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 |
|---|---|---|---|---|
114,340 | <p>Let $M\subset \mathbb C^2$ be a hypersurface defined by $F(z,w)=0$. Then for some point $p\in M$, I've
$$\text{ rank of }\left(
\begin{array}{ccc}
0 &\frac{\partial F}{\partial z} &\frac{\partial F}{\partial w} \\
\frac{\partial F}{\partial z} &\frac{\partial^2 F}{\partial ^ 2z} &\frac{\partial^2 F}{\partial z\partial w} \\
\frac{\partial F}{\partial w} &\frac{\partial^2 F}{\partial w\partial z} & \frac{\partial^2 F}{\partial w^2} \\
\end{array}
\right)_{\text{ at p}}=2.$$</p>
<p>What does it mean geometrically? Can anyone give a geometric picture near $p$? </p>
<p>Any comment, suggestion, please.</p>
<p>Edit: Actually I was reading about Levi flat points and Pseudo-convex domains. I want to understand the relation between these two concepts. A point p for which the rank of the above matrix is 2 is called Levi flat. If the surface is everywhere Levi flat then it is locally equivalent to $(0,1)\times \mathbb{C}^n$, so I have many examples....but what will happen for others for example take the three sphere in $\mathbb{C}^2$ given by $F(z,w)=|z|^2+|w|^2−1=0$. This doesn't satisfy the rank 2 condition. Can I have precisely these two situations?</p>
| Robert Israel | 8,508 | <p>Suppose $A$ is the adjacency matrix of graph $G_1$ and $B$ the adjacency matrix of graph $G_2$, where we consider both $G_1$ and $G_2$ to have the same vertices $1,\ldots,n$. Then $(AB)_{ij}$ is the number of ways to get from $i$ to $j$ by going first along an edge of $G_1$ and then along an edge of $G_2$.</p>
|
1,403,228 | <blockquote>
<p><strong>Question:</strong></p>
<p>Let <span class="math-container">$m, n, q, r \in \mathbb Z$</span>. If <span class="math-container">$m = qn + r$</span>, show that <span class="math-container">$\gcd(m, n) = \gcd(n, r)$</span>. Hence justify the Euclidean Algorithm.</p>
</blockquote>
<p>I found this question in a past test paper, but cannot seem to find a reference in my textbook that indicates how I can go about "proving" the above statement. Can anyone please point me in the right direction?</p>
| Community | -1 | <p>The following is a demonstration using notions of ring theory.</p>
<p>Let $m,n,q,r\in\mathbb{Z}$ such that $m=qn+r$.</p>
<p>Now consider the ideals $J=m\cdot \mathbb{Z}+n\cdot \mathbb{Z}$ and $J'=n\cdot\mathbb{Z}+r\cdot\mathbb{Z}$. The ring of integer is a principal ideal domain, therefore, since $J$ and $J'$ are ideals, there're $d,d'\in\mathbb{Z}$ such that $J=d\cdot\mathbb{Z}$ and $J'=d'\cdot \mathbb{Z}$. Recalling the definition of $\gcd$(greatest common divisor) it's easy to see that $d=\gcd(m,n)$ and $d'=\gcd(n,r)$ knowing that $d$ and $d'$ are the least positive integers, respectively, of $J$ and $J'$. Therefore to show that $\gcd(m,n)=\gcd(n,r)$ is sufficient to show that $J=J'$.</p>
<p>$j'\in J'\rightarrow \exists x',y'\in\mathbb{Z}: j'=nx'+ry'$</p>
<p>We know that $m=nq+r\Leftrightarrow r=m-nq$, so $j'=nx'+ry'=nx'+(m-nq)y'=my'+n(x'-qy')\in J\Rightarrow J'\subset J$.</p>
<p>$j\in J\rightarrow \exists x,y\in\mathbb{Z}: j=mx+ny$</p>
<p>We know that $m=qn+r$, so $j=mx+ny=(nq+r)x+ny=n(qx+y)+rx\in J'\Rightarrow J\subset J'$.</p>
<p>Since $J'\subset J$ and $J\subset J'$, then we have $J=J'$, which allows us to demonstrate what we want.</p>
|
3,170,368 | <p>The limit of <span class="math-container">$f(x,y)=\frac{xy^2}{x^2+y^4}$</span> as <span class="math-container">$(x,y) \longrightarrow(0,0)$</span> is doesn't exist, because if we take two paths:</p>
<p>1) Along the path, <span class="math-container">$x=0$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$</span>
2) Along the path, <span class="math-container">$x=y^2$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$</span></p>
<p>But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e.,
<span class="math-container">$x=r\cos(\theta),y=r\sin(\theta)$</span>, we know that <span class="math-container">$x^2+y^2=r^2$</span> and this indicates that <span class="math-container">$r\longrightarrow0$</span> as <span class="math-container">$(x,y)\longrightarrow(0,0)$</span>, therefore
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$</span>
<span class="math-container">$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$</span>
Please someone help me on this contradictory, why this is so happened?.</p>
| StackTD | 159,845 | <p>You need to be careful and consider all paths as well, not only paths with constant <span class="math-container">$\theta$</span>.</p>
<blockquote>
<p><span class="math-container">$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$$</span></p>
</blockquote>
<p><span class="math-container">$$=\lim_{r\longrightarrow0}\left(r\color{blue}{\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}}\right)$$</span></p>
<p>Your argument works if the expression in blue stays bounded (near the origin); for all <span class="math-container">$(r,\theta)$</span>.</p>
<hr>
<p>Inspired by your own path <span class="math-container">$x=y^2$</span>, check what happens in polar coordinates (<span class="math-container">$r \ne 0$</span>):
<span class="math-container">$$x=y^2 \longrightarrow r\cos\theta=r^2\sin^2\theta \implies r=\frac{\cos\theta}{\sin^2\theta}$$</span>
Along <span class="math-container">$r=\frac{\cos\theta}{\sin^2\theta}$</span> you can take <span class="math-container">$\theta\to\tfrac{\pi}{2}$</span> to get <span class="math-container">$r\to 0$</span>, but what happens with your function along that path?</p>
<p><span class="math-container">$$\lim_{r\to 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\stackrel{r=\frac{\cos\theta}{\sin^2\theta}}{\longrightarrow}\ldots$$</span></p>
|
2,877,174 | <p>Imagine that we roll two fair six-sided dice (i.e., all six sides have equal probability). Let X1 and
X2 be the random variables representing these outcomes.
Now, imagine we take one of the dice rolls, say X1, and add a (possibly negative) constant c to
the result. If this becomes less than zero, then we set it to zero; denote this by <br></p>
<p><em>(X + c)<sub>+</sub> = max(X + c, 0)</em></p>
<p>What is the expected value of <em>(X<sub>1</sub> - 2)<sub>+</sub></em> * <em>(X<sub>2</sub> + 1)<sub>+</sub></em>?</p>
<p>My answer <br>
E{<em>(X<sub>1</sub> - 2)<sub>+</sub></em> * <em>(X<sub>2</sub> + 1)<sub>+</sub></em>} = $\frac{1+2+3+4}{6}$ * $\frac{2+3+4+5+6+7}{6}$ = 7.515
<br></p>
<p>I don't know if my answer is right. For X<sub>2</sub>, Am I suppose to divide by 6 or 36? Can anyone correct me if I'm wrong?</p>
| José Carlos Santos | 446,262 | <p>Your argument is correct, but your notation is innapropriate. Where you wrote $\operatorname{rank}(\ker\cdots)$, you should have written $\dim(\ker\cdots)$.</p>
|
2,228,328 | <p>Is the function $ f(z)=\dfrac 1 {x^2+y^2} + i \dfrac 1 {x^2+y^2} $ is differentiable and holomorphic somewhere?</p>
<p>We have $z=x+iy$ and $f(z)= \dfrac{1+i}{|z|^2}$ . Now, $f(0)= \lim_{\delta z \rightarrow 0} \frac{f(0+\delta z) - f(0)}{\delta z}=$ undefined. So $f(z)$ is not differentiable at origin. Also since since $ f(z)= |z|^2$ is not differentiable anywhere except origin , so $f(z)$ is not not differentiable and hence it is not holomorphic. But I am not sure, please help me</p>
| JMJ | 295,405 | <p>Let the operators $D_1$ and $D_2$ be defined as
$$
D_1(u+iv) = u_x - v_y \ \text{ and } \ D_2(u+iv) = u_y + v_x.
$$
Then $u+iv$ is analytic at $z_0 = x_0 + i y_0$ if and only if $u(x_0,y_0)+iv(x_0,y_0)$ belongs to $\ker D_1 \cap \ker D_2$. </p>
|
189,618 | <p>I'm trying to plot 2D vectors in Mathematica. Built in functions don't really work for me because I want to plot vectors of matrices from the origin to the their coordinates with an arrow on their tips. I made a function </p>
<pre><code>plotMatrixVectors[mat_List] :=
Graphics[Table[Arrow[{origin, i}], {i, mat}], Axes -> True],
</code></pre>
<p>and it worked just fine, until it suddenly stopped working and instead of a graph started reporting this to me:</p>
<pre><code>{{plotMatrixVectors[1], plotMatrixVectors[2]}, {plotMatrixVectors[3],
plotMatrixVectors[1]}}
</code></pre>
<p>Note: these numbers in square bracket have nothing to the with coordinates in the lower example, this message came up when I tried to plot different vectors, but you get the point.</p>
<p>I rewrote the function because I thought I unintentionally messed it up somehow, but it still printed the same thing. However, when I copypasted the body of the function and just plugged in a 2x2 matrix, say this one </p>
<pre><code>plot2D1 = {{1, 5}, {-6, 4}};
Graphics[Table[Arrow[{origin, i}], {i, plot2D1}], Axes -> True]
</code></pre>
<p>it gives me this, which is good.
<a href="https://i.stack.imgur.com/VLyUf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VLyUf.png" alt="enter image description here"></a></p>
<p>I have 2 questions here.
1) Why did my function stopped working but the same thing works when it's typed outside of the function? I restarted Mathematica and it still didn't work.
2) Is there any way to strecth those axes so they go beyond the limits of matrix vectors coordinates?</p>
<p>Thanks a lot!</p>
| m_goldberg | 3,066 | <p>I don't know what went wrong with your code, but you might try this. It is simpler and more robust than your formulation.</p>
<pre><code>plotVectors[pts : {{_?NumberQ, _?NumberQ} __}] :=
Graphics[Arrow[{{0, 0}, #} & /@ pts], Axes -> True]
</code></pre>
<p>Tests</p>
<pre><code>plotVectors[{{1, 5}, {-6, 4}}
</code></pre>
<p><a href="https://i.stack.imgur.com/8e3ly.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8e3ly.png" alt="test1"></a></p>
<pre><code>SeedRandom[26];
With[{n = 15}, plotVectors[RandomReal[{-10, 10}, {n, 2}]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/rjYOo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rjYOo.png" alt="test2"></a></p>
|
478,620 | <p>Let $S$ be a set that satisfies the following property:</p>
<blockquote>
<p>For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$</p>
</blockquote>
<p>Does it follow that $S$ is finite?</p>
| Moataz | 316,923 | <p>(Hundreds) (Tens) (Units),
Units could be $(1, 3, 7, 9) \rightarrow 4$ numbers,
Tens could be $(0, 1, 2, 3, 4, 6, 7, 8, 9)\rightarrow 9$ numbers,
Hundreds could be $(1, 2, 3, 4, 6, 7, 8, 9) \rightarrow 8$ numbers,
(Hundreds) (Tens) (Units) $\rightarrow
(8) (9) (4) = 288$</p>
|
3,493,729 | <p>Although I looked up the answer on integral calculator com but I still have little to no idea as to how one would proceed to solve this integral.
Integrate <span class="math-container">$\dfrac{e^x(x^4+2)}{(1+x^2)^{5/2}}$</span> wrt <span class="math-container">$x.$</span> I initially tried to convert it to the form <span class="math-container">$e^x\cdot(f(x)+f'(x)).$</span> However, I wasn't successful in spite of struggling for more than half an hour :(</p>
<p>Any help would be appreciated :)</p>
| Adam Martens | 691,077 | <p>This is not possible. If <span class="math-container">$X,Y$</span> are topological spaces with <span class="math-container">$X$</span> connected and <span class="math-container">$f: X\to Y$</span> is continuous, then <span class="math-container">$f(X)\subset Y$</span> is connected. Since <span class="math-container">$\{0,1\}$</span> (with discrete topology) is disconnected, there exists no such surjective <span class="math-container">$f$</span></p>
|
380,696 | <p>Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.</p>
<p>My workings so far: I proceeded by induction. Obviously $1^2 \equiv 1 \mod 8$. Then assume that for a certain uneven number $k$ we have $k^2 \equiv 1 \mod 8$. Then the the next uneven number is $k+2$ and $(k+2)^2 = k^2 + 4k + 4 \equiv 4k + 5 \mod 8$. by the induction hypothesis. Now since $k$ is uneven we can write it as $k=2j+1$ and thus $4k+5 =8j+9 \equiv 1 \mod 8$ and we have shown what was asked for by induction.</p>
<p>However, in the case of a prime number $n$ I am not so certain how to proceed because I can't use induction. It comes down to proving that $24|(p^2-1)$. This is certainly the case for the first uneven prime $p \neq 3$, namely $p=5$ such that $p^2-1 =24$. How would I proceed from here, or how should I approach the problem differently? Many thanks in advance!</p>
| vonbrand | 43,946 | <p>If $n$ is odd, $n = 2 k + 1$ and $n^2 = 4 k^2 + 4 k + 1 = 4 k (k + 1) + 1$, and $4 k (k + 1)$ is divisible by 8. Thus $n^2 \equiv 1 \pmod{8}$.</p>
<p>If $p > 3$ is prime, it is odd, and so by the above $p^2 \equiv 1 \pmod{8}$. As it is prime, $p \equiv 1 \pmod{3}$ or $p \equiv 2 \pmod{3}$, so $p^2 \equiv 1 \pmod{3}$. Now, as $\gcd(3, 8) = 1$, combining the above congruences gives $p^2 \equiv 1 \pmod{24}$.</p>
|
306,178 | <p>Given
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.</p>
| Dominic Michaelis | 62,278 | <p>If you don't care about overkilling the Problem you can even do this with calculus, showing the derivative in n is negative.
The derivative is
$$ \frac{\left(\frac{1}{n}+1\right)^n (n+1) \left(n \log \left(\frac{1}{n}+1\right)-1\right)}{n^2}$$ and so we only need to show that
$$1> n \log(1+\frac{1}{n})$$
By substitution $n=\frac{1}{x}$ we have
$$\frac{\log(1+x)}{x}=\frac{\ln(1+x)-\ln(1)}{(x+1)-1}=\frac{1}{1+\xi}$$
with $\xi \in (0,x)$ (this is granted by the mean value theorem), and the last expression is less than 1.</p>
|
4,212,199 | <p><span class="math-container">$$\frac15\int_0^5(4+2^{0.1t^2})dt$$</span>
This question is from part (b) of question 1 in the AP Calculus BC 2019 FRQ. I've tried using u-substitution where <span class="math-container">$u = 0.1 t^2$</span>, but this won't work because of the <span class="math-container">$2t$</span> that will result when taking the derivative of <span class="math-container">$u$</span>. Integration by parts wouldn't work as the integral doesn't involve a product, and trigonometric substitution wouldn't work either. Is there a technique that would allow me to evaluate this?</p>
| Oğuzhan Kılıç | 481,167 | <p>There are many ways you can solve this problem, let's first do it in your way. Note that i'm not bothering with units. Because of feet, pound. :( not nice. Also let me note that we must talk about the <span class="math-container">$g$</span> (gravitation constant) because if we don't say anything about it answer wşll be no work is required.</p>
<p>Mass of hanging part of rope is <span class="math-container">$2(100-y)$</span>, force acting on this part of rope is <span class="math-container">$g.2(100-y)=gm$</span>. Work is given by <span class="math-container">$Force.Distance$</span>. Then the integral becomes</p>
<p><span class="math-container">$$ \int_0^{80}2g(100-y)\ dy = 9600$$</span></p>
<p>Another way to solve is to use energy which can be calculated easily, initial potential energy is</p>
<p><span class="math-container">$$ \int_0^{100} 2gy\ dy=10000 $$</span></p>
<p>Final potential energy</p>
<p><span class="math-container">$$\int_{80}^{100} 2gy \ dy + \int_0^{80} 200g dy=3600+16000 $$</span></p>
|
2,478,026 | <p><a href="https://i.stack.imgur.com/TyhHX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TyhHX.png" alt="enter image description here"></a></p>
<p>It doesn't make sense to me that |E[x] - m| = |E[X-m]|</p>
| bof | 111,012 | <p>$E[X-m]=E[X]-E[m]=E[X]-m$ because $m$ is a constant. For instance, if $m=5,$ then
$E[X-m]=E[X-5]=E[X]-E[5]=E[X]-5.$<br>
Similarly,<br>
$E[X-\mu]=E[X]-\mu=\mu-\mu=0,$ a familiar fact.</p>
|
2,916,168 | <p>I am a beginner in proofs and, unfortunately, I cannot wrap my mind around how to prove the simplest things, so I need a bit of help getting started. This is the proof that I am dealing with:</p>
<p>$\text{If }x< y< 0\text{, then }x^{2}> y^{2}\text{.}$</p>
<p>Thank you in advance.</p>
| user | 505,767 | <p>We have that</p>
<p>$$x<y<0 \iff 0<-y<-x $$</p>
<p>and since $f(a)=a^2$ is strictly increasing for $a>0$</p>
<p>$$0<-y<-x \implies 0<(-y)^2<(-x)^2$$</p>
<p>that is</p>
<p>$$x^2>y^2>0$$</p>
|
566,255 | <p>How to evaluate the following integral? </p>
<p>$\int_{0}^{\infty} \sin(kx)dx=\frac 1 k$ </p>
<p>The book Mathematical Physics by Butkov reads "The sequence $f_N(k)=\int_{0}^{N} \sin(kx) dx=\frac{(1-\cos kN)}k$diverges as N approaches $\infty$, but it is weakly convergent for a suitable chosen set of test functions $g(k)$ defined $0<k<\infty$."
The problem is how to choose the test function?</p>
| Chris Janjigian | 23,078 | <p>Take $\varphi \in C_c^\infty ((0,\infty))$ and call $f_n(k) = \int_0^n \sin(ky) dy = \frac{1-\cos(nk)}{k}$. We would like to show that
\begin{align*}
\lim_{n \to \infty} \int_{0}^\infty \varphi(k) f_n(k) dk = \int_{0}^\infty \varphi(k) \frac{1}{k} dk
\end{align*}
which would show that as distributions on $C_c^\infty ((0,\infty))$, $f_n(k) \to \frac{1}{k}$. It suffices to show that
\begin{align*}
\lim_{n \to \infty} \int_0^\infty \frac{\varphi(k)}{k}\cos(nk)dk = 0
\end{align*}
Notice that since $\varphi$ is a smooth function compactly supported away from $0$, $\frac{\varphi(k)}{k}$ is a continuous function of compact support on $\mathbb{R}$ and is therefore an $L^1$ function. The result now follows from the <a href="http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma" rel="nofollow">Riemann Lebesgue Lemma</a>.</p>
|
2,462,852 | <p>I want to prove that $f(x)=x^3+x+2$, $f: \mathbb R \rightarrow \mathbb R$ is bijective without calculus. My attempts at showing to prove that it' injective and surjective are written below:</p>
<p>$1)$ Injectivity: </p>
<p>I want to show that $\forall a,b \in \mathbb R$ $f(a)=f(b) \implies a=b$.</p>
<p>I started like this:
$$f(a)=f(b) \implies a^3+a+2=b^3+b+2$$
$$\implies a(a^2+1)=b(b^2+1)$$
$$\implies \frac{a}{b}=\frac{b^2+1}{a^2+1}$$
Then I said since $\frac{b^2+1}{a^2+1}>0$ $\forall a,b \in \mathbb R$ then either $a \land b < 0$ or $a \land b > 0$. (For the case when $b=0 \land a \in \mathbb R$ it would be easy to prove that $a=b$.) From there it seemed pretty obvious that $\frac{a}{b}=\frac{b^2+1}{a^2+1} \implies a=b$ so I couldn't really draw a logical argument to show that $a=b$.</p>
<p>$2)$ Surjectivity: </p>
<p>I want to show that $\forall b \in \mathbb R$ $\exists a \in \mathbb R$ s.t. $f(a)=b$. </p>
<p>I started like this:</p>
<p>Let $b \in \mathbb R$ and set $f(a)=b$ then we have:</p>
<p>$$a^3+a+2=b$$
$$\implies a^3+a=b-2$$
$$\implies a(a^2+1)=b-2$$</p>
<p>But then I couldn't find an expression for $a \in \mathbb R$ in terms of $b$.
