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 |
|---|---|---|---|---|
4,403,081 | <p><span class="math-container">$$\int \dfrac{dx}{x\sqrt{x^4-1}}$$</span></p>
<p>I need to solve this integration.
I solved and got <span class="math-container">$\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$</span>, however the answer given in my textbook is <span class="math-container">$\dfrac12\sec^{-1}(x^2) + C$</span></p>
<p>How can I prove that both quantities are equal? Is there something wrong with my answer?</p>
<p><strong>EDIT:</strong></p>
<p>Here's my work:
<span class="math-container">$$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$</span></p>
<p>Let <span class="math-container">$x^4 - 1 = t^2$</span>
<span class="math-container">$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$</span></p>
<p><span class="math-container">$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$</span></p>
| Community | -1 | <p>I think your problem is already solved with B. Goddard's answer, Here's an easy way to solve the given integral.</p>
<hr />
<p>We have,
<span class="math-container">$$\int\dfrac{dx}{x\sqrt{x^4-1}}$$</span></p>
<p>Let <span class="math-container">$x^2 = \sec(\theta) \implies 2x \, dx = \sec(\theta) \tan(\theta) \, d\theta$</span></p>
<p>Which further implies that, <span class="math-container">$dx = \dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{2}\, d\theta$</span></p>
<p>After substitution, the given integral changes to:
<span class="math-container">$$\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\sec^2(\theta) - 1}}\, d\theta$$</span></p>
<p><span class="math-container">$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\tan^2(\theta)}}\, d\theta$$</span></p>
<p><span class="math-container">$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\tan(\theta)}\, d\theta$$</span></p>
<p><span class="math-container">$$=\dfrac12\int d\theta$$</span></p>
<p><span class="math-container">$$=\dfrac12\theta + C$$</span></p>
<p><span class="math-container">$$=\boxed{\dfrac12\sec^{-1}(x^2) + C}$$</span></p>
|
815,432 | <p>I have sometimes seen notations like $a\equiv b\pmod c$. How do we define the notation? Have I understood correctly that $c$ must be an element of some ring or does the notation work in magmas in general?</p>
| Bill Dubuque | 242 | <p>In a commutative ring <span class="math-container">$\,R,\, $</span> <span class="math-container">$\, a\equiv b\pmod c\,$</span> means <span class="math-container">$\,c\mid a-b,\ $</span> i.e. <span class="math-container">$\,a-b = cr\,$</span> for some <span class="math-container">$\,r \in R,\,$</span> i.e. <span class="math-container">$\,a-b\in cR.\,$</span> More generally, <span class="math-container">$\,a\equiv b\pmod{ c_1,\ldots, c_n}\,$</span> means that <span class="math-container">$\,a-b\in (c_1,\ldots c_n)R := c_1 R+\cdots+ c_n R,\,$</span> the sum of the principal ideals <span class="math-container">$\,c_i R,\,$</span> i.e <span class="math-container">$\,a-b = c_1 r_1 + \cdots + c_n r_n\,$</span> for some <span class="math-container">$\,r_i\in R.\,$</span> If <span class="math-container">$\,S\subset R\,$</span> then <span class="math-container">$\,a\equiv b\pmod{\! S}$</span> means <span class="math-container">$\,a-b\in SR\,$</span> where <span class="math-container">$\,SR\,$</span> is the ideal <em>generated</em> by <span class="math-container">$S$</span> in <span class="math-container">$R$</span>.</p>
<p>Congruence arithmetic provides an "element-ary" way of working in quotient (residue) rings. See <a href="https://math.stackexchange.com/a/16157/242">here</a> for the general correspondence between congruences, ideals, & <span class="math-container">$R$</span>-subalgebras of a square <span class="math-container">$R^2$</span>.</p>
<p>Similarly for groups (and <a href="https://math.stackexchange.com/a/3925222/242">ideal-determined algebras</a>) we can normalize all equations <span class="math-container">$\,a = b\,$</span> to the form <span class="math-container">$\,a\!-\!b = \color{#c00}0,\,$</span> so a congruence is determined by a <span class="math-container">$\rm\color{#c00}{single}$</span> congruence class: all elements <span class="math-container">$\color{#c00}{\equiv 0}$</span> (as above for ideals in rings). In more general types of algebras we need to work directly with congruence (equivalence) relation operations.</p>
<p>As for magmas, or any other algebraic structure defined by purely equational axioms, the notion of <a href="http://en.wikipedia.org/wiki/Congruence_relation#Universal_algebra" rel="nofollow noreferrer">ring-congruence generalizes</a> in a straightforward way to that of an equivalence relation that is compatible with all of the operations of the structure, e.g. <span class="math-container">$\, A\equiv a,\ B\equiv b\,\Rightarrow\, A\oplus B = a\oplus b\,$</span> for all binary operations <span class="math-container">$\,\oplus\,$</span> of the algebra, and similarly for all other <span class="math-container">$\,n$</span>-ary operations of the structure. Just as for rings, there is a <a href="http://en.wikipedia.org/wiki/Quotient_algebra" rel="nofollow noreferrer">quotient algebra</a> that reifies the (modular) congruence arithmetic within an algebraic structure of the same type, just like even/odd parity arthmetic of integers mod <span class="math-container">$\,2\,$</span> is ring-theoretically reified as arithmetic in the quotient/residue ring <span class="math-container">$\,\Bbb Z/2.$</span></p>
|
4,474,806 | <p>I use the following method to calculate <span class="math-container">$b$</span>, which is <span class="math-container">$a$</span> <strong>increased</strong> by <span class="math-container">$x$</span> percent:</p>
<p><span class="math-container">$\begin{align}
a = 200
\end{align}$</span></p>
<p><span class="math-container">$\begin{align}
x = 5\% \text{ (represented as } \frac{5}{100} = 0.05 \text{)}
\end{align}$</span></p>
<p><span class="math-container">$\begin{align}
b = a \cdot (1 + x) \
= 200 \cdot (1 + 0.05) \
= 200 \cdot 1.05 \
= 210
\end{align}$</span></p>
<p>Now I want to calculate <span class="math-container">$c$</span>, which is also <span class="math-container">$a$</span> but <strong>decreased</strong> by <span class="math-container">$x$</span> percent.</p>
<p>My instinct is to preserve the method, but to use division instead of multiplication (being the inverse operation):</p>
<p><span class="math-container">$
\begin{align}
c = \frac{a}{1 + x} \
= \frac{200}{1 + 0.05} \
= \frac{200}{1.05} \
= 190.476190476 \
\end{align}
$</span></p>
<p>The result looks a bit off? But also interesting as I can multiply it by the percent and I get back the initial value (<span class="math-container">$190.476190476 \cdot 1.05 = 200$</span>).</p>
<p>I think the correct result should be 190 (without any decimal), using:</p>
<p><span class="math-container">$
\begin{align}
c = a \cdot (1 - x) \
= 200 \cdot (1 - 0.05) \
= 200 \cdot 0.95 \
= 190
\end{align}
$</span></p>
<p>What's the difference between them? What I'm actually calculating?</p>
| John Douma | 69,810 | <p>It is easier to translate these expressions from English to Math than to think in terms of multiplication and division. The latter way leads to memorizing an arbitrary rule which won't stay with you.</p>
<p>We are saying we want to increase <span class="math-container">$a$</span> by <span class="math-container">$x\%$</span> where it is understood that "by <span class="math-container">$x\%$</span>" means <span class="math-container">$x\%$</span> of <span class="math-container">$a$</span>. This gives us a straight forward translation <span class="math-container">$$a+\frac{x}{100}a=a(1+\frac{x}{100})$$</span></p>
<p>When <span class="math-container">$x=5$</span> we get <span class="math-container">$1.05a$</span> as you calculated.</p>
<p>If we decrease <span class="math-container">$a$</span> by <span class="math-container">$x\%$</span>, we get <span class="math-container">$$a-\frac{x}{100}a=a(1-\frac{x}{100})$$</span> and when <span class="math-container">$x=5$</span> we get <span class="math-container">$0.95a$</span>.</p>
|
4,474,806 | <p>I use the following method to calculate <span class="math-container">$b$</span>, which is <span class="math-container">$a$</span> <strong>increased</strong> by <span class="math-container">$x$</span> percent:</p>
<p><span class="math-container">$\begin{align}
a = 200
\end{align}$</span></p>
<p><span class="math-container">$\begin{align}
x = 5\% \text{ (represented as } \frac{5}{100} = 0.05 \text{)}
\end{align}$</span></p>
<p><span class="math-container">$\begin{align}
b = a \cdot (1 + x) \
= 200 \cdot (1 + 0.05) \
= 200 \cdot 1.05 \
= 210
\end{align}$</span></p>
<p>Now I want to calculate <span class="math-container">$c$</span>, which is also <span class="math-container">$a$</span> but <strong>decreased</strong> by <span class="math-container">$x$</span> percent.</p>
<p>My instinct is to preserve the method, but to use division instead of multiplication (being the inverse operation):</p>
<p><span class="math-container">$
\begin{align}
c = \frac{a}{1 + x} \
= \frac{200}{1 + 0.05} \
= \frac{200}{1.05} \
= 190.476190476 \
\end{align}
$</span></p>
<p>The result looks a bit off? But also interesting as I can multiply it by the percent and I get back the initial value (<span class="math-container">$190.476190476 \cdot 1.05 = 200$</span>).</p>
<p>I think the correct result should be 190 (without any decimal), using:</p>
<p><span class="math-container">$
\begin{align}
c = a \cdot (1 - x) \
= 200 \cdot (1 - 0.05) \
= 200 \cdot 0.95 \
= 190
\end{align}
$</span></p>
<p>What's the difference between them? What I'm actually calculating?</p>
| JonathanZ supports MonicaC | 275,313 | <p>I think there are two issues at play here.</p>
<p><strong>Issue #1</strong>: Increase vs. decrease. That determines whether you use <span class="math-container">$1 + \frac{p}{100}$</span> or <span class="math-container">$1-\frac{p}{100}$</span>, respectively.</p>
<p>Using <span class="math-container">$I$</span> for "Initial amount" and <span class="math-container">$F$</span> for "Final amount":</p>
<p><span class="math-container">$$F = I \times (1 \pm \frac{p}{100}) $$</span></p>
<p><strong>Issue #2</strong>: Initial vs. Final. Do you know the initial (i.e. before the change) value and want to find the final (after the change) value, or do you know what the value is after the change, and want to find the initial value? That determines whether you multiply or divide.</p>
<p>If you know the initial value, use the same equation above. If you know the final value, you will actually re-use the same equation again (there's really only one equation for percentages!), but it will be re-written as</p>
<p><span class="math-container">$$ I = \frac{F}{1 \pm \frac{p}{100}}$$</span></p>
<p>I think that last case is where you have seen solutions that use division. Just realize that it's not the difference between increase and decrease that determines that, it's the difference of whether you want to find the value before or after the change.</p>
|
1,296,420 | <p>I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?</p>
| Mikko Korhonen | 17,384 | <p>Take $$G = H \times H \times H \times \cdots$$ for $H$ any nontrivial group.</p>
|
3,165,937 | <p>Are orthogonal groups are lie groups? I think parameter space points corresponds to elements with determinant -1 break analytic property of lie groups , what is the general condition to check a group is lie group or not ?</p>
| José Carlos Santos | 446,262 | <p>Yes, orthogonal groups are Lie groups. Since, <span class="math-container">$O(n,\mathbb R)$</span> consists of two copoes of <span class="math-container">$SO(n,\mathbb R)$</span>, if <span class="math-container">$SO(n,\mathbb R)$</span> is a Lie group, then so is <span class="math-container">$O(n,\mathbb R)$</span>.</p>
|
1,011,718 | <p>$P1 : \sin(x/y)$ .</p>
<p>I tried using $y=mx$. $f$ becomes $\sin(1/m)$, so limit doesn't exist. But it is too easy. Am I right?</p>
<p>$P2 : F = x^2\log(x^2+y^2)$</p>
<p><img src="https://i.stack.imgur.com/pEaSs.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/CEI2K.jpg" alt="Attempt"></p>
<p>So function can be defined as O at origin in order to be continous . Am I right in these? Im learning this stuff on my own.</p>
| Przemysław Scherwentke | 72,361 | <p>P1 is easy indeed and you are correct.</p>
<p>In P2 the limit is 0, but your attempt shows only limits on lines $y=mx$. A correct version is similar to that from answers to your previous question:
$$
|x^2\log(x^2+y^2)|\leq|(x^2+y^2)\log(x^2+y^2)|.
$$
But you should know, that
$$
\lim_{x\to0^+}x\ln{x}=0,
$$
and this ends the proof.</p>
|
3,964,237 | <p>I am trying to understand the meaning of this exercise question in a logic textbook.</p>
<blockquote>
<p>For each of the following statement forms,find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters.<br/>
i. <span class="math-container">$A \rightarrow ( B \leftrightarrow \lnot C )$</span> <br/>
ii. <span class="math-container">$\lnot A \lor ( B \rightarrow C )$</span> <br/>
iii. <span class="math-container">$A \land (B \lor \lnot C)$</span> <br/></p>
</blockquote>
<p>But I keep failing. I don't understand the part "<em>find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters</em>". So I am unable to solve it.Can someone tell me what that means?</p>
| Neptune | 825,557 | <p>It means you are only using <span class="math-container">$\neg$</span> in front of <span class="math-container">$A$</span> <span class="math-container">$B$</span> or <span class="math-container">$C$</span>. So instead of writing <span class="math-container">$\neg(A \lor B)$</span>, for example, you would write <span class="math-container">$\neg A \land \neg B$</span></p>
|
2,658,195 | <p>I have the following problem with which I cannot solve. I have a very large population of birds e.g. 10 000. There are only 8 species of birds in this population. The size of each species is the same.</p>
<p>I would like to calculate how many birds I have to catch, to be sure in 80% that I caught one bird of each species.</p>
| SK19 | 509,159 | <p>If we assume that each die roll is done <strong>independent</strong> of the others and further assume that <strong>each side has the same probability</strong>, then this experiment is just the same as throwing a perfect coin until one party wins. There is no way to predict which of the eight predictions would be more likely because no one is.</p>
<p><strong>BUT</strong> we are dealing with a "real" die here (real in the Manga universe), and real dice don't really have each side with same probability (because they are not perfect, may have little cuts or a site weights more or something similar) AND our inability to predict how the die is thrown is bad due to too many variables (how often is the die shaken in the hand etc.). So if the manga characters would have some strong, supernatural sense of all the surroundings and how the one who throws the die does it, they could make predictions. But this is really, really far fetched.</p>
|
546,572 | <p><img src="https://i.stack.imgur.com/aJ2t5.jpg" alt="enter image description here"></p>
<p>Could anyone tell me how to solve 9b and 10? I've been thinking for five hours, I really need help.</p>
| detnvvp | 85,818 | <p>Hint: take $A=\mathbb N$ and $B=\{n+\frac{1}{n}\left|\right.n\in\mathbb N,n\geq 2\}$.</p>
|
2,473,951 | <p>The definition I have of a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ is that for every $x, y \in \mathbb{R}$ and every $\lambda \in [0, 1]$,
$$
f(\lambda x + (1-\lambda )y) \leq \lambda f(x) + (1- \lambda )f(y).$$</p>
<p>By proving that slopes increase I mean that for $x \leq y \leq z$, we get $$\frac{f(y) - f(x)}{y-x} \leq \frac{f(z) - f(x)}{z-x} \leq \frac{f(z) - f(y)}{z-y}. $$</p>
<p>Is there a simple proof of this which doesn't assume that such a convex function has a non-negative second derivative? It's difficult to see how the definition gets us here.</p>
| Community | -1 | <p><strong>Hint</strong>.</p>
<p>Note that if $x\leqslant y\leqslant z$, then we can write
$$
y=\lambda x+(1-\lambda)z
$$
for some $\lambda\in[0,1]$. <em>Explicitly</em>, you can find what is $\lambda$ in terms of $x,y$ and $z$. On the other hand, the definition of convexity tells you
$$
f(y)\leqslant \lambda f(x)+(1-\lambda)f(z).
$$
Now do some algebra to get what you want. </p>
|
2,399,842 | <p>I ran $\frac{d^n}{dx^n}[(x!)!]$ through <em><a href="https://www.wolframalpha.com/input/?i=nth%20derivative%20of%20(x!)!" rel="nofollow noreferrer">Wolfram|Alpha</a></em>, which returned</p>
<blockquote>
<p>$$\frac{\partial^n(x!)!}{\partial x^n} = \Gamma(1+x!)\,R(n,1+x!)$$
for</p>
<ul>
<li><p>$R(n,x)=\psi(x)\,R(-1+n,x)+R^{(0,1)}(-1+n,x)$</p></li>
<li><p>$R(0,x)=1$</p></li>
<li><p>$n\in\mathbb{Z}$</p></li>
<li><p>$n>0$</p></li>
</ul>
<p>where $\psi^{(n)}(x)$ is the $n$<sup>th</sup> derivative of the digamma function</p>
</blockquote>
<p>They define the <a href="http://mathworld.wolfram.com/PolygammaFunction.html" rel="nofollow noreferrer">polygamma function</a> as $$\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln[\Gamma(x)]$$</p>
<p>What on Earth is $R^{(0,1)}$, and how can I make sense of this $R(n,x)$ business?</p>
| leonbloy | 312 | <p>The superscript notation means (I think) the derivative: $$R^{(0,1)}(−1+n,x)=\frac{\partial R(n,x)}{\partial x}\Bigr|_{(−1+n,x)}$$</p>
<p>$R(n,x)$ is so ugly, it seems, that it cannot be expressed in some closed form but only as a recursion in $n$:</p>
<p>$$R(n,x)=\frac{d \ln[\Gamma(x)]}{dx} \,R(-1+n,x)+\frac{\partial R(n,x)}{\partial x}\Bigr|_{(−1+n,x)}$$</p>
|
1,612,220 | <p>This is an exercise page 7 from Sutherland's book Introduction to Metric and Topological Spaces.</p>
<blockquote>
<p>Suppose that <span class="math-container">$V,X,Y$</span> are sets with <span class="math-container">$V\subseteq X\subseteq Y$</span> and suppose that <span class="math-container">$U$</span> is a subset of <span class="math-container">$Y$</span> such that <span class="math-container">$X\setminus V=X\cap U$</span>.</p>
<p>Prove that <span class="math-container">$V=X\cap(Y\setminus U)$</span>.</p>
</blockquote>
<p>My attempt:</p>
<p>Let <span class="math-container">$x\in X\cap(Y\setminus U)$</span>. Then <span class="math-container">$x\in X$</span> and <span class="math-container">$x\in Y\setminus U$</span>. So, <span class="math-container">$x\in X$</span> and <span class="math-container">$x\notin U$</span>.</p>
<p>Here the solution given by Sutherland's book argues differently. So I am wondering if I can say: If an element <span class="math-container">$x$</span> is in the set <span class="math-container">$X$</span> then we can write <span class="math-container">$x\in V$</span> and <span class="math-container">$x\in X\cap V$</span>.</p>
<p>And continuing, we have <span class="math-container">$x\in V$</span> and <span class="math-container">$x\in X\cap V$</span> and <span class="math-container">$x\notin U$</span>.</p>
<p>The last two relations can be eliminated. And hence, <span class="math-container">$x\in V$</span>.</p>
<p>The second part of the proof is to prove conversely that <span class="math-container">$V\subseteq X\cap(Y\setminus U)$</span>.</p>
<p>I am wondering if the first part of my proof is valid, especially the second sentence.</p>
| DanielWainfleet | 254,665 | <p>Using $+$ to mean union, and multiplication to mean intersection,and $-$ for complement (That is $a-b=a\backslash b$) we have $$V=X-(X-V).$$ We also have $$X-V=X U \;\text { and } X=X Y.$$ Hence $$V=X-(X-V)=X-X U=( X Y) -( X Y)U=X(Y-Y U)=X(Y-U)=$$ $$=X\cap (Y\backslash U).$$ Note that the condition $U\subset Y$ was unnecessary.</p>
|
1,465,490 | <p>I have a set of target coordinates and a set of actually clicked coordinates which should be approximately the same, but not identical. The y coordinates are equal, however, the x-coordinates differ, such that negative coordinates are closer together than larger/positive coordinates.
e.g.:</p>
<pre><code>target_y actual_y
-691 -580
-675 -520
-650 -500
-638 -480
-588 - 420
-538 -320
-480 -260
-355 -60
-301 160
-301 360
-297 -560
-295 380
-222 100
-205 120
-203 120
-169 220
-103 300
-102 240
-41 360
-17 420
17 500
72 560
72 580
112 600
</code></pre>
<p>the difference is equal for all series, so I want to determine a function that tranforms the actual_y into the range of the target_y. I am thinking of something like target=0.2*actual-50. how can I find the correct function?</p>
<p>--
my edit was not needed, sorry</p>
| Hagen von Eitzen | 39,174 | <p>Make a sketch! In total you have two half circles and two line segments of length distance plus twice the radius.</p>
|
40,493 | <p>I am trying to use Mathematica to solve a relatively simple ODE involving parameter(s). I would like to use a set of conditions to solve for the particular solution of the ODE. I understand how to make Mathematica find values for the constants that arise during the process of solving the ODE, but what about solving for constants/coefficients already present in the original ODE? Here is a simple example involving Newton's Law of Cooling...
