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 |
|---|---|---|---|---|
4,242,765 | <p>Let <span class="math-container">$X_k = 1$</span> with probability <span class="math-container">$0.5$</span> and <span class="math-container">$X_k = -1$</span> with probability 0.5, and let <span class="math-container">$X_k$</span> be independent random variables <span class="math-container">$k = (1,2,...,n)$</span>.</p>
<p>I was able to prove that the characteristic function <span class="math-container">$\Omega_X (\omega)$</span> of the random variable <span class="math-container">$\sum_{i=1}^n \frac{X_i}{\sqrt{n}}$</span> equals <span class="math-container">$\left(\cos{\frac{\omega}{\sqrt{n}}}\right)^n$</span> without too much trouble.</p>
<p>However I'm struggling to prove the next part of the question, which asks me to prove that the limit of that characteristic function approaches <span class="math-container">$e^{-\frac{\omega^2}{2}}$</span>as n approaches infinity.</p>
<p>There is a hint provided which says that I should take the logarithm of the characteristic function and then apply L'Hopital's rule, but I'm struggling at the LHOP part. I'm sure it's a weakness in my calc skills, but I seem to be stuck unfortunately. Any help would be much appreciated!</p>
<p>Problem is from Schaum's Outline of Probability and Statistics, 4th edition. P 3.74.</p>
| Oliver Díaz | 121,671 | <p>Recall that
<span class="math-container">$$\cos z=\sum^\infty_{k=0}\frac{(-1)^kz^{2k}}{(2k)!}=1-\frac{z^2}{2}(1+g(z))$$</span>
where <span class="math-container">$\lim_{z\rightarrow0}g(z)=0$</span>.</p>
<p>Also, <a href="https://math.stackexchange.com/a/4128931/121671">recall</a> that for any sequence of complex numbers <span class="math-container">$c_n\xrightarrow{n\rightarrow\infty}c$</span>,
<span class="math-container">$$\lim_{n\rightarrow\infty}\Big(1+\tfrac{c_n}{n}\Big)^n=e^c$$</span></p>
<p>Then
<span class="math-container">$$\begin{align}
\Big(\cos\frac{\omega}{\sqrt{n}}\Big)^n&=\left(1-\frac{\omega^2}{2n}\Big(1+g\big(\frac{\omega^2}{n}\big)\Big)\right)^n\\
&=\left(1+\frac{-\tfrac{\omega^2}{2}(1+g\big(\tfrac{\omega^2}{n}\big))}{n}\right)^n\xrightarrow{n\rightarrow\infty}e^{-\tfrac{\omega^2}{2}}
\end{align}$$</span></p>
|
2,825,789 | <p>I struggle to understand the following theorem (not the proof, I can't even validate it to be true). Note: I don't have a math background.</p>
<blockquote>
<p>If S is not the empty set, then (f : T → V) is injective if and only if Hom(S, f) is injective.</p>
<p>Hom(S, f) : Hom(S, T) → Hom(T, V)</p>
</blockquote>
<p>As I understand, to prove</p>
<p><strong>f is injective ↔ Hom(S, f) is injective</strong></p>
<p>we can go two ways. We can either prove</p>
<ol>
<li><strong>f</strong> is injective → <strong>Hom(S, f)</strong> is injective AND</li>
<li><strong>f</strong> is not injective → <strong>Hom(S, f)</strong> is not injective</li>
</ol>
<p>Or we can prove</p>
<ol>
<li><strong>Hom(S, f)</strong> is injective → <strong>f</strong> is injective AND</li>
<li><strong>Hom(S, f)</strong> is not injective → <strong>f</strong> is not injective</li>
</ol>
<p>Both ways should give the same result, because biconditional is symmetric, right?!</p>
<p>Then I draw the following diagram:</p>
<p><a href="https://i.stack.imgur.com/1IRGM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1IRGM.png" alt="enter image description here" /></a></p>
<p>where I see <strong>f</strong> as injective but <strong>HOM(S, f)</strong> as not!</p>
<p>Where I'm wrong? How to visualize <strong>HOM(S, f)</strong> correctly?</p>
| Fred | 380,717 | <p>$f(x)=1/x$ is continuous on $[1, \infty)$, but $\int_1^{\infty} f(x) dx = \infty$.</p>
|
2,649,283 | <p>There are these two questions that my professor posted, and they absolutely stumped me:</p>
<p>$ \vdash (\exists x. \bot) \implies P $
and
$(\exists x. \top) \vdash (\forall x. \bot ) \implies P$.</p>
<p>What do I even do with the $(\exists x. \bot)$ part? It got me stuck for quite some time. Any help will be appreciated.</p>
| Peter Smith | 35,151 | <p>In a syntax that allows vacuous quantification (bad and un-Fregean, but now sadly usually permitted), $\exists x\,\phi$ is equivalent to plain $\phi$, where $\phi$ is closed, without free variables, and so $\exists x\,\bot$ is equivalent to plain $\bot$. And you know, presumably, about the ex falso principle $\vdash \bot \to P$. Put those two together to get the first result.</p>
<p>Can you now see why the second result similarly holds?</p>
|
1,819,841 | <p>Is it possible to find out what the "limit" is as the number of iterations of compositions a certain function with itself, like a trig function for example, tends to infinity? </p>
<ol>
<li><p>$\sin(\sin(\sin(\dots(\sin(x))\dots)))$ </p></li>
<li><p>$\cos(\cos(\cos(\dots(\cos(x))\dots)))$</p></li>
</ol>
<p>In the first one, I'm composing sine with itself infinitely many times and in the second, I'm composing cosine with itself. </p>
| E. Joseph | 288,138 | <p>Yes it is possible.</p>
<p>Just consider the sequence defined by induction:</p>
<p>$$\begin{cases} u_0=x \\ u_{n+1}=f(u_n). \end{cases}$$</p>
<p>where $f$ is the function you consider (here $\sin$ and $\cos$).</p>
<p>Then look at the limit</p>
<p>$$\lim_{n\to\infty} u_n$$</p>
<p>with the usuals theorems.</p>
|
1,819,841 | <p>Is it possible to find out what the "limit" is as the number of iterations of compositions a certain function with itself, like a trig function for example, tends to infinity? </p>
<ol>
<li><p>$\sin(\sin(\sin(\dots(\sin(x))\dots)))$ </p></li>
<li><p>$\cos(\cos(\cos(\dots(\cos(x))\dots)))$</p></li>
</ol>
<p>In the first one, I'm composing sine with itself infinitely many times and in the second, I'm composing cosine with itself. </p>
| Community | -1 | <p>Observe that $\sin(x) \in [-1,1]$ and for $x \in [-1,1]$ we have $\lvert \sin(x) \lvert \leq \lvert x \rvert$. This shows that it converges.</p>
<p>In fact, you could also just use Banachs Fixed Point theorem. And you'll see that the sequence will converge to the fixedpoints of $\sin$ and $\cos$ (in the interval $[-1,1]$).</p>
|
1,925,867 | <p>I can't find any. For saying $H$ is a subgroup of $G$ we have notation but it seems none exists for subrings.</p>
| D_S | 28,556 | <p>There doesn't seem to be any standard notation for "is a subring of." If $S, R$ are rings, and one writes $S \subseteq R$ (literally, $S$ is a subset of $R$), then it is tacitly assumed that the operations on $S$ making it into a ring are restricted from those of $R$. It follows from here that $0_S = 0_R$, but if $S$ and $R$ are rings with identity, it is not necessarily true that $1_S = 1_R$. Many authors tacitly assume this as well when they speak of subrings. </p>
|
3,792,135 | <p><strong>Question:</strong> Sum of the series <span class="math-container">$1-3x^2+5x^4 - ... + (-1)^{n-1} (2n-1)x^{2n-2} = \sum\limits_{n=0}^{\infty} (-1)^{n-1} (2n-1)x^{2n-2}$</span></p>
<p>My first idea is to integrate to get <span class="math-container">$\int f(x) dx = x -x^3 + x^5 - ... + (-1)^{ n-1}x^{2n-1} = \sum\limits_{n=1}^{\infty} (-1)^{n-1} x^{2n-1}$</span>. Now I am trying to modify this to a geometric form:</p>
<p><span class="math-container">$$\sum\limits_{n=1}^{\infty} (-1)^{n-1} x^{2n-1}$$</span></p>
| Community | -1 | <p><span class="math-container">$1-3x^2+5x^4-...=\sum_{n=1}^{\infty} (-1)^{n-1}(2n-1)x^{2n-2}=\big( \sum_{n=1}^{\infty} (-1)^{n-1}x^{2n-1} \big)'=\big(\frac{-1}{x}\sum_{n=1}^{\infty}(-x^2)^{n} \big)'=\big(\frac{-1}{x}\times (\frac{1}{1+x^2}-1) \big)'=\big(\frac{x}{1+x^2} \big)'=\frac{1-x^2}{(1+x^2)^2}$</span></p>
<p>when <span class="math-container">$|x|<1$</span></p>
|
3,792,135 | <p><strong>Question:</strong> Sum of the series <span class="math-container">$1-3x^2+5x^4 - ... + (-1)^{n-1} (2n-1)x^{2n-2} = \sum\limits_{n=0}^{\infty} (-1)^{n-1} (2n-1)x^{2n-2}$</span></p>
<p>My first idea is to integrate to get <span class="math-container">$\int f(x) dx = x -x^3 + x^5 - ... + (-1)^{ n-1}x^{2n-1} = \sum\limits_{n=1}^{\infty} (-1)^{n-1} x^{2n-1}$</span>. Now I am trying to modify this to a geometric form:</p>
<p><span class="math-container">$$\sum\limits_{n=1}^{\infty} (-1)^{n-1} x^{2n-1}$$</span></p>
| ratatuy | 812,151 | <p><span class="math-container">$x-x^3+x^5-x^7+\dots=\frac{x}{1+x^2}=\int f(x)\mathrm dx$</span>,
<span class="math-container">$f(x)=\left(\frac{x}{x^2+1}\right)'=\frac{1-x^2}{(1+x^2)^2}$</span></p>
|
2,062,706 | <p>I have the following function:</p>
<p>\begin{equation}
f(q,p) = q \sqrt{p} + (1-q) \sqrt{1 - p}
\end{equation}</p>
<p>Here, $q \in [0,1]$ and $p \in [0,1]$.</p>
<p>Now, given some value $q \in [0,1]$ what value should I select for $p$ in order to maximize $f(q,p)$? That is, I need to define some function $g(q)$ such that $f(q, g(q))$ is a local maximum.</p>
<p>I've been thinking about this problem for days and I don't know where to begin. Any help will be greatly appreciated.</p>
| Ahmed S. Attaalla | 229,023 | <p>Hint:</p>
<p>Now we take some constant $q$ so our function becomes a function with only $p$ varying:</p>
<p>$$\begin{equation}
f(p) = q \sqrt{p} + (1-q) \sqrt{1 - p}
\end{equation}$$</p>
<p>With the constraint $0 \leq p \leq 1$. To maximize one must consider critical points in our interval, and the endpoints. Keep in mind that the constant $q$ is somewhere between $0$ and $1$.</p>
|
2,789,002 | <p>How can I calculate the height of the tree? I am with geometric proportionality.</p>
<p><img src="https://i.stack.imgur.com/m4zMD.png"></p>
| Rhys Hughes | 487,658 | <p>Welcome to Math.SE. It is always better if you show us the work you've done on a problem, and where you got stuck. I'll answer your question but please bear this in mind for next time. </p>
<p>The length from the man to the base of the small tree is $16m$. The length from the man to the big tree is $26m(10m+16m)$
The ratio between these two is your geometric factor, $\frac{26}{16}=\frac{13}{8}=1.125$.
Simply multiply this factor by the height of the small tree to gain the height of the big tree. </p>
|
2,048,054 | <p>I need to find signed distance from the point to the intersection of 2 hyperplanes. I was quite sure that this is something that every mathematician do twice a week :) But not found any good solution or explanation for same problem.</p>
<p>In my case the hyperplanes is defined as $y = w'*x + x_0$, but it is ok to define it with the set of points if there is no other way to solve.</p>
<p>The only solution i found is method to find points of intersection from points from hyperplanes here: <a href="https://www.mathworks.com/matlabcentral/fileexchange/50181-affinespaceintersection-intersection-of-lines-planes-volumes-etc" rel="nofollow noreferrer">https://www.mathworks.com/matlabcentral/fileexchange/50181-affinespaceintersection-intersection-of-lines-planes-volumes-etc</a></p>
<p>But i stuck how to find signed distance after that.</p>
<p>I have strong feeling that there is easy solution, but i don't know correct keywords.</p>
<p>It will be great to see formulas and implementation on any language.
But for sure any help highly appreciated.</p>
<p>Thank you.</p>
| Ben Grossmann | 81,360 | <p>The answer is no. For example, $\{1/n:n\in \Bbb N\} \cup \{0\}$ is disconnected but compact.</p>
|
213,338 | <p>Suppose that $f$ is analytic in the unit disc D = {$z \in \mathbb{C}$ : |$z$| < 1} and $|$f($z$)$| \le 1/(1-|$z$|)$ for all $z\in D$.</p>
<p>Let $f($z$)= \sum _{n=0}^{\infty } a_nz^n$ be the power series expansion of f about $0$.</p>
<p>Prove that $$|a_n| \le (n+1)(1+1/n)^{n} < e(n+1)$$</p>
| Asaf Karagila | 622 | <p>We can rewrite $x^x=e^{x\ln x}$, now using continuity of exponentiation we know that $$\lim_{x\to 0}e^{x\ln x}=e^{\lim_{x\to 0} x\ln x}$$</p>
<p>Calculating $\lim\limits_{x\to0} x\ln x$ is simpler, and it is indeed $0$ (you can use L'Hospital to prove this limit), now we have: $$\lim_{x\to 0}x^x=\lim_{x\to0} e^{x\ln x}=e^0=1.$$</p>
|
3,320,193 | <blockquote>
<p>If given <span class="math-container">$P(B\mid A) =4/5$</span>, <span class="math-container">$P(B\mid A^\complement)= 2/5$</span> and <span class="math-container">$P(B)= 1/2$</span>, what is the probability of <span class="math-container">$A$</span>?</p>
</blockquote>
<p>I know I need to apply Bayes theorem here to figure this out, but I'm struggling a bit to understand how. </p>
<p>So far I've considered this formula:
<span class="math-container">$$P(B\mid A) = \dfrac{P (B \cap A) }{ P (B \cap A) + P(B^\complement \cap A)}$$</span></p>
<p>From this formula, I understand that <span class="math-container">$P(B \cap A) = P(A) \cdot P(B\mid A)$</span> so I plug in the given values but then only find that <span class="math-container">$P(B^\complement |A)$</span> is <span class="math-container">$2/25$</span>. But this does not get me any closer to my goal, <span class="math-container">$P(A)$</span>.</p>
<p>I imagine my understanding of this is quite backward. Any pointers would be helpful.</p>
<p>Thank you</p>
| Community | -1 | <p>You know that <span class="math-container">$$P(B)=P(B|A)P(A)+P(B|A^C)P(A^C)$$</span> and <span class="math-container">$P(A^C)=1-P(A)$</span>. From there, it's just plugging in and solving for <span class="math-container">$P(A)$</span>.</p>
|
2,895,284 | <blockquote>
<p>Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$</p>
</blockquote>
<p>I start by finding the derivative of the denominator, since I have to use the chain rule. </p>
<p>Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:</p>
<p>$$\frac{d}{dx}(x-1)^{2}=2(x-1)$$</p>
<p>After having found the derivative of the denominator I find the derivative of the numerator, which is $3x^2$. With the two derivatives found I apply the quotient rule, which states that </p>
<p>$$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v'u-vu'}{v^2}$$</p>
<p>and substitute in the numbers</p>
<p>$$\frac{d}{dx}\frac{x^3}{(x-1)^2}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}$$</p>
<p>Can I simplify this any further?Is the derivation correct?</p>
| Bernard | 202,857 | <p>You're mixing the product rule and the quotient rule. You can apply each of them, but not simultaneously.</p>
<ul>
<li>by the <em>product rule</em>: remember $\dfrac{\mathrm d}{\mathrm dx}\Bigl(\dfrac1{x^n}\Bigr)=-\dfrac n{x^{n+1}}$, so
\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)&=\dfrac{\mathrm d}{\mathrm dx}(x^3)\cdot\dfrac1{(1-x)^2}+x^3\dfrac{\mathrm d}{\mathrm dx}\dfrac1{(1-x)^2} \\
&= \frac{3x^2}{(1-x)^2}+\frac{2x^3}{(1-x)^3} =\frac{x^2\bigl(3(1-x)+2x\bigr)}{(1-x)^3}\\&
=\frac{x^2(3-x)}{(1-x)^3}.
\end{align}</li>
<li>by the <em>quotient rule</em>:
$$\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)=\frac{3x^2(1-x)^2+x^3\cdot2(1-x)}{(1-x)^4}=\frac{x^2\color{red}{(\not1-\not x)}\bigl(3(1-x)+2x\bigr)}{( 1-x)^{\color{red}{\not4}\,3}}=\dots$$</li>
<li>Since there are exponents, <em>logarithmic differentiation</em> may help make it shorter. It's simpler to use it with <em>Lagrange's notations</em>: set $f(x)=\dfrac{x^3}{(1-x)^2}$. Then
$$\frac{f'(x)}{f(x)}=\frac 3x+\frac2{1-x}=\frac{3-x}{x(1-x)}, \quad\text{so }\;f'(x)=\frac{f'(x)}{f(x)}\cdot f(x)=\dotsm $$</li>
</ul>
|
2,103,436 | <p>Suppose we have the vector space $V$ and the non-empty subspace $W$. I know there is a theorem that states that if $\bar{v}_1$ and $\bar{v}_2$ are vectors in a subspace $W$ then the vector $(\bar{v}_1 + \bar{v}_2)$ will also be in the subspace $W$. However is the converse true? Would having the vector $(\bar{v}_1 + \bar{v}_2)$ in $W$ imply that $\bar{v}_1$ and $\bar{v}_2$ are also in $W$?</p>
| Mark Fischler | 150,362 | <p>Of course not. Let $P_W(\vec{v})$ be the projection of $\vec{v}$ onto subspace $W$.</p>
<p>Then as long as $$\vec{v}_1- P_W(\vec{v}_1)=- (\vec{v}_2-P_W(\vec{v}_2))$$ their sum will be in subspace $W$.</p>
|
2,316,042 | <p><strong>Problem:</strong> Consider the set of all those vectors in $\mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?</p>
<p>In general, if $B$ is the set of all bases vectors then,
$$B=\{(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)\}.$$</p>
<p>There are $8(6\cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6\cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated. </p>
| Max Herrmann | 172,142 | <p>The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:</p>
<p><span class="math-container">$$
\begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \\
(0, 0, 1), (1, 0, 0), (1, 0, 1) \\
(0, 0, 1), (1, 1, 0), (1, 1, 1) \\
(0, 1, 0), (1, 0, 0), (1, 1, 0) \\
(0, 1, 0), (1, 0, 1), (1, 1, 1) \\
(0, 1, 1), (1, 0, 0), (1, 1, 1) \\
\end{eqnarray}
$$</span></p>
|
205,479 | <p>There are $K$ items indexed $X_1, X_2, \ldots, X_K$ in the pool. Person A first randomly take $K_A$ out of these $K$ items and put them back to the pool. Person B then randomly take $K_B$ out of these $K$ items. What is the expectation of items that was picked by B but not taken by A before?</p>
<p>Assuming $K_A \geq K_B$, the formula I get is,</p>
<p>\begin{equation}
E = \sum_{i=1}^{K_B} i \frac{{{K}\choose{K_A}}{{K_A}\choose{K_B - i}}{{K - K_A}\choose{i}}}{{{K}\choose{K_A}}{{K}\choose{K_B}}}
\end{equation}</p>
<p>When $K_B > K_A$, I can derive similar formulas. I am wondering if there is a way to simplify this formula? Thanks.</p>
| André Nicolas | 6,312 | <p>I would prefer to index the objects by the <em>numbers</em> $1,2,\dots,K$. </p>
<p>Define the indicator random variable $W_i$ by $W_i=1$ if item $i$ was taken by $B$ but not taken by $A$ before, and by $W_i=0$ otherwise. Let $p=\Pr(W_i=1)$. Then
$$p=\frac{K_B}{K}\left(1-\frac{K_A}{K}\right).$$
This is because the events "item labelled $i$ was taken by $B$" and "item $i$ was taken by $A$" are <em>independent</em>.
