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 |
|---|---|---|---|---|
48,864 | <p>I can't resist asking this companion question to the <a href="https://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking"> one of Gowers</a>. There, Tim Dokchitser suggested the idea of Grothendieck topologies as a fundamentally new insight. But Gowers' original motivation is to probe the boundary between a human's way of thinking and that of a computer. I argued, therefore, that Grothendieck topologies might be more natural to computers, in some sense, than to humans. It seems Grothendieck always encouraged people to think of an object in terms of the category that surrounds it, rather than its internal structure. That is, even the most lovable mathematical structure might be represented simply as a symbol $A$, and its special properties encoded in arrows $A\rightarrow B$ and $C\rightarrow A$, that is, a grand combinatorial network. I'm tempted to say that the idea of a Grothendieck topology is something of an obvious corollary of this framework. It's not something I've devoted much thought to, but it seems this is exactly the kind of reasoning more agreeable to a computer than to a woolly, touchy-feelly thinker like me.</p>
<p>So the actual question is, what other mathematical insights do you know that might come more naturally to a computer than to a human? I won't try here to define computers and humans, for lack of competence. I don't think having a deep knowledge of computers is really a prerequisite for the question or for an answer. But it would be nice if your examples were connected to substantial mathematics. </p>
<p>I see that this question is subjective (but not argumentative in intent), so if you wish to close it on those grounds, that's fine.</p>
<p>Added, 11 December: Being a faulty human, I had an inexplicable attachment to the past tense. But, being weak-willed on top of it all, I am bowing to peer pressure and changing the title.</p>
| Ivan Meir | 7,113 | <p>I think that many conjectures from number theory (which I think count as insights) might be more obvious to a computed than a human being since they would have access to a huge amount of empirical data from which to discover patterns and obtain estimates as to the probability that something is true or plausible. This is a very effective way to discover theorems in number theory. The method of discovery is along the lines described by Polya in his books on plausible reasoning in relation to Euler's discoveries.</p>
<p>To obtain the same level of confidence humans need insight and proof which in number theory is often really hard to obtain.</p>
<p>There are some mathematicians like Ramanujan, Euler, Gauss who had similar abilities but this is quite rare.</p>
<p>Also mathematical results that are accessible by humans must be true for a reason i.e. there must be a reasonably short deductive route from known theorems. Work of Chaitlin and others suggests that some theorems are not true for any reason i.e. they are not amenable to any deduction from a set of axioms of less complexity than themselves. On the borderline there must be profound mathematical results that are close to being empirically true in that sense. You would imagine that computers might have a better chance of understanding and perceiving these results since they might be able to reason more effectively from a much wider vantage point empirically speaking given their massive processing power.</p>
|
340,264 | <p>Given that</p>
<p>$L\{J_0(t)\}=1/(s^2+1)$</p>
<p>where $J_0(t)=\sum\limits^{∞}_{n=0}(−1)n(n!)2(t2)2n$,</p>
<p>find the Laplace transform of $tJ_0(t)$. </p>
<p>$L\{tJ_0(t)\}=$_<strong><em>_</em>__<em>_</em>__<em>_</em>___<em></strong>---</em>___?</p>
| Lubin | 17,760 | <p>There are so many ways of approaching this that one can get dizzy contemplating them all. You can use the Binomial series for $(1+z)^{1/2}$, plugging in $2/y^2-1$, for example, to get $\sqrt2/y$, then multiply that result by $y$.</p>
<p>Or you can use Newton-Raphson directly. I'll illustrate how it goes for $p=7$, where you can take $y=3$. You have a function $f(x)=x^2-2$, $f'(x)=2x$, and you want a root of $f$ that's close to $3$. Set $z_1=3$, and the first error is $f(z_1)=7$, and the first correction is $7/6$, so the second approximation is $z_2=3-7/6=11/6$. The second error is $f(z_2)=49/36$, so the second correction is $(49/36)\big/(22/6)=49/132$, and the third approximation is $z_3=z_2-49/132=193/132$. The third error is $f(z_3)=2401/17424$, in which the numerator is already $7^4$. Continue as long as you want.</p>
|
3,076,504 | <p>The problem states: </p>
<p>Right Triangle- perimeter of <span class="math-container">$84$</span>, and the hypotenuse is <span class="math-container">$2$</span> greater than the other leg. Find the area of this triangle. </p>
<p>I have tried different methods of solving this problem using Pythagorean Theorem and systems of equations, but cannot find any of the side lengths or the area of the right triangle. I looked for similar problems on StackExchange and around the internet, but could not find anything. </p>
<p>Does anyone know anything that could help find the side lengths of the triangle and the area as well?<br>
Method that I tried: </p>
<ol>
<li>Made a system with the values given.<br>
<span class="math-container">\begin{align}
a+b+c&=84 \\
c&=b+2
\end{align}</span></li>
<li>Substituted <span class="math-container">$c$</span> with <span class="math-container">$b+2$</span>.<br>
<span class="math-container">\begin{align}
a+b+b+2&=84 \\
a + 2b &= 82 & \text{subtracted $2$ from both sides}\\
a + a^2 - 4 &= 82
\end{align}</span></li>
<li><span class="math-container">$c^2$</span> is <span class="math-container">$(b+2)(b+2)$</span>, so I used Pythagorean Theorem to isolate one of the variables.<br>
<span class="math-container">\begin{align}
a^2+b^2 &=c^2\\
a^2 + b^2 &=(b+2)(b+2)\\
a^2+b^2 &=b^2+2b+4\\
a^2&=2b+4 & \text{ (Subtracted $b^2$ from both sides) }
\end{align}</span>
OR<br>
<span class="math-container">\begin{align}
a^2-4&=2b
\end{align}</span></li>
</ol>
<p>I do not know what to do after this point.</p>
| poetasis | 546,655 | <p><span class="math-container">$\\ \textbf{Finding triples, given perimeter using Euclid's formula}$</span> where <span class="math-container">$P=perimeter$</span></p>
<p><span class="math-container">$$P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn\implies n=\frac{P-2m^2}{2m}\quad where \quad \biggl\lceil\frac{\sqrt{P}}{2}\biggr\rceil\le m \le \biggl\lfloor\sqrt{\frac{P}{2}}\biggr\rfloor$$</span></p>
<p>Here, the lower limit ensures that <span class="math-container">$m>n$</span> and the upper limit insures that <span class="math-container">$n>0$</span>.</p>
<p>Example:
<span class="math-container">$$P=84\Rightarrow \biggl\lceil\frac{\sqrt{84}}{2} \biggr\rceil =5 \le m\le\biggl\lfloor\sqrt{\frac{84}{2}}\biggr\rfloor =6:\quad f(84,5)\notin\mathbb{N}\quad f(84,6)=1\Rightarrow
F(6,1)=(35,12,37)$$</span></p>
<p>There are no other solutions for <span class="math-container">$P=84$</span>.</p>
|
3,365,112 | <p>Pretty simple, for <span class="math-container">$a,b \in \mathbb R$</span>, show that <span class="math-container">$|a-b|<\frac{|b|}{2}$</span> implies <span class="math-container">$\frac{|b|}{2}<|a|$</span>. I can see this graphically on the number line, but I can't seem to show it algebraically.</p>
<p>I'm think it involves the triangle inequalities:</p>
<blockquote>
<p><span class="math-container">$$|x+y|\le|x|+|y|$$</span></p>
<p><span class="math-container">$$|x-y|\ge||x|-|y||.$$</span></p>
</blockquote>
| cmk | 671,645 | <p>Since <span class="math-container">$|a-b|<|b|/2,$</span> we know that <span class="math-container">$$-|b|/2<a-b<|b|/2,$$</span> or <span class="math-container">$$b-|b|/2<a<b+|b|/2,$$</span> from the definition of the absolute value. Try to consider what happens if you choose <span class="math-container">$b$</span> to have different signs (if <span class="math-container">$b\geq 0,$</span> look at what happens to the first <span class="math-container">$<$</span>, and if <span class="math-container">$b<0,$</span> do the same for the second <span class="math-container">$<$</span>).</p>
|
729,444 | <p>Let be two lists $l_1 = [1,\cdots,n]$ and $l_2 = [randint(1,n)_1,\cdots,randint(1,n)_m]$ where $randint(1,n)_i\neq randint(1,n)_j \,\,\, \forall i\neq j$ and $n>m$. How I will be able to found the number of elements $x\in l_1$, to select, such that the probability of $x \in l_2$ is $1/2$?. I'm trying using the birthday paradox but I cann't get.</p>
<p>$randint(x,y)$ pick a random number between $x$ and $y$.</p>
| Brian Fitzpatrick | 56,960 | <p>If you don't know what an eigenvalue is, and if you're not worried about elegance, then here is a more direct approach (assuming you're working over a field not of characteristic two).</p>
<p>There exist scalars $\lambda_{ij}$ for $1\leq i,j\leq n$ such that for every basis $\{v_1,\dotsc,v_n\}$ of $V$ we have
$$
\begin{array}{ccccccc}
T(v_1) & = & \lambda_{11}v_1 & + & \dotsb & + & \lambda_{n1}v_n \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
T(v_n) & = & \lambda_{1n}v_1 & + & \dotsb & + & \lambda_{nn}v_n
\end{array}\tag{1}
$$
Now, fix $v\in V$ and note that there exists a basis $\{v_1,\dotsc,v_i,\dotsc,v_n\}$ of $V$ such that $v_i=v$. Equation $(1)$ then implies
$$
T(v)=\lambda_{1i}v_1+\dotsb+\lambda_{ii}v_i+\dotsb+\lambda_{ni}v_n\tag{2}
$$
Next, since $$\{-v_1,\dotsc,-v_{i-1},v_i,-v_{i+1},\dotsc,-v_n\}$$ is also a basis for $V$, equation $(1)$ also implies
$$
T(v)=-\lambda_{1i}v_1-\dotsb-\lambda_{i-1,i}\cdot v_{i-1}+\lambda_{ii}v_i-\lambda_{i+1,i}\cdot v_{i+1}-\dotsb-\lambda_{ni}v_n\tag{3}
$$
Subtracting equation $(3)$ from equation $(2)$ gives
$$
\mathbf{0}=2\lambda_{1i}v_1+\dotsb+2\lambda_{i-1,i}\cdot v_{i-1}+2\lambda_{i+1,i}\cdot v_{i+1}+\dotsb+2\lambda_{ni}v_{n}\tag{4}
$$
Since $\{v_1,\dotsc,v_n\}$ are linearly independent, $(4)$ implies
$$
\lambda_{1i}=\dotsb=\lambda_{i-1,i}=\lambda_{i+1,i}=\dotsb=\lambda_{ni}=0\tag{5}
$$
Since our choice of $i$ was arbitrary, equation $(5)$ implies
$$
\lambda_{kl}=0\tag{6}
$$
whenever $k\neq l$. Moreover, equations $(2)$ and $(6)$ imply that $T(v)=\lambda_{kk}v=\lambda_{ll}v$ for all $k$ and $l$ so that
$$
\lambda_{kk}=\lambda_{ll}
$$
for all $k$ and $l$. That is, there exists a scalar $\lambda$ such that
$$
\lambda_{kl}=
\begin{cases}
0 & k\neq l \\
\lambda & k=l
\end{cases}\tag{7}
$$</p>
<p>Finally, we wish to show that there exists a scalar $\lambda$ such that $T(v)=\lambda v$ for every $v\in V$. To do so, let $v\in V$ and note that $(2)$ and $(7)$ imply $T(v)=\lambda_{ii} v=\lambda v$.</p>
|
2,155,652 | <p>I have a question regarding this proof my professor gave us. For the third property, I understand the proof up to the sentence "If $x \in E'$, i.e., x is a limit point of E."
Well, I also understand that if x is not in F, then x can't be a limit point since F is closed.
After that, I don't fully understand it. Could someone please give me an explanation?</p>
<p>Thank you in advance.</p>
<p><a href="https://i.stack.imgur.com/yECP4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yECP4.jpg" alt="enter image description here"></a></p>
| manofbear | 230,268 | <p>If $x$ is a limit point of $E$, that means there exists a sequence $(x_n)\subset E$ for which $x_n\neq x$ for all $n$, and $\lim x_n=x$. Notice that the same sequence $(x_n)\subset F$ satisfies the same conditions, so $x$ is a limit point of $F$. Since $F$ is closed, it contains its limit points. So $x\in F$.</p>
|
809,516 | <p>I need to calculate </p>
<p>$$\lim_{x \to \infty} \frac{((2x)!)^4}{(4x)! ((x+5)!)^2 ((x-5)!)^2}.$$</p>
<p>Even I used Striling Approximation and Wolfram Alpha, they do not help.</p>
<p>How can I calculate this?</p>
<p>My expectation of the output is about $0.07$.</p>
<p>Thank you in advance.</p>
| Claude Leibovici | 82,404 | <p>Using directly Stirling approximation of the factorial $$\begin{align}
\Gamma(n+1) \approx \sqrt{2 \pi} \ n^{n+1/2} \ e^{-n}
\end{align}$$ the expression becomes $$\frac{((2x)!)^4}{(4x)! ((x+5)!)^2 ((x-5)!)^2}\approx\sqrt{\frac{2}{\pi }} (x-5)^{9-2 x} x^{4 x+\frac{3}{2}} (x+5)^{-2 x-11}$$ which, for large values of $x$, can be approximated by $$\sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{x}}-50 \sqrt{\frac{2}{\pi }}
\left(\frac{1}{x}\right)^{3/2}+1275 \sqrt{\frac{2}{\pi }}
\left(\frac{1}{x}\right)^{5/2}+O\left(\left(\frac{1}{x}\right)^{7/2}\right)$$</p>
<p>For $x=100$, the exact value is $0.0484141$ while the approximation is $0.0500673$.<br>
For $x=1000$, the exact value is $0.0239969$ while the approximation is $0.0240019$.</p>
|
1,555,548 | <p>There are $8$ people and they want to sit in a bus which has $2$ single front seats and $4$ sets of $3$ seats with $1$ person that is always the designated driver. How many ways are there for the people to sit in the bus?</p>
<p>I solved it by using:</p>
<p>$6!*(\binom{9}{3}) - 4((6*5*4*3)*2(\binom{4}{2})+(6*5*4*3*2)(\binom{3}{2}) + 6!(\binom{2}{2})) + 7!*(\binom{10}{3})-4((7*6*5*4)*3!*(\binom{5}{2})+(7*6*5*4*3)*2!*(\binom{4}{2})+(7*6*5*4*3*2)(\binom{3}{2}) + 7!(\binom{2}{2}) = \boxed{233280}$</p>
<p>I did complementary counting and took out the cases where there were more than $3$ people in a set of rows. Can anyone tell me if my answer is right?</p>
| mathochist | 215,292 | <p>If the specific seat matters, it seems like there should be an $8$ way choice to choose the driver, then in the remaining $14$ seats, we have to choose $7$ for the remaining passengers, which gives $8{14 \choose 7}$.</p>
|
622,278 | <p>What is the relation between the convergence of $\sum a_{n}$ and $\prod (1+a_{n})$ where $a_{n} \in \mathbb{C} \ \forall n$ ?</p>
<p>Where can I find some references about this topic ?</p>
| benh | 115,596 | <p>Let $a_n\neq 0$ for all $n\in \Bbb N$, then $$\prod_{n=1}^\infty a_n \text{ converges} \Leftrightarrow \sum_{n=1}^\infty \log{a_n} \text{ converges}$$
Moreover, for $a_n \neq -1$, we have an equivalence $$\sum_{n=1}^\infty \log(1+a_n) \text{ converges absolutely} \Leftrightarrow \sum_{n=1}^\infty a_n \text{ converges absolutely}.$$
Thus, if $\sum_{n=1}^\infty a_n$ converges absolutely, $\prod_{n=1}^\infty (1+a_n)$ converges unconditionally.
Unfortunately, I have no reference for this in English language.</p>
|
1,811,812 | <p>I've to prove that $Q_8/Z(Q_8)$ is isomorphic to Klein's $4$-group. I know that
$Q_8/Z(Q_8)$ has order $4$. But, I'm not able to get the elements of $Q_8/Z(Q_8)$. Please help me in representing $Q_8/Z(Q_8)$ in Roster form.</p>
| carmichael561 | 314,708 | <p>I'm not sure what roster form is, but $1,i,j,k$ is a complete system of coset representatives for $Z(Q_8)=\{\pm 1\}$.</p>
|
1,811,812 | <p>I've to prove that $Q_8/Z(Q_8)$ is isomorphic to Klein's $4$-group. I know that
$Q_8/Z(Q_8)$ has order $4$. But, I'm not able to get the elements of $Q_8/Z(Q_8)$. Please help me in representing $Q_8/Z(Q_8)$ in Roster form.</p>
| rschwieb | 29,335 | <p>Here's another way to get at it. There is a popular group theory exercise that says <a href="https://math.stackexchange.com/q/546971/29335">if $G/Z(G)$ is cyclic, then $G$ is Abelian</a>.</p>
<p>You know that $Q_8$ is nonabelian, and that there are only two groups of order $4$ possible...</p>
|
65,658 | <p>Suppose $X_i$'s are i.i.d, with the density distribution $f(x) = e^{-x}$, $x \geq 0$. I was able to show that
$$P(\limsup X_n/\log{n} =1)=1$$ using Borel-Cantelli.</p>
<p>Define $M_n=\max \{X_1,\ldots,X_n\}$, can I claim $M_n/\log{n} \rightarrow 1$ a.s. in this case? Is it still true in general without knowing the distribution of $X_i$?</p>
| Robert Israel | 8,508 | <p>If $F(x) = 1 - e^{-x}$ for $x > 0$ is the CDF of each $X_i$, the CDF of $M_n$ is $F_{M_n}(x) = F(x)^n = (1 - e^{-x})^n$ for $x > 0$. Note that $\ln(F_{M_n}(x)) = n \ln(1 - e^{-x})$ and since $- t - t^2 < \ln(1-t) < -t$ for $0 < t < .683$, for any $c>0$ we have $-n^{1-c} - n^{1-2c} \le \ln(F_(M_n)(c \ln n)) \le -n^{1-c}$ for $n$ large enough. If $c < 1$, this says
$P\left( \frac{M_n}{\ln n} \le c\right) \le e^{-n^{1-c}}$, and $\sum_n e^{-n^{1-c}} < \infty$ so almost surely only finitely many $\frac{M_n}{\ln n} \le c$. If $c > 1$,
$P(\left( \frac{M_n}{\ln n} \le c \right) \ge e^{-n^{1-c} - n^{1-2c}} \to 1$ as $n \to \infty$, so almost surely infinitely many $\frac{M_n}{\ln n} \le c$. Thus $\lim \sup_n \frac{M_n}{\ln n} = 1$ almost surely. However, it's not so clear to me that $\lim \inf_n \frac{M_n}{\ln n} = 1$ almost surely. </p>
|
4,513,368 | <p>The following question seems to be quite simple, but I am having a hard time to prove it rigorously.</p>
<p>Consider <span class="math-container">$n\in\mathbb{N}$</span> vertices, for example <span class="math-container">$\{v_1,\ldots, v_n\}$</span>. I have some further information on these vertices, namely, that any of these vertices has at least one connection (by an edge) to itself or to any other of the vertices.</p>
<p>Now I would like to prove, that there must be a subset of <span class="math-container">$\{v_1,\ldots, v_n\}$</span>, which form a connected graph. Even the subgroup <span class="math-container">$\{v_1,\ldots, v_n\}$</span> would be allowed.</p>
<p>One more information to consider: I call a subset of <span class="math-container">$\{v_1,\ldots, v_n\}$</span>, which contains only one element, connected, if it is connected to itself by an edge.</p>
<p>In the best case, I would like to prove this through a contradiction: "Suppose, all subsets do not form a connected graph." However, I cannot find the contradictory statement, that follows from this false assumption.</p>
<p>Again, the statement itself seems very obvious, but I would like to prove it rigorously.</p>
<p>Any help is appreciated! Thank you in advance!</p>
| Prem | 464,087 | <p>Either I am missing something or the Question is missing something.</p>
<p>Let <span class="math-container">$V=\{a,b,c,u,v,w,x,y,z\}$</span> & Edges are <span class="math-container">$\{ab,bc,uu,uv,vw,ww,wx,wy,xy,zz\}$</span> where each Vertex is connected to something, maybe itself.</p>
<p>In set <span class="math-container">$V$</span>, take some "Starting Element", Eg <span class="math-container">$u$</span>, into new set <span class="math-container">$U=\{u\}$</span>.<br />
It is given that <span class="math-container">$u$</span> is connected to something (maybe itself), then take all other elements which are connected to <span class="math-container">$u$</span> (that is <span class="math-container">$u,v$</span>) & include these in <span class="math-container">$U$</span> to get <span class="math-container">$U=\{u,v\}$</span>.<br />
Then take all elements connected to new element <span class="math-container">$v$</span> (that is <span class="math-container">$w$</span>) & include in <span class="math-container">$U=\{u,v,w\}$</span>. Again take all new elements in <span class="math-container">$U$</span> (that is <span class="math-container">$w$</span>), & include all connected elements (that is <span class="math-container">$w,x,y$</span>) then <span class="math-container">$U=\{u,v,w,x,y\}$</span>.<br />
Again, include all elements which are connected to these new elements (that is <span class="math-container">$x,y$</span>) where we get no new elements, hence we terminate our algorithm and say that <span class="math-container">$U=\{u,v,w,x,y\}$</span> is Maximally Connected.</p>
<p>Maximally Connected means : Including <span class="math-container">$a,b,c,z$</span> will make it unconnected.</p>
<p>We will get same <span class="math-container">$U$</span> when "Starting Element" is some other element of this <span class="math-container">$U$</span>.<br />
With "Starting Element" <span class="math-container">$a$</span>, we will get <span class="math-container">$U=\{a,b,c\}$</span><br />
With "Starting Element" <span class="math-container">$z$</span>, we will get <span class="math-container">$U=\{z\}$</span></p>
<p>It seems very straight forward, hence Doubting whether I missed something !</p>
<p>In case we want <span class="math-container">$|U|$</span> to be Maximum, then we should try out all "Starting Elements" in the unconnected elements of <span class="math-container">$V$</span> & keep track of the Maximally Connected <span class="math-container">$U$</span> with Maximum Number of Elements.</p>
|
3,910,013 | <p>I'm preparing for a high school math exam and I came across this question in an old exam.</p>
<p>Let <span class="math-container">$f(x) = \dfrac{1}{2(1+x^3)}$</span>.</p>
<p><span class="math-container">$\alpha \in (0, \frac{1}{2})$</span> is the only real number such that <span class="math-container">$f(\alpha) = \alpha$</span>.</p>
<p><span class="math-container">$(u_n)$</span> is the series such that
<span class="math-container">$$
\begin{cases}
u_0 = 0 \\
u_{n+1} = f(u_n) \quad \forall n \in \mathbb{N}
\end{cases}
$$</span>
Prove that <span class="math-container">$|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha|$</span></p>
<p>The end goal is to prove that <span class="math-container">$(u_n)$</span> converges to <span class="math-container">$\alpha$</span>. But we have to do it this way instead of finding that both <span class="math-container">$(u_{2n})$</span> and <span class="math-container">$(u_{2n+1})$</span> converge to <span class="math-container">$\alpha$</span>.</p>
<p>It's supposed to be a very easy question. But I have been trying for the last two hours and I couldn't find it.</p>
<p>It's easy to prove that <span class="math-container">$u_n \in [0, \frac{1}{2}]$</span> by induction.</p>
<p>Here's what I tried:</p>
<ul>
<li>I tried separating the question into two cases <span class="math-container">$u_n \le \alpha$</span> and <span class="math-container">$u_n \ge \alpha$</span>.