<img src="https://i.stack.imgur.com/rivKO.jpg" alt="enter image description here"></p>
<p>Here is the code I tried:</p>
<pre><code> DSolve[
{
T'[t] == -k*(T[t] - Ta),
T[0] == 70,
T[1/2] == 110,
T[1] == 145
},
{T[t], t, k},
{t}
]
</code></pre>
<p>I feel like I need a two step process... first solve the ODE with the parameters, and then solve for the parameters afterwards. I'm just not sure where to start.</p>
<p>Thank you in advance!</p>
| Nasser | 70 | <pre><code>sol = T[t] /. First@DSolve[{T'[t] == -k*(T[t] - Ta)}, T[t], t]
</code></pre>
<p><img src="https://i.stack.imgur.com/wVWI5.png" alt="Mathematica graphics"></p>
<pre><code>sol = sol /. C[1] -> c
</code></pre>
<p><img src="https://i.stack.imgur.com/mba0H.png" alt="Mathematica graphics"></p>
<pre><code>eq1 = 70 == sol /. t -> 0;
eq2 = 110 == sol /. t -> 1/2;
eq3 = 145 == sol /. t -> 1;
Solve[{eq1, eq2, eq3}, {k, c, Ta}]
</code></pre>
<p><img src="https://i.stack.imgur.com/TVVKG.png" alt="Mathematica graphics"></p>
|
1,224,202 | <p>Does the following equation makes any sense at all?</p>
<p>$$
\frac{1}{|X|\cdot|Y|}\sum\limits_{x \in X}\sum\limits_{y \in Y}\begin{cases}
1 & \mathrm{if~} x > y\\
0.5 & \mathrm{if~} x = y\\
0 & \mathrm{if~} x < y
\end{cases}
$$</p>
<p>For every comparison of $x$ and $y$, I want to add 1, 0.5 or 0 according to the case statements, and then multiply by the left part of the equation. Is it correct the way it is written? Is there any more beautiful way of designing that equation?</p>
| Community | -1 | <p>$$\frac{1}{|X|\cdot|Y|}\left(|\{(x,y):x \in X,y\in Y:x>y\}|+\frac12|\{(x,y):x \in X,y\in Y:x=y\}|\right).$$</p>
<hr>
<p>Alternatively,
$$\frac1{|X|\cdot|Y|}\left(\sum_{\substack{x\in X \\ y \in Y}}(x>y)+\frac12\sum_{\substack{x\in X \\ y \in Y}}(x=y)\right)$$</p>
|
11,353 | <p>Thinking about the counterintuitive <em>Monty Hall Problem</em> (stick or switch?),
revisited in <a href="https://matheducators.stackexchange.com/a/11346/511">this ME question</a>,
I thought I would issue a challenge:</p>
<blockquote>
<p>Give in one (perhaps long) sentence a convincing explanation of why <em>switching</em> is twice as likely to lead to winning as <em>sticking</em>.</p>
</blockquote>
<p>Assume the game assumptions
are pre-stated and clear.</p>
<p>The probabilities are not even close, so there should be a convincing explanation after all
<a href="https://en.wikipedia.org/wiki/Monty_Hall_problem" rel="nofollow noreferrer">the discussion of this topic</a>,
even though "1,000 Ph.D."s got it wrong (in 1990 when it first went viral).</p>
| quid | 143 | <p>If you "stay" then you win when the prize is behind the <em>one</em> door your originally selected, yet when you "switch" you win when the prize is behind one of the <em>two</em> doors you originally did not select.</p>
|
231,773 | <p>Let $G=(V,E)$ be an undirected graph. We form a graph $H=(V',E')$ from $G$ such that </p>
<ul>
<li>$V' = V \cup \{ w_e \mid e \in E \}$, and </li>
<li>$E' = \{ aw_e, bw_e \mid ab = e \in E \} \cup \{ w_e w_f \mid e,f \text{ are adjacent edges in }G \}$. </li>
</ul>
<p>Informally, $H$ is built from $G$ by subdividing each edge, and by putting an edge between two newly created vertices iff the corresponding edges are adjacent in $G$.</p>
<p>The above construction feels quite natural. Does it have a name? It seems like some kind of a variation of the line graph.</p>
| assaferan | 74,819 | <p>The answer is true, using the following construction.
Let <span class="math-container">$B$</span> be the quaternion algebra of discriminant <span class="math-container">$p$</span> and let <span class="math-container">$O$</span> be a maximal order with an element <span class="math-container">$x$</span> satisfying <span class="math-container">$x^2 = -p$</span>. The reduced norm is a quadratic form on <span class="math-container">$O$</span>, with positive definite Gram matrix <span class="math-container">$A$</span> of determinant <span class="math-container">$p^2$</span>. The matrix <span class="math-container">$A^{-1}$</span> then represents the reduced norm on the dual lattice <span class="math-container">$O^{\sharp}$</span>. Recall that <span class="math-container">$\text{nrd}(\text{diff}(O)) = \text{discrd}(O) = p$</span> and since <span class="math-container">$O$</span> is maximal, <span class="math-container">$\text{diff}(O)$</span> is invertible and <span class="math-container">$O = \text{diff}(O) O^{\sharp}$</span> (see e.g.
<em>Voight, John</em>, <a href="http://dx.doi.org/10.1007/978-3-030-56694-4" rel="nofollow noreferrer"><strong>Quaternion algebras</strong></a>, <a href="https://zbmath.org/?q=an:07261776" rel="nofollow noreferrer">ZBL07261776</a>. Section 16.8), hence
<span class="math-container">$$
p O^{\sharp} \subseteq \text{diff}(O) O^{\sharp} = O
$$</span>
which implies that <span class="math-container">$p \cdot (O^{\sharp} / O) = 0$</span> (therefore <span class="math-container">$O^{\sharp} / O$</span> is an abelian group of exponent <span class="math-container">$p$</span> and size <span class="math-container">$p^2$</span>, so isomorphic to <span class="math-container">$(\mathbb{Z} / p \mathbb{Z})^2$</span>). Next, we note that the matrix <span class="math-container">$A^{-1}$</span> is also the change of basis matrix between the chosen basis of <span class="math-container">$O$</span> and its dual, therefore <span class="math-container">$pA^{-1}$</span> is integral.</p>
<p>This is not enough, but so far we have not used the element <span class="math-container">$x$</span>. The matrix <span class="math-container">$pA^{-1}$</span> is the matrix representing the norm form on the ideal <span class="math-container">$xO^{\sharp}$</span>, since <span class="math-container">$x^2 = -p$</span>. But <span class="math-container">$nrd(xO^{\sharp})=p$</span> is a maximal order with an element <span class="math-container">$x$</span> such that <span class="math-container">$x^2 = -p$</span>. By a theorem of Ibukiyama (see reference below), it is isomorphic (hence isometric as lattices) to <span class="math-container">$O$</span>.</p>
<p>This could also be seen directly (referring to some of the answers above - it is possible to do it using a finite number of cases) by explicitly writing down the maximal orders. I will list below explicit constructions, which are based on
<em>Ibukiyama, Tomoyoshi</em>, <a href="http://dx.doi.org/10.1017/S002776300002016X" rel="nofollow noreferrer"><strong>On maximal orders of division quaternion algebras over the rational number field with certain optimal embeddings</strong></a>, Nagoya Math. J. 88, 181-195 (1982). <a href="https://zbmath.org/?q=an:0473.12012" rel="nofollow noreferrer">ZBL0473.12012</a>. -</p>
<p>If <span class="math-container">$p \equiv 3 \bmod 4$</span>, then <span class="math-container">$B = (-p,-1)$</span>, and <span class="math-container">$O = \mathbb{Z}<(1+i)/2,j>$</span>. In this case, we have
<span class="math-container">$$
A = \left( \begin{array}{cccc}
2 & 1 & 0 & 0 \\
1 & \frac{p+1}{2} & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & \frac{p+1}{2}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc}
\frac{p+1}{2} & -1 & 0 & 0 \\
-1 & 2 & 0 & 0 \\
0 & 0 & \frac{p+1}{2} & -1 \\
0 & 0 & -1 & 2
\end{array} \right),
K = \left( \begin{array}{cccc}
-1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & -1 & 1 \\
0 & 0 & 1 & 0
\end{array} \right)
$$</span></p>
<p>If <span class="math-container">$p \equiv 1 \bmod 4$</span>, then choosing <span class="math-container">$q \equiv 3 \bmod 4$</span> such that
<span class="math-container">$\left( \frac{q}{p} \right) = -1$</span>, we can find <span class="math-container">$c$</span> such that
<span class="math-container">$c^2 \equiv -p \bmod q$</span>, and then <span class="math-container">$B = (-p,-q)$</span> and
<span class="math-container">$$
O = \mathbb{Z} \oplus \mathbb{Z} \frac{1+j}{2} \oplus
\mathbb{Z} \frac{i(1+j)}{2} \oplus \mathbb{Z} \frac{(c+i)j}{q}
$$</span>
In this case, we compute that
<span class="math-container">$$
A = \left( \begin{array}{cccc}
2 & 1 & 0 & 0 \\
1 & \frac{q+1}{2} & 0 & c \\
0 & 0 & \frac{p(q+1)}{2} & p \\
0 & c & p & \frac{2(p+c^2)}{q}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc}
\frac{(q+1)(c^2+p)}{2q} & -c-\frac{c^2+p}{q} & -c & \frac{c(q+1)}{2} \\
-c-\frac{c^2+p}{q} & 2(c^2 + \frac{c^2+p}{q}) & 2c & -c(q+1) \\
-c & 2c & 2 & -q \\
\frac{c(q+1)}{2} & -c(q+1) & -q & \frac{q(q+1)}{2}
\end{array} \right)
$$</span></p>
<p>and if <span class="math-container">$\{e_1, e_2, e_3, e_4 \}$</span> is the above basis for <span class="math-container">$O$</span>, then we see that <span class="math-container">$ \{e_2 e_4, ce_1 - e_4, e_1, j e_2 \} $</span> is a basis for the resulting module, which written in terms of the original basis yields the matrix
<span class="math-container">$$
K = \left( \begin{array}{cccc}
0 & c & 1 & -\frac{q+1}{2} \\
-c & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
\frac{q+1}{2} & -1 & 0 & 0
\end{array}
\right).
$$</span></p>
<p>Referring to Will Jagy's wondering in the first answer, the reason that this does not work for most lattices is that they do not correspond to a maximal order (as in the answer by few_reps, the quotient of the lattices is actually cyclic) and there are only one or two (depending on <span class="math-container">$p \bmod 4$</span> isomorphism classes of maximal orders which contain a root of <span class="math-container">$-p$</span>. For example, for <span class="math-container">$p = 37$</span>, there are only two isomorphism classes of maximal orders in the quaternion algebra, and only one of them contains a root of <span class="math-container">$-p$</span>.</p>
|
2,911,187 | <p>Lines (same angle space between) radiating outward from a point and intersecting a line:</p>
<p><a href="https://i.stack.imgur.com/52HY4.png" rel="noreferrer"><img src="https://i.stack.imgur.com/52HY4.png" alt="Intersection Point Density Distribution"></a></p>
<p>This is the density distribution of the points on the line:</p>
<p><a href="https://i.stack.imgur.com/CqbWF.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/CqbWF.jpg" alt="Plot line Graph"></a></p>
<p><sup>I used a Python script to calculate this. The angular interval is 0.01 degrees. X = tan(deg) rounded to the nearest tenth, so many points will have the same x. Y is the number of points for its X. You can see on the plot that ~5,700 points were between -0.05 and 0.05 on the line.</sup></p>
<p>What is this graph curve called? What's the function equation?</p>
| user | 505,767 | <p>It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form</p>
<p>$$y=\frac{d}{H+x^2}$$</p>
<p>where $d$ is related to the intensity and $H$ is the distance of the source point from the line.</p>
<p>Here is the <a href="https://www.desmos.com/calculator/162tozjwev" rel="noreferrer">plot</a> for the case $d=H=1$</p>
<p><a href="https://i.stack.imgur.com/g0Vls.png" rel="noreferrer"><img src="https://i.stack.imgur.com/g0Vls.png" alt="enter image description here"></a></p>
|
3,764,030 | <p>If <span class="math-container">$|z| = \max \{|z-1|,|z+1|\}$</span>, then:</p>
<ol>
<li><span class="math-container">$\left| z + \overline{z} \right| =1/2$</span></li>
<li><span class="math-container">$z + \overline{z} =1$</span></li>
<li><span class="math-container">$\left| z + \overline{z} \right| =1$</span></li>
<li><span class="math-container">$z - \overline{z} = 5$</span></li>
</ol>
<p>I am totally confused that how to calculate maximum of <span class="math-container">$|z-1|$</span> and <span class="math-container">$|z+1|$</span>, I think they represent line <span class="math-container">$x=1$</span> and <span class="math-container">$x=-1$</span> and if so then how to calculate maximum.</p>
<p>Please! Guide me how to proceed.</p>
| Kavi Rama Murthy | 142,385 | <p>There is no complex number <span class="math-container">$z$</span> such that <span class="math-container">$|z|$</span> is the maximum of <span class="math-container">$|z-1|$</span> and <span class="math-container">$|z+1|$</span>.</p>
<p>One can say that all four options are vacuously true!</p>
<p>Proof: write <span class="math-container">$z$</span> as <span class="math-container">$\frac {(z-1)+(z+1)} 2$</span>. This gives <span class="math-container">$|z| \leq \frac {|z-1|+|z+1|} 2 \leq \max \{|z-1|,|z+1|\}=|z|$</span>. Hence equality holds throughout. In particular this implies <span class="math-container">$|z-1|=|z+1|$</span> so <span class="math-container">$z$</span> must lies in the imaginary axis. But then <span class="math-container">$z=iy$</span> for some real number <span class="math-container">$y$</span> and the given equation fails.</p>
|
2,635,635 | <p>I have searched a lot and don't really understand the answers I've come across. So apologies in advance if I'm repeating a common question.</p>
<p>The problem is as follows: Distribute $69$ identical items across
$4$ groups where each groups needs to contain at least $5$ items.</p>
<p>The way I see the problem: $x_1 + x_2 + x_3 + x_4 = 69$, $x_i \geq 5$</p>
<p>Could this then be the same as: $x_1 + x_2 + x_3 + x_4 = 49$, $x_i \geq 0$, and therefore be solved using ${n + k -1} \choose {k - 1}$ = ${52}\choose{3}$ = $\frac{52!}{3!(52 - 3)!}$ = $\frac{52!}{3!49!}$</p>
<p>Or am I thinking completely wrong here?</p>
| AleksandrH | 287,621 | <p>Precisely, yes. What you did is correct. If you're interested, the "rigorous" mathematical approach would be the following:</p>
<p>$x_1 + x_2 + x_3 + x_4 = 69$, where $x_i \geq 5$ for all $i \in [1,4]$</p>
<p>Let's take $x_1$ as an example:</p>
<p>$x_1 \geq 5$</p>
<p>By simple subtraction from both sides, we see that:</p>
<p>$x_1 - 5 \geq 0$</p>
<p>Stars and bars is really easy to solve when $x_i \geq 0$ for all $i$. That's because it essentially translates to a scenario where you can pick $0$ or more of each item to place in each "bucket". Notice that the left-hand side of that inequality is indeed $\geq 0$. In that case, so we no longer have to refer to it as the "left-hand side of the inequality", let's call it $x_1'$.</p>
<p>$x_1' = x_1-5\geq0$</p>
<p>Now, let's just focus on the equality part:</p>
<p>$x_1' = x_1 - 5$</p>
<p>Rearranging:</p>
<p>$x_1 = x_1' +5$</p>
<p>And now, let's repeat this process for all other variables:</p>
<p>$x_2 = x_2'+5$</p>
<p>$x_3 = x_3' + 5$</p>
<p>$x_4 = x_4' + 5$</p>
<p>And substitute these all back into the original equation:</p>
<p>$x_1'+5+x_2'+5+x_3'+5+x_4'+5 = 69$</p>
<p>Then, subtracting all the $5$s from both sides, this simply becomes:</p>
<p>$x_1'+x_2'+x_3'+x_4' = 49$</p>
<p>Where all $x_i' \geq 0$. And there! You can now proceed solving this just as you did.</p>
<p>More on the stars and bars part, because I hate memorizing equations:</p>
<p>x1|x2|x3|x4</p>
<p>The variables indicate valid buckets where we can place stars. Let's call the leftmost bucket $x_1$, all the way to $x_4$ on the rightmost.</p>
<p>We have $49$ stars that we can put in these buckets to indicate "Hey, I would like this many of the variable $x_i$, please!" </p>
<p>And then the problem simply becomes a <a href="https://en.wikipedia.org/wiki/Permutation#Permutations_of_multisets" rel="nofollow noreferrer">permutation of a multiset</a>. If you've solved problems that ask you how many ways you can scramble the letters in the word "SORRY", for example, then this is precisely what you're doing—you translate the problem into an identical scenario. In this case, you're scrambling two "letters": the bars (|) and the stars (*).</p>
<p>So in your case, you've got $3$ bars and $49$ stars. You have $52$ "slots" in your "word". You choose $3$ of those to be the bars and the rest to be stars:</p>
<p>$52 \choose 3$ $49 \choose 49$ = $\frac{52!}{3!49!}$</p>
|
3,608,097 | <p>Three strings totaling a length <span class="math-container">$U= 3 a + 4b + 2 \pi r$</span> cut into three parts together enclose minimum total area</p>
<p><span class="math-container">$$ A= \frac{\sqrt3 a^2}{4} + b^2+\pi r^2,$$</span></p>
<p>when they are made into shapes of an equilateral triangle, square and circle respectively.</p>
<p>Please help determine <span class="math-container">$ (a,b,r)$</span> divisions in the case of these three polygons. </p>
<p>When there are only a square and a circle it is noted that the figures can be be drawn enclosed between parallels. <a href="https://math.stackexchange.com/questions/3607758/minimum-area-of-a-square-and-a-circle-of-total-1-unit-circumference-length/3607834#3607834">Geometrical arrangement</a>.</p>
<p>In the above it has been verbally indicated that first pairwise Lagrange Multiplier</p>
<p><span class="math-container">$$\dfrac{ \dfrac{\partial{U}}{\partial{b}}}{ \dfrac{\partial{U}}{\partial{r}}}= \dfrac{ \dfrac{\partial{A}}{\partial{b}}}{ \dfrac{\partial{A}}{\partial{r}}}=\lambda$$</span></p>
<p>could be taken to determine <span class="math-container">$(b,r)$</span> relation:</p>
<p><span class="math-container">$$ b=2r \tag1$$</span></p>
<p>Next <span class="math-container">$(a,r)$</span> pairwise relation is determined in a similar manner:</p>
<p><span class="math-container">$$ \frac{a}{\sqrt3 }=2r \tag2$$</span></p>
<p>Idea was that after determining <span class="math-container">$ (a,b,r)$</span> we could check if there would be such pattern/regularity here as well for three figures of minimum total area (equilateral triangle,square,circle) in parallel line packing.</p>
<p>However it turns out that for an equilateral triangle base <span class="math-container">$a$</span> the ratio of altitudes</p>
<p><span class="math-container">$$ \dfrac{\dfrac{a}{\sqrt3 }}{\dfrac{a \sqrt3 }{2} }= \dfrac32 $$</span></p>
<p><a href="https://i.stack.imgur.com/4hZhN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4hZhN.png" alt="String partitioning Min Area"></a></p>
<p>So no pattern is observed, it being seen that the arrangement with square/circle does not repeat with equilateral triangle and circle, further generalization for regular polygons with higher number of vertices <span class="math-container">$n=5,6,7..\infty \,$</span> &c..do not hold good on basis of total height of triangle. This conclusion has changed in view of the later observations.</p>
<p>On further examination the common top line for square and circle is found to match with the diameter of incircle <span class="math-container">$ 2 r_I=\dfrac{a}{\sqrt3}$</span> as shown and this common feature is further investigated for invariance finding for <em>general regular polygons set</em>... in the next answer area.</p>
| José Carlos Santos | 446,262 | <p>If <span class="math-container">$\lvert x\rvert\leqslant\frac13$</span>, then<span class="math-container">$$\left\lvert\frac{3^n}{2n^2+5}x^n\right\rvert\leqslant\frac1{2n^2+5},$$</span>and the series converges (apply the comparison test to this series and to the series <span class="math-container">$\sum_{n=1}^\infty\frac1{n^2}$</span>).</p>
<p>And if <span class="math-container">$\lvert x\rvert>\frac13$</span>, then<span class="math-container">\begin{align}\lim_{n\to\infty}\frac{\left\lvert\frac{3^{n+1}}{2(n+1)n^2+5}x^{n+1}\right\rvert}{\left\lvert\frac{3^n}{2n^2+5}x^n\right\rvert}&=\lim_{n\to\infty}\frac{2n^2+5}{2(n+1)^2+5}\times\frac3{\lvert x\rvert}\\&=\frac3{\lvert x\rvert}\\&>1,\end{align}</span>and so the series diverges then.</p>
|
3,488,226 | <p>We have <span class="math-container">$$\ln(v-1) - \ln(v+3) = \ln(x) + C$$</span></p>
<p>Multiplying through by e gives:</p>
<p><span class="math-container">$$(v-1)/(v+3) = x + e^C$$</span></p>
<p>But the answer in the textbook is:</p>
<p><span class="math-container">$$(v-1)/(v+3) = Dx$$</span></p>
<p>Where <span class="math-container">$$D = e^C$$</span></p>
<p>Question: why is the answer Dx and not x + D?</p>
<p>Thanks!</p>
| Community | -1 | <p><span class="math-container">$ln(\frac{v-1}{v+3})$</span> <span class="math-container">$=$</span> <span class="math-container">$ln(x)+C$</span> <span class="math-container">$\implies$</span> <span class="math-container">$\frac{v-1}{v+3} = Dx$</span>.</p>
|
2,867,521 | <p>I am interested in calculating the following double summation:</p>
<p>$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$</p>
<p>I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.</p>
<p>The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.</p>
| William Elliot | 426,203 | <p>Let the directed set be (0,1) and the net assign x to x in R.<br>
This net converges to 1 and its image along with 1 is (0,1]<br>
which is not compact.</p>
|
362,250 | <p>Let <span class="math-container">$\nu$</span> be a <em>finite</em> Borel measure on <span class="math-container">$\mathbb{R}^n$</span> and define the shift operator <span class="math-container">$T_a$</span> on <span class="math-container">$L^p_{\nu}(\mathbb{R}^n)$</span> by <span class="math-container">$f\to f(x+a)$</span> for some fixed <span class="math-container">$a\in \mathbb{R}^n-\{0\}$</span>. Suppose moreover that
<span class="math-container">$\nu$</span> is absolutely continuous wrt the Lebesgue measure <span class="math-container">$m$</span> and let
<span class="math-container">$
\frac{d \nu}{dm}(x)= h(x).
$</span></p>
<p>In this case, can we obtain a bound on <span class="math-container">$\|T_{a}\|_{\mathrm{op}}$</span> in terms of <span class="math-container">$h$</span> and of <span class="math-container">$a$</span>? </p>
<p>Usually when <span class="math-container">$\nu$</span> is the Lebesgue measure then this is commonly known to be <span class="math-container">$1$</span>, but here, in the finite and dominated case I can't seem to find such a result...</p>
| Yemon Choi | 763 | <p>This was intended as an extended comment but started to have too many formulas, so I thought that it would be more legible if posted as an answer.