Let random variable $Y$ be the number of objects taken by $B$ that had not been taken by $A$. Then $Y=W_1+W_2+\cdots+W_K$, and therefore by the linearity of expectation
$$E(Y)=E(W_1+W_2+\cdots+W_K)=E(W_1)+E(W_2)+\cdots+E(W_K)=Kp.$$
Note that the $W_i$ are in general <em>not</em> independent. But the linearity of expectation does not require independence.</p>
<p><strong>Remark:</strong> There is undoubtedly a way to obtain the expectation by manipulating a suitable expression that involves binomial coefficients. It is, however, very much the inefficient way to compute the expectation.</p>
|
2,155,071 | <p>You Throw Two Dice. One of the possible many outcomes that may occur is that you get a six on each die (this outcome is called a double six). How many times must you throw the two dice in order for the probability of getting a double six (on one of your throws) to be at least .50?</p>
<p>I completed this problem by setting up the equation .5 = (1/6)(1/6)N
N= 18 times. Is this correct? Can someone please provide me with the proper equation. Thanks so much. </p>
| PattuX | 390,101 | <p>Ask yourself the following:</p>
<p>$p(double \space six)=$</p>
<p>That means $p(no \space double \space six)=$</p>
<p>Chaining this: $p(no \space double \space six \space in \space n \space tries)=$</p>
<p>Again negating: $p(at \space least \space one \space double \space six \space in \space n \space tries)=$</p>
<p>Now just solve for what $n$ the last probability is $>0.5$</p>
<p>I can't quite tell what you mean by your equation...</p>
|
2,155,071 | <p>You Throw Two Dice. One of the possible many outcomes that may occur is that you get a six on each die (this outcome is called a double six). How many times must you throw the two dice in order for the probability of getting a double six (on one of your throws) to be at least .50?</p>
<p>I completed this problem by setting up the equation .5 = (1/6)(1/6)N
N= 18 times. Is this correct? Can someone please provide me with the proper equation. Thanks so much. </p>
| Doug M | 317,162 | <p>The probability of getting a double 6 on any throw is $\frac {1}{36}$</p>
<p>Your probability of getting not getting any 1 double 6s on N throws is $(1-\frac 1{36})^N$</p>
<p>Solve for N such that
$(1-\frac 1{36})^N < 0.5$</p>
<p>$N \log (\frac {35}{36}) < \log {\frac 12}\\
N > \frac {\log 2}{\log 36-\log 35}$</p>
|
244,433 | <p>I have a list:</p>
<pre><code>data = {{2.*10^-9, 0.0025}, {4.*10^-9, 0.0025}, {6.*10^-9, 0.0025}, {8.*10^-9, 0.0025}, {1.*10^-8, 0.0025}, {7.*10^-9, 0.0023}, {3.*10^-9, 0.0025},...}
</code></pre>
<p>And I wanted to remove every third pair and get</p>
<pre><code> newdata = {{2.*10^-9, 0.0025}, {4.*10^-9, 0.0025}, {8.*10^-9, 0.0025}, {1.*10^-8, 0.0025}, {3.*10^-9, 0.0025},...}
</code></pre>
| bill s | 1,783 | <p>Another way to construct the needed indices:</p>
<pre><code>data[[Union[Range[1, Length[data], 3], Range[2, Length[data], 3]]]]
{{2.*10^-9, 0.0025}, {4.*10^-9, 0.0025}, {8.*10^-9, 0.0025}, {1.*10^-8, 0.0025}, {3.*10^-9, 0.0025}}
</code></pre>
<p>Similarly:</p>
<pre><code>data[[Complement[Range[Length[data]], Range[3, Length[data], 3]]]]
</code></pre>
|
1,400,352 | <p>Confusion with the eccentricity of ellipse. On <a href="https://en.wikipedia.org/wiki/Ellipse#Directrix" rel="nofollow noreferrer">wikipedia</a> I got the following in the directrix section of ellipse.</p>
<blockquote>
<p>Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.
Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.</p>
</blockquote>
<p>It is given that $e=\frac fa=\frac ad$</p>
<p><a href="https://i.stack.imgur.com/pPLFX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pPLFX.png" alt="enter image description here"></a></p>
<p>In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).</p>
<p><strong>My question</strong></p>
<p>Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?</p>
| Prakhar Londhe | 261,867 | <p>The ellipse can also be defined as the locus of points whose distance from the focus is proportional to the horizontal distance from a vertical line known as the conic section directrix, where the ratio is $<1$. Letting $e$ be the ratio and $d$ the distance from the center at which the directrix lies, then in order for this to be true, it must hold at the extremes of the major and minor axes, so</p>
<p><a href="https://i.stack.imgur.com/lQ4oR.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lQ4oR.gif" alt="enter image description here"></a></p>
<p>$r=\frac{a-c}{d-a}=\frac{\sqrt{b^2+c^2}}{d}$</p>
<p>With a bit of solving, you get</p>
<p>$d= \frac{a^2}{\sqrt{a^2-b^2}} = \frac{a^2}{c}$
and</p>
<p>$e = \frac{\sqrt{a^2-b^2}}{a}=\frac{c}{a}$.</p>
<p>From the value of $d = \frac{a^2}{c}$</p>
<p>or $\frac{d}{a} = \frac{a}{c} = \frac{1}{e}$</p>
<p>Hence $e = \frac{a}{d}$</p>
<p>In this proof, you can see that we have derived d and e from the extremities, since their values are easily available. For the $\frac{PF}{PD}$ ratio, you can put $PF = a - c$ and $PD = d - a$ (similar to what we did for finding d and e).</p>
<p>Hope this helped.
Cheers! </p>
|
4,459,439 | <p>Suppose <span class="math-container">$G$</span> is an abelian finite group, and the number of order-2 elements in <span class="math-container">$G$</span> is denoted by <span class="math-container">$N$</span>.</p>
<p>I have found that <span class="math-container">$N= 2^n-1$</span> for some <span class="math-container">$n$</span> that satisfy <span class="math-container">$2^n| \ |G|$</span>. I write my proof. Would you tell me if this proof is correct? Moreover, Would you tell me can we say more about the number <span class="math-container">$N$</span>?</p>
<h2>My Proof</h2>
<p>I: The subset of all elements with order 2 union <span class="math-container">$\{e \}$</span> is a subgroup of <span class="math-container">$G$</span> because for all <span class="math-container">$x \ne y: (xy)^2=x^2y^2=e$</span>, and all elements are self inverse. Therefore, <span class="math-container">$N+1$</span> divides |G| by the Lagrange theorem.</p>
<p>II: If <span class="math-container">$N=1$</span> (i.e. there is only one element of order 2 in <span class="math-container">$G$</span>), namely <span class="math-container">$x$</span>, everything will be fine. However, if we have <span class="math-container">$x$</span> and <span class="math-container">$y$</span> as elements of order 2 in <span class="math-container">$G$</span>, then <span class="math-container">$xy$</span> has order 2 and <span class="math-container">$N=3$</span>. If there exists another element of order 2 in <span class="math-container">$G$</span>, namely <span class="math-container">$z$</span>, then <span class="math-container">$xz,\ yz,\ xyz$</span> have order 2 and <span class="math-container">$N=7$</span>. If there exists another element of order 2 in <span class="math-container">$G$</span>, namely <span class="math-container">$w$</span>, then <span class="math-container">$xw,\ yw,\ zw,\ xyw,\ xzw,\ yzw, \ xyzw$</span> have order 2 and <span class="math-container">$N=15$</span>. By induction, <span class="math-container">$N=$$n\choose{1}$$+\cdots +$${n}\choose{n}$</span> <span class="math-container">$= \ 2^n-1$</span> for some <span class="math-container">$n$</span>.</p>
<p>With I and II, <span class="math-container">$N= 2^n-1$</span> for some <span class="math-container">$n$</span> that satisfy <span class="math-container">$2^n| \ |G|$</span>.</p>
<p>Obviously, if |G| is odd, <span class="math-container">$N=0$</span>. Or if <span class="math-container">$|G|=36$</span>, <span class="math-container">$N=1$</span> or <span class="math-container">$N=3$</span>.</p>
<p>Is what I wrote correct?</p>
<p>Can we be more specefic about the number of elements of order 2 in <span class="math-container">$G$</span>?</p>
| Maxime Cazaux | 666,222 | <p>One can be very explicit using the following fact : a finite abelian group is isomorphic to <span class="math-container">$\bigoplus_{i=1}^n \frac{\mathbb{Z}}{n_i \mathbb{Z}}$</span>, so you only have to count the number of <span class="math-container">$n_i$</span>'s which are even.
If we call this number <span class="math-container">$m$</span>, then the number of order 2 elements should indeed be <span class="math-container">$2^m-1$</span>.</p>
|
145,303 | <p>Another question about the convergence notes by Dr. Pete Clark:</p>
<p><a href="http://alpha.math.uga.edu/%7Epete/convergence.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/convergence.pdf</a></p>
<p>(I'm almost at the filters chapter! Getting very excited now!)</p>
<p>On page 15, Proposition 4.6 states that for the following three properties of a topological space <span class="math-container">$X$</span>,</p>
<p><span class="math-container">$(i)$</span> <span class="math-container">$X$</span> has a countable base.</p>
<p><span class="math-container">$(ii)$</span> <span class="math-container">$X$</span> is separable.</p>
<p><span class="math-container">$(iii)$</span> <span class="math-container">$X$</span> is Lindelof (every open cover admits a countable subcover).</p>
<p>we always have <span class="math-container">$(i)\Rightarrow (ii)$</span> and <span class="math-container">$(i)\Rightarrow (iii)$</span>.</p>
<p>Also, we if <span class="math-container">$X$</span> is metrizable, we have <span class="math-container">$(iii)\Rightarrow (i)$</span>, and <em>thus all three are equivalent</em>.</p>
<p>This last part confuses me. We establish all the implications claimed in the proof, but there seems to be a missing link in the claim that all three are equivalent: namely <span class="math-container">$(ii)\Rightarrow (iii)$</span>.</p>
| Pete L. Clark | 299 | <p>The missing implication "separable metrizable implies second countable" is rather easy to prove -- as Carl's answer shows -- but the proof should still appear in the notes. </p>
<p>I have uploaded a new version taking this into account. Thanks for bringing this to my attention.</p>
|
2,853,401 | <p>Assume $E\neq \emptyset $, $E \neq \mathbb{R}^n $. Then prove $E$ has at least one boundary point. (i.e $\partial E \neq \emptyset $).</p>
<p>================= </p>
<p>Here is what I tried.<br>
Consider $P_0=(x_1,x_2,\dots,x_n)\in E,P_1=(y_1,y_2,\dots,y_n)\notin E $.<br>
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,\dots,ty_n+(1-t)x_n) $, $0\le t\le 1$.<br>
$ t_0=\sup\{t |P_t \in E\} $. And then I wanted to prove that $P_{t_0}\in \partial E$. </p>
<p>A. If $P_{t_0}\in E$. then $t\neq 1$ otherwise $P_{t_0}=P_1$. And by definition ,$P_t \notin E$ for $t_0 \lt t \leq 1 $. And choose $t_n$,such that $1\gt t_n\gt t_0$,$t_n \to t_0$, which makes $P_{t_n} \notin E$, but $P_{t_n} \to P_{t_0}$.Then $P_{t_o} \in \partial E$. </p>
<p>B. If $P_{t_0}\notin E$. then $t\neq 0$ otherwise $P_{t_0}=P_0$.And then choose $t_n$ such that $0\lt t_n\lt t_0$ , $t_n \to t_0$ ,therefore $P_{t_n} \to P_{t_0}$ and $P_{t_n} \in E$. Hence $P_{t_o} \in \partial E$.</p>
<p>Thus we have $\partial E \neq \emptyset$. </p>
<p>Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B).
$$P_{t_n} \to P_{t_0} \Rightarrow P_{t_o} \in \partial E$$
I can somewhat image this. But how to make this step strictly?</p>
| William Elliot | 426,203 | <p>Let S be a connected space.<br>
Assume A subset S and empty $\partial$A.<br>
Since $\partial$A = $\overline A$ $\cap$ $\overline{S-A},$
S = (S-A)$^o$ $\cup$ A$^o.$ </p>
<p>Thus either (S-A)$^o$ or A$^o$ is empty.<br>
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.<br>
If (S-A)$^o$ is empty, S = A$^o$. So S = A. </p>
<p>Since R^n is connected, I have shown the contraposit of your problem.<br>
Related to this is: S is connected iff the<br>
only clopen subsets are S and the empty set.<br>
Also, A is clopen iff boundary A is empty.</p>
|
3,583,778 | <p>This problem came to me when I was solving another binomial coefficient summation problem in this site.</p>
<p>I want to prove that <span class="math-container">$\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$</span></p>
| Z Ahmed | 671,540 | <p><span class="math-container">$$S=\sum_{i=0}^{p} \sum_{j=0}^{q+i} \sum_{k=0}^{r+j} {p \choose i} {q+i \choose j} {r+j \choose k}$$</span></p>
<p>First do the <span class="math-container">$k$</span> sum, then
<span class="math-container">$$S=\sum_{i=0}^{p} \sum_{j=0}^{q+i} {p \choose i} {q+i \choose j} 2^{r+j}$$</span>
Next do the <span class="math-container">$j$</span> sum, then
<span class="math-container">$$S=2^r 3^q\sum_{i=0}^{p} {p \choose i} 3^{i}=2^r 3^q 4^p $$</span>
Lastly do the <span class="math-container">$i$</span> sum
<span class="math-container">$$\implies S=2^r 3^q 4^p$$</span></p>
|
1,123,777 | <p><strong><span class="math-container">$U$</span> here represents the upper Riemann Integral.</strong></p>
<p><img src="https://i.stack.imgur.com/GbNm2.jpg" alt="enter image description here" /></p>
<p><img src="https://i.stack.imgur.com/KtRI4.jpg" alt="enter image description here" /></p>
<p><img src="https://i.stack.imgur.com/wlmAk.jpg" alt="enter image description here" /></p>
<p><img src="https://i.stack.imgur.com/pvKC3.jpg" alt="enter image description here" /></p>
<p><strong>I understand the vast majority of this proof</strong>, however the part underlined in orange states <span class="math-container">$\forall \varepsilon>0 $</span> should it not be <span class="math-container">$\forall \varepsilon\geq0 $</span> so that we have</p>
<p><span class="math-container">$U(f)\leq S(f,\Delta_\varepsilon ^1) \leq U(f)+\frac{\varepsilon}{2}$</span>?</p>
<p>For the green part , if the statement works <span class="math-container">$\forall\varepsilon>0$</span> surely it could work in the case <span class="math-container">$U(f+g)=50$</span>, <span class="math-container">$U(f)+U(g)=49$</span>, <span class="math-container">$\varepsilon=2$</span></p>
| Paul Vithayathil | 211,283 | <p>I noticed that you had for n = 2 the solutions as 1+1, 2+0 but 2+0 cannot be a solution based on the problem constraints.</p>
<p>Edit: My bad, as I was working through 2+0 did not make sense but 2 is clearly a solution. But I was able to prove this by thinking for each n, how many ways can you sum to n with k numbers (knowing the largest possible k is k =n where all the terms are 1. Example for n=5, k=3 there is only one possible way 1+2+2.</p>
|
2,064,984 | <p>I am learning about the student t-test. </p>
<p>I am struggling, however, to be given a reasonable explanation why the standard deviation of the standard normal distribution curve is 1. </p>
<p>It says "The Standard Normal Variable is denoted Z and has mean 0 and S.D 1..."</p>
<p>"... this is written as Z ~ N(0,1^2)</p>
<p>Can someone please explain?
Very much appreciate it.</p>
| Michael Hardy | 11,667 | <p>It's hard to be sure what your question means. Suppose $X$ is normally distributed and has expected value $\mu$ and standard deviation $\sigma$. Let $Z = \dfrac {X-\mu} \sigma,$ so that $X = \mu + \sigma Z.$ Then $Z$ has expected value $0$ and standard deviation $1$ and is normally distributed. The one member of this family of distributions that has expected value $0$ and standard deviation $1$ is the one we call the "standard" one.</p>
<p>The distribution of $X$ is
$$
\frac 1 {\sqrt{2\pi}} e^{-(1/2)\Big((x-\mu)/\sigma\Big)^2} \, \frac{dx} \sigma.
$$
If you let $z= \dfrac{x-\mu}\sigma$, then we have $dz=\dfrac{dx}\sigma$ and the measure becomes
$$
\frac 1 {\sqrt{2\pi}} e^{-(1/2) z^2} \, dz.
$$
The expected value of that is
$$
\int_{-\infty}^\infty z \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right)
$$
and this is clearly $0$ because an odd function is integrated over an interval symmetric about $0$. The variance takes some work. Let's call it $\tau^2:$
$$
\tau^2 = \operatorname{var}(Z) = \int_{-\infty}^\infty z^2 \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right).
$$
We have
$$
\operatorname{E}(X) = \operatorname{E}(\mu+\sigma Z) = \mu + \sigma\operatorname{E}(Z) = \mu+\sigma\cdot0 = \mu,
$$
and
$$
\operatorname{var}(X) = \operatorname{var}(\mu+\sigma Z) = \sigma^2 \operatorname{var}(Z) = \sigma^2\tau^2.
$$
If your question is how did we conclude that $\tau^2=1$, say so and maybe I'll post some more on that.</p>
|
4,112,958 | <p>This is a Number Theory problem about the extended Euclidean Algorithm I found:</p>
<p>Use the extended Euclidean Algorithm to find all numbers smaller than <span class="math-container">$2040$</span> so that <span class="math-container">$51 | 71n-24$</span>.</p>
<p>As the eEA always involves two variables so that <span class="math-container">$ax+by=gcd(a,b)$</span>, I am not entirely sure how it is applicable in any way to this problem. Can someone point me to a general solution to this kind of problem by using the extended Euclidean Algorithm?
Also, is there maybe any other more efficient way to solve this than using the eEA?</p>
<p>(Warning: I'm afraid I'm fundamentally not getting something about the eEA, because that section of the worksheet features a number of similiar one variable problems, which I am not able to solve at all.)</p>
<p>I was thinking about using <span class="math-container">$71n-24=51x$</span>, rearranging that into
<span class="math-container">$$71n-51x=24.$$</span> It now looks more like the eEA with <span class="math-container">$an+bx=gcd(a,b)$</span>, but <span class="math-container">$24$</span> isn‘t the <span class="math-container">$gcd$</span> of <span class="math-container">$71$</span> and <span class="math-container">$51$</span>...</p>
| Joffan | 206,402 | <p>This is asking you to find solutions to <span class="math-container">$71n-24\equiv 0 \bmod 51$</span>, which equivalently is <span class="math-container">$20n\equiv 24\bmod 51$</span>.</p>
<p>To do this you can proceed by finding the modular multiplicative inverse of <span class="math-container">$20 \bmod 51$</span> using the extended Euclidean algorithm , as you say.</p>
<p>So:<br />
<span class="math-container">$\begin {array}{rrc}
a&b&51a+20b\\ \hline
1 & 0 & 51\\
0 & 1 & 20\\
1 & -2 & 11\\
-1 & 3 & 9\\
2 & -5 & 2\\
-9 & 23 & 1 \\
\end{array}$</span></p>
<p>So the last line says: <span class="math-container">$-9\cdot 51 + 23\cdot 20 = 1$</span> (check this if nervous), so <span class="math-container">$23\cdot 20 \equiv 1\bmod 51$</span> and <span class="math-container">$23$</span> is the inverse of <span class="math-container">$20\bmod 51$</span>.</p>
<p>So then <span class="math-container">$23\cdot 20n\equiv 23\cdot 24\bmod 51$</span> giving <span class="math-container">$n\equiv 42\bmod 51$</span> and we have solutions <span class="math-container">$n\in\{42,93,144,\ldots\}$</span> up to the upper limit given.</p>
|
4,345,671 | <p>I have a series of cubic polynomials that are being used to create a trajectory. Where some constraints can be applied to each polynomial, such that these 4 parameters are satisfied.