So I'm wondering if anyone can tell me how I can proceed with my surjectivity and injectivity proofs.</p>
| Peter Szilas | 408,605 | <p>1) Injective:</p>
<p>$f(x) = x^3 +x+2$, $x \in \mathbb{R}$ is strictly monotonically increasing.</p>
<p>Let $x_1< x_2 $, then</p>
<p>$x_1^3 + (x_1+2) \lt x_2^3 + (x_2 +2)$,</p>
<p>is strictly monotonically increasing,</p>
<p>sum of $y_1= x^3$ and a linear function $y_2= x+2$, both strictly monotonically increasing.</p>
<p>2) Surjective.</p>
<p>A)$\lim_{x \rightarrow \infty} f(x) = \infty.$</p>
<p>B)$\lim_{x \rightarrow -\infty} f(x) = - \infty$.</p>
<p>Let $c \in \mathbb{R}$.</p>
<p>Choose $a$ with $f(a) < c$, and $b$ with $f(b) >c$, </p>
<p>(property A,B).</p>
<p>Consider the closed interval $ [a,b] .$</p>
<p>Since $f$ is continuos, and $f(a) < c < f(b)$,</p>
<p>there is a $p \in (a,b)$ such that $f(p) =c$.</p>
|
688,711 | <p>Assume the desity of air <span class="math-container">$\rho$</span> is given by</p>
<p><span class="math-container">$\rho(r)=\rho_0$$e^{-(r-R_0)/h_0}$</span> for <span class="math-container">$r\ge R_0$</span></p>
<p>where <span class="math-container">$r$</span> is the distance from the centre of the earth, <span class="math-container">$R_0$</span> is the radius of the earth in meters, <span class="math-container">$\rho_0=1.2kg/m^3$</span> and <span class="math-container">$h_0=10^4m$</span></p>
<p>Assuming the atmosphere extends to infinity, calculate the mass of the portion of the earth's atmosphere north of the equator and south of <span class="math-container">$30^\circ$</span>N latitude.</p>
<p>How do I even start this problem? Do I need to convert it into spherical coordinates? But then what limits do I use for the integration?</p>
| Julián Aguirre | 4,791 | <p>The limits of iteration in spherical coordinates could be</p>
<ul>
<li>$r>R_0$ (the exterior of the earth)</li>
<li>$0\le\theta<2\,\pi$ (all around the earth)</li>
<li>=$\pi/6<\phi<\pi/2$ (between 30ºN and the equator)</li>
</ul>
|
688,711 | <p>Assume the desity of air <span class="math-container">$\rho$</span> is given by</p>
<p><span class="math-container">$\rho(r)=\rho_0$$e^{-(r-R_0)/h_0}$</span> for <span class="math-container">$r\ge R_0$</span></p>
<p>where <span class="math-container">$r$</span> is the distance from the centre of the earth, <span class="math-container">$R_0$</span> is the radius of the earth in meters, <span class="math-container">$\rho_0=1.2kg/m^3$</span> and <span class="math-container">$h_0=10^4m$</span></p>
<p>Assuming the atmosphere extends to infinity, calculate the mass of the portion of the earth's atmosphere north of the equator and south of <span class="math-container">$30^\circ$</span>N latitude.</p>
<p>How do I even start this problem? Do I need to convert it into spherical coordinates? But then what limits do I use for the integration?</p>
| Phani | 545,039 | <p>To calculate the mass of the atmosphere above the earth's surface you do not need to know the density variation of air vertically nor the extent of the atmosphere above the earth. All one needs to know is the pressure at sea level and the value of acceleration due to gravity (g) at sea level (and an assumption that this does not vary over the depth of the atmosphere, which is about 50 km).</p>
<p>Use this formula P = m" g ; pressure is the weight/m^2 of the atmosphere at sea level.
p= 1.01325 N/m^2; g = 9.8 m/s^2, m" = 1.0339E4 kg/m^2
Radius of earth (R) = 6.372E3 km; Surface area of earth = 4 Pi R^2 = 5.1E8 km^2
Therefore mass of atmosphere over the earth = 1.0339E4 x 5.1E14 = 5.274E18 kg </p>
|
213,533 | <p>I am trying to play a sequence of sounds in Mathematica using <code>Table</code>. My real example is much more complicated so that is why I got to using <code>Table</code>, </p>
<p>Why doesn't the follow code work? </p>
<pre><code>frequencies = {{100, 200}, {200, 300}, {400, 600}};
Sound[
Table[
Sound[Play[Total[Sin[# 2 Pi t] & /@ frequencies[[i]]], {t, 0, 0.5}]],
{i, 1, 3, 1}]]
</code></pre>
<p>If I run the sounds as a list it works fine, but when using table it does not, I just get the following:</p>
<blockquote>
<pre><code>Sound[{Sound[
Sound[SampledSoundFunction[
Function[{Play`Time494},
Block[{t =
0. + 0.000125 Play`Time494}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] - 4.44089*10^-16) 0.568158]], 4000,
8000]]],
Sound[Sound[
SampledSoundFunction[
Function[{Play`Time495},
Block[{t =
0. + 0.000125 Play`Time495}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] + 1.88738*10^-15) 0.525731]], 4000,
8000]]],
Sound[Sound[
SampledSoundFunction[
Function[{Play`Time496},
Block[{t =
0. + 0.000125 Play`Time496}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] + 8.88178*10^-16) 0.525731]], 4000,
8000]]]}]
</code></pre>
</blockquote>
<p>Any help appreciated</p>
| Nasser | 70 | <p>I get something like this</p>
<pre><code>frequencies = {{100, 200}, {200, 300}, {400, 600}};
Sound[Table[
Evaluate@Sound[
Play[Total[Sin[#*2 Pi t] & /@ frequencies[[i]]], {t, 0,
0.5}]], {i, 1, 3, 1}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/8Ltau.png" alt="Mathematica graphics"></p>
<p>And it plays sound when clicking on the pay button. (just added <code>Evaluate</code>). See if this is what you expecting the sound to be. I am not good with music.</p>
|
2,448,745 | <p>I am trying to derive the general form of the equation of the circle given two points $(x_1,y_1)$, $(x_2,y_2)$and the angle $\theta$ subtended by the chord joining the two points. </p>
<p>So, $\theta$ is basically the angle between two lines
$\implies \tan \theta = |\dfrac{\dfrac{y-y_1}{x-x_1}-\dfrac{y-y_2}{x-x_2}}{1+( \dfrac{y-y_1}{x-x_1}\dfrac{y-y_2}{x-x_2})}| \\ \implies(x-x_1) (x-x_2)+(y-y_1)(y-y_2)= \pm \cot\theta[(y-y_1)(x-x_2)-(x-x_1)(y-y_2)]$</p>
<p>But this is the equation of the major arc right? How do I derive the equation of the circle from it?</p>
| M. Winter | 415,941 | <p>A <a href="https://en.wikipedia.org/wiki/Structure_(mathematical_logic)" rel="noreferrer">general algebraic object</a> is a set $X$ together with some specified</p>
<ul>
<li>relations $R_i\subseteq X^n,i\in I$,</li>
<li>functions $f_j:X^n\to X,j\in J$ and</li>
<li>constants $c_k\in X,k\in K$.</li>
</ul>
<p>As you can see, the functions can be of any arity. Such structures are mostly discussed in <a href="https://en.wikipedia.org/wiki/Algebraic_structure" rel="noreferrer">abstract setting</a> as in <em>model theory</em>. But there are examples which are used in applications, e.g. <a href="https://en.wikipedia.org/wiki/Planar_ternary_ring" rel="noreferrer">planar ternary rings</a> are used in incidence geometry. Another example which uses no functions but a <em>ternary relation</em> is a <a href="https://en.wikipedia.org/wiki/Cyclic_order" rel="noreferrer">cyclic order</a>.</p>
<hr>
<p><strong>Related</strong>:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/351369/why-dont-we-study-algebraic-objects-with-more-than-two-operations">Why don't we study algebraic objects with more than two operations?</a></li>
<li><a href="https://math.stackexchange.com/questions/94690/is-anybody-researching-ternary-groups">Is anybody researching "ternary" groups?</a></li>
</ul>
|
2,634,657 | <p>My question is: For <span class="math-container">$R>0$</span>, Can we choose a family of functions <span class="math-container">$\eta_R\in C_c^1(\mathbf{R}^N)$</span> satisfying
<span class="math-container">$0\leq\eta_R\leq 1$</span> in <span class="math-container">$\mathbf{R}^N$</span>, <span class="math-container">$\eta_R=1$</span> in <span class="math-container">$B_R(0)$</span> and
<span class="math-container">$\eta_R=0$</span> in <span class="math-container">$\mathbf{R}^N \setminus B_{2R}(0)$</span> with <span class="math-container">$|\nabla\eta|\leq\frac{C}{R}$</span> for <span class="math-container">$C>0$</span> a positive constant independent of <span class="math-container">$R$</span>.</p>
<p>I know these type of functions can be chosen for any <span class="math-container">$R$</span>, but don't know whether there is a choice of such functions for which the constant is independent of <span class="math-container">$R$</span>, as asked in the question.</p>
<p>Can anyone give a proper explanation to this question?</p>
<p>Thanks...</p>
| Montie | 517,270 | <p>Let $\eta$ be a smooth function on $\mathbb{R}^N$ satisfying $\eta \equiv 1$ in $B_1 (0)$ and $\eta (x) = 0$ for all $x \in \mathbb{R}^N \setminus B_2 (0)$. For $R > 0$, define $\eta_R$ by $\eta_R (x) := \eta (\frac{x}{R})$. Then $\eta_R$ is smooth, supported on the ball of radius $2R$, and we have by the chain rule
\begin{equation*}
\nabla \eta_R (x) = \nabla \eta_R (\frac{x}{R}) = \frac{1}{R} \nabla \eta (x).
\end{equation*}
Thus, we have your desired inequality with $C := \max_{B_2 (0)} |\nabla \eta|$, which does not depend on $R$.</p>
|
1,718,391 | <p>The sequence is </p>
<p>$$\sum_{n=1}^{\infty} \frac{n-1}{n!}$$</p>
<p>I could show that it converges using the Ratio test, but evaluating it seems to be hard. I'm trying to avoid power series and Taylor series and such. Is it possible to solve this some other, simpler way? </p>
<p>I tried writing out the first few terms, but since it's non-geometric, it doesn't seem to help. </p>
| Yiyuan Lee | 104,919 | <p>Hint: Note that</p>
<p>$$\frac{n - 1}{n!} = \frac{1}{(n - 1)!} - \frac{1}{n!}$$</p>
<p>A large amount of cancellations will occur between successive terms in the summation.</p>
|
1,718,391 | <p>The sequence is </p>
<p>$$\sum_{n=1}^{\infty} \frac{n-1}{n!}$$</p>
<p>I could show that it converges using the Ratio test, but evaluating it seems to be hard. I'm trying to avoid power series and Taylor series and such. Is it possible to solve this some other, simpler way? </p>
<p>I tried writing out the first few terms, but since it's non-geometric, it doesn't seem to help. </p>
| hamid kamali | 208,328 | <p>Consider the function $$f(x)=\frac{\mathbb e^x}{x}=\sum_{n=0}^\infty \frac{x^{n-1}}{n!}$$ And use derivativetion we get that: $$\frac{x\mathbb e^x-\mathbb e^x}{x^2}=\sum_{n=0}^\infty \frac{(n-1)x^{n-2}}{n!}$$ Or: $$\frac{x\mathbb e^x-\mathbb e^x}{x^2}+\frac1{x^2}=\sum_{n=1}^\infty \frac{(n-1)x^{n-2}}{n!}$$ Finally: $$1=\sum_{n=1}^\infty \frac{n-1}{n!}$$.</p>
|
72,098 | <p>This is my first time using Stack Exchange, but it looks like a good resource. I am here to ask a couple questions about my homework. We're working from the latest edition of <em>Abstract Algebra</em> by Herstein. This is problem #3 from p. 73 of that book, and it reads as follows:</p>
<blockquote>
<p>Let $G$ be any group and $A(G)$ the set of all 1-1 mappings of $G$, as a set, onto itself. Define $L_{a} \colon G \to G$ by $L_{a}(x)=xa^{-1}$. Prove that:<br>
(a) $L_{a} = A(G)$.<br>
(b) $L_{a}L_{b} = L_{ab}$.<br>
(c) The mapping $\psi:G \rightarrow A(G)$ defined by $\psi(a) = L_{a}$ is a monomorphism of G into A(G).</p>
</blockquote>
<p>I believe that I have answered parts b & c correctly, but have some concerns about the rigorousness of my approaches to all three problems. I will start out here by including my proposed solution to part a. Any advice or comments would be greatly appreciated.</p>
<p>(a) $A(G)$, as the set of all 1-1 mappings of $G$ onto itself, can be represented as the set of all operations on an element $x \in G$ such that the result is also in $G$. Since $G$ is a group, we know that this set can be represented as the set of all group multiplications $yx \mid x \in G, y \in G$ for a given element $x$. This is because any $x$ element can be fixed as the first parameter, while the $y$ elements are taken over every element of G. We have then that $yx \in G$, due to G's closure under group multiplication. Furthermore, any element $e \in G$ has an inverse element $e^{-1} \in G$, since $G$ is a group. This means that we can consider any element in $G$ as the inverse of its inverse: $e = (e^{-1})^{-1}$. This, when plugged in for our variable $x \in G$ above, gives us that $\forall x \in G, \forall y \in G, yx^{-1} \in G$. This is exactly our given mapping $L_{a}$ above, with the labels rearranged. This shows that $L_{a}$ is a mapping from $G \rightarrow G$, as required. In order to show why this set contains every such possible mapping, we will assume that there is a mapping $M_{a}(x) : G \rightarrow G$ such that $M_{a}(x) \notin L_{a}$. This mapping, as a mapping from G to G, must take the form of a group multiplication: $M_{a}(x) = x*a, x\in G$. However, since G is a group, we have again that $a = (a^{-1})^{-1}$, or, if we let $m = a^{-1}$, that $a = m^{-1}$, and so our mapping can be rewritten as $M_{a}(x) = x*m^{-1} m \in G$. However, this is exactly the same mapping as our above $L_{a}(x)$, showing that every mapping in $A(G)$ can indeed be written as a multiplication between some element $x \in G$ and another element's inverse $a^{-1} \in G$, and thus that $L_{a} \in A(G)$. $\blacksquare$</p>
<p>Thanks for taking the time to check this out, even if you don't feel you can offer any help.
<strong>EDIT</strong>: Thanks to those who pointed out that I had used =, not $\in$, above by mistake. Also appreciated is the edit to italicize my variables. Now I know how to as well. :) </p>
<p><strong>EDIT</strong>: The responses so far have been so helpful, I'd like to put the other two parts of my solution up to solicit feedback on them as well. Hopefully they aren't as muddled as the first part's was.<br>
<strong>changed a bit in response to feedback</strong><br>
(b)
$L_{a}(x) = xa^{-1}$<br>
$L_{b}(x) = xb^{-1}$<br>
$(L_{a}L_{b})(x) = L_{a}(L_{b}(x))$<br>
$L_{a}(L_{b}(x)) = L_{a}(xb^-1)$<br>
$L_{a}(xb^-1) = xb^{-1}a^{-1}$<br>
$xb^{-1}a^{-1} = x(ab)^{-1} = L_{ab}(x)$<br>
$L_{a}(x)L_{b}(x) = L_{ab}(x) \quad\blacksquare$ </p>
<p>(c)
$\psi(a) = L_{a}(x) = xa^{-1}$. To show that this mapping is a monomorphism of $G$ into $A(G)$, we will first rely on part (a) to state that $\psi(a) = L_{a}$ is indeed a mapping from $G$ to $A(G)$. Now we must show that $\psi$ is a monomorphism of $G$ into $A(G)$. First we will show that $\psi$ is a homomorphism of $G$ into $A(G)$. To this end, we will appeal to the results of our calculations in part (b) to state that $L_{a}L_{b} = L_{ab}$, which of course implies that $\psi(a)\psi(b) = L_{a}L_{b} = L_{ab} = \psi(ab)$, which proves that $\psi$ is a homomorphism. In order to continue and show that $\psi$ is a monomorphism, we must show that it is an injective (1-1) mapping. Let $\psi(a) = Z = \psi(b)$. This can be written as:
$\psi(a) = L_{a}(e) = ea^{-1} = Z$<br>
$\psi(b) = L_{b}(e) = eb^{-1} = Z$<br>
$Z = ea^{-1} = eb^{-1}$<br>
$a^{-1} = b^{-1} \Rightarrow a = b$<br>
The last line of the above follows from the uniqueness of inverse elements in $G$. This shows that the only way for two output values of $\psi$ to be equal is for their inputs to be equal as well, and thus $\psi$ is an injective homomorphism, or a monomorphism from $G$ to $A(G)$. $\blacksquare$</p>
| Dylan Moreland | 3,701 | <p>To be clear, $A(G)$ is the set of all <a href="http://en.wikipedia.org/wiki/Bijection" rel="nofollow">bijective</a> set maps $G \to G$. This $A(G)$ is even a group under composition, but an element of $A(G)$ doesn't have to be one of the $L_a$ or pay much attention at all to the group structure of $G$. For example, look at the cyclic group $G$ of order $3$, which I'll write as $\{1, x, x^2\}$. I can define a bijective map of sets $G \to G$ by swapping $x$ and $x^2$, fixing $1$. You can check that this is not of the form $L_a$.</p>
<p>Anyway, what you want to show for (a) is that each $L_a$ is bijective, and hence a member of $A(G)$. Perhaps you can find an inverse map: what operation will undo $x \mapsto xa^{-1}$? You shouldn't have to look very far.</p>
<p>And your solution for (b) looks good! I'm worried about (c) only because we seem to be mixing up functions and the formulas defining them. I would write this as: If $L_a$ and $L_b$ agree as maps of sets then in particular $L_a(1) = L_b(1)$, so $a^{-1} = b^{-1}$ and hence $a = b$, using the fact that $(a^{-1})^{-1} = a$. (Or whatever suits you.)</p>
|
2,054,338 | <p>I had this question on my exam and I thought I could solve it using the row echelon method. Well, I couldn't, and I still don't know how to solve it. We were asked to "determine $a$ such the equation system has a unique solution" and to determine $a$ itself. The system had three equations: </p>
<p>$2x - y + az = 3$ </p>
<p>$3x - 4y + 2az = 1$</p>
<p>$x + y - z = 2$.</p>
<p>Could you provide me with a step-by-step of how to solve this?</p>
<p>Thank you</p>
| DeepSea | 101,504 | <p><strong>hint</strong>: $1+x+x^2+\cdots+ x^k+\cdots = \dfrac{1}{1-x}, |x| < 1$. Differentiate both sides and plug $x = 1-a$ into the equation to get the answer.</p>
|
2,411,811 | <p>Determine the limit:</p>
<p>$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}$$</p>
<p>I know that the answer is $0$, I am just unsure of how to solve this. Thanks.</p>
| Claude Leibovici | 82,404 | <p>You have different ways for doing it.</p>
<p>The first one is to write $$\cos(2h)=1-2\sin^2(h)\implies 1-\cos(2h)=2\sin^2(h)$$ $$\frac{1-\cos(2h)} h=\frac{2\sin^2(h)}h=2h\frac{\sin^2(h)}{h^2}=2h \left(\frac{\sin(h)}{h} \right)^2$$</p>
<p>The second one would be Taylor expansion
$$\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right)$$ Make $t=2h$ to get $$\cos(2h)=1-2 h^2+\frac{2 h^4}{3}+O\left(h^6\right)$$
$$\frac{1-\cos(2h)} h=2 h-\frac{2 h^3}{3}+O\left(h^5\right)$$ which shows the limit and how it is approached.</p>
|
1,050,664 | <p>Can someone explain why the inverse $4$ modulo $9$ is $7$? What am I missing?
$$9 = 2\cdot4 + 1$$
$$1 = 9-4\cdot2$$
$$1 = -2\cdot4 + 1\cdot9$$</p>
<p>Isn't then $-2$ inverse of $4$ modulo $9$?</p>
| user141592 | 178,602 | <p>That is correct, but remember that $-2 \equiv 7 \pmod 9$.</p>
|
2,100,793 | <p>Let us say that I have a set $A=\{1,2, 3\}$. Now I need to access, say, the element $3$ of $A$. How do I achieve this?</p>
<p>I know that sets are unordered list of elements but I need to access the elements of a set. Can I achieve this with a tuple? Like $A=(1, 2, 3)$, should I write $A(i)$ to access the i-th element of $A$? Or is there any other notation? </p>
<blockquote>
<p>If I have a list of elements, what is the best mathematical object to represent it so that I can freely access its elements and how? In programming, I would use <code>arrays</code>.</p>
</blockquote>
| Noah Schweber | 28,111 | <p>In a general set, there is no indexing, except the trivial one: a set $A$ is indexed by itself, that is, $A=\{x_a: a\in A\}$ where $x_a=a$.</p>
<p>Indeed, in set theory without the axiom of choice, there can be sets which are impossible to "index nicely," although of course this takes work to be made precise. </p>
<p>So the general task you describe is either trivial (if you allow a set to index itself) or impossible.</p>
<p>That said, in restricted contexts we can do better. For instance, suppose we're looking exclusively at <em>finite sets of real numbers</em>. Then any such set is naturally ordered by cardinality, that is, we may speak of the $k$th <em>smallest</em> element of a set.</p>
|
1,897,163 | <p>Let $\frac{x^2}{y}\to c\neq 0$ as $(x,y)\to (0,0)^+$. Why does this mean that $y$ goes faster to $0$ as $x$ does?</p>
<p>For values of x and y that are small enough, we surely do not have $x^2/y=1$ so they cannot be identical, okay. Moreover, if the quotient would tend to 0, this would mean that $x$ would be faster decaying to 0 than y. </p>
| paf | 333,517 | <p>Let us fix $x\in \Bbb R^*$. We see that
$$f(x)=x\left(\dfrac{1}{1+x^2}\right)^n$$</p>
<p>Since $\left|\dfrac{1}{1+x^2}\right|<1$, we see that $\dfrac{f(x)}{x}$ is a geometric sequence converging to 0. </p>
|
4,232,796 | <p>Consider <span class="math-container">$\sqrt{x^2+y^2}+2\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-6x-8y+25}$</span>. I need to find global or local minima of this function, but Wolfram Alpha doesn't seem to find one; the answer is that <span class="math-container">$1 + 2\sqrt{5}$</span> is its global minimum</p>
<p>Am I doing something wrong? I use <a href="https://www.wolframalpha.com/input/?i=minimum+sqrt%28x%5E2%2By%5E2%29%2B2sqrt%28x%5E2%2By%5E2-2x%2B1%29%2Bsqrt%28x%5E2%2By%5E2-6x-8y%2B25%29" rel="nofollow noreferrer">this</a> input for the global minimum, and <a href="https://www.wolframalpha.com/input/?i=local+minimum+sqrt%28x%5E2%2By%5E2%29%2B2sqrt%28x%5E2%2By%5E2-2x%2B1%29%2Bsqrt%28x%5E2%2By%5E2-6x-8y%2B25%29" rel="nofollow noreferrer">this</a> input for local minima.</p>
| Hamza | 185,441 | <p>Since <span class="math-container">$f$</span> is entire, we deduce that <span class="math-container">$G=f^{(n)}$</span> is entire to. In particular
<span class="math-container">$$
G(\frac{1}{k+1})=0 \qquad k\ge 1.
$$</span>
and hence, <span class="math-container">$G(0)=G(\lim_k\frac{1}{k+1})=\lim_k G(\frac{1}{k+1})= 0$</span>, by Analytic continuation, <span class="math-container">$G(z)=0$</span>.
Since <span class="math-container">$G=f^{(n)}=0$</span>, we deduice that <span class="math-container">$f$</span> is a polynomial of degree at most <span class="math-container">$n-1$</span>.</p>
|
1,836,785 | <p>Following are the two theorems that Hardy and Wright state in their book</p>
<blockquote>
<p><strong>Theorem A:</strong> The number of primes not exceeding <span class="math-container">$x$</span> is given by <span class="math-container">$\pi(x) \sim \frac{x}{\log{x}}$</span>.</p>
<p><strong>Theorem B</strong>: The order of magnitude of <span class="math-container">$\pi(x)$</span> is <span class="math-container">$\pi(x) \asymp \frac{x}{\log{x}}$</span>.</p>
</blockquote>
<p>where,</p>
<ol>
<li><span class="math-container">$f \sim \phi$</span> iff <span class="math-container">$\; \frac{f}{\phi} \to 1\;$</span>, ie. two the two functions are asymptotically similar.</li>
<li><span class="math-container">$f \asymp \phi$</span> iff <span class="math-container">$\;A^{'}\phi < f < A \phi\;$</span> ie. <span class="math-container">$\;f\;$</span> is of same order of magnitude as <span class="math-container">$\phi$</span> ( <span class="math-container">$A$</span> and <span class="math-container">$A^{'}$</span> are constants ).</li>
</ol>
<p>How are the two theorems different ? Am I missing something trivial ?</p>
| tolanj | 349,700 | <p>As a concrete example think of f:x -> (sin x) + 2; and g:x -> 2;
Its clear that f/g does not converge anywhere, ie f !∼ g, but they are of the same order of magnitude as 0.4g < f < 2g.</p>
<p>In general 2. does not imply 1. x and 2x are a trivial example 2x/x is fixed at 2 but they clearly have the same magnitude.</p>
<p>1 does not imply 2 either <em>in the general case</em> (think of a function ϕ that approaches 0 at some place prior to its asymptotic behaviour where f is 1) but must imply it for some region (x,infinity)</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| Alexandre Eremenko | 110,120 | <p>The best advise I can share was given to me by my mother, (she was a researcher in medicine) when I was a first-year student (of mathematics):
find a good adviser and follow his/her advice.</p>
<p>As a beginner, you usually cannot judge yourself about research areas of mathematics, and what to do and what to learn. In all this you should rely on a good adviser, who must be a mathematician with well-established reputation, and a person you feel comfortable working with. So investigate carefully all potential advisers around and choose the best one.
Once you make your choice, follow his/her advises in everything.</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| Community | -1 | <p>Here is my answer in short</p>
<ol>
<li><p>Use YouTube videos .Some of them are good and explain basic content.</p>
</li>
<li><p>The Math Sorcerer channel is an excellent for math advice and<br />
inspiration</p>
</li>
<li><p>George Polya’s Mathematics and Plausible Reasoning vol 1 &2 These are great books to build mathematical skills</p>
</li>
<li><p>Circle around a particular problem to try solving it,instead of a
attacking it head on.</p>
</li>
<li><p>If you can’t solve a problem, go to another one and when you go to bed, think about it and forget it. Let “Deep mind”,
ie,subconscious mind ,find a way to solve it. You might be surprised it finds hints for your conscious mind to solve it</p>
</li>
<li><p>Use this site,it is wealth of info.</p>
</li>
</ol>
|
2,377,139 | <p>My question:</p>
<p>Characterize those power series $\sum_{k=0}^\infty a_{k}(x-c)^{k}$ that converge uniformly on ($-\infty, \,\infty$).