(You don't actually state if there are <span class="math-container">$\nu$</span>-null sets which are not Lebesgue null, so I'm going to build an example which is mutually absolutely continuous wrt Lebesgue measure.)</p>
<p>If <span class="math-container">$h$</span> is unbounded then it seems to me that your shift operator could be unbounded. You don't specify whether you want <span class="math-container">$p$</span> to be in the reflexive range, so let me take <span class="math-container">$p=2$</span> just to be sure, and take <span class="math-container">$n=1$</span>, <span class="math-container">$a=1$</span> for simplicity.
Take <span class="math-container">$h(x)=|x|^{-3/4}$</span> for <span class="math-container">$x\in [-1,1]$</span> and <span class="math-container">$h(x)=e^{-|x|}$</span> outside that interval, so that <span class="math-container">$h \in L^1_m({\bf R})$</span>. Put <span class="math-container">$d\nu = h\ dm$</span>, so that <span class="math-container">$\nu$</span> is a finite measure on <span class="math-container">${\bf R}$</span>.</p>
<p>Consider
<span class="math-container">$$ f(x) = \begin{cases}(x-1)^{-1/3} & \hbox{if $x\in (1,2]$} \\ 0 & \hbox{otherwise} \end{cases}$$</span>
This belongs to <span class="math-container">$L^2_\nu({\bf R})$</span> since
<span class="math-container">$$ \int_{\bf R} |f(x)|^2 h(x) \,dx = \int_1^2 (x-1)^{-2/3} e^{-x}\,dx <\infty $$</span>
On the other hand,
<span class="math-container">$$ (T_1f)(x) =f(x+1) = \begin{cases} x^{-1/3} & \hbox{if $x\in (0,1]$} \\ 0 & \hbox{otherwise} \end{cases}$$</span>
so
<span class="math-container">$$ \int_{\bf R} |T_1f(x)|^2 h(x) \,dx = \int_0^1 x^{-2/3} x^{-3/4}\,dx = +\infty $$</span></p>
|
1,862,807 | <blockquote>
<p>Show that for <span class="math-container">$x,y,z\in\mathbb{Z}$</span>, if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are coprime and <span class="math-container">$z$</span> is nonzero, then <span class="math-container">$\exists n\in\mathbb{Z}$</span> such that <span class="math-container">$z$</span> and <span class="math-container">$y+xn$</span> are coprime.</p>
</blockquote>
<p>Not sure where to start on this one. I understand that coprime indicates that their GCD is 1, but I am somewhat confused how to proceed.</p>
| Bill Dubuque | 242 | <p>This can be solved <em>intuitively</em> by using a slight twist on Euclid's idea for generating new primes. Euclid employed <span class="math-container">$\,1 + p_1\cdots p_n$</span> is coprime to <span class="math-container">$\,c = p_1\cdots p_n.\,$</span> Stieltjes noted the generalization that, furthermore, <span class="math-container">$\ \color{#c00}{p_1\cdots p_k} +\, \color{#0a0}{p_{k+1}\cdots p_n}\,$</span> is coprime to <span class="math-container">$\,c\,$</span> too, which motivates the following </p>
<p><strong>Key Idea</strong> <span class="math-container">$\, $</span> Coprimes to <span class="math-container">$\,c\,$</span> arise by partitioning into <span class="math-container">$\rm\color{#c00}{two}\ \color{#0a0}{summands}$</span> all prime factors of <span class="math-container">$\,c,\,$</span> i.e.</p>
<p><strong>Theorem</strong> <span class="math-container">$\,\ \ \color{#c00}a+\color{#0a0}b\ $</span> is coprime to <span class="math-container">$\ c\:$</span> if every prime factor of <span class="math-container">$\,c\,$</span> divides <span class="math-container">$\,a\,$</span> or <span class="math-container">$\,b,\,$</span> but not both.</p>
<p><strong>Proof</strong> <span class="math-container">$\ $</span> If not then <span class="math-container">$\,a+b\,$</span> and <span class="math-container">$\,c\,$</span> have a common prime factor <span class="math-container">$\,p.\,$</span> By hypothesis <span class="math-container">$\,p\mid a\,$</span> or <span class="math-container">$\,p\mid b.\,$</span> Wlog, say <span class="math-container">$\,p\mid b.\,$</span> Then <span class="math-container">$\,p\mid (a+b)-b = a,\,$</span> so <span class="math-container">$\,p\,$</span> divides both <span class="math-container">$\,a,b,\,$</span> contra hypothesis. <span class="math-container">$ $</span> <strong>QED</strong></p>
<p>We seek <span class="math-container">$\,\color{#c00}{y}+\color{#0a0}{xn}\,$</span> coprime to <span class="math-container">$\,z,\,$</span> so it suffices to choose <span class="math-container">$\,n\,$</span> such that each prime factor <span class="math-container">$\,p\,$</span> of <span class="math-container">$\,z\,$</span> divides exactly one of <span class="math-container">$\,y\,$</span> or <span class="math-container">$\,xn.\,$</span> Note <span class="math-container">$\,p\,$</span> can't divide both <span class="math-container">$\,x,y,\,$</span> since they are coprime. Hence it suffices to let <span class="math-container">$\,n\,$</span> be the product of primes in <span class="math-container">$\,z\,$</span> that do not occur in <span class="math-container">$\,x\,$</span> or in <span class="math-container">$\,y.\ \ $</span> <strong>QED</strong> </p>
<p><strong>Remark</strong> <span class="math-container">$\ $</span> Note how the solution becomes quite obvious after employing Stieltjes' idea, amounting to nothing but a trivial calculation of a difference of sets (of primes)</p>
|
2,550,365 | <p>I encountered a question regarding orthogonal projection. I only have some clues to approach this questions and am not sure whether my thoughts can be used to answer the question. Could anyone check my thoughts and correct it or figure out a better solution, if needed? Thanks in advance.</p>
<p>Suppose $W \subseteq ℝ^6$ is a subspace with basis ${\begin{bmatrix}1\\5\\-1\\6\\787\\0\end{bmatrix},\begin{bmatrix}-34\\4\\4\\4\\4\\8\end{bmatrix},\begin{bmatrix}2\\3\\4\\5\\6\\2\end{bmatrix}, \begin{bmatrix}-5\\123\\-4\\12\\-3\\1\end{bmatrix}}$, and let $P : ℝ^{\color{red}6}\toℝ^{\color{red}6}$ be the orthogonal projection onto $W$. The following do not require very much computation.
<ol>
<li>What is the rank of $P$? How do you know?</li>
<strong>My thought: the rank will be 4. As a vector in $ℝ^6$ will be projected onto a 4-dimensional subspace, 4 linearly independent columns are needed.</strong>
<li></p>
<p>What is the dimension of the $1$-eigenspace of $P$? How do you know?</li>
<strong>My thought: the dimension will be 4, too. As $P$ will transform a vector in $ℝ^6$ into the 4-dimensional subspace, the eigenvector should also be on the subspace first so that some linear combinations of the 4 vectors that form $W$ can transform the input vector into the vector itself. Therefore the vector can be represented by W and the dim of the eigenspace is 4.</strong>
<li></p>
<p>What is the dimension of the null space of $P$? How do you know?</li>
<strong>My thought: the dim of null space will be 2. As the dimension is 6 and the rank is 4, the dim of null space will therefore be 2. However, I got stuck here as I tried to image some form of Ax=b that transform a vector x into another b, what A can be so that the x in $ℝ^6$ can be transformed to a vector b that is the projection of A onto $W$.</strong>
<li></p>
<p>Explain why $P$ must be similar to a diagonal matrix, and find a diagonal matrix it is similar to (note: you are <i>not</i> being asked to find an invertible matrix $Q$ so that $P = QDQ^{-1}$).</p>
<p><strong>I have no clues for this one at all, could anyone give me the solution or hint to this one as well?</strong>
</ol></p>
<p>Thanks a lot! ^_^ </p>
<p>Bump!</p>
| amd | 265,466 | <p>You’ve found an eigenvalue with geometric multiplicity 4 and one with geometric multiplicity 2. The sum of their eigenspaces is six-dimensional, so must be all of $\mathbb R^6$. $P$ is therefore diagonalizable and in fact is similar to $\operatorname{diag}(1,1,1,1,0,0)$.</p>
|
3,074,668 | <p>Good evening,</p>
<p>Could someone please demonstrate why this property is valid?</p>
<blockquote>
<p>Given <span class="math-container">$\sigma\in S_n$</span></p>
<p><span class="math-container">$$\left|\prod_{i<j} \frac{\sigma(j)-\sigma(i)}{j-i}\right|=1$$</span></p>
</blockquote>
| Nick Peterson | 81,839 | <p>This is because <span class="math-container">$\sigma$</span> is a permutation, and therefore a one-to-one correspondence. </p>
<p>You can rewrite the product in terms of numerators and denominators by way of
<span class="math-container">$$
\begin{align*}
\left\lvert\prod_{i<j}\frac{\sigma(j)-\sigma(i)}{j-i}\right\rvert&=\frac{\prod\limits_{i<j}\left\lvert\sigma(j)-\sigma(i)\right\rvert}{\prod\limits_{i<j}(j-i)}.
\end{align*}
$$</span>
Re-index the product in the numerator by letting <span class="math-container">$h=\sigma^{-1}(i)$</span> and <span class="math-container">$k=\sigma^{-1}(j)$</span>. Note that we can't assume <span class="math-container">$h<k$</span>; however, we can still index the product in the numerator over all sets <span class="math-container">$\{h,k\}$</span> of two distinct integers in <span class="math-container">$[1,n]$</span>. </p>
<p>This reindexing yields
<span class="math-container">$$
\prod_{i<j}\lvert\sigma(j)-\sigma(i)\rvert=\prod_{\{h,k\}}\lvert k-h\rvert=\prod_{h<k}(k-h).
$$</span></p>
|
60,810 | <p>I am looking for a proof of the fact that if $f:\mathbb{R}\to \mathbb{R}$ is a group automorphism of $(\mathbb{R},+)$ that also preserves order, then there exists a positive real number $c$ s.t. $f(x)=cx$ for all $x\in \mathbb{R}$. If anyone can point to a reference, that will be great.</p>
| Dylan Moreland | 3,701 | <p>I don't have a reference, but here is an outline of how I would do this. Since we can approximate any real number from above and below by elements of $\mathbf Q$ and $f$ preserves the order, we can reduce to the case of rational $x$. Clearly we want $f(1) = c > 0$ to be our constant. Now, if $b$ is a positive integer then
$$
c = f(1) = f(b/b) = bf(1/b).
$$
It doesn't take much work to get from this to a proof that $f(x) = cx$ for each $x \in \mathbf{Q}$.</p>
|
1,989,950 | <p>I was doing proof of open mapping theorem from the book Walter Rudin real and complex analysis book and struck at one point. Given if $X$ and $Y$ are Banach spaces and $T$ is a bounded linear operator between them which is $\textbf{onto}$. Then to prove $$T(U) \supset \delta V$$ where $U$ is open unit ball in $X$ and $\delta V = \{ y \in Y : \|y\| < \delta\}$.</p>
<p>Proof- For any $y \in Y$ since map is onto, there exist an $x \in X$ such that $Tx = y$. It is also clear that if $\|x\| < k$, then $y \in T(kU)$ for any $k$. Clearly $$Y = \underset{k \in \mathbb{N}}{\cup} T(kU) $$ But as $Y$ is complete, by Baire category theorem it can't be written as countable union of nowhere dense sets. So there exist atleast one $k$ such that $ T(kU)$ is not nowhere dense. Thus this means $$(\overline{T(kU)})^0 \ne \emptyset$$ i.e. $ T(kU)$ closure has non empty interior. Let $W$ be open set contained in closure of $T(kU)$. Now for any $w \in W \implies w \in \overline{T(kU})$, so every point of $W$ is the limit of the sequence $\{Tx_i\}$, where $x_i \in kU$, Let us now fix $W$ and $k$.</p>
<p>Now choose $y_0 \in W$ and choose $\eta > 0$, so that $y_0+y \in W$ if $\|y\| < \eta$. This can be done as $W$ is open set, so every point of it has some neighborhood also there. Now as $y_0 , y_0+y \in W$ from above paragraph there exist sequences $\{x_i'\}$ and $\{x_i''\}$ in $kU$ such that $$T(x_i') \to y_0 \qquad T(x_i'') \to y_0+y \quad as \ i \to \infty$$ Set $x_i = x_i'-x_i''$. Then clearly $$\|x_i\| \leq \|x_i'\| + \|x_i''\| < 2k$$ and $T(x_i) \to y$. Since this holds for every $y$ with $\|y\|< \eta$. </p>
<p>Now it is written that, the linearity of $T$ shows that following is true for $\delta = \dfrac{\eta}{2k}$ </p>
<p>To each $y \in Y$ and to each $\epsilon > 0$ there corresponds an $x \in X$ such that $$\|x\| \leq \delta^{-1}\|y\| \quad \text{and} \quad \|Tx-y\| < \epsilon \quad (1)$$ How does this follows?</p>
<p>This proof is given in Walter rudin 3rd edition on page 112</p>
| Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>Assuming $\cos(\theta)\neq 0$,
$$\sec(\theta) + \tan(\theta) = \frac{3}{2}\implies 1+\sin(\theta)= \frac{3}{2}\cos(\theta) $$ Now, use the tangent half-angle substitution $t=\tan(\frac\theta 2)$ to get $$5 t^2+4 t-1=0\implies (t+1) (5 t-1)=0$$ One of the roots cannot be used (which one and why ?).</p>
<p>From the solution $\theta=2\tan^{-1}(t)$ and all other trigonometric functions.</p>
|
894,764 | <p>I am having problems understanding how to solve this question. </p>
<p>Find a linear function that satisfies both of the given conditions.</p>
<p>$f(-1) = 5, f(1) = 6$</p>
<p>Thanks,
<i>Note: i have the answer, just need help understanding</i></p>
| nchar | 132,757 | <p>A linear function (in one variable) is of the form $f(x) = ax + b$, where $a$ and $b$ are constants. The two conditions $f(-1) = 5$ and $f(1) = 6$ determine the value of $a$ and $b$. </p>
<p>To be explicit, using these two conditions, we have
$$ -a+b = 5,$$
$$ a+b = 6.$$
Now solve the equations above to get $a$ and $b$.</p>
|
376,484 | <p>My questions are motivated by the following exercise:</p>
<blockquote>
<p>Consider the eigenvalue problem
$$
\int_{-\infty}^{+\infty}e^{-|x|-|y|}u(y)dy=\lambda u(x), x\in{\Bbb R}.\tag{*}
$$
Show that the spectrum consists purely of eigenvalues. </p>
</blockquote>
<p>Let $A:L^2({\Bbb R})\to L^2({\Bbb R})$ be a linear operator such that
$$(Au)(x)=\int_{\Bbb R}k(x,y)u(y)dy$$
where $k(x,y)=e^{-|x|-|y|}$. Then $A$ is self-adjoint, since $\overline{k(x,y)}=k(y,x)$. Thus $\sigma(A)\subset{\Bbb R}$.</p>
<blockquote>
<p>My first <strong>question</strong>: is $A$ invertible?</p>
</blockquote>
<p>$A$ is a Hilbert-Schmidt operator since $k\in L^2(\Bbb{R}^2)$ and thus $A$ is compact. Then the answer should be NO since $L^2({\Bbb R})$ is an infinite-dimensional Hilbert space. It follows that $\lambda=0$ must be an eigenvalue of $A$ according to the conclusion in the exercise. But $\operatorname{ker}(A)={0}$ which implies that $\lambda=0$ is not an eigenvalue of $A$. </p>
<blockquote>
<p>My second <strong>question</strong>: what mistake do I make above?</p>
</blockquote>
| Hagen von Eitzen | 39,174 | <p>The function
$$f(n)=\begin{cases}1&\text{if }n\text{ odd}\\0&\text{if }n\text{ even}\end{cases}$$
is multiplicative and we have $\lim_{m\to\infty}f(p^m)=0$ if we let $p=2$. Since $\lim_{n\to\infty}f(n)$ does not exist, I guess that we may assume that <em>for all</em> primes $p$ we have $\lim_{m\to\infty}f(p^m)=0$.
But this is still not good enough:
A multiplicative function is uniqeuly determined by its values at prime powers. Even if we assume that $f$ should be <em>strongly</em> mutiplicative, we can let $f(p^m)=(1-\frac1p)^m$. Then indeed $f(p^m)\to0$ as $m\to\infty$ for any given $p$ (in fact, for any integer $p>1$), but since arbitraryly big primes exist, we have arbitrarily big numbers $n=p$ with $f(n)=(1-\frac1p)>\frac12$.</p>
<p>In summary: I cannot find an interpretation of the problem statement that would make the claim true.</p>
|
254,623 | <p>For example 100 is even and 100/2= 50 is also even</p>
<p>But 30 is also even but 30/2=15 is odd</p>
<p>Now let's say I have a number as large as 10^10000000000...</p>
<p>I want to know how many steps are involved in cutting this number in half. When the number is even, I divide it by 2. When it's odd, I subtract 1. I continue this process until I hit 0.</p>
<p>However, when the number is too high, I can't actually manipulate it directly (elaboration: too big to write out, and too big to fit into memory on a computer), so I am curious if there's a way for me to do this by just knowing the even/odd attributes along the chain.</p>
<p>I hope this makes sense!</p>
<p>Example:
If n=100, I have the following chain</p>
<p>100, 50, 25, 24, 12, 6, 3, 2, 1, 0</p>
<p>Which is a total of 9 "splitting steps" (10 if you count the original number)
And the following parities</p>
<p>Even, Even, Odd, Even, Even, Even, Odd, Even, Odd, Even</p>
<p>I am asking if, given n=10, there is a way to get this parity chain</p>
| N. S. | 9,176 | <p>Yes you can get the chain for $n=10$, but for large numbers it is not easy.</p>
<p>Write your number $n$ in binary. What you are doing is the following:</p>
<ul>
<li>If the last digit is 0, you erase it.</li>
<li>If the last digit is 1, you make it a zero.</li>
</ul>
<p>For $n=10$ in binary you have $n=1010$. Thus your string is</p>
<p>$$1010 \to 101 \to 100 \to 10 \to 1 \to 0$$</p>
<p>In total, the number of digits is exactly the number of digits in binary (which is exactly $\log_2 n$ rounded up) <strong>plus</strong> the number of 1 in the binary representation </p>
|
1,923,149 | <p>Preimage is defined as $X = \{x \in \mathbb{R}^n:Ax \in S\}$, where $A$ is a linear mapping $A \in \mathbb{R}^{m,n}$, and $S \subseteq \mathbb{R}^m$ is a sequentially-closed set.</p>
<p>The definition of sequential-closedness is that $S$ is sequentially-closed iff every convergent sequence in $S$ has its limit in $S$.</p>
<p>What I tried:
Suppose $X$ has a convergent sequence whose limit is not in $X$, then since the mapped sequence is in S, whose limit is not in $S$ (1). This is contradicted to what we assume that $S$ is a closed set, so all convergent sequences in $X$ has their limits in $X$, which proves that $X$ is a closed set.</p>
<p>Now my question is how can I prove (1) is true?</p>
| David K | 139,123 | <p>Interpret $f(z) = \binom zn$ as the number of combinations of $n$ objects that can be selected from a set of $z$ distinct objects
where $n, z \in \mathbb Z$ and $z \geq n \geq 0$.</p>
<p>For $n = 0$, we have $f(z) = 1$, which is a convex function.</p>
<p>For $n \geq 1$, consider a set of $z+1$ distinct objects,
$S_{z+1} = \{a_1, \ldots, a_z, a_{z+1}\}$.
Then $\binom {z+1}n$ is the number of combinations of $n$ objects
selected from $S_{z+1}$.
We can partition these combinations into two sets:</p>
<ul>
<li>The combinations in which $a_{z+1}$ is selected.</li>
<li>The combinations in which $a_{z+1}$ is <em>not</em> selected.</li>
</ul>
<p>The number of combinations in the first partition is the number of ways to select $a_{z+1}$ from $\{a_{z+1}\}$ (namely $1$) times the number of ways to select the remaining $n-1$ elements from the $z$-element set $S_{z+1} \setminus \{a_{z+1}\}$, which is $\binom z{n-1}$.
So there are $\binom z{n-1}$ combinations in this partition.</p>
<p>The number of combinations in the second partition is the number of ways to select $n$ objects from the $z$-element set $S_{z+1} \setminus \{a_{z+1}\}$, so there are $\binom zn$ combinations in this partition.</p>
<p>Adding together the size of the partitions gives the total number of combinations, so we have $$\binom z{n-1} + \binom zn = \binom {z+1}n.$$
(This is a well-known result but I wanted to make sure it was presented
with a combinatorial explanation.)
It follows that
$$
\binom {z+1}n - \binom zn = \binom z{n-1}
$$
and since $z+1 \geq n$ as well,
$$
\binom {z+2}n - \binom {z+1}n = \binom {z+1}{n-1}.
$$
</p>
<p>Since $n - 1 \geq 0$, a similar argument to the one above (including the case $n-1=0$ and the case $n - 1 \geq 1$) shows that
$$
\binom z{n-1} \leq \binom {z+1}{n-1},
$$
that is,
$$
\binom {z+1}n - \binom zn \leq \binom {z+2}n - \binom {z+1}n.
$$</p>
<p>We can extend this via induction to show that if $n \leq z_i < z_m < z_f$
then
$$
(z_f - z_m) \left( \binom{z_m}n - \binom{z_i}n \right)
\leq (z_m - z_i) \left( \binom{z_f}n - \binom{z_m}n \right).
$$</p>
<p>It follows that
$$
f(z_m) = \binom{z_m}n
\leq \frac{z_f - z_m}{z_f - z_i} \binom{z_i}n
+ \frac{z_m - z_i}{z_f - z_i} \binom{z_f}n
= (1-\lambda) f(z_i) + \lambda f(z_f)
$$
where $0 < \lambda = \dfrac{z_m - z_i}{z_f - z_i} < 1$,
which satisfies a reasonable definition of what it means for a
function over integers to be convex over the integers greater than or
equal to $n$.</p>
|
1,285,177 | <p>I know this is very simple and I'm missing something trivial here...</p>
<p>I'm having trouble converting this set of equations to polar form:</p>
<p>$$
\dot{x_1}=x_2-x_1 (x_1^2+x_2^2-1)\\
\dot{x_2}=-x_1-x_2 (x_1^2+x_2^2-1)
$$</p>
<p>where</p>
<p>$$
r= (x_1^2+x_2^2)^{1/2}\\
\theta=\arctan\left(\frac{x_2}{x_1}\right)
$$</p>
<p>The book I'm going through has these converted to the following equations:</p>
<p>$$
\dot{r}=-r(r^2-1)\\
\dot{\theta}=-1
$$</p>
<p>Here are the steps I've taken...</p>
<p>$$
\frac{dr}{dt}=(x_1\dot{x_1}+x_2\dot{x_2})(x_1^2+x_2^2)^{-1/2}\\
\dot{x_1}+\dot{x_2}=x_2-x_1-(x_1+x_2)(x_1^2+x_2^2-1)\\
\dot{x_1}+\dot{x_2}=x_2-x_1-(x_1+x_2)(r^2-1)
$$</p>
<p>Now I'm not sure what the next step to take would be... I've tried a few things and none of them got me to the correct result. Any help would be appreciated! :)</p>
| David Holden | 79,543 | <p>$$
A^\mathrm{T}=-A
$$
implies that $A$ has zeroes on the main diagonal, since they change sign on the RHS, but remain unchanged on the LHS. if you incorporate that into your representation, it should be easy to answer the question about dimension and basis.</p>
<blockquote>
as a related exercise you might like to consider the $2 \times 2$ matrices with complex entries. these are isomorphic to $\Bbb{R}^8$ as a real vector space, and to $\Bbb C^4$ as a complex vector space.
the <i>skew-hermitian</i> matrices satisfy:
$$
A^{\mathrm{T}}+A^* =0
$$
where $A^*$ is the complex conjugate of $A$. the skew-hermitian matrices are isomorphic to a $4$-dimensional subspace of $\Bbb R^8$. however in $\Bbb C^4$ they are only a subgroup, but do not form a subspace - the diagonal elements must be purely imaginary and this property is not preserved under multiplication by an arbitrary complex scalar.