-Initial Position
-final Position
-Initial Velocity
-final Velocity</p>
<p>The polynomials are pieced together such that the ends of one polynomial are identical to the beginnings of the next to preserve continuity.</p>
<p>I instead want to represent these polynomials as cubic Bézier curves.</p>
<p><strong>How would I find the x,y position of each control point for the cubic Bézier curves, such that it matches the curvature of the cubic polynomial.</strong></p>
<p>Here is what I have so far, made in desmos.</p>
<p><a href="https://www.desmos.com/calculator/agsywptfno" rel="nofollow noreferrer">https://www.desmos.com/calculator/agsywptfno</a></p>
<p>Currently the bezier curve is defined as a binomial, with a polynomial for X and or Y
e.g. Bezier = (X(t), Y(t))</p>
| Karl | 279,914 | <p>The problem is that your pieces are not triangles and don't "approach triangles in the limit" in the necessary way. For example, the proportion of shaded area above height <span class="math-container">$\frac h2$</span> in your figure doesn't approach <span class="math-container">$\frac14$</span> of the total shaded area as you cut the hemisphere into smaller pieces; it remains somewhat less. Your approach actually computes the surface area of a cone.</p>
<p>Another way to see that your pieces are not triangles: the angle between the base and the sides is always 90°. This means the sides are curved, and each piece is fatter at the bottom than a triangle. Cutting it into thinner pieces doesn't change this; the pieces always have more area than the corresponding triangle, by a fixed ratio that does not approach 1.</p>
|
764,905 | <p>Calculate $$\int_{D}(x-2y)^2\sin(x+2y)\,dx\,dy$$ where $D$ is a triangle with vertices in $(0,0), (2\pi,0),(0,\pi)$.</p>
<p>I've tried using the substitution $g(u,v)=(2\pi u, \pi v)$ to make it a BIT simpler but honestly, it doesn't help much.</p>
<p>What are the patterns I need to look for in these problems so I can get an integral that's viable to calculate? Everything I try always leads to integrating a huge function and that's extremely error prone.</p>
<p>I mean, I can obviously see the $x-2y$ and $x+2y$ but I don't know how to use it to my advantage. Also, when I do my substitution, I get $\sin(2\pi(u+v))$ and in the triangle I have, $u+v$ goes from 0 to 1, so the $\sin$ goes full circle. Again, no idea if that helps me.</p>
<p>Any help appreciated.</p>
| Disintegrating By Parts | 112,478 | <p>Your integral can be evaluated as an iterated integral and bunch of integration-by-parts:
$$
\int_{0}^{\pi}\int_{0}^{2\pi-2y}(x-2y)^{2}\sin(x+2y)dxdy \\
= \int_{0}^{\pi}\int_{2y}^{2\pi}(x-4y)^{2}\sin(x)\,dx dy \\
= \int_{0}^{\pi}\left(\left.-(x-4y)^{2}\cos(x)\right|_{x=2y}^{x=2\pi}+\int_{2y}^{2\pi}2(x-4y)\cos(x)\,dx\right)\,dy \\
= \int_{0}^{\pi}\left(-(2\pi-4y)^{2}+4y^{2}\cos(2y)+\int_{2y}^{2\pi}2(x-4y)\cos(x)\,dx\right)\,dy \\
= \int_{0}^{\pi}\left(\left.-(2\pi-4y)^{2}+4y^{2}\cos(2y)+2(x-4y)\sin(x)\right|_{x=2y}^{2\pi}-\int_{2y}^{2\pi}2\sin(x)\,dx\right)\,dy \\
=\int_{0}^{\pi}\left(-(2\pi-4y)^{2}+4y^{2}\cos(2y)+4y\sin(2y)+(2-2\cos(2y))\right)\,dy \\
= \left.\left[\frac{1}{12}(2\pi -4y)^{3}+2y^{2}\sin(2y)+(2y-\sin(2y))\right] \right|_{y=0}^{\pi} \\
= -\frac{4\pi^{3}}{3}+2\pi.
$$
I get the same answer as wolf.</p>
|
7,647 | <p>Given a polyhedron consists of a list of vertices (<code>v</code>), a list of edges (<code>e</code>), and a list of surfaces connecting those edges (<code>s</code>), how to break the polyhedron into a list of tetrahedron?</p>
<p>I have a convex polyhedron.</p>
| Joseph Malkevitch | 1,618 | <p>There are polyhedra which are homeomorphic to a sphere with the property that every edge which is not already an edge of the polyhedron lies completely in the exterior of the polyhedron. Sometimes these polyhedra are called Lennes Polyhedra. These polyhedra can not be subdivided into tetrahedra using existing vertices of the polyhedron. In the plane, any simple plane polygon can be triangulated. This forms the basis for Steve Fisk's elegant result about "guarding" plane simple polygons. The analogue of this can not be carried out for 3-dimensional polyhedra. For a brief discussion of Lennes Polyhedra see pages. 253 and 254 of J. O'Rourke, Art Gallery Theorems and Algorithms.</p>
<ul>
<li>Joe Malkevitch</li>
</ul>
|
4,059,420 | <p>If f is holomorphic at every point on the open disc <span class="math-container">$$D=\{z:|z|\lt1\}$$</span>
I want to show that f is constant</p>
| qwfwq | 899,490 | <p><span class="math-container">$\infty$</span> can't be the biggest number, because it isn't a number. It's just a symbol used in limits. Ok, you can consider compactifications of real or complex numbers, but that's certainly not what you had in mind, because both are very far from "counting".</p>
|
4,059,420 | <p>If f is holomorphic at every point on the open disc <span class="math-container">$$D=\{z:|z|\lt1\}$$</span>
I want to show that f is constant</p>
| Oscar Lanzi | 248,217 | <p>Set ordinalities:</p>
<p><span class="math-container">$0,1,2,...,\omega,\omega+1,\omega+2,...,\omega^n,...,\omega^{\omega},...,\omega^{\omega^{\omega^{...}}}=\epsilon_0,...$</span></p>
<p>One might say it's hugely, vastly, mind-bogglingly big.</p>
|
399,804 | <p>The Question was:</p>
<blockquote>
<p>Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.</p>
</blockquote>
<p>I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.</p>
| Kris Williams | 38,143 | <p>Given the equation $$2 \cos(x) = \sin(x)$$ and the instruction to write solely in terms of $\sin(x)$, I would begin by looking for an identity that involves $\cos(x)$, the term we want to transform and $\sin(x)$ the term we want to write everything in. This leaves us with the identity $$\cos(x)^2 + \sin(x)^2 =1.$$ We may then subtract $\sin(x)^2$ from both sides to have have $$\cos(x)^2 = 1 - \sin(x)^2.$$ As $\cos(x)^2$ is non-negative, we may take the square root of both sides and are left with $$\cos(x) = \pm \sqrt{1 - \sin(x)^2}.$$ We then replace the $\cos(x)$ in our equation with $\pm\sqrt{1 - \sin(x)^2}$ and thus we have two equations $$2 \sqrt{1 - \sin(x)^2} = \sin(x)$$ and $$-2 \sqrt{1 - \sin(x)^2} = \sin(x)$$ </p>
|
399,804 | <p>The Question was:</p>
<blockquote>
<p>Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.</p>
</blockquote>
<p>I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.</p>
| Kushashwa Ravi Shrimali | 42,058 | <p>As already mentioned by the users , you must use the identity : </p>
<p>$$\large \sin^2 x + \cos^2 x = 1 ........... \boxed{1} $$ </p>
<p>Here is , how to start solving the problem : </p>
<p>$$\large \text{Given : } \quad 2\cos x = \sin x ......... \boxed{2}$$
Now, simplifying equation 1 further :</p>
<p>$$\large \cos^2 x = 1 - \sin^2 x $$
Solve for $\cos x $ further by taking square root both sides and put that value in equation 2 , you will get your answer.</p>
|
399,804 | <p>The Question was:</p>
<blockquote>
<p>Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.</p>
</blockquote>
<p>I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.</p>
| robjohn | 13,854 | <p>Of course, since $\sin^2(x)+\cos^2(x)=1$, if $2\cos(x)=\sin(x)$, squaring and adding $4\sin^2(x)$ to both sides yields
$$
4=5\sin^2(x)\tag{1}
$$
Of course, $(1)$ also has solutions where $2\cos(x)=-\sin(x)$. Knowledge of
$\sin(x)$ fully determines $|\cos(x)|$, but says nothing about the sign of $\cos(x)$. For this reason, the equation $2\cos(x)=\sin(x)$ cannot be fully characterized in terms of $\sin(x)$ alone.</p>
<p>On the other hand, $2\cos(x)=\sin(x)$ is equivalent to $\tan(x)=2$.</p>
|
506,720 | <p>Hi how can I find the dimension of a vector space? For example :
$V = \mathbb{C} , F = \mathbb{Q}$
what is the dimension of $V$ over $\mathbb{Q}$?</p>
| Mhenni Benghorbal | 35,472 | <p><strong>Hint:</strong> Check this basis for $\mathbb{C}$ </p>
<blockquote>
<p>$$ \left\{ 1, i\right\}. $$</p>
</blockquote>
<p>Now, you need to know what the definition of dimension is? </p>
|
4,183,263 | <p>If Tychonoff's theorem is true, why closed ball in <span class="math-container">$\mathbb{R}^n$</span> is not compact?</p>
<p>The theorem says that if <span class="math-container">$X_i$</span> is compact, for every <span class="math-container">$i\in I$</span>, so <span class="math-container">$\prod_{i\in I}X_i$</span> is compact. Then take <span class="math-container">$n\in\mathbb{N}$</span> and we have <span class="math-container">$\prod_{n\in\mathbb{N}}[-1,1]_i$</span> in <span class="math-container">$\mathbb{R}^\infty$</span> is not compact. But, what??</p>
| Hoang Nguyen | 934,337 | <p><span class="math-container">$\prod_{n \in \mathbb{N}}[-1,1]$</span> is indeed compact under product topology. I believe you misunderstand the structure of a closed ball, that is a closed ball <span class="math-container">$B_{d}(x, \epsilon)$</span> is defined to be the set <span class="math-container">$\{y: d(x,y) \leqslant \epsilon\}$</span> for some <span class="math-container">$\epsilon$</span>, not <span class="math-container">$\prod_{n \in \mathbb{N}}[-1,1]$</span> as you claimed. And the set <span class="math-container">$B_{d}(x, \epsilon)$</span> is clearly not compact.</p>
<p><strong>Proof</strong> Let <span class="math-container">$\gamma < \epsilon$</span>. Consider the set <span class="math-container">$B = \{ (\gamma,0,0,...), (0,\gamma,0,...), (0,0,\gamma,0,...) \} \subset B_{d}(x, \epsilon)$</span> is closed in <span class="math-container">$B_{d}(x, \epsilon)$</span> and has discrete topology, therefore it can not has limit point in <span class="math-container">$B_{d}(x, \epsilon)$</span>. This contradicts with the face that compactness implies limit point compactness.</p>
<p>Hope this helps!</p>
|
1,599,886 | <p>What is the proper way of proving : the density operator $\hat{\rho}$ of a pure state has exactly one non-zero eigenvalue and it is unity, i.e,</p>
<p>the density matrix takes the form (after diagonalizing):
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 \\
\end{bmatrix}}
\end{equation}</p>
<p>For mixed state: $\hat{\rho}=\sum \limits_{i}P_{i}|\psi_{i}\rangle\langle\psi_{i}|$ </p>
<p>For any state: $Tr(\hat{\rho})=\sum\limits_{i}P_{i}=1$</p>
<p>For pure state: $\hat{\rho}=|\psi\rangle\langle\psi|$</p>
<p>$|\psi\rangle$ is the statevector of the system</p>
<p>$P_{i}$ is the probability to be in the state $|\psi_{i}\rangle$, which are the eigenvalues of the density operator.</p>
| Martin Argerami | 22,857 | <p>Since $\langle \psi|\psi\rangle=1$, we have
$$
(|\psi\rangle\langle\psi|)^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|
$$
Then $|\psi\rangle\langle\psi|$ is a projection, and its eigenvalues are $0$ and $1$. Since we know that the trace of $|\psi\rangle\langle\psi|$ is one we conclude that, counting multiplicities, the eigenvalues of $|\psi\rangle\langle\psi|$ are $1,0,\ldots,0$. So its diagonal form is
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 \\
\end{bmatrix}}
\end{equation}</p>
|
752,045 | <p>Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$</p>
<p>First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).</p>
<p>Here is what I did for the first question:</p>
<p>The area $D$ can be written as:</p>
<p>$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$</p>
<p>Where: </p>
<p>$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$
$$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$
$$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$
$$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$
$$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$</p>
<p>First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?</p>
<p><img src="https://i.stack.imgur.com/frEJp.jpg" alt="enter image description here"></p>
<p>P.S. This is a homework.</p>
| mookid | 131,738 | <p>You should write
$$
D =\{(x,y): -1<x<1, -1<y<1+x^2
\}
-
\{ (x,y): 0<x<1, -\sqrt{x}<y<\sqrt{x}
\}
$$</p>
|
752,045 | <p>Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$</p>
<p>First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).</p>
<p>Here is what I did for the first question:</p>
<p>The area $D$ can be written as:</p>
<p>$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$</p>
<p>Where: </p>
<p>$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$
$$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$
$$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$
$$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$
$$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$</p>
<p>First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?</p>
<p><img src="https://i.stack.imgur.com/frEJp.jpg" alt="enter image description here"></p>
<p>P.S. This is a homework.</p>
| Ellya | 135,305 | <p>It will take a while but to write as unions you need to break things down I.e. for the top left section you can write as $\{-1\leq x\leq 1, 0\leq y\leq 1\}\cup\{-1\leq x\leq 1,1\leq y\leq 1+x^2\}$. And continue this for the other areas.</p>
<p>Btw $R$ refers to the whole region you have written as unions, and A just refers to the fact that you have an area integral.</p>
|
1,634,325 | <blockquote>
<p><strong>Problem</strong>:
Is there sequence that sublimit are $\mathbb{N}$? If it's eqsitist prove this.</p>
</blockquote>
<p>I try to solve this problem by guessing what type of sequence need to be. <br>For example:
$a_n=(-1)^n$ has two sublimit $\{1,-1\}$.
<br>
$a_n=n
\times\sin(\frac{\pi}{2})$ has 0 sublimit because $\lim _{x\to \infty}{n\times\sin(\frac{\pi}{2})}=\infty$. <br>
But I don't know how to solve this so please give me a hint.</p>
| Community | -1 | <p>There is such a sequence. For example make it by counting. Fix a natural $n$, first count to $n$, then start again and count to $n+1$, then to $n+2$ etc. </p>
|
887,200 | <p>So I have the permutations:
$$\pi=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 3 & 7 & 1 & 6 & 5 & 4 & 9 & 8
\end{array} \right)$$
$$\sigma=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
9 & 5 & 6 & 8 & 7 & 1 & 2 & 4 & 3
\end{array} \right)$$</p>
<p>I found $\pi\sigma$ and $\sigma\pi$ to be</p>
<p>$$\pi\sigma=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
5 & 6 & 2 & 9 & 1 & 7 & 8 & 3 & 4
\end{array} \right)$$</p>
<p>$$\sigma\pi=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
8 & 6 & 5 & 9 & 4 & 2 & 3 & 1 & 7
\end{array} \right)$$.</p>
<p>So my question is: did I do these correctly or did I mix them up (i.e, the permutation matrix I have for $\pi\sigma$ is actually $\sigma\pi$)?</p>
| amWhy | 9,003 | <p>You've done just fine - EXCEPT: the first is $\sigma \pi$ and the second is $\pi\sigma$.</p>
<p>$$\sigma \pi = \sigma\circ \pi = \sigma(\pi(i))\quad i = 1, 2, \ldots, 9\, $$ So we can only find $\sigma(\pi(i))$ after finding $\pi(i)$.</p>
|
179,230 | <p>I draw <a href="http://reference.wolfram.com/language/ref/Cos.html" rel="nofollow noreferrer"><code>Cos</code></a> function using the code line : </p>
<pre><code>GraphicsColumn[
{
Plot[Cos[0.0625*Pi x], {x, 0, 40*Pi}, Axes -> False],
Plot[Cos[0.0625*Pi x], {x, 0, 40*Pi}, Axes -> False],
Plot[Cos[0.0625*Pi x], {x, 0, 40*Pi}, Axes -> False],
Plot[Cos[0.0625*Pi x], {x, 0, 40*Pi}, Axes -> False]
}
,ImageSize -> Large
]
</code></pre>
<p>and manually add circles and arrows as you can see in the attached picture , how can I export the figure . export doesn't work and I can't save it or copy it please help<br>
<a href="https://i.stack.imgur.com/eSZ4E.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eSZ4E.jpg" alt="see image"></a></p>
| user7739 | 7,739 | <p>I know this post could be shorter, but I wanted to outline a methodology, as in teach a man to fish..., so please bear with me. The first point to make is that the geometric scale of x vs y is easiest to manage from within the Graphics[] environment. Each Plot[] command creates its own coordinate context and so you can introduce distortions you don't intend unless you apply a great deal of care. Secondly, mixing different graphical elements is easiest to do in Graphics[] as well.</p>
<p>First of all, decide how to draw the 4 cosine curves. I noticed that your diagrams simply have 4 cycles and no axes so I dispensed with the frequency and duration you used. Here we have a quick set of 4 cosines in a column:</p>
<pre><code>With[{baseline = -2.5 {0, 1, 2, 3}, t = Range[-2, 2, 0.01]},
Module[{s},
s = Cos[2 Pi t];
Graphics[{
Table[Line@Thread@{t, s + baseline[[k]]}, {k, 1, 4}]
}]
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/o4JjL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o4JjL.png" alt="enter image description here"></a></p>
<p>Notice I separated the curves by a little more than their peak to peak height.</p>
<p>Now we want to draw the dots in a systematic way. Imagine each set of dots is drawn with respect to a point on the baseline associated with the cosine curve. We can use the baseline index to also choose that time position.</p>
<pre><code>With[{baseline = -2.5 {0, 1, 2, 3}, t = Range[-2, 2, 0.01], tdots = {-1.5, -0.5, 0.5, 1.5}},
Module[{s, dots},
s = Cos[2 \[Pi] t];
dots[k_] := With[{td = tdots[[k]], y0 = baseline[[k]] + 0.15, dy = 0.4, r = 0.16},
{Red, Disk[{td, y0 + dy}, r], Disk[{td, y0}, r], Disk[{td, y0 - dy}, r]}
];
Graphics[{
Table[Line@Thread@{t, s + baseline[[k]]}, {k, 1, 4}],