What does it mean to characterize a power series?</p>
<p>Let {$a_{k}$} be a sequence of coefficients for a power series. By definition, the radius of convergence $R = \infty$ if $\limsup\sqrt[k]{|a_{k}|} = 0$. </p>
| Hans Engler | 9,787 | <p>These power series are precisely the polynomials. </p>
<p>The partial sums of a uniformly convergent power series must be a Cauchy sequence in the sup-norm. Let $S_n(x)$ be the $n$-th partial sum. Uniform convergence implies<br>
$$
\forall \epsilon > 0 \, \exists N > 0 \, \ni \, \forall n > N \, \forall x \in \mathbb{R} \,
|S_{n+1}(x) - S_n(x)| = |a_{n+1}(x-c)^{n+1}| < \epsilon
$$
Now this is only possible if $a_{n+1} = 0$. Therefore the condition becomes
$$
\exists N > 0 \, \ni \, \forall n > N \, a_{n+1} = 0
$$
i.e. this is a polynomial. </p>
|
979,947 | <p>If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?</p>
<p>UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$</p>
<p>My further work: </p>
<p>Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$</p>
<p>Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$</p>
<p>Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.</p>
<p>Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.</p>
<p>Case $1$: $k<C$</p>
<p>Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$ </p>
<p>By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$</p>
<p>This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.</p>
<p>If $k=1$ we have $A+B=C$ which violates the condition $A>C$.</p>
<p>If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$</p>
<p>However, this leads to the result that $A=B$ which violates that stated conditions.</p>
<p>Case $1$, $k<C$, fails.</p>
<p>Case $2$: $k=C$</p>
<p>Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$</p>
<p>This later equation is clearly false and thus Case $2$, $k=C$ fails.</p>
<p>Case $3$: $k>C$</p>
<p>Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.</p>
<p>Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.</p>
<p>Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$ </p>
<p>Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$</p>
<p>By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$</p>
<p>Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.</p>
<p>By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$</p>
<p>and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$</p>
<p>From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.</p>
<p>Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$</p>
<p>Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$</p>
<p>Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.</p>
<p>Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.</p>
<p>Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.</p>
| Nameless | 68,482 | <p>If points are awarded as you say, then indeed the expected points you can gain from choosing one of the five answers randomly is
$$Ep=1/5*1-4/5*1/4=4/20-4/20=0.$$
However, you are missing the fact that this happens only in expectation, so it is <em>not</em> the case that</p>
<blockquote>
<p>guessing (blind or otherwise) is <em>always</em> at least as good as leaving it blank.</p>
</blockquote>
<p>You can put it another way: in 4 out of 5 cases guessing is worse than leaving it blank. Though fair enough, in 1 out of 5 you are better off.</p>
<p>Another point: people are usually risk averse, that is, would rather have 0 for sure than 0 in expectation, because the latter implies you sometimes get a negative score. But this is a matter of your preferences: if you are generally a gambler, then you might as well guess and gamble that you got the right answer.</p>
<p>Finally, as noted in the comments, as soon as you can reasonably rule out some options or can at least order the answers with respect to likelihood of being correct, then you should guess (but not blindly, i.e., pick the answer you think is most likely).</p>
|
2,146,571 | <p>I solved this problem in my textbook but noticed their solution was different than mine. <br/></p>
<p>$1. \ 9e^{-2x}=1$ </p>
<p>$2. \ e^{-2x}=\frac{1}{9}$</p>
<p>$3. -2x=\ln(\frac{1}{9})$</p>
<p>$4. \ x=-\ \frac{1}{2}\ln(\frac{1}{9})$</p>
<p>However, the answer that my textbook gives is $\frac{\ln(9)}{2}$ </p>
<p>I plugged these expressions into my calculator and they are indeed equivalent, however I don't see what properties I could use to get from my messy answer to the textbook's much cleaner one. Any help would be greatly appreciated. Thank you.</p>
| Hushus46 | 240,398 | <p>There exists the following property for logarithms:</p>
<p>$$n \ln{x} = \ln{x^n}$$</p>
<p>So for your problem you have:</p>
<p>$$ -\frac{1}{2} \ln{\left(\frac{1}{9}\right)}=\frac{1}{2}\ln{\left(\left(\frac{1}{9}\right)^{-1}\right)}=\frac{1}{2}\ln{9}= \frac{\ln9}{2}$$</p>
<p>I hope this is sufficient as an explanation.</p>
|
718,609 | <p>This theorem is the converse of Wilson's theorem:</p>
<blockquote>
<p>If $n$ is composite and $n>4$, then $(n-1)! \equiv 0 \pmod n$</p>
</blockquote>
<p>The question holds up for all the composites I have tried but I'm struggling to form a proof for all composites greater than $4$.</p>
| Olivier Bégassat | 11,258 | <p>If $n$ is composite, then there are $1<a<b<n$ with $ab=n$, unless $n=p^2$ is the square of a prime. In the first case, $a$ and $b$ appear in the product $(n-1)!$ thus $n|(n-1)!$. In the second case, your hypothesis $n>4$ implies $p\geq 3$ and thus $1<p<2p<n-1$ so that $2n=p\times 2p$ divides $(n-1)!$.</p>
|
718,609 | <p>This theorem is the converse of Wilson's theorem:</p>
<blockquote>
<p>If $n$ is composite and $n>4$, then $(n-1)! \equiv 0 \pmod n$</p>
</blockquote>
<p>The question holds up for all the composites I have tried but I'm struggling to form a proof for all composites greater than $4$.</p>
| ajotatxe | 132,456 | <p>If $n=pq$, try to prove that $pq$ divides $(n-1)!$</p>
|
631,348 | <p>Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?</p>
<p>P.S. using Kummer's test doesn't count</p>
| user127.0.0.1 | 50,800 | <p>The series even converges <em>absoute</em>, thus set $a_n:=\left|a_n\right|$</p>
<p>Now (if $n$ is big enough)
$$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right) \Longleftrightarrow \frac{a_{n+1}}{a_n}\leq\frac{n-L}{n}$$</p>
<p>This is equivalent to
$$\left(L-1\right)a_n \leq \left(n-1\right)a_n-na_{n+1}$$</p>
<p>Because of $L>1$ the left side is bigger than zero, so</p>
<p>$$0\leq \left(n-1\right)a_n-na_{n+1}\Longleftrightarrow na_{n+1}\leq \left(n-1\right)a_n$$</p>
<p>That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum</p>
<p>$$\sum b_n = \sum \left(n-1\right)a_n-na_{n+1}$$</p>
<p>Which is a telescop-sum and thus it converges</p>
<p>But that implies that the sum over $a_n$
$$\sum a_n \leq \sum \left(L-1\right)a_n \leq \sum b_n$$
is convergent as well by the comparison test.</p>
|
2,226,386 | <p>Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?</p>
| Doug M | 317,162 | <p>Here is how I approached it.</p>
<p>Translate the whole thing such that the vertex opposite M is the origin.</p>
<p>represent the other vertexes of the triangle with complex numbers $2a,$ and $2b$</p>
<p>Point $M$ is then $a+b$</p>
<p>The center of one square is at $a(1-i)$
and the other is at $b(1+i)$</p>
<p>We can represent the two lines with: $a(1-i) - (a+b) = -b-ia$ and $b(1+i) - (a+b) = a+ib$</p>
<p>if they are equal then $|-b-ia| = |a+ib|$
and if they are perpendicular $Re[(-b-ia)(a+ib)] = 0$</p>
|
2,553,470 | <p>The equations of the two chords of the parabola y^2=4ax are to be found such that the pass through the point (-6a,0) and subtends an angle of 45 degree at the vertex of the parabolas. Tried it in many ways including using parametric equations but could not get the two equations</p>
<p>(note. Let the chord intersect the parabola at points p and q. And let the vertex be point r then prq..=45 degree) </p>
| Shreyas Pimpalgaonkar | 477,919 | <p>You shouldn't complicate things at this level.</p>
<p>What you are trying to accomplish is the exact definition of the binomial coefficient.</p>
<p>It is the number of ways of choosing n places from a total of m places.</p>
<p>In this case, choose 4 places from 8 places, and for every such selection of places permuting A, B, C, D in those places is exactly 1, as you want it in alphabetical order.</p>
<p>So there you have it, the answer is equal to mCn</p>
|
512,960 | <p>Let's say I have a statement: if p then q.<br>
The converse would be: if q then p.<br>
The inverse would be: if not p then not q.<br>
The contraposition would be: if not q then not p.<br>
What would you call the following? if not p then q.<br>
Thanks.</p>
| Glen O | 67,842 | <p>There's no name for it, because there's no real connection between them. The inverse and converse exist because they still assert a direct correlation between p and q; it's just that, as opposed to the regular and contrapositive forms, the condition for failure is reversed.</p>
<p>On the other hand, "if not p then q" is a completely different assertion.</p>
|
245,789 | <p>Let $f$ be a function defined on the interval $(−1, 1)$ such that for all $x, y \in (−1, 1)$, $$f(x + y) = \frac{{f(x) + f(y)}}{{1 - f(x)f(y)}}.$$ Suppose that $f$ is differentiable at $x = 0$, show that $f$ is differentiable on $(−1, 1)$.</p>
<p>Need your kind guidance and help. Thanks in advance!</p>
| DonAntonio | 31,254 | <p>Since $\,f(0)=0\,$, we get:</p>
<p>$$f'(0)=\lim_{x\to 0}\frac{f(x)}{x}\Longrightarrow \,\forall w\in (-1,1)\setminus\{0\}:$$</p>
<p>$$f'(w)=\lim_{h\to 0}\frac{f(w+h)-f(w)}{h}=\lim_{h\to 0}\frac{\frac{f(w)+f(h)}{1-f(w)f(h)}-f(w)}{h}=$$</p>
<p>$$=\lim_{h\to 0}\frac{\left(f(w)^2+1\right)f(h)}{h\left[1-f(w)f(h\right]}=\lim_{h\to 0}\frac{f(h)}{h}\cdot\lim_{h\to 0}\frac{f(w)^2+1}{1-f(w)f(h)}$$</p>
<p>where the last step is justified since both limits on the right hand side exist finitely...</p>
|
947 | <p>I'm looking for the algorithm that efficiently locates the "loneliest person on the planet", where "loneliest" is defined as:</p>
<p>Maximum minimum distance to another person — that is, the person for whom the closest other person is farthest away.</p>
<p>Assume a (admittedly miraculous) input of the list of the exact latitude/longitude of every person on Earth at a particular time.</p>
<p>Also take as provided a function <span class="math-container">$d(p_1, p_2)$</span> that returns the distance on the surface of the earth between <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> - I know this is not trivial, but it's "just spherical geometry" and not the important (to me) part of the question.</p>
<p>What's the most efficient way to find the loneliest person?</p>
<p>Certainly one solution is to calculate <span class="math-container">$d(\ldots)$</span> for every pair of people on the globe, then sort every person's list of distances in ascending order, take the first item from every list and sort those in descending order and take the largest. But that involves <span class="math-container">$n(n-1)$</span> invocations of <span class="math-container">$d(\ldots)$</span>, <span class="math-container">$n$</span> sorts of <span class="math-container">$n-1$</span> items and one last sort of <span class="math-container">$n$</span> items. Last I checked, <span class="math-container">$n$</span> in this case is somewhere north of six billion, right? So we can do better?</p>
| Reid Barton | 126,667 | <p>The paper <em>Vaidya, Pravin M.</em>, <a href="https://doi.org/10.1007/BF02187718" rel="noreferrer"><strong>An <span class="math-container">$O(n \log n)$</span> algorithm for the all-nearest-neighbors problem</strong></a>, Discrete Comput. Geom. 4, No. 2, 101-115 (1989), <a href="https://zbmath.org/?q=an:0663.68058" rel="noreferrer">ZBL0663.68058</a> gives an <span class="math-container">$O(n \log n)$</span> algorithm for the "all-nearest-neighbors" problem: given a set of points <span class="math-container">$S$</span>, find all the values <span class="math-container">$m(p)$</span> where <span class="math-container">$p$</span> is a point of <span class="math-container">$S$</span> and <span class="math-container">$m(p)$</span> is the minimum distance from <span class="math-container">$p$</span> to a point of <span class="math-container">$S \setminus \{p\}$</span>. Then the "loneliest point" is the point <span class="math-container">$p$</span> which maximizes <span class="math-container">$m(p)$</span>. So your problem can be solved in <span class="math-container">$O(n \log n)$</span> time, which is pretty good.</p>
<p>(In case it's not clear, I'm applying their algorithm to the set of points viewed as living inside <span class="math-container">$\mathbb{R}^3$</span>, using the fact that there's an order-preserving relationship between distance along the sphere and straight-line distance in <span class="math-container">$\mathbb{R}^3$</span>.)</p>
|
947 | <p>I'm looking for the algorithm that efficiently locates the "loneliest person on the planet", where "loneliest" is defined as:</p>
<p>Maximum minimum distance to another person — that is, the person for whom the closest other person is farthest away.</p>
<p>Assume a (admittedly miraculous) input of the list of the exact latitude/longitude of every person on Earth at a particular time.</p>
<p>Also take as provided a function <span class="math-container">$d(p_1, p_2)$</span> that returns the distance on the surface of the earth between <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> - I know this is not trivial, but it's "just spherical geometry" and not the important (to me) part of the question.</p>
<p>What's the most efficient way to find the loneliest person?</p>
<p>Certainly one solution is to calculate <span class="math-container">$d(\ldots)$</span> for every pair of people on the globe, then sort every person's list of distances in ascending order, take the first item from every list and sort those in descending order and take the largest. But that involves <span class="math-container">$n(n-1)$</span> invocations of <span class="math-container">$d(\ldots)$</span>, <span class="math-container">$n$</span> sorts of <span class="math-container">$n-1$</span> items and one last sort of <span class="math-container">$n$</span> items. Last I checked, <span class="math-container">$n$</span> in this case is somewhere north of six billion, right? So we can do better?</p>
| bhwang | 239 | <p>This is called the "largest empty circle problem" in Computational Geometry, and has an <span class="math-container">$O(n \log n)$</span> solution provided that you are given the convex hull of the points and the corresponding Voronoi diagrams. Here is a very readable paper on the problem:</p>
<blockquote>
<p>Schuster, M. (2008). <a href="https://www.cs.swarthmore.edu/%7Eadanner/cs97/s08/papers/schuster.pdf" rel="nofollow noreferrer">The largest empty circle problem</a>. In: Proceedings of the Class of 2008 Senior Conference, Computer Science Department, Swarthmore College (pp. 28-37).</p>
</blockquote>
|
2,198,293 | <p>Having a bit of trouble in what way I am suppose to go about solving this problem. Any guidance would be great.</p>
<p>Show that there is at least one real solution to:$$x^5 - x^2 - 4 = 0 $$</p>
<p>Thanks in advance.</p>
| Jaideep Khare | 421,580 | <p>Since at $x \rightarrow \infty $ function tends to $\infty$ and at $x \rightarrow -\infty$ function tends to $-\infty$ . Hence it will cut $x$-axis at some point for sure since function is well-continuous.This is true for all polynomials with odd degree.</p>
|
3,790,932 | <p>If <span class="math-container">$f : \bar{\mathbb{Q}} \to \mathbb{Q}$</span> is a continuous function, where <span class="math-container">$\bar{\mathbb{Q}}$</span> denotes the set of algebraic numbers, does such function have to be constant?</p>
| Jyrki Lahtonen | 11,619 | <p>The claim is actually false. Proffering the following counterexample
<span class="math-container">$$f(x)=\begin{cases}1,&\ \text{if the real part of $x$ is greater than $\pi$, and}\\
0,&\ \text{otherwise.}\end{cases}$$</span></p>
|
3,178,648 | <blockquote>
<p>We assign to every element <span class="math-container">$i$</span> from <span class="math-container">$N=\{1,2,...,n\}$</span> a positive integer <span class="math-container">$a_i$</span>. Suppose <span class="math-container">$$a_1+a_2+...+a_n = 2n-2$$</span> then prove that map <span class="math-container">$T: \mathcal{P}(N) \to \{1,2,...,2n-2\}$</span> defined with <span class="math-container">$$T(X) = \sum _{i\in X}a_i$$</span> is surjective. </p>
</blockquote>
<hr>
<p>We can assume that <span class="math-container">$a_1\leq a_2\leq ...\leq a_n$</span>. </p>
<p>Clearly, <span class="math-container">$a_1 = a_2 = 1$</span> and thus <span class="math-container">$1,2,2n-3,2n-4$</span> are in a range. </p>
<p>Also, if <span class="math-container">$a_i=2$</span> for some <span class="math-container">$i$</span> then we could easily apply induction. </p>
<p>Say <span class="math-container">$b_1< b_2<...<b_k$</span> are all different values that appear among <span class="math-container">$a_i$</span>. </p>
<p>Then we have <span class="math-container">$n _1\cdot b_1+n_2\cdot b_2+...+n_k \cdot b_k = 2n-2$</span> and <span class="math-container">$n_1+n_2+..+n_k = n$</span>. We have to prove that for each <span class="math-container">$l\leq 2n-2$</span> we have <span class="math-container">$$n' _1\cdot b_1+n'_2\cdot b_2+...+n'_k \cdot b_k = l$$</span></p>
<p>for some <span class="math-container">$n'_i\leq n_i$</span>. And here it stops. I have no idea how to find all those <span class="math-container">$n_i'$</span>. Any ideas?</p>
| Calvin Lin | 54,563 | <p>Here's a direct solution (which I believe gets to the heart of the problem). (Sorry, I just noticed that OP requested a solution based on graph theory in the bounty. I ignored that.)</p>
<p>In <span class="math-container">$ \{ a_i \}$</span>, let there be <span class="math-container">$k$</span> 1's, and let <span class="math-container">$a_n$</span> be the largest value.</p>
<p><strong>Claim 1:</strong> <span class="math-container">$k \geq a_n$</span>. </p>
<p>If <span class="math-container">$ k = n-1$</span>, then <span class="math-container">$a_n = (2n-2) - (n-1) = n-1$</span> so <span class="math-container">$k \geq a_n$</span> as desired.<br>
If <span class="math-container">$ k < n-1$</span>, then <span class="math-container">$ 2n-2 = \sum a_i \geq k + 2 (n-k-1) + a_n \Rightarrow k \geq a_n$</span> as desired.</p>
<p><strong>Claim 2:</strong> For any <span class="math-container">$ 1 \leq i \leq 2n-1 $</span>, we can find a subset of <span class="math-container">$\{a_i\}$</span> that sums to these values. </p>
<p>If <span class="math-container">$ i \leq 2n-2 - k $</span>, then ignore the <span class="math-container">$k$</span> values of 1 for now. Take the largest possible subset that gives a sum <span class="math-container">$ \leq i $</span>. Note that the difference between <span class="math-container">$i$</span> and this sum is strictly less than any non-1 element, hence strictly less than <span class="math-container">$k$</span> from claim 1. As such, we can use additional 1's to get the sum to <span class="math-container">$i$</span>.<br>
If <span class="math-container">$i > 2n-2 - k$</span>, then take all the non-1 values, and enough 1 values.</p>
<hr>
<p>Bonus: Prove that the statement still holds true if <span class="math-container">$\sum a_i = 2n-1$</span>. Use the same argument (modifying the first claim to <span class="math-container">$ k \geq a_n - 1$</span>. The second claim is fine as is). </p>
<p>Bonus: Prove that the statement still holds true if <span class="math-container">$\sum a_i \in [n+1, 2n-1]$</span> using a similar argument.</p>
|
3,178,648 | <blockquote>
<p>We assign to every element <span class="math-container">$i$</span> from <span class="math-container">$N=\{1,2,...,n\}$</span> a positive integer <span class="math-container">$a_i$</span>. Suppose <span class="math-container">$$a_1+a_2+...+a_n = 2n-2$$</span> then prove that map <span class="math-container">$T: \mathcal{P}(N) \to \{1,2,...,2n-2\}$</span> defined with <span class="math-container">$$T(X) = \sum _{i\in X}a_i$$</span> is surjective. </p>
</blockquote>
<hr>
<p>We can assume that <span class="math-container">$a_1\leq a_2\leq ...\leq a_n$</span>. </p>
<p>Clearly, <span class="math-container">$a_1 = a_2 = 1$</span> and thus <span class="math-container">$1,2,2n-3,2n-4$</span> are in a range. </p>
<p>Also, if <span class="math-container">$a_i=2$</span> for some <span class="math-container">$i$</span> then we could easily apply induction. </p>
<p>Say <span class="math-container">$b_1< b_2<...<b_k$</span> are all different values that appear among <span class="math-container">$a_i$</span>. </p>
<p>Then we have <span class="math-container">$n _1\cdot b_1+n_2\cdot b_2+...+n_k \cdot b_k = 2n-2$</span> and <span class="math-container">$n_1+n_2+..+n_k = n$</span>. We have to prove that for each <span class="math-container">$l\leq 2n-2$</span> we have <span class="math-container">$$n' _1\cdot b_1+n'_2\cdot b_2+...+n'_k \cdot b_k = l$$</span></p>
<p>for some <span class="math-container">$n'_i\leq n_i$</span>. And here it stops. I have no idea how to find all those <span class="math-container">$n_i'$</span>. Any ideas?</p>
| PkT | 435,267 | <p>Here's a solution based on graph theory. The idea is to remove a "leaf" from the tree and proceed by induction. </p>