</blockquote>
|
1,176,435 | <p>Consider $G(4,p)$ - the random graph on 4 vertices. What is the probability that vertex 1 and 2 lie in the same connected component?</p>
<p>So far, I have considered the event where 1 and 2 do not lie in the same component. Then vertex 1 must lie in a component of order 1, 2 or 3 that doesn't contain vertex 2. However, I am unsure about how to compute these probabilities. For 1 to be in a component if order 1, I think this has probability $(1-p)^3$. </p>
| Siméon | 51,594 | <p>Let $E_{ij} = E_{ji}$ denote the event that the edge $\{i,j\}$ is added in the graph. The event <em>"1 is connected to 2"</em> is the disjoint union of the five following events:</p>
<ul>
<li>$E_{12}$, with probability $p$</li>
<li>$\overline{E_{12}} \wedge E_{13} \wedge E_{32}$, with probability $ p^2(1-p)$</li>
<li>$\overline{E_{12}} \wedge E_{13} \wedge \overline{E_{32}} \wedge E_{34} \wedge E_{42}$, with probability $p^3(1-p)^2$</li>
<li>$\overline{E_{12}} \wedge \overline{E_{13}} \wedge E_{14} \wedge E_{42}$, with probability $p^2(1-p)^2$</li>
<li>$\overline{E_{12}} \wedge \overline{E_{13}} \wedge E_{14} \wedge \overline{E_{42}} \wedge E_{43} \wedge E_{32}$, with probability $p^3(1-p)^3$</li>
</ul>
<p>Therefore, the probability that vertex 1 and 2 lie in the same connected component is simply
$$
p + p^2(1-p) + p^3(1-p)^2 + p^2(1-p)^2 + p^3(1-p)^3
$$</p>
|
2,540,007 | <p>I have the following question. Find the matrix representation of the transormation $T:\mathbb{R}^3\to\mathbb{R}^3$ that rotates any vector by $\theta=\frac{\pi}{6}$ along the vector $v=(1,1,1)$.</p>
<p>A hint is given to find the rotation matrix about the $z-axis$ by $\frac{\pi}{6}$which is $$
\begin{bmatrix}
\frac{\sqrt{3}}{2} &\frac{-1}{2} & 0 \\
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
0 & 0 &1
\end{bmatrix}
$$
and then find an orthogonal basis for $\mathbb{R}^3$ that has $v$ as one of its vector then finally rewrite the above matrix using the new basis and that should be my answer.</p>
<p>My questions are </p>
<ol>
<li>Do I pick two random vectors and join them to $v$ and use the Gram-Schmidt process to get an orthogonal basis.</li>
<li>Why does that work?</li>
</ol>
| Lee Mosher | 26,501 | <p>The Gram-Schmidt process is a rather tedious way to do such a simple geometric task.</p>
<p>First, the plane $P$ orthogoal to $v = (1,1,1)$ is $x+y+z=0$.</p>
<p>Second, pick any vector $P$ in that plane, say $w = (1,-1,0)$.</p>
<p>Third, let $u$ be the cross product of $v,w$:
$$u = v \times w = (1,1,1) \times (1,-1,0) = (1,1,-2)
$$
It follows that $\{v,w,u\}$ is an orthogonal basis.</p>
<p>Finally, a note of caution: before proceeding to rewrite the rotation matrix using this new basis, what you really want is an ortho<strong>normal</strong> basis, hence you should first normalize the basis by replacing $v$ with $\frac{v}{|v|}$ and similarly for $w$ and $u$.</p>
|
362,895 | <p>I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.</p>
<p>Proof: Notice that $$\langle 1 - i \rangle\\
\Rightarrow 1 = i\\
\Rightarrow 2 = 0.$$
Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus
$$
\mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}.
$$</p>
<p>This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$</p>
| davidlowryduda | 9,754 | <p>Your answer is great, but I'd like to give a different view as well.</p>
<p>A standard first or second example of a Euclidean Domain is the Gaussian integers $\mathbb{Z}[i]$, so that in particular the Gaussian integers form a principal ideal domain. We also know that in PIDs, nonzero prime ideals are maximal. So if we were to show that $1 - i$ is a Gaussian prime, then $\langle 1 - i \rangle$ would be a prime ideal, and thus a maximal ideal. Thus, quotienting by it would give a field.</p>
<p>So how do we show that $1 - i$ is prime? Well, compute its norm (from the Euclidean Domain norm, where $|x + iy| = x^2 + y^2$. Its norm is $2$. Norms are multiplicative, so if $1-i = ab$, then $2 = |a||b|$. But its norm is also an integer, and $2$ is a prime (in the reals). Thus $1-i$ is a prime.</p>
<p>And so we have it.</p>
|
2,551,683 | <blockquote>
<p>A jet has a $5\%$ chance of crashing on any given test flight. Once it crashes the program will be halted. Find the probability that the program lasts less than three flights.</p>
</blockquote>
<p>The correct answer to this question is $0.1426$, but I can't figure out how to get it.</p>
<p>Here's my attempt at the question:
$p=0.05$</p>
<p>$q=0.95$</p>
<p>$x=0,1,2$</p>
<p>$$(0.05)(0.95)^{-1} + (0.05)(0.95)^0 + (0.05)(0.95)^1 = 0.1501$$ </p>
<p>I used the geometric probability method by using the formula $pq^{x-1}$ where $p$ is the probability of successes, $q$ is the probability of failures, and $x$ is the waiting time/how many events occurred before the success occurred.</p>
| Dean | 393,411 | <p>The probability that the jet survives after three tests: $(0.95)^3$. The probability for the other cases is $1-(0.95)^3=0.1426$.</p>
|
3,632,097 | <p>Given a sheaf <span class="math-container">$F$</span> on a topological space <span class="math-container">$X$</span> and <span class="math-container">$U$</span> is an open subset of <span class="math-container">$X$</span>. Denote <span class="math-container">$F|_U$</span> be the restricted sheaf of <span class="math-container">$F$</span>. Then to any <span class="math-container">$y\in U$</span>. Do we have <span class="math-container">$F_y=(F|_U)_y$</span>?</p>
| Bueggi | 683,835 | <p>Yes we have.
To sections define the same germ if they agree on an open neighborhood of a point p. So they will also agree on a smaller neighborhood.</p>
|
2,252,317 | <p>I have a rather challenging question on my assignment and I have put in my best effort for now. I think I just need a tiny nudge to set me in the right direction to finish this proof. If you could have a look, that would be great!</p>
<hr>
<p><strong>Background on Cosets and Operations Defined on $V/W$</strong></p>
<p>Let $W$ be a subspace of a vector space $V$ over $\mathbb{F}$. For some fixed $\mathbf{v} \in V$, a coset of $W$ is defined to be the set</p>
<p>$$
\{\mathbf{v}\} + W = \{ \mathbf{v} + \mathbf{w} \,\,|\,\, \mathbf{w} \in W \}.
$$
We usually denote this as $\mathbf{v} + W$, though.</p>
<p>The set of all cosets of $W$ is denoted by
$$
V/W=\{ \mathbf{v} + W \,\,|\,\, \mathbb{v} \in V \}.
$$</p>
<p>Addition and scalar multiplication are defined on $V/W$ by
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1} + \mathbf{v_2}) + W, \quad \text{and} \\
k(\mathbf{v} + W) = k\mathbf{v} + W, \quad k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>The Question</strong></p>
<p>I am required to prove that these operations are well-defined, i.e. if $\mathbf{v_1} + W = \mathbf{v_1'} + W$ and $\mathbf{v_2} + W = \mathbf{v_2'} + W$, then
$$
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W), \quad \text{and}\\
k(\mathbf{v_1} + W) = k(\mathbf{v_1'} + W), \quad \forall k \in \mathbb{F}.
$$</p>
<hr>
<p><strong>My Problem</strong></p>
<p>I am having great difficulty proving the first part. So far I have this.
$$
\begin{align*}
\mathbf{v_1} + W = \mathbf{v_1'} + W &\implies \mathbf{v_1} - \mathbf{v_1'} = \mathbf{w_1} \in W \implies \mathbf{v_1} = \mathbf{v_1'} + \mathbf{w_1}, \quad \text{and}\\
\mathbf{v_2} + W = \mathbf{v_2'} + W &\implies \mathbf{v_2} - \mathbf{v_2'} = \mathbf{w_2} \in W \implies \mathbf{v_2} = \mathbf{v_2'} + \mathbf{w_2}.
\end{align*}
$$</p>
<p>Then,
$$
\begin{align*}
(\mathbf{v_1} + W) + (\mathbf{v_2} + W) &= (\mathbf{v_1} + \mathbf{v_2}) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{w_1}) + (\mathbf{v_2'} + \mathbf{w_2}) \right), \quad \mathbf{w_1}, \mathbf{w_2} \in W\\
&\quad \,\, \text{Let } \mathbf{w_1} + \mathbf{w_1} = \mathbf{w_3} \in W.\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + \mathbf{w_3} \right) + W\\
&=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + W \right) + (\mathbf{w_3} + W)\\
&= (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W) + (\mathbf{w_3} + W), \quad \mathbf{w_3} \in W.
\end{align*}
$$</p>
<p>As you can see, this is frustratingly close to the result I wanted to prove. The only thing in the way is the extra $+ (\mathbf{w_3} + W)$. Is there some way I could make that disappear though? Or is my whole proof just wrong?</p>
<p>The only thing I know about the coset $\mathbf{w} + W$ where $\mathbf{w} \in W$ is that it is a subspace of the vector space $V$. Could I somehow use this fact?</p>
| Sarvesh Ravichandran Iyer | 316,409 | <blockquote>
<p>$w_3 + W = W$!</p>
</blockquote>
<p>Proof : $x \in w_3 + W \implies x = w_3 + w$ for some $w \in W $, which means $x \in W$ since $W$ is a subspace.</p>
<p>On the other hand, $x \in W \implies x = w_3 + (x - w_3) \in w_3 + W$, since $W$ is a subspace, so $x-W_3 \in W$.</p>
<p>Hence, continuing from your line of thought:
$$
(v_1'+W) + (v_2' + W) +(w_3'+W) = (v_1'+W) + (v_2'+W) + W \\ = (v_1' + W) + (v_2' + W + W) = (v_1' + W) + (v_2' + W)
$$</p>
<p>Which completes the proof.</p>
<p>Use a similar logic to prove that $kW = W$ for any non-zero scalar $k$. This will be helpful for your second part.</p>
|
902,522 | <p>How would I simplify a fraction that has a radical in it? For example:</p>
<p>$$\frac{\sqrt{2a^7b^2}}{{\sqrt{32b^3}}}$$</p>
| FWE | 170,600 | <p>See for example <a href="http://rads.stackoverflow.com/amzn/click/0521356539" rel="nofollow"><em>Lambek and Scott: Introduction to higher order categorical logic</em></a>, ch 0.1 (unfortunately not in the net but for sure in ur university library). There first is defined a <strong>graph</strong>, then a <strong>deductive system</strong> as a graph with</p>
<ul>
<li>For each object $A$ an <strong>identity</strong> arrow $1_A:A\rightarrow A$</li>
<li>For each pair of arrows $f:A\rightarrow B$ and $g:B\rightarrow C$ the
<strong>composition</strong> $gf:A\rightarrow C$ </li>
</ul>
<p>The idea of a logician might be that of objects as <strong>formulas</strong> and arrows as <strong>deductions</strong>, the composition in this context becoming a <strong>rule of inference</strong>
$$\frac{f:A\rightarrow B\space\space\space g:B\rightarrow C}{gf:A\rightarrow C}$$</p>
<p>However this is just an interpretation and letting this aside one can take that abstract definition of a deductive system to come out with the usual category:</p>
<p>A <strong>category</strong> is a deductive system with the ususal equations for <strong>identity</strong> and <strong>associativity</strong> - i.e. for $f:A\rightarrow B$, $g:B\rightarrow C$ and $h:C\rightarrow D$
$$f1_a=f=1_Bf,\space\space (hg)f=h(gf)$$</p>
|
813,395 | <p>I have read that linear independence occurs when:</p>
<p>$$\sum_{i=1}^n a_i v_i =0$$</p>
<p>Has only $a_i=0$ as a solution, but what if all $v_i$ were $0$ then $a_i$ could vary and still yield $0$. Does that mean that such a vector set is not linearly independent?</p>
<p>What if I have:</p>
<p>Let $\{c_0,c_1,c_2,\dots,c_n\}$ denote a set of $n+1$ distinct elements in $\mathbb{R}$. Define the set of $n+1$ polynomials.</p>
<p>$$f_j(x)=\prod_{k=0,k\ne j}^n \frac{x-c_k}{c_j - c_k} $$</p>
<p>Note that $f_j(x) \in P_n(\mathbb{R})$ with the property
$$f_j(c_l) = \left\{ \begin{align} 0&& \text{if}&& j\ne l\\ 1&& \text{if}&& j= l \end{align} \right.$$</p>
<p>And $\alpha = \{f_0(x),f_1(x),\dots,f_n(x)\}$, then this is or isn't linearly independent based on my $x$ value. Is there something here that forces $x$ to equal one of my $c_j$? For I am told that this $\alpha$ is linearly independent.</p>
| DeepSea | 101,504 | <p>By definition of linearly independent vectors the space $X$ itself is supposed to consist a non-zero vector $v$, thus not all of the $v_i$'s are zero vectors. </p>
|
3,115,090 | <p>This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function. </p>
<p><span class="math-container">$$\lim_{n\to\infty}{\frac{1}{n} {\sum_{k=3}^{n}{\frac{3}{k^2-k-2}}}}$$</span></p>
<p>Well, even the fact that <span class="math-container">$\frac{3}{k^2-k-2} = \frac{1}{k-1}-\frac{1}{k+2}$</span> doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not. </p>
<p>Is that a Riemann sum at all? </p>
| Romeo | 28,746 | <p>Isn't it <span class="math-container">$k^2 -k -2 = (k+1)(k-2)$</span>? In that way,
<span class="math-container">$$
\frac{3}{k^2-k-2} = \frac{1}{k-2} - \frac{1}{k+1}
$$</span>
and this is likely to be telescopic. </p>
|
1,988,731 | <p>Can one of you math this for me?</p>
<p>You've got a deck of 88 cards. Let's call the first card "1", the next "2", and so on.</p>
<p>What are the odds of pulling "1", "2", and "3" out of the deck, in that order?</p>
| Chas Brown | 167,790 | <p>If you mean "what are the odds that the first card drawn is $1$, the second card drawn is $2$, and the third card drawn is $3$?": There are $88!$ different possible shuffles of the deck of $88$ cards. There are "only" $(88-3)!$ shuffles that have the first three cards being $1,2,3$ in that order. So the probability is $\frac{85!}{88!} = \frac{1}{88\cdot 87\cdot 86}$.</p>
<p>If you mean "what are the odds that $1$ is drawn before either $2$ or $3$, and also $2$ is drawn before $3$?", then that is $\frac{1}{6}$.</p>
|
951,332 | <p>the integer 220, 251 304 represent three consecutive perfect squares in base b. Determine the value of b.</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Use the identity, $$(a+2)^2+a^2-2(a+1)^2=2$$</p>
<p>Here $(a+2)^2=(304)_b=3b^2+4$ and so on</p>
|
3,290,839 | <p>I want to prove that that given <span class="math-container">$f:R^2 \rightarrow R$</span> which is continuous with compact support s.t the integral of <span class="math-container">$f$</span> for every straight line <span class="math-container">$l$</span> is zero (<span class="math-container">$\int f(l(t))\mathrm{d}t=0$</span>) then <span class="math-container">$f$</span> is almost everywhere <span class="math-container">$0.$</span></p>
<p>Well I know how to proof it in case that <span class="math-container">$l = 1_{B(x,r)}$</span> is measurable and bounded with compact support (from other thread) , but that's not the case here.
Any idea? thanks!</p>
| Daniele Tampieri | 317,063 | <p>The problem you pose is the weak version of a classical result in harmonic analysis, which does not require the continuity nor the compactness of the support of the datum <span class="math-container">$f$</span>: in order to prove this stronger result, first note that
<span class="math-container">$$
\begin{align}
\int\limits_{\Bbb R} f\big(l(t)\big)\mathrm{d}t=0\:\text{ for every}&\text{ straight line } l\subsetneq\Bbb R^2\label{1}\tag{1}\\
\Updownarrow&\\
\int\limits_{\sigma(l_0)} f(l)\,\mathrm{d}l=0\:\text{ for every rigid }&\text{motion }\sigma\in\mathsf{Aut}(\Bbb R^2)\label{2}\tag{2}
\end{align}
$$</span>
where by <em>rigid motion</em> we mean a bijection of <span class="math-container">$\Bbb R^2$</span> in itself which is the composition of a rotation and a translation and <span class="math-container">$l_0\subsetneq\Bbb R^2$</span> is a given straight line: this is obvious, since substituting a dilatation of a given straight line <span class="math-container">$l$</span> in \eqref{1} does not change the value of the integral, thus these linear transformations do not contribute to it and it is not necessary to consider them.
Now we can state and prove the following:</p>
<p><strong>Theorem (Cramér & Wald, Newman, Besicovitch)</strong>. If <span class="math-container">$f\in L^1(\Bbb R^2)$</span> is such that, for an arbitrarily given straight line <span class="math-container">$l_0\subsetneq\Bbb R^2$</span>, equation \eqref{2} holds, then <span class="math-container">$f\equiv 0$</span> a.e..<br>
<strong>Proof</strong>. Let <span class="math-container">$\xi=(\xi_1,\xi_2)\neq 0$</span> so that <span class="math-container">$\Vert \xi\Vert\neq0$</span> and define <span class="math-container">$(x,y) \mapsto \sigma_{\Vert \xi\Vert}(x,y)=(u,v)$</span> as
<span class="math-container">$$
\begin{pmatrix}
u\\
v
\end{pmatrix}=
\begin{pmatrix}
\frac{\xi_1}{\Vert \xi\Vert}& \frac{\xi_2}{\Vert \xi\Vert}\\
-\frac{\xi_2}{\Vert \xi\Vert}& \frac{\xi_1}{\Vert \xi\Vert}
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}\quad\forall (x,y)\in\Bbb R^2
$$</span>
The Jacobian of <span class="math-container">$\sigma_{\Vert \xi\Vert}$</span> is equal to one therefore, by applying the Fourier transform to <span class="math-container">$f$</span>, we get
<span class="math-container">$$
\begin{split}
\hat{f}(\xi)&=\iint\limits_{\Bbb R^2} e^{-i\langle\xi,(x,y)\rangle}f(x,y)\,\mathrm{d}x\mathrm{d}y\\
&=\int\limits_{\Bbb R} e^{-i\Vert\xi\Vert u} \left[\,\int\limits_{\Bbb R}f\big(\sigma_{\Vert \xi\Vert}^{-1}(u,v)\big)\,\mathrm{d}v\right] \mathrm{d}u=0
\end{split}
$$</span>
since the integral inside the square brackets in the right side of the above equality is equal to \eqref{2} for some <span class="math-container">$\sigma=\sigma_{\Vert \xi\Vert}(u,\cdot)$</span>. By the arbitrariness of <span class="math-container">$\xi\in\Bbb R^2\setminus\{0\}$</span>, <span class="math-container">$f\equiv 0$</span> a.e.. <span class="math-container">$\blacksquare$</span></p>
<p><strong>Final notes</strong></p>
<ul>
<li>When I saw the question and the comment by Christian Blatter, I though this was a classical problem for the Radon transform. However, I saw the comment of mathworker21, I remembered a counterexample and the fact that this is in reality a result related to the Pompeiu problem: I found the above simple and beautiful theorem in reference [1] (Theorem 1, p. 26) while looking for that example. </li>
<li>Finally, <em>Chakalov's example</em> ([1], p. 27) <em>shows that disks have not the Pompeiu property</em>, i.e. there exists (even very smooth) functions <span class="math-container">$f$</span> such that, for many (even infinitely many) <span class="math-container">$r>0$</span>,
<span class="math-container">$$
\int\limits_{B(x_0,r)} f(y)\,\mathrm{d}y=0\quad \forall x_0\in\Bbb R^2,\; \text{ does not imply }f\equiv 0\, \text{a.e.}
$$</span>
where <span class="math-container">$x_0=(x_{0,1},x_{0,2})$</span> and <span class="math-container">$y=(y_1,y_2)$</span> are points in <span class="math-container">$\Bbb R^2$</span>.<br>
<strong>Edit</strong>. Following the comment of mathworker21, I checked Chakalov's example as reported in [1], and I found that integral formula (2) in the paper is flawed by typos. The correct one is given by Garofalo and Segala in [2], p. 137, and for the completeness I report it here. If we choose <span class="math-container">$f(x,y)=\sin(ax)$</span>, <span class="math-container">$a>0$</span> then we have
<span class="math-container">$$
\int\limits_{B(x_0,r)} f(x,y)\,\mathrm{d}x\mathrm{d}y=\frac{2\pi r}{a}\sin(ax_{0,1})J_1(ar)
$$</span>
where <span class="math-container">$J_1$</span> is the first kind Bessel function of order <span class="math-container">$1$</span>. Choosing <span class="math-container">$a$</span> such that <span class="math-container">$ar$</span> is a zero of <span class="math-container">$J_1$</span> makes the above integral zero for any <span class="math-container">$x_0\in\Bbb R^2$</span>. Note that the example works also in dimension <span class="math-container">$n>2$</span>, for the higher dimensional analogue of the Pompeiu problem.</li>
</ul>
<p><strong>References</strong></p>
<p>[1] Nicola Garofalo (1989), "<a href="https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwihwsi6_7njAhUKsaQKHQQJBHQQFjAAegQIBRAC&url=http%3A%2F%2Fwww.seminariomatematico.unito.it%2Frendiconti%2Fcartaceo%2F47-3%2F25.pdf&usg=AOvVaw1UhOoa6ZLtacWktd3UnkTw" rel="nofollow noreferrer">A new result on the Pompeiu problem</a>",
Rendiconti del Seminario Matematico, Torino, Fascicolo Speciale "PDE and Geometry", vol. 46, 25-38 (1989), <a href="http://www.ams.org/mathscinet-getitem?mr=MR1086204" rel="nofollow noreferrer">MR1086204</a>, <a href="https://zbmath.org/?q=an%3A0737.35145" rel="nofollow noreferrer">Zbl 0737.35145</a></p>
<p>[2] Nicola Garofalo and Fausto Segala (1994), "<a href="https://doi.org/10.1090/S0002-9947-1994-1250819-4" rel="nofollow noreferrer">Univalent functions and the Pompeiu problem</a>", Transactions of the American Mathematical Society, 346, 137-146, <a href="https://www.ams.org/mathscinet-getitem?mr=1250819" rel="nofollow noreferrer">MR1250819</a>, <a href="https://zbmath.org/?q=an%3A0823.30027" rel="nofollow noreferrer">Zbl 0823.30027</a>.</p>
|
2,956,744 | <p><a href="https://i.stack.imgur.com/lhDyB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lhDyB.png" alt="https://i.imgur.com/vvQDuVa.png" /></a></p>
<p>I'll provide a quickly-drawn representation of what the problem is.