Table[dots[k], {k, 1, 4}]
}]
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/eDczd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eDczd.png" alt="enter image description here"></a></p>
<p>Now let's add the arrows. We will call them hop's. Here we will also assign the output to the variable g for export below.</p>
<pre><code>g = With[{baseline = -2.5 {0, 1, 2, 3}, t = Range[-2, 2, 0.01],
tdots = {-1.5, -0.5, 0.5, 1.5}},
Module[{s, dots, hop},
s = Cos[2 \[Pi] t];
dots[k_] :=
With[{td = tdots[[k]], y0 = baseline[[k]] + 0.15, dy = 0.4, r = 0.16},
{Red, Disk[{td, y0 + dy}, r], Disk[{td, y0}, r], Disk[{td, y0 - dy}, r]}
];
hop[k1_, k2_] := With[{y = baseline[[k1]] + 1, t1 = tdots[[k1]], t2 = tdots[[k2]]},
Arrow[BezierCurve[
{{t1, y}, {0.9 t1 + 0.1 t2, y + 0.5}, {0.1 t1 + 0.9 t2, y + 0.5}, {t2, y}}]
] ];
Graphics[{
Table[Line@Thread@{t, s + baseline[[k]]}, {k, 1, 4}],
Table[dots[k], {k, 1, 4}],
hop[1, 2], hop[2, 3], hop[3, 4], hop[4, 1]
}]
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/4h63w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4h63w.png" alt="enter image description here"></a></p>
<p>You want to export to a pixel oriented format like png as opposed to a photographic image format like jpg because the jpg image compression can introduce artifacts. This command will save the png file in the same directory as the notebook containing this code.</p>
<pre><code>Export[FileNameJoin[{NotebookDirectory[], "dot_hops.png"}],g,ImageResolution->200]
</code></pre>
|
2,883,023 | <p>Find the number of zeros of $f(z)=z^6-5z^4+3z^2-1$ in $|z|\leq1$. </p>
<p>My attempts have not gotten far. </p>
<p>I know we can examine the related equation $f(w)=w^3-5w^2+3w-1$ in $|w|\leq1$, letting $w=z^2$.</p>
<p>It is clear that $f(w)=0$ for $|w|=1$ if and only if $w=-1$. </p>
<p>My main problem is that this seems obviously like an application of Rouché's theorem, where we let $g(w)=5w^2$, whereupon we have the relation $|f(w)|< |g(w)|$ for all $|w|=1$, with the sole exception that equality holds when $w=-1$. </p>
<p>I would like to use Rouché to equate the number of zeros of $f$ and $g$, but my understanding of Rouché is that the inequality should be strict with no exceptions. </p>
| Doug M | 317,162 | <p>Rouche's theorem.</p>
<p>if $|g(z)| > |h(z)|$ for all $z$ along some closed contour. $g(z) + h(z)$ has as many zeros inside the contour as $g(z)$ has inside the contour.</p>
<p>But what about when $|g(z)| = |h(z)|$? Then there is the possibility that some zeros lie on the contour.</p>
<p>In this case
let $g(z) = 5z^4, h(z) = z^6 + 3z^2 - 1$
on the contour $|z| = 1$</p>
<p>$|g(z)| = 5, |h(z)| \le 5$</p>
<p>There will be $4$ zeros in the disk $|z|\le 1$ with some possibly on the contour.</p>
|
96,080 | <p>The empty clause is a clause containing no literals and by definition is false.</p>
<p>c = {} = F</p>
<p>What then is the empty set, and why does it evaluate to true?</p>
<p>Thanks!</p>
| Isnax | 554,236 | <p>You can get an intuition why this is so by observing that:</p>
<ol>
<li>A disjunction is true iff there exists a member which is true. In an empty disjunction (empty clause) there is no such member, so it is always false.</li>
<li>A conjunction is true iff no member which is false exists. An empty conjunction (empty set in cnf problems) has no member (and a fortiori no member which is false) so it is always true.</li>
</ol>
<p>It is a similar intuition to the idea of universal quantification being true on the empty domain.</p>
|
974,656 | <p><img src="https://i.stack.imgur.com/LyqzL.jpg" alt="enter image description here"></p>
<p>One way to solve this and my book has done it is by : </p>
<p><img src="https://i.stack.imgur.com/2wYSn.jpg" alt="enter image description here"></p>
<hr>
<p>This is a well known way, but I have a different method, and it seems logical to me (but I don't know what the mistake is). And yes it's wrong, but I don't understand what's wrong with my following method :</p>
<p>For 1 component the mean is $E(X)=2.5$ so for 5 components it's : </p>
<p>$$E(5X)=5E(X)=5(2.5)=12.5$$</p>
<p>So for 5 items we can say : </p>
<p>$$\lambda= 1/E(5X)= 1/12.5$$</p>
<p>$$X ~ Expo (1/12.5)$$</p>
<p>$$P(T \geq 3)=1-e^{-3/12.5}$$
$$P(T \geq3)=0.21$$</p>
<p>Which is not the same as in the book. Please help me, what is wrong with my method.</p>
| Petite Etincelle | 100,564 | <p>Suppose $a_0 < 0$ and $a_n >0$</p>
<p>Firstly we have $P(0) = a_0 < 0$.</p>
<p>$P$ is of even degree, which means $\lim_{x \to +\infty} P(x) = \lim_{x\to -\infty}P(x) = +\infty$, since $a_n > 0$</p>
<p>Apply IVT on $[0, +\infty)$ and $(-\infty, 0]$, we find two roots for $P(x)$.</p>
<p>If $a_0 > 0$ and $a_n <0$ we can argue in a similar way.</p>
|
28,532 | <p><code>MapIndexed</code> is a very handy built-in function. Suppose that I have the following list, called <code>list</code>:</p>
<pre><code>list = {10, 20, 30, 40};
</code></pre>
<p>I can use <code>MapIndexed</code> to map an arbitrary function <code>f</code> across <code>list</code>:</p>
<pre><code>{f[10, {1}], f[20, {2}], f[30, {3}], f[40, {4}]}
</code></pre>
<p>where the second argument to <code>f</code> is the part specification of each element of the list.</p>
<p>But, now, what if I would like to use <code>MapIndexed</code> only at certain elements? Suppose, for example, that I want to apply <code>MapIndexed</code> to only the second and third elements of <code>list</code>, obtaining the following:</p>
<pre><code>{10, f[20, {2}], f[30, {3}], 40}
</code></pre>
<p>Unfortunately, there is no built-in "<code>MapAtIndexed</code>", as far as I can tell. What is a simple way to accomplish this? Thanks for your time.</p>
| Michael E2 | 4,999 | <p>Another way, using <code>MapIndexed</code>'s functionality, like <a href="https://mathematica.stackexchange.com/users/5">rm-rf</a>'s:</p>
<pre><code>mapAtIndexed[f_, expr_, pos_, levelspec_: 1, opts : OptionsPattern[MapIndexed]] :=
Module[{f0},
f0[x_, p : Alternatives @@ pos] := f[x, p];
f0[x_, _] := x;
MapIndexed[f0, expr, levelspec, opts]
]
</code></pre>
<p>OP's example:</p>
<pre><code>mapAtIndexed[g, list, {{2}, {3}}]
(* {10, g[20, {2}], g[30, {3}], 40} *)
</code></pre>
<p>Multiple levels (pay attention to the <code>{2}</code>):</p>
<pre><code>mapAtIndexed[f, Table[10 i + j, {i, 4}, {j, 3}],
{{1, 2}, {1, 3}, {2}, {2, 1}, {3, 2}, {4, 3}}, 2]
(* {{11, f[12, {1, 2}], f[13, {1, 3}]},
f[{f[21, {2, 1}], 22, 23}, {2}],
{31, f[32, {3, 2}], 33},
{41, 42, f[43, {4, 3}]}} *)
</code></pre>
<p>Heads, too:</p>
<pre><code>mapAtIndexed[f, list, {{2}, {4}, {0}}, Heads -> True]
(* f[List, {0}][10, f[20, {2}], 30, f[40, {4}]] *)
</code></pre>
<hr>
<p><strong>Alternate solution</strong></p>
<p>In this we have to ensure <code>f</code> is mapped at the lower levels first, using <code>Reverse @ Sort @ ...</code>, since we're not using <code>Map</code>. However it's quite a bit faster.</p>
<pre><code>mapAI2[f_, expr_, pos_] := Module[{e0 = expr},
(e0[[Sequence @@ #]] = f[e0[[Sequence @@ #]], #]) & /@ Reverse @ Sort @ pos;
e0
]
</code></pre>
<p>Example</p>
<pre><code>mapAI2[f, Table[10 i + j, {i, 4}, {j, 3}],
{{1, 2}, {3, 0}, {1, 3}, {2}, {2, 1}, {3, 2}, {4, 3}}]
(* {{11, f[12, {1, 2}], f[13, {1, 3}]},
f[{f[21, {2, 1}], 22, 23}, {2}],
f[List, {3, 0}][31, f[32, {3, 2}], 33],
{41, 42, f[43, {4, 3}]}} *)
</code></pre>
|
113,797 | <p>I'm trying to extract every 21st character from this text, s (given below), to create new strings of all 1st characters, 2nd characters, etc.</p>
<p>I have already separated the long string into substrings of 21 characters each using</p>
<pre><code> splitstring[String : str_, n_] :=
StringJoin @@@ Partition[Characters[str], n, n, 1, {}]
</code></pre>
<p>giving:</p>
<pre><code> {"ALJJJEAQZJMZKOZDKEHBL", "XLPXNEHZCSEJVVLWHTUDJ", \
"WFYXKKMWNNTNPHDTMGIOP", "OOSYPXGTLOHOPHTDHBHWO", \
"MWGSKXSTNNEYSQHRSGPKP", "CJBNVIYCZHIVPFSWCKFPJ", \
"OZQLNGPTLCIALHMBIGUOP", "ESYNDGACTURTALHLSGFBR", \
"LPRMYKFQFXTEEZQHIUMOC", "CLSUIKWYLRPRRJZWCKUOW", \
"WJLLVYEJXNIEMDQYQTDFC", "HJQLYKFQJTKUCICIRKOWX", \
"PHLWYCCDRKVPAHCYPZFRL", "CNYBIOEWWGHQQBCDHGPRW", \
"WHWIUOQTYOJLHGLFRTTVL", "CQCIUEHZYJKJEWHOVMYOM", \
"JOBTIHCOSGCZVZJFEYHAC", "JQCQFSTWLOHYPXZDHBHTW", \
"TLGIUJIWEXSJGKLTMKAFF", "PYGYICGSPRPLPOZNKPUSH", \
"NSMQCGPWHILVXLZARZLIS", "POBDNEXJYMESQDTAQWKWL", \
"WPBIDCCJXKHQCOTAJXBVZ", "WWDNCCCCWACQHZJREEHRO", \
"LQSSRCVPFHCJCGLURBHFM", "POYKFKWRHKVGLLSYRTISC", \
"ESYKXCEYFNSAHUHYIBJVL", "WZLKPJHQJEQQVHNHSEHTO", \
"TNFJCRMZCCIEXKSALTMFP", "YTLGCORTMRIAILOISWKRJ", \
"DTMRXGVYHEHRSIPRIWKOX", "SLBQVJHSZTMZMIHRAFFTJ", \
"CTMSYGADQEHQXUZIQHLZJ", "OOGRZGVHLKPTIUEHHKHKT", \
"CHCNCMMYLIMBWWHNKXBQP", "DEWQCIVZRNETVVPFCVYST", \
"RTYZYEPWZTMQHLNHSGPRO", "TNFJCRRLNNPRHGYARRJVW", \
"HJBIFTAPWRVUCZODYYLGF", "COBDWKMDTGJCEBDTVRDRO", \
"HJQLPLVHJYKNJSLGEBZDL", "OOWKROWOOOJRXKRAMYMMM", \
"FOBFTGTSLHIGLQLWVVLTL", "TDYBEGVNJLEAQDPRQXKRH", \
"WOGIUCPLCJEAJBYGLTSCY", "OCUDECCQCURAELOAPEHKP", \
"YJBIVOPWZTEVHJHNEYNMX", "CFSHYKPPWCGUMEWYKNHZW", \
"JQSWCRANSOALVIJLPRZDL", "YOFDJVCDHTAVAHTRMTBHP", "RJZBIOEOSCR"}
</code></pre>
<p>How can I make strings of all the first characters, second characters, etc.?</p>
<pre><code> s=ALJJJEAQZJMZKOZDKEHBLXLPXNEHZCSEJVVLWHTUDJWFYXKKMWNNTNPHDTMGIOPOOSYPXG\
TLOHOPHTDHBHWOMWGSKXSTNNEYSQHRSGPKPCJBNVIYCZHIVPFSWCKFPJOZQLNGPTLCIALH\
MBIGUOPESYNDGACTURTALHLSGFBRLPRMYKFQFXTEEZQHIUMOCCLSUIKWYLRPRRJZWCKUOW\
WJLLVYEJXNIEMDQYQTDFCHJQLYKFQJTKUCICIRKOWXPHLWYCCDRKVPAHCYPZFRLCNYBIOE\
WWGHQQBCDHGPRWWHWIUOQTYOJLHGLFRTTVLCQCIUEHZYJKJEWHOVMYOMJOBTIHCOSGCZVZ\
JFEYHACJQCQFSTWLOHYPXZDHBHTWTLGIUJIWEXSJGKLTMKAFFPYGYICGSPRPLPOZNKPUSH\
NSMQCGPWHILVXLZARZLISPOBDNEXJYMESQDTAQWKWLWPBIDCCJXKHQCOTAJXBVZWWDNCCC\
CWACQHZJREEHROLQSSRCVPFHCJCGLURBHFMPOYKFKWRHKVGLLSYRTISCESYKXCEYFNSAHU\
HYIBJVLWZLKPJHQJEQQVHNHSEHTOTNFJCRMZCCIEXKSALTMFPYTLGCORTMRIAILOISWKRJ\
DTMRXGVYHEHRSIPRIWKOXSLBQVJHSZTMZMIHRAFFTJCTMSYGADQEHQXUZIQHLZJOOGRZGV\
HLKPTIUEHHKHKTCHCNCMMYLIMBWWHNKXBQPDEWQCIVZRNETVVPFCVYSTRTYZYEPWZTMQHL\
NHSGPROTNFJCRRLNNPRHGYARRJVWHJBIFTAPWRVUCZODYYLGFCOBDWKMDTGJCEBDTVRDRO\
HJQLPLVHJYKNJSLGEBZDLOOWKROWOOOJRXKRAMYMMMFOBFTGTSLHIGLQLWVVLTLTDYBEGV\
NJLEAQDPRQXKRHWOGIUCPLCJEAJBYGLTSCYOCUDECCQCURAELOAPEHKPYJBIVOPWZTEVHJ\
HNEYNMXCFSHYKPPWCGUMEWYKNHZWJQSWCRANSOALVIJLPRZDLYOFDJVCDHTAVAHTRMTBHP\
RJZBIOEOSCR
</code></pre>
| Edmund | 19,542 | <p>The last substring does not contain 21 characters but you can use the padded form of <code>Partition</code> to get sublist of equal length with padding. Then it is just a <code>Transpose</code> and <code>StringJoin</code> to the result. I use <code>"&"</code> for padding which you can substitute for any character.</p>
<pre><code>StringJoin@Flatten@Transpose@Partition[Characters[s], 21, 21, {1, 1}, "&"]
</code></pre>
<p>Hope this helps.</p>
|
172,894 | <p>Suppose $A$ is an integral domain with integral closure $\overline{A}$ (inside its fraction field), $\mathfrak{q}$ is a prime ideal of $A$, and $\mathfrak{P}_1,\ldots,\mathfrak{P}_k$ are the prime ideals of $\overline{A}$ lying over $\mathfrak{q}$. Show that $\overline{A_\mathfrak{q}} = \bigcap\overline{A}_\mathfrak{P_i}$ (note that the LHS is the integral closure of a localization, whereas the RHS is the intersection of localizations of integral closures of $A$).</p>
<p>If it would help, I suppose we could assume that $A$ has dimension 1, so that $\overline{A}$ is Dedekind, though I don't think that assumption is required.</p>
<p>(Geometrically, we're comparing the integral closure of a local ring at a singular point Q of some variety with the intersection of local rings at points in the normalization mapping to Q).</p>
<p>Thanks.</p>
| Cantlog | 37,134 | <p>First notice that without extra hypothesis, the integral closure needs not be finite over $A$, and there might be infinitely many prime ideals lying over a given one in $A$. Neverthless, the equality holds. </p>
<p>It is easy to see that the LHS is contained in the RHS because the latter is integrally closed and contained in the field of fractions of $A_q$. Let $f$ be an element of the RHS (possibly infinite intersection). Denote by $B=\overline{A}$. Consider the denominator ideal
$$ I=\{ b\in B \mid bf\in B\}$$
of $f$. I claim that $I\cap A$ is not contained in $q$. Admitting this, let $s\in I\cap A\setminus q$, then $sf\in B$ and $f$ is integral over $A_q$. </p>
<p>Now we prove the claim. Suppose that $I\cap A\subseteq q$. Then, geometrically, $q\in V(I\cap A)\subseteq \mathrm{Spec}(A)$. As $A/(I\cap A)\to B/I$ is injective and integral, $\mathrm{Spec}(B/I)\to \mathrm{Spec}(A/(I\cap A))$ is surjective. Identify $V(I)$ with $\mathrm{Spec}(B/I)$, this implies that there exists $p\in V(I)$ such that $p\cap A=q$. So $p$ is a prime ideal of $B$ lying over $q$ and containing $I$. This contradicts the hypothesis $f\in B_p$.</p>
|
464,426 | <p>Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$</p>
<p>How should I approach it? I tried to use L'Hopital's Rule but it's just keep giving me 0/0.</p>
| André Nicolas | 6,312 | <p>L'Hospital's Rule works just fine, and in one step. Remember that you are taking the limit as $x$ approaches $1$.</p>
|
464,426 | <p>Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$</p>
<p>How should I approach it? I tried to use L'Hopital's Rule but it's just keep giving me 0/0.</p>
| Michael Hardy | 11,667 | <p>If $u^{15}= x$ then $x^{1/5}=(u^{15})^{1/5}=u^3$, and similarly $x^{1/3}=u^5$. Therefore
$$
\frac{x^{1/5}-1}{x^{1/3}-1} = \frac{u^3-1}{u^5-1}= \frac{(u-1)(u^2+u+1)}{(u-1)(u^4+u^3+u^2+u+1)}.
$$
Do the obvious cancelation. Then ask: as $x\to1$, then $u\to\text{what}$? Then finding the limit is easy.</p>
<p>You could also apply L'Hopital's rule to $\dfrac{u^3-1}{u^5-1}$ and the answer emerges quite fast.</p>
|
51,555 | <p>Consider the set $(s_1, ..., s_N) \in S$, where all $s_i$ are positive integers selected from some interval $[M, L]$ and the sum of any $k$ integers in $S$ is required to be unique and to have a distance of at least $d$ from all other possible sums of $k$ integers in the set. For example, if $k = 2$, the sum of any two randomly chosen elements, i.e. $s_a$ and $s_b$, must generate a unique sum that has a difference of at least $d$ from all other possible sums of two elements in $S$. </p>
<p>As a function of $k$, the minimum difference between sums $d$, and the size of the interval, $(L - M)$, what is the largest possible size of the set $S$, and what is the most efficient way to compute its elements?</p>
<p>Edit - I'm of course happy to hear any $d = 1$ solutions. </p>
<p>Edit 2 - For the $d=1$ case, is this problem NP-complete, similar to the subset-sum problem?</p>
| Gerry Myerson | 8,269 | <p>Here's a start. The number of choices of $k$ elements is $N\choose k$ (or a little more, if repeats are allowed), and each sum is between $kM$ and $kL$, so you must have $${N\choose k}\lt kL-kM+2$$ which gives you an upper bound on $N$. It doesn't tell you whether you can achieve that bound, nor an efficient way to try. </p>
<p>EDIT: Ah, you've changed the problem, so my answer is now relevant only to your case $d=1$. But if you divide the right side of the displayed inequality by $d$, it should apply to your more general question. </p>
|
51,555 | <p>Consider the set $(s_1, ..., s_N) \in S$, where all $s_i$ are positive integers selected from some interval $[M, L]$ and the sum of any $k$ integers in $S$ is required to be unique and to have a distance of at least $d$ from all other possible sums of $k$ integers in the set. For example, if $k = 2$, the sum of any two randomly chosen elements, i.e. $s_a$ and $s_b$, must generate a unique sum that has a difference of at least $d$ from all other possible sums of two elements in $S$. </p>
<p>As a function of $k$, the minimum difference between sums $d$, and the size of the interval, $(L - M)$, what is the largest possible size of the set $S$, and what is the most efficient way to compute its elements?</p>
<p>Edit - I'm of course happy to hear any $d = 1$ solutions. </p>
<p>Edit 2 - For the $d=1$ case, is this problem NP-complete, similar to the subset-sum problem?</p>
| Julián Aguirre | 4,791 | <p>This is an algorithm that I am convinced is optimal, although I have not tried to prove it. First of all, we may assume that $L=0$ (consider ${s_1-L,\dots,s_N-L}$). For the algorithm I will describe, we may also asume that $d=1$.</p>
<p>The key is to observe that if $\{s_1,\dots,s_N\}$ is such that all sums of subsets of $k$ elements are different, so must be the sums of subsets of $m$ elements for $1\le m\le k$. This implies that $\{s_1,\dots,s_k\}$ must be the first $k$ terms of the sequence <a href="http://oeis.org/A005255" rel="nofollow">A005255</a>. Let $n>k$ and suppose that $\{s_1,\dots,s_n\}$ is such that the sums of the subsets of $m$ elements are all different for $1\le m\le k$. Then we choose $s_{n+1}$ in such a way that for each $m$, $1\le m\le k$, the smallest possible sum of $m$ elements, one of them beeing $s_{n+1}$, is equal to the largest possible sum of $m$ terms in $\{s_1,\dots,s_n\}$ plus $1$. If the $s_j$'s are ordered in increasing order, this means
$$
s_{n+1}=\max_{1\le m\le k}\Bigl(\,\sum_{j=n-m+1}^ns_j-\sum_{j=1}^{m-1}s_j\Bigr)+1.
$$
It is easy to show that for all $n>k$
$$
\sum_{j=n-k}^{n-1}s_j-\sum_{j=1}^{k-1}s_j\le s_n\le2\,s_{n-1}.