I got that I just need to prove that <span class="math-container">$u_{n+1} + \frac{1}{2} u_n \ge \frac{3}{2} \alpha$</span>. But I didn't know how to go from here. Substituting <span class="math-container">$u_{n+1}$</span> with <span class="math-container">$\dfrac{1}{2(1+u_n^3)}$</span> seems to only make the problem more complicated.</li>
<li>I tried squaring both sides. It only made the expression more complicated.</li>
</ul>
<p>The problem vaguely reminds me of the epsilon-delta definition of <span class="math-container">$\lim_{x \to \alpha} f(x) = \alpha$</span>. But that seems like it won't lead anywhere.</p>
| AlanD | 356,933 | <p>Hint: Since <span class="math-container">$f(\alpha)=\alpha$</span>, you are trying to prove
<span class="math-container">$$
|u_{n+1}-\alpha|\leq \frac 12|u_n-\alpha|,
$$</span>
which is equivalent to showing
<span class="math-container">$$
|f(u_n)-f(\alpha)|\leq \frac 12|u_n-\alpha|
$$</span>
or
<span class="math-container">$$
\frac{|f(u_n)-f(\alpha)|}{|u_n-\alpha|}\leq \frac 12
$$</span>
By the mean value theorem, what does the left hand side equal and how can you show it is bounded above by <span class="math-container">$1/2$</span>?</p>
|
133,604 | <p>"A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line." - Wikipedia</p>
<p><img src="https://i.stack.imgur.com/Rl1WS.gif" alt="cycloid animation"></p>
<p>In many calculus books I have, the cycloid, in parametric form, is used in examples to find arc length of parametric equations. This is the parametric equation for the cycloid:</p>
<p>$$\begin{align*}x &= r(t - \sin t)\\
y &= r(1 - \cos t)\end{align*}$$</p>
<p>How are these equations found in the first place?</p>
| Ross Millikan | 1,827 | <p>The center of the circle moves along a horizontal line at constant velocity. If we want the cusps to be at $y=0$, that means the center should be $(x_c,y_c)=(rt,r)$. Then we add on the location of the point on the rim relative to the center. This will be something like $(r\cos t, r\sin t)$ but we still need to get the phase right. If we start with the point on the rim at $(0,0)$ at $t=0$ the rim point is at an angle of $\frac {-\pi}2$ at $t=0$, that is, pointing straight down. A little fiddling with the phases gets the expression you quote. The scale between the center motion and rotation is set by the requirement that there be no slippage, which means the velocity of the point on the road must be $0$.</p>
|
133,604 | <p>"A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line." - Wikipedia</p>
<p><img src="https://i.stack.imgur.com/Rl1WS.gif" alt="cycloid animation"></p>
<p>In many calculus books I have, the cycloid, in parametric form, is used in examples to find arc length of parametric equations. This is the parametric equation for the cycloid:</p>
<p>$$\begin{align*}x &= r(t - \sin t)\\
y &= r(1 - \cos t)\end{align*}$$</p>
<p>How are these equations found in the first place?</p>
| J. M. ain't a mathematician | 498 | <p>Here is a cartoon depiction of what Robert and Ross were showing, courtesy of <a href="http://books.google.com/books?hl=en&id=EbVrWLNiub4C&pg=PA78" rel="noreferrer">Stan Wagon</a>:</p>
<p><img src="https://i.stack.imgur.com/Yd6sv.gif" alt="rolling penny"></p>
|
133,604 | <p>"A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line." - Wikipedia</p>
<p><img src="https://i.stack.imgur.com/Rl1WS.gif" alt="cycloid animation"></p>
<p>In many calculus books I have, the cycloid, in parametric form, is used in examples to find arc length of parametric equations. This is the parametric equation for the cycloid:</p>
<p>$$\begin{align*}x &= r(t - \sin t)\\
y &= r(1 - \cos t)\end{align*}$$</p>
<p>How are these equations found in the first place?</p>
| louie mcconnell | 131,577 | <p>This book is a great resource. See pdf page 599, actual page 567.</p>
<p><a href="http://www.marystarhigh.com/apps/download/7vb7ETI4n4RtLFWDnZw0xNfQRUSB1swoBHQpP7i1l9pXZS1Y.pdf/Precalculus%20Book.pdf" rel="nofollow">http://www.marystarhigh.com/apps/download/7vb7ETI4n4RtLFWDnZw0xNfQRUSB1swoBHQpP7i1l9pXZS1Y.pdf/Precalculus%20Book.pdf</a></p>
<p>You should go to the page before reading on and while reading the rest of the post.</p>
<p>In it, it explains everything very coherently and breaks down the derivation into 4 steps: finding an equation for the location of the center of the circle (x and y coordinates), and then finding the equation for the point P in <strong>in reference</strong> to the center.</p>
<p>We will start off by trying to find where the center of the circle is at angle $\theta$. The x coordinate is going to be equal to the distance traveled, which is the same thing as the length of the sector of the circle we have already covered. The sector is equal to the radius times the central angle, so the center will be at $x = a \theta$</p>
<p>The y coordinate of the center at any time is really easy because the center is always the height of the radius, which is $a$. Therefore, the center is at coordinates $(a\theta, a)$ at angle $\theta$.</p>
<p>Now, let's try and find the location of point P <strong>in reference</strong> to the center. We will start with the x coordinate. </p>
<p>At angle $\theta$, P will start by lagging behind, then jumping ahead, then going back to where it started. Therefore, we want to start by subtracting $0a$, then $1a$, then $0a$, then -$1a$, then going back to $0$ again. This behavior is exhibited by $a \sin \theta$, so our x coordinate is now complete: $x = a\theta - a \sin \theta = a(\theta - \sin \theta)$</p>
<p>Now for the y coordinate. To get the height of point P at angle $\theta$, we notice that it starts out below the center, then goes above the center, then back below. Therefore, we want to subtract $1a$, then $0a$, then $-1a$ (add $1a$), then go back to $0a$ again. The pattern of $(1, 0, -1, 0, 1)$ is exhibited by $a \cos \theta$, so we want to subtract this from the center, giving us $y = a - a \cos \theta$ , or $y = a(1 - \cos \theta)$.</p>
<p>Now, we are done. Our two equations are $$x = a(\theta - \sin \theta)$$ $$y = a(1 - \cos \theta)$$.</p>
|
138,091 | <p>I am trying to compute an explicit formula using Mathematica for the following multinomial expression:</p>
<blockquote>
<p>\begin{equation} \sum_{n_{1}+n_{2}+...+n_{M}=N}^{M} {N \choose
n_{1},n_{2},...,n_{M }} \cdot n_{i} = ? \end{equation}</p>
</blockquote>
<p>where $i={1,2,...,M}$ and using </p>
<pre><code>multinomial[n__] := (Plus @@ {n})!/Times @@ (#! & /@ {n})
</code></pre>
<p>but I don't know how make the sumatoria over all the index $n_{k}$.</p>
<p>In fact I know the following:</p>
<blockquote>
<p>\begin{equation} \sum_{n_{1}+n_{2}+...+n_{M}=N}^{M} {N \choose
n_{1},n_{2},...,n_{M }} = M^{N} \end{equation}</p>
</blockquote>
<p>but I think that this previous results can not be used in order to obtain the result at the first equation, it's the reason why I am asking for a code in <em>Mathematica</em> that tries to compute this thing in an analytical way.</p>
<p>example:</p>
<p>Taking for example M=N=2 and $i=1$ then I have to obtain:</p>
<blockquote>
<p>\begin{equation} \sum_{n_{1}+n_{2}=2}^{2} {2 \choose n_{1},n_{2}} \cdot n_{1} = {2 \choose 2,0} \cdot 2+{2 \choose 0,2} \cdot 0+{2 \choose 1,1} \cdot 1\end{equation}</p>
</blockquote>
<p>In fact I I take $i=2$ i would obtain the same:</p>
<blockquote>
<p>\begin{equation} \sum_{n_{1}+n_{2}=2}^{2} {2 \choose n_{1},n_{2}} \cdot n_{2} = {2 \choose 2,0} \cdot 0+{2 \choose 0,2} \cdot 2+{2 \choose 1,1} \cdot 1\end{equation}</p>
</blockquote>
| Mr.Wizard | 121 | <p>Following Carl Woll's comment, correcting $m$ and $n$, and providing a more efficient form:</p>
<pre><code>n = 17;
m = 9;
p = IntegerPartitions[n, {m}, Range[0, n]];
Sum[Total[Permutations[x][[All, 1]] * Multinomial @@ x], {x, p}]
n m^(n - 1)
</code></pre>
<blockquote>
<pre><code>31501343210481297
31501343210481297
</code></pre>
</blockquote>
<p>Solved, it would seem.</p>
|
1,201,900 | <p>This is a rather soft question to I will tag it as such.</p>
<p>Basically what I am asking, is if anyone has a good explanation of what a homomorphism is and what an isomorphism is, and if possible specifically pertaining to beginner linear algebra.</p>
<p>This is because, in my courses we have talked about vector spaces, linear transformations, etc., but we have always for some reason skipped the sections on isomorphisms and homomorphisms.</p>
<p>And yes I have tried to look on wikipedia and such, but it just isn't really clicking for me what it is and what it represents/use of it.</p>
<p>I am under the impression that two spaces with bijection are isomorphic to one another, but that is about it.</p>
<p>Any ideas/opinions?</p>
<p>Thanks!</p>
| Community | -1 | <p>A <em>homomorphism</em> is a structure-preserving mapping.<br>
An <em>isomorphism</em> is a bijective homomorphism.</p>
<p>"Structure" can mean many different things, but in the context of linear algebra, almost exclusively means the vectorial structure -- i.e. all those rules about addition and scalar multiplication. So a homomorphism is just a mapping from one vector space to another vector space that "<a href="https://math.stackexchange.com/questions/1159339/what-does-is-mean-for-a-transformation-to-preserve-an-operation">preserves addition and scalar multiplication</a>". I.e. a homomorphism in linear algebra is just a linear transformation. A typical linear transformation in linear algebra is $\mathbf x \mapsto A\mathbf x$, where $\mathbf x \in \Bbb R^m$ and $A$ is an $n\times m$ matrix.</p>
<p>An isomorphism is not only a homomorphism -- it's bijective. So these are the invertible linear transformations. So the mapping $\mathbf x \mapsto A\mathbf x$ is an isomorphism if $A$ is an <em>invertible</em> matrix.</p>
|
2,410,517 | <p>I feel like I'm missing something very simple here, but I'm confused at how Rudin proved Theorem 2.27 c:</p>
<p>If <span class="math-container">$X$</span> is a metric space and <span class="math-container">$E\subset X$</span>, then <span class="math-container">$\overline{E}\subset F$</span> for every closed set <span class="math-container">$F\subset X$</span> such that <span class="math-container">$E\subset F$</span>. Note: <span class="math-container">$\overline{E}$</span> denotes the closure of <span class="math-container">$E$</span>; in other words, <span class="math-container">$\overline{E} = E \cup E'$</span>, where <span class="math-container">$E'$</span> is the set of limit points of <span class="math-container">$E$</span>.</p>
<p>Proof: If <span class="math-container">$F$</span> is closed and <span class="math-container">$F \supset E$</span>, then <span class="math-container">$F\supset F'$</span>, hence <span class="math-container">$F\supset E'$</span>. Thus <span class="math-container">$F \supset \overline{E}$</span>.</p>
<p>What I'm confused about is how we know <span class="math-container">$F \supset E'$</span> from the previous facts?</p>
| Trevor Gunn | 437,127 | <p>If $x$ is a limit point of $E$ then $x = \lim x_n$ for some sequence $x_n \in E \setminus \{x\}$. If $E \subseteq F$ then $x_n \in F \setminus \{x\}$ so we can also say that $x$ is a limit point of $F$. Therefore</p>
<p>$$ E' \subseteq F' \subseteq F. $$</p>
|
894,159 | <p>I was assigned the following problem: find the value of $$\sum_{k=1}^{n} k \binom {n} {k}$$ by using the derivative of $(1+x)^n$, but I'm basically clueless. Can anyone give me a hint?</p>
| Community | -1 | <p>Note that $$(1+x)^n= \sum_{k=0}^{n} \binom {n} {k} x^k$$ </p>
<blockquote class="spoiler">
<p> $$f'(x)=n \cdot (1+x)^{n-1}= \sum_{k=1}^{n} k \binom {n} {k} x^{k-1}$$
Whence, we have:
$$\sum_{k=1}^{n} k \binom {n} {k} = f'(1) = n \cdot 2^{n-1}$$</p>
</blockquote>
|
2,809,686 | <p>Let S={1,2,3,...,20}. Find the probability of choosing a subset of three numbers from the set S so that no two consecutive numbers are selected in the set.
"I am getting problem in forming the required number of sets."</p>
| Graham Kemp | 135,106 | <p>Count the total ways to select any three numbers from the set of twenty.</p>
<p>Count the ways to select a number, its immediate successor, and any other number <em>that is not the first number's immediate predecessor</em> (to avoid overcounting). There will be two cases to consider: first select $1$ or first select from $\{2..19\}$</p>
<p>Subtract and divide as appropriate to get the probability.</p>
<hr>
<p>Alternately: </p>
<p>Count ways to select any three numbers from $\{2..19\}$, so you can then subtract one from the least and add one to the greatest to ensure they are not consecutive <em>and</em> come from $\{1..20\}$. </p>
<p>Count the total ways to select any three numbers from the set of twenty.</p>
<p>Divide and answer.</p>
|
112,226 | <p>Prove that there are exactly</p>
<p>$$\displaystyle{\frac{(a-1)(b-1)}{2}}$$ </p>
<p>positive integers that <em>cannot</em> be expressed in the form </p>
<p>$$ax\hspace{2pt}+\hspace{2pt}by$$</p>
<p>where $x$ and $y$ are non-negative integers, and $a, b$ are positive integers such that $\gcd(a,b) =1$.</p>
| Brian M. Scott | 12,042 | <p>Call an integer <em>bad</em> if it cannot be represented as an integer combination of $a$ and $b$ with non-negative coefficients. There are $(a-1)(b-1)$ non-negative integers less than $(a-1)(b-1)$, and you know that all of the bad integers are among them. Take a look at a simple example. Suppose that $a=4$ and $b=7$, so that the bad integers must lie between $0$ and $17$ inclusive. </p>
<p>$$\begin{array}{r|c}
\text{Good}:&0&16&15&14&4&12&11&7&8\\ \hline
\text{Bad}:&17&1&2&3&13&5&6&10&9
\end{array}$$</p>
<p>Notice that the numbers in each column add up to $17=(a-1)(b-1)-1$. A little experimentation will suggest that this is a general phenomenon: if $b=(a-1)(b-1)-1$ is the largest bad integer, and $0\le n\le b$, then exactly one of $n$ and $b-n$ is bad. If you can prove this, you’ll have as an immediate consequence that $\frac12(b+1)=\frac12(a-1)(b-1)$ integers are bad.</p>
<p>It’s easy to see that $n$ and $b-n$ cannot both be good: that would make $b$ good. Thus, you want to show that they also cannot both be bad. For this you’ll probably want to use what you know about the general solution to the Diophantine equation $c=ax+by$. If you get stuck, take a look at the proof of Lemma 3 in <a href="http://math.pugetsound.edu/~mspivey/FrobFinal.pdf" rel="nofollow noreferrer">this paper</a> by Mike Spivey; it isn’t exactly what you want, but it’s very close and should give you the ideas that you need to complete the argument.</p>
|
2,909,480 | <p>Please notice the following before reading: the following text is translated from Swedish and it may contain wrong wording. Also note that I am a first year student at an university - in the sense that my knowledge in mathematics is limited.</p>
<p>Translated text:</p>
<p><strong>Example 4.4</strong> Show that it for all integers $n$ is true that $n^3 - n$ is evenly divisible by $3$.<sup>1</sup> </p>
<p>Here we are put in front of a situation with a statement for <em>all integers</em> and not all positive. But it is enough that we treat the cases when $n$ is non-negative, for if $n$ is negative, put $m = -n$. Then $m$ is positive, $n^3 - n = -(m^3 - m)$ and if $3$ divides $a$, then $3$ also divides $-a$.</p>
<p>Now here also exists a statement for $n = 0$ so that we have a sequence $p_0, p_1, p_2, \; \ldots$ of statements, but that the first statement has the number $0$ and not $1$ is of course not of any higher meaning. Statement number $0$ says that $0^3 - 0$, which equals $0$, is evenly divisible by $3$, which obviously is true. If the statement number $n$ now is true, id est $n^3 - n = 3b$ for some integer $b$, then the statement number $n+1$ also must be true for</p>
<p>$
\begin{split}
(n + 1)^3 - (n + 1) & = n^3 - n + 3n^2 + 3n \\
& = 3b + 3n^2 + 3n \\
& =3(b + n^2 + n)
\end{split}
$</p>
<p>and $b + n^2 + n$ is an intege. What we was supposed to show now follows from the induction principle. $\square$</p>
<p><p>1. That an integer $a$ is "evenly divisible by 3" are everyday language rather than mathematical. The precise meaning is that it exists another integer $b$ such that $a = 3b$.</p></p>
<p><strong>In the above written text</strong>, I understanding everything (or I at least think so) except </p>
<p>$
\begin{split}
(n + 1)^3 - (n + 1) & = n^3 - n + 3n^2 + 3n \\
& = 3b + 3n^2 + 3n \\
& =3(b + n^2 + n).