<p>Assuming <span class="math-container">$A_n = \{a_1, \dots, a_n\}$</span>, <span class="math-container">$a_1 = 1$</span> and <span class="math-container">$a_2 > 1$</span>, consider the set <span class="math-container">$A_{n-1} = \{a_2-1, a_3, \dots a_n\}$</span> which has <span class="math-container">$n-1$</span> elements whose sum is <span class="math-container">$2n-4$</span> (i.e. remove a leaf). From the induction hypothesis the map <span class="math-container">$T^{(n-1)}$</span> covers <span class="math-container">$\{1, \dots, 2n-4\}$</span>. Now for <span class="math-container">$i \in \{1, \dots , 2n-4\}$</span>, there is some <span class="math-container">$X \subset A_{n-1}$</span> such that <span class="math-container">$T^{(n-1)}(X) = i$</span>.</p>
<ul>
<li>If <span class="math-container">$a_2-1 \in X$</span> , define <span class="math-container">$Y :=(X - \{a_2-1\}) \cup \{a_2\}\subset A_n$</span> so that <span class="math-container">$T^{(n)}(Y) = i+1$</span>.</li>
<li>If <span class="math-container">$a_2-1 \notin X$</span>, define <span class="math-container">$Y :=X \cup \{a_1\} \subset A_n$</span> so that <span class="math-container">$T^{(n)}(Y) = i+1$</span>.</li>
</ul>
<p>So <span class="math-container">$T^{(n)}$</span> covers <span class="math-container">$2, \dots, 2n-3$</span>. But clearly it also covers <span class="math-container">$1 (=a_1)$</span> and <span class="math-container">$2n-2 (=\sum_{a\in A_n} a)$</span>, so we're done.</p>
|
1,842,781 | <p>Suppose we have an alphabet of the following allowed characters: </p>
<ul>
<li>the lowercase letters $a$ through $z$ (26)</li>
<li>the uppercase letters $A$ through $Z$ (26)</li>
<li>the numerals $0$ through $9$ (10)</li>
<li>the common punctuation marks: $.,;:'"()!?-$ ( 11)</li>
<li>the white space (1)</li>
</ul>
<p>Making a total of 74 options. </p>
<p>Each character $c$ in our 10-character sequence can be a member of the set of the allowed characters, $C$. </p>
<p>Suppose we have a given ten-character sequence, $S$. (For the sake of example, let $S = Abc123.,!$)</p>
<p>What is the probability that a randomly constructed sequence, $S'$, (with each character being randomly selected from $C$) matches $S$?</p>
| Théophile | 26,091 | <p>Starting off with @heropup's approach, we can speed up the calculations by appealing to symmetry:</p>
<p>\begin{array}{ccccc} & a & & b & \\ 0 & & c & & 1000 \\ & d & & e & \end{array}</p>
<p>Observe that we must have $a=d, b=e$, and $a = 1000-b$. This eliminates all variables but $b$ and $c$ (we might as well work with $b$ rather than $a$, since the question asks for the greatest among the five values, which will clearly be $b$), so the system of equations is:</p>
<p>\begin{align*} 3b &= (1000-b) + c + 1000 \\ 6c &= 2(1000-b) + 2b + 1000. \end{align*}</p>
<p>That is,</p>
<p>\begin{align*} 4b &= 2000 + c \\ 6c &= 3000. \end{align*}</p>
<p>This gives $c=500$, from which we see that $b=625$.</p>
|
4,081,029 | <p>I've been reading <a href="https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec03.pdf" rel="nofollow noreferrer">this pdf</a> about vector transformations and I don't quite understand how to implement it in a computer program. On page 10, it shows you how to transform a vector from one coordinate system to another, exactly how would this look on paper? Say I had two 3d vectors that represented directions and I wanted to get the relative direction of <span class="math-container">$vector A$</span> to <span class="math-container">$vector B$</span>. So if <span class="math-container">$A = (0,0,0)$</span> and <span class="math-container">$B = (0,1,0)$</span>, <span class="math-container">$C$</span> would also be <span class="math-container">$(0,1,0)$</span>. If <span class="math-container">$A = (0,1,0)$</span>, <span class="math-container">$C$</span> would then be <span class="math-container">$(0,0,-1)$</span>. Would I make matrices of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and just multiply them together? Other places have talked about getting the dot product? What would the <span class="math-container">$i$</span> and <span class="math-container">$i'$</span> vectors be in this case? Thank you.</p>
<p>I'm pretty sure this is all the info I need but in case there is an obviously better way of doing it I will explain my project. It's a simple voxel 3D raycaster. I'm actually using Unity and rendering to a texture using a compute shader.</p>
<p>The way I am calculating the rays is I am centering the pixel coordinates as if they were 3D space coordinates, moving them a little forward and getting the directions from 0,0,0 to each of those points (and shrinking the dimensions for field-of-view). Then, I transform those directions relative to the player's transform every frame to get the ray directions for the GPU. Obviously, this also needs to be calculated on the GPU so I can't use the handy Unity method I used just to test if it would work (Transform.TransformDirection). So I guess I could also use a relative point transform function too and just send those points and the player's transform to the GPU.</p>
| Glärbo | 907,918 | <p>Voxel tracing differs from ordinary ray tracing in that you want to use a coordinate system where the voxels are defined for each unit cell, i.e. for unit (integer) coordinates.</p>
<p>You can use one vector, say <span class="math-container">$\vec{o}$</span> for observer (eye) location (in the voxel coordinates), and an unit quaternion <span class="math-container">$\mathbf{q} = q_r + q_x \mathbf{i} + q_y \mathbf{j} + q_z \mathbf{k} = (q_x, q_y, q_z; q_r)$</span> to describe the observer orientation; it corresponds to rotation matrix <span class="math-container">$\mathbf{R}$</span> via
<span class="math-container">$$\mathbf{R} = \left[ \begin{matrix}
1-2( q_y^2 + q_z^2 ) & 2( q_x q_y- q_z q_r) & 2( q_x q_z+ q_y q_r) \\
2( q_x q_y+ q_z q_r) & 1-2( q_x^2 + q_z^2 ) & 2( q_y q_z- q_x q_r) \\
2( q_x q_z- q_y q_r) & 2( q_y q_z+ q_x q_r) & 1-2( q_x^2 + q_y^2 ) \\
\end{matrix} \right] = \left[ \begin{matrix}
u_x & v_x & w_x \\ u_y & v_y & w_y \\ u_z & v_z & w_z \\
\end{matrix} \right] = \left[ \begin{matrix} \hat{u} & \hat{v} & \hat{w} \end{matrix} \right]$$</span>
where <span class="math-container">$\hat{w}$</span> is the "forward" unit vector in voxel coordinates, <span class="math-container">$\hat{v}$</span> is "up", and <span class="math-container">$\hat{u}$</span> is "right". For the identity unit quaternion <span class="math-container">$\mathbf{q} = 1 = (0, 0, 0; 1)$</span>, <span class="math-container">$\hat{u} = (1, 0, 0)$</span>, <span class="math-container">$\hat{v} = (0, 1, 0)$</span>, and <span class="math-container">$\hat{w} = (0, 0, 1)$</span>.</p>
<p>The reason for using a quaternion (or a bivector, which yields the same math) is that the orientation can be safely normalized by dividing each of the four components by <span class="math-container">$\sqrt{q_r^2 + q_x^2 + q_y^2 + q_z^2}$</span>, because for unit quaternions, <span class="math-container">$q_r^2 + q_x^2 + q_y^2 + q_z^2 = 1$</span>. To apply a rotation <span class="math-container">$\mathbf{t}$</span> to the orientation, you use Hamilton product:
<span class="math-container">$$\left\lbrace\begin{aligned}
q_r^\prime &= t_r q_r - t_x q_x - t_y q_y - t_z q_z \\
q_x^\prime &= t_r q_x + t_x q_r + t_y q_z - t_z q_y \\
q_y^\prime &= t_r q_y - t_x q_z + t_y q_r + t_z q_x \\
q_z^\prime &= t_r q_z + t_x q_y - t_y q_x + t_z q_r \\
\end{aligned} \right.$$</span>
If the projection plane (view in voxel coordinates) is <span class="math-container">$2 \chi$</span> wide and <span class="math-container">$2 \gamma$</span> tall, at distance <span class="math-container">$d$</span> from the observer, each ray <span class="math-container">$(i, j)$</span> (<span class="math-container">$0 \le i \le i_\max$</span>, <span class="math-container">$0 \le j \le j_\max$</span>) has unnormalized ray direction vector <span class="math-container">$\vec{s}$</span>,
<span class="math-container">$$\vec{s} = d \hat{w} + \left(\frac{i}{i_\max} - \frac{1}{2}\right) \chi \hat{u} + \left(\frac{j}{j_\max} - \frac{1}{2}\right) \gamma \hat{v}$$</span>
and starts at <span class="math-container">$\vec{p}$</span>,
<span class="math-container">$$\vec{p} = \vec{o} + \vec{s}$$</span>
The interesting thing is that if you calculate the unit ray vector <span class="math-container">$\hat{r}$</span>,
<span class="math-container">$$\hat{r} = \frac{\vec{s}}{\lVert\vec{s}\rVert} = \left[\begin{matrix} r_x \\ r_y \\ r_z \end{matrix} \right]$$</span>
its reciprocal components tell you the interval at which the ray steps one voxel unit along each axis,
<span class="math-container">$$\left\lbrace \begin{aligned}
L_x &= \frac{1}{\lvert r_x \rvert}, \\
L_y &= \frac{1}{\lvert r_y \rvert}, \\
L_z &= \frac{1}{\lvert r_z \rvert}, \\
\end{aligned} \right. \quad \begin{aligned}
S_x &= \operatorname{sign}(r_x) = \frac{r_x}{\lvert r_x \rvert} \\
S_y &= \operatorname{sign}(r_y) = \frac{r_y}{\lvert r_y \rvert} \\
S_z &= \operatorname{sign}(r_z) = \frac{r_z}{\lvert r_z \rvert} \\
\end{aligned}$$</span>
where <span class="math-container">$S_x$</span>, <span class="math-container">$S_y$</span>, <span class="math-container">$S_z$</span> are <span class="math-container">$-1$</span>, <span class="math-container">$0$</span>, or <span class="math-container">$+1$</span> depending on the direction of the ray, and division by zero yields either <span class="math-container">$\infty$</span> or a value larger than the voxel map size.</p>
<p>The initial distance <span class="math-container">$d_x$</span> where the ray intersects a voxel <span class="math-container">$x$</span> face, is
<span class="math-container">$$d_x = \begin{cases}
( p_x - \lfloor p_x \rfloor ) L_x, & S_x = -1 \\
\infty, & S_x = 0 \\
( \lceil p_x \rceil - p_x ) L_x, & S_x = +1 \\
\end{cases}$$</span>
<span class="math-container">$d_x \ge 0$</span>; and similarly for the <span class="math-container">$y$</span> and <span class="math-container">$z$</span> axes.</p>
<p>Instead of progressing the ray at fixed length steps, you examine where the ray next intersects a voxel face. You do this by choosing the smallest of <span class="math-container">$d_x$</span>, <span class="math-container">$d_y$</span>, and <span class="math-container">$d_z$</span>.</p>
<p>If <span class="math-container">$d_x$</span> is the smallest of the three, the ray length <span class="math-container">$d$</span> is set to <span class="math-container">$d_x$</span>, then <span class="math-container">$d_x$</span> increased by <span class="math-container">$L_x$</span> (corresponding to the distance where the ray intersects the next <span class="math-container">$x$</span> face), and the ray intersects a voxel <span class="math-container">$x$</span> face at <span class="math-container">$\vec{p} + d \hat{r}$</span>. Similarly for the other two coordinates.</p>
<p>You may need to handle the rare cases <span class="math-container">$d_x = d_y$</span>, <span class="math-container">$d_x = d_z$</span>, etc. where the ray intersects a voxel edge, and <span class="math-container">$d_x = d_y = d_z$</span> where the ray intersects a voxel vertex, separately.</p>
<p>The voxel is identified by the integer part of coordinates of <span class="math-container">$\vec{p} + d \hat{r}$</span>. The fractional parts correspond to the fractional coordinates within the voxel, remembering that the above method only examines voxel <em>faces</em>; at least one of the coordinates is always an integer. This allows you to use textures for each voxel face, if you want.</p>
<hr />
<p>Consider the following 2D scenario for a single ray starting at lower left, progressing up right:
<img src="https://i.stack.imgur.com/RgX3e.png" alt="2D lattice wall intersections" />
The distance along the ray between the blue tabs is <span class="math-container">$L_x$</span>, and between the red tabs is <span class="math-container">$L_y$</span>. We walk in increasing distance.</p>
<p>If we call vertical/blue faces <span class="math-container">$x$</span> (because they occur at integer <span class="math-container">$x$</span> coordinates), and horizontal/red faces <span class="math-container">$y$</span>, the order in which the ray encounters these faces is <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$x$</span>, <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$x$</span>, <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$x$</span>, <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, and so on.</p>
<hr />
<p>Nothing beats an example, I guess. Here is a horrible-crude-simple tracer in Python 3:</p>
<pre><code># SPDX-License-Identifier: CC0-1.0
# -*- coding: utf-8 -*-
from math import inf as _inf, sqrt as _sqrt, copysign as _copysign, floor as _floor
class Vector(tuple):
"""Immutable, compact 3D vector"""
def __new__(cls, *args):
"""Create a new 3D vector
Usage:
v = Vector(x, y, z)
v = Vector((x, y, z))
v = Vector([x, y, z])
v = Vector(v)
"""
if len(args) == 1:
if isinstance(args[0], (tuple, list)):
args = args[0]
else:
raise TypeError("Cannot construct a Vector from a %s." % str(type(args[0])))
if len(args) != 3:
raise ValueError("Vector() needs three components; %d given." % len(args))
return tuple.__new__(cls, (float(args[0]), float(args[1]), float(args[2])))
def __init__(self, *args):
"""Vectors are immutable"""
pass
def __str__(self):
return "(%.6f, %.6f, %.6f)" % self
@property
def x(self):
"""X coordinate"""
return self[0]
@property
def y(self):
"""Y coordinate"""
return self[1]
@property
def z(self):
"""Z coordinate"""
return self[2]
@property
def normsqr(self):
"""Squared norm, i.e. Euclidean length squared"""
return self[0]*self[0] + self[1]*self[1] + self[2]*self[2]
@property
def length(self):
"""Euclidean length"""
return _sqrt(self[0]*self[0] + self[1]*self[1] + self[2]*self[2])
@property
def normalized(self):
"""Vector normalized to unit length"""
n = _sqrt(self[0]*self[0] + self[1]*self[1] + self[2]*self[2])
if n == 1:
return self
elif n > 0:
return tuple.__new__(self.__class__, (self[0]/n, self[1]/n, self[2]/n))
else:
raise ValueError("Cannot normalize a zero vector to unit length!")
def dot(self, other):
"""Vector dot product"""
if not isinstance(other, Vector):
other = Vector(other)
return self[0]*other[0] + self[1]*other[1] + self[2]*other[2]
def cross(self, other):
"""Vector cross product"""
if not isinstance(other, Vector):
other = Vector(other)
return tuple.__new__(self.__class__, ( self[1]*other[2] - self[2]*other[1],
self[2]*other[0] - self[0]*other[2],
self[0]*other[1] - self[1]*other[0] ))
def proj(self, other):
"""Vector projection to another vector"""
if not isinstance(other, Vector):
other = Vector(other)
n = ( self[0]*other[0] +
self[1]*other[1] +
self[2]*other[2] ) / ( other[0]*other[0] +
other[1]*other[1] +
other[2]*other[2] )
return tuple.__new__(self.__class__, ( n*other[0], n*other[1], n*other[2] ))
def perp(self, other):
"""The part of current vector perpendicular to the other vector"""
unit = Vector(other).normalized
return self - unit*self.dot(unit)
def __add__(self, other):
"""Vector addition"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( self[0] + float(other[0]),
self[1] + float(other[1]),
self[2] + float(other[2]) ))
else:
return NotImplemented
def __iadd__(self, other):
"""Vector addition"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( self[0] + float(other[0]),
self[1] + float(other[1]),
self[2] + float(other[2]) ))
else:
return NotImplemented
def __radd__(self, other):
"""Vector addition"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( float(other[0]) + self[0],
float(other[1]) + self[1],
float(other[2]) + self[2] ))
else:
return NotImplemented
def __sub__(self, other):
"""Vector subtraction"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( self[0] - float(other[0]),
self[1] - float(other[1]),
self[2] - float(other[2]) ))
else:
return NotImplemented
def __isub__(self, other):
"""Vector subtraction"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( self[0] - float(other[0]),
self[1] - float(other[1]),
self[2] - float(other[2]) ))
else:
return NotImplemented
def __rsub__(self, other):
"""Vector subtraction"""
if isinstance(other, (tuple, list)) and len(other) == 3:
return tuple.__new__(self.__class__, ( float(other[0]) - self[0],
float(other[1]) - self[1],
float(other[2]) - self[2] ))
else:
return NotImplemented
def __mul__(self, other):
"""Vector-scalar multiplication"""
if isinstance(other, (int, float)):
return tuple.__new__(self.__class__, ( self[0]*other,
self[1]*other,
self[2]*other ))
else:
return NotImplemented
def __imul__(self, other):
"""Vector-scalar multiplication"""
if isinstance(other, (int, float)):
return tuple.__new__(self.__class__, ( self[0]*other,
self[1]*other,
self[2]*other ))
else:
return NotImplemented
def __rmul__(self, other):
"""Scalar-vector multiplication"""
if isinstance(other, (int, float)):
return tuple.__new__(self.__class__, ( other*self[0],
other*self[1],
other*self[2] ))
else:
return NotImplemented
def __truediv__(self, other):
"""Vector-scalar division"""
if isinstance(other, (int, float)):
return tuple.__new__(self.__class__, ( self[0]/other,
self[1]/other,
self[2]/other ))
else:
return NotImplemented
def __itruediv__(self, other):
"""Vector-scalar division"""
if isinstance(other, (int, float)):
return tuple.__new__(self.__class__, ( self[0]/other,
self[1]/other,
self[2]/other ))
else:
return NotImplemented
def __rtruediv__(self, other):
"""Division by vector not implemented"""
return NotImplemented
def __neg__(self):
"""Unary - (Vector negation)"""
return tuple.__new__(self.__class__, ( -self[0], -self[1], -self[2] ))
def __pos__(self):
"""Unary + (No effect)"""
return self
def __abs__(self):
"""abs() (Euclidean length)"""
return _sqrt(self[0]*self[0] + self[1]*self[1] + self[2]*self[2])
def __invert__(self):
"""Unary ~ (Vector scaled to reciprocal length)"""
n = self[0]*self[0] + self[1]*self[1] + self[2]*self[2]
return tuple.__new__(self.__class__, ( self[0]/n, self[1]/n, self[2]/n ))
def __bool__(self):
"""Zero vectors are False, nonzero vectors True"""
return (self[0]*self[0] + self[1]*self[1] + self[2]*self[2]) > 0.0
def _step_start_length(direction, start):
one = abs(direction)
try:
len_x = one / abs(direction.x)
sgn_x = int(_copysign(1.0, direction.x))
except ZeroDivisionError:
len_x = _inf
sgn_x = 0
try:
len_y = one / abs(direction.y)
sgn_y = int(_copysign(1.0, direction.y))
except ZeroDivisionError:
len_y = _inf
sgn_y = 0
try:
len_z = one / abs(direction.z)
sgn_z = int(_copysign(1.0, direction.z))
except ZeroDivisionError:
len_z = _inf
sgn_z = 0
if sgn_x > 0:
dist_x = (_floor(start.x + 1) - start.x) * len_x
elif sgn_x < 0:
dist_x = (start.x - _floor(start.x)) * len_x
else:
dist_x = len_x
if sgn_y > 0:
dist_y = (_floor(start.y + 1) - start.y) * len_y
elif sgn_y < 0:
dist_y = (start.y - _floor(start.y)) * len_y
else:
dist_y = len_y
if sgn_z > 0:
dist_z = (_floor(start.z + 1) - start.z) * len_z
elif sgn_z < 0:
dist_z = (start.z - _floor(start.z)) * len_z
else:
dist_z = len_z
return (sgn_x, sgn_y, sgn_z), Vector(dist_x, dist_y, dist_z), Vector(len_x, len_y, len_z)
class View:
"""2D view into a voxel lattice"""
def __init__(self, **kwargs):
self._voxel = None
self._position = Vector(0, 0, -50)
self._basis = (Vector(1,0,0), Vector(0,1,0), Vector(0,0,1))
self._picture_distance = 10
self._picture_halfsize = 8
self._xsize = 320
self._ysize = 160
self._colors = [ b'\x00\x00\x00', # Background #000000
b'\x33\x99\xff', # X faces #3399ff
b'\x33\xff\x99', # Y faces #33ff99
b'\x33\xcc\xcc', # XY edges #33cccc
b'\xff\x99\x33', # Z faces #ff9933
b'\xcc\x66\xcc', # XZ edges #cc66cc
b'\xcc\xcc\x66', # YZ edges #cccc66
b'\xcc\xcc\xcc', # XYZ vertx #cccccc
]
def renderPPM(self, output, voxel):
output.write(b'P6\n%d %d 255\n' % (self._xsize, self._ysize))
picx = (self._picture_halfsize / self._xsize) * self._basis[0]
picy = (-self._picture_halfsize / self._xsize) * self._basis[1]
pic0 = self._position + self._picture_distance * self._basis[2] - picx * (0.5 * self._xsize) - picy * (0.5 * self._ysize)
for y in range(0, self._ysize):
for x in range(0, self._xsize):
start = pic0 + x*picx + y*picy
delta = (start - self._position).normalized
step, nextdist, length = _step_start_length(delta, start)
dist = 0
while True:
# Blank wall at distance 200
if dist >= 200:
face = 0
break
elif nextdist.x < nextdist.y and nextdist.x < nextdist.z:
dist = nextdist.x
nextdist += (length.x, 0, 0)
face = 1
elif nextdist.y < nextdist.x and nextdist.y < nextdist.z:
dist = nextdist.y
nextdist += (0, length.y, 0)
face = 2
elif nextdist.z < nextdist.x and nextdist.z < nextdist.y:
dist = nextdist.z
nextdist += (0, 0, length.z)
face = 4
elif nextdist.x == nextdist.y and nextdist.x < nextdist.z:
dist = nextdist.x
nextdist += (length.x, length.y, 0)
face = 1 + 2
elif nextdist.x == nextdist.z and nextdist.x < nextdist.y:
dist = nextdist.x
nextdist += (length.x, 0, length.z)
face = 1 + 4
elif nextdist.y == nextdist.z and nextdist.y < nextdist.x:
dist = nextdist.y
nextdist += (0, length.y, length.z)
face = 2 + 4
else:
dist = nextdist.x
nextdist += (length.x, length.y, length.z)
face = 1 + 2 + 4
if voxel(start + dist * delta, face):
break
output.write(self._colors[face])
print("Row %d of %d rendered" % (y + 1, self._ysize))
def voxel(point, face):
# Voxel map has a spheroid-thingy of radius 6 at (10,4,3)
if (point - (10,4,3)).normsqr <= 36:
return True
# and a bigger one at (-20,0,10)
if (point - (-20,0,10)).normsqr <= 144:
return True
return False
if __name__ == '__main__':
view = View()
with open('out.ppm', 'wb') as out:
view.renderPPM(out, voxel)
print("Saved 'out.ppm'.")
