math-container">$H$</span> is a normal subgroup of <span class="math-container">$G$</span>.</p>
</blockquote>
|
79,317 | <p>This concerns one of those "well known" facts, referred to in a recent preprint I've been looking at. In principle it's elementary, but I can't pin down an explicit textbook reference for it. Start with two finite groups $A,B$ and their product $G:=A \times B$, working over a splitting field $K$ for the groups involved with prime characteristic dividing $|G|$. Let $S_1, \dots, S_m$ and $T_1, \dots, T_n$ be respective sets of representatives of isomorphism classes of simple modules for the group algebras $KA, KB$. In turn let the projective covers (=injective hulls) be respectively $P_i, Q_j$. These are the PIMs or
indecomposable projective modules for the two group algebras.</p>
<p>It's a standard observation (found in some books) that there is an obvious isomorphism between $KG$ and the tensor product algebra $KA \otimes_K KB$, while each group algebra splits into the direct sum (as a left module over itself) of the various PIMs taken with multiplicity equal to the dimension of the corresponding simple module. It's also a standard fact (found in some books) that each $S_i \otimes T_j$ is a simple module for $KG$. From these ingredients one can conclude that
$P_i \otimes Q_j$ is the corresponding PIM, thereby exhausting all isomorphism classes for $KG$. </p>
<blockquote>
<p>Is all of this written down in a self-contained way somewhere?</p>
</blockquote>
| Geoff Robinson | 14,450 | <p>I'm not sure what Fein does, but can't you do everything with Brauer characters, at least in the splitting field case. Use the fact that the Brauer characters of the PIMs are the unique class funcstion $\theta_i$ such that $\langle \theta_i, \phi_j \rangle = \delta_{ij}$where the $\phi_j$ are the Brauer characters of the simple modules. Since the simple modules of a direct product are easily determined, this uniquely determines the Brauer characters of the PIMs of the same direct product, and it's clear that they are the pairwise products of the Brauer characters of the PIMs for the two direct factors. </p>
|
1,419,105 | <p>I had a course in the construction of numbers last semester. I understand the potencial of most of the proofs, for example: I guess I can answer decently why commutativity is important.</p>
<p>But when it comes to the proof of uniqueness of $0,1$, I have no idea why that is important. For $\Bbb{N}$, I guess it's important because using peano axioms, we could have:</p>
<p>$$0\to 1 \to 2 \to \dots$$</p>
<p>$$0'\to 1' \to 2' \to \dots$$</p>
<p>Inside $\Bbb{N}$, but I'm not sure if that is the reason, nor what problem it would represent. </p>
| Zubin Mukerjee | 111,946 | <p>It's nice to make sure we aren't working in the <a href="http://mathworld.wolfram.com/TrivialRing.html" rel="nofollow">trivial ring</a>. Ideally the axioms we assume for $\mathbb{N}$ should eliminate the possibility that $\mathbb{N}$ is the trivial ring; a proof that $0 \neq 1$ is one way to do that.</p>
|
9,302 | <p>Say I have a symmetric matrix. I have the concept of 2-norm as defined on wikipedia. Now I want to prove (disprove?) that the norm of a symmetric matrix is maximum absolute value of its eigenvalue. I would really appreciate if this can be done only using simple concepts of linear algebra.</p>
<p>I am quite new to mathematics. </p>
| Ross Millikan | 1,827 | <p>Given a symmetric matrix, you have a complete set of eigenvalues and orthogonal eigenvectors. Any vector can be represented as a linear combination of the eigenvectors. Multiply your matrix by an arbitrary unit vector decomposed into the eigenvectors. Then note that the maximum length of the resultant vector is achieved when the input vector is along the eigenvector associated with the largest eigenvalue in absolute value.</p>
|
1,934,268 | <p>Let's say I have an arbitrary matrix $A$. </p>
<p>An eigenvector of $A$ would be a vector $\vec{x}$, such that </p>
<ol>
<li>$A\vec{x} = \lambda \vec{x}$</li>
<li>$\vec{x} \in N(A-\lambda I)$ i.e. $(A - \lambda I)\vec{x} = \vec{0}$</li>
</ol>
<p>where $\lambda$ is an eigenvalue of $\vec{x}$</p>
<hr>
<p>Now are all $\vec{x} \in N(A-\lambda I)$ eigenvectors of $A$, i.e. do all $\vec{x}$ in the nullspace of $A - \lambda I$ also automatically satisfy condition $(1)$, that $A\vec{x} = \lambda \vec{x}$?</p>
<p>Written more concretely is the following statement true:</p>
<p>$$\vec{x} \in N(A-\lambda I) \implies A\vec{x} = \lambda \vec{x} \ \ \ \ \ \forall \vec{x} \in N(A-\lambda I)$$</p>
<p>If that was the case then the eigenvectors, would $span$ the entire nullspace. </p>
<p>Furthemore what relation does $N(A-\lambda I)$ have to $N(A)$ and the eigenvectors? If $\vec{x} \in N(A-\lambda I)$, does that imply that $\vec{x} \in N(A)$? Can we deduce anything about the eigenvectors of $A$ from $N(A)$ or can we only deduce them from $N(A - \lambda I)$?</p>
| Josu Etxezarreta Martinez | 360,329 | <p>The first statement is correct by the definition of nullspace, as $x\in N(A-\lambda I)$, then all those vectors should fulfill the relationship $(A-\lambda I)x=0$, from where we can derive easily the relation $Ax=\lambda x$.</p>
<p>For the statement you make that if $x\in N(a-\lambda I)$ then $x\in N(A)$, it is false as it can be seen in the next counterexample. Consider matrix A:
$\left( \begin{array}{ccc}
2 & 0 \\
0 & 1 \\
\end{array} \right)$, it can be clearly seen that the nullspace of this matrix is zero dimensional. However, it has eigenvectors $a=(1,0),b=(0,1)$, which form the nullspace of the matrix $A-\lambda I$, that is the zero matrix, so being in that nullspace does not imply to be in the nullspace of $A$</p>
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