\end{split}
$</p>
<p>Could someone please explain what happened, because I am totally lost?</p>
| Hagen von Eitzen | 39,174 | <p>First line: Just expand the third power:
$$ (n+1)^3=(n+1)(n+1)(n+1)=n^3+3n^2+3n+1$$
(and then subtract $n+1$, of course).</p>
<p>Second line: We know by induction hypothesis that $n^3-n=3b$ (for some $b$)</p>
<p>Third line: extract the common factor $3$.</p>
|
3,154,407 | <p>I would like to define a function whose domain is any multiset of real numbers and image is a real number.</p>
<p>To my understanding, the domain of a function that can be applied on any set of real numbers is the power set <span class="math-container">$\mathcal{P}(\mathbb{R})$</span>. Is it correct? If yes, is there an equivalent "power multiset" and what is its notation?</p>
| Gary Moon | 477,460 | <p>I think you'd probably be best served by defining your own notation. You can write multisets as sets using various conventions (see Wikipedia page), and so you could perhaps view the function as having domain <span class="math-container">$\mathcal{P}(\mathbb{R}) \times \mathcal{P}(\mathbb{R}^2)$</span> (I believe). In the end though, it seems to me that you'd probably be better off with your own notation.</p>
|
3,006,511 | <p>Let <span class="math-container">$c = \{ x = \{x_k\}_{k=1}^{\infty} \in l^\infty \vert \exists \lim_{k \to \infty} x_k \in \mathbb{C} \}$</span>. </p>
<p>Let <span class="math-container">$x_n \in c$</span>, with <span class="math-container">$x_n \to x = \{x_k\}$</span> with the sup norm. </p>
<p>I want to prove that <span class="math-container">$ x \in c$</span>.</p>
<p>So I am a bit stuck here, I want to use the fact that <span class="math-container">$x_n$</span> belongs to c, so it has a limit that I call <span class="math-container">$l_n$</span>. But from now on I do not know how to keep going. Any help?</p>
| alepopoulo110 | 351,240 | <p>As said in the comments, you should improve your notation. It will help.</p>
<p>Let <span class="math-container">$(x_n)\subset c$</span> with <span class="math-container">$x_n=(x_n^1,x_n^2,\dots)$</span> for any <span class="math-container">$n$</span>.</p>
<p>Let <span class="math-container">$x=(b_1,b_2,\dots)$</span>. It suffices to prove that <span class="math-container">$(b_k)$</span> is a Cauchy sequence.</p>
<p>We have that <span class="math-container">$\|x_n-x\|_\infty\to0$</span>, therefore <span class="math-container">$\displaystyle{\lim_{n\to\infty}\sup_{k\in\mathbb{N}}|x_n^k-b_k|=0}$</span>. Let <span class="math-container">$\varepsilon>0$</span>; then there exists <span class="math-container">$n_0\in\mathbb{N}$</span> such that for all <span class="math-container">$n\geq n_0$</span> and for all <span class="math-container">$k\in\mathbb{N}$</span> it is <span class="math-container">$|x_n^k-b_k|<\varepsilon$</span>. Now since <span class="math-container">$x_{n_0}\in c$</span> it is a Cauchy sequence therefore for this <span class="math-container">$\varepsilon$</span> there exists <span class="math-container">$N\geq 0$</span> such that for all <span class="math-container">$m,n\geq N$</span> it is <span class="math-container">$|x_{n_0}^n-x_{n_0}^m|<\varepsilon$</span>.</p>
<p>Now we have for <span class="math-container">$m,n\geq N$</span>: <span class="math-container">$|b_n-b_m|\leq|b_n-x_{n_0}^n|+|x_{n_0}^n-x_{n_0}^m|+|x_{n_0}^m-b_m|<3\varepsilon$</span>.</p>
<p>Since for a random <span class="math-container">$\varepsilon>0$</span> we have found an integer <span class="math-container">$N$</span> such that for all <span class="math-container">$m,n$</span> greater than <span class="math-container">$N$</span> it is <span class="math-container">$|b_n-b_m|<3\varepsilon$</span>, by the definition of a Cauchy sequence we have that <span class="math-container">$(b_n)$</span> is Cauchy.</p>
|
760,767 | <p>I don't understand the last part of this proof:</p>
<p><a href="http://www.proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Sylow_P-Subgroup" rel="nofollow">http://www.proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Sylow_P-Subgroup</a></p>
<p>where they say: $p \nmid \left[{N : P \cap N}\right]$, thus, $P \cap N$ is a Sylow p-subgroup of $N$. I don't see why this implicaction is true. On the other hand, I understand that $P$ being a Sylow p-subgroup of $G$ implies that $p \nmid [G : P]$, for $[G:P]=[G:N_G(P)][N_G(P):P]$ and $p$ does not divide any of these two factors. So, what I don't understand why is true is the inverse implication, that is, if $p \nmid [G : P]$ then $P$ is a Sylow p-subgroup of $G$.</p>
| ayuanx | 246,849 | <p>Interesting topic.</p>
<p>First to @mathguy, $\int r d\theta = r θ$ is wrong. Because usually $r$ is a function of $\theta$. so $\int r d\theta =\int r(\theta)d\theta$.</p>
<p>The reason why the concept of "<em>curve length</em> $r d\theta$" can apply to area intergration in polar coordinates $Area=\int \frac{1}{2} r (r d\theta)$ is that the radial component of the length of the curve doesn't contribute to the area at all, so it is safe to use $r d\theta$ as "<em>effective curve length</em>" while the radial component of curve length is completely dropped off in area integration.</p>
|
2,498,628 | <p>This was a question in our exam and I did not know which change of variables or trick to apply</p>
<p><strong>How to show by inspection ( change of variables or whatever trick ) that</strong></p>
<p><span class="math-container">$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx \tag{I} $$</span></p>
<p>Computing the values of these integrals are known as routine. Further from their values, the equality holds. But can we show equality beforehand?</p>
<blockquote>
<p><strong>Note</strong>: I am not asking for computation since it can be found <a href="https://math.stackexchange.com/questions/187729/evaluating-int-0-infty-sin-x2-dx-with-real-methods?noredirect=1&lq=1.">here</a>
and we have as well that,
<span class="math-container">$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx =\sqrt{\frac{\pi}{8}}$$</span>
and the result can be recover here, <a href="https://math.stackexchange.com/questions/187729/evaluating-int-0-infty-sin-x2-dx-with-real-methods?noredirect=1&lq=1">Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?</a>.</p>
</blockquote>
<p>Is there any trick to prove the equality in (I) without computing the exact values of these integrals beforehand?</p>
| Quanto | 686,284 | <p>Note that
<span class="math-container">$\int_0^\infty(\mathcal{L} g)(x)f(x)\,dx = \int_0^\infty g(x)(\mathcal{L}f)(x)\,dx$</span>
with <span class="math-container">$f(x) =\sin x -\cos x$</span>, <span class="math-container">$g(x)=\frac1{\sqrt x}$</span> and their transformations <span class="math-container">$\mathcal{L} f=\frac{x-1}{x^2+1}$</span>, <span class="math-container">$
\mathcal{L} g=\sqrt{\frac \pi x}$</span>
leads to
<span class="math-container">$$\sqrt\pi\int_0^\infty\frac{ \sin x -\cos x}{\sqrt x}dx=
\int_0^\infty\frac{ x-1}{\sqrt x(x^2+1)}dx
\overset{x\to \frac1x}= \int_0^\infty\frac{1- x}{\sqrt x(x^2+1)}dx=0
$$</span>
where the vanishing LHS with <span class="math-container">$x\to x^2$</span> yields <span class="math-container">$\int_0^\infty \sin x^2 dx= \int_0^\infty \cos x^2dx$</span>.</p>
|
44,771 | <p>A capital delta ($\Delta$) is commonly used to indicate a difference (especially an incremental difference). For example, $\Delta x = x_1 - x_0$</p>
<p><strong>My question is: is there an analogue of this notation for ratios?</strong></p>
<p>In other words, what's the best symbol to use for $[?]$ in $[?]x = \dfrac{x_1}{x_0}$?</p>
| Amad | 625,606 | <p>The greek letter <span class="math-container">$\Delta$</span> stands for <em>Difference</em>
So the symbol for the multiplicative increase of variable should be related to letter <span class="math-container">$Q$</span> in greek which is the ancient Qoppa letter and stands for <em>Quotient</em>
<a href="https://i.stack.imgur.com/scmlW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/scmlW.png" alt="enter image description here" /></a></p>
|
1,053,506 | <p>I had thought that the ultra-metric property was just a rule that someone made up, that if applied shows some bizarre behavior. I however came across these notes: <a href="http://www.math.harvard.edu/~thorne/all.pdf" rel="nofollow">Lecture notes</a> and it seems that the ultra-metric property is actually derived from a p-adic norm. i however don't understand what they are doing (Note Theorem 0.4 which leads to Definition 0.5).</p>
<p>So it seems as if the ultra-metric or non-Archimedean metric is actually derived from the Archimedean metric as applied to p-adic numbers?</p>
<p>Thanks,</p>
<p>Brian</p>
| Slade | 33,433 | <p>This has nothing to do with the usual Archimedean/Euclidean metric. We can define a norm (hence a metric) on $\mathbb{Q}$ as follows: let $\| a \|_p$ be $c^{-\nu_p (a)}$. Here $\nu_p (a)$ is the largest number of powers of $p$ dividing $a$ (possibly negative, or $\infty$ for $a=0$). So things are small exactly when they are highly divisible by $p$</p>
<p>This makes $\mathbb{Q}$ into an example of an ultrametric space, because the number of powers of $p$ dividing $a+b$ is always at least the the number of powers of $p$ that divide both $a$ and $b$.</p>
<hr>
<p>Ultrametric spaces are more common than you might think, though we are not used to noticing them. For example, it is a good idea to think of genetic distance as an ultrametric. Fractals, or data that is tree-like, will tend to have ultrametric qualities.</p>
<p>Again, <em>nothing</em> to do with the Archimedean norm—though it turns out that, up to scaling, the only possible norms on $\mathbb{Q}$ are the Archimedean one, and the $p$-adic ones, a fact of great interest to number theorists.</p>
<hr>
<p>Normed fields that do <em>not</em> have the ultrametric property are actually quite rare: they are just the subfields of $\mathbb{C}$. On the other hand, there are ultrametric fields of every infinite cardinality.</p>
<hr>
<p>I think the $p$-adics are not such a great first example for an ultrametric space, or even for a normed field. For the former, any discrete space will do. For the latter, we could look instead at the ring $\mathbb{C}(T)$ of rational functions, and measure divisibility by $T$ instead of $p$ (and otherwise mimic the $p$-adics, as $\mathbb{C}[T]$ and $\mathbb{Z}$ are much more similar than they first appear...).</p>
<p>I think this might be a little more intuitive, because it's about measuring the order of poles and zeros of meromorphic functions, which feels more natural, at least at first, than measuring the number of powers of $p$ dividing a number*.</p>
<hr>
<p>*My dear departed teacher, Paul Sally, claimed to have accidentally referred to this, during a lecture, as "the $p$-ness of a number".</p>
|
3,712,128 | <p>I've been reading about combinatorial games, specifically about positions in such games can be classified as either winning or losing positions. However, what I'm not sure about now is how I can represent draws using this: situations where neither player wins or loses. Do I use a winning position or losing position to represent a position where a draw has occurred? Why is such a representation correct?</p>
<p>Thanks in advance. </p>
| Z Ahmed | 671,540 | <p><span class="math-container">$(a)$</span>: two parallel non-intersecting lines, (b) intesecting lines.</p>
<p>In the case of three planes (equations) when non two are parallel and no two are idendtical: Take <span class="math-container">$z=k$</span> and solve for %x, y$ punt these in the third equation.
One of the three things will happen:</p>
<p>(1) <span class="math-container">$k$</span> gets determined, the planes meet in a unique point form a an open tetrahedron.</p>
<p>(2) <span class="math-container">$k$</span> disappears giving a true statement e,g,, <span class="math-container">$6=6$</span> so many solutions, three planes meet in a line
like three pages of a book.</p>
<p>(3) <span class="math-container">$k$</span> disappears giving leaving a wrong statement <span class="math-container">$6=5$</span>, so no solution, planes meeting an open prism.</p>
|
295,618 | <p>Problem A: Please fill each blank with a number such that all the statements are true:</p>
<p>0 appears in all these statements $____$ time(s)<br>
1 appears in all these statements $____$ time(s)<br>
2 appears in all these statements $____$ time(s)<br>
3 appears in all these statements $____$ time(s)<br>
4 appears in all these statements $____$ time(s)<br>
5 appears in all these statements $____$ time(s)<br>
6 appears in all these statements $____$ time(s)<br>
7 appears in all these statements $____$ time(s)<br>
8 appears in all these statements $____$ time(s)<br>
9 appears in all these statements $____$ time(s) </p>
<p>Note: they are treated as numbers, not digits. e.g. 11 counts as occurrence of 11, but not two 1.</p>
<p><strong>EDIT</strong><br>
How do number of solutions behave, with respect to number of statements there are? I need a sketch of the proof.</p>
| Ross Millikan | 1,827 | <p>Douglas Hofstadter wrote about these problems years ago. He suggested a useful approach is to fill in the blanks with something, then just count and refill, iterating to convergence. On the first one, I got trapped in a loop between $1741111121$ and $1821211211$ and for the second a loop between $254311311150$ and $262323111160$ where the largest is taken from the current iteration instead of the last.</p>
|
101,098 | <p>I apologize in advance because I don't know how to enter code to format equations, and I apologize for how elementary this question is. I am trying to teach myself some differential geometry, and it is helpful to apply it to a simple case, but that is where I am running into a wall.</p>
<p>Consider $M=\mathbb{R}^2$ as our manifold of interest. I believe that the tangent space is also $\mathbb{R}^2$. From linear algebra, we know that a basis set for $\mathbb{R}^2$ is $$\left\{\left[\matrix{1\\0}\right], \left[\matrix{0\\ 1}\right] \right\}\;.$$</p>
<p>Now, from differential geometry, we are told that basis vectors are $\frac{d}{dx}$ and $\frac{d}{dy}$ where the derivatives are partial deravatives.</p>
<p>So my question is how does one obtain a two-component basis vector of linear algebra from a simple partial derivative?</p>
<hr>
<p>EDIT: Thanks to everyone for the replies. They have been very helpful, but thinking as a physicist, I would like to see how the methods of differential geometry could be used to derivive the standard basis of linear algebra. It seems that there must be more to it than saying that there is an isomorphism between the space of derivatives at a point and R^n which sets up a natural correspondence between the basis vectors.</p>
<p>I may be completely off-the-wall wrong, but somehow I think that the answer involves partial derivatives of a local orthognal coordinate system at point p.</p>
| Jeremy Mann | 23,200 | <p>Yes, the fact that the tangent space at a point isn't actually R^2 cannot be emphasized enough. Another way to think of the tangent space at a point is through equivalence classes of differentiable curves through the given point, with the relation being that two curves are equivalent if the curve composed with your coordinate chart (in this case your chart is trivial) have the same velocity at this point. So in order to obtain a basis for the tangent space all you need to do is take two curves through your given point, and then differentiate them and evaluate. These curves then act on real-valued differentiable functions, say f:M->R^2
so then tangent vector is a function $ \alpha'(0):D\rightarrow\mathbb{R} $
$$
\alpha'(0)f=\frac{d}{dt}(f\circ\alpha) |_{t=0}
$$
Do Carmo's book has a good explanation of this.</p>
|
3,852,952 | <p>Given a projective space <span class="math-container">$\mathbb{P}^n(\mathbb{C})$</span>, I can consider the Grasmannian of lines <span class="math-container">$G(2,n+1)$</span>, which has a structure of projective variety inside <span class="math-container">$\mathbb{P}^N$</span>, where <span class="math-container">$N=\binom{2+n+1}{2}-1$</span> thanks to the Pl<span class="math-container">$\ddot{u}$</span>cker embedding.</p>
<p>It has been told me -literally, I have no references, it was a speech- that while a point <span class="math-container">$p\in G(2,n+1)$</span> obviously represents a line in <span class="math-container">$\mathbb{P}^n$</span> by definition, a <em>line</em> inside <span class="math-container">$G(2,n+1)$</span> -viewd as a projective variety in <span class="math-container">$\mathbb{P}^N$</span>-corresponds to a pencil of planes.</p>
<p>Unfortunately I'm still trying to properly understand why this work -because unfortunately it is not crystal for me -, but I was also wondering if it makes sense -that is, if there is a geometrical meaning-, also for quadrics contained in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Therefore I'd like to have an idea of what happens for quadrics, and moreover even a little help for lines in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Since this question comes primarily from my curiosity, I apologize for the vagueness: any comment, reference or answer would be much appreciate!</p>
| mathma | 270,091 | <p>I don't know if I'm understanding your question right, but here are some ideas. The Grassmanian <span class="math-container">$G(1,n)$</span> can be seen as the projective <span class="math-container">$n-1$</span> dimensional space, because lines are seen as points. Then a line in projective space comes from a plane in <span class="math-container">$n$</span>-dimensional space, perhaps that's what they meant? That plane can be understood as a pencil of lines.</p>
<p>What is unclear to me is your claim that the embeding is in <span class="math-container">$\mathbb{P}^N$</span> with <span class="math-container">$N={n+3\choose 2}-1$</span>. The embedding should go <span class="math-container">$i: G(1,n)\rightarrow \mathbb{P}(\Lambda^1 k^n)\simeq \mathbb{P}^{n-1}$</span>, so the Plücker embedding would be just the isomorphism of the Grassmanian <span class="math-container">$G(1,n)$</span> with <span class="math-container">$\mathbb{P}^{n-1}$</span>.</p>
|
359,277 | <p>Can you find function which satisfies $f(ab)=\frac{f(a)}{f(b)}$? For example $log(x)$ satisfies condition $f(ab)=f(a)+f(b)$ and $x^2$ satisfies $f(ab)=f(a)f(b)$?</p>
| Luke Mathieson | 35,289 | <p>You could take a relatively trivial function like $f(x) = 1$. Or a slightly more general version that takes everything to the identity.</p>
|
359,277 | <p>Can you find function which satisfies $f(ab)=\frac{f(a)}{f(b)}$? For example $log(x)$ satisfies condition $f(ab)=f(a)+f(b)$ and $x^2$ satisfies $f(ab)=f(a)f(b)$?</p>
| baharampuri | 50,080 | <p>Let us reformulate the question as classify all maps $f : G \rightarrow H$ which need not be group morphism that satisfies the condition $f(ab)=f(a)f(b)^{-1}$</p>
<p>A simple calculation shows that $f(e)=f(x)f(x^{-1})^{-1}= f(x^{-1})f(x)^{-1}$ or we have $f(x)=f(e)^{-1}f(x^{-1})=f(e)f(x^{-1})$ or $f(e)^{-1}=f(e)$ Now $f(x)=f(ex)=f(e)f(x)^{-1}=f(e)^{-1}f(x)^{-1}=(f(x)f(e))^{-1}=(f(x)f(e)^{-1})^{-1}=f(xe)^{-1}=f(x)^{-1}$ so even if we don't assume a group morphism we have the image involutive. And hence it's a group morphism $f(ab)=f(a)f(b)^{-1}=f(a)f(b)$</p>
|
3,534,566 | <p>I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?</p>
<p><span class="math-container">$$x^2y'+ 2xy = 5y^3$$</span></p>
<p><span class="math-container">$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$</span></p>
<p><span class="math-container">$n = 3$</span></p>
<p><span class="math-container">$v = y^{-2}$</span></p>
<p><span class="math-container">$-\frac{1}{2}v'=y^{-3}$</span></p>
<p><span class="math-container">$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$</span></p>
<p><span class="math-container">$$v'+\frac{4}{x}v=\frac{-10}{x^2}$$</span></p>
<p>this is now a first order linear ODE where:</p>
<p><span class="math-container">$$y(x)=\frac{1}{u(x)}\int u(x)q(x)$$</span></p>
<p><span class="math-container">$u(x)=e^{4\int\frac{1}{x}}=x^4$</span></p>
<p><span class="math-container">$q(x) = \frac{-10}{x^2}$</span></p>
<p><span class="math-container">$$\frac{1}{x^4}\int x^4 \frac{-10}{x^2}=\frac{1}{x^4}\frac{-10x^{2+1}}{2+1}+C$$</span></p>
<p>which leaves us with :</p>
<p><span class="math-container">$$\frac{1}{y^2} = \frac{-10}{3x}+x^{-4}C$$</span></p>
<p>naturally </p>
<p><span class="math-container">$$y^2= \frac{1}{\frac{-10}{3x}+x^{-4}C}$$</span></p>
<p>The book states the answer as being:</p>
<p><span class="math-container">$$y^2= \frac{x}{2+Cx^5}$$</span> </p>
| user577215664 | 475,762 | <p>You made a sign mistake here:
<span class="math-container">$$\frac{-1}{2}v'+\frac{2}{x}v = \frac{5}{x^2}$$</span>
And also
<span class="math-container">$$\left (\frac 1 {y^2} \right )'=-2\frac {y'}{y^3}$$</span>
<span class="math-container">$$\implies v'=-2\frac {y'}{y^3}$$</span>
Another way:
<span class="math-container">$$x^2dy+ 2xydx - 5y^3dx=0$$</span>
<span class="math-container">$$d(x^2y) - 5y^3dx=0$$</span>
<span class="math-container">$$\frac {dx^2y}{(x^2y)^3} - 5\frac {dx}{x^6}=0$$</span>
<span class="math-container">$$- \frac {1}{2(x^2y)^2} +\frac {1}{x^5}=C$$</span></p>
|
2,871,949 | <p>Let $X_1, X_2, X_3, X_4$ be independent Bernoulli random variables. Then
\begin{align}
Pr[X_i=1]=Pr[X_i=0]=1/2.
\end{align}
I want to compute the following probability
\begin{align}
Pr( X_1+X_2+X_3=2, X_2+X_4=1 ).
\end{align}
My solution: Suppose that $X_1+X_2+X_3=2$ and $X_2+X_4=1$. Then $(X_2, X_4)=(0,1)$ or $(1,0)$. The probabilities of $(X_2, X_4)=(0,1)$ and $(1,0)$ are both $1/2 \times 1/2 = 1/4$. If $(X_2, X_4)=(0,1)$, then $X_1+X_2+X_3=X_1+X_3=2$. Therefore $X_1 = X_3 =1$. This happens with probability $1/4$. If $(X_2, X_4)=(1,0)$, then $X_1+X_2+X_3=X_1+1+X_3=2$. Therefore $(X_1, X_3) \in \{(1,0), (0,1)\}$. This happens with probability $1/4$. Therefore
\begin{align}
Pr( X_1+X_2+X_3=2, X_2+X_4=1 ) = 1/16+2/16=3/16.
\end{align}
Is this correct? Thank you very much.</p>
| BGM | 297,308 | <p>Just add as a supplementary technique:</p>
<p>You may also try to use the law of total probability with conditioning on $X_2$, since only $X_2$ are in common of the two events, then make use of the independence:</p>
<p>$$ \begin{align} &\Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1\} \\
=& \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\
&+ \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 1\}\Pr\{X_2 = 1\} \\
=& \Pr\{X_1 + X_3 = 2, X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\
&+ \Pr\{X_1 + X_3 = 1, X_4 = 0|X_2 = 1\}\Pr\{X_2 = 1\} \\
=& \Pr\{X_1 + X_3 = 2\}\Pr\{X_4 = 1\}\Pr\{X_2 = 0\} + \\
&+ \Pr\{X_1 + X_3 = 1\}\Pr\{X_4 = 0\}\Pr\{X_2 = 1\} \\
=& \frac {1} {4}\times \frac {1} {2} \times \frac {1} {2} + \frac {1} {2} \times \frac {1} {2} \times \frac {1} {2} \\
=& \frac {3} {16}
\end{align}$$</p>
<p>So essentially you have counted the cases, just like what you have did.</p>
|
63,525 | <p>I asked this question in math.stackexchange but I didn't have much luck. It might be more appropiate for this forum. Let $z_1,z_2,…,z_n$ be i.i.d random points on the unit circle ($|z_i|=1$) with uniform distribution on the unit circle. Consider the random polynomial $P(z)$ given by
$$
P(z)=\prod_{i=1}^{n}(z−z_i).
$$
Let $m$ be the maximum absolute value of $P(z)$ on the unit circle $m=\max\{|P(z)|:|z|=1\}$.</p>
<p>How can I estimate $m$? More specifically, I would like to prove that there exist $\alpha>0$ such that the following holds almost surely as $n\to\infty$
$$
m\geq \exp(\alpha\sqrt{n}).
$$
Or at least that for every $\epsilon>0$ there exists $n$ sufficiently large such that
$$
\mathbb{P}(m\geq\exp(\alpha\sqrt{n}))>1-\epsilon
$$
for some $\alpha$ independent on $n$.</p>
<p>Any idea of what can be useful here?</p>
| Joe Silverman | 11,926 | <p>I have a couple of comments regarding Chandru's answer, but they're too long to fit in the comment box, so I'm making this a separate answer. First, the quantity
$$
F(P) = \int_0^{2\pi} \log|P(e^{i\phi})| d\phi
$$
is not non-negative without some further assumption. The easiest is to assume that the polynomial $P$ is <em>monic</em>. But as a trivial counterexample in general, one can take $P(z)=c$
with $|c|<1$. For Chandru's argument, somewhere one needs to use the assumption that $P(0)=1$ to make the conclusion that $F(P)\ge0$.</p>
<p>Second, the quantity $F(P)$ is essentially the logarithm of the classical <em>Mahler measure</em>, which is defined by
$$
\log M(P) = \frac{1}{2\pi} \int_0^{2\pi} \log|P(e^{i\phi})| d\phi.
$$
Factoring $P$ as
$$
P(z) = c(z-z_1)\cdots(z-z_n),
$$
it is an elementary classical computation to show that
$$
\log M(P) = \log|c|+\sum_{i=1}^n \log\max(1,|z_i|).