Basically, there is a line <span class="math-container">$$l: y=-x+b$$</span> and there are 2 known points on it: <span class="math-container">$$A = (-6,8)$$</span> and <span class="math-container">$$B = (-2,4)$$</span> The line in question (let's name it k, it's the red one) passes through point A. Additionally, the distance between l and point C, located on k, as shown on the image, is 2. Is it possible to get the equation of k with just this amount of information?</p>
| Eleven-Eleven | 61,030 | <p>Generally, the alternate definition of the derivative is </p>
<p><span class="math-container">$$f'(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}$$</span></p>
<p>Note that this definition requires a limit. Thus we must have that </p>
<p><span class="math-container">$$\lim_{x\rightarrow c^{\color{red}{-}}}\frac{f(x)-f(c)}{x-c}=\lim_{x\rightarrow c^{\color{red}{+}}}\frac{f(x)-f(c)}{x-c}$$</span></p>
<p>Take left and right limits using the definition here. What happens?</p>
|
2,561,125 | <p>Hey having trouble finishing this question.</p>
<p>Prove by induction that $n^3 \le 2^n$ for all natural numbers $n\ge 10$.</p>
<p>This is what I have so far:</p>
<p>Base step: For $n = 10$ </p>
<p>$1000 \le 1024$</p>
<p>Assumption Step: For $n = k$</p>
<p>Assume $k^3 \le 2^k$</p>
<p>Induction step: For $n = (k+1)$</p>
<p>$(k+1)^3 \le 10^{k+1}$</p>
<p>$k^3 +3k^2 + 3k +1 \le 10^k*10$</p>
<p>Not really sure where to go from here</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>We have to show that $(k+1)^3<2^{k+1}$. Note that:</p>
<p>$$(k+1)^3=k^3\cdot\frac{(k+1)^3}{k^3}=k^3\left(\frac{k+1}k\right)^3\;.$$</p>
<p>Since $k\ge 10$, $\frac{k+1}k=1+\frac1k\le 1+\frac1{10}=\frac{11}{10}$, and therefore </p>
<p>$$\left(\frac{k+1}k\right)^3\le \,\, ??$$</p>
|
4,132,998 | <p>I figure it has something to do with minimal polynomials and the Jordan canonical form, I just can't piece it together. I would really appreciate a hint on how to start it.</p>
| lhf | 589 | <p>Your hunch is correct: The minimal polynomial must be divide <span class="math-container">$t^m$</span> and have degree at most <span class="math-container">$n$</span>.</p>
|
274,249 | <p>I'd like to write the current data, but the Monitor and PutAppend are being written in such a way that the function <strong>pot</strong> does not match the current variables <strong>(a and b)</strong>. How can I fix this?</p>
<pre><code>f[a_, b_] = Exp[-Sqrt[b*x + Cos[c]]] + a*y;
PP[a_, b_] := NIntegrate[f[a, b], {c, 0, 2*Pi}, {x, 0, 10}, {y, 0, 10}];
P = Monitor[Table[{{a, b}, pot = PP[a, b]}, {a, 1, 5, 1}, {b, 1, 5, 1}], {{" a =", a, " b =", b, " p =", pot},
PutAppend[{" a =", a, " b =", b, " p =", pot}, "C:/Users/Public/P1.txt"]}]
<span class="math-container">```</span>
</code></pre>
| Ulrich Neumann | 53,677 | <p>Start with</p>
<pre><code>NonlinearModelFit[data, a Exp[b (x - 1990)], {a, b }, x]
</code></pre>
|
4,263,784 | <p>I'm trying to find the multiplicity of <span class="math-container">$z=0$</span> on <span class="math-container">$f(z)=z\cos(z)-\sin(z)$</span> using complex analysis.</p>
<p>I'm new to complex analysis and the argument principle/Rouché's theorem so I'm not quite sure where to start. I can prove how many zero's this function has but I'm not quite sure what theorem would help me determine the multiplicity of this root, can someone point me in the right direction or provide a proof?</p>
<p>Thank you</p>
| kelalaka | 338,051 | <p>One Time Pad (OTP) requires uniform random key bits to be information-theoretically secure. Once the key is transmitted securely with a trusted carrier, than the messages transmitted between the two sides are secure as long as the keystream is never used again ( keywords: two-time pad, many-time pad, and crib-dragging).</p>
<p>In the correct usage of OTP, the attacker cannot learn anything about the message other than the message length of the transmitted message.</p>
<p>Once an eavesdropper obverserved an <span class="math-container">$\ell$</span> length OTP message, then all possible English text of size <span class="math-container">$\ell$</span> are possible candidates. The question is how can you distinguish?. For example, you expected that your enemy will attack tomorrow but don't know the time. You received a message of size 14 characters (Well, omit the encoding). Is it <span class="math-container">$\texttt{ATTACK AT DOWN}$</span> or <span class="math-container">$\texttt{ATTACK AT NOON}$</span>? How you can decide which one? Why do we limit these two messages, because it is educational. What about <span class="math-container">$\texttt{WAIT FOR ORDER}$</span>?. Usually the (military) messages are not this short long, so there will be numerous possibilities for each message. Even for two possible messages you fail to distinguish, how you can distinguish if there is more than one possibility?</p>
<p>You may some candidates of possible messages, however, this is your prejudice about the messages. If you look at the <a href="https://en.wikipedia.org/wiki/One-time_pad#Exploits" rel="nofollow noreferrer">historical exploits</a>, you will see that either weak random, or two time-pad for the exploits!</p>
<p>Now, why do you limit the message to meaningfull English? Maybe they are just coordinates? Maybe some other binary data!</p>
<p>To make it more interesting, let they sent a single bit! <code>1</code> for <em>attack</em> <code>0</code> for <em>wait</em> in every minute. Can you distinguish?</p>
<p>No matter what computainally power you have, you cannot distinguish the two messages!</p>
|
3,322,492 | <p><strong>Prove:</strong> <span class="math-container">$A \cap (B - C) = (A \cap B) − (A \cap C)$</span></p>
<p>I can understand this using Venn Diagrams, however I am struggling to translate this into a formal proof. </p>
| Community | -1 | <p>The most common strategy for proving that sets <span class="math-container">$S=T$</span> is to show that <span class="math-container">$S\subset T$</span> and <span class="math-container">$T\subset S$</span>. And the most common strategy for proving <span class="math-container">$S\subset T$</span> is to prove that <span class="math-container">$(x\in S) \implies (x\in T)$</span> for any element <span class="math-container">$x$</span> in the universe of the problem. So let's do that!</p>
<p>First, let <span class="math-container">$x\in A\cap (B-C)$</span> be given. This means that <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B-C$</span>, meaning that <span class="math-container">$x\in B$</span> and <span class="math-container">$x\notin C$</span>. In other words, <span class="math-container">$x\in A\cap B$</span> but <span class="math-container">$x\notin A\cap C$</span>, so <span class="math-container">$x\in (A\cap B)-(A\cap C)$</span>. </p>
<p>Conversely, let <span class="math-container">$x\in(A\cap B)-(A\cap C)$</span> be given. Then <span class="math-container">$x\in A\cap B$</span> and <span class="math-container">$x\notin A\cap C$</span>. Since <span class="math-container">$x\in A\cap B$</span>, <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B$</span>. But <span class="math-container">$x\notin A\cap C$</span>, so it must be that <span class="math-container">$x\notin C$</span>. Since <span class="math-container">$x\in B$</span> and <span class="math-container">$x\notin C$</span>, <span class="math-container">$x\in B-C$</span>. Therefore, <span class="math-container">$x\in A\cap (B-C)$</span>.</p>
|
3,322,492 | <p><strong>Prove:</strong> <span class="math-container">$A \cap (B - C) = (A \cap B) − (A \cap C)$</span></p>
<p>I can understand this using Venn Diagrams, however I am struggling to translate this into a formal proof. </p>
| Cornman | 439,383 | <p>To show that two sets are equal you normaly show the two relations <span class="math-container">$\subseteq$</span> and <span class="math-container">$\supseteq$</span>.</p>
<p>This is how it is done, where I write it down explanatory and not how I would do so in a mathematical proof.</p>
<p>So we first want to show that <span class="math-container">$A\cap (B-C)\subseteq (A\cap B)-(A\cap C)$</span></p>
<p>So let <span class="math-container">$x\in A\cap (B-C)$</span>.
By definition of <span class="math-container">$\cap$</span> this means that <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B-C$</span>.
By definition of <span class="math-container">$-$</span> we have that <span class="math-container">$B-C$</span> implies <span class="math-container">$x\in B$</span> and <span class="math-container">$x\notin C$</span>.</p>
<p>Now we have <span class="math-container">$x\in A$</span> and <span class="math-container">$x\in B$</span>. So <span class="math-container">$x\in A\cap B$</span>. (By definitino of <span class="math-container">$\cap$</span>)
Since <span class="math-container">$x\in A$</span> and <span class="math-container">$x\notin C$</span>, it is <span class="math-container">$x\notin A\cap C$</span>. (By definition of <span class="math-container">$\cap$</span>)</p>
<p>So <span class="math-container">$x\in (A\cap B)-(A\cap C)$</span> (By defintion of <span class="math-container">$-$</span>)</p>
<p>Now show <span class="math-container">$A\cap (B-C)\supseteq (A\cap B)-(A\cap C)$</span>.</p>
<p>Start like this: Let <span class="math-container">$x\in (A\cap B)-(A\cap C)$</span>. Then ...</p>
<p>The proof is done (exactly) like above.</p>
|
3,322,492 | <p><strong>Prove:</strong> <span class="math-container">$A \cap (B - C) = (A \cap B) − (A \cap C)$</span></p>
<p>I can understand this using Venn Diagrams, however I am struggling to translate this into a formal proof. </p>
| user0102 | 322,814 | <p>Here it is an approach based on the definition of difference and the De Morgan Laws:</p>
<p><span class="math-container">\begin{align*}
(A\cap B) - (A\cap C) & = (A\cap B)\cap(\overline{A\cap C}) = (A\cap B)\cap(\overline{A}\cup\overline{C})\\
& = (A\cap B\cap\overline{A})\cup(A\cap B\cap\overline{C})\\
& = A\cap B\cap\overline{C} = A\cap(B - C)
\end{align*}</span></p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Kevin O'Bryant | 935 | <p>Elementary means that I know how to reduce it to "counting on fingers", college algebra, and linear algebra. That doesn't mean that other areas are to be avoided at all, though. </p>
<p>This definition works differently for different people, as that "I know how" expression is vague. In my book, probability is elementary but the Stone-Cech compactification $\beta N$ of the naturals is not.</p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| user717 | 717 | <p>I refine my answer: I think that problems in elementary number theory can be characterized as problems in number theory for which both the problem and its solution can be understood in a fair amount of time by someone with undergraduate mathematical knowledge. The condition that also the solution is elementary and does not involve advanced techniques should not be dropped because there are a lot of elementary problems (Fermat's last theorem) which are very easy to state but are not that easy to solve (that's a slight understatement of course).</p>
<p>And I think that it is wrong to exclude the complex numbers. For example the basic material on cyclotomic fields is what I would definitely call elementary number theory. Even quadratic reciprocity involves complex numbers (okay, perhaps one of the 10^18 proofs does not, I don't know).</p>
<p>I want to add that "elementary" should not be translated as "easy"!</p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Thomas Riepe | 451 | <p>I would regard as 'elementary number theory' that what needs no previous mathematical knowledge, esp. no abstract algebra, no Galois theory and no (complex) analysis. 'Elementary number theory' could include what Euler and Gauss did and of course "elementary" means not "simple". I would regard the reducability of statements like the four colour theorem to diophantine ones as belonging to elementary number theory too. BTW, if I remember correctly, Euler wrote an algebra textbook containing the Fermat for exp.=3 case with the help of an uneducated pupil to guarantee it's elementary nature. </p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Ilya Nikokoshev | 65 | <p>To me, "elementary" = <strong>in principle can be understood by a person who only knows high school math</strong>. </p>
<p>Notes:</p>
<ul>
<li>The person can be assumed to be extremely smart or a genius.</li>
<li>What to mean by <em>high school math</em> can vary from person to person, but derivatives are certainly in and algebraic curves are certainly out. </li>
</ul>
<p>Actually, this definition essentially coincides with the one in Wikipedia:</p>
<blockquote>
<p>Several important discoveries of this field are Fermat's little theorem, Euler's theorem, the Chinese remainder theorem and the law of quadratic reciprocity. The properties of multiplicative functions such as the Möbius function and Euler's φ function, integer sequences, factorials, and Fibonacci numbers all also fall into this area.</p>
</blockquote>
<p>The examples they give are exactly things people sometimes teach at schools — that happens only rarely, but the point is that none of these formally uses any "college material".</p>
|
4,463 | <p>It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs." </p>
<p>I have two closely related questions.</p>
<ol>
<li>Is my understanding of the usage of "elementary" correct?</li>
<li>It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid? </li>
</ol>
<p>NB: I'm asking about how "elementary" usually <strong>is</strong> defined and why, not how it <strong>should be</strong> defined.</p>
| Konstantinos Gaitanas | 38,851 | <p>In a book i saw a nice answer:<br>
''Elementary Number Theory'' is Number Theory which is based on mathematics you can find in Euclid's ''Elements''</p>
|
194,421 | <p>This is homework. The problem was also stated this way: </p>
<p>Let A be a dense subset of $\mathbb{R}$ and let x$\in\mathbb{R}$. Prove that there exists a decreasing sequence $(a_k)$ in A that converges to x.</p>
<p>I know:</p>
<p>A dense in $\mathbb{R}$ $\Rightarrow$ every point in $\mathbb{R}$ is either in A or a limit point of A.</p>
<p>If x is a limit point of A, then there is a sequence in A that converges to x. </p>
<p>What if $x\in A$?</p>
<p>Also, how can I know if the sequence is increasing or decreasing?</p>
| Shankara Pailoor | 39,210 | <p>Hint: Recursively define $a_n \in A$ such that $a_n \in (x, a_{n-1})$. </p>
|
3,516,754 | <p>I got stuck while doing exercise of the Apostol's Calculus, the exercise 28 of Section 5.5.</p>
<p>Here's the question</p>
<hr>
<p>Given a function <span class="math-container">$f$</span> such that the integral <span class="math-container">$A(x) = \int_a^xf(t)dt$</span> exists for each <span class="math-container">$x$</span> in an interval <span class="math-container">$[a, b]$</span>. Let <span class="math-container">$c$</span> be a point in the open interval <span class="math-container">$(a, b)$</span>. Consider the following ten statements about this <span class="math-container">$f$</span> and this A:</p>
<hr>
<p>And there are five (a) ~ (e) statements on the left, and five (<span class="math-container">$\alpha$</span>) ~ (<span class="math-container">$\epsilon$</span>) statements on the right. The author asks the reader to decide the implicative relation from statements on the left to statements on the right. I thought I answered correctly but the solution at the end tells different. I don't know why this is wrong.</p>
<p>(d) <span class="math-container">$f'(c)$</span> exists. <span class="math-container">$\implies$</span> (<span class="math-container">$\epsilon$</span>) <span class="math-container">$A'$</span> is continuous at c.</p>
<p>This is my argument:
By the Example 7 of Section 4.4, the differentiability of <span class="math-container">$f$</span> at c implies the continuity of <span class="math-container">$f$</span> at c. Since <span class="math-container">$f$</span> is differentiable at c, <span class="math-container">$f$</span> is continuous at c, so that <span class="math-container">$A'$</span>, which equals to <span class="math-container">$f$</span>, should continuous at c. </p>
<p>But the solution at the end says (d) does not implies (<span class="math-container">$\epsilon$</span>).</p>
<p>Sorry for the partializing the problem, it maybe tough to point out what is wrong. </p>
| symplectomorphic | 23,611 | <p>Paramanand gives an explicit counterexample, which proves that your <em>claim</em> is false. But I want to add this further answer to expose the flaw in your <em>reasoning</em>.</p>
<p>Your argument appears to rely on the following missing assumption: if <span class="math-container">$g$</span> and <span class="math-container">$h$</span> are two functions defined on a set containing <span class="math-container">$c$</span>, <span class="math-container">$g(c)=h(c)$</span>, and <span class="math-container">$g$</span> is continuous at <span class="math-container">$c$</span>, then <span class="math-container">$h$</span> is continuous at <span class="math-container">$c$</span>. (If this were true, your conclusion would follow. In your case, <span class="math-container">$g=f$</span> and <span class="math-container">$h=F’$</span>.)</p>
<p>This assumption is false. Here’s a very simple counterexample. Take <span class="math-container">$g(x)=0$</span> on the real line and let <span class="math-container">$h(x)=1$</span> for all <span class="math-container">$x$</span> except zero, where <span class="math-container">$h$</span> jumps to zero. Then <span class="math-container">$g(0)=0=h(0)$</span> and <span class="math-container">$g$</span> is continuous at zero, but <span class="math-container">$h$</span> is not continuous at zero.</p>
|
189,266 | <p><strong>Q1:</strong> If a Morse function on a smooth closed $n$-manifold $X$ has critical points of only index $0$ and $n$, does it follow that $X\approx \mathbb{S}^n\coprod\ldots\coprod\mathbb{S}^n$?</p>
<p>I think the following question is essential in regard to the one above:</p>
<p><strong>Q2:</strong> If $f$ is a Morse function on a closed connected smooth $n$-manifold $X$ that has critical points of only index $0$ and $n$ and $f(X)\!=\![a,b]$, can a critical point of index $0$ or $n$ be mapped into $(a,b)$?</p>
| André Nicolas | 6,312 | <p>There are many choices, particularly if you do not ask that the "distance" function satisfy the triangle inequality.</p>
<p>For example, if the vectors are $(x_1,x_2,\dots,x_n)$ and $(y_1,y_2,\dots,y_n)$ we can use $\sum (x_i-y_i)^2$ (the square of the Euclidean distance). This has some very nice properties that have made it the general favourite ever since the time of Gauss. </p>
<p>Or else you can use the "taxicab" distance $\sum|x_i-y_i|$.</p>
<p>You might wish also to experiment with expressions like $\sum |x_i-y_i|^p$ for various $p$, to see wheether a $p$ other than $2$ gives behaviour that you prefer. </p>
|
2,846,114 | <p>$$\int\frac{2}{x(3x-8)}dx=P\cdot \ln\left|x\right|+Q\cdot \ln\left|3x-8\right|$$</p>
<p>Find out what P and Q are equal to.</p>
<p>This is what I worked out:</p>
<p>$$\frac{A}{x}+\frac{B}{3x-8}=\frac{2}{x(3x-8)}$$
$$-\frac{1}{4}=A,\ \ \ \frac{3}{4}=B$$
$$P=A, Q=B$$</p>
<p>why is the answer $P=-\frac{1}{4}, Q=\frac{1}{4}$?</p>
| Key Flex | 568,718 | <p>You have the integral $\int\dfrac{2}{x(3x-8)}=2\int\dfrac{1}{x(3x-8)}$</p>
<p>Now if you solve $\dfrac{1}{x(3x-8)}$ using partial fractions then you will get
$$\dfrac{1}{x(3x-8)}=\frac{3}{8(3x-8)}-\frac{1}{8x}$$
Now,
$$2\int\dfrac{1}{x(3x-8)}=2\int\frac{3}{8(3x-8)}-\frac{1}{8x}$$
$$=2\left(\frac18\ln|3x-8|-\frac18\ln|x|\right)=\frac14\ln|3x-8|-\frac14\ln|x|$$
Therefore, $P=-\dfrac14$ and $Q=\dfrac14$</p>
<p><strong>Edit:</strong>
$$2\int\frac{3}{8(3x-8)}dx=\frac34\int\frac{1}{3x-8}dx$$
Apply $u=3x-8$ and we get $$=\frac34\int\frac{1}{3u}=\frac14\int\frac1udu=\frac14\ln|u=\frac14\ln|3x-8|$$</p>
<p>And</p>
<p>$$2\int\frac{1}{8x}=\frac{2}{8}\ln|x|=\frac14\ln|x|$$</p>
|
261,031 | <p>i hope some of you can support to solve my problem, i need to work on data in the following way, where the length of each of the lists or sublists is equal. As an example i want to share the data-pattern with you:</p>
<pre><code>list1={a,b,c};
list2={{d,e,f},{g,h,i},......} (in reality the number of sublists in list2 is about 30)
data=list2[[1]]
</code></pre>
<p>the goal is now to combine these lists in the following form
<code>{ {{a,d},{b,e},{c,f}}, {{a,g},...} ...}</code> .</p>
<hr />
<p>The next step to plot it or to create a fit-formula. The values of list1 should be always plotted as the x-Data.