$$
In particular $s_n$ is <em>almost</em> like $2^n$.</p>
<p>For $d>1$, the algorithm will give $d$ times the solution with $d=1$.</p>
<p>The algorithm is implemented in the following Mathematica ce; you have to provide the values of $k$ and $M$. </p>
<pre><code>k = 5;
M = 1000000;
s[1] = 0;
s[2] = 1;
s[n_Integer] :=
s[n] = Max[
Table[Sum[s[j], {j, n - m, n - 1}] - Sum[s[j], {j, m - 1}] +
1, {m, n - 1}]];
sol = Table[s[j], {j, k}];
If[Last[sol] <= S,
While[next =
Max[Table[
Total[Take[sol, -m]] - Total[Take[sol, m - 1]] + 1, {m, k}]];
next <= S, AppendTo[sol, next]]; Print[sol],
Print["M is too small; try again with a larger M"]]
{0,1,2,4,7,13,24,46,88,172,337,661,1298,2550,5012,9852,19367,38073,74848,147146,289280,568708}
</code></pre>
|
3,999,996 | <p>I know how you can show this geometrically, but is there any way to prove this algebraically?</p>
| paulinho | 474,578 | <p>Since <span class="math-container">$\arcsin x$</span> is bounded above by <span class="math-container">$\pi / 2$</span>, by monotonicity of the integral we have
<span class="math-container">$$\int_0^1 x \arcsin x dx \leq \frac \pi 2 \int_0^1 x dx = \frac \pi 4$$</span></p>
|
1,501,876 | <blockquote>
<p>I want to prove $A_n$ has no subgroups of index 2. </p>
</blockquote>
<p>I know that if there exists such a subgroup $H$ then $\vert H \vert = \frac{n!}{4}$ and that $\vert \frac{A_n}{H} \vert = 2$ but am stuck there. I have tried using the proof that $A_4$ has no subgroup of order 6 to get some ideas but am still stuck. Sorry I don't have much else to add at this point. Thanks a bunch.</p>
| MooS | 211,913 | <p>This elementary argument works for all $n$ at once and does not need simplicity of the alternating group: Let $N$ be of index $2$ in $A_n$ and let $x$ be a $3$-cycle. We have $xN=x^4N=(xN)^4=N$, where the latter holds since the factor group is of order $2$. We derive that $N$ contains all $3$-cycles. But it is a well known (and easy to show) fact that the alternating group is generated by $3$-cycles.</p>
<p>Of course we implictly used that a subgroup of index $2$ is always normal.</p>
|
2,869,898 | <p>I want to prove that <span class="math-container">$$
f(x,y)=
\begin{cases} \frac{xy^2}{x^2+y^2} &\text{ if }(x,y)\neq (0,0)\\
0 &\text{ if }(x,y)=(0,0)
\end{cases}
$$</span>
is not differentiable at <span class="math-container">$(0,0)$</span>.</p>
<p>I thought that I can prove that it is not continuous around <span class="math-container">$(0,0)$</span> but it certainly is!</p>
<p>So how can I prove that it is not differentiable?</p>
| Fred | 380,717 | <p>Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) \in \mathbb R^2$ with $u^2+w^2=1$ we then have</p>
<p>$$(*) \quad \frac{\partial f}{\partial v}(0,0)=v \cdot gradf(0,0).$$</p>
<p>Since </p>
<p>$$ (**) \quad\frac{\partial f}{\partial v}(0,0)= \lim_{t \to 0}\frac{f(tv)-f(0,0)}{t}=\frac{uw^2}{u^2+w^2}=uw^2,$$</p>
<p>we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$</p>
<p>$$\frac{\partial f}{\partial v}(0,0)=0.$$</p>
<p>But if $u=w= \frac{1}{\sqrt{2}}$, then by $(**)$</p>
<p>$$\frac{\partial f}{\partial v}(0,0) \ne 0.$$</p>
<p>A contradiction !</p>
<p>Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.</p>
|
2,869,898 | <p>I want to prove that <span class="math-container">$$
f(x,y)=
\begin{cases} \frac{xy^2}{x^2+y^2} &\text{ if }(x,y)\neq (0,0)\\
0 &\text{ if }(x,y)=(0,0)
\end{cases}
$$</span>
is not differentiable at <span class="math-container">$(0,0)$</span>.</p>
<p>I thought that I can prove that it is not continuous around <span class="math-container">$(0,0)$</span> but it certainly is!</p>
<p>So how can I prove that it is not differentiable?</p>
| Rene Schipperus | 149,912 | <p>Look at the directional derivative:</p>
<p>$$d_{\mathbf{v}}f=\lim\limits_{t\to 0}\frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$</p>
<p>in your case you have for $\mathbf{x}=(0,0)$</p>
<p>$$d_{(a,b)}f=\lim\limits_{t\to 0}\frac{\frac{t^3ab^2}{t^2(a^2+b^2)}}{t}=\frac{ab^2}{a^2+b^2}$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable. </p>
|
86,422 | <p>In his notes in Algebraic Number theory, J S Milne gives the following as an example of an unramified Abelian extension :</p>
<p>$ K = \mathbb Q (\sqrt{-5})$ having a quadratic extension $L = \mathbb Q (\sqrt{-1}, \sqrt{-5})$.
Then, $L/K$ has discriminant a unit, so it ramifies. </p>
<p>My question is, considering the simple extension $L = K(i)$ gives the discriminant to be $-4$, which clearly isn't a unit in $\mathcal O_K$. Am I committing any mistake? </p>
<p>Can you suggest other examples of unramified extensions?</p>
<p>Since I am a beginner in Class Field theory, related examples (of Abelian / non-Abelian extensions), counter-examples and other insights are more than welcome. </p>
| David Loeffler | 2,481 | <p>You are slipping up because $i$ does not generate the ring of integers of $L$ as an $\mathcal{O}_K$-algebra: we have $\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]$, but $\mathcal{O}_L = \mathbb{Z}\left[i, \frac{1 + \sqrt{5}}{2}\right] \ne \mathbb{Z}[i, \sqrt{-5}]$. Hence the discriminant of $L/K$ is not the same as the discriminant of the order $\mathbb{Z}[i, \sqrt{-5}]$, which is what you've calculated.</p>
|
86,422 | <p>In his notes in Algebraic Number theory, J S Milne gives the following as an example of an unramified Abelian extension :</p>
<p>$ K = \mathbb Q (\sqrt{-5})$ having a quadratic extension $L = \mathbb Q (\sqrt{-1}, \sqrt{-5})$.
Then, $L/K$ has discriminant a unit, so it ramifies. </p>
<p>My question is, considering the simple extension $L = K(i)$ gives the discriminant to be $-4$, which clearly isn't a unit in $\mathcal O_K$. Am I committing any mistake? </p>
<p>Can you suggest other examples of unramified extensions?</p>
<p>Since I am a beginner in Class Field theory, related examples (of Abelian / non-Abelian extensions), counter-examples and other insights are more than welcome. </p>
| Franz Lemmermeyer | 3,503 | <p>A quick way of seeing what's going on is using the fact that
$L = K(i) = K(\sqrt{5})$; the fact that the different above (say of $L/{\mathbb Q}(i)$ divides the different below (e.g. of ${\mathbb Q}(\sqrt{5})/{\mathbb Q}$) shows (take norms) that the discriminant of $L/K$ divides both $-4$ and $5$, hence is trivial. This works for a lot of other examples, too.</p>
|
4,139,141 | <p>If the conditions of the theorem are met for some ordinary differential equation, then we are guaranteed that a solution exists. However, I don't fully understand what it means for a solution to exist. If we can show that a solution exists, does that mean that it can be found explicitly using known methods? Or, are there some differential equations, that we know exist because of the theorem, but for which we can not find a general solution, and are thus forced to use numerical methods for an approximation?</p>
| Christopher K | 101,768 | <p>Here's some guidance to each of your questions. Hopefully this helps!</p>
<ol>
<li>For all <span class="math-container">$a \in A$</span>, there exist disjoint open sets <span class="math-container">$V_{a} \ni a$</span> and <span class="math-container">$U_{a} \ni x$</span>. Since <span class="math-container">$A = \cup_{a \in A} \{a\} \subseteq \cup_{a \in A} V_{a}$</span> is a union of open sets that covers <span class="math-container">$A$</span>, it is an open cover.</li>
<li>Since <span class="math-container">$A$</span> is compact, all open covers (including this one) admit a finite subcover <span class="math-container">$\{V_{a_{j}}\}_{j=1}^{n}$</span> (that is, <span class="math-container">$A \subseteq \cup_{j=1}^{n} V_{a_{j}} = V$</span>) for some <span class="math-container">$\{a_{j} \}_{j=1}^{n} \subseteq A$</span>.</li>
<li>Note that <span class="math-container">$U = \cap_{j=1}^{n} U_{a_{j}}$</span> is a finite intersection of open sets containing <span class="math-container">$\{x \}$</span> and so <span class="math-container">$x \in U \neq \emptyset$</span> is open.</li>
<li>Since <span class="math-container">$V$</span> is an open cover of <span class="math-container">$A$</span>, <span class="math-container">$A \subseteq V$</span>. Since <span class="math-container">$x \in U_{a_{j}}$</span> for each <span class="math-container">$a_{j}$</span>, it follows that <span class="math-container">$x \in U$</span>.</li>
<li>The set <span class="math-container">$U$</span> is disjoint from <span class="math-container">$V$</span> since <span class="math-container">$V_{a_{j}} \cap U \subseteq V_{a_{j}} \cap U_{a_{j}} = \emptyset$</span> for each <span class="math-container">$a_{j}$</span>. Note that <span class="math-container">$V \cap U = \cup_{j=1}^{n} (V_{a_{j}} \cap U) = \emptyset$</span>. This is just de Morgan.</li>
<li>Finally <span class="math-container">$A \cap U \subseteq V \cap U = \emptyset$</span> and so <span class="math-container">$A, U$</span> are disjoint.</li>
<li>For <span class="math-container">$x \in X \setminus A$</span>, <span class="math-container">$\exists U \ni x$</span> such that <span class="math-container">$U \subseteq X \setminus A$</span> (or equivalently <span class="math-container">$U \cap A = \emptyset$</span>).</li>
<li>Hence, since <span class="math-container">$x$</span> is arbitrarily chosen from <span class="math-container">$X \setminus A$</span>, it follows that <span class="math-container">$X \setminus A$</span> is open and <span class="math-container">$A$</span> is closed in <span class="math-container">$X$</span>.</li>
</ol>
|
772,315 | <p>Statement: $\limsup\limits_{n\to\infty} c_n a_n = c \limsup\limits_{n\to\infty} a_n$</p>
<p>Please help find a counterexample to this statement if $c<0$.</p>
<p>Edit: also suppose $c_n \to c$ and $\limsup a_n$ is finite</p>
| Community | -1 | <p>Let $c_n = -1 \to -1 = c $, and $a_n = (-1)^n$. then</p>
<p>$$ limsup (c_na_n) = 1$$</p>
<p>$$ -1 ( \limsup(a_n) ) = -1 $$</p>
|
2,173,918 | <p>Let $f(z)=\sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}$. I want to show that this series represents a holomorphic function in the unit disk. I'm, however, quite confused. For example, is $f(z)$ even a power series? It doesn't look as such. Here's what I have so far come up with.</p>
<blockquote>
<p>Proof:</p>
</blockquote>
<p>$$ \sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}=-\sum\limits_{k=1}^\infty\frac{1}{1-z^{-k}}=-\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty z^{kn}$$</p>
<p>[If the above expression is correct then we have a Laurent series of a power series].</p>
<p>Now, $|c_n|=1$ for all $n$. By Parseval's formula,</p>
<p>$$2\pi\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}=\sum\limits_{k=-\infty}^{-1} \int\limits_0^{2\pi}\left|g(\rho^ke^{it})\right|^2dt=^? \int\limits_0^{2\pi} \sum\limits_{k=1}^{\infty} \left|g(\rho^{-k}e^{it})\right|^2dt$$
where $0\le\rho\le 1$, and $g$ is some function (holomorphic) on the unit disk. So we know that this integral (above in the middle) exists.</p>
<p>Now, I believe, what remains to be proved is that the infinite series of this integral also exists. Do you think this approach is OK, or did I make some mistakes in it? How can we prove that the series $\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}$ converges? And then, since it converges, does this imply that $G$ is holomorphic?</p>
<p>Thanks a lot.</p>
| reuns | 276,986 | <blockquote>
<p>Theorem : if $\sum_{n=0}^\infty a_n z^n$ converges absolutely for $|z| < R$ then it is analytic/holomorphic for $|z| < R$.</p>
</blockquote>
<p>For $|z| < 1$ : $$\sum_{k=1}^\infty\frac{z^k}{1-z^k}=\sum_{k=1}^\infty\sum_{n=1}^\infty z^{kn} = \sum_{m=1}^\infty z^{m} \tau(m)$$ where $\tau(m)=\sum_{d | m} 1$ is the number of divisors. Since $\tau(m) <m$ it means that $\sum_{m=1}^\infty z^{m} \tau(m)$ converges absolutely for $|z| < 1$. Hence changing the order of summation doesn't matter, and $\sum_{m=1}^\infty z^{m} \tau(m)=\sum_{k=1}^\infty\frac{z^k}{1-z^k}$ is analytic/holomorphic on $|z| < 1$.</p>
|
4,041,758 | <p>If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve?</p>
<p>More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles)</p>
<p>I saw it in a debate thread and it got me curious now.</p>
| dbmag9 | 73,894 | <p>As you identify in your question, the real point of contention is the definition of <em>angle</em>. As the other answers have indicated, if your definition includes angles at the intersection of two curves then a semicircle certainly has two right angles.</p>
<p>However, as the controversy on Twitter (a few days ago at time of posting, started because this question was in someone's daughter's primary school homework) indicates, <em>angle</em> as commonly used is often restricted to the intersections of (straight) lines. Under this definition there is no angle at the vertices of a semicircle.</p>
<p>Which definition is more common is an empirical question (and the consensus seems to be that the wider definition is common among professional mathematicians, while the stricter definition might be something a layperson would describe), but to my mind the fact that there exist differing definitions makes your question ill-formed. I think the educational value is to expose the fact that we should be careful with how we think about our definition.</p>
|
964,989 | <p>Question: Find the order of $(1/2)^{1/2}$, $(1/e)^{1/e}$, $(1/3)^{1/4}$ without using calculator.</p>
<p>Extra constraint: You only have about 150 seconds to do it, failing to do so will eh... make you run out of time on the exam which affects the chance of admitting into graduate school!</p>
<p>Back to the question, when I first saw it I tried to use $(1/2.7)^{1/3}$ to approximate $(1/e)^{1/e}$, compare the $12^{th}$ power of each number and end up in getting the wrong result.</p>
<p>Later(too late!) I find out $f(x)=(1/x)^{1/x}$ has minimum at $x=e$ by taking the derivative $f'(x)=f(x)(ln(x)-1)/x^2$. In addition, second derivative or some kind of feeling is required to make sure it's minimum not maximum. However I feel this approach might unnecessarily take too long.</p>
<p>Had you seen this question, what would have been your first instinct/approach? Is there any faster way to do it?</p>
| Community | -1 | <p>$$\left(\frac13\right)^{\frac14}>\left(\frac14\right)^{\frac14}=\left(\frac12\right)^{\frac12}.$$</p>
<p>Knowing there is a minimum at $x=e$,</p>
<p>$$\left(\frac1e\right)^\frac1e<\left(\frac12\right)^{\frac12}<\left(\frac13\right)^{\frac14}.$$</p>
<p>Hint:</p>
<p>Take $\log(x^x)=x\log x$, derivative $\log x+1$, second derivative $1/x>0$: it has a minimum at $1/e$.</p>
|
4,492,104 | <p>I'm looking for closed-form expressions for the integral <span class="math-container">$$I:=\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx .$$</span>
Some related integrals that I've found <a href="https://math.stackexchange.com/questions/2354232/is-it-possible-to-evaluate-int-0-infty-log1-coshx-fracx2ex-dx?rq=1">include</a>: <span class="math-container">$$J:=-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx = 6 + 2\ln(2) - {2\pi^{2} \over 3} - 4\zeta(3) - 2\pi i,$$</span>
and <a href="https://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosx-mathrmdx?noredirect=1&lq=1">this</a> one: <span class="math-container">$$K:= \int_0^{\frac{\pi}{2}}\log\Big{(}\cos(x)\Big{)}dx = - \frac{\pi}{2} \log(2) ,$$</span> <a href="https://www.researchgate.net/profile/Mundher-Khaleel/post/Can_anybody_solve_the_integration/attachment/59d62aad79197b80779890e3/AS%3A340147343249408%401458108854707/download/Table_of_Integrals_Series_and_Products_Tablicy_Integralov_Summ_Rjadov_I_Proizvedennij_Engl._2%282%29.pdf" rel="nofollow noreferrer">and</a> (p. 530) <span class="math-container">$$L:=\int_{0}^{\infty} \big{(} \ln(1-e^{-x}) \big{)} dx = - \frac{\pi^{2}}{6} .$$</span></p>
<p>Moreover, I've encountered various integrals of the <a href="https://en.wikipedia.org/wiki/Lobachevsky_integral_formula" rel="nofollow noreferrer">sinc</a> function, <a href="https://www.wolframalpha.com/input?i=integrate+ln%28sin%28x%29%2Fx%29+from+x%3D0+to+pi%2F2" rel="nofollow noreferrer">including</a>: <span class="math-container">$$M:= \int_{0}^{\pi/2} \ln \bigg{(} \frac{\sin(x)}{x} \bigg{)} dx = \frac{\pi}{2} \Big{(} 1 - \ln(\pi) \Big{)} .$$</span> However, I haven't found any closed-form expressions for <span class="math-container">$I$</span> yet, only an approximation: <span class="math-container">$$I \approx -5.75555891011162780816$$</span> and I'm not sure how to proceed with the integral.</p>
| metamorphy | 543,769 | <p>More precise computation of the given integral: <span class="math-container">$$I=-5.031379591902842520548271636746403412607399342991051\cdots$$</span></p>
<p>This is computed using <a href="https://pari.math.u-bordeaux.fr/" rel="nofollow noreferrer">PARI/GP</a>. The integrand is oscillating, so <code>intnum</code> cannot handle it directly, and PARI/GP user's guide "suggests" expanding it into Fourier series. Instead, using <span class="math-container">$$\int_0^\infty f(x)\,dx+\sum_{n=1}^\infty\int_0^\pi\big(f(2n\pi-x)+f(2n\pi+x)\big)\,dx$$</span> for our integrand <span class="math-container">$f(x)=\log(1-\frac{\sin x}{x})$</span>, we have <span class="math-container">$$f(2n\pi-x)+f(2n\pi-x)=\log\left[1-\left(\frac{x-\sin x}{2n\pi}\right)^2\right]-\log\left[1-\left(\frac{x}{2n\pi}\right)^2\right]$$</span> and, recognizing the infinite product for the sine function, we obtain <span class="math-container">$$I=\int_0^\pi\log\frac{\sin\big((x-\sin x)/2\big)}{\sin(x/2)}\,dx=\color{blue}{\int_0^\pi\log\left(2\sin\frac{x-\sin x}{2}\right)dx}.$$</span> I compute this as follows.</p>
<pre><code>sinctail(x)=if(x>1e-3,(1-sinc(x))/x^2,suminf(n=0,(-x^2)^n/(2*n+3)!));
kernel(x)=log(sinctail(x))+log(sinc((x-sin(x))/2));
answer()=3*Pi*(log(Pi)-1)+intnum(x=0,Pi,kernel(x));
</code></pre>
<p>An amusing consequence: using Bessel functions, we have <span class="math-container">$\color{blue}{I=-\pi\sum_{n=1}^\infty J_n(n)/n}$</span>.</p>
<hr />
<p>Another line of thoughts, thanks to the observation by @Tyma Gaidash: <span class="math-container">$$y=y(\lambda,x)=2\sum_{n=1}^\infty J_n(n\lambda)\frac{\sin nx}{n}\qquad(|\lambda|\leqslant 1)$$</span> is <a href="https://en.wikipedia.org/wiki/Kapteyn_series" rel="nofollow noreferrer">the solution</a> of <span class="math-container">$y=\lambda\sin(x+y)$</span>. Writing <span class="math-container">$y(x):=y(1,x)$</span>, we get <span class="math-container">$$-I=\int_0^\infty\frac{y(x)}{x}\,dx=\frac12\int_0^\pi y(x)\cot\frac x2\,dx$$</span> similarly to the above. Now let's get rid of the implicitly defined <span class="math-container">$y(x)$</span> by an inverse substitution. Namely, for <span class="math-container">$0<x<\pi/2-1$</span>, <span class="math-container">$y(x)$</span> grows from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, and we have <span class="math-container">$x=\arcsin y-y$</span>; similarly, for <span class="math-container">$\pi/2-1<x<\pi$</span>, <span class="math-container">$y(x)$</span> goes from <span class="math-container">$1$</span> to <span class="math-container">$0$</span>, and we have <span class="math-container">$x=\pi-\arcsin y-y$</span>. Thus
<span class="math-container">\begin{align*}
-I&=\frac12\int_0^1y\left(\frac1{\sqrt{1-y^2}}-1\right)\cot\frac{\arcsin y-y}2\,dy\\&+\frac12\int_0^1y\left(\frac1{\sqrt{1-y^2}}+1\right)\tan\frac{\arcsin y+y}2\,dy\\\implies I&=\color{blue}{\int_0^1\frac{(1-y^2-\cos y)\ y}{(y-\sin y)\sqrt{1-y^2}}\,dy}.