</code></pre>
<p>Near the end, the <code>voxel()</code> function determines what is visible. It saves a NetPBM P6 image (Portable Pixmap format) <code>out.ppm</code>, which looks like:
<img src="https://i.stack.imgur.com/rjNGW.png" alt="out.ppm" /></p>
|
3,122,612 | <p>I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:</p>
<p><span class="math-container">$\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n$</span></p>
<p>I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).</p>
<p><a href="http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf" rel="nofollow noreferrer">http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf</a></p>
<p>EDIT: The full equation is</p>
<p><span class="math-container">$\frac{\partial A}{\partial z}+\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n \frac{\partial^{n} A}{\partial t^{n}}=-i \frac{\chi^{(2)}\omega}{2nc}AA^{*}e^{-i\Delta \textbf{k}\cdot \textbf{z}}$</span></p>
<p>so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..</p>
| Community | -1 | <p><span class="math-container">$$\sum\limits_{n=1}^\infty \frac{(-i)^{n-1}}{n!}k^n=\sum\limits_{n=1}^\infty \frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=\frac{1}{-i}\sum\limits_{n=1}^\infty\frac{(-i)^n}{n!}k^n=i(e^{-ik}-1).$$</span> So I would say yes it is known and there is a name for it -- exponential function.</p>
|
1,696,713 | <p>I am solving exact differential equation, but I am stuck on the step on how to simplify this term or how to rewrite it. </p>
<p>$e^{-2\ln{\sin{x}}}$</p>
| Twiffy | 153,233 | <p>I think the mathematical concept you want is <em>implication</em>.</p>
<p>For example, think about mathematical equations that only work sometimes. How about $x^2=x$. If you plug in a random value of $x$, that equation isn't true. But it happens to work when $x=1$. Written mathematically,</p>
<p>$$ x = 1 \quad \Longrightarrow \quad x^2 = x$$</p>
<p>The symbol $\Longrightarrow$ is read as <em>implies</em>, or as the <em>then</em> part of an if-then statement. You can also think about it as meaning mathematical causation -- e.g. the fact that $x=1$ causes the equation $x^2=x$ to be true. This arrow doesn't always go both ways. If it's true that $x^2=x$, that doesn't imply that $x=1$. (It could instead be true that $x=0$.)</p>
<p>Warning: there are some problems with this perspective, namely that in mathematical logic, any false statement implies any statement you want. This is worth thinking about some more!</p>
|
4,014,022 | <p>In example <span class="math-container">$1.5$</span> of <em>Cracking the GRE Subject Test</em>, the authors make the following calculation in one step with no additional commentary:</p>
<blockquote>
<p>we interchange <span class="math-container">$x$</span> and <span class="math-container">$y$</span> and solve for <span class="math-container">$y$</span>:
<span class="math-container">\begin{align}
\vdots\\
xy^2 + y-x &= 0\\
y &= \frac{-1 \pm \sqrt{1+4x^2}}{2x}
\end{align}</span></p>
</blockquote>
<p>What technique allows such a breezy solution?</p>
| Jessie | 794,754 | <p><span class="math-container">$$y = \frac{x}{1-x^2}$$</span></p>
<p>Replace <span class="math-container">$x$</span> with <span class="math-container">$y$</span></p>
<p><span class="math-container">$$x = \frac{y}{1-y^2}$$</span></p>
<p>and solve <span class="math-container">$y,x=\frac{y}{1-y^2}$</span></p>
<p><span class="math-container">$$-\frac{1\pm\sqrt{4x^2+1}}{2x}$$</span></p>
|
73,112 | <blockquote>
<p>Find <span class="math-container">$e^{At}$</span>, where <span class="math-container">$$A = \begin{bmatrix} 1 & -1 & 1 & 0\\ 1 & 1 & 0 & 1\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$$</span></p>
</blockquote>
<hr />
<p>So, let me just find <span class="math-container">$e^{A}$</span> for now and I can generalize later. I notice right away that I can write</p>
<p><span class="math-container">$$A = \begin{bmatrix} B & I_{2} \\ 0_{22} & B \end{bmatrix}$$</span></p>
<p>where</p>
<p><span class="math-container">$$B = \begin{bmatrix} 1 & -1\\ 1 & 1\\ \end{bmatrix}$$</span></p>
<p>I'm sort of making up a method here and I hope it works. Can someone tell me if this is correct?</p>
<p>I write:</p>
<p><span class="math-container">$$A = \mathrm{diag}(B,B) + \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$$</span></p>
<p>Call <span class="math-container">$S = \mathrm{diag}(B,B)$</span>, and <span class="math-container">$N = \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$</span>. I note that <span class="math-container">$N^2$</span> is <span class="math-container">$0_{44}$</span>, so</p>
<p><span class="math-container">$$e^{N} = \frac{N^{0}}{0!} + \frac{N}{1!} + \frac{N^2}{2!} + \cdots = I_{4} + N + 0_{44} + \cdots = I_{4} + N$$</span></p>
<p>and that <span class="math-container">$e^{S} = \mathrm{diag}(e^{B}, e^{B})$</span> and compute:</p>
<p><span class="math-container">$$e^{A} = e^{S + N} = e^{S}e^{N} = \mathrm{diag}(e^{B}, e^{B})\cdot[I_{4} + N]$$</span></p>
<p>This reduces the problem to finding <span class="math-container">$e^B$</span>, which is much easier.</p>
<p>Is my logic correct? I just started writing everything as a block matrix and proceeded as if nothing about the process of finding the exponential of a matrix would change. But I don't really know the theory behind this I'm just guessing how it would work.</p>
| hmakholm left over Monica | 14,366 | <p>A different, but rather specific, strategy would be to use the ring homomorphism
$${a+bi\in\mathbb C \mapsto \pmatrix{a&-b \\ b&a}\in\mathbb R^{2\times 2}}$$in the block decomposition. Then your problem is equivalent to finding
$$e^{t\pmatrix{1+i & 1\\ 0 & 1+i}}=e^{\pmatrix{t+ti & t\\ 0 & t+ti}}=e^{t+ti}e^{\pmatrix{0&t\\0&0}}=(e^{t+ti})\pmatrix{1&t\\0&1}$$
which unfolds to
$$\pmatrix{e^t\cos t & -e^t\sin t & t e^t \cos t & -t e^t \sin t \\ e^t \sin t & e^t \cos t & t e^t \sin t & t e^t \cos t \\ 0 & 0 & e^t\cos t & -e^t\sin t \\ 0&0& e^t\sin t & e^t\cos t }$$</p>
|
3,748,879 | <p><span class="math-container">$g : \mathbb R \to [0,1]$</span> is a non-decreasing and right continuous step function such that <span class="math-container">$g(x)=0$</span> for all <span class="math-container">$x \leq 0$</span> and <span class="math-container">$g(x)=1$</span> for all <span class="math-container">$x \geq 1$</span>. Let us define <span class="math-container">$g^{-1}(y) = \inf { \{x : x \geq 0, \ g(x) \geq y\} }$</span></p>
<p>Then, is it a continuous function, right/left continuous or neither ?</p>
<p>Where I'm specifically having a problem is the [<span class="math-container">$g(x) \geq y$</span>] part. I do not understand what this means in this context.</p>
<p>Edit: it has been pointed out to me that the function <span class="math-container">$g$</span> is not defined in <span class="math-container">$(0,1)$</span> so the question is incorrect. So please just assume that function is well defined but however many steps the question says exists, exist between <span class="math-container">$(0,1)$</span>.</p>
| N. S. | 9,176 | <p><strong>Hint</strong>
<span class="math-container">$$ \int_0^\infty \frac{1}{(x^p＋2020)^q} \,dx$$</span>
converges if and only if
<span class="math-container">$$\int_1^\infty \frac{1}{(x^p＋2020)^q} \,dx$$</span></p>
<p>Now apply the Limit Comparison Theorem for <span class="math-container">$$\int_1^\infty \frac{1}{(x^p＋2020)^q} \,dx \, \mbox{ and } \, \int_1^\infty \frac{1}{x^{pq}} \,dx$$</span></p>
|
2,657,571 | <p>Apologies if this is a duplicate. I searched and didn't find anything quite like it.</p>
<p>Suppose I have a drawer with an equal number of N black socks and N white socks. They're all mixed up. So, my chances of picking a matching pair in the first two selections is (N-1)/(2N-1), right? Well, what if, <em>before</em> I pick the first sock, I randomly (so I don't know the colors of the socks I'm moving) <em>partition</em> the drawer so that there are N socks on each side, and I draw one sock from each side. <em>Do the chances of drawing a matching pair change</em>?</p>
<p>On the one hand, we can see that selection from one side doesn't <em>change</em> the composition of socks on the other side of the partition. However, whichever color I choose from the "first" side, it's likely that there are more of that color on that side. On other words, if I draw a black sock from one side, it's more likely that that side had N-1 blacks and 1 white than it is that that side had 1 black and N-1 whites.</p>
<p>My suspicion is that I need to do some kind of hypothesis testing, where I consider the chances of every possible partitioning, but that's way above my skill level.</p>
| Bram28 | 256,001 | <p>Of course it shouldn't change!</p>
<p>Let's see if the math works out for a simple example, say $N=3$</p>
<p>First, the 'normal' calculation says that the chances of getting a match $M$ is:</p>
<p>$$P(M)= \frac{N-1}{2N-1}=\frac{2}{5}$$</p>
<p>Now let's see what happens when you randomly partition them.</p>
<p>Let's say you first pick from the left side. After randomly splitting the socks into two groups, there can be $0$, $1$, $2$, or $3$ socks there, with respective probabilities of:</p>
<p>$$P(0)= \frac{{3 \choose 0}\cdot{3 \choose 3}}{6 \choose 3} = \frac{1}{20}$$</p>
<p>$$P(1)= \frac{{3 \choose 1}\cdot{3 \choose 2}}{6 \choose 3} = \frac{9}{20}$$</p>
<p>$$P(2)= \frac{{3 \choose 2}\cdot{3 \choose 1}}{6 \choose 3} = \frac{9}{20}$$</p>
<p>$$P(3)= \frac{{3 \choose 3}\cdot{3 \choose 0}}{6 \choose 3} = \frac{1}{20}$$</p>
<p>Now, getting a match $M$ in the first and last situation is impossible, so $$P(M|0)=P(M|3)=0$$</p>
<p>When there is $1$ white sock on the left, the chances of getting a match are the chance of getting that white sock times the chances of getting one of the two socks on the right side, plus the chances of getting one of the two black ones on the left and the one black one on the right. And, with $2$ white socks on the left it's all symmetrical. So:</p>
<p>$$P(M|1)=P(M|2)=\frac{1}{3}\cdot \frac{2}{3} + \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{9}$$</p>
<p>In sum:</p>
<p>$$P(M)=P(0)\cdot P(M|0) + P(1)\cdot P(M|1) +P(2)\cdot P(M|2) +P(3)\cdot P(M|3) =$$</p>
<p>$$ \frac{1}{20} \cdot 0 + \frac{9}{20} \cdot \frac{4}{9} + \frac{9}{20} \cdot \frac{4}{9} + \frac{1}{20} \cdot 0 = \frac{8}{20} = \frac{2}{5}$$</p>
<p>OK, so yes, same chance!</p>
<p>Now, I'm sure you can generalize this for any $N$ ... thus evaluating the series:</p>
<p>$$P(M)=\sum_{i=0}^N \frac{{N \choose i} \cdot {N \choose {N-i}}}{2N \choose N} \cdot 2 \cdot \frac{i}{N} \cdot \frac{N-i}{N}$$</p>
<p>[... Insert some math that's over my head ....] </p>
<p>... and you'll see that this ends up being $$\frac{N-1}{2N-1}$$</p>
|
3,080,294 | <blockquote>
<p>A cone has its guiding curve to the circle <span class="math-container">$x^2+y^2+2ax+2by=0$</span> and passes through a fixed point <span class="math-container">$(0,0,c)$</span>. If the section of the cone by plane <span class="math-container">$y=0$</span> is a rectangular hyperbola. Prove that the vertex lies on fixed circle <span class="math-container">$x^2+y^2+z^2+2ax+2by=0$</span> and <span class="math-container">$2ax+2by+cz=0$</span>.</p>
</blockquote>
<p>Attempt:</p>
<p>Using equation of circle and fixed point, equation of cone can be found out, it comes:</p>
<p><span class="math-container">$$cx^2+cy^2-2axz-2byz+2acx+2bcy=0$$</span></p>
<p>Its section by <span class="math-container">$y=0$</span> plane comes out to be
<span class="math-container">$$cx^2-2axz+2ax=0$$</span></p>
<p>I am unable to proceed further. Please help</p>
| Intelligenti pauca | 255,730 | <p>Let <span class="math-container">$V=(x_0,y_0,z_0)$</span> be the vertex of the cone. The cone is composed of all lines passing through <span class="math-container">$V$</span> and a point of circle <span class="math-container">$\gamma$</span> of equation <span class="math-container">$x^2+y^2+2ax+2by=0$</span> in the <span class="math-container">$xy$</span> plane; the equation of the cone is then:
<span class="math-container">$$
(z_0x-x_0z)^2+(z_0y-y_0z)^2+2a(z_0x-x_0z)(z_0-z)+2b(z_0y-y_0z)(z_0-z)=0.
$$</span>
Intersecting this with plane <span class="math-container">$y=0$</span> gives the equation of a conic in the <span class="math-container">$xz$</span> plane:
<span class="math-container">$$
(z_0x-x_0z)^2+y_0^2z^2+2a(z_0x-x_0z)(z_0-z)-2by_0z(z_0-z)=0.
$$</span>
This represents a rectangular hyperbola if the coefficients of <span class="math-container">$x^2$</span> and <span class="math-container">$z^2$</span> in that equation are opposite, which leads to the equation:
<span class="math-container">$$
\tag{1}
x_0^2+y_0^2+z_0^2+2ax_0+2by_0=0.
$$</span>
This is the equation of the sphere having <span class="math-container">$\gamma$</span> as a great circle.</p>
<p>We know, on the other hand, that point <span class="math-container">$P=(0,0,c)$</span> lies on the cone, implying that <span class="math-container">$V$</span> belongs to the cone having <span class="math-container">$P$</span> as vertex and <span class="math-container">$\gamma$</span> as guiding curve. Hence the coordinates of <span class="math-container">$V$</span> must satisfy the equation:
<span class="math-container">$$
\tag{2}
cx_0^2+cy_0^2+2ax_0(c-z_0)+2by_0(c-z_0)=0.
$$</span>
Combining equations <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> we obtain the equation of a plane, to which <span class="math-container">$V$</span> must then belong:
<span class="math-container">$$
\tag{3}
2ax_0+2by_0+cz_0=0.
$$</span>
Vertex <span class="math-container">$V$</span> must then lie on the intersection of sphere <span class="math-container">$(1)$</span> and plane <span class="math-container">$(3)$</span>, which is exactly the circle we were required to find.</p>
<p><a href="https://i.stack.imgur.com/E793w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E793w.png" alt="enter image description here"></a></p>
|
2,575,156 | <p>I thought of this question through an excercise in algebraic geometry where the rings were $\Gamma(V)$ and $\mathcal{O}_P(V)$, although my question is more general. If $R\subset S$ two (commutative with 1) rings (R subring of S) and I prime ideal in R, then consider I'=(I) the ideal in S, the ideal generated by I. Is that ideal prime? I would think that the answer is positive even in this general case but i can't seem to find any obvious and fast proof. <br/>Any help would be appreciated. <br/>Just for reference the question came up from excercise 2.18 in fulton algebraic curves in which long story short i had to prove that there is a 1-1 correspondence between prime ideals in $\mathcal{O}_P(V)$ and prime ideals in $\Gamma(V)$.</p>
| Bernard | 202,857 | <p>As mentioned in @Asquire's answer, you can have counter-examples taking for $R=\mathbf Z$ and for $S$ the ring of integers of some quadratic extension. It is know that an odd prime number $p$ remains prime in $S$ (one says it is <em>inert</em> in this case)if and only if the discriminant of the extension is <em>not</em> a square modulo $p$. In the other cases, it is either the product of two prime ideals or the squre of a prime ideal.</p>
<p>Nevertheless the property is true in the case that $S$ is a polynomial ring $R[X_1,\dots,X_n]$.</p>
<p>It is also true in the case that $s$ is a ring of fractions $\Sigma^{-1} R$, for prime ideals which do not intersect the multiplicative system $\Sigma$.</p>
|
3,438,043 | <p><strong>Definition:</strong></p>
<p>Let <span class="math-container">$X$</span> be a set.</p>
<p>A set <span class="math-container">$\tau \subset P(X)$</span> is called a topology on <span class="math-container">$X$</span> if:</p>
<p>(a) <span class="math-container">$\emptyset , X\in \tau$</span></p>
<p>(b) <span class="math-container">$A,B\in \tau$</span> implies <span class="math-container">$A\cap B\in \tau$</span></p>
<p>(c) If <span class="math-container">$\alpha \in \tau$</span> then <span class="math-container">$\underset{A\in \alpha}\bigcup A\in \tau$</span>.</p>
<p>I have been given <span class="math-container">$$\tau :=\left \{ U\subset \mathbb{R} : \text{ For every } x\in U \text{ exists } \varepsilon >0 \text{ with } (x-\varepsilon, x+\varepsilon)\subset U\right \}$$</span>
Show that<span class="math-container">$(\mathbb{R}, \tau)$</span> is a topological space</p>
<p><strong>My attempt:</strong></p>
<p>In order to show that <span class="math-container">$\emptyset, \mathbb{R}\in \tau$</span> , I'd say that if <span class="math-container">$\varepsilon \to 0$</span>, we have <span class="math-container">$(x,x)=\emptyset$</span> and if <span class="math-container">$\varepsilon \to \infty$</span> we have <span class="math-container">$U\subset \mathbb{R}\in \tau$</span>.</p>
<p>In oder to show (b) I'd take to arbitrary intervalls and add them together - but how do I do it formally?</p>
<p>Sadly, I don't really know how to show (c).</p>
| José Carlos Santos | 446,262 | <p>The empty set belongs to <span class="math-container">$\tau$</span> trivially.</p>
<p>The set <span class="math-container">$\mathbb R$</span> belongs to <span class="math-container">$\tau$</span> because is <span class="math-container">$x\in U$</span> and you take <span class="math-container">$\varepsilon=1$</span>, then <span class="math-container">$(x-\varepsilon,x+\varepsilon)\subset\mathbb R$</span>.</p>
<p>If <span class="math-container">$A,B\in\tau$</span>, take <span class="math-container">$x\in A\cap B$</span>. There are <span class="math-container">$\varepsilon_1,\varepsilon_2>0$</span> such that <span class="math-container">$(x-\varepsilon_1,x+\varepsilon_1)\subset A$</span> and <span class="math-container">$(x-\varepsilon_2,x+\varepsilon_2)\subset B$</span>. So,<span class="math-container">\begin{align}\bigl(x-\min\{\varepsilon_1,\varepsilon_2\},x+\min\{\varepsilon_1,\varepsilon_2\}\bigr)&=(x-\varepsilon_1,x+\varepsilon_1)\cap(x-\varepsilon_2,x+\varepsilon_2)\\&\subset A\cap B.\end{align}</span></p>
<p>And if <span class="math-container">$\alpha\subset\tau$</span> and <span class="math-container">$x\in\bigcup_{A\in\alpha}A$</span>, then <span class="math-container">$x\in A_0$</span> for some <span class="math-container">$A_0\in\alpha$</span>. So, there is a <span class="math-container">$\varepsilon>0$</span> such that <span class="math-container">$(x-\varepsilon,x+\varepsilon)\subset A_0$</span>. So,<span class="math-container">$$(x-\varepsilon,x+\varepsilon)\subset\bigcup_{A\in\alpha}A.$$</span></p>
|
3,438,043 | <p><strong>Definition:</strong></p>
<p>Let <span class="math-container">$X$</span> be a set.</p>
<p>A set <span class="math-container">$\tau \subset P(X)$</span> is called a topology on <span class="math-container">$X$</span> if:</p>
<p>(a) <span class="math-container">$\emptyset , X\in \tau$</span></p>
<p>(b) <span class="math-container">$A,B\in \tau$</span> implies <span class="math-container">$A\cap B\in \tau$</span></p>
<p>(c) If <span class="math-container">$\alpha \in \tau$</span> then <span class="math-container">$\underset{A\in \alpha}\bigcup A\in \tau$</span>.</p>
<p>I have been given <span class="math-container">$$\tau :=\left \{ U\subset \mathbb{R} : \text{ For every } x\in U \text{ exists } \varepsilon >0 \text{ with } (x-\varepsilon, x+\varepsilon)\subset U\right \}$$</span>
Show that<span class="math-container">$(\mathbb{R}, \tau)$</span> is a topological space</p>
<p><strong>My attempt:</strong></p>
<p>In order to show that <span class="math-container">$\emptyset, \mathbb{R}\in \tau$</span> , I'd say that if <span class="math-container">$\varepsilon \to 0$</span>, we have <span class="math-container">$(x,x)=\emptyset$</span> and if <span class="math-container">$\varepsilon \to \infty$</span> we have <span class="math-container">$U\subset \mathbb{R}\in \tau$</span>.</p>
<p>In oder to show (b) I'd take to arbitrary intervalls and add them together - but how do I do it formally?</p>
<p>Sadly, I don't really know how to show (c).</p>
| Root | 466,361 | <p>Your proof of a) is not right. Note that interval size tending to zero is different from it being zero.
a) is vacuuously true for null set to put it differently there is no x the 'for all x'statement is always true. For <span class="math-container">$\mathbb{R}$</span> it clear that as for any real number say interval with size 1 is real. </p>
|
3,438,043 | <p><strong>Definition:</strong></p>
<p>Let <span class="math-container">$X$</span> be a set.</p>
<p>A set <span class="math-container">$\tau \subset P(X)$</span> is called a topology on <span class="math-container">$X$</span> if:</p>
<p>(a) <span class="math-container">$\emptyset , X\in \tau$</span></p>
<p>(b) <span class="math-container">$A,B\in \tau$</span> implies <span class="math-container">$A\cap B\in \tau$</span></p>
<p>(c) If <span class="math-container">$\alpha \in \tau$</span> then <span class="math-container">$\underset{A\in \alpha}\bigcup A\in \tau$</span>.</p>
<p>I have been given <span class="math-container">$$\tau :=\left \{ U\subset \mathbb{R} : \text{ For every } x\in U \text{ exists } \varepsilon >0 \text{ with } (x-\varepsilon, x+\varepsilon)\subset U\right \}$$</span>
Show that<span class="math-container">$(\mathbb{R}, \tau)$</span> is a topological space</p>
<p><strong>My attempt:</strong></p>
<p>In order to show that <span class="math-container">$\emptyset, \mathbb{R}\in \tau$</span> , I'd say that if <span class="math-container">$\varepsilon \to 0$</span>, we have <span class="math-container">$(x,x)=\emptyset$</span> and if <span class="math-container">$\varepsilon \to \infty$</span> we have <span class="math-container">$U\subset \mathbb{R}\in \tau$</span>.</p>
<p>In oder to show (b) I'd take to arbitrary intervalls and add them together - but how do I do it formally?</p>
<p>Sadly, I don't really know how to show (c).</p>
| Mark | 470,733 | <p>I'm not sure you really understood the definition of <span class="math-container">$\tau$</span>. A set <span class="math-container">$U$</span> is in <span class="math-container">$\tau$</span> if for each <span class="math-container">$x\in U$</span> there is some <span class="math-container">$\epsilon>0$</span> (which might depend on <span class="math-container">$x$</span>) for which <span class="math-container">$(x-\epsilon,x+\epsilon)\subseteq U$</span>. This has nothing to do with taking <span class="math-container">$\epsilon$</span> to <span class="math-container">$0$</span> or to <span class="math-container">$\infty$</span>. </p>
<p>An example: for each <span class="math-container">$x\in\mathbb{R}$</span> we have <span class="math-container">$(x-1,x+1)\subseteq\mathbb{R}$</span>. So for each <span class="math-container">$x\in\mathbb{R}$</span> we can take <span class="math-container">$\epsilon=1$</span>. Hence <span class="math-container">$\mathbb{R}\in\tau$</span>. </p>
<p>Now the empty set: well, if we suppose it isn't in <span class="math-container">$\tau$</span> then there must be some <span class="math-container">$x\in\emptyset$</span> such that for all <span class="math-container">$\epsilon>0$</span> we have that <span class="math-container">$(x-\epsilon,x+\epsilon)$</span> is not contained in <span class="math-container">$\emptyset$</span>. But this is a contradiction because there can't be any elements <span class="math-container">$x\in\emptyset$</span>. </p>
<p>Now let's show <span class="math-container">$\tau$</span> is closed to unions. Let's say <span class="math-container">$\{A_i\}_{i\in I}$</span> is a collection of sets in <span class="math-container">$\tau$</span> and we want to show their union is in <span class="math-container">$\tau$</span>. Let <span class="math-container">$x\in\cup_{i\in I} A_i$</span>. Then there is some <span class="math-container">$j\in I$</span> such that <span class="math-container">$x\in A_j$</span>. Since <span class="math-container">$A_j\in\tau$</span> we know there is some <span class="math-container">$\epsilon>0$</span> for which <span class="math-container">$(x-\epsilon,x+\epsilon)\subseteq A_j\subseteq\cup_{i\in I} A_i$</span>. So we proved exactly what we wanted, the union is in <span class="math-container">$\tau$</span> as well.</p>
<p>Finally, we have to show that <span class="math-container">$\tau$</span> is closed to finite intersections. Let <span class="math-container">$A,B\in\tau$</span>. We want to show that their intersection is in <span class="math-container">$\tau$</span>. Let <span class="math-container">$x\in A\cap B$</span>. Since <span class="math-container">$x\in A$</span> and <span class="math-container">$A\in\tau$</span> there is <span class="math-container">$\epsilon_1>0$</span> for which <span class="math-container">$(x-\epsilon_1,x+\epsilon_1)\subseteq A$</span>. Similarly, there is <span class="math-container">$\epsilon_2>0$</span> such that <span class="math-container">$(x-\epsilon_2,x+\epsilon_2)\subseteq B$</span>. Let <span class="math-container">$\epsilon=\min\{\epsilon_1,\epsilon_2\}$</span>. Then <span class="math-container">$(x-\epsilon,x+\epsilon)\subseteq A\cap B$</span>. </p>
|
1,351,458 | <p>Can someone please show me the steps (all of them… yeah, even the obvious ones)
to go from</p>
<p>$$\begin{align}\frac{y+1}{y-1} = 10^{x^2}\end{align}$$</p>
<p>to</p>
<p>$$\begin{align}y=\frac{10^{x^2}+1}{10^{x^2}-1}\end{align}$$</p>
| Mythomorphic | 152,277 | <p>For $y\neq1$
$$\frac{y+1}{y-1}=10^{x^2}$$ $$y+1=y\cdot10^{x^2}-10^{x^2}$$ $$1+10^{x^2}=y(10^{x^2}-1 )$$</p>
<p>Then?</p>
|
1,351,458 | <p>Can someone please show me the steps (all of them… yeah, even the obvious ones)
to go from</p>
<p>$$\begin{align}\frac{y+1}{y-1} = 10^{x^2}\end{align}$$</p>
<p>to</p>
<p>$$\begin{align}y=\frac{10^{x^2}+1}{10^{x^2}-1}\end{align}$$</p>
| wythagoras | 236,048 | <p>$$\frac{y+1}{y-1} = 10^{x^2}$$</p>
<p>$$y+1 = 10^{x^2}(y-1)$$</p>
<p>$$y+1 = 10^{x^2}y-10^{x^2}$$</p>
<p>$$y+1+10^{x^2} = 10^{x^2}y$$</p>
<p>$$1+10^{x^2} = (10^{x^2}-1)y$$</p>
<p>$$y=\frac{10^{x^2}+1}{10^{x^2}-1}$$</p>
|
2,056,979 | <p>I am attempting:
<a href="https://i.stack.imgur.com/FM80Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FM80Z.png" alt="enter image description here"></a></p>
<p>My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.