I created some of the most interesting datasets as follows: `</p>
<pre><code>plottedList=Table[{list1[[k]], data[[k]]}, {k, 1, Length[data]}]
ListPlot[plottedList]
</code></pre>
<p>My Problem is that i now would need to plot all of the data-pairs and combine the data to create the lists which i can find a linear or nonlinear fit.</p>
<hr />
<p>I hope you can give me some advice,
Best Chris!</p>
| Syed | 81,355 | <p>A minor variation based on @kglr's answer.</p>
<pre><code>list1 = {a, b, c};
list2 = {{d, e, f}, {g, h, i}};
MapThread[List, {list1, #}] & /@ list2
</code></pre>
<hr />
<p>Using Transpose:</p>
<pre><code>Transpose[{list1, #}] & /@ list2
</code></pre>
<hr />
<p>Result:</p>
<blockquote>
<p>{{{a, d}, {b, e}, {c, f}}, {{a, g}, {b, h}, {c, i}}}</p>
</blockquote>
|
4,422,824 | <p><strong>Edit: This question involves derivatives, please read my prior work!</strong></p>
<p>This question has me stumped.</p>
<blockquote>
<p>A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.</p>
</blockquote>
<p>First, from the instructions, I believe...</p>
<p><span class="math-container">$f'(x)=(5280/60)-ax=88-ax$</span></p>
<p><span class="math-container">$f''(x)=-a$</span></p>
<p>I also believe <span class="math-container">$f(x)={\int}f'(x)dx=88x-a{\int}x=88x-a\frac{x^2}{2}$</span>, because <code>a</code> is known to be constant.</p>
<p>Where I'm lost is what comes next. I can compute <code>a</code> and <code>x</code> in terms of each other at <code>f(x)=450</code>, but this doesn't seem to get me closer to the answer. Neither does the fact that <span class="math-container">$f^{-1}(450)=x$</span>. What am I missing here? Thank you!</p>
| VERNON HEZRON | 1,002,019 | <p>Well, I believe the solution is rather simple. We have however been given not a fixed distance but an upper constraint on the distance to be travelled before it stops. Therefore, I will derive a simple equation of motion and use it to compute the dece;eration required. First of all, we are farmiliar with the fact that <span class="math-container">$a = \frac{v - u}{t}$</span> and <span class="math-container">$s = \frac{v + u}{2}$</span>. From the first relation, <span class="math-container">$t = \frac{v - u}{a}.$</span> Substituting this into the second relation yields <span class="math-container">$s = \frac{v + u}{2}\frac{v - u}{a} = \frac{v^{2} - u^{2}}{2a}$</span>. Therefore, <span class="math-container">$v^{2} = u^{2} + 2as$</span>. Substituting the given values into our derived relation with <span class="math-container">$v = 0$</span>, <span class="math-container">$u^{2} = -2as$</span>
<span class="math-container">$3600 = -2(450)\cdot a$</span>. <span class="math-container">$a = -4$</span>. Hence, the deceleration for it to travel the <span class="math-container">$450$</span>m would be <span class="math-container">$4ms^{-2}$</span>. However, to stop at less than <span class="math-container">$450$</span>m, it would have to decelerate at a higher rate. Hence the solution is <span class="math-container">$a > 4ms^{-2}$</span>.</p>
|
180,296 | <p>I need an algorithm to decide quickly in the worst case if a 20 digit integer is prime or composite.</p>
<p>I do not need the factors.</p>
<p>Is the fastest way still a prime factorization algorithm? Or is there a faster way given the above relaxation?</p>
<p>In any case which algorithm gives the best worst case performance for a 20 digit prime?</p>
<p><strong>Update:</strong></p>
<p>Here is the simple method I started with:</p>
<pre><code> int64 x = 981168724994134051LL; // prime
int64 sq = int64(ceil(sqrt(x)));
for(int64 j = 2; j <= sq; j++)
{
if (x % j == 0)
cout << "fail" << endl;
}
</code></pre>
<p>It takes 9 seconds on my 3.8Ghz i7 3930K. I need to get it down by a factor of about 1000. Going to try a low end "primorial" sieve and see what that does.</p>
<p><strong>Update 2:</strong></p>
<p>I created a prime sieve using $2.3.5.7.11.13.17 = 510510 = c$ entries. And then searched for factors in blocks of 510510, disregarding factors that are divisible by one of the 7 mentioned primes by a lookup table. It actually made running time worst (11 seconds), I suspect because the memory access time is not worth it compared to the density of numbers cooprime to $(2,3,5,..,17)$</p>
| M Turgeon | 19,379 | <p>Write down what it means for an integer to have an inverse modulo $n$, and recall <a href="http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" rel="nofollow">Bézout's identity</a>.</p>
|
157,074 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/155685/prove-binomp-1k-equiv-1k-pmod-p">Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$</a> </p>
</blockquote>
<p>The question is as follows:</p>
<blockquote>
<p>Let $p$ be prime. Show that ${p \choose k}\bmod{p}=0$, for $0 \lt k \lt p,\space k\in\mathbb{N}$. What does this imply about the binomial co-efficients ${p-1 \choose k}$?</p>
</blockquote>
<p>By the definition of binomial coefficients:</p>
<p>$${p \choose k}=\frac{p!}{k!(p-k)!}$$</p>
<p>Now if $0 \lt k \lt p$, then we have $p\mid{p\choose k}$, therefore ${p \choose k}\equiv0\pmod{p}, \space 0 \lt k \lt p. \space \blacksquare$</p>
<p>Note that we can write: ${p \choose k}={p-1 \choose k}+{p-1 \choose k-1}$, and therefore:</p>
<p>$${p-1 \choose k}={p \choose k}-{p-1 \choose k-1}=\frac{p!}{k!(p-k)!}-\frac{(p-1)!}{(k-1)!(p-k)!}=\frac{(p-1)!}{(k-1)!(p-k)!}\left(\frac{p}{k}-1\right)$$</p>
<p>However, I am unsure how to proceed with this question, the book I am working from states that:</p>
<p>$${p-1 \choose k}\equiv(-1)^{k}\pmod{p}, \space 0 \le k \lt p$$</p>
<p>But I am unsure how the authors have derived this congruence, so I'd appreciate any hints. </p>
<p>Thanks in advance.</p>
| André Nicolas | 6,312 | <p>It turns out that we do not even need Wilson's Theorem. Note the identity
$$\binom{p-1}{k+1}(k+1)=\binom{p-1}{k}(p-k-1).$$
This is easily obtained from the fact that $\binom{n}{m}=\frac{n!}{m!(n-m)!}$.
Now note that $p-k-1\equiv -(k+1)\pmod p$. Thus
$$\binom{p-1}{k+1}(k+1)\equiv -\binom{p-1}{k}(k+1)\pmod p.$$
If $0\le k \lt p-1$, then $k+1$ is not divisible by $p$, so we can cancel and obtain
$$\binom{p-1}{k+1}\equiv -\binom{p-1}{k}\pmod p.\tag{$1$}$$
Thus $\binom{p-1}{k}$ changes sign modulo $p$ every time that we increment $k$.
But $\binom{p-1}{0}=1$, and the result follows.</p>
|
2,579,572 | <blockquote>
<p>In an election, $10\%$ voters did not participate and $1200$ votes are found invalid. The winner gets $68\%$ of total voting list and he won by $56400$ votes. Find the votes polled in favor of losing candidate.</p>
</blockquote>
<p>I can't understand what should I do with the number of invalid votes.</p>
<hr>
<p>Edit:</p>
<p><a href="https://i.stack.imgur.com/EtIOr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EtIOr.jpg" alt="enter image description here"></a></p>
<p>Can anybody tell why 1200 has been subtracted from 56400 in</p>
<p>$$46units=56400-1200$$</p>
<p>Shouldn't $$46 units =56400$$</p>
<p>Because 1200 votes are not counted in voting, how can it be a part of the votes by which the candidate won?</p>
| N. F. Taussig | 173,070 | <p>Let $x$ be the number of voters on the list. The total number of votes is equal to the number of votes received by the winner plus the number of votes received by the loser plus the number of voters who did participate in the election plus the number of voters whose votes were invalidated. </p>
<p>The winner receives $0.68x$.</p>
<p>Since the loser receives $56~400$ fewer votes, the loser receives $0.68x - 56~400$. </p>
<p>The number of voters who did participate in the election is $0.1x$. </p>
<p>There were $1200$ invalid votes. </p>
<p>Hence,
\begin{align*}
0.68x + 0.68x - 56~400 + 0.1x + 1200 & = x\\
1.46x - 55~200 & = x\\
0.46x & = 55~200\\
46x & = 5~520~000\\
x & = 120~000
\end{align*}
Therefore, the loser receives
$$0.68(120~000) - 56~400 = 81~600 - 56~400 = 25~200$$
votes.</p>
|
1,127,596 | <p>I am trying to show that the value of $\int^\infty_0$$\int^\infty_0$ sin($x^2$+$y^2$) dxdy is $\frac{\pi}{4}$ using Fresnel integrals. I'm having trouble splitting apart the integrand in order to actually be able to use the Fresnel integrals. Any help is appreciated. </p>
<p>Answer:
$\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) = $\int^\infty_0$ $\int^\infty_0$ sin($x^2$)cos($y^2$)+cos($x^2$)sin($y^2$)dxdy</p>
<p>= $\int^\infty_0$ $\frac{\sqrt2\pi}{4}$ cos$(y^2)$+$\frac{\sqrt2\pi}{4}$ sin$(y^2)$ dy (the $\frac{\sqrt2\pi}{4}$ comes from established Fresnel integrals values)</p>
<p>= do the same thing for dy and then you get $\frac{\pi}{4}$ as desired.</p>
<p>Now I'm working on doing this with polar coords.</p>
| user 1591719 | 32,016 | <p>Does it work like that? $$\int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2}) \,dx\ dy=\Im\left\{\int_{0}^{\infty}\int_{0}^{\infty} e^{i(x^2+y^2)} \ dx \ dy\right\}=\Im\left\{\int_{0}^{\infty}e^{i x^2} \ dx\int_{0}^{\infty} e^{i y^2} \ dy\right\}=\frac{\pi}{4}$$</p>
|
3,015,596 | <p>A short introduction: The independence number <span class="math-container">$\alpha(G)$</span> of a graph <span class="math-container">$G$</span> is the cardinality of the largest independent vertex set. Independent vertex set is made only of vertices with no edges between them. </p>
<p><a href="https://i.stack.imgur.com/VtuKm.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/VtuKm.jpg" alt="enter image description here"></a></p>
<p>For example, the set shown above has <span class="math-container">$\alpha(G)=3$</span> because in set <span class="math-container">$\{A,C,E\}$</span> all vertices are independent and it's impossible to construct a set with four independent elements.</p>
<p>I have stumbled upon a theorem of Caro and Wei who independently provided the following estimate for the lower bound:</p>
<p><span class="math-container">$$\alpha(G)\ge\sum_{i=1}^{n}\frac1{d_i+1}\tag{1}$$</span></p>
<p>My naive attempts to prove the statement fell short so I started to look for a proof elsewhere. Despite the fact that this famous result of Caro and Wei was quoted in many places (Google returned thousands of pages on the first try), it was not easy to find a full proof. This one turned out be my personal favorite (copied from <a href="http://www.cs.utexas.edu/~panni/lec5.pdf" rel="noreferrer">here</a>):</p>
<p><a href="https://i.stack.imgur.com/rC4Sx.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rC4Sx.jpg" alt="enter image description here"></a></p>
<p>All other proofs that I have found so far are more or less just a copy of it, most often with fewer words. Another copy can be found on <a href="https://math.stackexchange.com/questions/1407478/caro-wei-theorem-proof">MSE</a> (please note that this post is <em>not</em> a copy of the same question).</p>
<p>The proof starts with some combinatorics and then jumps to probabilities and expectations. It's perfectly readable and I have enjoyed it immensely but...</p>
<p>Imagine that you have to explain the theorem to someone (like my son) who is very good at combinatorics but has zero experience with probabilities and expectations.</p>
<p>My question is: Is there a proof of (1) that is <strong>not</strong> based on a probabilistic argument? Can we replace the probabilistic part of the proof with something that is "more basic", if you understand what I mean. My attemts to find such proof were unsuccessful. </p>
| Yair Caro | 1,067,865 | <p>Induction on the order of <span class="math-container">$G$</span>. It's true for <span class="math-container">$|G| = 1$</span>. Now assume it's true for <span class="math-container">$|G| \leq n$</span> and let's prove it for <span class="math-container">$|G| = n+1$</span>.</p>
<p>Choose a vertex <span class="math-container">$v$</span> of minimum degree. Consider <span class="math-container">$H = G - N[v]$</span>.
Clearly
<span class="math-container">$$
\begin{align*}
\alpha(G) & \geq 1 + \alpha(H) \\
& \geq 1 + \sum_{u \in V(H)} \frac{1}{\mathrm{deg}_H(u)+1} \\
& \geq \sum_{w \in N[v]} \frac{1}{\mathrm{deg}_G(w)+1} + \sum_{u \in V(H)} \frac{1}{\mathrm{deg}_H(u)+1} \tag{*} \\
& = \sum_{w \in N[v]} \frac{1}{\mathrm{deg}_G(w)+1} + \sum_{u \in V(G) \setminus N[v]} \frac{1}{\mathrm{deg}_H(u)+1} \\
& \geq \sum_{w \in V(G)} \frac{1}{\mathrm{deg}_G(w)+1}
\end{align*}
$$</span>
where <span class="math-container">$\mathrm{deg}_G(w)$</span> is the degree of <span class="math-container">$w$</span> in <span class="math-container">$G$</span> and <span class="math-container">$\mathrm{deg}_H(u)$</span> is the degree of <span class="math-container">$u$</span> in <span class="math-container">$H$</span>.</p>
<p>To show <span class="math-container">$(^*)$</span>, observe
<span class="math-container">$$
1 = \sum_{w \in N[v]} \frac{1}{\mathrm{deg}(v)+1} \geq \sum_{w \in N[v]} \frac{1}{\mathrm{deg}_G(w)+1}
$$</span>
since <span class="math-container">$v$</span> has minimum degree.</p>
|
3,206,138 | <p>I have the matrix <span class="math-container">$A=\begin{pmatrix}
1 & -1\\
1 &1
\end{pmatrix}$</span> and <span class="math-container">$A^{n}=\begin{pmatrix}
x_{n} & -y_{n}\\
y_{n} &x_{n}
\end{pmatrix}$</span></p>
<p>At first exercise I found that <span class="math-container">$2A-A^{2}=2I_{2}$</span>.At the second exercise I found that <span class="math-container">$A^{48}=2^{24}I_{2}$</span>.</p>
<p>I got stuck at the last exercise where I need find the value of <span class="math-container">$$\frac{x_{10}^{2}+y_{10}^2}{x_{8}^{2}+y_{8}^2}=?$$</span></p>
<p>I tried to multiplicate these 2 matrix <span class="math-container">$\begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}\cdot \begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}$</span> but I don't get something useful.How to approach this exercise?</p>
| E.H.E | 187,799 | <p><span class="math-container">$$x=\frac{i}{1+i}=\frac{i(1-i)}{(1+i)(1-i)}=\frac{i+1}{2}$$</span></p>
|
3,206,138 | <p>I have the matrix <span class="math-container">$A=\begin{pmatrix}
1 & -1\\
1 &1
\end{pmatrix}$</span> and <span class="math-container">$A^{n}=\begin{pmatrix}
x_{n} & -y_{n}\\
y_{n} &x_{n}
\end{pmatrix}$</span></p>
<p>At first exercise I found that <span class="math-container">$2A-A^{2}=2I_{2}$</span>.At the second exercise I found that <span class="math-container">$A^{48}=2^{24}I_{2}$</span>.</p>
<p>I got stuck at the last exercise where I need find the value of <span class="math-container">$$\frac{x_{10}^{2}+y_{10}^2}{x_{8}^{2}+y_{8}^2}=?$$</span></p>
<p>I tried to multiplicate these 2 matrix <span class="math-container">$\begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}\cdot \begin{pmatrix}
x_{10} & -y_{10}\\
y_{10} &x_{10}
\end{pmatrix}$</span> but I don't get something useful.How to approach this exercise?</p>
| DanielWainfleet | 254,665 | <p><span class="math-container">$[1].$</span> Suppose <span class="math-container">$A\implies B\implies C \implies x\in \{0,1\}.$</span> You cannot conclude, from this, that <span class="math-container">$x\in \{0,1\}\implies A.$</span> This does not mean that it isn't useful to use one-way implications. It's very useful but it's not the whole story. You have <span class="math-container">$A\implies x\in \{0,1\}$</span> and then you may find </p>
<p>(i). <span class="math-container">$x=0\implies \neg A$</span> and <span class="math-container">$x=1\implies \neg A,$</span> and conclude that <span class="math-container">$A$</span> is false for all <span class="math-container">$x,$</span> OR</p>
<p>(ii). <span class="math-container">$x=0 \implies A$</span> and <span class="math-container">$x=1\implies \neg A,$</span> and conclude that <span class="math-container">$A\iff x=0,$</span> OR</p>
<p>(iii). <span class="math-container">$x=0\implies \neg A$</span> and <span class="math-container">$x=1\implies A,$</span> and conclude that <span class="math-container">$A\iff x=1,$</span> OR</p>
<p>(iv) <span class="math-container">$x=0\implies A$</span> and <span class="math-container">$x=1\implies A,$</span> and conclude that <span class="math-container">$A\iff (x=0 \lor x=1).$</span></p>
<p><span class="math-container">$[2].$</span> Let <span class="math-container">$a,b \in \Bbb R.$</span> Then <span class="math-container">$a+ib \ne 0$</span> iff <span class="math-container">$a,b$</span> are not both <span class="math-container">$0.$</span> And if <span class="math-container">$a+ib\ne 0 $</span> then <span class="math-container">$a+ib$</span> has a multiplicative inverse <span class="math-container">$(a/r-ib/r)=(a-ib)/r$</span> where <span class="math-container">$r=a^2+b^2.$</span> Because <span class="math-container">$$(a+ib)(a-ib)/r=(a^2-i^2b^2)/r=(a^2+b^2)/r=1.$$</span> So if <span class="math-container">$x,y,z \in \Bbb C$</span> and <span class="math-container">$z\ne 0$</span> then <span class="math-container">$x=y\iff xz=yz.$</span> (Just as when <span class="math-container">$x,y,z\in \Bbb R$</span> and <span class="math-container">$z\ne 0$</span>). </p>
<p>If <span class="math-container">$a=b=1$</span> then <span class="math-container">$r= a^2+b^2=2$</span> so <span class="math-container">$1/(a+ib) =(a-ib)/r=(1-i)/2.$</span></p>
<p>Therefore <span class="math-container">$$x+ix= i\iff x(1+i)=i\iff$$</span> <span class="math-container">$$\iff x\cdot (1+i)[1/(1+i)]=i\cdot[1/(1+i)] \iff$$</span> <span class="math-container">$$\iff x\cdot 1=i\cdot [(1-i)/2]\iff$$</span> <span class="math-container">$$\iff x=[i-i^2]/2=[i-(-1)]/2=(i+1)/2. $$</span></p>
<p>Notice the implications are two-way so there is nothing more to do.</p>
<p>In brief, solving a linear equation in one variable <span class="math-container">$(x)$</span> in <span class="math-container">$\Bbb C$</span> is done just the way you do it in <span class="math-container">$\Bbb R.$</span></p>
|
3,197,683 | <p>Here is the theorem that I need to prove</p>
<blockquote>
<p>For <span class="math-container">$K = \mathbb{Q}[\sqrt{D}]$</span> we have</p>
<p><span class="math-container">$$\begin{align}O_K = \begin{cases}
\mathbb{Z}[\sqrt{D}] & D \equiv 2, 3 \mod 4\\
\mathbb{Z}\left[\frac{1 + \sqrt{D}}{2}\right] & D \equiv 1 \mod 4
\end{cases}
\end{align}$$</span></p>
</blockquote>
<p>The theorem we need to use is this one that can be found in any generic number theory textbook.</p>
<blockquote>
<p>an element <span class="math-container">$\alpha\in K$</span> is an algebraic integer if and only if its minimal polynomial has coefficients in <span class="math-container">$\mathbb{Z}$</span>.</p>
</blockquote>
<p>I tried many avenues of attack but it is extremely hard to prove. How do I prove it?</p>
| J. W. Tanner | 615,567 | <p>Let <span class="math-container">$\alpha=p+q\sqrt D$</span> with <span class="math-container">$p,q\in\mathbb Q$</span>. The other root of its minimal polynomial is <span class="math-container">$\bar \alpha=p-q\sqrt D.$</span> </p>
<p>The minimal polynomial is <span class="math-container">$(x-\alpha)(x-\bar\alpha)=x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha=x^2-2px+(p^2-q^2D). $</span></p>
<p>For <span class="math-container">$-2p$</span> to be an integer, say <span class="math-container">$-m$</span>, <span class="math-container">$p=m/2$</span>, where <span class="math-container">$m \in \mathbb Z$</span>. </p>
<p>For <span class="math-container">$p^2-q^2D$</span> to be an integer, say <span class="math-container">$n$</span>, this means also <span class="math-container">$q=l/2$</span> for some <span class="math-container">$l \in \mathbb Z$</span>.</p>
<p>So <span class="math-container">$p^2-q^2D=(\frac m 2)^2-(\frac l 2)^2D$</span> is an integer so <span class="math-container">$m^2-l^2D$</span> is a multiple of <span class="math-container">$4$</span>. </p>
<p>Modulo <span class="math-container">$4$</span>, <span class="math-container">$m^2$</span> and <span class="math-container">$l^2$</span> could be <span class="math-container">$\equiv0$</span> or <span class="math-container">$1$</span>.</p>
<p>Can you take it from here?</p>
|
1,375,958 | <p>I am looking for a bounded funtion $f$ on $\mathbb{R}_+$ satisfying $f(0)=0$, $f'(0)=0$ and with bounded first and second derivatives. My intitial idea has been to consider trigonometric functions or compositions of them, but I still haven't found an adequate one. Any ideas would be greatly appreciated.</p>
| David C. Ullrich | 248,223 | <p>f=0.</p>
<p>Now it turns out a reply must be at least thirty characters. Hmm. Humdee hum...</p>
|
1,888,187 | <p>I am working on algebraic functions and I am stuck on this problem:</p>
<p>$f(x) = a * r^x$<br>
$(2,1),(3,1.5)$<br></p>
<p>This would be a simple problem if it weren't for that $1.5$ -</p>
<p>So, I have plugged in $2$ and $1$ into the function and this is what I got:
<a href="https://i.stack.imgur.com/MVbxv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MVbxv.png" alt="My Problem Solving"></a></p>
<p>[I also solved for $3,1.5$]<br>
So, this is where I get stuck every time. </p>
<p>I have tried solving it multiple ways but it always ends up to the same effect, where $r$ is always $=$ to something funny that is unsolvable - at least for me.</p>
<p>Could someone explain how you got your answer so that I can understand the process? I know how to do this type of problem, but it is just <em>this</em> one that I can't get!!!!</p>
<p>Thanks for your help and time!</p>
<p><b><em>EDIT</em></b>: I need this simplified, that is why I cant leave $r^2 = 1.5$</p>
| Salech Alhasov | 25,654 | <p>So if we have the function f(x)=ar^x, then by substituting the first point given, namely $(2,1)$, we'll get:</p>
<p>$$1=ar^2$$</p>
<p>Similarly, substituting $(3, \frac{3}{2})$, we end up with</p>
<p>$$\frac{3}{2}=ar^3$$</p>
<p>Now, to find $a\neq 0$ and $r\neq 0 $, you should divide this two equations, what you get then?</p>
|
1,914,686 | <p>A fleet of nine taxis is dispatched to three airports, in such a way that three go to airport A, five go to airport B and one goes to airport C.</p>
<p>If exactly three taxis are in need of repair. What is the probability that every airport receives one of the taxis requiring repairs.</p>
<p>My method was total number of ways is (9!)/(3!*5!*1!)= 504 (this is correct)</p>
<p>Now you have three groups and of every group one position is already filled. So: ....R, ..R, R
So the numbers of ways to distribute the rest is: (6!)/(2!*4!)=15</p>
<p>But according to answer model of my professor, it should be 90. So I assume he says the number of ways of distrubiting these 3 broken cabs over the 3 spots is 3*2*1.