\end{align*}</span>
This gives an alternative way(s) of computation of <span class="math-container">$I$</span>.</p>
|
4,492,104 | <p>I'm looking for closed-form expressions for the integral <span class="math-container">$$I:=\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx .$$</span>
Some related integrals that I've found <a href="https://math.stackexchange.com/questions/2354232/is-it-possible-to-evaluate-int-0-infty-log1-coshx-fracx2ex-dx?rq=1">include</a>: <span class="math-container">$$J:=-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx = 6 + 2\ln(2) - {2\pi^{2} \over 3} - 4\zeta(3) - 2\pi i,$$</span>
and <a href="https://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosx-mathrmdx?noredirect=1&lq=1">this</a> one: <span class="math-container">$$K:= \int_0^{\frac{\pi}{2}}\log\Big{(}\cos(x)\Big{)}dx = - \frac{\pi}{2} \log(2) ,$$</span> <a href="https://www.researchgate.net/profile/Mundher-Khaleel/post/Can_anybody_solve_the_integration/attachment/59d62aad79197b80779890e3/AS%3A340147343249408%401458108854707/download/Table_of_Integrals_Series_and_Products_Tablicy_Integralov_Summ_Rjadov_I_Proizvedennij_Engl._2%282%29.pdf" rel="nofollow noreferrer">and</a> (p. 530) <span class="math-container">$$L:=\int_{0}^{\infty} \big{(} \ln(1-e^{-x}) \big{)} dx = - \frac{\pi^{2}}{6} .$$</span></p>
<p>Moreover, I've encountered various integrals of the <a href="https://en.wikipedia.org/wiki/Lobachevsky_integral_formula" rel="nofollow noreferrer">sinc</a> function, <a href="https://www.wolframalpha.com/input?i=integrate+ln%28sin%28x%29%2Fx%29+from+x%3D0+to+pi%2F2" rel="nofollow noreferrer">including</a>: <span class="math-container">$$M:= \int_{0}^{\pi/2} \ln \bigg{(} \frac{\sin(x)}{x} \bigg{)} dx = \frac{\pi}{2} \Big{(} 1 - \ln(\pi) \Big{)} .$$</span> However, I haven't found any closed-form expressions for <span class="math-container">$I$</span> yet, only an approximation: <span class="math-container">$$I \approx -5.75555891011162780816$$</span> and I'm not sure how to proceed with the integral.</p>
| Quanto | 686,284 | <p>Expand as follows
<span class="math-container">$$\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx
=-\sum_{n\ge1}\frac1n \int_0^\infty \left( \frac{\sin x}{x} \right)^n dx $$</span>
where <a href="https://math.stackexchange.com/q/307510/686284"><span class="math-container">$\int_0^\infty \left( \frac{\sin x}{x} \right)^n dx =
\frac{\pi}{2^n (n-1)!}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1} $</span></a>.</p>
|
4,492,104 | <p>I'm looking for closed-form expressions for the integral <span class="math-container">$$I:=\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx .$$</span>
Some related integrals that I've found <a href="https://math.stackexchange.com/questions/2354232/is-it-possible-to-evaluate-int-0-infty-log1-coshx-fracx2ex-dx?rq=1">include</a>: <span class="math-container">$$J:=-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx = 6 + 2\ln(2) - {2\pi^{2} \over 3} - 4\zeta(3) - 2\pi i,$$</span>
and <a href="https://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosx-mathrmdx?noredirect=1&lq=1">this</a> one: <span class="math-container">$$K:= \int_0^{\frac{\pi}{2}}\log\Big{(}\cos(x)\Big{)}dx = - \frac{\pi}{2} \log(2) ,$$</span> <a href="https://www.researchgate.net/profile/Mundher-Khaleel/post/Can_anybody_solve_the_integration/attachment/59d62aad79197b80779890e3/AS%3A340147343249408%401458108854707/download/Table_of_Integrals_Series_and_Products_Tablicy_Integralov_Summ_Rjadov_I_Proizvedennij_Engl._2%282%29.pdf" rel="nofollow noreferrer">and</a> (p. 530) <span class="math-container">$$L:=\int_{0}^{\infty} \big{(} \ln(1-e^{-x}) \big{)} dx = - \frac{\pi^{2}}{6} .$$</span></p>
<p>Moreover, I've encountered various integrals of the <a href="https://en.wikipedia.org/wiki/Lobachevsky_integral_formula" rel="nofollow noreferrer">sinc</a> function, <a href="https://www.wolframalpha.com/input?i=integrate+ln%28sin%28x%29%2Fx%29+from+x%3D0+to+pi%2F2" rel="nofollow noreferrer">including</a>: <span class="math-container">$$M:= \int_{0}^{\pi/2} \ln \bigg{(} \frac{\sin(x)}{x} \bigg{)} dx = \frac{\pi}{2} \Big{(} 1 - \ln(\pi) \Big{)} .$$</span> However, I haven't found any closed-form expressions for <span class="math-container">$I$</span> yet, only an approximation: <span class="math-container">$$I \approx -5.75555891011162780816$$</span> and I'm not sure how to proceed with the integral.</p>
| Tyma Gaidash | 905,886 | <p>The final sum looks like a <a href="https://www.wolframalpha.com/input?i=Kapteyn+Series" rel="nofollow noreferrer">Kapteyn Series</a>. Maybe it can be simplified or is the solution to a transcendental equation. From @metamorphy’s answer with <a href="https://www.wolframalpha.com/input?i=besselj" rel="nofollow noreferrer">Bessel J</a>:</p>
<p><span class="math-container">$$-\frac I\pi=\sum_{n=1}^\infty\frac{\text J_n(n)}n= \sum_{n=1}^\infty\frac{\text J_n(n)}n \cos\left(2\pi n\right)$$</span></p>
<p><a href="https://www.wolframalpha.com/input?i=inversebetaregularized" rel="nofollow noreferrer"><strong>Inverse Beta Regularized <span class="math-container">$\text I^{-1}_s(a,b)$</span></strong></a> we have a <a href="https://www.wolframalpha.com/input?i=inversehaversine%28inversebetaregularized%28x%2Fpi%2C3%2F2%2C1%2F2%29%29+vs+x-sin%28x%29+" rel="nofollow noreferrer">partial closed form</a> and a <a href="https://math.stackexchange.com/posts/1170733/revisions"><strong>Kepler equation series solution</strong></a>:</p>
<p><span class="math-container">$$x-\sin(x)=y\implies x=\boxed{\text{hav}^{-1}\left(\text I^{-1}_\frac y\pi\left(\frac32,\frac12\right)\right)=y+2\sum_{n=1}^\infty \frac{\text J_n(n)}n\sin(y n)}\tag{1}$$</span></p>
<p><a href="https://www.wolframalpha.com/input?i=-inversehaversine%28inversebetaregularized%28e%2Fpi%2C3%2F2%2C1%2F2%29%29%2Be%2B2+sum%5Bbesselj%28n%2Cn%29%2Fn+sin%282.71828+n%29%2C%7Bn%2C1%2C700%7D%5D" rel="nofollow noreferrer"><strong>which is shown here</strong></a> with the <a href="https://www.wolframalpha.com/input?i=inversehaversine" rel="nofollow noreferrer">inverse haversine</a>.
<span class="math-container">$$\sum_{n=1}^\infty\frac{\text J_n(n)}n \sin\left(\frac\pi2n\right)= \sum_{n=1}^\infty\frac{\text J_{4n-3}(4n-3)}{4n-3}-\sum_{n=1}^\infty\frac{\text J_{4n-1}(4n-1)}{4n-1} $$</span></p>
<p><span class="math-container">$$\frac 2{\sqrt3}\sum_{n=1}^\infty \frac{\text J_n(n)}n \sin\left(\frac\pi3n\right)= \sum_{n=1}^\infty\frac{\text J_{6n-5}(6n-5)}{6n-5}+ \sum_{n=1}^\infty\frac{\text J_{6n-4}(6n-4)}{6n-4}-\sum_{n=1}^\infty\frac{\text J_{6n-2}(6n-2)}{6n-2}-\sum_{n=1}^\infty\frac{\text J_{6n-1}(6n-1)}{6n-1}$$</span>
<span class="math-container">$$\vdots$$</span></p>
<p>All we need to do now is find something like <span class="math-container">$\sin(yn)=1,n\in\Bbb N$</span>, but no value of <span class="math-container">$y$</span> satisfies this condition, so we must add multiple sines together or manipulate the sum. Then we remember the Fourier series:</p>
<p><span class="math-container">$$\sum_{n=1}^\infty\frac{\text J_n(n)}n \cos\left(2\pi n\right)=\sum_{n=1}^\infty \sum_{m=1}^\infty a_m \frac{\text J_n(n)}n\sin\left(\frac{\pi m n }L\right)\mathop=^? \sum_{m=1}^\infty a_m \sum_{n=1}^\infty \frac{\text J_n(n)}n\sin\left(\frac{\pi m n }L\right),a_m= \frac 2L \int _0^L \cos(2\pi x)\sin\left(\frac{\pi m x}L\right)dx=\frac{2m((-1)^m\cos(2\pi L)-1)}{\pi(4L^2-m^2)}$$</span></p>
<p>where <span class="math-container">$a_m$</span> are series coefficients. If the series converges on <span class="math-container">$[-L,L]$</span>, then we get a sum made of <span class="math-container">$\text{hav}^{-1}\left(\text I^{-1}_\frac mL\left(\frac32,\frac12\right)\right)$</span> terms using <span class="math-container">$(1)$</span></p>
<p><span class="math-container">$$\text{hav}^{-1}\left(\text I^{-1}_\frac mL\left(\frac32,\frac12\right)\right),0\le \frac mL\le \pi$$</span></p>
<p><span class="math-container">$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-1}\left(\frac12,\frac32\right)\right)+\pi,\pi\le \frac mL\le 2\pi$$</span></p>
<p><span class="math-container">$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-2}\left(\frac32,\frac12\right)\right)+2\pi,2\pi\le \frac mL\le 3\pi$$</span></p>
<p><span class="math-container">$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-3}\left(\frac12,\frac32\right)\right)+3\pi,3\pi\le \frac mL\le 4\pi$$</span></p>
<p><span class="math-container">$$\text{hav}^{-1}\left(\text I^{-1}_{\frac mL-4}\left(\frac32,\frac12\right)\right)+4\pi,4\pi\le \frac mL\le 5\pi$$</span>
<span class="math-container">$$\vdots$$</span></p>
<p>Does this method work?</p>
|
1,383,380 | <p>On page 12 of Stein, Shakarchi textbook 'Complex analysis', the authors state that the <em>Cauchy-Riemann equations link complex and real analysis</em>. I have completed courses on real and complex analysis, but I feel that this is somewhat of an over-statement. But perhaps it is just me which doesnt have a good enough overview.</p>
<p>$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}
$$</p>
<p>If anyone with a clear insight is able to concisely explain how one could justify writing something like this -- then that insight would be most valuable.</p>
| David C. Ullrich | 248,223 | <p>Why is there not a fixed point $f(x)=x$? Nobody said there wasn't.</p>
<p>To show there does exist an $x$ with $f(x)=2x$, let $g(x)=f(x)-2x$. Then $g(0)=f(0)\ge 0$, while $g(1)=f(1)-2\le0$. So the intermediate value theorem shows there exists $x$ with $g(x)=0$.</p>
<p>(Or, if you have a theorem saying any map from $[0,1]$ to itself has a fixed point, consider $f/2$.)</p>
|
464,489 | <p>We are given $H = \{(1),(13),(24),(12)(34),(13)(24),(14)(23),(1234),(1432)\}$ is a subgroup of $S_4$. Also assume $K = \{(1),(13)(24)\}$ is a normal subgroup of $H$. Show $H/K$ isomorphic to $Z_2\oplus Z_2$. </p>
<p>This is just a practice question (not assignment). So I have tried finding $H/K$ explicitly.</p>
<p>$H/K = \{\{(1),(13)(24)\},\{(13),(24)\},\{(14)(23),(12)(34)\},\{(1234),(1423)\}\}$. We know there are only $2$ groups of order $4$. One of the elements in $H/K$ we see is $(1234)K$, doesn't this element have a order of $4$, making $H/K$ cyclic and hence not isomorphic to $Z_2\oplus Z_2$? </p>
| user77404 | 77,404 | <p>After Tobias comment. I realized the order of (1234)K is actually 2 and not 4. Since
(1234)K(1234)K = (1234)(1234)K = (13)(24)K = K. We know there are only 2 groups of order 4. That is $Z_4$ and $Z_2 \bigoplus Z_2$. Since we see $H/K$ does not have any elements of order 4 it is not cyclic and cannot be isomorphic to $Z_4$, hence $H/K$ must be isomorphic to $Z_2 \bigoplus Z_2$.</p>
|
1,017,965 | <p>Just as the title, my question is what is the matrix representation of Radon transform (Radon projection matrix)? I want to have an exact matrix for the Radon transformation. </p>
<p>(I want to implement some electron tomography algorithms by myself so I need to use the matrix representation of the Radon transformation.)</p>
<p>(An introduction of the Radon transformation: <a href="http://en.wikipedia.org/wiki/Radon_transform" rel="nofollow noreferrer">click here</a>.)</p>
| AnonSubmitter85 | 33,383 | <p>A couple of years ago I was looking for the same answer. You can see that thread <a href="https://stackoverflow.com/questions/12166562/radon-transform-matrix-representation">here</a>. However, I don't think my solution is the only way to do it nor necessarily the best way.</p>
<p>We know it's a linear transform, so if you can take the transform of the right identity matrix, you'll get the matrix representation. As an easy example using matlab notation, if we wanted to get the matrix representation of a column-wise DFT matrix using the <code>fft()</code> function, we could do it with <code>A = fft(eye(N))</code>. Now, it's a little more complicated with the radon transform to do it efficiently, but the same idea applies.</p>
<p>If the image is $\mathbf{X} \in \mathbb{C}^{M \times N}$ and we wanted to find the DRT matrix $\mathbf{R} : \mathbb{C}^{M \times N} \rightarrow \mathbb{C}^{L\times P}$, then we know that $\mathbf{R} \in \mathbb{C}^{LP \times MN}$ and that the DRT is given by $\mathbf{Y} = \mathbf{R} \cdot \operatorname{vec}(\mathbf{X}) \in \mathbb{C}^{LP \times 1}$, where $\operatorname{vec} : \mathbb{C}^{A\times B} \rightarrow \mathbb{C}^{AB \times 1}$. The typical 2-D display of the DRT would then be had by simply reshaping the $LP \times 1$ vector into a $L \times P$ array.</p>
<p>We don't know $\mathbf{R}$, but we have access to a function that will efficiently compute the DRT, call it <code>radon()</code>, such that <code>Y = radon(X)</code> is equivalent to $\mathbf{Y} = \mathbf{R} \cdot \operatorname{vec}(\mathbf{X})$. Since $\mathbf{R} \cdot \mathbf{I} \cdot \operatorname{vec}(\mathbf{X}) = \mathbf{R} \cdot \operatorname{vec}(\mathbf{X})$, where $\mathbf{I}$ is the identity matrix, if we can used <code>radon()</code> in such a way that <code>A = radon(I)</code>, then we have that <code>A * vec(X) = radon(X)</code>.</p>
<p>The problem is that I don't see how to use <code>radon</code> to efficiently compute the above. Most <code>radon()</code> implementations take a 2-D input and output a 2-D array. It's simple enough to reshape the input/output into vectors, but the problem is that you would have to do that $MN$ times. Even for modest image sizes, the above will become computationaly prohibitive rather fast.</p>
<p>I hope the above helps. If you come up with an efficient solution, please post it here or at the stackoverlow thread linked above.</p>
|
3,725,007 | <p>Consider the rings, <span class="math-container">$Z_2[x]/(1 + x^2)$</span> and <span class="math-container">$ Z_2[x]/(1 + x + x^2)$</span>, despite having different polynomial as divisor, I have been told that -</p>
<p><span class="math-container">$$Z_2[x]/(1 + x^2) = \{0, 1, x, 1 + x\}$$</span></p>
<p>and</p>
<p><span class="math-container">$$Z_2[x]/(1 + x + x^2)= \{0, 1, x, 1 + x\}$$</span></p>
<p>Then what is the difference between <span class="math-container">$Z_2[x]/(1 + x^2)$</span> and <span class="math-container">$ Z_2[x]/(1 + x + x^2)$</span>?</p>
| Preston Lui | 556,396 | <p>The way I will approach it is to prove that <span class="math-container">$\lim_{x \to \infty}G(x)\le G(\pi)$</span>, where <span class="math-container">$G(x) is \int_{0}^{x}\frac{sin(y)}y dy $</span>. The reason being if <span class="math-container">$f(y)=\frac{sin(y)}y$</span>, then <span class="math-container">$\int_{r \pi}^{(r+2)\pi}f(y)dy\leq0$</span> for odd <span class="math-container">$r$</span></p>
<p>The rest is trivial</p>
|
891,370 | <p>I got the function $8.513 \times 1.00531^{\Large t} = 10$. The task is to solve $t$. The correct answer is $t = 31$. How do I get there ?.</p>
| beep-boop | 127,192 | <p>Hint: $$8.513 \cdot 1.00531^t=10 \iff 1.00531^t=\frac{10}{8.513} \iff \ln[1.00531^t]=\ln\left[\frac{10}{8.513}\right] $$
$$\iff t\ln[1.00531]=\ln\left[\frac{10}{8.513}\right] \iff t= \quad?$$</p>
|
891,370 | <p>I got the function $8.513 \times 1.00531^{\Large t} = 10$. The task is to solve $t$. The correct answer is $t = 31$. How do I get there ?.</p>
| amWhy | 9,003 | <ul>
<li><p>Isolate $ 1.00531^{\large t}$ on the left side of the equation.</p></li>
<li><p>Take the $\ln$ of each side of that equation.</p></li>
<li><p>And use the fact that $\ln a^b = b \ln a$.</p></li>
<li><p>Solve for $t$ as you would any first degree polynomial.</p></li>
</ul>
|
16,795 | <p>Consider a finite simple graph $G$ with $n$ vertices, presented in two different but equivalent ways:</p>
<ol>
<li>as a logical formula $\Phi= \bigwedge_{i,j\in[n]} \neg_{ij}\ Rx_ix_j$ with $\neg_{ij} = \neg$ or $ \neg\neg$ </li>
<li>as an (unordered) set $\Gamma = \lbrace [n],R \subseteq [n]^2\rbrace$ </li>
</ol>
<p>In each case the complement $G'$ of $G$ is easily presented and is of course <em>not</em> isomorphic to $G$ (in the usual sense) generally:</p>
<ol>
<li>$ \Phi' = \bigwedge_ {i,j} \neg \neg_{ij}\ R x_i x_j $ </li>
<li>$\Gamma' = \lbrace [n],[n]^2 \setminus R\rbrace$</li>
</ol>
<p>Let's state for the moment that the presentation as a logical formula is the more "flexible" one: we can easily omit single literals, leaving it open whether $Rx_ix_j$ or not. But this can be mimicked for set presentation by making it from a pair to a triple $\lbrace[n],R,\neg R \subseteq [n]^2 \setminus R\rbrace$. </p>
<p>Let's call a presentation <em>complete</em>, if it leaves nothing open, i.e. no omitted literal and $\neg R = [n]^2 \setminus R$, resp.</p>
<p>Now, let a graph be given in complete set presentation $\lbrace[n],R,\neg R = [n]^2 \setminus R\rbrace$. Since order in this set should not matter, any sensible definition of "graph isomorphism" should make any graph isomorphic to its complement.</p>
<blockquote>
<p>Where and how do I run into trouble when I
assume - following this line of
reasoning, contrary to the usual line of thinking - that every (finite) graph is
isomorphic to its complement?</p>
</blockquote>
| Tony Huynh | 2,233 | <p>The <em>set</em> { $[n], R, \neg R$ } does not actually specify a graph, since we cannot distinguish between edges and non-edges. The <em>triple</em> ($[n], R, \neg R$ ) does specify a graph since by convention we can say that the second coordinate specifies the edges and the third specifies the non-edges. Thus, ($[n], R, \neg R$ ) is not isomorphic to ($[n], \neg R, R$ ). </p>
|
382,526 | <p>I can't calculate the Integral:</p>
<p>$$
\int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^{6}}}dx
$$</p>
<p>any help would be great!</p>
<p>p.s I know it converges, I want to calculate it.</p>
| Shuhao Cao | 7,200 | <p>Doing the substitution first $x^3 = \cos\theta$ , then $\theta/2 = t$, would lead us to a simpler integral:
$$
\int^{\frac{\pi}{2}}_0 \frac{1}{3\sqrt{\cos\theta}} d\theta = \int^{\frac{\pi}{4}}_0 \frac{2}{3\sqrt{1 - 2\sin^2t}} dt
$$
and this is <a href="http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html" rel="nofollow">elliptic integral of the first kind</a>, hence no closed form. For the numerical value please refer to vadim123's comment.</p>
|
958,099 | <p>I have the following question:</p>
<blockquote>
<p>For real $x$, $f(x) = \frac{x^2-k}{x-2}$ can take any real value. Find the range of values $k$ can take.</p>
</blockquote>
<p>Here is how I commenced:</p>
<p>$$
y(x-2) = x^2-k \\
-x^2 + xy - 2y + k = 0\\
$$</p>
<p>So we have $a=-1$, $b=y$, $c=(-2y+k)$. In order to find out the range of $f(x)$, i.e. the nature of its roots, we need the discriminant:</p>
<p>$$
y^2 - 4(-1)(-2y+k) \ge 0 \\
y^2 -8y + 4k \ge 0
$$</p>
<p>So $k$ must equal something to make that equation true - but the only way I understand "true" here is if I can factorise the equation after substituting a possible value of $k$. There are a few values which settle this criteria, like $k=4$, $k=3$ and $k=\frac{7}{4}$. The listed answer says that $k \ge 4$, which makes sense because the latter values produce $y$ values less than 0.</p>
<p>Is there a better way to find out the values of k in this situation? Or do I need to rely on factorising with trial and error?</p>
| Yulia V | 86,800 | <p>$$y^2 -8y + 4k = (y-4)^2+4(k-4)$$</p>
<p>For $y=4$, we need $k \geq 4$ to ensure that the discriminant is positive.</p>
|
3,040,110 | <p>What is the Range of <span class="math-container">$5|\sin x|+12|\cos x|$</span> ?</p>
<p>I entered the value in desmos.com and getting the range as <span class="math-container">$[5,13]$</span>.</p>
<p>Using <span class="math-container">$\sqrt{5^2+12^2} =13$</span>, i am able to get maximum value but not able to find the minimum.</p>
| Makina | 603,955 | <p>Another possible approach.</p>
<p>For the first quadrant: <span class="math-container">$5\sin(x) + 12\cos(x) = 13\sin(x + \arccos(\frac{5}{13}))$</span>. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of <span class="math-container">$x$</span>.</p>
|
2,279,120 | <p>How do we prove that for any complex number $z$ the minimum value of $|z|+|z-1|$ is $1$ ?
$$
|z|+|z-1|=|z|+|-(z-1)|\geq|z-(z-1)|=|z-z+1|=|1|=1\\\implies|z|+|z-1|\geq1
$$</p>
<p>But, when I do as follows
$$
|z|+|z-1|\geq|z+z-1|=|2z-1|\geq2|z|-|1|\geq-|1|=-1
$$
Since LHS can never be less than 0, $|z|+|z-1|\geq0$</p>
<p>Why do I seem to get different result compared to the first method ?</p>
<p>ie.