<a href="https://i.stack.imgur.com/9onWu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9onWu.jpg" alt="enter image description here"></a></p>
| Community | -1 | <p>Better consider the expression: $$\delta(k+1) = 4^{k+2} + 5^{2k+1} = 4\times 4^{k+1} + 5^2\times 5^{2k-1} = 4[7m-5^{2k-1}] +5^{2k-1}(5^2) = [5^{2k-1} \times (5^2-4) +4(7m)] \mid 7$$ We considered $\delta(k) = 4^{k+1} +5^{2k-1} = 7m$ , since we know $7\mid \delta(k)$. Hope it helps.</p>
|
4,126,527 | <p>Suppose <span class="math-container">$w_1$</span>, <span class="math-container">$w_2$</span>, ..., <span class="math-container">$w_{n-1}$</span> are the complex roots not equal to <span class="math-container">$1$</span> of <span class="math-container">$z^n-1=0$</span>, where <span class="math-container">$n$</span> is odd.</p>
<p>Show: <span class="math-container">$\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}$</span>.</p>
<p>Hello, I am struggling with this complex number proof. Any help is welcomed; thank you in advance.</p>
<p>I have tried expanding the LHS and comparing it to the RHS but am not able to make them equal; I got this far:
<span class="math-container">$\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}$</span></p>
<p>Part 2 of the question:
Hence show: <span class="math-container">$$\sum_{k=1}^{n-1} \frac{1-\bar{w_k}}{1+w_k} = n$$</span></p>
<p>Any suggestions?</p>
| Community | -1 | <p>From <span class="math-container">$w^n=1$</span> we deduce <span class="math-container">$\bar w^n=1$</span> and <span class="math-container">$(w\bar w)^n=1$</span>. Then as <span class="math-container">$w\bar w$</span> is real and <span class="math-container">$n$</span> is odd, <span class="math-container">$w\bar w=\sqrt[n]1=1$</span>.</p>
<p>The given identity holds for <span class="math-container">$w=1$</span>. Now assuming <span class="math-container">$w\ne1$</span>, we multiply both sides of the equation by <span class="math-container">$(1+w)(1+\bar w)=2-w-\bar w\ne0$</span> and get</p>
<p><span class="math-container">$$2-w^2-\bar w^2=4-w^2-2w\bar w-\bar w^2,$$</span> which completes the proof.</p>
|
4,126,527 | <p>Suppose <span class="math-container">$w_1$</span>, <span class="math-container">$w_2$</span>, ..., <span class="math-container">$w_{n-1}$</span> are the complex roots not equal to <span class="math-container">$1$</span> of <span class="math-container">$z^n-1=0$</span>, where <span class="math-container">$n$</span> is odd.</p>
<p>Show: <span class="math-container">$\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}$</span>.</p>
<p>Hello, I am struggling with this complex number proof. Any help is welcomed; thank you in advance.</p>
<p>I have tried expanding the LHS and comparing it to the RHS but am not able to make them equal; I got this far:
<span class="math-container">$\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}$</span></p>
<p>Part 2 of the question:
Hence show: <span class="math-container">$$\sum_{k=1}^{n-1} \frac{1-\bar{w_k}}{1+w_k} = n$$</span></p>
<p>Any suggestions?</p>
| Toby Mak | 285,313 | <p>Since <span class="math-container">$\omega$</span> lies on the unit circle, more generally, let <span class="math-container">$\omega = \cos \theta + i\sin\theta$</span> and <span class="math-container">$\overline{\omega} = \cos \theta-i\sin\theta$</span>:</p>
<p><span class="math-container">$$\text{LHS} = \frac{2-\omega^2-\overline{\omega}^2}{2+\omega+\overline{\omega}}$$</span>
<span class="math-container">$$= \frac{2 - (\cos^2 \theta - \sin^2 \theta + 2i \cos \theta \sin \theta)-(\cos^2 \theta - \sin^2 \theta- 2i\cos \theta \sin \theta)}{2 + 2\cos\theta}$$</span>
<span class="math-container">$$= \frac{2 + 2\sin^2 \theta- 2 \cos^2 \theta}{2 + 2 \cos \theta} = \frac{1 + (1 - \cos^2 \theta) - \cos^2 \theta}{1 + \cos \theta} = \frac{2(1 + \cos \theta)(1 - \cos \theta)}{1 + \cos \theta} = 2 - 2 \cos \theta$$</span>
<span class="math-container">$$= 2 - (\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta) = 2 - \omega - \overline{\omega} = \text{RHS}$$</span></p>
|
3,114,663 | <p>Let there be <span class="math-container">$m$</span> indistinguishable balls, <span class="math-container">$k$</span> bins, <span class="math-container">$C$</span> capacity.
Let <span class="math-container">$X_j$</span> denote the total balls in bin <span class="math-container">$j$</span>.
I've seen ways to calculate the total number of combinations, but I'm not sure how to go about calculating the mean and variance of <span class="math-container">$X_j$</span>.
It is understood that the ball is always thrown into one of the empty bins with uniform probability. edit: nonfull bins not empty bins sorry.</p>
| saulspatz | 235,128 | <p>This seems computationally very difficult to me. You might do better with simulation.</p>
<p>It <span class="math-container">$C\ge m$</span> then the constraint has no effect, and we are just looking for the mean and variance of the number of balls in bin <span class="math-container">$j.$</span> This has a binomial distribution, and the answer is well known.</p>
<p>When <span class="math-container">${m\over k}\leq C< M,$</span> the situation is much more complicated, because the result depends on the sequence in which the balls were thrown into the bins. Before any bin fills up, a ball has probability <span class="math-container">$1/k$</span> of landing in bin <span class="math-container">$j.$</span> After some other ball fills up, it has probability <span class="math-container">$1/(k-1)$</span> of landing in bin <span class="math-container">$j$</span>. The situation is even more complicated if it's possible for more than one ball to fill up.</p>
<p>Consider the case <span class="math-container">$k=3, j=1.$</span> Let <span class="math-container">$P(x_1,x_2,x_3)$</span> be the probablity that at some point there are <span class="math-container">$x_i$</span> balls in bin <span class="math-container">$i$</span> for <span class="math-container">$i=1,2,3.$</span> To compute the expectation, we have to sum up terms of the form <span class="math-container">$aP(a,b,c)$</span> where <span class="math-container">$a+b+c=m.$</span> If <span class="math-container">$\max\{a,b,c\}<C,$</span> we have simply <span class="math-container">$$P(a,b,c)={m\choose a,b,c}3^{-n}\tag{1}$$</span> But suppose <span class="math-container">$c=C.$</span> Then <span class="math-container">$$P(a,b,c)=P(a,b,C)=\frac12P(a-1,b,C)+\frac12P(a,b-1,C)+\frac13P(a,b,C-1)\tag{2}$$</span> taking account of all bins the last ball might have been thrown into.</p>
<p>The last term on the right-hand side of <span class="math-container">$(2)$</span> can be computed directly from <span class="math-container">$(1),$</span> but the first two have to be computed recursively from <span class="math-container">$(2)$</span>. It won't take many bins, or many balls, before this computation becomes unwieldy, if the capacity constraint is effective. </p>
<p>There are so many possible sequences of throws that memoization is unlikely to be effective, so far as I can see. I've been tying to think of ways to simplify the calculations, but so far, I don't have a glimmer. </p>
|
2,990,400 | <p>I need to find all the continuous functions <span class="math-container">$f$</span> such that
<span class="math-container">$$
[f(t)]^2=\int_0^t f(s) ds
$$</span>
<strong>Attempt:</strong></p>
<p>Since
<span class="math-container">$$
f(t)\leq [f(t)]^2+c, \, \, \forall c \geq \frac14$$</span>
we have</p>
<p><span class="math-container">$$
f(t)\leq c+ \int_0^t f(s)ds $$</span>
Now, Gronwall's Lemma gives
<span class="math-container">$$
f(t) \leq c+ \int_0^t c \,e^{t-\tau}d\tau \iff$$</span>
<span class="math-container">$$ f(t) \leq c \, e^t, \, \,\forall c\geq \frac14$$</span></p>
<p>Given that these steps were correct, is this inequality the final answer?</p>
| Lutz Lehmann | 115,115 | <p>Just observe that <span class="math-container">$f(0)=0$</span> and then take the derivative
<span class="math-container">$$
2ff'=f\implies f=0\lor 2f'=1,~f=\frac t2.
$$</span>
As per a hint of Ian in a prior comment, you can also combine a segment where <span class="math-container">$f=0$</span> and then the nullity of <span class="math-container">$f(t)(2f'(t)-1)=0$</span> switches to the second factor generating a solution <span class="math-container">$f(t)=\frac12\max(0, t-a)$</span>.</p>
|
4,264,808 | <p>So we know that <span class="math-container">$\frac{n^2}{2} \geq \frac{n}{2}$</span>, but I'm stuck proving that <span class="math-container">$\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$</span>. Am I missing something?</p>
| Alann Rosas | 743,337 | <p>Notice that the expression <span class="math-container">$\frac{x^2}{2^\sqrt{\ln(x)}}$</span> is well-defined if and only if <span class="math-container">$x\geq1$</span>. Division by <span class="math-container">$x$</span> then shows that your inequality is equivalent to</p>
<p><span class="math-container">$$\frac{x}{2^\sqrt{\ln(x)}}\geq\frac{1}{2}$$</span></p>
<p>We can prove this inequality with calculus:</p>
<p>Consider the function <span class="math-container">$f:(1,\infty)\to\mathbb{R}$</span> defined by</p>
<p><span class="math-container">$$f(x)=\frac{x}{2^\sqrt{\ln(x)}}$$</span></p>
<p>This function is differentiable, so any maxima or minima in its domain will have derivative <span class="math-container">$0$</span>.</p>
<p>Using the quotient rule and the chain rule, we find that for every <span class="math-container">$x>1$</span>, the derivative of <span class="math-container">$f$</span> at <span class="math-container">$x$</span> is</p>
<p><span class="math-container">\begin{align}
f'(x) &= \frac{2^\sqrt{\ln(x)}-x\cdot2^\sqrt{\ln(x)}\ln(2)\cdot\frac{1}{2\sqrt{\ln(x)}}\cdot\frac{1}{x}}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{2^\sqrt{\ln(x)}\left(1-\frac{\ln(2)}{2\sqrt{\ln(x)}}\right)}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}
\end{align}</span></p>
<p>We now solve the equation <span class="math-container">$f'(x)=0$</span> to find <span class="math-container">$f$</span>'s critical points, if any exist.</p>
<p><span class="math-container">\begin{align}
f'(x)=0 &\iff 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}=0\\
&\iff \frac{\ln(2)}{2\sqrt{\ln(x)}}=1\\
&\iff \frac{\ln(2)}{2}=\sqrt{\ln(x)}\\
&\iff \ln(x)=\frac{\ln^2(2)}{4}\\
&\iff x=\exp\left(\frac{\ln^2(2)}{4}\right)
\end{align}</span></p>
<p>Thus, <span class="math-container">$e^{\frac{\ln^2(2)}{4}}$</span> is the only critical point of <span class="math-container">$f$</span>. This actually corresponds to a minimum of <span class="math-container">$f$</span>, which we can prove by showing that <span class="math-container">$f'(x)$</span> is negative for <span class="math-container">$1<x<e^{\frac{\ln^2(2)}{4}}$</span> and positive for <span class="math-container">$x>e^{\frac{\ln^2(2)}{4}}$</span>.</p>
<p><span class="math-container">\begin{align}
1<x<e^{\frac{\ln^2(2)}{4}} &\implies 0<\ln(x)<\frac{\ln^2(2)}{4}\\
&\implies 0<\sqrt{\ln(x)}<\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}>1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}<0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}<0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}</span></p>
<p>This proves that <span class="math-container">$f'(x)$</span> is negative for <span class="math-container">$1<x<e^{\frac{\ln^2(2)}{4}}$</span>.</p>
<p><span class="math-container">\begin{align}
x>e^{\frac{\ln^2(2)}{4}} &\implies \ln(x)>\frac{\ln^2(2)}{4}>0\\
&\implies \sqrt{\ln(x)}>\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}<1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}>0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}>0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}</span></p>
<p>This proves that <span class="math-container">$f'(x)$</span> is positive for <span class="math-container">$x>e^{\frac{\ln^2(2)}{4}}$</span>. We infer that <span class="math-container">$f$</span> has an absolute minimum at <span class="math-container">$e^{\frac{\ln^2(2)}{4}}$</span>, so for every <span class="math-container">$x>1$</span>,</p>
<p><span class="math-container">\begin{align}
f(x) &\geq f\left(e^{\frac{\ln^2(2)}{4}}\right)\\
&= \frac{e^{\frac{\ln^2(2)}{4}}}{2^\sqrt{\ln\left(e^{\frac{\ln^2(2)}{4}}\right)}}\\
&= \frac{e^{\frac{\ln(2)}{4}\cdot\ln(2)}}{2^\sqrt{\frac{\ln^2(2)}{4}}}\\
&= \frac{2^\frac{\ln(2)}{4}}{2^\frac{\ln(2)}{2}}\\
&= 2^{\frac{\ln(2)}{4}-\frac{\ln(2)}{2}}\\
&= 2^{-\frac{\ln(2)}{4}}
\end{align}</span></p>
<p>All that remains to be shown is that <span class="math-container">$2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$</span>:</p>
<p><span class="math-container">\begin{align}
2^{-\frac{\ln(2)}{4}}>\frac{1}{2} &\iff e^{-\frac{\ln(2)}{4}\cdot\ln(2)}>\frac{1}{2}\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>\ln\left(\frac{1}{2}\right)\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>-\ln(2)\\
&\iff \frac{\ln(2)}{4}<1\\
&\iff \ln(2)<4
\end{align}</span></p>
<p>The last of these is true because <span class="math-container">$\ln(2)<1$</span>, so we have that <span class="math-container">$2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$</span>. Thus, for every <span class="math-container">$x>1$</span>,</p>
<p><span class="math-container">$$f(x)>\frac{1}{2}$$</span></p>
<p>or</p>
<p><span class="math-container">$$\frac{x}{2^\sqrt{\ln(x)}}>\frac{1}{2}$$</span></p>
|
3,245,854 | <p>Hi I am trying to solve an integral problem that involves trig substitution. First I tried completing the square, which gave me <span class="math-container">$1/\sqrt{(x+3)^2+2^2}$</span>. I know I am supposed to use <span class="math-container">$x = \arctan(\theta)$</span>. Does that mean it should be: </p>
<p><span class="math-container">$x + 3 = \arctan(\theta)$</span></p>
<p><span class="math-container">$x = \arctan(\theta) - 3$</span></p>
<p>and then Integrate from there? I am not sure if this is a good way of thinking about this problem. I would appreciate any help!
<a href="https://i.stack.imgur.com/N2IV6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N2IV6.png" alt="enter image description here"></a></p>
| Mohammad Riazi-Kermani | 514,496 | <p>Note that <span class="math-container">$$x^2+6x+13 = (x+3)^2+4$$</span></p>
<p>Your correct substitution is<span class="math-container">$$(x+3)=2\tan(\theta)$$</span></p>
<p><span class="math-container">$$dx=2\sec^2(\theta)d\theta$$</span></p>
<p><span class="math-container">$$(x+3)^2+4=4\tan ^2(\theta)+4 =4(1+\tan^2 (\theta)= 4\sec^2(\theta)$$</span></p>
<p><span class="math-container">$$\sqrt {x^2+6x+13}=2\sec (\theta)$$</span></p>
<p>Will take care of the radical and the integral is manageable from here on. </p>
|
3,245,854 | <p>Hi I am trying to solve an integral problem that involves trig substitution. First I tried completing the square, which gave me <span class="math-container">$1/\sqrt{(x+3)^2+2^2}$</span>. I know I am supposed to use <span class="math-container">$x = \arctan(\theta)$</span>. Does that mean it should be: </p>
<p><span class="math-container">$x + 3 = \arctan(\theta)$</span></p>
<p><span class="math-container">$x = \arctan(\theta) - 3$</span></p>
<p>and then Integrate from there? I am not sure if this is a good way of thinking about this problem. I would appreciate any help!
<a href="https://i.stack.imgur.com/N2IV6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N2IV6.png" alt="enter image description here"></a></p>
| clathratus | 583,016 | <p>So you are trying to find
<span class="math-container">$$I=\int\frac{dx}{\sqrt{x^2+6x+13}}=\int\frac{dx}{\sqrt{(x+3)^2+2^2}}$$</span>
And your goal is to find the right substitution <span class="math-container">$$x+3=a\cdot\tan\theta$$</span>
such that <span class="math-container">$$(x+3)^2+2^2=(a\tan\theta)^2+2^2=2^2(\tan^2\theta+1)=2^2\sec^2\theta.$$</span>
It does not take much to show that <span class="math-container">$a=2$</span>. Can you take it from here?</p>
|
3,441,647 | <p>I was asked to find the domain of <span class="math-container">$$\arcsin[\frac{x^2+1}{2x}]$$</span>
My first step was <span class="math-container">$$-1\leq\frac{x^2+1}{2x} \leq1$$</span></p>
<p>What I don't understand is why I cannot cross multiply to get <span class="math-container">$$-2x\le {1+x}^{2} \le 2x$$</span> and then solve the inequality? I tried doing this and got the wrong answer.</p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$x=\tan y$</span></p>
<p><span class="math-container">$z=\arcsin\dfrac{1+x^2}{2x}=\arcsin(\csc2y)$</span></p>
<p><span class="math-container">$\implies\sin z=\csc2y$</span> which is either <span class="math-container">$\ge1$</span> or <span class="math-container">$\le-1$</span></p>
<p>But <span class="math-container">$-1\le\sin z\le1$</span></p>
<p>So, <span class="math-container">$\dfrac{1+x^2}{2x}=\pm1$</span></p>
|
1,435,718 | <blockquote>
<p>Find all pairs of prime numbers $p , q$ for which:
$$p^2 \mid q^3 + 1 \tag{A}$$
and
$$q^2 \mid p^6 − 1 \tag{B}$$</p>
</blockquote>
<p>The question is from the Bulgaria National Olympiad 2014.</p>
<p><em>I'm looking for any solution I may have missed, and generally any alternative method that might reduce the case work (could combine cases 2.1 and 3.1, I suppose).</em></p>
<hr>
<p>I will split the work into three cases:</p>
<ol>
<li>$p=q\ge2$. </li>
<li>$p>q\ge2$.</li>
<li>$q>p\ge2$.</li>
</ol>
<h2>Case 1</h2>
<p>It is clear that neither (A) nor (B) are met.</p>
<h2>Case 2</h2>
<p>First consider two subcases:</p>
<ol>
<li>$p>q$ and $q\in\{2,3\}$.</li>
<li>$p>q\ge5$.</li>
</ol>
<h3>Case 2.1</h3>
<p>$$\begin{align}
q=2 &\implies p^2\mid9 &\implies p^2=9 &\implies p=3 \\
q=3 &\implies p^2\mid28 &\implies p^2=4 &\implies\text{ no solution} \\
\end{align}$$</p>
<p>So $\boxed{(p,q)=(3,2)}$ is the only solution for this case.</p>
<h3>Case 2.2</h3>
<p>(A) factorises as $p^2 \mid (q+1)(q^2-q+1)$. Now</p>
<p>$$q^2-q+1=(q+1)(q-2)+3 \implies \gcd(q+1,q^2-q+1)=
\begin{cases} 3,\quad\text{if }3\mid q+1\\1,\quad\text{otherwise}\end{cases}$$</p>
<p>Since $p>5$ is prime, we must have either $p^2\mid q+1$ or $p^2\mid q^2-q+1$. But this is impossible because $p>q\ge5 \implies p^2>q+1\text{ and }p^2>q^2>q^2-q+1$. So there are no solutions here.</p>
<h2>Case 3</h2>
<p>Consider two subcases:</p>
<ol>
<li>$q>p$ and $p\in\{2,3\}$.</li>
<li>$q>p\ge5$.</li>
</ol>
<h3>Case 3.1</h3>
<p>$$\begin{align}
p=2 &\implies q^2\mid63 &\implies q^2=9 &\implies q=3 \\
p=3 &\implies q^2\mid728 &\implies q^2=4 &\implies\text{ no solution} \\
\end{align}$$</p>
<p>So $\boxed{(p,q)=(2,3)}$ is the only solution for this case.</p>
<h3>Case 3.2</h3>
<p>(B) factorises as $q^2 \mid (p^3+1)(p^3-1)$. Now</p>
<p>$$p^3+1=(p^3-1)+2 \implies \gcd(p^3+1,p^3-1)=
\begin{cases} 2,\quad\text{if }p\text{ is odd}\\1,\quad\text{otherwise}\end{cases}$$</p>
<p>Since $q>5$ is prime, we must have either $q^2\mid p^3+1$ or $q^2\mid p^3-1$. If $q^2\mid p^3+1$, the the same arguments as in case 2.2 can be applied to show that $q^2\mid p+1$ or $q^2\mid p^2-p+1$ neither of which is possible when $q>p$. </p>
<p>So the only remaining possibility is $q^2\mid p^3-1$. This factorises as $q^2\mid (p-1)(p^2+p+1)$. Now</p>
<p>$$p^2+p+1=(p-1)(p+2)+3 \implies \gcd(p-1,p^2+p+1)=
\begin{cases} 3,\quad\text{if }3\mid p-1\\1,\quad\text{otherwise}\end{cases}$$</p>
<p>Since $q>5$ is prime, we must have either $q^2\mid p-1$ or $q^2\mid p^2+p+1$. But this is impossible because $q>p\ge5 \implies q^2>p-1\text{ and }q^2\ge (p+1)^2=p^2+2p+1>p^2+p+1$. So there are no solutions here either.</p>
| Keith Backman | 29,783 | <p>The insight to simplify is that $p$ and $q$ must have different parities; they can't both be odd. At least one of them must be odd, because there are not two even primes. Then the other one must be even, because the addition or subtraction of $1$ makes one of the compound expressions even. Hence one of them must be 2. If we choose $p=2$, then $p^6-1=63$ and $q=3$. This checks out because $4\mid28$. If $q=2$, then $q^3+1=9$ and $p=3$. This checks out because $4\mid728$. $(p,q)=(2,3)$ in either order.</p>
|
390,662 | <p>Let <span class="math-container">$X$</span> be a compact Riemann surface. I would like to find a somehow complete reference for the proof of the so called non-Abelian Hodge correspondence relating Dolbeaut, Betti and Higgs bundle moduli spaces.</p>
<p>I've tried to read the original articles by Hitchin (1987) or Simpson (1990) but it seems to me that I've not found somehow a complete reference showing a precise proof of all the statements involved (for the case of a curve).</p>
<p>Does anyone know a book or notes related to this?</p>
| Niels | 11,682 | <p>I recently attented a nice online talk by Pengfei Huang and he indicated two sources:</p>
<ul>
<li><p>the first chapter of his own phd <a href="https://tel.archives-ouvertes.fr/tel-03134917" rel="noreferrer">Non-abelian Hodge theory and some specializations - TEL - Thèses en ligne</a></p>
</li>
<li><p><a href="https://arxiv.org/abs/1406.1693" rel="noreferrer">Introduction to Nonabelian Hodge Theory: flat connections, Higgs bundles and complex variations of Hodge structure</a> by Alberto Garcia-Raboso, Steven Rayan (see <a href="https://link.springer.com/chapter/10.1007%2F978-1-4939-2830-9_5" rel="noreferrer">Introduction to Nonabelian Hodge Theory | SpringerLink</a> for the published version)</p>
</li>
</ul>
<p>I am not sure to which extent they can be called complete references, but as general references on the subject, I think they certainly deserve to be mentionned.</p>
|
12,047 | <p>How to prove the following trigonometric identities ?</p>
<p>1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$
</p>
<p>I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?</p>
<p>2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $
</p>
<p>For this one I tried substituting them with the sides of the triangle, but not successful to the final result.</p>
<p>These are not my homework, I am trying to learn maths almost on my own, so ...</p>
| user02138 | 2,720 | <p>When I need to prove trigonometric identities, I tend to use <em>complex exponentials</em>. For instance, to prove your first identity observe that
\begin{align}
\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} \quad \text{and} \quad \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},
\end{align}
so
\begin{align}
\tan \theta = \frac{1}{i} \frac{e^{i\theta} - e^{-i\theta} }{e^{i \theta} + e^{-i \theta}}.