$$
In particular, if $|c|=1$, then $\log M(P)\ge0$.</p>
|
3,919,284 | <p>Let <span class="math-container">$S^1 \subseteq \mathbb{R}^2$</span> be the unit circle circle <span class="math-container">$\mathbb{V}(X^2+Y^2-1)$</span>. Let <span class="math-container">$S$</span> be an infinite subset of <span class="math-container">$S^1$</span>. I want to show that
<span class="math-container">$$\overline{S}^{\text{Zariski}}= \mathbb{V}(\mathbb{I}(S))= S^1$$</span></p>
<p>Is there an easy way to see this?</p>
<p>Clearly <span class="math-container">$S^1$</span> itself is Zariski closed, so it contains the Zariski-closure of <span class="math-container">$S$</span>. However, the other inclusion seems more delicate. Any ideas how I can show this?</p>
| Sabino Di Trani | 82,009 | <p>The ring <span class="math-container">$\mathbb{R}[x,y]$</span> has dimension 2 and consequently a prime ideal containing <span class="math-container">$x^2+y^2-1$</span> must be a maximal one.</p>
<p>Moreover, the ring <span class="math-container">$\mathbb{R}[x,y]$</span> is Noetherian and every ideal has a finite number of associated primes.</p>
<p>Geometrically this two facts mean exactly what you want: a Zarisky closed proper subset of the Circle can only be a finite union of (closed) points.</p>
|
3,919,284 | <p>Let <span class="math-container">$S^1 \subseteq \mathbb{R}^2$</span> be the unit circle circle <span class="math-container">$\mathbb{V}(X^2+Y^2-1)$</span>. Let <span class="math-container">$S$</span> be an infinite subset of <span class="math-container">$S^1$</span>. I want to show that
<span class="math-container">$$\overline{S}^{\text{Zariski}}= \mathbb{V}(\mathbb{I}(S))= S^1$$</span></p>
<p>Is there an easy way to see this?</p>
<p>Clearly <span class="math-container">$S^1$</span> itself is Zariski closed, so it contains the Zariski-closure of <span class="math-container">$S$</span>. However, the other inclusion seems more delicate. Any ideas how I can show this?</p>
| KReiser | 21,412 | <p>Sabino Di Trani's answer is good. Here is a solution which is perhaps more hands-on.</p>
<p>Suppose <span class="math-container">$f\in \Bbb R[x,y]/(x^2+y^2-1)$</span> vanishes on an infinite subset of <span class="math-container">$S^1$</span>. We'll show it's actually zero, which shows that the Zariski closure of any infinite subset of <span class="math-container">$S^1$</span> is all of <span class="math-container">$S^1$</span>: <span class="math-container">$I(S)=(0)$</span> which implies <span class="math-container">$V(I(S))=V(0)=S^1$</span>.</p>
<p>We can write <span class="math-container">$f$</span> as <span class="math-container">$p+yq$</span> for polynomials <span class="math-container">$p,q\in\Bbb R[x]$</span> by rewriting every instance of <span class="math-container">$y^2$</span> as <span class="math-container">$1-x^2$</span>. Then <span class="math-container">$f\cdot(p-yq)=p^2-y^2q^2=p^2+(x^2-1)q^2$</span> is a polynomial in <span class="math-container">$x$</span> vanishing on an infinite set, so it's zero. As <span class="math-container">$\Bbb R[x,y]/(x^2+y^2-1)$</span> is a domain, we must have that either <span class="math-container">$f=p+yq$</span> or <span class="math-container">$p-yq$</span> is zero, both of which give <span class="math-container">$f=0$</span>.</p>
|
1,530,848 | <p>Let $F(\mathbb{R})$ be the set of all functions $f : \mathbb{R} → \mathbb{R}$. Define pointwise addition and multiplication as follows. For any $f$ and $g$ in $F(\mathbb{R})$ let:</p>
<p>(i) $(f + g)(s) = f(x) + g(x)$ for all $x \in \mathbb{R}$</p>
<p>(ii) $(f · g)(s) = f(x) · g(x)$ for all $x \in \mathbb{R}$</p>
<p>Prove that $F(\mathbb{R})$ forms a ring under these two operations.</p>
<p>I know the axioms of rings that I need to show hold for this to be true. My question is, when I am showing closure for instance,</p>
<p>$(f + g)(s) +(f + g)(t) = f(x) + g(x)+ f(y) + g(y) \in \mathbb{R}$ and</p>
<p>$(f · g)(s)·(f · g)(t) = f(x) · g(x) · f(y) · g(y) \in \mathbb{R}$</p>
<p>is this the correct way to apply the axioms? My book was very general about the definition of a ring. The whole pointwise notation versus the definition is confusing me. Any help would be appreciated.</p>
| Slade | 33,433 | <p>This is not really an answer, but it was getting too long to be a comment.</p>
<p>Mathematics draws much of its power from deep, sometimes mysterious dualities between geometry and algebra, so I do not think there is any way to understand the relationship between geometric intuition and symbol manipulation in general.</p>
<p>Diophantine equations are one of the least geometrically intuitive areas of mathematics, though it is certainly worth mentioning that geometric insights (albeit very difficult ones) are behind some of the deepest results in the field, like Fermat's Last Theorem.</p>
<p>One can sometimes make arguments from intuitive principles that give partial answers. To take your example of $(xy-7)^2 = x^2 + y^2$, one can observe immediately that this is the intersection in the plane of two curves, one of degree $2$ and one of degree $4$, that have no components in common. By <a href="https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem" rel="nofollow">Bézout's theorem</a>, there are at most $8$ intersection points (in fact, we can conclude that there are exactly $8$ solutions, if we know how to count them).</p>
<p>This tells us that the <em>integer</em> solutions, in particular, are a finite, possibly empty set. But it doesn't really give a good a priori method for finding them. That requires some symbol pushing.</p>
<p>To some extent, we can gain a lot of intuition for pushing symbols around. But this can be a lifelong process, and I think any mathematician will tell you that there are advantages to being able to manipulate equations without completely understanding what you are doing. There is some truth to the famous quotation that mathematics is about getting used to things, and for myself I can say that diligent practice has been far more valuable than theory when it comes to developing comfort with difficult problems.</p>
<p>The theory and intuition comes eventually, and with specific examples you may get little boosts to your insight here and there from helpful mentors, but there is no singular approach that will allow you to "plan".</p>
|
3,450,713 | <p>Let <span class="math-container">$a+3b=7$</span> and <span class="math-container">$c=3$</span>. Then value of <span class="math-container">$a+3(b+c)$</span> is</p>
<p>A) <span class="math-container">$10$</span></p>
<p>B) <span class="math-container">$16$</span> </p>
<p>C) <span class="math-container">$21$</span></p>
<p>D) <span class="math-container">$30$</span></p>
<p>Answer is B, but how?</p>
| fleablood | 280,126 | <p>Method 1: Distribute first then substitute what you recognize:</p>
<p><span class="math-container">$a+3(b+c) =$</span></p>
<p><span class="math-container">$a + 3b + 3c = $</span></p>
<p><span class="math-container">$(a + 3b) + 3(c)= $</span> (and we know <span class="math-container">$c=3$</span> and <span class="math-container">$a+3b = 7$</span> so..)</p>
<p><span class="math-container">$7 + 3*3 = 7 + 9 = 16$</span>.</p>
<p>Method 2: Substitute what we know first.</p>
<p><span class="math-container">$a+3(b+c) =$</span></p>
<p><span class="math-container">$a+3(b+3) =$</span></p>
<p><span class="math-container">$a+3b + 9 =$</span></p>
<p><span class="math-container">$(a+3b) +9 =$</span></p>
<p><span class="math-container">$7+9 = 16$</span>.</p>
<p>Method 3: (perhaps overkill) Reduce the unknowns:</p>
<p><span class="math-container">$a +3b = 7$</span> so <span class="math-container">$a = 7 - 3b$</span>. </p>
<p>Replace <span class="math-container">$a$</span> with <span class="math-container">$7-3b$</span> and <span class="math-container">$c$</span> with 3$.</p>
<p><span class="math-container">$a + 3(b+c) = $</span></p>
<p><span class="math-container">$(7-3b) + 3(b + 3) =$</span></p>
<p><span class="math-container">$(7-3b) + 3b + 9 =$</span></p>
<p><span class="math-container">$7 +(-3b+3b) + 9=$</span></p>
<p><span class="math-container">$7+0 + 9 = 16$</span>.</p>
<p>3<span class="math-container">$\frac 12$</span>. (the wrong choice but all ways will work;)</p>
<p><span class="math-container">$a +3b = 7$</span> so <span class="math-container">$3b = 7-a$</span> and <span class="math-container">$b = \frac {7-a}3$</span>.</p>
<p>Replace every <span class="math-container">$b$</span> with <span class="math-container">$\frac {7-a}3$</span> and every <span class="math-container">$c$</span> with <span class="math-container">$3$</span>.</p>
<p><span class="math-container">$a + 3(b + c) =$</span></p>
<p><span class="math-container">$a + 3(\frac {7-a}3 + 3)=$</span></p>
<p><span class="math-container">$a + (7-a) + 9 =$</span></p>
<p><span class="math-container">$(a-a) + 7 + 9 = 0 + 7+9 = 16$</span>.</p>
|
3,450,713 | <p>Let <span class="math-container">$a+3b=7$</span> and <span class="math-container">$c=3$</span>. Then value of <span class="math-container">$a+3(b+c)$</span> is</p>
<p>A) <span class="math-container">$10$</span></p>
<p>B) <span class="math-container">$16$</span> </p>
<p>C) <span class="math-container">$21$</span></p>
<p>D) <span class="math-container">$30$</span></p>
<p>Answer is B, but how?</p>
| suhbell | 592,879 | <p>For questions like these where you cannot solve for each value to then find the solution, you are supposed to substitute some form of the given equation in for the variable. You are given a number for c so substitute that into the equation <span class="math-container">$ a+3(b+c) $</span> first.
<span class="math-container">$$ a+3(b+3) $$</span>
We cant to find a way to substitute <span class="math-container">$ a+3b $</span> into the equation since we know the value of <span class="math-container">$ a+3b $</span> is 7. First, let's try expanding the formula to see if the equation appears:
<span class="math-container">$$ a+3b+9 $$</span>
We see <span class="math-container">$a+3b$</span> in the first part of the equation, so we can substitute 7 in for <span class="math-container">$a+3b$</span>:
<span class="math-container">$$ 7+9 $$</span>
Now, we can solve!
<span class="math-container">$$ 16 $$</span>
For problems like these, find a way to rearrange the final equation in a way that will allow you to substitute the given equations into it.</p>
|
1,478,314 | <p>In this particular case, I am trying to <strong>find all points $(x,y)$ on the graph of $f(x)=x^2$ with tangent lines passing through the point $(3,8)$</strong>. </p>
<p>Now then, I know the <a href="http://www.meta-calculator.com/online/?panel-102-graph&data-bounds-xMin=-10&data-bounds-xMax=10&data-bounds-yMin=-7.28&data-bounds-yMax=7.28&data-equations-0=%22y%3Dx%5E2%22&data-rand=undefined&data-hideGrid=false" rel="nofollow">graph</a> of $x^2$. What now?</p>
| Ian Miller | 278,461 | <p>A point on the graph is $(x,x^2)$. The slope from that point to $(3,8)$ is given by: $\frac{x^2-8}{x-3}$. This has to be equal to the derivative at the point for it to be a tangent. So:
$$\frac{x^2-8}{x-3}=2x$$
$$x^2-8=2x^2-6x$$
$$0=x^2-6x+8$$
$$0=(x-2)(x-4)$$
So $x=2$ or $x=4$.
So the points are $(2,4)$ and $(4,16)$.</p>
|
27,965 | <p>I'm looking at <a href="https://math.stackexchange.com/questions/2669893/calculating-the-sums-of-series">this question</a>. I gave the answer that was accepted. Please bear in mind that, when I answered this question, it was a different edit. In particular, there were more parts to the question.</p>
<p>The reason I'm here is because my question got a few up votes as well as a few down votes, attaining a score of $-1$ at its nadir. I was wondering, was I justified in giving this answer?</p>
<p>I'm aware of the guidelines about good questions for the site, and also aware that this question fails them. I've also seen <a href="https://math.meta.stackexchange.com/questions/27259/is-it-acceptable-to-answer-a-poor-quality-question">this meta post</a>.</p>
<p>The reason I gave my answer is because I believed the asker really did have little clue about how to go about these questions. I figured that all the asker needed was one good, fully justified worked example, and they could do the rest on their own. The asker's comment at the end vindicated this view, and they changed the question so that my answer would more comfortably fit it!</p>
<p>I understand that it's generally not preferable to reward poor quality questions on the site. I don't want the quality of questions to (further) drop. But, I do think that, for many people, these first steps are the most daunting. It's very hard for a new student to produce the work they've done, when they haven't done any. First and foremost, I want to help people on this site, as I'm sure most other people here do as well, and part of me finds it hard to accept that these people, who are genuinely confused at the first hurdle, cannot get help.</p>
<p>I know there are ways around it too. I know that a more savvy user of the site knows that they can mitigate this, for example, by copying out the relevant formulae they know. However, this is not something that new users pick up easily. I don't think it necessarily indicates laziness, or a lack of due consideration of a problem, merely an unfamiliarity with the subtler workings of this site.</p>
<p>Bear in mind, I only answered one part of the question too. I wasn't giving out all the answers for all the parts, for the asker to present as their work. The asker would still be required to do most of the work in order to answer the question. I was just using one part as an example to impart the necessary tools for the asker to answer the question.</p>
<p>So, with all that said, was I justified in giving this answer? Or were the downvotes justified instead?</p>
| hardmath | 3,111 | <p>I think better of <em>thorough</em> answers to poorly stated Questions than of fragmentary "hint" posts.</p>
<p>I suspect there is a temptation for many to give hasty replies to substandard Questions, accompanied by frequent rationalizations that "I didn't want to give the OP a full solution that can be copy-pasted in their homework".</p>
<p>A good Answer is apt to provide value to future Readers, so I prefer that to cryptic replies in the Answer box to "make the OP think for themselves.'</p>
|
730,929 | <p>Let $E$ be an extension field of a finite field $F$ , where $F$ has $q$ elements. Let $a \in E$ be algebraic over $F$ of degree $n$. Prove that $F(a)$ has $q^n$ elements.</p>
<p>I am not sure how to do this one, but furthermore, what does $a$ being algebraic over $F$ of degree $n$ mean? Does it mean the polynomial $a$ solves in $F$ is of degree $n$?</p>
| user134824 | 134,824 | <p>The statement "$a$ is algebraic over $F$ of degree $n$" means two things together:</p>
<ol>
<li>$a$ is the root of some polynomial in $F[x]$ (that is, the coefficients of the polynomial lie in $F$) that has degree $n$.</li>
<li>Every other nonzero polynomial in $F[x]$ for which $a$ is a root has degree at least $n$.</li>
</ol>
<p>In other words, the statement means that the <em>minimal polynomial of $a$ over $F$</em> has degree $n$. For example, convince yourself that the following claims are true:</p>
<ul>
<li>$\sqrt[4]{2}$ has degree $4$ over $\mathbb Q$.</li>
<li>$\sqrt[4]{2}$ has degree $2$ over $\mathbb Q(\sqrt{2})$.</li>
<li>$\sqrt[4]{2}$ has degree $1$ over $\mathbb Q(\sqrt[4]{2})$.</li>
</ul>
<p>Here's a second characterization of degree, one which you'll find useful in solving your problem. If $a$ is algebraic over $F$ of degree $n$ then the set $\{1,a,a^2,\dots,a^{n-1}\}$ forms a basis for the <em>vector space</em> $F(a)$. In other words, we can think of $F(a)$ as being a vector space over $F$ and the dimension of the vector space is the degree of $a$. For vector spaces over finite fields, what is the relationship between dimension and cardinality?</p>
|
24,318 | <p>I have an expression as below:</p>
<pre><code>Equations = 2.0799361919940695` x[1] + 3.3534325557330327` x[1]^2 -
4.335179297091139` x[1] x[2] + 1.1989715511881491` x[2]^2 -
3.766597877399148` x[1] x[3] - 0.33254815073371535` x[2] x[3] +
1.9050048836042945` x[3]^2 + 1.1386715715291826` x[1] x[4] +
2.802846492104668` x[2] x[4] - 0.6210244597295915` x[3] x[4] +
4.943369095158792` x[4]^2
</code></pre>
<p>I want to write it in an output file. So I use the below code:</p>
<pre><code>removebracketvar[x_] :=
StringReplace[
StringReplace[
ToString[x], {"[" -> "", "]" -> "", "," -> "", "*^" -> "e",
".*" -> ".0*"}], Whitespace -> ""];
SetDirectory["C:\\folder"];
WriteString["eqfile.txt",
removebracketvar[
ToString[Equations , InputForm, NumberMarks -> False]] ];
Close["eqfile.txt"]
</code></pre>
<p>The slight problem with the code for me is that it inserts the floating point numbers up to 16 digits of precision. I just want them to around up to 10 digits of precision.
When I use <code>SetPrecision[Equations,10]</code>, it weirdly changes <code>x[1]</code> etc. to <code>x[1.0000000]</code>, etc.! I want to leave the variables as they are but want to change the floating points to less number of digits after the decimal point. What would be the best way of doing this?</p>
| bill s | 1,783 | <p>How about:</p>
<pre><code>apply[func_] := Module[{}, xvalues = Range[0, 500, 2.5];
points1 = Map[func1, xvalues];
Do[If[points1[[i]] < 0, points1[[i]] = 0], {i, 1, Length[points1], 1}];
table1 = Transpose[{xvalues, points1}]];
</code></pre>
<p>Now you call the function apply with your desired funcX as an argument</p>
<pre><code>apply[func1]
</code></pre>
<p>Or you can automate this by defining</p>
<pre><code>allFuncs = {func1, func2, func3, func4};
</code></pre>
<p>and then </p>
<pre><code>apply/@allFuncs
</code></pre>
<p>will run them all.</p>
|
4,316,780 | <p>Let <span class="math-container">$X$</span> be a real Banach space. Let <span class="math-container">$J \colon X \to 2^{X^*}$</span> be its (normalized) duality map,
<span class="math-container">$$ J(x) = \{ x^* \in X^* \colon \langle x^* , x \rangle =||x|| \ ||x^*||, \ || x^* ||=||x|| \} , \ x \in X.$$</span>
Assume that <span class="math-container">$X$</span> is smooth, so that <span class="math-container">$J(x)= \{j(x)\}$</span> is a singleton. Fix <span class="math-container">$x,y \in X$</span>. It is known that, if
<span class="math-container">$$||x|| \le ||x+ry||, \tag 1$$</span> for every <span class="math-container">$r \in \mathbb R$</span>, then <span class="math-container">$\langle j(x),y \rangle =0$</span>. What happens if we only take <span class="math-container">$r \ge0$</span> in <span class="math-container">$(1)$</span>? I expect that <span class="math-container">$ \langle j(x),y \rangle \ge 0$</span>, since that is the case for Hilbert spaces. Indeed, if <span class="math-container">$X$</span> is a Hilbert space with inner product <span class="math-container">$(\cdot,\cdot) $</span>, then <span class="math-container">$J$</span> is just the identity map, and <span class="math-container">$(1)$</span> implies that
<span class="math-container">$$ r^2 ||y||^2 + 2r (x,y) \ge 0, $$</span>
for every <span class="math-container">$r \ge 0$</span>. Dividing by <span class="math-container">$r$</span> and then letting <span class="math-container">$r \to 0$</span> we obtain that <span class="math-container">$(x,y) \ge 0$</span>.</p>
| Evangelopoulos Phoevos | 739,818 | <p>Here is a different approach. For <span class="math-container">$r >0$</span> consider the unit functional <span class="math-container">$y^*_r = j(x+r^{-1}y)/\|j(x+r^{-1}y)\|$</span>. Then
<span class="math-container">\begin{align*}
\|x\| &\le \|x+r^{-1}y\| = \frac{1} {\| j(x+r^{-1}y) \|}\langle j(x+r^{-1}y),x+r^{-1}y\rangle = \langle y^*_r, x +r^{-1}y\rangle
\\
&=\langle y_r^* ,x \rangle + \tfrac 1r\langle y_r^* ,y\rangle \le \|x\| + \tfrac 1r \langle y_r^* ,y\rangle.