1st method $\implies (|z|+|z-1|)_{min}=1\\$</p>
<p>2nd method$\implies (|z|+|z-1|)_{min}=0\\$</p>
<p>What is going wrong in the second approach ?</p>
| Vim | 191,404 | <p>By the triangle inequality:
$$|z|+|z-1|\ge |z-(z-1)|=1$$
And $1$ is attained when $z=0$, say.</p>
|
3,846,339 | <p>Suppose I have the inequality <span class="math-container">$(\frac{A}{B})^X < (\frac{C}{D})\cdot(\frac{E}{F})^Y$</span> and I want X by itself.</p>
<p>Can I do this <span class="math-container">$X\cdot \log(\frac{A}{B}) < \log(\frac{C}{D})\cdot(Y\cdot \log(\frac{E}{F}))$</span>?
Am I breaking any rules on the right-hand side?</p>
| KingLogic | 503,528 | <p>One of the logarithm rules is <span class="math-container">$\log (ab)=\log a + \log b$</span>. Therefore, on the right, it should be <span class="math-container">$\log (\frac{C}{D}) + Y \log (\frac{E}{F})$</span>.</p>
|
275,785 | <p>Let $a_{1}, a_{2}, \ldots, a_{n}$, $n \geq 3$. Prove that at least one of the number $(a_{1}+a_{2}\ldots +a_{n})^{2}-(n^2-n+2)a_{i}a_{j}$ is greater or equal with $0$ for $1 \leq i < j \leq n$.</p>
<p>I don't know at least how to catch this problem .
Thanks :)</p>
| pre-kidney | 34,662 | <p>Suppose for contradiction that the condition didn't hold. This gives you a set of ${n \choose 2}$ sharp inequalities in the $(i,j)$ which you can multiply together, to yield a sharp inequality of homogeneous polynomials of degree $n(n-1)$. However, the reverse inequality can be proven via "bashing" techniques, so it shows that your hypothesis was false - meaning that the condition holds for some $(i,j)$ pair, which is what we were trying to show.</p>
<p>The reason I am being vague is because I don't think this is your own problem, and you haven't given an attribution - so I am unsure if it is an open contest problem, for instance.</p>
|
1,393,154 | <p><span class="math-container">$4n$</span> to the power of <span class="math-container">$3$</span> over <span class="math-container">$2 = 8$</span> to the power of negative <span class="math-container">$1$</span> over <span class="math-container">$3$</span></p>
<p>Written Differently for Clarity:</p>
<p><span class="math-container">$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$</span></p>
<hr />
<blockquote>
<p><strong>EDIT</strong></p>
<p>Actually, the problem should be solving <span class="math-container">$4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$</span>. Another user edited this question for clarity, but they edited it incorrectly to add parentheses around the right hand side, as can be seen above.</p>
</blockquote>
| haqnatural | 247,767 | <p>$${ \left( 4n \right) }^{ \frac { 3 }{ 2 } }={ \left( 8 \right) }^{ -\frac { 1 }{ 3 } }\\ \left( { \left( 4n \right) }^{ \frac { 3 }{ 2 } } \right) ^{ 2/3 }=\left( { \left( { 2 }^{ 3 } \right) }^{ -\frac { 1 }{ 3 } } \right) ^{ 2/3 }\\ 4n=2^{ -\frac { 2 }{ 3 } }\\ n=\frac { 2^{ -\frac { 2 }{ 3 } } }{ 4 } =\frac { 1 }{ 4\sqrt [ 3 ]{ 4 } } $$</p>
|
2,348,909 | <p>Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$. This confuses me. As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{n=0}^{\infty}x^n,
\end{align}
and such a series converges to $\frac{1}{1 - x}$ if $x \in [0,1)$. Now, it is certainly true that if we put $y = \frac{x^2}{1 + x^2}$, then $\frac{1}{1 - y} = 1 + x^2$. But in this case, shouldn't the series we're discussing be
\begin{align}
\sum_{n=0}^{\infty}\left[\frac{x^2}{1 + x^2}\right]^n?
\end{align}
What am I missing?</p>
| farruhota | 425,072 | <p>You should write the first few terms to understand its behavior:
$$\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\frac{x^2}{(1+x^2)^3}+\cdots=$$
$$x^2\left(1+\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}+\cdots\right)\stackrel{x\ne 0}=$$
$$x^2\cdot \frac{1}{1-\frac{1}{1+x^2}}=1+x^2.$$</p>
|
2,843,560 | <p>If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, then $x$ is equal to</p>
<p>(a) $y$</p>
<p>(b) $y/2$</p>
<p>(c) $2y$</p>
<p>(d) $y/6$</p>
<p>I expanded the first equation to reach $2\sin x(2+\cos x-2\sin x)= \sin y$, but I doubt it leads me anywhere. A little hint would be appreciated. Thanks!</p>
| lab bhattacharjee | 33,337 | <p><strong>Generalization</strong> :</p>
<p>Using <a href="https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro">How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?</a>,</p>
<p>$$f(n)=\dfrac{\sum_{k=0}^{n-1}\sin (a+k \cdot d) }{\sum_{k=0}^{n-1}\cos (a+k \cdot d)}= \tan \frac{2 a + (n-1)d}2$$</p>
<p>If $a=d=x,$ $$f(n)=\tan\dfrac{x(n+1)}2$$</p>
<p>Here $n=3$</p>
|
81,267 | <p>I have the following problem: I have a (a lot)*3 table, meaning that I have 3 columns, say X, Y and Z, with real values. In this table some of the rows have the same (X,Y) values, but with different value of Z. For instance</p>
<pre><code>{{12.123, 4.123, 513.423}, {12.123, 4.123, 33.43}}
</code></pre>
<p>have the same (X,Y) but different Z. This is a case of multiplicity=2, but in principle it could be higher. What I want to do is to take all the unique rows, AND in case they have multiplicity >1 (i.e. repeated (X,Y)), pick the one with minimum Z value. In the previous example it would be the second one.</p>
<p>I hope I have been clear! Thank you very much indeed!</p>
| Bob Hanlon | 9,362 | <pre><code>data = Table[{x, y, RandomReal[]},
{x, 3}, {y, 3}, {3}] // Flatten[#, 2] &
</code></pre>
<blockquote>
<p>{{1, 1, 0.987008}, {1, 1, 0.682772}, {1, 1, 0.923863}, {1, 2,
0.464991}, {1, 2, 0.963954}, {1, 2, 0.829995}, {1, 3, 0.773942}, {1, 3, 0.550081}, {1, 3,
0.0821332}, {2, 1, 0.466804}, {2, 1, 0.964716}, {2, 1, 0.679577}, {2, 2,
0.820486}, {2, 2, 0.10238}, {2, 2, 0.810867}, {2, 3, 0.87029}, {2, 3,
0.530354}, {2, 3, 0.91676}, {3, 1, 0.345255}, {3, 1, 0.939562}, {3, 1,
0.456195}, {3, 2, 0.497948}, {3, 2, 0.13677}, {3, 2, 0.544248}, {3, 3,
0.271007}, {3, 3, 0.539826}, {3, 3, 0.927424}}</p>
</blockquote>
<pre><code>First /@ (SortBy[#, Last] & /@ GatherBy[data, Most])
</code></pre>
<blockquote>
<p>{{1, 1, 0.682772}, {1, 2, 0.464991}, {1, 3, 0.0821332}, {2, 1,
0.466804}, {2, 2, 0.10238}, {2, 3, 0.530354}, {3, 1, 0.345255}, {3, 2, 0.13677}, {3, 3,
0.271007}}</p>
</blockquote>
<p>However, since {x,y} are identical in each grouping, then <code>SortBy[#,Last]&</code> is equivalent to <code>Sort</code></p>
<pre><code>First /@ (Sort /@ GatherBy[data, Most])
</code></pre>
<blockquote>
<p>{{1, 1, 0.682772}, {1, 2, 0.464991}, {1, 3, 0.0821332}, {2, 1,
0.466804}, {2, 2, 0.10238}, {2, 3, 0.530354}, {3, 1, 0.345255}, {3, 2, 0.13677}, {3, 3,
0.271007}}</p>
</blockquote>
<pre><code>% == %%
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
353 | <p>One of the challenges of undergraduate teaching is logical implication. The case by case definition, in particular, is quite disturbing for most students, that have trouble accepting "false implies false" and "false implies true" as true sentences. </p>
<blockquote>
<p><strong>Question:</strong> What are good point of view, methods and tips to help students grasp the concept of logical implication?</p>
</blockquote>
<p>To focus the question, I would like to restrict to math majors, although the question is probably equally interesting for other kind of students.</p>
| Tim Seguine | 208 | <p>One of the things that really helped me (not to learn it, but to appropriately apply it), back when I originally learned this myself was the equivalence between $p\implies q$ and $\neg p \vee q$. I constantly reminded myself of it when manipulating logic statements. Make sure the students are aware of this equivalence.</p>
<p>What it did was give me a basis of comparison to something I already understood well. It gave me something to check in case I was worried I have made a mistake. It made negation of implication a lot easier to understand as well.</p>
|
246,808 | <p>Trying to solve the following PDE with BC <code>T==1</code> on a spherical cap of a unit sphere and <code>T==0</code> at infinity (approximated as <code>r==(x^2 + y^2 + z^2)^0.5==40^0.5</code>) and the flux over the remaining surfaces taken to be zero (only half domains has been specified due to symmetry reasons):</p>
<pre><code>p = 0.2;
Pe = 20;
<< NDSolve`FEM`
boundaries = {-x^2 - y^2 - z^2 + 1, x^2 + y^2 + z^2 - 40, -y,
z - p + 1};
\[CapitalOmega] =
ToElementMesh[
ImplicitRegion[And @@ (# <= 0 & /@ boundaries), {x, y, z}]];
Show[\[CapitalOmega][
"Wireframe"["MeshElement" -> "MeshElements", Boxed -> True]],
Axes -> True, AxesLabel -> {"x", "y", "z"},
PlotRange -> {{-7, 7}, {-0.5, 7}, {-7, 1}}]
Show[\[CapitalOmega]["Wireframe"], Axes -> True,
AxesLabel -> {"x", "y", "z"},
PlotRange -> {{-7, 7}, {-0.5, 7}, {-7, 1}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/VrfMr.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/VrfMr.gif" alt="dominio" /></a></p>
<p>The free-surface is located at <code>z==-0.8</code> in the chosen reference system.
The differential equation is:</p>
<pre><code>sol = NDSolveValue[{D[T[x, y, z], x] ==
1/Pe Laplacian[T[x, y, z], {x, y, z}], {DirichletCondition[
T[x, y, z] == 1., boundaries[[1]] == 0.],
DirichletCondition[T[x, y, z] == 0., boundaries[[2]] == 0.]}},
T, {x, y, z} \[Element] \[CapitalOmega]]
</code></pre>
<p>I noticed a non-smooth behavior of the solution <code>sol</code>, as confirmed by the density plots:</p>
<pre><code>z1 = -0.8;
DensityPlot[sol[x, y, z1], {x, -4, 4}, {y, 0, 2}, PlotRange -> All,
PlotPoints -> 100, AspectRatio -> 1/2]
DensityPlot[sol[x, 0, z], {x, -4, 4}, {z, -0.8, -2}, PlotRange -> All,
PlotPoints -> 100, AspectRatio -> 1/2]
</code></pre>
<p><a href="https://i.stack.imgur.com/S4c63.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/S4c63.gif" alt="fig1" /></a>
<a href="https://i.stack.imgur.com/5P7Dm.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/5P7Dm.gif" alt="fig2" /></a></p>
<p>and by some plots of the solution</p>
<pre><code>Plot[sol[x, 0, -0.8], {x, 0.6, 6.1}, Frame -> True,
PlotRange -> {{-0.1, 7}, {-0.1, 1.2}}]
Plot[sol[x, 0, -0.8], {x, -6.1, -0.6}, Frame -> True,
PlotRange -> {{-7, 0.1}, {-0.1, 1.2}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/NRs8c.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/NRs8c.gif" alt="fig3" /></a>
<a href="https://i.stack.imgur.com/Sw4XW.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/Sw4XW.gif" alt="fig4" /></a></p>
<p>The derivatives of the solution show an even worse behavior. I report here as example just the derivative with respect to <code>x</code>:</p>
<pre><code>Dr[x_, y_, z_] = D[sol[x, y, z], x]
Plot[Dr[x, 0, -0.8], {x, 0.6, 8}, Frame -> True]
Plot[Dr[x, 0, -0.8], {x, -8, -0.6}, Frame -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/I0bTo.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/I0bTo.gif" alt="fig5" /></a>
<a href="https://i.stack.imgur.com/6EKBK.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/6EKBK.gif" alt="fig6" /></a></p>
<p>I am interested in the derivative because I am trying to calculate the total flux of the gradient of <code>sol</code> through the two portion of the spherical cap, <code>SC1</code> and <code>SC2</code>, which border the domain. As known, this is proportional to the heat flow through <code>SC1</code> and <code>SC2</code>. The definition of <code>SC1</code> and <code>SC2</code> is:</p>
<pre><code>SC1 = ImplicitRegion[
x^2 + y^2 + z^2 == 1 && z <= p - 1 && y >= 0 && x >= 0, {x, y, z}];
SC2 = ImplicitRegion[
x^2 + y^2 + z^2 == 1 && z <= p - 1 && y >= 0 && x <= 0, {x, y, z}];
Show[DiscretizeRegion[SC1, {{-5, 5}, {-5, 5}, {-3, 3}},
MaxCellMeasure -> 0.0001], Axes -> True,
AxesLabel -> {"x", "y", "z"},
PlotRange -> {{-1, 1}, {-0.2, 1}, {-0.8, -1}}]
Show[DiscretizeRegion[SC2, {{-5, 5}, {-5, 5}, {-3, 3}},
MaxCellMeasure -> 0.0001], Axes -> True,
AxesLabel -> {"x", "y", "z"},
PlotRange -> {{-1, 1}, {-0.2, 1}, {-0.8, -1}}]
</code></pre>
<p>and the heat flow through <code>SC1</code> and <code>SC2</code> should be:</p>
<pre><code>NIntegrate[#, {x, y, z} \[Element] SC1] & /@ (Grad[
sol[x, y, z], {x, y, z}].{x, y, z})
NIntegrate[#, {x, y, z} \[Element] SC2] & /@ (Grad[
sol[x, y, z], {x, y, z}].{x, y, z})
</code></pre>
<p>Unfortunately, I get a lot of errors from the last calculation, I don't know if this happens because of the non-smooth behavior of the solution and of the derivative or for other mistakes that I am doing. Thank you in advance.</p>
| user21 | 18,437 | <p>Here are a couple of different ideas for generating the meshes. This works in 12.3</p>
<pre><code>Needs["NDSolve`FEM`"]
Pe = 15;
p = 2/10;
reg = RegionDifference[Cuboid[{-2.5, 0, -3.8}, {7.5, 4, -0.8}],
Ball[]];
bmesh = ToBoundaryMesh[reg, AccuracyGoal -> 3.5];
</code></pre>
<p>The key here is the <code>AccuracyGoal</code>. In version 12.3 the 3D default boundary mesh generator is OpenCascade. OpenCascade will not refine flat boundaries if the accuracy goal is increased.</p>
<pre><code>bmesh["Wireframe"]
</code></pre>
<p><a href="https://i.stack.imgur.com/NdHrk.png" rel="noreferrer"><img src="https://i.stack.imgur.com/NdHrk.png" alt="enter image description here" /></a></p>
<p>Visualize the boundary mesh and it's boundary markers:</p>
<pre><code>groups = bmesh["BoundaryElementMarkerUnion"];
temp = Most[Range[0, 1, 1/(Length[groups])]];
colors = ColorData["BrightBands"][#] & /@ temp;
Show[
bmesh["Edgeframe"],
bmesh["Wireframe"["MeshElement" -> "BoundaryElements",
"MeshElementStyle" -> (Directive[EdgeForm[], FaceForm[#]] & /@
colors)]]
, Epilog -> Inset[LineLegend[colors, groups], Scaled[{0.85, 0.6}]]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/x66qY.png" rel="noreferrer"><img src="https://i.stack.imgur.com/x66qY.png" alt="enter image description here" /></a></p>
<p>The solution can then be found with</p>
<pre><code>mesh = ToElementMesh[bmesh];
sol = NDSolveValue[{D[T[x, y, z], x] ==
1/Pe Laplacian[T[x, y, z], {x, y, z}], {DirichletCondition[
T[x, y, z] == 1., ElementMarker == 7],
DirichletCondition[T[x, y, z] == 0., ElementMarker == 1]}},
T, {x, y, z} \[Element] mesh];
</code></pre>
<p>Another idea is to split the cap. This can be done with the OpenCascadeLink. The idea is to create a left and right part of the geometry, take the faces and merge them back together.</p>
<pre><code>Needs["OpenCascadeLink`"]
rleft = RegionDifference[Cuboid[{-2.5, 0, -3.8}, {0, 4, -0.8}],
Ball[]];
rright = RegionDifference[Cuboid[{0, 0, -3.8}, {7.5, 4, -0.8}],
Ball[]];
left = OpenCascadeShape[rleft];
right = OpenCascadeShape[rright];
shape = OpenCascadeShapeUnion[
Flatten[{OpenCascadeShapeFaces[left],
OpenCascadeShapeFaces[right]}]]
OpenCascadeShapeSurfaceMeshToBoundaryMesh[shape]["Wireframe"]
</code></pre>
<p><a href="https://i.stack.imgur.com/mUZrV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/mUZrV.png" alt="enter image description here" /></a></p>
<p>Note, that we have a split along the x=0 plane. Next, we generate a boundary mesh</p>
<pre><code>(bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[shape,
"ShapeSurfaceMeshOptions" -> {(*"LinearDeflection"->0.00025*)
"AngularDeflection" -> 0.025
}])["Wireframe"]
</code></pre>
<p>You can tell OpenCascade to either use a linear deflection or an angular deflection (See documentation <a href="https://reference.wolfram.com/language/OpenCascadeLink/tutorial/UsingOpenCascadeLink.html#295318819" rel="noreferrer">here</a>)</p>
<p>Inspect the boundary mesh and it's markers</p>
<pre><code>groups = bmesh["BoundaryElementMarkerUnion"];
temp = Most[Range[0, 1, 1/(Length[groups])]];
colors = ColorData["BrightBands"][#] & /@ temp;
Show[
bmesh["Edgeframe"],
bmesh["Wireframe"["MeshElement" -> "BoundaryElements",
"MeshElementStyle" -> (Directive[EdgeForm[], FaceForm[#]] & /@
colors)]]
, Epilog -> Inset[LineLegend[colors, groups], Scaled[{0.85, 0.6}]]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/e6AEF.png" rel="noreferrer"><img src="https://i.stack.imgur.com/e6AEF.png" alt="enter image description here" /></a></p>
<p>Inspect the detail of the cap:</p>
<pre><code>bmesh["Wireframe"[ElementMarker == 7 || ElementMarker == 9,
"MeshElement" -> "BoundaryElements"]]
</code></pre>
<p><a href="https://i.stack.imgur.com/fuLuQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fuLuQ.png" alt="enter image description here" /></a></p>
<p>Generate the mesh:</p>
<pre><code>mesh = ToElementMesh[bmesh];
mesh["Wireframe"]
</code></pre>
<p><a href="https://i.stack.imgur.com/bAGYC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bAGYC.png" alt="enter image description here" /></a></p>
<p>Solve the PDE</p>
<pre><code>sol = NDSolveValue[{D[T[x, y, z], x] ==
1/Pe Laplacian[T[x, y, z], {x, y, z}], {DirichletCondition[
T[x, y, z] == 1., ElementMarker == 7 || ElementMarker == 9],
DirichletCondition[T[x, y, z] == 0., ElementMarker == 1]}},
T, {x, y, z} \[Element] mesh];
z1 = -0.8;
DensityPlot[sol[x, y, z1], {x, -4, 4}, {y, 0, 2}, PlotRange -> All,
PlotPoints -> 100, AspectRatio -> 1/2]
DensityPlot[sol[x, 0, z], {x, -4, 4}, {z, -0.8, -2}, PlotRange -> All,
PlotPoints -> 100, AspectRatio -> 1/2]
</code></pre>
<p><a href="https://i.stack.imgur.com/sCHut.png" rel="noreferrer"><img src="https://i.stack.imgur.com/sCHut.png" alt="enter image description here" /></a></p>