But to me this seems wrong?
The broken taxis don't differ so R(1)R(2)R(3) is same as R(2)R(1)R(3).
If you standing at the airport A, it doesn't matter which of the broken cabs stands there?</p>
| Henry | 6,460 | <p>A probability is usually between $0$ and $1$. </p>
<p>Try a simpler question: two airports, namely D with two taxis and E with two. Suppose the taxis are $T_1$, $T_2$, $T_3$, $T_4$ and the even numbers are broken.</p>
<p>The four equally probably cases with a broken taxi at each airport are </p>
<ul>
<li>$\{T_1,T_2\}, \{T_3,T_4\}$</li>
<li>$\{T_1,T_4\}, \{T_2,T_3\}$</li>
<li>$\{T_2,T_3\}, \{T_1,T_4\}$</li>
<li>$\{T_3,T_4\}, \{T_1,T_2\}$</li>
</ul>
<p>The two equally probably cases with both broken taxis at the same airport are</p>
<ul>
<li>$\{T_1,T_3\}, \{T_2,T_4\}$</li>
<li>$\{T_2,T_4\}, \{T_1,T_3\}$</li>
</ul>
<p>Your method would give $\dfrac{4!}{2!\,2!}=6$ ways of distributing the four taxis and $\dfrac{2!}{1!\,1!}=2$ ways with one broken taxi at each airport, making the probability $\dfrac26=\dfrac13$. Too small. </p>
<p>If instead we multiplied by $2!$ because there are two broken taxis, that would make the probability $\dfrac46=\dfrac23$. Just right.</p>
|
82,765 | <p><strong>Bug introduced in 9.0 and persisting through 12.2</strong></p>
<hr />
<p>I get the following output with a fresh Mathematica (ver 10.0.2.0 on Mac) session</p>
<pre><code>FullSimplify[Exp[-100*(i-0.5)^2]]
(* 0. *)
Simplify[Exp[-100*(i-0.5)^2]]
(* E^(-100. (-0.5+i)^2) *)
</code></pre>
<p><code>FullSimplify</code> seems to be a bit overambitious and kills the expression completely. Is there anything that explains this behavior or is this simply a bug?</p>
<p>As suggested in the comments, I did an additional test:</p>
<pre><code>FullSimplify[Exp[-100*(i-1/2)^2]]
(* E^(-25 (1-2 i)^2) *)
</code></pre>
<p>Apparently, the float point math causes the problem.</p>
| Silvia | 17 | <p>Although confirmed by Wolfram officially, I will still hesitate to call this a bug.</p>
<p>The expression DOES very close to zero (though not a constant) everywhere except in a small neighborhood around $i=0.5$:</p>
<pre><code>LogPlot[E^(-25 (1 - 2 i)^2), {i, -1, 2}, PlotRange -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/LQbtf.png" alt="neighborhood around 0.5"></p>
<p>There is no way $10^{-80}$ can be handled correctly with a machine precision method.</p>
<p>One of the properer ways is do the math near $0.5$:</p>
<pre><code>FullSimplify[E^(-100 (i - .5)^2) /. i -> δ + .5] /. δ -> i - .5
</code></pre>
<blockquote>
<pre><code>E^(-100 (-0.5 + i)^2)
</code></pre>
</blockquote>
<h1>Edit</h1>
<p>To answer OP's comment, I still think it's actually a problem with using machine-precision number at an inappropriate place. Try change <code>0.5</code> to an arbitrary-precision number, say <code>0.5`2</code>, then the result will be fine:</p>
<pre><code>FullSimplify[Exp[-100*(i - 0.5`2)^2]] // InputForm
</code></pre>
<blockquote>
<pre><code>E^(-100.`10.*(-0.5`10. + i)^2)
</code></pre>
</blockquote>
|
3,830,231 | <p>I'm trying to prove the following 'covariance inequality'
<span class="math-container">$$
|\text{Cov}(x,y)|\le\sqrt{\text{Var}(x)}\sqrt{\text{Var}(y)}\,,
$$</span>
where covariance and variance are defined using discrete values,
<span class="math-container">$$
\text{Cov}(x,y) = \frac{1}{n-1}\sum_{i=1}^n \big[(x_i-\bar{x})(y_i-\bar{y})\big]\,,
$$</span>
<span class="math-container">$$
\text{Var}(x) = \frac{\sum_{i=1}^n(x_i-\bar{x})}{n-1}\,,
$$</span>
<span class="math-container">$$
\text{Var}(y) = \frac{\sum_{i=1}^n(y_i-\bar{y})}{n-1}\,.
$$</span></p>
<p>There are plenty of proofs of to be found online (such as <a href="https://www.probabilitycourse.com/chapter6/6_2_4_cauchy_schwarz.php" rel="nofollow noreferrer">this one</a>), however, they all either seem to be for continuous random variables, or just refer me to the Cauchy-Schwarz inequality, which I am aware of, but not sure how to apply to this particular proof. Basically, I am wondering if there is a way to prove this inequality using those above definitions.</p>
<p>I've tried substituting these definitions into the inequality above, but after expanding these summations and ridding of the <span class="math-container">$1/(n-1)$</span> on both sides, I'm left with a mess (as you can imagine) with summation terms on both sides, some in the absolute value, and some in the square root. I'm not sure if there's some algebraic mistake I'm making, some summation property I'm missing, or if substitution is just the wrong way to go about this proof.</p>
| tommik | 791,458 | <p>Without considering continuous or discrete, the following is a very elementary proof.</p>
<p>Let's consider the rv <span class="math-container">$Z=Y+aX$</span> and let's calculate its variance.</p>
<p><span class="math-container">$$V(Z)=V(Y)+a^2V(X)+2a cov(X,Y)\geq0$$</span></p>
<p>As per the fact that variance cannot be negative, the above expression is a 2nd degree inequality in <span class="math-container">$a$</span> and as per the fact that the 2nd degree term has positive coefficient the <span class="math-container">$\Delta$</span> must be <span class="math-container">$\leq0$</span></p>
<p>Thus</p>
<p><span class="math-container">$$4cov^2(X,Y)-4V(X)V(Y)\leq0$$</span></p>
<p><span class="math-container">$$cov^2\leq V(X)V(Y)$$</span></p>
|
3,882,214 | <p>I have a question that involves an Argand diagram. The complex number <strong>u = 1 + 1i</strong> is the center of that circle, and the radius is one. In other words, <span class="math-container">$$|z - (1 + 1i)| = 1$$</span></p>
<p>Now my issue is the following: I need to <strong>calculate the least value of |z| for the points on this locus using the diagram</strong>. Here's the sketch:</p>
<p><a href="https://i.stack.imgur.com/RPa9e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RPa9e.png" alt="enter image description here" /></a></p>
<p>So how do I find that least |z|? I understand that it'll involve a tangent to the circle, and I assume it's on the bottom right side of the circle, closest to the origin, but I'm not sure how to go about doing this.</p>
| Karan Elangovan | 497,101 | <p>The closest point is the point on the line through the origin and <span class="math-container">$(1 + i)$</span>.</p>
<p>The distance between the origin and the centre is <span class="math-container">$|1+i| = \sqrt{2}$</span></p>
<p>The distance between the centre and the point is 1 (radius is 1)</p>
<p>So the distance we want is <span class="math-container">$\sqrt2 - 1$</span>.</p>
<p>From a visual standpoint, it should be clear why this is the closest point. However we can make this rigorous by noting for any <span class="math-container">$z$</span> on the circle:
4</p>
<p><span class="math-container">$$ 1 = |z - (i+1)| = |(i+1) - z| \geq |i+1| - |z| $$</span></p>
<p><span class="math-container">$$ |z| \geq |i+1| - 1 = \sqrt{2} - 1$$</span></p>
<p>So at the shortest we can get is <span class="math-container">$\sqrt2 - 1$</span>, which we've shown is possible.</p>
|
76,505 | <p>In the eighties, Grothendieck devoted a great amount of time to work on the foundations of homotopical algebra. </p>
<p>He wrote in "Esquisse d'un programme": "[D]epuis près d'un an, la plus grande partie de mon énergie a été consacrée à un travail de réflexion sur les <em>fondements de l'algèbre (co)homologique non commutative</em>, ou ce qui revient au même, finalement, de l'<em>algèbre homotopique</em>." (Beginning of section 7. English version <a href="http://matematicas.unex.es/~navarro/res/esquisseeng.pdf" rel="noreferrer">here</a>: "Since the month of March last year, so nearly a year ago, the greater part of my energy has been devoted to a work of reflection on the foundations of non-commutative (co)homological algebra, or what is the same, after all, of homotopic[al] algebra.) </p>
<p>In <a href="http://www.math.jussieu.fr/~maltsin/groth/ps/lettreder.pdf" rel="noreferrer">a letter to Thomason</a> written in 1991, he states: "[P]our moi le “paradis originel” pour l’algèbre topologique n’est nullement la sempiternelle catégorie ∆∧ semi-simpliciale, si utile soit-elle, et encore moins celle des espaces topologiques (qui l’une et l’autre s’envoient dans la 2-catégorie des topos, qui en est comme une enveloppe commune), mais bien la catégorie Cat des petites caégories, vue avec un œil de géomètre par l’ensemble d’intuitions, étonnamment riche, provenant des topos." [EDIT 1: Terrible attempt of translation, otherwise some people might miss the reason why I have asked this question: "To me, the "original paradise" for topological algebra is by no means the never-ending semi-simplicial category ∆∧ [he means the simplex category], for all its usefulness, and even less is it the category of topological spaces (both of them imbedded in the 2-category of toposes, which is a kind of common enveloppe for them). It is the category of small categories Cat indeed, seen through the eyes of a geometer with the set of intuitions, surprisingly rich, arising from toposes."]</p>
<p>If $Hot$ stands for the classical homotopy category, then we can see $Hot$ as the localization of $Cat$ with respect to functors of which the topological realization of the nerve is a homotopy equivalence (or equivalently a topological weak equivalence). This definition of $Hot$ still makes use of topological spaces. However, topological spaces are in fact not necessary to define $Hot$. Grothendieck defines a <em>basic localizer</em> as a $W \subseteq Fl(Cat)$ satisfying the following properties: $W$ is weakly saturated; if a small category $A$ has a terminal object, then $A \to e$ is in $W$ (where $e$ stands for the trivial category); and the relative version of Quillen Theorem A holds. This notion is clearly stable by intersection, and Grothendieck conjectured that classical weak equivalences of $Cat$ form the smallest basic localizer. This was proved by Cisinski in his thesis, so that we end up with a categorical definition of the homotopy category $Hot$ without having mentionned topological spaces. (Neither have we made use of simplicial sets.) </p>
<p>I personnally found what Grothendieck wrote on the subject quite convincing, but of course it is a rather radical change of viewpoint regarding the foundations of homotopical algebra. </p>
<p>A related fact is that Grothendieck writes in "Esquisse d'un programme" that "la "<em>topologie générale</em>" <em>a été développée</em> (dans les années trente et quarante) <em>par des analystes et pour les besoins de l'analyse</em>, non pour les besoins de la topologie proprement dite, c'est à dire l'étude des <em>propriétés topologiques de formes géométriques</em> diverses". ("[G]eneral topology” was developed (during the thirties and forties) by analysts and in order to meet the needs of analysis, not for topology per se, i.e. the study of the topological properties of the various geometrical shapes." See the link above.) This sentence has already been alluded to on MO, for instance in Allen Knutson's answer <a href="https://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/14354#14354">there</a> or Kevin Lin's comment <a href="https://mathoverflow.net/questions/14314/algebraic-topologies-like-the-zariski-topology">there</a>. </p>
<p>So much for the personal background of this question.</p>
<p>It is not new that $Top$, the category of all topological spaces and continuous functions, does not possess all the desirable properties from the geometric and homotopical viewpoint. For instance, there are many situations in which it is necessary to restrict oneself to some subcategory of $Top$. I expect there are many more instances of "failures" of $Top$ from the homotopical viewpoint than the few I know of, and I would like to have a list of such "failures", from elementary ones to deeper or less-known ones. I do not give any example myself on purpose, but I hope the question as stated below is clear enough. Here it is, then: </p>
<blockquote>
<p>In which situations is it noticeable that $Top$ (the category of general topological spaces and continuous maps) is not adapted to geometric or homotopical needs? Which facts "should be true" but are not? And what do people usually do when encountering such situations? </p>
</blockquote>
<p>As usual, please post only one answer per post so as to allow people to upvote or downvote single answers.</p>
<p>P.S. I would like to make sure that nobody interpret this question as "why should we get rid of topological spaces". This, of course, is not what I have in mind! </p>
| Omar Antolín-Camarena | 644 | <p>The smash product of pointed topological spaces is not associative, i.e., $(X \wedge Y)\wedge Z$ need not be homeomorphic to $X \wedge (Y \wedge Z)$. (It fails, for example, for $X = Y = \mathbb{Q}$ and $Z = \mathbb{N}$.)</p>
|
2,550,568 | <p>Suppose we have an alphabet of $a$ letters and a word $w$ of length $r$. What is the probablity that $w$ will appear in a sequence of $n$ letters drawn at random from the given alphabet?</p>
<p>I have posted a general question since there seem to be a few of these questions appearing, and this is intended as a general form of question to which general answers can be given. Anyone who wants to add to what I have written - asymptotic of solutions in more detail, for example, or alternative methods - that would be great.</p>
<p>And are there any good references for this kind of problem.</p>
| Francisco José Letterio | 482,896 | <p>Well, the possible number of words of length n using a characters is $a^n$</p>
<p>On the other side, the odds of w showing up is the different ways you can choose r consecutive positions in a string of n positions, which is (if I'm not mistaken) $n-r+1$, </p>
<p>So the odds are $$\frac {n-r+1}{a^n}$$</p>
<p>Edit: I'll try to fix this later since it's obviously wrong and I'll be counting multiple cases if I fix it now</p>
|
3,902,418 | <p>Ok, so I know that the mean value of a function, <span class="math-container">$f(x)$</span>, on the interval <span class="math-container">$[a,b]$</span> is given by (or defined by?)