\end{align*}</span>
Since the closed unit ball of <span class="math-container">$X^*$</span> is weak-star compact, <span class="math-container">$(y_r)_{r>0}$</span> admits a weak-star convergent subnet. If <span class="math-container">$y^*$</span> is its limit point, then it is true that
<span class="math-container">$$ \|y^*\| \le 1, ~~~ \langle y^*,y\rangle \ge 0 ~~~\text{and} ~~~\langle y^*,x\rangle \ge \|x\|.$$</span>
Since <span class="math-container">$\langle y^*,x\rangle \le \|x\|,$</span> we obtain that
<span class="math-container">$\langle y^*,x \rangle=\|x\|$</span> and thus <span class="math-container">$\|y^*\|=1$</span>. It is easy to see that <span class="math-container">$j(x) = \|x\|y^*$</span> and so
<span class="math-container">$$\langle j(x),y\rangle= \|x\| \langle y^*,x \rangle \ge 0.$$</span></p>
<blockquote>
<p>Note: The converse is also true. That is, if <span class="math-container">$\langle j(x),y)\ge 0$</span> then for any <span class="math-container">$r>0$</span>
<span class="math-container">$$ \|x\| \le \|x+ry\|.$$</span>
The proof is similar to the case of Hilbert spaces: We estimate
<span class="math-container">$$ \|x\|^2 = \langle j(x),x \rangle \le \langle j(x), x \rangle +r \langle j(x),y \rangle = \langle j(x) , x+ry\rangle \le \|x\| \|x+ry\|.$$</span></p>
</blockquote>
|
122,503 | <p>It seems that the current state of quantum Brownian motion is ill-defined. The best survey I can find is <a href="http://arxiv.org/pdf/1009.0843v1.pdf">this one</a> by László Erdös, but the closest the quantum Brownian motion comes to appearing is in this conjecture (p. 30):</p>
<blockquote>
<p>[<b>Quantum Brownian Motion Conjecture</b>]: For small [disorder] $\lambda$ and [dimension] $d \ge 3$, the location of the electron is governed by a heat equation in a vague sense: $$\partial_t \big|\psi_t(x)\big|^2 \sim \Delta_x \big|\psi_t(x)\big|^2 \quad \Rightarrow \quad \langle \, x^2 \, \rangle_t \sim t, \quad t \gg 1.$$
The precise formulation of the first statement requires a scaling limit. The second
statement about the diffusive mean square displacement is mathematically precise, but
what really stands behind it is a diffusive equation that on large scales mimics the
Schrödinger evolution. Moreover, the dynamics of the quantum particle converges to
the Brownian motion as a process as well; this means that the joint distribution of
the quantum densities $\big|\psi_t(x)\big|^2$ at different times $t_1 < t_2 < \dots < t_n$ converges to the corresponding finite dimensional marginals of the Wiener process.</p>
</blockquote>
<p>This is the Anderson model in $\mathbb R^d$ with disordered Hamiltonian $H = -\Delta + \lambda V$. The potential $V$ is disordered, and is generated by i.i.d. random fields; the parameter $\lambda$ controls the scale of the disorder.</p>
<hr>
<p><a href="http://en.wikipedia.org/wiki/Brownian_motion">Classical Brownian motion</a> admits many characterizations and generalizations. For example, <a href="http://en.wikipedia.org/wiki/Wiener_measure">Wiener measure</a> leads to the construction of an <a href="http://en.wikipedia.org/wiki/Abstract_Wiener_space">abstract Wiener space</a>, which is the appropriate setting for the powerful <a href="http://en.wikipedia.org/wiki/Malliavin_calculus">Mallivin calculus</a>. The <a href="http://en.wikipedia.org/wiki/Structure_theorem_for_Gaussian_measures">structure theorem of Gaussian measures</a> says that <i>all</i> <a href="http://en.wikipedia.org/wiki/Gaussian_measure#Gaussian_measures_on_infinite-dimensional_spaces">Gaussian measures</a> are abstract <a href="http://ncatlab.org/nlab/new/Wiener+measure">Wiener measures</a> in this way. I would love to know what all this theory looks like in the language of <a href="http://www.mitchener.staff.shef.ac.uk/free.pdf">non-commutative probability theory</a>.</p>
<p>The QBM Conjecture states roughly that a quantum particle in a weakly disordered environment should behave like a quantum Brownian motion. This is an important open problem, but it doesn't quite capture what a QBM <i>is</i>, nor what different <i>types</i> of QBM may exist. Thus my question:</p>
<blockquote>
<p>What kind of precise mathematical object is a quantum Brownian motion? </p>
</blockquote>
| Jess Riedel | 5,789 | <p>(In words explained below:) <strong>Quantum Brownian motion (QBM) is a class of possible dynamics for an open, quantum, continuous degree of freedom in which the reduced dynamics are specified by a quadratic Hamiltonian and linear Lindblad operators in the phase-space variables $x$ and $p$.</strong></p>
<p>Consider the arbitrary time-evolution of a system's density matrix when it is in contact with an environment:
\begin{align}
\rho = \rho_{\mathcal{S}}(t) = \mathrm{Tr}_{\mathcal{E}} [U_t \sigma^0_{\mathcal{SE}} U_t^\dagger],
\end{align}
where $\sigma^0_{\mathcal{SE}}$ is the initial global state (both $\mathcal{S}$ and $\mathcal{E}$) and $U_t$ is the unitary governing the global evolution. Then the system is said to evolve according to a special case of quantum Brownian motion — a QBM <a href="https://en.wikipedia.org/wiki/Quantum_dynamical_semigroup" rel="nofollow">quantum dynamical semigroup</a> — when the evolution of its density matrix obeys a <a href="http://en.wikipedia.org/wiki/Lindblad_equation" rel="nofollow">Lindblad master equation</a>
\begin{align}
\partial_t \rho = -i [\hat{H},\rho] + \sum_i \left(V_i \rho V_i^\dagger - \frac{1}{2} \{V_i^\dagger V_i, \rho\} \right),
\end{align}
generated by a time-independent Hamiltonian that is a quadratic polynomial in $x$ and $p$
\begin{align}
\hat{H} = \frac{1}{2m}\hat{p}^2 + \frac{\mu}{2} \{\hat{x},\hat{p}\} + \frac{m\omega^2}{2} \hat{x}^2,
\end{align}
with $\mu$, $m$, and $\omega^2$ real, and by time-independent Lindblad operators that are linear polynomials in the same
\begin{align}
V_i = a_i \hat{p} + b_i \hat{x}, \qquad (i=1,2)
\end{align}
with $a_i$ and $b_i$ complex. The master equation can be re-written as
\begin{align}
\partial_t \rho = -i &[\hat{H},\rho] + i (\lambda/2) [\hat{p},\{\hat{x},\rho\}] - i (\lambda/2) [\hat{x},\{\hat{p},\rho\}] \\
&- D_{pp}[\hat{x},[\hat{x},\rho]] - D_{xx}[\hat{p},[\hat{p},\rho]] + D_{xp}[\hat{p},[\hat{x},\rho]] + D_{px}[\hat{x},[\hat{p},\rho]]
\end{align}
with coefficients
\begin{align}
D_{xx} &= \frac{\vert a_1 \vert^2 + \vert a_2 \vert^2}{2} \quad , \quad & D_{pp} &= \frac{\vert b_1 \vert^2 + \vert b_2 \vert^2}{2},\\
D_{xp} &= D_{px} = -\mathrm{Re} \frac{a_1^* b_1 + a_2^* b_2}{2} \quad , \quad & \lambda &= \mathrm{Im} (a_1^* b_1 + a_2^* b_2),
\end{align}</p>
<p><strong>More generally, we say a system undergoes <em>quantum Brownian motion</em> when it evolves according to the above master equation</strong>, regardless of whether it forms a quantum dynamical semigroup. If it obeys the master equation with time-independent coefficients then the QBM is <em>time-homogeneous</em> (in the sense of a Markov process); otherwise it is <em>time-inhomogeneous</em>. The class of all possible instantaneous QBM dynamics is parameterized by $\mu$, $m$, $\omega^2$, $a_i$, and $b_i$.</p>
<p>The resulting dynamics take a particularly beautiful form in the <a href="https://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution" rel="nofollow">Wigner representation</a>. The above master equation for $\rho$ is equivalent to the following dynamical equation for the Wigner function $W(x,p)$:
\begin{align}
\partial_t W = -\frac{p}{m}\partial_x W + m\omega^2 & x \partial_p W + (\lambda - \mu)\partial_x (x W) + (\lambda + \mu)\partial_p (p W)\\
&+D_{pp} \partial^2_x W + D_{xx} \partial^2_p + (D_{xp}+D_{px}) \partial_x \partial_p W.
\end{align}
More compactly:
\begin{align}
\partial_t W (\alpha) &= \left[ F_{ab} \partial_a \alpha_b + D_{ab} \partial_a \partial_b \right] W(\alpha)
\end{align}
where
\begin{align}
F_{ab} = \left( \begin{array}{cc} \lambda - \mu & -1/m \\ m \omega^2 & \lambda+\mu \end{array} \right) \quad, \quad D_{ab} = \left( \begin{array}{cc} D_{xx} & D_{xp} \\ D_{px} & D_{pp} \end{array} \right)
\end{align}
are matrices with real elements. Above, the phase-space indices $a,b$ take the values $x,p$, with Einstein summation assumed, so that $\alpha_a$ is a vector in phase space. (The directional derivative $\partial_a$ is just shorthand for $\partial_{\alpha_a}$.)</p>
<p>This is <em>identical</em> in form to a Klein-Kramers equation (more generally a <a href="https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation" rel="nofollow">Fokker-Planck</a>-type equation) for the phase-space probability distribution of a <em>classical</em> point particle undergoing Brownian motion. </p>
<p>This is remarkable because such equations were originally derived for true probability distribution, but they also apply to the Wigner function. As a bonus, this gives us an immediate and simple physical interpretation for each of the terms in the QBM master equation</p>
<p>The best comprehensive modern statement of the above definition is probably</p>
<ul>
<li>Isar et al. "<a href="http://www.worldscientific.com/doi/abs/10.1142/S0218301394000164" rel="nofollow">Open Quantum Systems</a>"</li>
</ul>
<p>which includes comparisions to important special cases discussed by other authors. Here are some more references I found useful in compiling the above: </p>
<ul>
<li>Alicki and Lendi, <a href="http://arxiv.org/abs/quant-ph/0205188" rel="nofollow">arXiv:quant-ph/0205188</a>.</li>
<li>Alicki, <a href="http://rads.stackoverflow.com/amzn/click/354070860X" rel="nofollow">Quantum Dynamical Semigroups and Applications</a>.</li>
<li>Lindblad, G. (1976). "<a href="http://link.springer.com/article/10.1007%2FBF01608499" rel="nofollow">On the generators of quantum dynamical semigroups</a>". Commun. Math. Phys. 48 (2) 119.</li>
<li>Breuer and Petruccione, <a href="http://rads.stackoverflow.com/amzn/click/0198520638" rel="nofollow">The Theory Open Quantum Systems</a>.</li>
<li>C. Caves, <a href="http://info.phys.unm.edu/~caves/reports/lindblad.pdf" rel="nofollow">Completely positive maps, positive maps, and the Lindblad form</a>.</li>
<li>G. Lindblad, "<a href="http://www.sciencedirect.com/science/article/pii/003448777690029X" rel="nofollow">Brownian motion of a quantum harmonic oscillator</a>".</li>
<li>H. Dekker, "<a href="http://journals.aps.org/pra/abstract/10.1103/PhysRevA.16.2126" rel="nofollow">Quantization of the linearly damped harmonic oscillator</a>"</li>
<li>A. Sandulescu and H. Scutaru, "<a href="http://www.sciencedirect.com/science/article/pii/000349168790162X" rel="nofollow">Open quantum systems and the damping of collective modes in deep inelastic collisions</a>".</li>
<li>A.O. Caldeira and A.J. Leggett, "<a href="http://www.sciencedirect.com/science/article/pii/0378437183900134" rel="nofollow">Path integral approach to quantum Brownian motion</a>".</li>
<li>B. Hu, J. Paz, and Y. Zhang, "<a href="http://journals.aps.org/prd/abstract/10.1103/PhysRevD.45.2843" rel="nofollow">Quantum Brownian motion in a general environment: Exact master equation with nonlocal dissipation and colored noise</a>".</li>
<li>J. Halliwell and T. Yu, "<a href="http://journals.aps.org/prd/abstract/10.1103/PhysRevD.53.2012" rel="nofollow">Alternative derivation of the Hu-Paz-Zhang master equation of quantum Brownian motion</a>".</li>
<li>W Zurek, "<a href="http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.75.715" rel="nofollow">Decoherence, einselection, and the quantum origins of the classical</a>".</li>
</ul>
|
4,130,809 | <p>I have questions regarding the proof that I made about the following statement: "Let <span class="math-container">$(X,\tau_{X})$</span> be a topological space and <span class="math-container">$\lbrace \infty\rbrace$</span> an object that doesn't belong to X. Define <span class="math-container">$Y=X\cup\lbrace\infty\rbrace$</span> and
<span class="math-container">\begin{equation}
\tau_{\infty}=\lbrace U\subset Y|U\in \tau_X\text{ or }Y-U\text{ is compact and closed in } X\rbrace
\end{equation}</span>
which is a topology on <span class="math-container">$Y$</span>. Show that <span class="math-container">$(Y,\tau_{\infty})$</span> is compact.</p>
<p>Ok, here is my attempt of the proof: Let <span class="math-container">$C=\lbrace C_\alpha \rbrace_{\alpha \in \Lambda}$</span> be an arbitrary open cover of <span class="math-container">$Y$</span>. Since <span class="math-container">$C_\alpha \in \tau_\infty$</span> for every <span class="math-container">$\alpha$</span>, then <span class="math-container">$Y-C_\alpha$</span> is compact and closed in <span class="math-container">$X$</span> for every <span class="math-container">$\alpha$</span>. Since <span class="math-container">$Y-C_\alpha$</span> is compact, let <span class="math-container">$C_x=\lbrace C_{\alpha_x}\rbrace_{\alpha_x\in \Lambda_x}$</span> be any open cover of <span class="math-container">$Y-C_\alpha$</span>. Then there is a finite subcover <span class="math-container">$C_x^{\prime}=\lbrace C_{\alpha_{x},i}\rbrace_{i=1}^n$</span>, with
<span class="math-container">\begin{equation*}
Y-C_\alpha \subset \cup_{i=1}^nC_{\alpha_{x},i}.
\end{equation*}</span></p>
<p>Now, select <span class="math-container">$\alpha_\infty$</span> such that <span class="math-container">$C_{\alpha_\infty}$</span> in <span class="math-container">$C$</span> contains the object <span class="math-container">$\lbrace \infty \rbrace$</span>. Then,
<span class="math-container">\begin{equation}
Y-C_{\alpha_\infty}\subset \cup_{i=1}^nC_{\alpha_{x},i}
\end{equation}</span>
Then clearly <span class="math-container">$\lbrace C_{\alpha_{x},i}\rbrace_{i=1}^n \cup \lbrace C_{\alpha_\infty}\rbrace$</span> is a finite subcover of <span class="math-container">$Y$</span>. Hence, <span class="math-container">$(Y,\tau_\infty)$</span> is compact.</p>
<p>I just want to know if my proof is correct. I would appreciate any corrections or other suggestions in case if it is incorrect. Thank you!</p>
| SolubleFish | 918,393 | <blockquote>
<p>Since <span class="math-container">$C_\alpha \in \tau_\infty$</span> for every <span class="math-container">$α$</span>, then <span class="math-container">$Y−C_\alpha$</span> is compact and closed in <span class="math-container">$X$</span> for every <span class="math-container">$\alpha$</span>.</p>
</blockquote>
<p>This is not true : if <span class="math-container">$C_\alpha \in \tau_X$</span>, you don't yet know that the complement is compact.</p>
<p>Fortunately, since <span class="math-container">$\infty \in C_{\alpha_\infty}$</span>, you have <span class="math-container">$C_{\alpha_\infty} \notin \tau_X$</span> and therefore its complement is closed and compact in <span class="math-container">$X$</span>. The rest of your proof works from this point on.</p>
|
774,209 | <p>I got stuck to find a fair formula to calculate the average ranking of the items that I found after consecutive searches, look:</p>
<p><img src="https://i.stack.imgur.com/5LazY.jpg" alt="enter image description here" /></p>
<p>If I calculate the simple average of the item2 for example I get 1,33 as a result, not even nearly close to an "average" ranking :P</p>
<p>any ideas?</p>
| Kathystats | 700,749 | <p>The arithmetic mean is a parametric descriptive statistic that presumes normality in the data. Ordinal data (rank-ordered data) does not meet the assumptions of normality, and you cannot use an arithmetic mean to describe the central tendency—only the median.</p>
|
2,348,131 | <p>In our class, we encountered a problem that is something like this: "A ball is thrown vertically upward with ...". Since the motion of the object is rectilinear and is a free fall, we all convene with the idea that the acceleration $a(t)$ is 32 feet per second square. However, we are confused about the sign of $a(t)$ as if it positive or negative. </p>
<p>Now, various references stated that if we let the upward direction to be positive then $a$ is negative and if we let downward to be the positive direction, then $a$ is positive. The problem in their claim is that they did not explain well how they arrived with that conclusion. </p>
<p>My question now is that, why is the acceleration $a$ negative if we choose the upward direction to be positive. Note: I need a simple but comprehensive answer. Thanks in advance. </p>
| crankk | 202,579 | <p>You can model this on a linear one dimensional space, the "heigth" $x$ of the particle. Different forces are acting on the ball, on the one hand the gravitation, which is directed to the "ground", on the other hand in the beginning a force is applied towards the "top" as the ball is thrown in that direction. You can derive the equations of motion through Newtons formular $F(t)=m\ddot{x}(t)$ and the superposition of forces, the resulting force is the sum of each force.</p>
|
2,348,131 | <p>In our class, we encountered a problem that is something like this: "A ball is thrown vertically upward with ...". Since the motion of the object is rectilinear and is a free fall, we all convene with the idea that the acceleration $a(t)$ is 32 feet per second square. However, we are confused about the sign of $a(t)$ as if it positive or negative. </p>
<p>Now, various references stated that if we let the upward direction to be positive then $a$ is negative and if we let downward to be the positive direction, then $a$ is positive. The problem in their claim is that they did not explain well how they arrived with that conclusion. </p>
<p>My question now is that, why is the acceleration $a$ negative if we choose the upward direction to be positive. Note: I need a simple but comprehensive answer. Thanks in advance. </p>
| Crazy | 449,016 | <p>The acceleration due to gravity is always downward. The convention is so. </p>
<p>Case I:<strong>Downward is the positive direction</strong>.</p>
<p>Let's us examine your case where the ball is falling vertically from the sky. Normally, we take the direction of initial motion of the ball to be positive. This happens in physics. Since the ball is accelerating downward due to gravity and it occurs in the direction of the initial motion then the acceleration is positive since we have choose downward as the positive direction. If there is a air resistance force acts in the ball. We know that air resistance occurs in the other direction of the object. So the air resistance must be negative!</p>
<p>The differential equation for an object undergoes acceleration can be modeled as follows.(We assume that mass is a constant variable)</p>
<p>$$m\frac{dv}{dt}=F_R$$</p>
<p>Where $F_R$ is a resultant force.</p>
<p>$$m\frac{dv}{dt}=F+kv$$</p>
<p>In our case</p>
<p>$F$ will be the $mg$
$k$ is the proportionality of the force</p>
<p>$kv$ or sometimes we use $kv^2$ instead to model the resistance acting on the object. We know that the resistance acts on the object opposes the direction of the object. Then we attribute it as</p>
<p>$$m\frac{dv}{dt}=F-kv$$</p>
<p>Case II <strong>Upward as the positive direction</strong>!</p>
<p>Again consider the object is falling downwards.We model our differential equation as. The weight of the object is acting downwards</p>
<p>$$m\frac{dv}{dt}=-F+kv$$</p>
<p>What we say about negative and positive of acceleration. Since acceleration are vectors. Vectors make up of magnitude and direction!</p>
<p>Can you see why case II is negative compared to case I.</p>
<p>I always think of $$ma=Upward+downward$$</p>
|
3,612,016 | <p>Consider <span class="math-container">$(\Bbb R\setminus\{-1\},*)$</span>, where
<span class="math-container">$$a*b:=ab+a+b \qquad a,b \in \Bbb R\setminus\{-1\}.$$</span>
We have to prove that it's an Abelian group. While it's easy to show how the properties of associativity, identity element, inverse element and commutativity hold, how do we prove the <strong>closure</strong>?</p>
| Ivo Terek | 118,056 | <p>You have to show that <span class="math-container">$ab+a+b$</span> is not equal to <span class="math-container">$-1$</span>, if <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are both not equal to <span class="math-container">$-1$</span> to begin with. This can be shown by assuming that <span class="math-container">$ab+a+b=-1$</span> and showing that necessarily <span class="math-container">$a$</span> or <span class="math-container">$b$</span> equals <span class="math-container">$-1$</span>. </p>
<p>Reorganize things as <span class="math-container">$a(b+1)+b=-1$</span>, which leads to <span class="math-container">$a(b+1)+b+1=0$</span>. Factor this to <span class="math-container">$(a+1)(b+1)=0$</span>. Conclude.</p>
|
3,612,016 | <p>Consider <span class="math-container">$(\Bbb R\setminus\{-1\},*)$</span>, where
<span class="math-container">$$a*b:=ab+a+b \qquad a,b \in \Bbb R\setminus\{-1\}.$$</span>
We have to prove that it's an Abelian group. While it's easy to show how the properties of associativity, identity element, inverse element and commutativity hold, how do we prove the <strong>closure</strong>?</p>
| Akash Yadav | 474,986 | <p>Hint : </p>
<p>If <span class="math-container">$ab+a+b=-1$</span>, then either <span class="math-container">$a=-1$</span> or <span class="math-container">$b=-1$</span>.</p>
|
1,076,292 | <p>I wish to use two points say $(x_1$,$y_1)$ and $(x_2$,$y_2)$ and obtain the coefficients of the line in the following form: $$ Ax + By + C = 0$$</p>
<p>Is there any direct formula to compute.</p>
| DeepSea | 101,504 | <p>$\text{slope} = m = \dfrac{y_2-y_1}{x_2-x_1} \Rightarrow y - y_1 = m(x-x_1) = \dfrac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \Rightarrow (x_2-x_1)y - y_1(x_2-x_1) = (y_2-y_1)x - x_1(y_2-y_1) \Rightarrow -(y_2-y_1)x + (x_2-x_1)y -y_1(x_2-x_1) +x_1(y_2-y_1)=0$. This gives the formula:</p>
<p>$A = y_1-y_2$</p>
<p>$B = x_2-x_1$</p>
<p>$C = x_1y_2-x_2y_1$</p>
|
1,076,292 | <p>I wish to use two points say $(x_1$,$y_1)$ and $(x_2$,$y_2)$ and obtain the coefficients of the line in the following form: $$ Ax + By + C = 0$$</p>
<p>Is there any direct formula to compute.</p>
| Community | -1 | <p>Here are a few ways of finding values of $A$, $B$, and $C$.</p>
<p><strong>Method 1:</strong> A line is the set of points $\vec{x}$ such that
$$\vec{a}\cdot \vec{x_0} = \vec{a} \cdot \vec{x}$$
where $x_0$ is an arbitrary point on the line and $\vec{a}$ is a nonzero vector perpendicular to the line. We know that the line will be parallel to the vector
$$\vec{d} = \langle x_2-x_1, y_2-y_1\rangle,$$
so $\vec{a}$ will be a vector perpendicular to $\vec{d}$. One such vector is
$$\vec{a} = \langle y_1-y_2, x_2 - x_1\rangle$$
Let's suppose that $\vec{x} = \langle x, y\rangle$ is a point on the line. Then, $\vec{x}$ satisfies
$$\vec{a} \cdot \vec{x_0} = \vec{a} \cdot \vec{x}$$
$$\langle y_1-y_2, x_2 - x_1\rangle \cdot \langle x_1, y_1\rangle = \langle y_1-y_2, x_2 - x_1\rangle \cdot \langle x, y \rangle$$
$$y_2x_1 - x_2 y_1 = (y_1 - y_2)x + (x_2 - x_1)y$$
$$(y_1 - y_2)x + (x_2 - x_1) y + x_2y_1 - y_2x_1 = 0$$
So
$$A = y_1 - y_2$$
$$B = x_2 - x_1$$
$$C = x_2y_1 - y_2x_1$$</p>
<p><strong>Method 2:</strong> Just plug your two points into the equation $Ax + By + C = 0$:
$$Ax_1 + By_1 + C = 0 \tag{1}$$
$$Ax_2 + By_2 + C = 0 \tag{2}$$
Subtracting (1) from (2) gives us
$$A(x_2 - x_1) + B(y_2 - y_1) = 0$$
Of course, $A = B = 0$ is a solution, but it will not give us a line. Another solution is $B = (x_2 - x_1)$, $A = (y_1 - y_2)$. Since $(x_1,y_1)$ and $(x_2, y_2)$ were distinct points, $A$ and $B$ are not both zero. Now let us solve for $C$ using (1):
$$(y_1-y_2)x_1+(x_2-x_1)y_1 + C = 0$$
$$y_1x_1 - y_2x_1 + x_2y_1 - x_1y_1 + C = 0$$
$$C = y_2x_1 - x_2y_1$$</p>
<hr>
<p>It's worth pointing out that there is no <em>unique</em> solution to your problem. If $$Ax + By + C = 0$$
is one solution, then
$$\lambda Ax + \lambda By + \lambda C = 0 \qquad \lambda \not= 0$$
is another way of expressing the same line. </p>
|
3,738,508 | <p>If <span class="math-container">$G$</span> is order <span class="math-container">$p^2q$</span>, where <span class="math-container">$p$</span>, <span class="math-container">$q$</span> are primes, prove that either a Sylow <span class="math-container">$p$</span>-subgroup or a Sylow <span class="math-container">$q$</span>-subgroup must be normal in <span class="math-container">$G$</span>.</p>
| quasi | 400,434 | <p>Assume <span class="math-container">$x^4+2$</span> factors in <span class="math-container">$\mathbb{Z}_5[x]$</span> as
<span class="math-container">$$x^4+2=(x^2+ax+b)(x^2+cx+d)$$</span>
If the <span class="math-container">$\text{RHS}$</span> is expanded,</p>
<ul>
<li>The coefficient of the <span class="math-container">$x^3$</span> term is <span class="math-container">$a+c$</span>, hence we must have <span class="math-container">$c=-a$</span>.<span class="math-container">$\\[4pt]$</span>
<li>The constant term is <span class="math-container">$bd$</span>, hence we must have <span class="math-container">$d={\large{\frac{2}{b}}}$</span>.