\end{align}
We calculate
\begin{align}
1 = \left( \frac{1}{i} \frac{e^{i\alpha} - e^{-i\alpha} }{e^{i \alpha} + e^{-i \alpha}} \right) \left(\frac{1}{i} \frac{e^{i\beta} - e^{-i\beta} }{e^{i \beta} + e^{-i \beta}} \right) = - \frac{(e^{2 i \alpha} - 1)(e^{2 i \beta} - 1)}{(e^{2 i \alpha} + 1)(e^{2 i \beta} + 1)}
\end{align}
Hence,
\begin{align}
(e^{2 i \alpha} + 1)(e^{2 i \beta} + 1) = - (e^{2 i \alpha} - 1)(e^{2 i \beta} - 1).
\end{align}
or $e^{2 i (\alpha + \beta)} + 1 = 0$, which implies that $\alpha + \beta = \frac{\pi}{2}$ by Euler's equation $e^{\pi i} + 1 = 0$, provided that we consider only angles in the fundamental region $[-\frac{\pi}{2}, \frac{\pi}{2}]$. A similar calculation works for your second identity.</p>
|
2,254,694 | <p>I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:</p>
<p>Attempt 1:</p>
<p>Write out the terms of $b_n$:</p>
<p>$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$</p>
<p>Adding up the terms you get:</p>
<p>$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$</p>
<p>But a dead end here.</p>
<p>Attempt 2:</p>
<p>For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$</p>
<p>Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. </p>
<p>For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. </p>
| Paramanand Singh | 72,031 | <p>Let $b_n=a_{n+1}-(a_{n}/2)$ so that $b_{n} \to 0$ as $n\to\infty$. Now we have $a_{2}=b_{1}+(a_{1}/2),a_{3}=b_{2}+b_{1}/2+a_{1}/4$ and continuing in this manner we get $$a_{n+1}=b_{n}+b_{n-1}/2+\cdots+b_{1}/2^{n-1}+a_{1}/2^{n}$$ Let $\epsilon>0$ be given then we can see that there is a positive integer $m$ such that $|b_n|<\epsilon$ for all $n\geq m$. Thus if $n>m$ then we have $$|a_{n+1}|\leq |b_{n} |+|b_{n-1}|/2+\cdots+|b_m|/2^{n-m}+|b_{m-1}|/2^{n-m+1}+\cdots +|b_{1}|/2^{n-1}+|a_{1}|/2^{n}$$ First $(n-m+1) $ terms on right are together less than $2\epsilon$ and the rest of $m$ terms tend to $0$. Thus taking $\limsup$ on both sides we see that $\limsup|a_{n+1}|\leq 2\epsilon$ and since $\epsilon$ was arbitrary this means that $a_{n} \to 0$ as $n\to\infty$. </p>
|
1,248,668 | <p>I do have a system of n equations with m variables where m > n with integer coefficients. I wish to find a set of integer solutions to this system (In my case n = 2 and m = 4). Could somebody tell me how I can do it? I already solved this system with Mathematica but I would like to redo these calculations by hand to understand how their were obtained.</p>
<p>The system is:
$\left\{
\begin{array}{l l}
4u - 3v + 4w + 3z = 1\\
-4v - 3u - 4z + 3w = 0
\end{array} \right.$</p>
| k1.M | 132,351 | <p>There is no way, as I know, to determine all integer solutions to the system. You should at first find all real solutions to the system, this gives you a subspace $V\subset\mathbb R^m$ of dimension at least $m-n$. Now for finding all integer solutions you should find the lattice points of the subspace $V$.</p>
|
495,064 | <p>I'm currently tackling the following problems:</p>
<p>a) Let $a,b \in \mathbb{Z}$ and let $m$ be a nonnegative integer. Prove that $(a,b)=1$ if and only if $(a^m,b)=1$.</p>
<p>If $(a^m,b)=1$ then $pa^m+qb=1$ for some integers $p$ and $q$. It follows that $(pa^{m-1})a+qb=1$, which shows that $(a,b)=1$. This took care of the ''$\Leftarrow$'' direction. I can't think of a way for the ''$\Rightarrow$'' direction. Can you help me here?</p>
<p>b) In addition to the above, let now $n$ be a nonnegative integer as well. Prove that $(a,b)=1$ if and only if $(a^m,b^n)=1$.</p>
<p>The ''$\Leftarrow$'' direction is completely analogous. How about the ''$\Rightarrow$'' direction? </p>
<p><strong>Edit:</strong> Wow, this really wasn't that hard. Here is what I ended up with:</p>
<p>(a) We prove this by contradiction. First assume that $(a,b)=1$ and suppose that $(a^m,b)=d>1$. Then there is a prime $p$ in the prime factorization of $d$ such that $p \mid a^m$ and $p \mid b$. But if $p \mid a^m$, then $p \mid a$, which implies that $p \mid a$ and $p \mid b$, which contradicts the fact that $(a,b)=1$. Thus, $(a^m,b)=1$.</p>
<p>Now assume that $(a^m,b)=1$ and $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and therefore $p \mid (a^m,b)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
<p>(b) This proof is analogous to the one in (a). Assume that $(a,b)=1$ and suppose that $(a^m,b^n)=d>1$. Then we have a prime $p$ such that $p \mid b^n$ and $p \mid a^n$. But then $p \mid a$ and $p \mid b$, from which it follows that $p \mid (a,b)$, which contradicts our initial assumption. Thus, $(a^m,b^n)=1$.</p>
<p>Now assume that $(a^m,b^n)=1$ and that $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and $p \mid b^n$. It follows that $p \mid (a^m,b^n)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
| user66733 | 66,733 | <p>HINT: How about thinking this way: $(a,b) \neq 1 \iff (a^m,b) \neq 1$?</p>
<p>So, for example, if $(a^m,b)=d>1$ then you can find a prime number that $p \mid d$, and this is guaranteed by the prime factorization theorem in integers. Now see what happens in this case.</p>
|
2,008,243 | <p>How to find the exact value of the following limit?</p>
<p>$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>$$\dfrac{e^{x\cos x^2}-e^x}{x^5}=e^x\cdot\dfrac{e^{x(\cos x^2-1)}-1}{x^5}$$</p>
<p>$$\dfrac{e^{x(\cos x^2-1)}-1}{x^5}=\dfrac{e^{x(\cos x^2-1)}-1}{x(\cos x^2-1)}\cdot\dfrac{x(\cos x^2-1)}{x^5}$$</p>
<p>$$\dfrac{\cos x^2-1}{x^4}=-\left(\dfrac{\sin x^2}{x^2}\right)^2\cdot\dfrac1{(1+\cos x^2)}$$</p>
<p>Now use $\lim_{h\to0}\dfrac{e^h-1}h=1$ and $\lim_{y\to0}\dfrac{\sin y}y=1$</p>
|
2,303,339 | <p>How does this proof work?</p>
<p><strong>Theorem.</strong>$\quad$Let $G$ be a group. Then $G$ has a unique identity. </p>
<p><strong>Proof.</strong>$\quad$Assume that $e$ and $f$ are two identities in $G$. Since $e$ is
an identity, $ef=f$; and since $f$ is an identity, $ef=e$. Thus $e=ef=f$. </p>
<p>I think need to get my understanding of variables sorted out, because when I read the first line of the proof I picture $e$ as an object different from $f$ and it's confusing to then read the conclusion that $e$ and $f$ are equal. Also, how does this show that $G$ has a unique identity?</p>
| HK Lee | 37,116 | <p>We use Caley-Hamilton theorem, i.e. $X^2-{\rm tr}\ X \cdot X+{\rm
det}\ XI=0$, for $A+B$ :</p>
<p>For direct computation we have a claim $$ {\rm }
{\rm tr }A \ {\rm tr}B- {\rm tr} (AB) =
{\rm det} (A+B)- {\rm det}A- {\rm det} B $$</p>
<p>If $A=\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right),\ B=\left(
\begin{array}{cc}
x & y \\
z & w \\
\end{array}
\right)$ then $$
{\rm det} (A+B)- {\rm det}A- {\rm det} B =aw +dx-bz-cy$$</p>
<p>Remaining thing is also followed from direct computation. </p>
|
2,033,834 | <blockquote>
<p>Evaluate the work integral where $F(x,y)=\langle-y,x\rangle$ over a triangle with vertices $A(-2,-2)$, $B(2,-2)$, $C(0,1)$.</p>
</blockquote>
<p>I am not sure how to approach this problem.
I tried setting $AB(4,0)$, $BC(-2,3)$ and $CA(-2,-3)$ but I am not sure how to proceed.</p>
<p>Without using Green's theorem</p>
| Akarsh Verma | 392,733 | <p>You will want to split this line integral into three. The first being the line integral from A to B, second from B to C, and third from C to A. The first line integral is along the line $y=-2$ and it is from $ x =-2$ to $x=2$. Thus, it can be parametrized as $x=t$ and $y=-2$ where $-2 \le t \le 2$. From this, we get the integral: $$\int_{-2 }^2 <2,t> . <1,0> dt$$ (The dot means dot product)
Using this, I think you can do the rest! Comment if you need more information/help :)</p>
|
644,325 | <p>Let $M$ be an $R$-module and $N$ be a submodule of $M$. Can we always write $M \cong N \oplus M/N$ as $R$-modules ? If not, then under what conditions on $M$, the direct sum holds ?</p>
| zcn | 115,654 | <p>(This question has almost certainly been asked before, but it seemed easier to just type this up than search for a previous answer). </p>
<p>There is always an exact sequence</p>
<p>$$0 \to N \to M \to M/N \to 0$$</p>
<p>I will state when this sequence is split exact (as pointed out in the comments, this is not the only possibility for which $M \cong N \oplus M/N$). Some cases when it is split:</p>
<p>i) $M/N$ is projective<br>
ii) $N$ is injective</p>
<p>Beyond these two cases, there is not much more that can be said in general (edit: as Pete L. Clark mentions, these are both special cases of $\operatorname{Ext}^1(M/N,N) = 0$). Notice that this is not really a condition on $M$, but rather on the embedding $N \hookrightarrow M$ (or the quotient $M/N$). </p>
|
644,325 | <p>Let $M$ be an $R$-module and $N$ be a submodule of $M$. Can we always write $M \cong N \oplus M/N$ as $R$-modules ? If not, then under what conditions on $M$, the direct sum holds ?</p>
| Wei Zhou | 106,010 | <p>No, it is not true. Consider the cyclic of order 4. It is a Z-module. Let N be the subgroup of order 2, then you get a counter example.</p>
|
999,000 | <p>Theorem 3.29 If $p >1$ then $\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p}$ converges; if $p \leq 1$, the series diverges. </p>
<p>Proof:</p>
<p>The monotonicity of the logarithmic function implies that $\{\log n\}$ increases. Hence $\frac{1}{n\log n}$ decreases, and we can apply theorem 3.27; this leads us to the series </p>
<p>$\sum_{k=1}^{\infty} 2^k \frac{1}{2^k(\log 2^k)^p}= \sum_{k=1}^{\infty} \frac{1}{(k\log 2)^p}= \frac{1}{(\log 2)^p} \sum_{k=1}^{\infty} \frac{1}{k^p}$ and the conclusion follows from Theorem 3.28. </p>
<p>I understand the proof thus far, but Rudin goes on to say that this procedure may evidently be continued. For instance,</p>
<p>$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)}$ diverges, whereas</p>
<p>$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^2}$ converges. </p>
<p>I want to continue the procedure to show for which $p$ does $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converge. </p>
<p>Here are the relevant theorems:</p>
<p>Theorem 3.27 Suppose $a_1 \geq a_2 \geq ... \geq 0$. Then the series $\sum_{k=1}^{\infty} a_n$ converges if and only if $\sum_{k=1}^{\infty} 2^n a_{2^n}$ converges. </p>
<p>Theorem 3.28 $\sum \frac{1}{n^p}$ converges if $p>1$ and diverges if $p \leq 1$. </p>
<p>Here is what I have so far:</p>
<p>By Theorem 3.27 $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges if and only if $\sum_{k=2}^{\infty}2^n \frac{1}{2^n\log 2^n(\log\log 2^n)^p} = \frac{1}{\log 2} \sum_{k=2}^{\infty} \frac{1}{n(\log\log 2^n)^p}$ </p>
<p>Now I'm stuck. How may the procedure be continued? Please note it is my goal to show a result similar to Theorem 3.29. It is not my goal to simply show that $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges for some $p$. Any hint would be greatly appreciated. Thank you. </p>
| Dap | 467,147 | <p>(This is more of a "sketch proof" at the moment.)</p>
<p>The pairs are exactly <span class="math-container">$(t,n)$</span> with <span class="math-container">$t\leq n.$</span> The necessity was posted in a comment which I'll quote here:</p>
<blockquote>
<p>For <span class="math-container">$t>n$</span> there is no such set: For all sufficiently large <span class="math-container">$k$</span> the set <span class="math-container">$S$</span> must contain a point <span class="math-container">$x_k$</span> with <span class="math-container">$\|x_k\|=k^{1/t},$</span> by connectedness. Since <span class="math-container">$N_t(x_k)=k,$</span> the points <span class="math-container">$x_k$</span> must be uniformly separated: there exists <span class="math-container">$\delta>0$</span> such that <span class="math-container">$\|x_k−x_j\|\geq\delta$</span> for all <span class="math-container">$k\neq j.$</span> For large <span class="math-container">$R,$</span> the ball <span class="math-container">$B(0,R)$</span> contains over <span class="math-container">$R^t/2$</span> points <span class="math-container">$x_k.$</span> Since the balls <span class="math-container">$B(x_k,\delta/2)$</span> are disjoint, it follows that <span class="math-container">$R_n\geq(R^t/2)(\delta/2)^n,$</span> which yields a contradiction when <span class="math-container">$R$</span> is large enough. – <strong>user147263</strong></p>
</blockquote>
<p>For <span class="math-container">$t=n$</span> there's no contradiction, but the balls <span class="math-container">$B(x_k,\delta/2)$</span> end up covering a positive proportion of <span class="math-container">$B(0,R).$</span> So the set isn't quite space-filling, but close: a <span class="math-container">$\delta/2$</span>-neighborhood of the set covers a positive fraction of the space.</p>
<p>I'll just consider the <span class="math-container">$t=n$</span> case, because I want to reuse the letter <span class="math-container">$t.$</span>
I'll assume <span class="math-container">$n\geq 2.$</span> (Perhaps interestingly, for <span class="math-container">$n\geq 3$</span> the construction can take place inside any cone with non-empty interior, in particular in <span class="math-container">$\mathbb R_{>0}\times \mathbb R^{n-1},$</span> whereas for <span class="math-container">$n=2$</span> unless I'm mistaken there is no unbounded connected subset of the upper half plane <span class="math-container">$\mathbb R_{>0}\times\mathbb R$</span> on which <span class="math-container">$N_2$</span> is continuous.)</p>
<p>Let <span class="math-container">$\mathbb T$</span> denote the metric space <span class="math-container">$\mathbb R/\mathbb 2\pi Z$</span> with the metric <span class="math-container">$d(x,y)=\min_{n\in\mathbb Z}|x+2\pi n-y|.$</span>
There is a bilipschitz embedding <span class="math-container">$\phi$</span> of <span class="math-container">$\mathbb T\times [0,1]^{n-2}$</span> into the sphere <span class="math-container">$S^{n-1}.$</span> For <span class="math-container">$n=2$</span> this is just embedding a circle in a circle; for <span class="math-container">$n>2$</span> embed <span class="math-container">$\mathbb T\times [0,1]\to\mathbb R^2$</span> as an annulus, pass through the other coordinates to get an embedding <span class="math-container">$\mathbb T\times [0,1]^{n-2}\to\mathbb R^{n-1},$</span> then use stereographic projection <span class="math-container">$\mathbb R^{n-1}\to S^{n-1}.$</span></p>
<p>Start with the curve <span class="math-container">$\gamma(t)=(t,0,\dots,0)$</span> for <span class="math-container">$t>1$</span> - we'll modify this. For each odd integer <span class="math-container">$m>3,$</span> in the segment <span class="math-container">$m<t<m+1,$</span> the curve lives in a <span class="math-container">$(n-1)$</span>-dimensional cube <span class="math-container">$[m,m+1]\times [0,1]^{n-2}.$</span> We can think of this as being a path through an <span class="math-container">$(n-1)$</span>-dimensional <span class="math-container">$m\times m\times \dots\times m$</span> grid graph, with step size <span class="math-container">$1/(m-1),$</span> where the path starts in one corner and exits at another.
Replace the curve within this cube by a Hamiltonian path with the same start and end - this is quite easy to construct. Smooth the corners. The important feature is that there are no significant "short cuts" smaller than <span class="math-container">$\Theta(1/m)$</span>; specifically, for any two points at Euclidean distance <span class="math-container">$L<1/2m$</span> the arc length of the curve between these points is at most <span class="math-container">$L/2.$</span></p>
<p>Do the modification for each odd <span class="math-container">$m>3.$</span> Let <span class="math-container">$s(t)$</span> denote the following modified arc length:
<span class="math-container">$$s(t)=\int_1^t \tau|\gamma'(\tau)|\;d\tau.$$</span></p>
<p>Then <span class="math-container">$s(t)$</span> increases by <span class="math-container">$\Theta(m^{n-1})$</span> during <span class="math-container">$m<t<m+2,$</span> which gives <span class="math-container">$s(t)=\Theta(t^n).$</span> Set <span class="math-container">$S_n=\{s(t)^{1/n}\phi(\gamma(t))\mid t>1\}.$</span> It should be clear that <span class="math-container">$S_n$</span> is unbounded and path-connected. I claim that <span class="math-container">$N_n$</span> is uniformly continuous on <span class="math-container">$S_n,$</span> in fact satisfying a kind of Lipschitz property on small scales.