<pre><code>FEMNBoundaryIntegrate[
Grad[sol[x, y, z], {x, y, z}] . {x, y, z}, {x, y, z}, mesh,
ElementMarker == 7]
-2.98032
FEMNBoundaryIntegrate[
Grad[sol[x, y, z], {x, y, z}] . {x, y, z}, {x, y, z}, mesh,
ElementMarker == 9]
-0.946663
</code></pre>
<p>You can still use a <code>MeshRefinementFunction</code> to refine the mesh away from the cap.</p>
<p>As a side note, check out these new (12.3) OpenCascade tutorials: <a href="https://reference.wolfram.com/language/OpenCascadeLink/tutorial/BookShelfBracket.html" rel="noreferrer">Book Shelf Bracket</a> and <a href="https://reference.wolfram.com/language/OpenCascadeLink/tutorial/HelicalBevelGear.html" rel="noreferrer">Helical Bevel Gear</a></p>
|
468,487 | <p>I have calculated the likelihood of an event to be $1$ in $1.07 \times 10^{2867}$.</p>
<p>I'm looking for a way to describe to a layperson how unlikely this event is to occur, but the number is so mind boggling large I can't find a way to put it into words.</p>
<p>Any suggestions would be appreciated</p>
| user5402 | 81,595 | <p>There's about $n=10^{80}$ atoms in the universe and about $p=4\times 10^{27}$ atoms in the human body. If you choose arbitrarily $p$ atoms in the universe, what is the probability that these atoms are exactly the atoms you are composed of at the instant you finish this sentence?</p>
<p>It's $\binom{n}{p}^{-1}$</p>
|
3,283,606 | <p>Good Evening,</p>
<p>I know this is a basic question, but I haven't been able to find a clear explanation for how to simplify the follow equation:
<span class="math-container">$$n\log_2n=10^6$$</span>
Solving this equation is part of the solution for Problem 1-1 from the Intro. to Algorithms book by CLRS:
<a href="http://atekihcan.github.io/CLRS/P01-01/" rel="nofollow noreferrer">http://atekihcan.github.io/CLRS/P01-01/</a></p>
<p>The author there simplifies the above to:
<span class="math-container">$$n=62746$$</span>
But I can't see how to do this. Thank you.</p>
| Ross Millikan | 1,827 | <p>You are doing one dimensional <a href="https://en.wikipedia.org/wiki/Root-finding_algorithm" rel="nofollow noreferrer">root finding</a> on the function <span class="math-container">$f(n)=10^6-n \log n$</span>. This is a large subject, a chapter in every numerical analysis book. The simplest algorithm to describe is bisection. We note that <span class="math-container">$f(1) \gt 0, f(10^6) \lt 0$</span> and check the midpoint. We replace the endpoint of the same sign with the midpoint, which cuts the interval in half. We do this as many times as needed to get the interval short enough that the error is acceptable. There are many fancier algorithms that may converge more rapidly.</p>
|
251,028 | <p>I want to make <span class="math-container">$S[\{,\cdots, \}]$</span> as follows</p>
<p>First input of <span class="math-container">$S$</span> is given list <span class="math-container">$\{1,2,3,\cdots, n\}$</span> and it produces <span class="math-container">$s_{123\cdots n}$</span></p>
<p>Further, if the ordering of the list is given differently, still gives increasing order. i.e.,</p>
<p><span class="math-container">$S[{1,3,2}] = s_{123}$</span></p>
<p><span class="math-container">$S[{1,2,4,3}] = s_{1234}$</span></p>
<p>and so on.</p>
| Andrzej | 79,522 | <p>If you want an input to be a <code>List</code> then e.g.:</p>
<pre><code>S[list_] := Subscript[S, StringDelete[ToString[Sort[list]], {",", " ","{", "}"}]]
</code></pre>
<p>If a <code>Sequence</code> then e.g.:</p>
<pre><code>R[seq__] := Subscript[R, StringDelete[ToString[{seq}], {",", " ", "{", "}"}]]
</code></pre>
<p>I'm sure there are other (better) ways, but these might need to be tailored to what you need.</p>
|
210,849 | <p>Let $p$ be an odd prime and $a$ be an integer with $\gcd(a, p) = 1$. Show that $x^2 - a \equiv 0 \mod p$ has either $0$ or $2$ solutions modulo $p$</p>
<p>I am clueless with this one. Hints please.</p>
| N. S. | 9,176 | <p>If $x^2-a=0 \mod p$ has some solution $b$, it means that $b^2=a$, and hence you original question becomes </p>
<p>$$x^2-b^2 \equiv 0 \mod p$$</p>
<p>Can you prove now that this equation has exactly two solutions?</p>
|
210,849 | <p>Let $p$ be an odd prime and $a$ be an integer with $\gcd(a, p) = 1$. Show that $x^2 - a \equiv 0 \mod p$ has either $0$ or $2$ solutions modulo $p$</p>
<p>I am clueless with this one. Hints please.</p>
| DonAntonio | 31,254 | <p>For $\,p=2\,$ the claim fails, as $\,x^2-1=x^2+1=(x+1)^2=0\pmod 2\,$ has one unique solution.</p>
<p>If $\,p\,$ is odd and $\,x^2-a=0\pmod p\,$ has a solution $\,b\,$, then</p>
<p>$$x^2-a=x^2-b^2=(x-b)(x+b)=0\pmod p\Longleftrightarrow$$</p>
<p>$$ x=b\pmod p\,\,or\,\,x=-b\pmod p$$</p>
<p>And now you only have to justify why the two solutions above are actually different.</p>
|
2,780,043 | <p>In the Navy we had bunk beds with a locker and lock both built in. You could gain access with a combination lock on the side that you could reprogram the code for. The lock was nothing more than several push buttons. Ive wondered for a long time now how many possible combinations there were. I believe the lock had $n=4$ buttons, but I would like to generalize to all $n\in\Bbb N$.</p>
<p>The system is easy enough to understand. Here are the rules:</p>
<ul>
<li>there are $n$ push buttons</li>
<li>each button can be pressed no more than once, and clearly you needn't push any button at all.</li>
<li>the locker combination can comprise of any number of distinct pressings.</li>
<li>order of pressing is relevant, so its a permutation problem</li>
<li>any grouping of buttons can be pressed simultaneously (wherein concurrently pressed buttons have no order).</li>
</ul>
<p>So, for example, if there are $n=3$ buttons and we are pressing all three buttons then $(1)(2)(3)$ is a viable combination, which is distinct from $(2)(1)(3)$ and from $(3)(2)(1)$, et. cetera, totaling 6 possible permutations when pressed separately. These are three button, three pressing combinations.</p>
<p>But those combinations are also distinct from three button, two pressing combinations such as $(1)(23)$ and $(12)(3)$ and three button, one pressing combinations like $(123)$. These were an additional three when (some) pressings are done concurrently, but since the order of the pressings are relevant then $(3)(12)$ and $(23)(1)$ are two more. Note that $(12)=(21)$ since concurrently pressed buttons have no order. But $(1)(23)\ne (23)(1)$ since the non-concurrently pressed groups $(1)$ and $(23)$ are ordered with respect to one another.</p>
<p>Naturally, one can press but two of the three, one of the three, or none of the three, for a multitude of additional possible combinations.</p>
<p>Id like to solve this problem, but in particular for the $n=4$ case but also in the general case.</p>
<p>This is a problem Ive been trying to solve for a decade.</p>
<p>For three buttons $1,2,3$ the combinations are $(), (1), (2), (3), (12), (13), (23), (123), (1)(2), (2)(1), (1)(3), (3)(1), (2)(3), (3)(2), (1)(23), (23)(1), (13)(2), (2)(13), (12)(3), (3)(12), (1)(2)(3), (1)(3)(2), (2)(1)(3), (2)(3)(1), (3)(1)(2), (3)(2)(1)$</p>
<p>26 combinations unless I missed some.</p>
<p>I shouldnt have to specify that there is no explicit limit on the number of pressings a combination can comprise, but you are implicitly restricted by the number of buttons and the fact that each can be pressed no more than once.</p>
| Ray Chou | 560,580 | <p>There are $\dbinom{n}{k}$ ways to choose k objects from a set of $n$ buttons, and there are $S(k,j)$ (Stirling numbers of the second kind) ways to partition these $k$ labeled objects into $j$ different (unlabeled) sets, and $j!$ ways to order these sets. In total, that makes</p>
<p>$$\sum_{k=0}^{n}\left[\dbinom{n}{k} \sum_{j=0}^{k} S(k,j) \: j!\right]$$</p>
<p>ways. Letting $a_{k}$ denoting the $k$th ordered Bell Number, there are</p>
<p>$$\sum_{k=0}^n\dbinom{n}{k} a_{k}$$</p>
<p>different combinations.</p>
|
623,703 | <blockquote>
<p>Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator. </p>
</blockquote>
<p>I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$</p>
<p>So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$. </p>
<p>The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you. </p>
| Archis Welankar | 275,884 | <p>We have to find the exact value of <span class="math-container">$\;\tan\left(\sin^{-1}\dfrac{1}{\sqrt{2}}\right)$</span>.</p>
<p>Since <span class="math-container">$\;\sin(45)=\dfrac{1}{\sqrt{2}}\;,\;x=45\;$</span> so <span class="math-container">$\;\tan(45)=1\;,\;$</span> hence we are done. Now <span class="math-container">$\;\sin(45)=\cos(45)\;,\;$</span> thus <span class="math-container">$\;\tan(45)=1\;.$</span></p>
|
10,666 | <p>My question is about <a href="http://en.wikipedia.org/wiki/Non-standard_analysis">nonstandard analysis</a>, and the diverse possibilities for the choice of the nonstandard model R*. Although one hears talk of <em>the</em> nonstandard reals R*, there are of course many non-isomorphic possibilities for R*. My question is, what kind of structure theorems are there for the isomorphism types of these models? </p>
<p><b>Background.</b> In nonstandard analysis, one considers the real numbers R, together with whatever structure on the reals is deemed relevant, and constructs a nonstandard version R*, which will have infinitesimal and infinite elements useful for many purposes. In addition, there will be a nonstandard version of whatever structure was placed on the original model. The amazing thing is that there is a <em>Transfer Principle</em>, which states that any first order property about the original structure true in the reals, is also true of the nonstandard reals R* with its structure. In ordinary model-theoretic language, the Transfer Principle is just the assertion that the structure (R,...) is an elementary substructure of the nonstandard reals (R*,...). Let us be generous here, and consider as the standard reals the structure with the reals as the underlying set, and having all possible functions and predicates on R, of every finite arity. (I guess it is also common to consider higher type analogues, where one iterates the power set ω many times, or even ORD many times, but let us leave that alone for now.) </p>
<p>The collection I am interested in is the collection of all possible nontrivial elementary extensions of this structure. Any such extension R* will have the useful infinitesimal and infinite elements that motivate nonstandard analysis. It is an exercise in elementary mathematical logic to find such models R* as ultrapowers or as a consequence of the Compactness theorem in model theory. </p>
<p>Since there will be extensions of any desired cardinality above the continuum, there are many non-isomorphic versions of R*. Even when we consider R* of size continuum, the models arising via ultrapowers will presumably exhibit some saturation properties, whereas it seems we could also construct non-saturated examples. </p>
<p>So my question is: what kind of structure theorems are there for the class of all nonstandard models R*? How many isomorphism types are there for models of size continuum? How much or little of the isomorphism type of a structure is determined by the isomorphism type of the ordered field structure of R*, or even by the order structure of R*? </p>
| Joel David Hamkins | 1,946 | <p>Let me offer one counterpoint to John's excellent answer.</p>
<p>Under the Continuum Hypothesis, the ultrapower version of R* will be saturated in any countable language. That is, it will realize all finitely realizable countable types with countably many parameters. Thus, by the usual back-and-forth construction, if we take the reduct to any countable part of the language, such as the ordered field structure and more, then under CH there will be only one <em>saturated</em> model of size continuum. </p>
<p>I'm not sure if John's construction can produce saturated models, but if so, then under CH this observation will answer his question at the end about whether one can have non-isomorphic R*'s that are isomorphic orders or even as ordered fields.</p>
|
467,574 | <p>Using permutation or otherwise, prove that $\displaystyle \frac{(n^2)!}{(n!)^n}$ is an integer,where $n$ is a positive integer.</p>
<p>I have no idea how to prove this..!!I am not able to even start this Can u give some hints or the solution.!cheers.!!</p>
| Davood | 477,916 | <p><strong>Remark</strong> : For every (prime) number $p$:
$$
n
\cdot
\lfloor \dfrac{n}{p^{\alpha}} \rfloor
\leq
\lfloor \dfrac{n^2}{p^{\alpha}} \rfloor
\
.
$$
<strong>Proof</strong> :
Let's denote by $m$ and $p$ the integral part and fractional part of
$\dfrac{n}{p^{\alpha}}$ </p>
<p>$$
m=\lfloor \dfrac{n}{p^{\alpha}} \rfloor
\ \ \ \ \
\text{and}
\ \ \ \ \
p=\lfloor \dfrac{n}{p^{\alpha}} \rfloor
\
;
$$
then one can see easilly that :
$$
\dfrac{n}{p^{\alpha}}=m+p
\Longrightarrow
\dfrac{n^2}{p^{\alpha}}=n \cdot m + n \cdot p
\Longrightarrow
\\
n \cdot \lfloor \dfrac{n}{p^{\alpha}} \rfloor
=
n \cdot m
=
\lfloor n \cdot m \rfloor
\leq
\lfloor \dfrac{n^2}{p^{\alpha}} \rfloor
\
.
$$ </p>
<hr>
<p><strong>Remark (II)</strong> : For every prime number $p$:
$$
n
\cdot
\sum
\lfloor \dfrac{n}{p^{\alpha}} \rfloor
\leq
\sum
\lfloor \dfrac{n^2}{p^{\alpha}} \rfloor
\ \
\Longrightarrow
\\
\ \ \ \ \
n
\cdot
v_p(n!)
\leq
v_p\Big((n^2)!\Big)
\Longrightarrow
\\
\ \ \
v_p(n! ^n)
\leq
v_p\Big((n^2)!\Big)
\
.
$$ </p>
<hr>
<p><strong>Remark (III)</strong> :
$$
n! ^n \mid (n^2)!
\
.
$$ </p>
|
18,895 | <p>I know that the answer is $C(8,2)$, but I don't get, why.
Can anyone, please, explain it?</p>
| ThomasMcLeod | 4,735 | <p>$C[8, 2] = \dfrac{8!}{2!(8-2)!} = (8!/6!)/2! = (8*7)/2 = 28$. Think of it this way. The $8$ is the choice of the first bit, and the $7$ is the choice of the second bit (it's only $7$ because there are only $7$ bits available after the first bit is decided). The $2$ represents the number of permutations of the chosen bits. We divide by $2$ because we don't want to count each ordering of the chosen bits separatetly. </p>
|
894,909 | <p>Given </p>
<p>$(x+3)(y−4)=0 $</p>
<p>Quantity $A = xy $</p>
<p>Quantity $B = -12 $</p>
<p>A Quantity $A$ is greater.<br>
B Quantity $B$ is greater.<br>
C The two quantities are equal.<br>
D The relationship cannot be determined from the information given. </p>
<p>How is the answer D and not C ? </p>
<p>Please help.</p>
| Vikram | 11,309 | <p>$(x+3)(y-4)=0 ...(I)\Rightarrow$ either $x=-3$ <strong>OR</strong> $y=4$ <strong>OR</strong> $ x=-3$ <strong>AND</strong> $y=4$.</p>
<p>If $x=-3$, then $y$ can assume any value and when $y=4$, $x$ can assume any value to satisfy eqn (I)</p>
<p>In both the cases $A=xy$ can be anything, so we can not be sure about $A$ being greater than or less than -$12$.</p>
<p><strong>Only</strong> when $x=-3$ <strong>AND</strong> also, $y=4$, we get $A=-12$, but such condition is not specified in the problem.</p>
|
1,098,438 | <p>This problem is really bothering me for some time, I appreciate if you have some idea and insight.</p>
<blockquote>
<p>Prove that</p>
<p>$$2^{2^n}+5^{2^n}+7^{2^n}$$</p>
<p>is divisible by $39$ for all natural numbers $n$.</p>
</blockquote>
<p>There was a suggestion that this should be done by mathematical induction, however, with a twist, but I could not see what the twist is.</p>
| wendy.krieger | 78,024 | <p>You first find $2^2$, $5^2$ and $7^2$, $\mod 39$, to be $4$, $25$ and $10$ respectively. These add to $39$.</p>
<p>Their squares are $16$, $1$, and $22$, which also add to $39$. This is power $2^2$.</p>
<p>The squares of these are $22$, $1$, $16$, which is the same as before, Thus if it is true for $2^n$, it's true for $2^{n + 1}$.</p>
<p>therefore $2^a + 5^a + 7^a$ is a multiple of $39$ if $a = 2^n$ (it's actually a multiple if $a \mod 12 = 4, 8$.)</p>
|
1,098,438 | <p>This problem is really bothering me for some time, I appreciate if you have some idea and insight.</p>
<blockquote>
<p>Prove that</p>
<p>$$2^{2^n}+5^{2^n}+7^{2^n}$$</p>
<p>is divisible by $39$ for all natural numbers $n$.</p>
</blockquote>
<p>There was a suggestion that this should be done by mathematical induction, however, with a twist, but I could not see what the twist is.</p>
| N. S. | 9,176 | <p>Here is an alternate approach. </p>
<p>As $3$ and $13$ are prime, it follows that for any $a$ not divisible by $3$ and $13$ we have
$$a^2 \equiv 1 \pmod{3}$$
$$a^{12} \equiv 1 \pmod{13} $$</p>
<p>From the first one you get immediately that $a^{2^n} \equiv 1 \pmod{3}$ for $a \in \{ 2,5,7 \}$.</p>
<p>The second one implies that
$$a^{16} \equiv a^{4} \pmod{ 13} \, \mbox{ for } a \in \{ 2,5,7 \} \,.$$</p>
<p>Therefore, for all $n \geq 1$ we have
$$(a^{16})^{2^{n-2}} \equiv (a^{4})^{2^{n-2}} \pmod{ 13} \, \mbox{ for } a \in \{ 2,5,7 \} \,.$$
or
$$a^{2^{n+2}} \equiv a^{2^{n}} \pmod{ 13} \, \mbox{ for } a \in \{ 2,5,7 \} \,.$$</p>
<p>Hence</p>
<p>$$2^{2^n}+5^{2^n}+7^{2^n} \equiv 2^{2^{n+2}}+5^{2^{n+2}}+7^{2^{n+2}} \pmod{13}$$</p>
<p>This is the inductive step $P(n) \Rightarrow P(n+2)$, you need to check $P(1), P(2)$.</p>
|
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