</ul>
<p>hence the factorization can be rewritten as
<span class="math-container">$$x^4+2=(x^2+ax+b)(x^2-ax+\frac{2}{b})$$</span>
Expanding, we get
<span class="math-container">$$x^4+2=x^4+\left(b+\frac{2}{b}-a^2\right)x^2+\left(\frac{2a}{b}-ab\right)x+2$$</span>
hence we must have
<span class="math-container">$$
\left\lbrace
\begin{align*}
&\frac{2a}{b}-ab=0&&(\text{eq}1)\\[4pt]
&b+\frac{2}{b}-a^2=0&&(\text{eq}2)\\[4pt]
\end{align*}
\right.
$$</span>
If <span class="math-container">$a\ne 0$</span>, <span class="math-container">$(\text{eq}1)$</span> yields <span class="math-container">$b^2=2$</span>, and if <span class="math-container">$a=0$</span>, <span class="math-container">$(\text{eq}2)$</span> yields <span class="math-container">$b^2=-2$</span>.</p>
<p>
Either way we have a contradiction since <span class="math-container">$2$</span> and <span class="math-container">$-2$</span> are not squares in <span class="math-container">$\mathbb{Z}_5$</span>.
|
1,301,476 | <p>(Cross-posted in <a href="https://matheducators.stackexchange.com/q/8173/"><strong>MESE 8173</strong></a>.) </p>
<p>I want to start to do mathematical Olympiad type questions but have absolutely no knowledge on how to solve these apart from my school curriculum. I'm $16$ but know maths up to the $18$ year old level. I think I will start learning the theory of the topics (Elementary Number Theory, Combinatorics, Euclidean Plane Geometry) then going on to trying the questions, but I need help in knowing what books to use to learn the theory. I have seen several list (such as <a href="http://www.artofproblemsolving.com/wiki/index.php/Olympiad_books" rel="nofollow noreferrer">http://www.artofproblemsolving.com/wiki/index.php/Olympiad_books</a>) but does anyone know which ones are the best for my level of knowledge.</p>
<p>P.S. I live in UK if that matters.</p>
| Keith | 244,951 | <p>I understand your reluctance to focus solely on contest books. Often some of the "tricks" used are really applications of some more general theory in disguise. Greater clarity is sometimes brought by studying math at a higher level or at least in a more orderly way (from books where the exposition is directed at a greater purpose).</p>
<p>The books in <a href="http://www.maa.org/publications/ebooks/anneli-lax-new-mathematical-library" rel="noreferrer">this series</a> tend to be good and are aimed at a high school audience. I particularly like the two by Niven.</p>
<p>You could also study some number theory (for example from the book by Stark). That tends to be closely related to a lot of contest material.</p>
<p>In parallel, you shouldn't hesitate to study math at university level now - particularly rigorous calculus (such as in Spivak's book) and algebra (e.g. from Artin's book). When I was in high school, my contest-type ability went up a lot in the space of a year when I started studying analysis and abstract algebra on my own. </p>
<p>Plane geometry needs to be studied in its own right for contests, however. One doesn't get that knowledge from typical university-level reading. <em>Geometry Revisited</em>, which appears in the series I linked to, is excellent in that respect. So too are Yaglom's books.</p>
|
2,280,133 | <p><em>I need help to understand, some steps of the proof of this theorem.</em> </p>
<p><strong>(Kolmogorov-M. Riesz-Fréchet)</strong> Let $\mathcal{F}$ be a bounded set in $L^p(\mathbb{R}^N)$ with $1\leq p < \infty$. Assume that </p>
<p>\begin{equation}
\lim\limits_{|h|\longrightarrow 0 }\|\tau_hf-f\|_p=0 \ uniformly
\; in \, f \in \mathcal{F},
\end{equation}</p>
<p>i.e
$\forall \varepsilon >0 \; \exists \delta >0$ such that $\|\tau_hf-f\|_p<\varepsilon \; \forall f \in \mathcal{F}, \; \forall h \in \mathbb{R}^N$ with $|h|<\delta$.
Then the closure of $\mathcal{F}_{|\Omega}$ is compact for any measurable set in $\Omega \subset \mathbb{R}^N$ with finite measure. </p>
<p><em>Well, you can find it in Haim Brezis, Functional Analysis, Sobolev Spaces and PDE. page 111</em></p>
<p>In step 1: We claim that
\begin{equation}
\|(\rho_n*f)-f\|_{L^p(\mathbb{R}^N)}\leq \varepsilon \ \forall f \in \mathcal{F}, \ \forall n > 1/n. \label{4.26_kolmogorov_23}
\end{equation}</p>
<p>And we have,
\begin{equation*}
\begin{split}
|(\rho_n*f)(x)-f(x)| &\leq \int |(f(x-y)-f(x))\rho_n(y)dy|\\
&\leq \left[ \int |(f(x-y)-f(x))|^p\rho_n(y)dy \right]^{1/p}
\end{split}
\end{equation*}
By Hölder´s inequality </p>
<p><em>So, i can´t understand, how use the Hölder´s inequality in the last inequality, please help me!!</em></p>
| Felix B. | 445,105 | <p>If $\rho_n$ is a probability density function and $\mu_n(B)=\int_B\rho_n(x)dx$, then
\begin{align}\int |f(x-y)-f(x)|\rho_n(y)dy&=\int|f(x-y)-f(x)|d\mu_n\\
&\le \left(\int|f(x-y)-f(x)|^{p}d\mu_n\right)^{1/p}\left(\int |1|^{q}d\mu_n\right)^{1/q}\\
&=\left(\int|f(x-y)-f(x)|^{p}\rho_n(y)dy\right)^{1/p}
\end{align}</p>
<p>I know that you didn't mention that $\rho_n$ was a density but given that $\rho_n$ is not written as to the power of p after Hölder and the q seems to be missing this is my best guess of what happened.</p>
|
2,433,174 | <p>I'm struggling with the following (<em>is it true?</em>):</p>
<blockquote>
<p>Let <span class="math-container">$X$</span> be a set and denote <span class="math-container">$\aleph(X)$</span> the <em><strong>cardinality</strong></em> of <span class="math-container">$X$</span>. Suppose that <span class="math-container">$\aleph(X)\geq \aleph_0$</span>, the cardinality <span class="math-container">$\aleph_0:=\aleph(\Bbb N)$</span>, with <span class="math-container">$\Bbb N$</span> the set of natural numbers.</p>
<p>Consider two subsets <span class="math-container">$A_1,A_2\subset X$</span> with <span class="math-container">$\aleph(A_i)<\aleph(X)$</span>, <span class="math-container">$i=1,2$</span>. Show that <span class="math-container">$\aleph(\bigcup_1^2 A_i)<\aleph(X)$</span>.</p>
</blockquote>
<p>From this, it would follow by induction that <span class="math-container">$\aleph(\bigcup_1^n A_i)<\aleph(X)$</span>, for any sets <span class="math-container">$A_1,\dots, A_n\subset X$</span> with <span class="math-container">$\aleph(A_i)<\aleph(X)$</span>, <span class="math-container">$i=1,\dots,n$</span>.</p>
<p>With this result, I would be able to solve <em>Dugundji's Exercise 1-b (Chapter III - Topological Spaces)</em>, which asserts:</p>
<blockquote>
<p>1.b) If <span class="math-container">$\aleph(X)\geq \aleph_0$</span>, then <span class="math-container">$\scr{A}_1$$=\{\emptyset\}\cup \{A\,|\, \aleph(X-A)<\aleph(X)\}$</span> is a topology on <span class="math-container">$X$</span>.</p>
</blockquote>
<p>The fact that <span class="math-container">$\emptyset$</span> and <span class="math-container">$X$</span> are in <span class="math-container">$\scr A_1$</span> is easy. Arbitrary unions of elements of <span class="math-container">$\scr A_1$</span> is in <span class="math-container">$\scr A_1$</span>, since the complement in <span class="math-container">$X$</span> of such an union is an <em><strong>intersection</strong></em> of complements, which have, each of of them, cardinality strictly smaller than <span class="math-container">$\aleph(X)$</span> (intersections decrease cardinality). But to show that finite intersections of elements of <span class="math-container">$\scr A_1$</span> are in <span class="math-container">$\scr A_1$</span>, I need the fact above...</p>
<p>It is easy to see that it is true if <span class="math-container">$\aleph(X)=\aleph_0$</span>, since in this case <span class="math-container">$A_1$</span> and <span class="math-container">$A_2$</span> must be finite. The problem arises if <span class="math-container">$\aleph_0\leq \aleph(A_i)<\aleph(X)$</span>...</p>
| Peter Szilas | 408,605 | <p>$T:\mathbb{R^2} \rightarrow \mathbb{R^2}$.</p>
<p>$T(x,y) \mapsto (x',y')$,</p>
<p>$x'=4x-y$; $y'=3x-2y.$</p>
<p>Solving for $x,y$ in terms of $x',y'$.</p>
<p>$2x'-y'=5x$; and $ 3x' -4y' =5y$.</p>
<p>The line $x+2y = 6$ is transformed:</p>
<p>$5x +10y =30 \rightarrow$</p>
<p>$(2x'-y' ) + 2 (3x'-4y') =30.$</p>
<p>$8x' - 9y' = 30.$</p>
<p>Slope of the image line: $m' = 8/9.$</p>
|
2,604,825 | <p>So I have a problem (two problems, actually) that a friend helped me out with, I'm able to work out the components of this problem but get lost when I have to bring it all together... so what I have is this</p>
<p>$f(t)=t^{2}e^{-2t}+e^{-t}\cos(3t)+5$</p>
<p>Simple enough. I got:</p>
<p>$$\mathcal{L}[t^2]=\frac{2}{s^2}$$</p>
<p>$$\mathcal{L}[e^{-2t}]=\frac{1}{s+2}$$</p>
<p>$$\mathcal{L}[e^{-t}]=\frac{1}{s+1}$$</p>
<p>$$\mathcal{L}[\cos(3t)]=\frac{s}{s^2+9}$$</p>
<p>$$\mathcal{L}[5]=\frac{5}{s}$$</p>
<p>Now the problem is bringing it all together... Apparently,</p>
<p>$$\mathcal{L}[t^2e^{-2t}]=\frac{2}{(s+2)^2}$$</p>
<p>And</p>
<p>$$\mathcal{L}[e^{-t}\cos(3t)]=\frac{s}{(s+1)^2+9}$$</p>
<p>and I just don't seem to get why...? how do you bring those guys together like that? whats the reason behind it?</p>
<p>the other problem is much simpler and I'm sure I have the correct answer... it is:</p>
<p>$$t^3-5\cos(5t)$$</p>
<p>I get:</p>
<p>$$\mathcal{L}[t^3]=\frac{6}{s^4}$$</p>
<p>And</p>
<p>$$\mathcal{L}[5\cos(5t)]=\frac{s}{s^2+25}$$</p>
<p>which together is supposedly,</p>
<p>$$\mathcal{L}[t^3-5\cos(5t)]=\frac{6}{s^4}-\frac{5s}{s^2-25}$$</p>
<p>Is it correct? Any help at all is much appreciated. Thank you.</p>
| Olivier Oloa | 118,798 | <p>One may use the <a href="https://math.stackexchange.com/questions/980551/how-to-show-sin-x-geq-frac2x-pi-x-in-0-frac-pi2">standard identity</a>
$$
\sin x\ge \frac{2x}{\pi},\qquad x \in \left[0,\frac \pi2 \right],
$$ giving
$$\int_0^{\frac{\pi}{2}}\sqrt{\sin x}\:dx >\int_0^{\frac{\pi}{2}}\sqrt{ \frac{2x}{\pi}}\:dx=\frac{\pi}{3}$$ as wanted.</p>
|
4,214,474 | <p>Consider <span class="math-container">$\mathbb{R}^\omega$</span> (countably infinite product of <span class="math-container">$\mathbb{R}$</span>) with the uniform metric.</p>
<p>Let <span class="math-container">$A$</span> be the set of infinite bounded sequences of <span class="math-container">$\mathbb{R}$</span>, i.e. <span class="math-container">$$A = \{x = (x_1, x_2, \dots ) \mid |x_k| \leq M \ \text{for all $k$ , for some positive number $M$}\}$$</span></p>
<p>Consider the metric <span class="math-container">$d(x,y) = \mathop{\text{sup }}_k \{\text{min}\{|x_k-y_k|, 1\}\}$</span>.</p>
<p>I want to prove <span class="math-container">$A$</span> is closed in the topology generated by <span class="math-container">$d$</span> (uniform topology).</p>
<p>My attempt:</p>
<p>We want to prove <span class="math-container">$A^c$</span> is open. Let <span class="math-container">$x = (x_k) \in A^c$</span>, so <span class="math-container">$x = (x_k)$</span> is a unbounded sequence. Suppose there exist a <span class="math-container">$M$</span> such that <span class="math-container">$|x_k| > M$</span> for infinitely many <span class="math-container">$k$</span>. I want to show there exist some <span class="math-container">$r>0$</span> such that <span class="math-container">$B_d(x,r) \subset A^c$</span>. Take <span class="math-container">$r = \frac{M}{2}$</span>, and <span class="math-container">$B_d(x, \frac{M}{2}) \subset A^c$</span>. Let <span class="math-container">$y = (y_k) \in B_d(x, \frac{M}{2})$</span>.</p>
<p>But I have no idea how to proceed further?</p>
<p>Actually, I want to prove <span class="math-container">$y = \{y_k\}$</span> is unbounded.</p>
<p>Please help me.</p>
| Huy Nguyen | 922,437 | <p>You are almost there. Suppose <span class="math-container">$y=(y_k) \in B_d(x, \frac{M}{2})$</span> is bounded which means that there exsits <span class="math-container">$N > 0$</span> such that <span class="math-container">$|y_k| \leq N$</span> for all <span class="math-container">$k$</span>. But since <span class="math-container">$y \in B_d(x, \frac{M}{2}) $</span>, one has : <span class="math-container">$|x_k| < \frac{M}{2} + |y_k| \leq \frac{M}{2} +N $</span>. This is the contradiction since <span class="math-container">$(x_k)$</span> is assumed to be unbounded.</p>
|
1,238,783 | <p>I am currently in high school where we are learning about present value. </p>
<p>I struggle with task like these: Say you get 6% interest each year, how much interest would that be each month?</p>
| Alberto Debernardi | 140,199 | <p>In order to compute the monthly interest, you know that in a year (which has 12 months) you get $6\%$ interest. Call $a$ the (decimal) monthly interest. Thus,
$$
(1+a)^{12}=1.06.
$$
Then you just solve the equation for $a$.</p>
|
58,947 | <p>Let $X$ be a non-compact holomorphic manifold of dimension $1$. Is there a compact Riemann surface $\bar{X}$ suc that $X$ is biholomorphic to an open subset of $\bar{X}$ ?</p>
<p><strong>Edit:</strong> To rule out the case where $X$ has infinite genus, perhaps one could add the hypothesis that the topological space $X^{\mathrm{end}}$ (is it a topological surface?), obtained by adding the <em>ends</em> of $X$, has finitely generated $\pi_1$ (or $H_1$ ). Would the new question make sense and/or be of any interest?</p>
<p><strong>Edit2:</strong> What happens if we require that $X$ has finite genus? (the <em>genus</em> of a non-compact surface, as suggested in a comment below, can be defined as the maximal $g$ for which a compact Riemann surface $\Sigma_g$ minus one point embeds into $X$)</p>
| Sylvain Bonnot | 15,673 | <p>You should probably check the following article:
Migliorini, Luca, "On the compactification of Riemann surfaces".
Here is the Mathscinet review about it:
"In this paper the author studies some questions concerning the compactifications of Riemann
surfaces. It is proved that if $X$ is an open connected Riemann surface then X has finite genus if and only if there exists a holomorphic injection $i: X \hookrightarrow \tilde{X}$
(with $\tilde{X}$ a compact Riemann surface), $i(X)$ being dense in $\tilde{X}$..."</p>
|
624,715 | <p>Find this follow ODe solution
$$y''-y'+y=x^2e^x\cos{x}$$</p>
<p>I konw solve this follow three case
$$y''-y'+y=x^2\cos{x}$$
$$y''-y'+y=e^x\cos{x}$$
$$y''-y'+y=x^2e^x$$</p>
<p>But for $f(x)=x^2e^x\cos{x}$, I can't.</p>
<p>Thank you very much!</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
In general, we want to solve
$\ds{{\rm y}''\pars{x} - {\rm y}'\pars{x} + {\rm y}\pars{x} = \fermi\pars{x}}$ where $\fermi$ is a given function. Let's assume, for simplicity, we know the 'boundary conditions'
$$
{\rm y_{p}}\pars{0}\quad\mbox{and}\quad{\rm y_{p}}'\pars{0}\,,\qquad
\mbox{( other cases can be handled in a similar way )}\tag{1}
$$
The solution is given by:
$$
{\rm y}\pars{x}
={\rm y_{p}}\pars{x} + \int_{-\infty}^{\infty}{\rm G}\pars{x,t}\fermi\pars{t}\,\dd t
\tag{2}
$$
where ${\rm y_{p}}\pars{x}$ satisfies
$\ds{{\rm y_{p}}''\pars{x} - {\rm y_{p}}'\pars{x} + {\rm y_{p}}\pars{x} = 0}$ and the 'boundary conditions' $\pars{1}$. ${\rm G}\pars{x,t}$ satisfies:
$$
\pars{\totald[2]{}{x} - \totald{}{x} + 1}{\rm G}\pars{x,t} = \delta\pars{x - t}\,,
\qquad
\left\vert%
\begin{array}{rcl}
{\rm G}\pars{0,t} & = & 0
\\[2mm]
\left.\partiald{{\rm G}\pars{x,t}}{x}\right\vert_{x\ =\ 0} & = & 0
\end{array}\right.\tag{3}
$$
The $\ds{{\rm G}\pars{x,t}}$ differential equation in $\pars{2}$ is equivalent to:
$$\left\lbrace%
\begin{array}{rcl}
\pars{\totald[2]{}{x} - \totald{}{x} + 1}{\rm G}\pars{x,t} = 0
& \mbox{if} & x \not= t
\\[2mm]
\left.\partiald{{\rm G}\pars{x,t}}{x}\right\vert_{x\ =\ t^{-}}^{x\ =\ t^{+}}
= 1&&
\end{array}\right.\tag{4}
$$
Solutions of $\ds{{\rm y}''\pars{x} - {\rm y}'\pars{x} + {\rm y}\pars{x} = 0}$ has the form $\ds{\expo{x/2}\bracks{A\sin\pars{kx} + B\cos\pars{kx}}}$ with
$\ds{k = \root{3}/2}$. Then
$$
{\rm G}\pars{x,t}
=
\left\lbrace%
\begin{array}{lcl}
\expo{x/2}\bracks{A\sin\pars{kx} + B\cos\pars{kx}} & \mbox{if} & x < t
\\
\expo{x/2}\bracks{C\sin\pars{kx} + D\cos\pars{kx}} & \mbox{if} & x > t
\end{array}\right.
$$
The ${\rm G}\pars{x,t}$ 'boundary conditions' in $\pars{3}$ leads to $A = B = 0$.