<span class="math-container">$$\frac{1}{b-a}\int_a^bf(x)~dx$$</span>
but I have <span class="math-container">$2$</span> basic questions about this:</p>
<p><strong><span class="math-container">$1$</span></strong>: From a purely mathematical point of view, does this have any practical use? Does it give us any extra weapon to add to our mathematical arsenal?</p>
<p><strong><span class="math-container">$2$</span></strong>: How excatly can I interpret this as a mean value? I've seen the geometrical interpretation (looking at the area under the graph within the given interval), but I still don't understand how it links to a mean value. For a finite set of values I divide the sum of the values by however many values I have to obtain the mean, but in this case there are infinitely many values, so how is this a mean value?</p>
<p>Thanks for your help.</p>
| Lee Mosher | 26,501 | <p>You wrote:</p>
<blockquote>
<p>For a finite set of values I divide the sum of the values by however many values I have to obtain the mean...</p>
</blockquote>
<p>Let's look at a Riemann sum for the expression <span class="math-container">$\frac{1}{b-a} \int_a^b f(x) \, dx$</span>, namely:
<span class="math-container">$$\frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \Delta x = \frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \frac{b-a}{n} = \frac{\sum_{i=1}^n f(x_i)}{n}
$$</span>
Surprise! It's a finite sum of values divided by the number of values.</p>
<p>Now the amazing thing is that if you study what happens to the numerical values of these <em>finite</em> mean expressions as <span class="math-container">$n \to \infty$</span>, the absolutely amazing thing is that it approaches a limit (assuming <span class="math-container">$f : [a,b] \to \mathbb R$</span> is continuous). That limit has its own special notation
<span class="math-container">$$\frac{1}{b-a} \left(\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \, \Delta x \right) = \frac{1}{b-a} \int_a^b f(x) \, dx
$$</span>
And for this reason, that last expression is called the mean value of <span class="math-container">$f$</span> itself, rather than just the mean value of a selected finite number of values of <span class="math-container">$f$</span>.</p>
|
4,522,097 | <p><span class="math-container">$$
X = \begin{pmatrix}
1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\
1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\
b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\
b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\
b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\
b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6}
\end{pmatrix}$$</span></p>
<p>The Schur complement w.r.t. the first and last row/column gives</p>
<p><span class="math-container">$$S = \begin{pmatrix}
1+b_1 & \frac{1}{a_6}\\
b_6 & 1 - \frac{a_5}{a_6}
\end{pmatrix}
-\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 &1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}^{-1}
\begin{pmatrix}1+b_2 & -\frac{a_1}{a_6} \\ b_3 & -\frac{a_2}{a_6}\\ b_4 & -\frac{a_3}{a_6}\\ b_5 & 1-\frac{a_4}{a_6}\end{pmatrix}.$$</span></p>
<p><span class="math-container">$$S = \begin{pmatrix} b_1 - b_2 + b_4 - b_5 & - \frac{-1-a_1+a_3-a_4+a_6}{a_6}\\-1-b_2+b_3-b_5+b_6 & \frac{a_1-a_2 + a_4 - a_5}{a_6}\end{pmatrix}$$</span></p>
<p>Then <span class="math-container">$\det(X) = \det\Biggr(\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}\Biggl). \det(S)$</span>.</p>
<p>How matrix <span class="math-container">$S$</span> is obtained? I am not sure why and how to take the blocks here. How <span class="math-container">$S$</span> is derived? I can see that in matrix <span class="math-container">$M$</span> but how to approach it in matrix <span class="math-container">$X$</span>?</p>
<p>Suppose <span class="math-container">$M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}$</span>. The Schur complement of <span class="math-container">$D
$</span> w.r.t <span class="math-container">$M$</span> is given by <span class="math-container">$M/D = A - B D^{-1} C$</span>. It is easy to see when there are contiguous blocks. But when they are not contiguous, how we apply the formula? Like how we get matrix <span class="math-container">$S$</span>. But I am not sure how to connect this with matrix <span class="math-container">$X$</span>.</p>
<p>How matrix <span class="math-container">$X$</span> will be different from <span class="math-container">$Y$</span> below, by clubbing the blocks together the blocks used to construct matrix <span class="math-container">$S$</span>.</p>
<p><span class="math-container">$$
Y = \begin{pmatrix}
1+b_1 & \frac{1}{a_6} & 1 & 0 & 0 & 0 \\
b_6 & 1-\frac{a_5}{a_6} & 0 & 0 & 0 & 1 \\
1+b_2 & -\frac{a_1}{a_6} & 1 & 1 & 0 & 0 \\
b_3 & -\frac{a_2}{a_6} & 1 & 1 & 1 & 0 \\
b_4 & -\frac{a_3}{a_6} & 0 & 1 & 1 & 1\\
b_6 & 1-\frac{a_4}{a_6} & 0 & 0 & 1 & 1
\end{pmatrix}$$</span></p>
| quarague | 169,704 | <p>They key difference between your two examples is the dimension of the space. The manifold is a 2-dimensional surface in both cases. In the first case this surface is embedded into the 3-dimensional space <span class="math-container">$\mathbb{R}^3$</span>.</p>
<p>In the second case you consider the parametrizations which map [parts of] the 2-dimensional space <span class="math-container">$\mathbb{R}^2$</span> to [parts of] the surface. On these parts the parametrization maps or charts are bijections.</p>
|
692,582 | <p>When I was reading a paper, I found an strange derivation like
$$\int^{+\infty}_{-\infty}\mathrm{ln}(1+e^w)f(w)dw\\=\int^0_{-\infty}\ln(1+e^w)f(w)+\int^\infty_0[\ln(1+e^{-w})+w]f(w)dw$$
when $w$ is the normal random variable and $f(w)$ is the normal density.</p>
<p>Why is that natural log integration broken up into that?</p>
<p>I think it just need a simple principle.</p>
<p>Thank you.</p>
| Gerry Myerson | 8,269 | <p>$$x^2+y(y+1)^2=0$$ gives something like what you want. </p>
<p>See also <a href="https://math.stackexchange.com/questions/276933/graph-of-an-infinitely-extending-rollercoaster-loop">Graph of an infinitely extending rollercoaster loop</a></p>
|
3,840,253 | <blockquote>
<p>How to show that <span class="math-container">$\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$</span>?</p>
</blockquote>
<p>My attempt:<br />
<span class="math-container">\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}</span>
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.</p>
| lab bhattacharjee | 33,337 | <p>If <span class="math-container">$\csc3x=\csc3y\iff \sin3y=\sin3x$</span></p>
<p><span class="math-container">$$3y=3x+(-1)^nn\pi$$</span> where <span class="math-container">$n$</span> is any integer</p>
<p><span class="math-container">$y=x+\dfrac{2n\pi}3; n=-1,0,1$</span></p>
<p>Now as <span class="math-container">$\sin3y=3\sin y-4\sin^3y,$</span></p>
<p>the roots of <span class="math-container">$$4\sin^3y-3\sin y+\sin3x=0$$</span> are</p>
<p><span class="math-container">$\sin\left(x-\dfrac{2\pi}3\right)=\sin\left(-\pi+\left(x+\dfrac\pi3\right)\right)=-\sin\left(x+\dfrac\pi3\right)=x_1,$</span></p>
<p><span class="math-container">$\sin x=x_2,$</span></p>
<p><span class="math-container">$\sin\left(x+\dfrac{2\pi}3\right)=\sin\left(\pi-\left(\dfrac\pi3-x\right)\right)=\left(\dfrac\pi3-x\right)=x_3$</span></p>
<p>By Vieta's formula,</p>
<p><span class="math-container">$$\dfrac1{x_1}+\dfrac1{x_2}++\dfrac1{x_3}=\dfrac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=\dfrac{\dfrac{-3}4}{-\dfrac{\sin3x}4}$$</span></p>
|
2,388,738 | <blockquote>
<p>I'm messing around with doing a visualization that has nothing to do with the primes and in order to execute it correctly I need an ordered list of all point in the order that the Ulam Spiral crosses them. I've tried some of my work but have only run in to abundantly complicated paths to solution. Also is there a name for looking for pattern that generally occur in spiral, whether they are related to primes or not?</p>
</blockquote>
<p>E.g.: Starting with this image:</p>
<p><a href="https://i.stack.imgur.com/90A3p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/90A3p.png" alt="enter image description here"></a></p>
<p>if $1$ is at the origin, then the list would be as follows: $$(0,0),(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1),(2,-1),(2,0),...$$</p>
| sOvr9000 | 186,910 | <p>A piecewise, explicit function is provided by the accepted answer of <a href="https://math.stackexchange.com/questions/3157030/parametrizing-the-square-spiral?rq=1">a similar question</a>.</p>
|
3,566,469 | <p>I am confused by a discussion with a colleague. The discussion is about the period of a periodic function.</p>
<p>For example, the periodic function <span class="math-container">$$f(x)=\sin(x), \quad x\in (0,\infty)$$</span> has period <span class="math-container">$2\pi$</span>. If I change the scale and build the function, <span class="math-container">$$g(x)=\sin(\ln x),\quad x\in (0,\infty)$$</span> is this new function, g, periodic? If it is, what is the period?</p>
<p><strong>EDIT</strong></p>
<p>I will clarify my point. If I change the scale of the function <span class="math-container">$g$</span>, let's say, <span class="math-container">$\ln x =u$</span> then I will have function <span class="math-container">$$h(u)=\sin u, \quad u\in \mathbb R$$</span> and now <span class="math-container">$h$</span> is periodic on <span class="math-container">$u\in \mathbb R $</span>. </p>
<p>So, my point is can I say that <span class="math-container">$g$</span> is not periodic in <span class="math-container">$x$</span>-domain but it is in <span class="math-container">$\log$</span>-domain?</p>
| J.G. | 56,861 | <p>Functions such as <span class="math-container">$\sin(\ln x)$</span> are called <a href="https://arxiv.org/abs/1002.1010" rel="nofollow noreferrer"><em>log-periodic</em></a> rather than periodic.</p>
|
373,068 | <p>For a real number $a$ and a positive integer $k$, denote by $(a)^{(k)}$ the number $a(a+1)\cdots (a+k-1)$ and $(a)_k$ the number
$a(a-1)\cdots (a-k+1)$. Let $m$ be a positive integer $\ge k$. Can anyone show me, or point me to a reference, why the number
$$ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}= \frac{2^{2k}(m)^{(k)}(m)_k}{(2k)!}$$ is always an integer?</p>
| vonbrand | 43,946 | <p>Using Knuth's notation:
$$
x^{\underline{k}} = x (x - 1) \ldots (x - k + 1) = (x)_{(k)} \qquad
x^{\overline{k}} = x (x + 1) \ldots (x + k - 1) = (x)^{(k)}
$$
What you are looking for is:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
$$
As if $k > m$ then $m^{\underline{k}} = 0$, while all other factors are positive, the range of interest is $0 \le k \le m$. We also have:
$$
\frac{m^{\underline{k}}}{k!} = \binom{m}{k}
$$
which is an integer, we still need:
$$
m^{\overline{k}} = (m + k - 1)^{\underline{k}}
$$
$$
\begin{align*}
(1/2)^{\overline{k}} &= (1/2) \cdot (3 / 2) \cdot \ldots \cdot (1/2 + k - 1) \\
&= \frac{1 \cdot 3 \cdot \ldots \cdot (2 k - 1)}{2^k} \\
&= \frac{1 \cdot 2 \cdot \ldots \cdot 2k}
{2^k \cdot 2^k \cdot k!} \\
&= \frac{(2 k)!}{2^{2 k} k!}
\end{align*}
$$
Thus:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
= \frac{m (m + k - 1)^{\underline{2 k-1}} 2^{2 k} }{(2 k)!}
$$</p>
|
2,036,301 | <p>Can someone please help me prove that this series is convergent? <br></p>
<p>The problem is I don't know what to do with sin.<br> </p>
<p>$$\sum_{n=1}^{\infty} 2^n \sin{\frac{\pi}{3^n}} $$</p>
| user389056 | 389,056 | <p>Consider the root test. </p>
<p>Let $L = \lim_{n \to \infty} |2^n\sin(\frac{\pi}{3^n})|^{\frac{1}{n}}.$
Then $L = \lim_{n \to \infty} (2^n\sin(\frac{\pi}{3^n}))^{\frac{1}{n}} = \lim_{n \to \infty} 2\sin(\frac{\pi}{3^n})^{\frac{1}{n}} = 0$, so the series converges by root test. </p>
|
139,934 | <p>Suppose I want to solve an equation for the matrix elements of $\bar{W}$:
$$\alpha W_{ba}+\beta W_{bb}=x; \alpha W_{aa}+\beta W_{ab}=y$$</p>
<p>Using the syntax <code>Subscript[W, ij]</code> for my matrix element (on the $i$th row and $j$ th column), I get the following message:</p>
<p>Set::write: Tag Times in 2 x is Protected.</p>
<p>Is is possible at all to write such double subscript in Mathematica?</p>
| Alexey Popkov | 280 | <pre><code>file = "--------------
SCF ITERATIONS
--------------
ITER Energy Delta-E Max-DP RMS-DP [F,P] Damp
*** Starting incremental Fock matrix formation ***
0 -8693.9185205626 0.000000000000 0.02365877 0.00022580 0.2453968 0.8500
1 -8694.4485310565 -0.530010493916 0.02170668 0.00021389 0.1786600 0.8500
***Turning on DIIS***
2 -8694.7699232947 -0.321392238231 0.01563958 0.00015675 0.1186396 0.8500
3 -8694.9659372250 -0.196013930265 0.00723160 0.00008969 0.0765966 0.8500
4 -8695.1123473361 -0.146410111134 0.00420004 0.00005953 0.0579101 0.8500
5 -8695.2253413821 -0.112994046020 0.00296590 0.00004350 0.0478892 0.8500
6 -8695.3172110457 -0.091869663593 0.00244669 0.00003472 0.0405933 0.8500
**** Energy Check signals convergence **** ";
start = StringPosition[file, "***Turning on DIIS***" ~~ " " ... ~~ "\n"][[1, 2]] + 1;
end = StringPosition[file,
"\n" ~~ " " ... ~~ "**** Energy Check signals convergence ****"][[1, 1]] - 1;
table = ImportString[StringTake[file, {start, end}], "Table"]
</code></pre>
<blockquote>
<pre><code>{{2, -8694.7699232947`, -0.321392238231`, 0.01563958`, 0.00015675`, 0.1186396`, 0.85`},
{3, -8694.965937225`, -0.196013930265`, 0.0072316`, 0.00008969`, 0.0765966`, 0.85`},
{4, -8695.1123473361`, -0.146410111134`, 0.00420004`, 0.00005953`, 0.0579101`, 0.85`},
{5, -8695.2253413821`, -0.11299404602`, 0.0029659`, 0.0000435`, 0.0478892`, 0.85`},
{6, -8695.3172110457`, -0.091869663593`, 0.00244669`, 0.00003472`, 0.0405933`, 0.85`}}
</code></pre>
</blockquote>
<p>If your file contains some heads/comments between <code>start</code> and <code>end</code>, you can filter them out if you know the structure of your file, for example as <a href="https://mathematica.stackexchange.com/questions/139914/reading-a-text-file-and-parsing-only-the-text-between-two-delimiters/139916?noredirect=1#comment377724_139915">Michael E2 suggests</a>:</p>
<pre><code>table2 = Cases[table, {_Integer, Repeated[_Real, {6}]}]
</code></pre>
<p>Another approach:</p>
<pre><code>table2 = DeleteCases[table, {a_String, ___, a_} /; StringMatchQ[a, "*" ..]]
</code></pre>
|
3,442,173 | <p>Give a counter example to each of the following:<br>
(a) G is a connected graph with a cut-vertex, then G contains a bridge.
(b) G is a tree if and only if it contains no cycle.</p>
| egreg | 62,967 | <p>Consider <span class="math-container">$g(x)=f(x)-cx$</span>. Then <span class="math-container">$g’(x)=f’(x)-c=c-c=0$</span>.</p>
<p>Therefore <span class="math-container">$g$</span> is constant, say <span class="math-container">$g(x)=d$</span>.</p>
<p>Then <span class="math-container">$f(x)=cx+d$</span>.</p>
|
3,059,857 | <p>We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it. </p>
<p><span class="math-container">$ x = A + k \times 2\pi$</span></p>
<p>and</p>
<p><span class="math-container">$x = \pi - A + k \times 2\pi$</span> </p>
<p>where <span class="math-container">$k$</span> is supposed to be a random integer? and A in this case is <span class="math-container">$\frac{11}{9}\pi$</span> </p>
| KM101 | 596,598 | <p>Note that you have the arcsine function has a range of <span class="math-container">$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$</span>, but sine is negative in both quadrants <span class="math-container">$3$</span> and <span class="math-container">$4$</span>, so only the quadrant <span class="math-container">$4$</span> angle is returned. (The opposite is also true: sine is positive in both quadrant <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, but only the quadrant <span class="math-container">$1$</span> angle is returned.) Also note that <span class="math-container">$\sin(\pi-\theta) = \sin(\theta)$</span>.</p>
<p>Hence, we have</p>
<p><span class="math-container">$$\sin (x) = y \implies x = \begin{cases}\ \arcsin y+2\pi k \\ \pi-\arcsin y+2\pi k\end{cases}; \quad k\in \mathbb{Z}$$</span></p>
<p>So, if <span class="math-container">$$\sin(x) = \sin\left(\frac{11}{9}\pi\right)$$</span></p>
<p>then there are two cases:</p>
<p><span class="math-container">$$x = \begin{cases}\ \frac{11}{9}\pi+2\pi k \\ \pi-\frac{11}{9}\pi+2\pi k\end{cases}$$</span></p>
<p>with the first being for angles in quadrant <span class="math-container">$3$</span> and the second being for angles in quadrant <span class="math-container">$4$</span>.</p>
<hr>
<p>As for why we have <span class="math-container">$2\pi k$</span> for <span class="math-container">$k \in \mathbb{Z}$</span> (<span class="math-container">$k$</span> must be an integer), this is because trig functions are periodic. Note that since <span class="math-container">$2\pi$</span> (or <span class="math-container">$360°$</span>) represents one full revolution, you return back to where you started. For example, <span class="math-container">$\frac{9\pi}{4}$</span> radians would mean <span class="math-container">$2\pi+\frac{\pi}{4}$</span>, so it’s represents the same angle as <span class="math-container">$\frac{\pi}{4}$</span>. So, any two angles separated by <span class="math-container">$2\pi$</span>, <span class="math-container">$4\pi$</span>, <span class="math-container">$6\pi$</span>, ... radians actually represent the same angle. (In degrees, that would be <span class="math-container">$360°$</span>, <span class="math-container">$720°$</span>, <span class="math-container">$1080°$</span>, etc.) So, we generalize this as all integer multiples of <span class="math-container">$2\pi$</span> radians or <span class="math-container">$360°$</span>.</p>
<p>As a simple example, assume we have <span class="math-container">$\sin(x) = 1$</span>. Clearly, the answer must be <span class="math-container">$\pi$</span>. But <span class="math-container">$\pi+2\pi$</span> also works. So does <span class="math-container">$\pi-2\pi$</span>. So does any angle <span class="math-container">$2\pi k$</span> radians apart from <span class="math-container">$\pi$</span>. So we would generalize the answer and give the full solution as <span class="math-container">$\pi+2\pi k$</span>, as they all represent the same angle, a right angle. In other words, there are infinite solutions. </p>
<p>However, you may occasionally be given a restricted domain such as <span class="math-container">$0 \leq x \leq 2\pi$</span>. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the <span class="math-container">$2\pi k$</span> for both cases, and in the example I gave, the answer simply becomes <span class="math-container">$\pi$</span>.)</p>
|
2,054,175 | <p>This problem is giving me loads of confusion. I just need someone to walk through it because I have the answer and I can't get to it to save my life. I have been on it for days. Please help.</p>
<p>$$\frac{x + 3}{x - 4}\le 0$$ </p>
| jameselmore | 86,570 | <p>Hint:<br>
Let "$+$" be any positive real number, and "$-$" be any negative real number. It should be clear that:</p>
<p>$$\frac{+}{+} > 0;\ \ \frac{-}{-} > 0$$
$$\frac{+}{-} < 0;\ \ \frac{-}{+} < 0$$</p>
<p>If you can figure out for which values of $x$ has $(x+3)$ positive\negative, and which values of $x$ make $(x-4)$ positive\negative, can you use the above rules to figure out where the fraction is negative?</p>
|
93,274 | <p>Wielandt wrote a paper titled "Remarks on diagonable matrices".</p>
<p>According to Mathematische Werke - Mathematical Works : Linear Algebra and Analysis
by Helmut Wielandt, Hans Schneider, Bertram Huppert (Editor) page 260 this paper from Wielandt remained unpublished (at least from the 1950s to the 1980s).</p>
<p>Does anyone have a copy of it or an idea of the proof on non defective pencils?</p>
<p>The main theorem states that for $A,B \in \mathcal{M}_n(\mathbb C)$, if in the pencil $\lambda A+ \mu B$ all matrices are diagnosable ($\forall \lambda. \mu \in \mathbb{C}$), then $AB=BA$.</p>
<p>Motzkin and Taussky proved that result (MR0086781 (19,242c)), using algebraic geometry, Kato proved it differently (MR1335452 (96a:47025)), using theory of complex functions in one variable.
Wielandt seemed to have given another proof, hence my request.</p>
<p>Thanks</p>
| Federico Poloni | 1,898 | <p>Now that the server is back up, I am posting this as a real answer.</p>
<p>With some work, you might be able to find the proof in Wielandt's notebooks, which were TeXxed and put online <a href="http://www3.math.tu-berlin.de/numerik/Wielandt/index_en.html" rel="nofollow">here</a>.</p>
<p>The TeX source files are also published, so you can download them and use an automated search tool. Nevertheless, there's lots of material there, so it is not an easy task if you have no idea which period to look at.</p>
|
3,340,686 | <p>The <span class="math-container">$7$</span>th floor of a building is <span class="math-container">$23$</span>m above street level and <span class="math-container">$13$</span>th floor is <span class="math-container">$41$</span>m above street level. What is the height (above street level) of the first floor and what is the height of one floor?</p>
<p>My working out is this:</p>
<p><span class="math-container">$t_7=23$</span></p>
<p><span class="math-container">$t_{13}=41$</span></p>
<p><span class="math-container">$t_n=t_1 + (n-1) d$</span></p>
<p><span class="math-container">$23= t_1 + 6d (1)$</span></p>
<p><span class="math-container">$41 = t_1 + 12d (2)$</span></p>
<p><span class="math-container">$18= 6d$</span></p>
<p><span class="math-container">$d=3$</span></p>
<p>Sub <span class="math-container">$3$</span> into equation <span class="math-container">$2$</span></p>
<p><span class="math-container">$41=t_1 + 12(3)$</span></p>
<p><span class="math-container">$41= t_1 + 36$</span></p>
<p><span class="math-container">$5 = t_1$</span></p>
<p>Can someone verify if I am doing it correct?</p>
| Siddhant | 687,664 | <p>Yes it is perfectly correct. The first floor is <strong><span class="math-container">$5m$</span></strong> above the street level above which every floor with a height of <strong><span class="math-container">$3m$</span></strong>.</p>
|
1,893,168 | <p>$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$</p>
<p>I can solve this by using L'Hopital's rule but how would I do this without this?</p>
| Marco Cantarini | 171,547 | <p>$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{1}{2}\frac{\log\left(1-\sin^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)}=-\frac{\sin^{2}\left(x\right)+O\left(\sin^{4}\left(x\right)\right)
}{2\sin^{2}\left(x\right)}\stackrel{x\rightarrow0}{\rightarrow}-\frac{1}{2}.$$</p>
|
1,893,168 | <p>$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$</p>
<p>I can solve this by using L'Hopital's rule but how would I do this without this?</p>
| paf | 333,517 | <p>Note that $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2(\frac x2)}{x^2}=\lim_{x\to 0}\frac 14 \frac{2\sin^2(\frac x2)}{\frac{x^2}{4}}=\frac{1}{2}$$
Now,
$$\lim_{u\to 1}\frac{\ln u}{u-1}=\ln'(1)=1$$
Hence, since $u=\cos x$ tends to 1 when $x$ tends to 0:
$$\lim_{x\to 0}\frac{\ln(\cos x)}{\cos x-1}=1$$
So
$$\lim_{x\to 0}\frac{\ln(\cos x)}{x^2} =\lim_{x\to 0}\frac{\ln(\cos x)}{\cos x-1}\lim_{x\to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2}$$</p>
<p>Finally, since $$\lim_{x\to 0} \left(\frac{\sin x}{x}\right)^2 = 1^2=1$$
,we have
$$\lim_{x\to 0}\frac{\ln(\cos x)}{\sin^2x}=\lim_{x\to 0}\frac{\ln(\cos x)}{x^2}\lim_{x\to 0} \left(\frac{x}{\sin x}\right)^2=-\frac{1}{2}$$</p>
<p><strong>Edit:</strong> with Taylor expansions, life becomes easier!
Near 0, we have
$$\frac{\ln(\cos x)}{\sin^2x}=\frac{\ln\left(1-\frac{x^2}{2}+o(x^2)\right)}{x^2+o(x^2)}$$</p>
<p>Since $\ln(1+u)=u+o(u)$ near 0, we obtain (letting $u=-\frac{x^2}{2}+o(x^2)$ above):
$$\frac{\ln(\cos x)}{\sin^2x}=\frac{-\frac{x^2}{2}+o(x^2)}{x^2+o(x^2)}$$
near 0. So the limit is $-\frac 12$. </p>
|
2,668,839 | <blockquote>
<p>Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$</p>
</blockquote>
<p>Try: put $\sin x=t$ and $-1\leq t\leq 1$</p>
<p>So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$</p>
<p>$$2yt^2+8yt+8y=t^2+4t+5$$</p>
<p>$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$</p>
<p>For real roots $D\geq 0$</p>
<p>So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$</p>
<p>$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$</p>
<p>$y\geq 0.5$</p>
<p>Could some help me where I have wrong, thanks</p>
| Community | -1 | <p>You can simplify the expression as</p>
<p>$$\frac12+\frac1{2(\sin x+2)^2}$$ and the extreme values are</p>
<p>$$\frac12+\frac1{2\cdot3^2},\\\frac12+\frac1{2\cdot1^2}.$$</p>
|
3,894,437 | <p>Let <span class="math-container">$\mathcal{A}$</span> be a finite set and consider the set of all sequences <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> on <span class="math-container">$\mathbb{Z}$</span> with values in <span class="math-container">$\mathcal{A}$</span>. This set has a cardinality of <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> which is not countable, right? (is there actually a simple explanation?)</p>
<p>My actual question however is if there is at least a countable and dense subset (w.r.t. to the product topology when endowing each <span class="math-container">$\mathcal{A}$</span> with the discrete topology), i.e. is <span class="math-container">$\mathcal{A}^{\mathbb{Z}}$</span> separable?</p>
<p>Sadly, I couldn't come up with anything myself. I ask this to answer another question from ergodic theory, namely if the two-sided full shift over a finite alphabet is transitive.</p>
| JunderscoreH | 632,927 | <p>For the side question, as long as <span class="math-container">$|\mathcal{A}|\ge 2$</span> we have <span class="math-container">$2^{\aleph_0}\le|\mathcal{A}^{\mathbb{Z}}|$</span> so that it's uncountable.</p>
<p>For the actual question, I'll consider <span class="math-container">$\mathcal{A}^{\mathbb{N}}$</span>, but the same idea works for <span class="math-container">$\mathbb{Z}$</span>. We can see that the product topology on such a set has a basis of <em>cones</em> <span class="math-container">$\{x\in\mathcal{A}^{\mathbb{N}}:\sigma\text{ is contained in }x\}$</span> for <span class="math-container">$\sigma\in\mathcal{A}^{<\mathbb{N}}$</span>. As a result, if we consider (say) the sequences which are almost always some <span class="math-container">$a\in\mathcal{A}$</span> (in the sense that there are only finitely many entries not <span class="math-container">$a$</span>), then this set will be dense of size <span class="math-container">$\mathcal{A}^{<\mathbb{N}}=\aleph_0$</span>.</p>
|
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