Consider <span class="math-container">$3<t<t'.$</span> We want to show <span class="math-container">$$|s(t)-s(t')|\leq C\|s(t)^{1/n}\phi(\gamma(t))-s(t')^{1/n}\phi(\gamma(t'))\|\tag{*}$$</span>
for some large constant <span class="math-container">$C,$</span> whenever the right-hand-side is smaller than some small constant <span class="math-container">$c>0.$</span></p>
<p>The nice way for the right-hand-side to be small is when <span class="math-container">$|t-t'|<2$</span> and the quantity <span class="math-container">$L=|\gamma(t)-\gamma(t')|$</span> is at most <span class="math-container">$1/4t.$</span> Then because there are no "short cuts" we must have <span class="math-container">$|s(t)-s(t')|<2Lt.$</span>
Using <span class="math-container">$s(t)^{1/n}=\Theta(t),$</span> the right-hand-side of (*) is <span class="math-container">$\Theta(Lt),$</span> which is perfect.</p>
<p>We need to rule out the possibility that the right-hand-side is small after looping around <span class="math-container">$\mathbb T$</span> a number of times. But this would only occur if <span class="math-container">$|t-t'|>2,$</span> which means <span class="math-container">$|s(t)-s(t')|>\Theta(t^{n-1}),$</span> which makes <span class="math-container">$|s(t)^{1/n}-s(t')^{1/n}|>\Theta(1).$</span> So the right-hand-side cannot be small this way.</p>
|
2,963,560 | <blockquote>
<p>Given <span class="math-container">$a,b,c\in \Bbb Z$</span>, pairwise distinct, and <span class="math-container">$n\in \Bbb N\setminus\{0\}$</span> prove that
<span class="math-container">$$S(n)=\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)}\in \Bbb Z.$$</span></p>
<p>Source: tagged as Kurshar 1959 in a text with problems from math contests</p>
</blockquote>
<p><strong>My attempt</strong>: I approached the problem trying first to solve a particular instance, such as <span class="math-container">$n=1$</span>, to get some insight. </p>
<p>In this particular case (<span class="math-container">$n=1$</span>), the proof is straightforward:
<span class="math-container">$$S(1)=\frac{a(b-c)-b(a-c)+c(a-b)}{(a-b)(a-c)(b-c)}$$</span>
<span class="math-container">$$S(1)=\frac{ab-ac-ab+bc+ac-bc}{(a-b)(a-c)(b-c)}=0\in \Bbb Z$$</span>
Then, I tried the avenue of an induction proof, considering that the proposition is true for <span class="math-container">$S(1)$</span> so that assuming that it is also true for <span class="math-container">$S(n-1)$</span> it would imply it is true for <span class="math-container">$S(n)$</span>. But I couldn't make this step to work. </p>
<p>Hints and answers, not necessarily with induction will be appreciated. But if possible with induction, that would be nice. Sorry if this is a dup.</p>
| metamorphy | 543,769 | <p>For <span class="math-container">$n > 1$</span>, <span class="math-container">$S(n)=\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$</span> (not the most elegant, but nice I think).</p>
<p>For a motivation: the <a href="https://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial" rel="nofollow noreferrer">complete homogeneous symmetric polynomial</a> <span class="math-container">$h_k(x_1, \ldots, x_n)$</span>, when <span class="math-container">$x_1, \ldots, x_n$</span> are pairwise distinct, is equal to <span class="math-container">$\displaystyle\sum_{i=1}^{n}\frac{x_i^{n+k-1}}{\prod_{j\neq i}(x_i-x_j)}$</span> (proven by induction on <span class="math-container">$n$</span>, using the identity <span class="math-container">$h_k(x_1,\ldots,x_{n+1})=\displaystyle\sum_{d=0}^k h_d(x_1,\ldots,x_n)x_{n+1}^{k-d}$</span> that holds by the definition of <span class="math-container">$h_k$</span>).</p>
|
28,586 | <p>Imagine I have two lists:</p>
<pre><code>List1 = {0, 1, 0.6, 0.5, 1.2, 0.4};
List2 = {"a", "b", "c", "d", "e", "f"};
</code></pre>
<p>How can I use <code>Pick</code> to return a third list consisting of items in List2 that have a array position corresponding to an array position with a value $\leq N$ in the first list? Here, for example, for $N = 0.5$ we should have:</p>
<pre><code>List3 = {"a", "d", "f"}
</code></pre>
| Mike Honeychurch | 77 | <p>Several ways to create the stencil. Perhaps most intuitive is just to run a test:</p>
<pre><code>stencil = # <= 0.5 & /@ List1
Pick[List2, stencil]
(* {"a", "d", "f"} *)
</code></pre>
<p><strong>Edit</strong></p>
<p>To answer the comment below, <code>Pick</code> "picks out those elements of <em>list</em> for which the corresponding element of <em>sel</em> is <code>True</code>" so the second list as argument to <code>Pick</code> has to be a list of <code>True</code>/<code>False</code> unless a specific pattern is specified.</p>
<p>So the "default" approach above creates a new list of <code>True</code>/<code>False</code> from <code>List1</code>. An alternative is to specify a pattern:</p>
<pre><code>Pick[List2, List1, x_ /; x <= 0.5]
(* {"a", "d", "f"} *)
</code></pre>
<p>or</p>
<pre><code>Pick[List2, List1, _?(# <= 0.5 &)]
(* {"a", "d", "f"} *)
</code></pre>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| Ali Shadhar | 432,085 | <p>From <a href="https://math.stackexchange.com/questions/3263927/prove-frac-partial-partial-m-textbn-m-textbn-m-sum-k-0n-1-f/3271177#3271177">here</a> we have </p>
<p><span class="math-container">$$\displaystyle\int_0^1 x^{n-1}\ln^2(1-x)\ dx=\frac1n\left({H_n^2}+H_n^{(2)}\right)$$</span></p>
<p>dividing both sides by <span class="math-container">$n$</span> then summing w.r.t <span class="math-container">$n$</span> from <span class="math-container">$n=1$</span> to <span class="math-container">$\infty$</span> we get
<span class="math-container">\begin{align*}
\sum_{n=1}^{\infty}\frac1{n^2}\left({H_n^2}+H_n^{(2)}\right)&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^{\infty}\frac{x^n}{n}\ dx=-\int_0^1\frac{\ln^3(1-x)}{x}\ dx\\
&=-\int_0^1\frac{\ln^3(x)}{1-x}\ dx=6\sum_{n=1}^{\infty}\frac{1}{n^4}=6\zeta(4)
\end{align*}</span></p>
<p>we have, using <span class="math-container">$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^{(a)}}{n^a}=\frac12\left(\zeta(2a)+\zeta^2(a)\right)$</span> that <span class="math-container">$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}=\frac12\left(\zeta(4)+\zeta^2(2)\right)=\frac74\zeta(4)$</span></p>
<p>finally <span class="math-container">$$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}=6\zeta(4)-\frac74\zeta(4)=\frac{17}4\zeta(4)$$</span></p>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| Ali Shadhar | 432,085 | <p><strong>I think this is the shortest solution</strong> </p>
<p>Using the generating function</p>
<p><span class="math-container">$$\frac12\ln^2(1-x)=\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$$</span></p>
<p>Multiply both sides by <span class="math-container">$\frac{\ln(1-x)}{x}$</span> then <span class="math-container">$\int_0^1$</span> and use the fact that <span class="math-container">$\int_0^1 x^{n-1}\ln(1-x)=-\frac{H_n}{n}$</span> we get</p>
<p><span class="math-container">$$\frac12\int_0^1\frac{\ln^3(1-x)}{x}\ dx=-3\zeta(4)=\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(-\frac{H_n}{n}\right)=\sum_{n=1}^\infty\frac{H_n}{n^3}-\sum_{n=1}^\infty\frac{H_n^2}{n^2}$$</span></p>
<p>Substituting <span class="math-container">$\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$</span> gives <span class="math-container">$\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^2}=\frac{17}4\zeta(4)$</span></p>
<p>Note that the fuction used above follows from integrating both sides of <span class="math-container">$\sum_{n=1}^\infty x^n H_n=-\frac{\ln(1-x)}{1-x}$</span></p>
|
13,467 | <p>I will present two problems alongside solutions, student is doing problems of type I like a cakewalk but has several issues with the problems of type 2;</p>
<p><strong>Type I</strong></p>
<blockquote>
<p>Consider an experiment of rolling two dice:</p>
<p>Sample space $$ S = \{(1,1),(1,2),(1,3), \cdots, (6,6) \} $$</p>
<p>Let $A$ be the event of getting 6 as sum on two dice:</p>
<p>Event $$A = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$ </p>
<p>Let $B$ be the event of getting 4 on first die:</p>
<p>Event $$B = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$$ </p>
<p>Now, the probability of getting sum of 6 on two dice given that first
die appears as 4 is given by</p>
<p>$$p(A \mid B) = \dfrac{p(A\cap B)}{p(B)} = \dfrac{p\{(4,2)\}}{p(B)} =
\dfrac{1}{6}$$</p>
</blockquote>
<p><strong>Type II</strong></p>
<blockquote>
<p>Let $C$ be the event of having cancer and $p(C) = 1/2$</p>
<p>The probability of having both tumor and cancer is $p(C\cap T) = 1/6$</p>
<p>Then the probability of having tumor given cancer is given by</p>
<p>$$p(T\mid C) = \dfrac{p(T\cap C)}{p(C)} = \dfrac{1}{3}$$</p>
</blockquote>
<p>Students are understanding type I problems and type 2 problems well, but some students are asking to present $C, T, C\cap T$ in terms of sets. I tried to convince them using Venn diagrams. But they are asking for either roster form (for discrete sets) or set builder form(for any set).</p>
<p>I am trying to do like follows </p>
<p>\begin{align}
\text{Sample space} = S &= \{x \mid \text{$x$ is a living being} \} \\
C &= \{x \mid \text{$x$ has cancer}\} \\
C\cap T &= \{x \mid \text{$x$ has both tumor and cancer}\} \\
\end{align}</p>
<p>Is it right way to do? Students are facing difficulty and they are asking every event inform of set, since definition of event is that it is a subset of sample space (which is a set).</p>
| guest | 9,298 | <p>The question is very confusingly stated. (Hopefully it is more clearly done in native language.) I can't help with the logic barrier for the students, but if they are having a hard time with the logic and then you add on confusing language, this makes things worse. Or may be more the problem than the logic barrier itself. </p>
<p>"Let T be the event of having cancer and p(C)=1/2". </p>
<p>Is T tumor or cancer??? And why have you combined such separate ideas (definition of T tumor and probability of C cancer) into one joined compound sentence? [You are lacking a comma, too. Wouldn't normally nit about this--make mistakes myself--but it adds to the bafflement.] For that matter what does "cancer" and "tumor" mean? (Note how I add the term "detectable" below to help with the huh factor of people equating C and T.)</p>
<p>Let me try:</p>
<ol>
<li>C is the fraction of the population that has cancer. </li>
<li>T is the fraction with a detectable tumor.* </li>
</ol>
<p>Given:</p>
<p>A. Probability of cancer C in the population is 0.5
B. Probability of having both cancer C and a detectable tumor T is 1/6.</p>
<p>Question: What is the probability of having a detectable tumor T, given having cancer C?</p>
<p>Answer: (your equation) </p>
<p>P.s. This is just a student math problem but mistakes in the medical literature are easy to make and common. In some cases even affecting patient treatment. So try to be super precise. I'm not perfect either probably have some mistakes myself. But I do know the importance of clear communication in medical work! </p>
<p>*Not clear to me if you are considering benign detectable tumors (i.e. possible to have T sans C). This has implications to how to think about the problem in real world. For instance, you could have a person who has C (cancer) and has T (detectable tumor) but the detected tumor is benign. Thus examination, biopsy, surgery etc. might not detect or treat the person's cancer. I don't think this affects your math problem, but just be careful.</p>
|
4,141,812 | <p>Considering the vector space <span class="math-container">$\mathbb{R}^3$</span>, find all values of <span class="math-container">$k$</span> such that
<span class="math-container">\begin{align*}
\begin{bmatrix}
2k^2 \\
-3k \\
1
\end{bmatrix} \in \text{ span }
\left\{
\begin{bmatrix}
1 \\
1 \\
3
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix},
\begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
\right\}.
\end{align*}</span></p>
<p>Here is my approach:<br />
By creating the system of equations
<span class="math-container">\begin{align*}
\begin{cases}
x_1 + x_3 &= 2k^2\\
x_1 + x_2 + 2x_3 &= -3k\\
3x_1 + x_2 + 4x_3 &= 1
\end{cases}
\end{align*}</span>
And then, I acquired its RREF by performing elementary row operations.
<span class="math-container">\begin{align*}
\begin{bmatrix}
1 & 0 & 1 & 2k^2 \\
0 & 1 & 1 & -2k^2-3k \\
0 & 0 & 0 & -4k^2 + 3k + 1
\end{bmatrix}
\end{align*}</span>
Since we get a pivot point at the augmented column,
<span class="math-container">\begin{align*}
-4k^2+3k+1 &= 0\\
k &= 1, -\dfrac{1}{4}
\end{align*}</span>
I'm not really sure with what I did in the last part. Is this right?</p>
| gfppoy | 390,474 | <p>By the definition of a vertical asymptote it suffices to find <span class="math-container">$a$</span> such that <span class="math-container">$e^x+x \rightarrow 0$</span> as <span class="math-container">$x \rightarrow a^+$</span> or <span class="math-container">$x \rightarrow a^-$</span>. But since <span class="math-container">$e^x+x$</span> is continuous we simply need to find <span class="math-container">$a$</span> such that <span class="math-container">$e^a+a=0$</span>. We can see such an <span class="math-container">$a$</span> exists in a number of ways, such as either graphically or by usage of the <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert W function</a>.</p>
|
216,040 | <p>A model for the movement of a stock price supposes that if the present price is S then after one period, say one second, it will either go up to uS with probability p or go down to dS with probability q = 1 - p. Assuming that successive movements are independent,approximate the probability that the stock will be up by at least 5% after the next 1000 periods for u = 1.02, d = 0.95 and p = 0.6</p>
| André Nicolas | 6,312 | <p>Let the random variable $W$ be our net gain in one throw of the three dice. </p>
<p>With probability $\dfrac{1}{6^3}$, our number comes up $3$ times, and $W=3$.</p>
<p>With probability $\dbinom{3}{2}\dfrac{1}{6^2}\dfrac{5}{6}$ our number comes up twice, and $W=2$.</p>
<p>With probability $\dbinom{3}{1}\dfrac{1}{6}\dfrac{5^2}{6^2}$ our number comes up once, and $W=1$.</p>
<p>Finally, with probability $\dfrac{5^3}{6^3}$ our number comes up $0$ times, and $W=-1$.<br>
Use the above numbers to find $E(W)$. We have
$$E(W)=3\cdot \frac{1}{216}+2\cdot\frac{15}{216}+1\cdot \frac{75}{216}+(-1)\cdot\frac{125}{216}.$$
That gives us the expected amount of money we "win" per game. (This is a realistic problem. The number $E(W)$ is negative. On average, the more we play the more we lose.)</p>
<p><strong>Another way:</strong> We could instead let $Y$ be the amount of money the casino <strong>hands over</strong>. Then in the various cases $Y$ is $4$, $3$, $2$, or $0$. Find $E(Y)$ using the same basic method, and at the end subtract the $1$ dollar we had to pay to play the game. The result will be the same.</p>
|
4,619,164 | <p>So one way to solve this ordinary differential equation is by computing the integral of both sides of <span class="math-container">$$\frac{y'}{y} = 1$$</span>
However, I did it another way and I think I made a mistake, but where is it? What I did is:
<span class="math-container">$$y' = y $$</span>
<span class="math-container">$$ \int y'dx = \int y dx $$</span>
<span class="math-container">$$ \int \frac{dy}{dx}dx = \int y dx $$</span>
<span class="math-container">$$ \int dy = \int y dx $$</span>
<span class="math-container">$$ y = yx + C $$</span>
<span class="math-container">$$ y = \frac{C}{1-x}$$</span></p>
| Trinity-Slifer | 1,139,301 | <p>If <span class="math-container">$y'=\frac{dy}{dx}$</span> as you wrote <span class="math-container">$y$</span> is a function in the variable <span class="math-container">$x$</span> so <span class="math-container">$\int y(x) dx$</span> is not <span class="math-container">$yx$</span>. I think the only way is to integrate <span class="math-container">$\frac{y'}{y}$</span>.</p>
|
1,148,674 | <p>I have always asked myself why this happens. </p>
<p>If $x = 4$, then $\sqrt{x} = 2$, but if I search for the $\sqrt{4}$, I get $2$ & $-2$.</p>
| Akshay Bodhare | 213,722 | <p>There is a difference.
$$\sqrt {x^2}=|x|$$</p>
<p>And,If
$$x^2=k \implies x=\pm\sqrt{k}$$</p>
|
1,342,951 | <p>I was thinking what function I should compare it to. If I say whether a function is smaller or bigger than this one, then I must prove that. I was thinking of (x+1)^2 but I realized that this converges so that won't prove the integral diverges. So, what function should I use?</p>
| Jahan Claes | 247,555 | <p>You define conditional probability by the Kolmogorov definition. From the Kolmogorov definition, you can prove $P(A\cap B) = P(A\mid B)P(B)$. Then you just have to define what you mean when you write $P(A,B)$. You've defined it to mean $P(A,B)=P(A\cap B)$. Thus, from the definition of conditional probability and the definition of $P(A,B)$, you can prove the relation $P(A,B)=P(B\mid A)P(A)$. There's nothing circular here, we're just using definitions!</p>
<p>Note that we don't need a "rigorous definition" of $P(A\cap B)$. We already have a rigorous definition of what $P(C)$ means for any set $C$. $A\cap B$ is just another set to plug into that definition: set $C=A\cap B$, and you've got your rigorous definition.</p>
|
1,284,086 | <p>Given a state $(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$, is it possible to factorise out the $|0\rangle$ in the middle of both of them?</p>
<p>Alternatively written as $|101\rangle + |001\rangle$ for convenience.</p>
<p>For a state such as the following it is possible to factorise out:</p>
<p>$|101\rangle + |100\rangle = |1\rangle (|01\rangle + |00\rangle)$</p>
<p>as shown above. Is there any way to factorise out the middle tensor in such a form?</p>
| Fallen Apart | 201,873 | <p>If you consider the definition of tensor product by <a href="http://en.wikipedia.org/wiki/Tensor_product#Universal_property" rel="nofollow">universal property</a>, which is unavoidably equal to one you have in mind, then you have two natural isomorphisms $V\otimes W\cong W\otimes V$ and $(V_1\otimes V_2)\otimes V_3\cong V_1\otimes (V_2\otimes V_3).$ This implies that you can simply say that the position on which we write particular factor of tensor is just a matter of convinience. I bet you start with three qubits. The order of them depends on you. If someone fix some order, then these isomorphisms tells you that you can make a permutation. But after you done a permutation and you want put these cubits in some gate then you also have to change the order in this gate (in imput and output). The point is that the tensor product gives you a flexibility in making permutations, in sucha a way, that if you permute the tensors and perumute the gates (i.e. imputs and outputs) then result will be the same. Which in fact is all we care about.</p>
|
83,236 | <p>I need a <code>compare-operation</code> for two lists of same length (usually > 100000) which does the following:</p>
<blockquote>
<p>{2,3,5,4,1,8,7} <code>compare-operation</code> {1,4,6,3,2,8,8} = {2,4,6,4,2,8,8}</p>
</blockquote>
<p>The resulting list has at each position the greater (or equal element in case both are equal) of the two lists (because 2>1,4>3,6>5,4>3,2>1,8=8,8>7).</p>
<p>How can that be done?</p>
| yode | 21,532 | <p>If you are in 11.1 or later verion</p>
<pre><code>ThreadingLayer[Max][{{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}}]
</code></pre>
<blockquote>
<p><code>{2.,4.,6.,4.,2.,8.,8.}</code></p>
</blockquote>
<p>Of course,we have build-in method if you are in old verion</p>
<pre><code>Internal`MaxAbs[{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}]
</code></pre>
<blockquote>
<p><code>{2,4,6,4,2,8,8}</code></p>
</blockquote>
<p>Or</p>
<pre><code>Random`Private`MapThreadMax[{{2, 3, 5, 4, 1, 8, 7}, {1, 4, 6, 3, 2, 8, 8}}]
</code></pre>
<p>will give a same result</p>
|
4,340,560 | <p>I have to teach limits to infinity of real functions of one variable.
I would like to start my course with a beautiful example, not simply a basic function like <span class="math-container">$1/x.$</span> For instance, I thought of using the functions linked to the propagation of covid-19 and show that, under the basic model, the number of contaminations will go to <span class="math-container">$0$</span> when time goes to <span class="math-container">$+\infty.$</span> However, this is a bad idea because the model is not so easy to explain and moreover students are sick of covid-subjects.</p>
<p>Hence, I ask you some help to find interesting examples from physics, geography, etc ... I suppose that an example with "time" going to <span class="math-container">$+\infty$</span> would be nice.</p>
| Paul Sinclair | 258,282 | <p>The problem with "real-life" applications of limits at infinity is that at a fundamental level, there are none. Every place you see something relating to the infinite in the real world, it is actually not the real world, but a mathematical model of the real world. And, without fail, if you examine the model closely, either the model breaks down - that is, fails to represent the real world situation - as you get close to the infinite; or else, the infinity exists in a part of the model that cannot be tested. (Is the universe infinite? It is impossible to prove it.) So any infinities you have "in the real world" are actually just mathematical infinities that do not relate to anything actually in the real world.</p>
<p>Instead I suggest you introduce infinite mathematical limits that the students are already used to dealing with, but may have never realized it. And the grandaddy of them all is non-terminating decimal notation. In particular, I suggest <span class="math-container">$0.999\ldots = 1$</span>.</p>
<p>For any finite number of <span class="math-container">$9$</span> digits, <span class="math-container">$1 - 0.9\ldots 9 > 0$</span>. But given any <span class="math-container">$x > 0$</span>, you can add enough <span class="math-container">$9$</span> digits for <span class="math-container">$1 - 0.9\ldots 9 < x$</span> for any <span class="math-container">$x$</span>. (Archimedean principle: There has to be an integer <span class="math-container">$n > \frac 1x$</span>, and <span class="math-container">$10^n > n$</span>.) So <span class="math-container">$1 - 0.999\ldots$</span> is <span class="math-container">$\ge 0$</span>, but must be smaller and any <span class="math-container">$x > 0$</span>. This leaves <span class="math-container">$0$</span> as the only possibility. In other words <span class="math-container">$$\lim_{n \to \infty} \frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^n} = 1$$</span></p>
|
3,009,351 | <p>Consider the vectors <span class="math-container">$q_1=(1,1,1)$</span> and <span class="math-container">$q_3=(1,1,-2)$</span>. I need to find a third vector <span class="math-container">$q_2$</span> such that <span class="math-container">$\{q_1,q_2,q_3\}$</span> is a arthogonal basis for <span class="math-container">$\mathbb{R}^3$</span>. </p>
<p>My problem is the following: I did take <span class="math-container">$v=(1,0,0)$</span> and I did verify that <span class="math-container">$\{q_1,q_3,v\}$</span> is a basis for <span class="math-container">$\mathbb{R}^3$</span>. Then I did take <span class="math-container">$$q_2=v-\langle v|q_1\rangle q_1-\langle v|q_3\rangle q_3=(-1,-2,1)$$</span></p>
<p>And, by Gram-Schmidt process, <span class="math-container">$q_2$</span> must be orthogonal to <span class="math-container">$q_1$</span> and <span class="math-container">$q_3$</span>. But, as we can see, it does not happen. So, where is my mistake?</p>
| MathIsNice1729 | 274,536 | <p>None of <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> are normalized. Hence, the formula would be
<span class="math-container">$$ q_2=v-{\langle v|q_1\rangle \over \langle q_1|q_1\rangle} q_1-{\langle v|q_3\rangle \over \langle q_3|q_3\rangle} q_3 = ({1 \over 2},-{1 \over 2},0)$$</span>.</p>
|
1,238,887 | <p>A special case of the Lagrange multiplier theorem may be stated as: Let $S, T \subset \mathbb{R}^{n}$ be open. Let $f: S \to \mathbb{R}$ be differentiable on $S$ and $g: T \to \mathbb{R}$ differentiable on $T$ such that $S \cap g^{-1}\{ 0 \}$ is not empty. If $f(x)$ is an extremum of $f$ for some $x \in S \cap g^{-1}\{ 0 \}$ and $\nabla g (x) \neq 0,$ then there is a $\lambda \in \mathbb{R}$ such that
$$\nabla f (x) = \lambda \nabla g(x).$$</p>
<p>However, since $f(x)$ is an extremum and $S \cap g^{-1} \{ 0 \} \subset S,$ should not we have $\nabla f(x) = 0$? </p>
| SiXUlm | 58,484 | <p>Not true.</p>
<p>Take $a_n = n * (\frac{e}{2})^{n}$. It's easy to verify that $\lim \frac{a_n}{e^n} = 0$</p>
<p>On the other hand, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n}$, where i runs from 1 to n</p>
<p>Now consider the denominator:</p>
<p>$\sum \frac{a_i}{a_n} = \sum \frac{i}{n}*(\frac{e}{2})^{i-n} $ < $\sum (\frac{e}{2})^{i-n}$</p>
<p>Let $k = \frac{e}{2}$. Then, the RHS can be written as: $\frac{1}{k^n}*\sum k^i$ = $\frac{k-1/k^n}{k-1} < \frac{k}{k-1} = \frac{e}{e-2}$</p>
<p>Thus, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n} > \frac{e-2}{e} = constant > 0 $. </p>
<p>Then it can't go to $0$.</p>
|
1,238,887 | <p>A special case of the Lagrange multiplier theorem may be stated as: Let $S, T \subset \mathbb{R}^{n}$ be open. Let $f: S \to \mathbb{R}$ be differentiable on $S$ and $g: T \to \mathbb{R}$ differentiable on $T$ such that $S \cap g^{-1}\{ 0 \}$ is not empty. If $f(x)$ is an extremum of $f$ for some $x \in S \cap g^{-1}\{ 0 \}$ and $\nabla g (x) \neq 0,$ then there is a $\lambda \in \mathbb{R}$ such that
$$\nabla f (x) = \lambda \nabla g(x).$$</p>
<p>However, since $f(x)$ is an extremum and $S \cap g^{-1} \{ 0 \} \subset S,$ should not we have $\nabla f(x) = 0$? </p>
| Yiorgos S. Smyrlis | 57,021 | <p>Not true.</p>
<p>Take the sequence $\{a_n\}$ of the form:
$$
a_n=b_j,\quad \text{for}\,\,\,k_j\le n< k_{j+1},
$$
such that</p>
<p>a. The $b_j$ are positive and form an increasing sequence,</p>
<p>b. $b_j=a_1+\cdots+a_{k_j-1}$, and hence $\dfrac{a_{k_j}}{a_1+\cdots+a_{k_j}}=\dfrac{1}{2}$,</p>
<p>c. $k_j$ is chosen so that
$\dfrac{a_{k_j-1}}{a_1+\cdots+a_{k_j-1}}<\dfrac{1}{2^j}$.</p>
<p>d. Such sequence is: $1,1,2,2,2,8,8,8,8,8,8,8,64,\cdots,64,2^{10},\cdots$,
which satisfies
$$
a_{2^n} \le 2^{n(n+1)/2}\quad \text{and}\quad a_m\le m^{\log m},
$$
and hence
$$
0<\frac{a_m}{\mathrm{e}^{\delta m}}\le\frac{m^{\log m}}{\mathrm{e}^{\delta m}}\to 0, \quad\text{for all $\delta>0$.}
$$</p>
|
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