${\rm G}\pars{t^{-},t} = {\rm G}\pars{t^{+},t}$ and the 'boundary condition' in $\pars{4}$ lead to:
\begin{align}
&\expo{t/2}\bracks{C\sin\pars{kt} + D\cos\pars{kt}} = 0
\\[3mm]&
\half\,\expo{t/2}\bracks{C\sin\pars{kt} + D\cos\pars{kt}}
+
k\expo{t/2}\bracks{C\cos\pars{kt} - D\sin\pars{kt}} = 1
\end{align}
which are equivalent to:
$$
\left.
\begin{array}{rcrcl}
\sin\pars{kt}C & + & \cos\pars{kt}D & = & 0
\\
\cos\pars{kt}C & - & \sin\pars{kt}D & = & {\expo{-t/2} \over k}
\end{array}\right\rbrace\
\imp\
\left\lbrace%
\begin{array}{rcl}
C & = & \phantom{-}{\expo{-t/2}\cos\pars{kt} \over k}
\\[2mm]
D & = & -\,{\expo{-t/2}\sin\pars{kt} \over k}
\end{array}\right.
$$
and
$$
{\rm G}\pars{x,t}
=\left\lbrace%
\begin{array}{lcl}
0 & \mbox{if} & x < t
\\[2mm]
{\expo{\pars{x - t}/2}\sin\pars{k\bracks{x - t}} \over k} & \mbox{if} & x > t
\end{array}\right.
$$
$\ds{{\rm y}\pars{x}
=
{\rm y_{p}}\pars{x} + {\expo{x/2} \over k}
\int_{-\infty}^{x}\expo{-t/2}\sin\pars{k\bracks{x - t}}\fermi\pars{t}\,\dd t
=
{\rm y_{p}}\pars{x} -
{1 \over k}\int_{-\infty}^{0}\expo{-t/2}\sin\pars{kt}\fermi\pars{t + x}\,\dd t}$
$$\color{#0000ff}{\large%
{\rm y}\pars{x}
=
{\rm y_{p}}\pars{x} + {1 \over k}
\int_{0}^{\infty}\expo{t/2}\sin\pars{kt}\fermi\pars{x - t}\,\dd t\,,
\qquad k = {\root{3} \over 2}}
$$
$\tt\mbox{Just plug your function}\ \fermi\pars{x - t}\ \mbox{in this expression !!!}$</p>
|
142,127 | <p>A simplex is regular if its all edges have the same length.</p>
<p>How to test in Mathematica whether a <code>Simplex</code> is regular or not, without checking all the edges manually? I'm not really familiar with loops in Mathematica. I also can't find in the documentation how to access the vertices of a <code>Simplex</code>.</p>
| kglr | 125 | <pre><code>ClearAll[regSimplexQ]
regSimplexQ = Equal @@ PropertyValue[{MeshRegion[#, Simplex[{1, 2, 3, 4}]] & @@ #, 1},
MeshCellMeasure] &;
regSimplexQ@Simplex[{{0, 1, 0}, {1, 0, 0}, {0, 0, 1}, {1, 1, 1}}]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>regSimplexQ@Simplex[{{0, 0, 1}, {1, 0, 0}, {1, 0, 1}, {1, 1, 1}}]
</code></pre>
<blockquote>
<p>False</p>
</blockquote>
|
1,573,262 | <p>What's an example of a convergent, yet unbounded sequence?</p>
<p>I'm having trouble of thinking how to do this. I want to use a piecewise function, but I feel like there might be something easier than this.</p>
| Adhvaitha | 228,265 | <p>Consider the sequence of functions on the interval $(0,1)$ given by $f_n(x) = nx^n$. Note that $f_n(x)$ converges to $0$ point wise on $(0,1)$. However, $f_n(x)$ is unbounded on $(0,1)$.</p>
|
1,573,262 | <p>What's an example of a convergent, yet unbounded sequence?</p>
<p>I'm having trouble of thinking how to do this. I want to use a piecewise function, but I feel like there might be something easier than this.</p>
| MC989 | 292,147 | <p>Rise is correct. I can't believe I didn't see this sooner. all convergent sequences are bounded. I was confusing this with convergent functions, which are different. (i.e. $f(x)= \frac {1}{x}$)</p>
|
3,704,633 | <p>Evaluate: <span class="math-container">$$\lim_{n \to \infty} \sqrt[n]{\frac{n!}{\sum_{m=1}^n m^m}}$$</span>
In case it's hard to read, that is the n-th root. I don't know how to evaluate this limit or know what the first step is... I believe that: <span class="math-container">$$\sum_{m=1}^n m^m$$</span> doesn't have a closed form so I suppose there must be some identity or theorem that must be applied to this limit. According to the answer key, the limit evaluates to <span class="math-container">$\frac{1}{e}$</span>.</p>
| Fnacool | 318,321 | <p>Let <span class="math-container">$a_n= (n! / \sum_{m=1}^n m^m)^{1/n}$</span>. </p>
<p>Observe that <span class="math-container">$n^n \le \sum_{m=1}^n m^m \le n n^n$</span>. </p>
<p>Then <span class="math-container">$$(\frac{n!}{n^n})^{1/n} \frac{1}{n^{1/n}} \le a_n \le (\frac{n!}{n^n})^{1/n}.$$</span> </p>
<p>Since <span class="math-container">$n^{1/n}\to 1$</span>, we need to find the limit of <span class="math-container">$(n! / n^n)^{1/n}$</span>. Take the logarithm of this expression to obtain the Riemann sum </p>
<p><span class="math-container">$$ \frac{1}{n} \sum_{j=1}^n \ln (j/n) \to \int_0^1 \ln x dx = -1.$$</span> </p>
<p>Therefore <span class="math-container">$a_n \to e^{-1}$</span>. </p>
|
1,620,686 | <p>Prove that this works for all $x$ and and only some $y$
$$\sqrt{(x-1)^2-(y+2)^2}=0.$$</p>
<p>This is as far as I got so far</p>
<p>Difference of squares:</p>
<p>$\sqrt{(x-1-y-2)(x-1+y+2)}=0$<br>
$\sqrt{x-y-3}\sqrt{x+y+1}=0$</p>
<p>Therefore $x-y-3=0 \implies y=x-3$ </p>
<p>$x+y+1=0$ and $y=-1-x$</p>
<p>I just don't know where to go from here?</p>
| Ross Millikan | 1,827 | <p>What you must mean (and it <em>really</em> helps to write it clearly) is that for all $x$ you can find a $y$ that makes the statement true and you can find another $y$ that makes the statement false. So given an $x$ you can exhibit a $y$ that makes the statement true-just solve for $y$<br>
$$(x-1)^2=(y+2)^2\\y+2=\pm(x-1)\\ y=\begin {cases}x-3\\-1-x \end{cases}$$
Either of these $y$'s will work, which you can show by plugging back in. Now let $y=x-4$, for example, to show that there is at least one $y$ that doesn't work. </p>
<p>Note that if you choose $y$ first, the same thing happens. For all $y$ there are some $x$ that work and some $x$ that do not work.</p>
|
1,285,273 | <p>Looking for hints to find the orthnormal basis for the null space/range of the following matrix</p>
<p>$A = \frac{1}{3}\left( \begin{array}{ccc}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \end{array} \right)$</p>
| BruceET | 221,800 | <p>This is an extended Comment, not an answer. Actually, I'm not
sure there <em>is</em> an answer along the lines you seek. First, although
the proof of the CLT is in terms of MGFs, I'm not sure MGFs
are going to be helpful judging how well sums or averages of
iid observations match the normal distribution. Second, the
$n \ge 30$ rule is often quoted and, in my experience, seldom helpful.</p>
<p>Roughly speaking the CLT often works for surprisingly small $n,$
especially if the distribution of the $X_i$ is (close to) symmetrical
and without 'fat' tails.</p>
<p><strong>Example 1.</strong> Sums of variables $X_i \sim Unif(0,1)$ are very
close to normal for very small $n.$ In the early days of computer
simulation $Z = \sum_1^{12} X_i - 6$ was often used to simulate
standard normal data. Such values have $E(Z) = 0$ and $V(Z) = 1.$
Although they necessarily have support $(-6,6),$ that does not
matter much because standard normal has only minuscule probability
outside that interval. (Nowadays when trig and log functions
are quickly executed, an exact method using such functions is
used to simulate standard normal variates; Google 'Box-Muller transformation', if interested.) </p>
<p>The brief R program below generates 1000 'standard normals' in this way. As expected the average is near 0 and the sample variance is near 1.
A Shapiro-Wilk test of normality for our first couple of runs could
not distinguish the sequence from a normal sample (P-values about .73 and about .12.) </p>
<pre><code> m = 1000; u = runif(12*m)
DTA = matrix(u, nrow=m) # 1000 x 12 matrix
z = rowSums(DTA) - 6
mean(z); var(z)
## 0.01051165
## 1.025977
</code></pre>
<p>However, agreement can be quite bad even for relatively large $n$ if the $X_i$ are skewed and or have heavy tails. Of course, the CLT does not apply to Student t random variables with DF = 1 or 2 (nonexistent variance). And convergence is extremely slow for $T(3)$ because of heavy tails.</p>
<p><strong>Example 2.</strong> If the $X_i$ are exponential, we have an extremely
skewed distribution with a fat right tail. For such variables,
the CLT does not 'work' satisfactorily even for $n = 100.$
Sums of 100 exponentials are gamma with shape parameter 100
and thus skewness 0.2. Sums from the program below (with $X_i \sim Exp(rate=1)$) typically
give histograms that show noticeable skewness and results from
almost all our runs failed the Shapiro-Wilk test for normality.</p>
<pre><code> m = 1000; x = rexp(100*m)
DTA = matrix(x, nrow=m) # 1000 x 100 matrix
z = rowSums(DTA)
</code></pre>
<p><strong>Example 3.</strong> You asked specifically about what the CLT has
to say about Poisson distributions. The sum of $n$ Poisson variates with mean $\lambda$ is Poisson with mean $n\lambda.$ Agreement with a normal distribution
depends on $n\lambda.$ If $n=100$ and $\lambda = .1,$ then agreement will be horrible. If $n = 10$ and $\lambda = 25,$ then agreement
will be pretty good for most applications. Clearly, no rule-of-thumb about $n$ makes
any sense unless you specify $\lambda.$</p>
<p>Specifically, if $X \sim Pois(225),$ then $P(X \le 250) = 0.9535$ and the appropriate normal approximation with continuity correction is $0.9554.$ But if $X \sim Pois(10),$ then $P(7 \le X \le 9) = 0.3278,$ whereas the normal approximation is only $0.3030.$</p>
<p>In practical applications, exact
computations with statistical software have pretty much replaced
CLT approximations.</p>
|
126,901 | <p>How to evaluate this determinant $$\det\begin{bmatrix}
a& b&b &\cdots&b\\ c &d &0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots& 0\\c&0&\cdots&0&d
\end{bmatrix}?$$</p>
<p>I am looking for the different approaches.</p>
| Davide Giraudo | 9,849 | <p>If the dimension of the matrix is $2$, it's only $ad-bc$. If it's $\geq 3$, and $d=0$ the determinant is $0$. If $d\neq 0$, do $C_1\leftarrow C_1-\frac cdC_j$, $2\leq j\leq n$. The first column becomes $\pmatrix{a-(n-1)\frac{bc}d\\\ 0\\\ \vdots\\\ 0}$, and the determinant is $d^{n-1}\left(a-(n-1)\frac{bc}d\right)=d^{n-2}(ad-(n-1)bc)$. </p>
|
2,517,469 | <p>Let $P$ be a projective module and $P=P_1+N$, where $P_1$ is a direct summand of $P$ and $N$ is a submodule. Show that there is $P_2\subseteq N$ such that $P=P_1\oplus P_2$. </p>
<p>I know that there is a submodule $P'$ of $P$ such that $P=P_1\oplus P'$. I wanted to consider the projection from this to $P_1$ and use the definition of being projective. But I would also need a map from $P=P_1+N$ to $P_1$ and I don't know how to get a well defined map there because it is not a direct sum. </p>
| Guest | 90,271 | <p>The condition $P = P_1 + N$ implies that the natural map $N \to P/P_1$ is surjective. As $P$ is projective this means the quotient map $P \to P/P_1$ factors through $N$, so there is a homomorphism $P \to N$ such that the composition $P \to N \to P/P_1$ is the quotient map. Since the quotient map gives an isomorphism $P' \simeq P/P_1$ we can define $P_2 \subseteq N$ to be the image of $P'$ under the map $P \to N$. Now $P_1$ and $P_2$ don't intersect because the quotient map was injective on $P'$ and $P = P_1 + P_2$ because $P'$ surjects onto $P/P_1$. Thus $P = P_1 \oplus P_2$.</p>
|
2,284,451 | <blockquote>
<p><span class="math-container">$A$</span> and <span class="math-container">$B$</span> alternately throw a pair of coin. The player who throws head two times first will win.</p>
<p>A has the first throw. The find chance of winning <span class="math-container">$A$</span> is</p>
</blockquote>
<p>Attempt: Let <span class="math-container">$\displaystyle P(A) = \frac{1}{2}$</span> (Probability of occuring head when <span class="math-container">$A$</span> throw coin) and</p>
<p><span class="math-container">$\displaystyle P(B) = \frac{1}{2}$</span> (Probability of occuring head when <span class="math-container">$A$</span> throw coin)</p>
<p>So chance of winning <span class="math-container">$A$</span> when he throw head 1st time is <span class="math-container">$$\displaystyle P(A)+P(A)P(\bar{B})P(A)+P(\bar{A})P(\bar{B})P(\bar{A})P(\bar{B})P(A)+\cdots \cdots$$</span></p>
<p>could some help me how to go for original question, thanks</p>
| true blue anil | 22,388 | <p>Assuming that each throw <strong>two coins</strong> alternately (the more difficult formulation), one way to solve is to consider the <strong>odds</strong> for the first two rounds.<br>
[ Subsequent rounds of $2$ will only add some common multiplier, <strong>odds</strong> won't change]</p>
<p>$\Bbb P$(A wins on first round) $=\frac14$</p>
<p>$\Bbb P$(A doesn't win on first round and B wins on second round) $=\frac34\cdot\frac14 = \frac3{16}$</p>
<p>Odds in favor of A = $\frac14 : \frac3{16} = 4:3$</p>
<p>Thus $\Bbb P$(A wins) $= \frac4{4+3} = \frac47$ </p>
|
140,358 | <p>Let $X$ and $Y$ be two topological spaces with $C(X) \cong C(Y)$ (where $C(X)$ is the ring of all continuous real valued functions on $X$). I know that we can not conclude that $X$ and $Y$ are homeomorphic. But I wonder how independent $X$ and $Y$ could be ? For example is there any forced relation between their cardinality ?</p>
| Joseph Van Name | 22,277 | <p>Since we are talking about rings of continuous functions, I will only talk about completely regular spaces in this problem. It is well known that for completely regular spaces $X$, $C(X)\simeq C(Y)$ if and only if $\upsilon X\simeq\upsilon Y$ where $\upsilon X$ denotes the Hewitt-realcompactification of a space $X$. Of course, for information on rings of continuous functions, obviously one should consult the book Rings of Continuous Functions by Gillman and Jerison.</p>
<p>If $X$ is a completely regular space, then the Hewitt realcompactification $\upsilon X$ is defined to be the subset of the Stone-Cech compactification where $x\in\upsilon X$ if and only if whenever $f:X\rightarrow[0,1)$ is continuous and $\overline{f}:\beta X\rightarrow[0,1]$ is the unique continuous extension, then $f(x)<1$. </p>
<p>Of course, since $|\upsilon X|\leq 2^{2^{|X|}}$ for all spaces $X$, if $C(X)\simeq C(Y)$, then $|X|\leq 2^{2^{|Y|}}$ and $|Y|\leq 2^{2^{|X|}}$.</p>
<p>Assuming the existence of certain large cardinals, there are discrete spaces with $X$, $|\upsilon X|=2^{2^{|X|}}$. For example, if $\kappa$ is a strongly compact cardinal, and $X$ is a discrete space of cardinality $\kappa$, then $|\upsilon X|=2^{2^{|X|}}$, but
$C(X)\simeq C(\upsilon X)$. This is because the points in $\upsilon X$ are in a one-to-one correspondence with the $\sigma$-complete ultrafilters on $X$ for discrete spaces $X$, and if there is a compact cardinal $\kappa$, then there are many $\sigma$-complete ultrafilters on a set $X$ of cardinality $\kappa$. If $\kappa$ is a measurable cardinal, and $|X|=\kappa$ then $|\upsilon X|\geq 2^{|X|}$ and $C(\upsilon X)\simeq C(X)$. There are probably examples of spaces $X,Y$ of different cardinality that do not involve large cardinals where $C(X)\simeq C(Y)$, but $|X|\neq|Y|$, but in order to have a discrete space as such a counterexample, one needs the existence of measurable cardinals.</p>
<p>$\textbf{Added some time later}$ I will now give an example of a space $X$ with $|X|<|\upsilon X|$ without resorting to large cardinal hypotheses. In particular, we have $C(X)\simeq C(\upsilon X)$, but $|X|<|\upsilon X|$.
Let $\mathbb{N}$ denote the natural numbers. For each unbounded function $f:\mathbb{N}\rightarrow[0,\infty)$, let $\overline{f}:\beta\mathbb{N}\rightarrow[0,\infty]$ be the unique extension of the function $f$. Then there is some point $x_{f}\in\beta\mathbb{N}$ with $\overline{f}(x_{f})=\infty$. Let $X=\mathbb{N}\cup\{x_{f}|f:\mathbb{N}\rightarrow[0,\infty)\,\textrm{is unbounded}\}$. Then $X$ is a space of cardinality continuum. Furthermore, the space $X$ is pseudocompact. If $f:X\rightarrow[0,\infty)$ is continuous and unbounded, then $f|_{\mathbb{N}}$ is unbounded as well, so $f(x_{f|_{\mathbb{N}}})=\infty$, a contradiction. It is well known that a space $Y$ is pseudocompact if and only if $\upsilon Y=\beta Y$. Therefore, we have $\upsilon X=\beta X=\beta\mathbb{N}$. In particular, $|\upsilon X|=2^{2^{\aleph_{0}}}$ while $|X|\leq 2^{\aleph_{0}}$.</p>
|
2,309,721 | <p>The problem is: Prove that $7|x^2+y^2$ only if $7|x$ and $7|y$ for $x,y∈Z$.</p>
<p>I found a theorem in my book that allows to do the following transformation:
if $a|b$ and $a|c$ -> $a|(b+c)$</p>
<p>So, can I prove it like this: $7|x^2+y^2 =>7|x^2, 7|y^2 => 7|x*x, 7|y*y => 7|x, 7|y$ ?</p>
<p>I am not really sure because I have this simple example in my head that even if $6|18$ => $6|14+4$, $6 ∤ 14$ and $6 ∤ 4$.</p>
<p>Any help will be appreciated, thank you!</p>
| Bernard | 202,857 | <p><strong>Hint:</strong></p>
<p>Work in the field $\;\mathbf Z/7\mathbf Z=\{0,\pm1,\pm2,\pm3\}$, compute all squares, then all sums of squares (a <em>Pythagoras' table</em> will be useful) and check $$x^2+y^2\equiv 0\mod7\iff (x\equiv 0)\wedge (y\equiv 0)\mod 7.$$</p>
|
155,429 | <p>Consider a square skew-symmetric $n\times n$ matrix $A$. We know that $\det(A)=\det(A^T)=(-1)^n\det(A)$, so if $n$ is odd, the determinant vanishes.</p>
<p>If $n$ is even, my book claims that the determinant is the square of a polynomial function of the entries, and Wikipedia confirms this. The polynomial in question is called the <a href="http://en.wikipedia.org/wiki/Pfaffian">Pfaffian</a>.</p>
<p>I was wondering if there was an easy (clean, conceptual) way to show that this is the case, without mucking around with the symmetric group. </p>
| mike stone | 252,564 | <p>By continuity, we can assume that $A$ can always be reduced to block diagonal form with blocks
$$
\left(\begin{matrix} 0 &\lambda_i\cr -\lambda_i &0 \end{matrix} \right)
$$
on the diagonal. In this case computing the determinant gives $\prod \lambda^2_i$ and computing the Pfaffian gives $\prod \lambda_i$, so the determinant is the square of the Pfaffian.</p>
|
3,785,967 | <p>Let <span class="math-container">$E(R)_X$</span> denote the expected return of asset <span class="math-container">$X$</span>.</p>
<p>Given a market with only 3 assets; <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and <span class="math-container">$C$</span>, the following three things can happen at the next timepoint:</p>
<p>With probability <span class="math-container">$ p = 0.5$</span>: <span class="math-container">$$E(R)_A = 0.4, E(R)_B = 0.02, E(R)_C = 0.3$$</span></p>
<p>With probability <span class="math-container">$p = 0.4$</span>: <span class="math-container">$$E(R)_A = 0.2, E(R)_B = 0.02, E(R)_C = 0.25$$</span></p>
<p>With probability <span class="math-container">$p = 0.1$</span>: <span class="math-container">$$E(R)_A = 0.1, E(R)_B = 0.02, E(R)_C = 0.15$$</span></p>
<p>Now, I'm trying to find the standard deviation for each asset.</p>
<p>Hence, I've found the expected return for each asset, via the following (for example, asset <span class="math-container">$A$</span>): <span class="math-container">$$E(R)_A = 0.5(0.4) + 0.4(0.2) + 0.1(0.1) = 0.29$$</span></p>
<p>Now, how do I find the standard deviation?</p>
<p>Do I simply do: <span class="math-container">$$ \sigma = \sqrt{\frac{(0.4-0.29)^2 + (0.2-0.29)^2 + (0.1-0.29)^2}{3}}$$</span></p>
<p>or do I need to somehow incorporate the probabilities in to this formula?</p>
| tommik | 791,458 | <blockquote>
<p>or do I need to somehow incorporate the probabilities in to this formula?</p>
</blockquote>
<p>Yes, You do need.</p>
|
162,655 | <p>Does there exist a Ricci flat Riemannian or Lorentzian manifold which is geodesic complete but not flat? And is there any theorm about Ricci-flat but not flat? </p>
<p>I am especially interset in the case of Lorentzian Manifold whose sign signature is (- ,+ ,+ , + ). Of course, the example is not constricted in Lorentzian case.</p>
<p>I know there are many Ricci flat case in General Relativiy which is the vacuum solution to Einstein's equation. But what I know, such as Kerr solution, are all geodesic incomplete. So I want a geodesic complete example and reference. Thanks!</p>
| Dietrich Burde | 32,332 | <p>For an example, let $N$ be a compact complex hypersurface of degree $m+1$ of the complex projective space $\mathbb{CP}^m$ with complex dimension $m\ge 3$ (for $m=3$ this is a complex $K3$ surface). The first Chern class of $N$ vanishes, and hence $N$ admits a Ricci-flat but <em>nonflat</em> Riemannian metric, by a theorem of Yau.</p>
|
73,383 | <p>The problem is:
$$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2}.$$</p>
<p>The tutor guessed it didn't exist, and he was correct. However, I'd like to understand why it doesn't exist.</p>
<p>I think I have to turn it into spherical coordinates and then see if the end result depends on an angle, like I've done for two variables with polar coordinates. I don't know how though.</p>
<p>I know $\rho = \sqrt{x^2+y^2+z^2}$ and $\theta = \arctan \left(\frac{y}{x} \right)$ and $\phi = \arccos \left( \frac{z}{\rho} \right)$, but how on earth do I break this thing up?</p>
| Chris Taylor | 4,873 | <p>Let's take the limit $z\to0$ first, getting</p>
<p>$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2}$$</p>
<p>Now consider what happens if you take the limit along $y=x$:</p>
<p>$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2} = \lim_{x\to0} \frac{x^2}{5x^2} = \frac{1}{5}$$</p>
<p>and along $y=2x$:</p>
<p>$$\lim_{x,y\to 0} \frac{xy}{x^2+4y^2} = \lim_{x\to 0} \frac{2x^2}{17x^2} = \frac{2}{17}$$</p>
<p>That's enough to tell you that the limit depends on direction.</p>
|
733,280 | <p>I cannot understand why $\log_{49}(\sqrt{ 7})= \frac{1}{4}$. If I take the $4$th root of $49$, I don't get $7$.</p>
<p>What I am not comprehending? </p>
| Mr.Fry | 68,477 | <p>Solve: $49^x = \sqrt{7} \Rightarrow 49^{2x} = 7 \Rightarrow \frac{\log7}{\log 49} = 2x \Rightarrow \frac{1}{2}=2x \Rightarrow x = \frac{1}{4}$</p>
|
2,813,595 | <p>which of the following can be expressed by exact length but not by exact number?</p>
<p>(i) $ \sqrt{10} $</p>
<p>(ii) $ \sqrt{7} $</p>
<p>(iii) $ \sqrt{13} \ $</p>
<p>(iv) $ \ \sqrt{11} \ $</p>
<p><strong>Answer:</strong></p>
<p>I basically could not understand th question.</p>
<p>What is meant by expressing by exact length ?</p>
<p><strong>Does we need to satisfy Pythagorean law?</strong></p>
<p>Help me with hints</p>
| Peter Szilas | 408,605 | <p>Correct me if wrong .</p>
<p>$√n$ is constructible, $n \in \mathbb{Z^+}$.</p>
<p>$n=1$, ok.</p>
<p>Assume $√n$ is constructible.</p>
<p>Step: </p>
<p>Show that $\sqrt{n+1}$ is constructible.</p>
<p>Pythagorean Theorem:</p>
<p>$(\sqrt{n})^2+1= n+1= (\sqrt{n+1})^2$, i.e.</p>
<p>construct a right triangle with leg lengths, $1$ and $√n$,</p>
<p>then the length of the hypotenuse is $\sqrt{n+1}$.</p>
|
57,988 | <p>I am a programmer/analyst with limited (and pretty rusty) knowledge of math.</p>
<p>"Just for the heck of it" I have decided to try my hand at <a href="http://spectrum.ieee.org/automaton/robotics/artificial-intelligence/you-you-can-take-stanfords-intro-to-ai-course-next-quarter-for-free" rel="nofollow">Stanford's introductory course on Artificial Intelligence</a> and according to their course description:</p>
<blockquote>
<p><strong>Prerequisites:</strong></p>
<p>A solid understanding of probability and linear algebra
will be required.</p>
</blockquote>
<p>Can someone please point me to concise introductory texts on these two topics? I don't know if what I'd like even exists (i.e. is there anything like "Linear Algebra for dummies"?) but my main requisites would be:</p>
<ul>
<li>Concise, as in "I doubt I can go through a 500 pages book"</li>
<li>Easy to approach (If the book itself has its own list of prerequisites, I doubt I can make much use of it, either).</li>
</ul>
<p>I can't be much more specific than this (the course introduction doesn't tell much more, unless they have recently updated it). </p>
<p>Thanks in advance!</p>
| Juan S | 2,219 | <p>For Linear Algebra I think that the (free) book by <a href="http://joshua.smcvt.edu/linearalgebra/" rel="nofollow">Heffron</a> is pretty good. </p>
<p>For probability I don't know too much, but maybe the first half of <a href="http://rads.stackoverflow.com/amzn/click/0521540364" rel="nofollow">Tijms</a> book</p>
|
3,109,001 | <p>I want to compute the Picard group of <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span>, which is not a Dedekind domain. The problem is that I don't even know where to begin.</p>
<p>Any ideas would be helpful.</p>
<p>Thanks</p>
<p>Edit- Since there's some confusion by what I mean by the "Picard group".</p>
<p><span class="math-container">$\textit{Definition}$</span></p>
<p>Let A be an integral domain. Let K denote it's field of fractions. Consider the set of all invertible fractional ideals F(A). Then F(A) forms a group. Consider the subgroup of non-zero principal fractional ideals P(A). So any element in P(A) looks like xA where x <span class="math-container">$\in$</span> K<span class="math-container">$^*$</span>. Then the Picard group of A is defined as F(A)/P(A).</p>
<p>When A is a Dedekind domain, the Picard group coincides with the class group.</p>
<p>Edit 2 - So here's why I'm having a hard time with this problem.</p>
<p>Had <span class="math-container">$\mathbb{Z}\sqrt{-19}$</span> been integrally closed, we could proceed using the Minkowski bound and Kummer-Dedekind theory. Unfortunately, <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span> is <span class="math-container">$\textit{not}$</span> integrally closed.</p>
<p>The integral closure in it's field of fractions <span class="math-container">$\mathbb{Q}(\sqrt{-19})$</span> is actually strictly bigger, given by <span class="math-container">$\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$</span>.</p>
<p>And thus, I can't use routine methods to try and understand this group.</p>
| DanLewis3264 | 480,329 | <p>You want to use the Kummer-Dedekind Theorem to factor the prime ideals of norm at most the Minkowski bound. </p>
<p>I recommend you follow 2.6.1 of <a href="https://jrsijsling.eu/notes/ant-notes.pdf" rel="nofollow noreferrer">https://jrsijsling.eu/notes/ant-notes.pdf</a></p>
|
3,109,001 | <p>I want to compute the Picard group of <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span>, which is not a Dedekind domain. The problem is that I don't even know where to begin.</p>
<p>Any ideas would be helpful.</p>
<p>Thanks</p>
<p>Edit- Since there's some confusion by what I mean by the "Picard group".</p>
<p><span class="math-container">$\textit{Definition}$</span></p>
<p>Let A be an integral domain. Let K denote it's field of fractions. Consider the set of all invertible fractional ideals F(A). Then F(A) forms a group. Consider the subgroup of non-zero principal fractional ideals P(A). So any element in P(A) looks like xA where x <span class="math-container">$\in$</span> K<span class="math-container">$^*$</span>. Then the Picard group of A is defined as F(A)/P(A).</p>
<p>When A is a Dedekind domain, the Picard group coincides with the class group.</p>
<p>Edit 2 - So here's why I'm having a hard time with this problem.</p>
<p>Had <span class="math-container">$\mathbb{Z}\sqrt{-19}$</span> been integrally closed, we could proceed using the Minkowski bound and Kummer-Dedekind theory. Unfortunately, <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span> is <span class="math-container">$\textit{not}$</span> integrally closed.</p>
<p>The integral closure in it's field of fractions <span class="math-container">$\mathbb{Q}(\sqrt{-19})$</span> is actually strictly bigger, given by <span class="math-container">$\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$</span>.</p>
<p>And thus, I can't use routine methods to try and understand this group.</p>
| pisco | 257,943 | <p>In case of imaginary quadratic field, there is a nice correspondence between the Picard group of orders with discriminant <span class="math-container">$D$</span> and reduced primitive quadratic form with the same discriminant. More precisely, the form <span class="math-container">$ax^2+bxy+cy^2$</span> corresponds to the ideal class of <span class="math-container">$(a,(-b+\sqrt{D})/2)$</span>. </p>
<p>For more information and proof, consult chapter 7 of the well-known <em>Primes of the Form <span class="math-container">$x^2+ny^2$</span></em> by David Cox.</p>
<hr>
<p>The order <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span> has discriminant <span class="math-container">$-56$</span>, and there are three primitive reduced forms:
<span class="math-container">$$x^2+19y^2 \qquad 4x^2-2xy+5y^2 \qquad 4x^2+2xy+5y^2$$</span>
they respectively translate to the trivial ideal, <span class="math-container">$(4,1+\sqrt{-19})$</span> and <span class="math-container">$(4,-1+\sqrt{-19})$</span>.</p>
<p>Hence the Picard group of <span class="math-container">$\mathbb{Z}[\sqrt{-19}]$</span> is cyclic of order <span class="math-container">$3$</span> generated by <span class="math-container">$(4,1+\sqrt{-19})$</span>.</p>
|
307,701 | <p>Show that if $G$ is a finite group with identity $e$ and with an even number of elements, then there is an $a \neq e$ in $G$, such that $a \cdot a = e$.</p>
<p>I read the solutions here <a href="http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf" rel="nofollow">http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf</a></p>
<p>Why do they say $D = \{a, a^\prime\}$? Isn't $D$ not a group? There is no identity and if they include the identity they get 3 elements, which means $|D| = 3 = $ odd.</p>
| i.a.m | 60,195 | <p>let $e,a_1,a_2,...,a_n$ be the elements of the group since the number of these elements is even we get the number of these elemnts $a_1,a_2,...,a_n$ is odd, now start putting every element with its inverse in a set lets say $\{a_i,a_j\}$ since the number is odd you will be left with one element call it $a$ and you have $a.a=e$.</p>
|
4,132,402 | <p>Can this be solved without trigonometry?</p>
<blockquote>
<p><span class="math-container">$AB$</span> is the base of an isosceles <span class="math-container">$\triangle ABC$</span>. Vertex angle <span class="math-container">$C$</span> is <span class="math-container">$50^\circ$</span>. Find the angle between the altitude and the median drawn from vertex <span class="math-container">$A$</span> to the opposite side.</p>
<p><a href="https://i.stack.imgur.com/YwI7Dm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YwI7Dm.png" alt="enter image description here" /></a></p>
</blockquote>
<p>I think I know how to do this using the Law of Cosines.</p>
<p>Call the side of the isosceles triangle <span class="math-container">$x$</span>, <span class="math-container">$CA=CB=x$</span>, and let <span class="math-container">$E$</span> be the midpoint of <span class="math-container">$BC$</span>. Then in the triangle formed by the median, <span class="math-container">$\triangle CAE$</span>, using the Law of Cosines:</p>
<p><span class="math-container">$$AE = \sqrt{x^2 + \frac{x^2}4-2\frac{x^2}{2}\cos50^\circ}$$</span></p>
<p>From this you can find <span class="math-container">$AE$</span> in terms of <span class="math-container">$x$</span>.</p>
<p>Then apply the Law of Cosines again to find <span class="math-container">$\angle CAE$</span> using
<span class="math-container">$$CE=\frac{x}{2}= \sqrt{x^2 + AE^2-2xAE\cos\angle CAE}$$</span>
and from <span class="math-container">$\cos\angle CAE$</span> you find <span class="math-container">$\angle CAE$</span>, and the angle we want is <span class="math-container">$40^\circ-\angle CAE$</span>.</p>
<p>But is it possible w/o trig?</p>
| Blue | 409 | <p>As <a href="https://math.stackexchange.com/a/4132518/409">@quasi's answer</a> suggests, the target angle almost-certainly isn't rational, so avoiding trig is unlikely.</p>
<p>That said, there's a pretty quick trigonometric approach to the target:</p>
<p><a href="https://i.stack.imgur.com/d7cfi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d7cfi.jpg" alt="enter image description here" /></a></p>
<p>Let <span class="math-container">$s$</span> be the triangle's half-leg, <span class="math-container">$M$</span> the midpoint of <span class="math-container">$\overline{BC}$</span>, and <span class="math-container">$N$</span> the foot of the altitude from <span class="math-container">$A$</span>. Then, in right <span class="math-container">$\triangle ACN$</span> we have
<span class="math-container">$$|AN|=2s\sin C \qquad |NC|=2s\cos C$$</span>
Thus,</p>
<blockquote>
<p><span class="math-container">$$\tan\theta =\frac{|MN|}{|AN|}=\frac{|CN|-|CM|}{|AN|} = \frac{2s\cos C-s}{2s\sin C} = \frac{2\cos C-1}{2\sin C} $$</span></p>
</blockquote>
<p>For <span class="math-container">$C=50^\circ$</span>, this gives <span class="math-container">$\theta=10.558\ldots^\circ$</span>.</p>
|
947,618 | <p>For T: V2->V2</p>
<p>T maps each point with polar coordinate (r.theta) to each point with polar coordinate (r,2theta) and T maps 0 onto itself.</p>
<p>Hi,</p>
<p>I was trying to do this by letting r= square root of x^2 + y^2 and theta=arctan(y/x) </p>
<p>but I failed.</p>
<p>can anybody please explain it? </p>
| Slade | 33,433 | <p>$T$ is surjective but not injective. For a linear transformation from a finite-dimensional vector space to itself, this is impossible.</p>
|
947,618 | <p>For T: V2->V2</p>
<p>T maps each point with polar coordinate (r.theta) to each point with polar coordinate (r,2theta) and T maps 0 onto itself.</p>
<p>Hi,</p>
<p>I was trying to do this by letting r= square root of x^2 + y^2 and theta=arctan(y/x) </p>
<p>but I failed.</p>
<p>can anybody please explain it? </p>
| Ben Grossmann | 81,360 | <p>We can deduce that $T$ is not linear because
$$
T[(1,0)] + T[(1,\pi)] = [(1,0) + (1,0)] = (2,0)
$$
But
$$
T[(1,0) + (1,\pi)] = T[0] = 0
$$</p>
|
1,663,838 | <p>Show that a positive integer $n \in \mathbb{N}$ is prime if and only if $\gcd(n,m)=1$ for all $0<m<n$.</p>
<p>I know that I can write $n=km+r$ for some $k,r \in \mathbb{Z}$ since $n>m$</p>
<p>and also that $1=an+bm$. for some $a,b \in \mathbb{Z}$</p>
<p>Further, I know that $n>1$ if I'm to show $n$ is prime.</p>
<p>I'm not sure how I would go about showing this in both directions though.</p>
| ThisIsNotAnId | 24,567 | <p>What you've written on the right of the iff. is the definition of a prime number. You have just stated it using more concise notation.</p>
<p>According to the <a href="http://mathworld.wolfram.com/PrimeNumber.html" rel="nofollow">wolfram page</a> for <em>Prime Number</em>,</p>
<blockquote>
<p>A prime number (or prime integer, often simply called a "prime" for short) is a positive integer p>1 that has no positive integer divisors other than 1 and p itself. </p>
</blockquote>
|
3,354,990 | <p>I have points and limits of a function and even the shape of the function and I'm looking for the function, something that very interesting for me how could I control the curve of the function?</p>
<p>(1) <span class="math-container">$\lim\limits_{x \to inf} f(x) = 1 $</span></p>
<p>(2) <span class="math-container">$f(\frac{1}{c}) = 1 $</span></p>
<p>(3) <span class="math-container">$0\lt x$</span></p>
<p>(4) <span class="math-container">$0\lt c \leq 1$</span></p>
<p><a href="https://i.stack.imgur.com/DqYRT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DqYRT.jpg" alt="Shape of function"></a></p>
| Kavi Rama Murthy | 142,385 | <p>One such function is <span class="math-container">$1-(1-c^{2}x^{2})^{+}$</span> where <span class="math-container">$y^{+}$</span> denotes <span class="math-container">$\max \{y,0 \}$</span>. </p>
<p>Edit: this function is convex on <span class="math-container">$(0,\frac 1 c)$</span>. To make it concave consider <span class="math-container">$\min \{1, \frac {1-e^{-cx}} {1-e^{-1}} \}$</span>. </p>
|
2,930,413 | <p>The problem is as shown. I tried using gradient and Hessian but can not make any conclusions from them. Any ideas?</p>
<p><span class="math-container">$$\max x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$$</span></p>
<p>subject to</p>
<p><span class="math-container">$$\sum_{i=1}^nx_i=1,\quad x_i\geq 0,\quad i=1,2,\ldots,n,$$</span></p>
<p>where <span class="math-container">$a_i$</span> are given positive scalars. Find a global maximum and show that it is unique.</p>
| Ahmed S. Attaalla | 229,023 | <p>The potential energy of the spring is $\frac{1}{2}x^2$ so,</p>
<p>$$\frac{1}{2}x^2+\frac{1}{2} \left(\frac{dx}{dt} \right)^2=E$$</p>
<p>It follows,</p>
<p>$$(\frac{dx}{dt})^2=2E-x^2$$</p>
<p>Supposing that $\frac{dx}{dt} \geq 0$ (ie spring is moving in positive direction), we get:</p>
<p>$$\frac{dx}{dt}=\sqrt{2E-x^2}$$</p>
<p>$$dt=\frac{dx}{\sqrt{2E-x^2}}$$</p>
<p>$$t+c=\arcsin(\frac{x}{\sqrt{2E}})$$</p>
|
323,783 | <p>How do I evaluate this definite integral?
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$
I know this is a trig. function. </p>
| Community | -1 | <p>From <a href="https://math.stackexchange.com/questions/20397/striking-applications-of-integration-by-parts/20481#20481">here</a>, we have
$$\int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdots \dfrac34 \cdot \dfrac12 \cdot\dfrac{\pi}2$$
Hence,
$$\int_0^{\pi/2} \sin^4(x) \cos^4(x) dx = \dfrac1{16} \int_0^{\pi/2} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi} \sin^4(t) dt = \dfrac2{32} \int_0^{\pi/2} \sin^4(t) dt$$
Hence, the answer is
$$\dfrac1{16} \cdot \dfrac34 \cdot \dfrac12 \cdot \dfrac{\pi}2 = \dfrac{3 \pi}{256}$$</p>
<hr>
<p>If the integral is from $0$ to $\pi/12$, then, we get that
\begin{align}
\int_0^{\pi/{12}} \sin^4(x) \cos^4(x) dx & = \dfrac1{16} \int_0^{\pi/12} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi/6} \sin^4(t) dt\\
& = \dfrac1{32} \int_0^{\pi/6} \dfrac{(1-\cos(2t))^2}{4} dt\\
& = \dfrac1{128} \int_0^{\pi/6} (1+\cos^2(2t) - 2 \cos(2t)) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(1+\dfrac{1+\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(\dfrac32+\dfrac{\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sin(4 \pi/6)}{8}-\sin(2 \pi/6)\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sqrt3}{16} - \dfrac{\sqrt3}2\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 - \dfrac{7\sqrt3}{16} \right) = \dfrac{4 \pi - 7 \sqrt3}{2048}
\end{align}</p>
|
743,473 | <p>A long Weierstrass equation is an equation of the form
$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
Why are the coefficients named $a_1, a_2, a_3, a_4$ and $a_6$ in this manner, corresponding to $xy, x^2, y, x$ and $1$ respectively? Why is $a_5$ absent?</p>
| Alex | 38,873 | <p>OK, here's my suggestion: for 1) use the law of total probability
$$
P_M=P_{M|L}P_L + P_{M|L'}P_{L'}
$$
Once you've done that, use $P_{M'}=1-P_M$ for 2):
$$
P_{L|M'}=\frac{P_{M'|L}P_L}{P_{M'}}
$$</p>
|
2,798,598 | <p>We have the series $\sum\limits_{n=1}^{\infty} \frac{(-1)^n n^3}{(n^2 + 1)^{4/3}}$. I know that it diverges, but I'm having some difficulty showing this. The most intuitive argument is perhaps that the absolute value of the series behaves much like $\frac{n^3}{\left(n^2\right)^{4/3}} = \frac{1}{n^{-1/3}}$, which diverges, though this doesn't seem like it would disprove thee fact that we could be dealing with a conditionally-convergent series. The computation of the limit, even of the absolute value of the general term, also seems nearly impossible to do by hand, as successive applications of L'Hospital's Rule seem to produce a result just as disorderly as what I started with. Limit comparison also doesn't quite seem to work, especially with the alternating-factor.</p>
<p>Thanks in advance for any insights on this. </p>
| user | 505,767 | <p>Note that</p>
<p>$$ \left|\frac{(-1)^n n^3}{(n^2 + 1)^{4/3}}\right|\sim {n^\frac13}\to \infty$$</p>
<p>thus the given series diverges since each terms in the limit diverges.</p>
<p>Recall indeed for convergence we always need that, as necessary condition, that $|a_n|\to 0$.</p>
|
283,824 | <p>Let $x$ be a random vector uniformly distributed on the unit sphere $\mathbb{S}^{n-1}$. Let $V$ be a linear subspace of dimension $k$ and let $P_V(x)$ be the orthogonal projection of $x$ onto $V$.
I have seen quoted in the literature that
\begin{align}
\mathbb{P}[|\left\| P_V(x)\right\|_2 - \sqrt{k/n} | \le \epsilon] \ge 1 -2\exp(-n\epsilon^2/2). \, \, \, \, \, \, \, (1)
\end{align} However, i can still not find a concrete proof. What i do understand is that for a $1$-Lipschitz function $f:\mathbb{S}^{n-1} \rightarrow \mathbb{R}$ such as $x \mapsto |\left\| P_V(x)\right\|_2$, we have that
\begin{align}
\mathbb{P}[|f - M_f | \le \epsilon] \ge 1 -2\exp(-n\epsilon^2/2), \, \, \, \, \, \, \, (2)
\end{align} where $M_f$ is the median of $f$. (2) mostly follows from the isoperimetric inequality on the sphere. The issue though with (1) is that $\sqrt{k/n}$ does not seem to be the median of $x \mapsto |\left\| P_V(x)\right\|_2$. Is anyone able to provide a clean argument for (1) or a self-contained reference in the literature? Many thanks. </p>
| Dirk | 9,652 | <p>Actually, the answer is almost in the paper you linked. There the author refers to </p>
<blockquote>
<p>Duality theorems for marginal problems, Hans G. Kellerer, Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete, November 1984, Volume 67, Issue 4, pp 399–432, <a href="https://link.springer.com/article/10.1007/BF00532047" rel="nofollow noreferrer">https://link.springer.com/article/10.1007/BF00532047</a></p>
</blockquote>
<p>One dual problem is already stated on page one. If you are looking for a "Kantorovich-Rubinstein" dual problem with Lipschitz-functions: I am not sure if such a thing exists - one uses the metric in an essential ways and I don't see anything that replaces this here…</p>
|
73,629 | <p>I want to use <code>Listplot</code> with <code>Tooltip</code>that displays <code>Position</code>of the element I'm hovering over.</p>
<pre><code>data={{0,1},{1,3},{2,2}};
ListPlot[Tooltip[data]]
</code></pre>
<p>This is displaying the value of the element. Can I use the <code>Position</code>function in the tooltip?</p>
| kglr | 125 | <pre><code>data = Sort@RandomInteger[10, {10, 2}];
tts = {Directive[Red, 16, Bold], CellFrame -> 3, CellFrameMargins -> 5};
ListPlot[MapIndexed[Tooltip[#, First@#2, TooltipStyle -> tts] &, data],
PlotStyle -> PointSize[Large], Frame -> True, AxesOrigin -> {0, 0}]
</code></pre>
<p><img src="https://i.stack.imgur.com/IssyY.gif" alt="enter image description here"></p>
|
4,358 | <p>I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:</p>
<p>If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.</p>
<p>Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.</p>
<p>My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.</p>
| Dan Ramras | 1,392 | <p>For the first question, let $S$ be the set of all rationals less than $\pi$. Then $S$ has no least upper bound in $\mathbb{Q}$. On the other hand, if $T = S\cup {4}$, then 4 is the least upper bound of $S$ inside $T$.</p>
<p>For the second question, let $S$ be the set of all rationals strictly less than 1, and let $T = \mathbb{Q}\cap \left([-\infty, 1)\cup [2, \infty)\right)$. Then the least upper bound of $S$ inside $T$ is 2. But of course the least upper bound of $S$ in $\mathbb{Q}$ is 1. </p>
<p>This shows the answer to your second question is "Yes."</p>
|
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