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 |
|---|---|---|---|---|
2,210,372 | <p>I need to identify which of the curves $A_1,...,A_5$ are related to algorithms whose run times are proportional to $n, \log(n), n^2, n^3$ and $1.1^n$:</p>
<p>(Mentioning that the first figure of the $A_5$ column should be $0.015$ (not $0.025$)</p>
<p><a href="https://i.stack.imgur.com/70I2d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/70I2d.png" alt="enter image description here"></a></p>
| Marcelo Fornet | 427,329 | <p>It can be done at sight I think, but you can try to adjust each function to each curve and pick the one that minimize the error using minimum square deviation. Let's say you want to check if curve A1 is proportional to $f(n)=n^2$. You try to minimize the function $g(a,b)=\Sigma(a f(x_i) + b - y_i)^2$ Here $x_i$ is the size of the input and $y_i$ is the time it takes.</p>
|
674,598 | <p>(a) $f(x) = |x|; -\infty <x < \infty $</p>
<p>The way i approach is to take the derivative for x from 0 to infinity which gives me 1 and take the derivative for -x from 0 to negative infinity which gives me negative 1...therefore, since there is not constant can be 1 and -1 at the same time ....my answer is there is no Lipschitz constant which satisfies this condition...is this a right approch?</p>
| Ben Grossmann | 81,360 | <p>Not quite. A function can be Lipschitz without being differentiable, and it can be differentiable without being Lipschitz.</p>
<p>For this problem, $1$ will work as a Lipschitz constant.</p>
<p>Why? Note that for any $x,y$, we have
$$
\left||x| - |y|\right| \leq |x-y|
$$</p>
|
1,359,743 | <p>If $\tan \theta = 3\cfrac{15}{16}$, then find $\sin \theta$.</p>
| SF Math | 254,277 | <p>Note that $\tan\theta=\frac{opp}{adj}$. Thus you have a right triangle with side lengths $\text{adj}=16$ and $\text{opp}=63$. I think you can solve it from here.</p>
|
1,403,228 | <blockquote>
<p><strong>Question:</strong></p>
<p>Let <span class="math-container">$m, n, q, r \in \mathbb Z$</span>. If <span class="math-container">$m = qn + r$</span>, show that <span class="math-container">$\gcd(m, n) = \gcd(n, r)$</span>. Hence justify the Euclidean Algorithm.</p>
</blockquote>
<p>I found this question in a past test paper, but cannot seem to find a reference in my textbook that indicates how I can go about "proving" the above statement. Can anyone please point me in the right direction?</p>
| John Douma | 69,810 | <p>$m-qn=r$ so if $c$ divides $m$ and $n$ it also divides r. Since this is true for all divisors of $m$ and $n$ it is true for $gcd(m,n)$ and so $gcd(m,n)\le gcd(n,r)$. If $d$ divides $n$ and $r$ then $d$ divides $qn+r$ which equals $m$ so $d$ is a common divisor of $m$ and $n$. Therefore $gcd(n,r)\le gcd(m,n)$ and $gcd(m,n) = gcd(n,r)$</p>
|
2,871,084 | <p>Consider that we have a circle drawn a round the <strong>origin (0,0)</strong>. That circle has some points drawn on its circumference. Each of those points has range and azimuth $(r,\theta)$, the <strong>r</strong> and <strong>$\theta$</strong> values of these points are calculate with responding to the <strong>origin (0,0)</strong>. </p>
<p>I want to move or translate that circle and its (on-circumference) points to a new center (x,y), the new center can be in any quadrant. Here is an image for more demonstration (consider that it was translated to the first quadrant):<a href="https://i.stack.imgur.com/slaiD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slaiD.png" alt="enter image description here"></a></p>
<p><strong>The question is what is the new $(r, \theta)$ of each point with responding to the origin (0,0) after translation to the new center?</strong></p>
| Quantic_Solver | 581,446 | <p>The map isn't surjective. A counterexample is the <a href="https://en.wikipedia.org/wiki/Weierstrass_function" rel="nofollow noreferrer">Weierstrass function</a> which is continuous.</p>
|
478,620 | <p>Let $S$ be a set that satisfies the following property:</p>
<blockquote>
<p>For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$</p>
</blockquote>
<p>Does it follow that $S$ is finite?</p>
| qaphla | 85,568 | <p>There are four digits that the number can end with and be odd, not $\frac{9}{2}$, which is what your calculation uses -- that is, there are more even numbers without a five than odd numbers without a five.</p>
<p>More correctly:</p>
<p>$8 * 9 * 4 = 72 * 4 = 288$, as the first digit can be any of $1,2,3,4,6,7,8,9$, the second any but $5$, and the third must be $1,3,7,$ or $9$.</p>
|
478,620 | <p>Let $S$ be a set that satisfies the following property:</p>
<blockquote>
<p>For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$</p>
</blockquote>
<p>Does it follow that $S$ is finite?</p>
| André Nicolas | 6,312 | <p>There is no reason that there are just as many odd integers that do not contain $5$ as there are even integers that do contain 5. The proper fraction is $\dfrac{4}{9}$.</p>
|
478,620 | <p>Let $S$ be a set that satisfies the following property:</p>
<blockquote>
<p>For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$</p>
</blockquote>
<p>Does it follow that $S$ is finite?</p>
| Zulqarnain Ansari | 228,238 | <p>out of the nine digits 0,1,2,3,4,6,7,8, and 9. The digit at hundred place may be any digit other than 0, any of the nine digits can occupy tens place and the unit place can be occupied by 1,3,7 and 9. Thus the required number of three digits odd numbers will be 8*9*4=288</p>
|
1,450,669 | <p>Find the value of $$\lim\limits_{x\to 0^+}[1+[x]]^{\frac2x}$$where $[x]$ denotes greatest integer function less than or equal to $x$. <br></p>
<hr>
<p><strong>My attempt:</strong> <br>
I calculated $[1+[x]]$ to be $1$ as $x\to 0^+$. <br> Now I am stuck. <br> Please help me.</p>
| Landon Carter | 136,523 | <p>Recall that if $f(x)\to 1$ and $g(x)\to\infty$ as $x\to0$ then $\lim_{x\to0}f(x)^{g(x)}=e^{\lim_{x\to0}g(x)(f(x)-1)}$.</p>
<p>Here $f(x)=1+[x]\to1$ and $g(x)=2/x\to\infty$ as $x\to0+$. Thus the limit is $e^{\lim_{x\to0+}(2/x)[x]}$</p>
<p>Now as $x$ goes to $0$, $x<1$ must be true after some stage. Then for all such $x$, $[x]=0$ so $\lim_{x\to0+}\dfrac{[x]}{x}=0$.</p>
<p>So the answer is $e^0=1$.</p>
|
3,493,729 | <p>Although I looked up the answer on integral calculator com but I still have little to no idea as to how one would proceed to solve this integral.
Integrate <span class="math-container">$\dfrac{e^x(x^4+2)}{(1+x^2)^{5/2}}$</span> wrt <span class="math-container">$x.$</span> I initially tried to convert it to the form <span class="math-container">$e^x\cdot(f(x)+f'(x)).$</span> However, I wasn't successful in spite of struggling for more than half an hour :(</p>
<p>Any help would be appreciated :)</p>
| Henno Brandsma | 4,280 | <p><strong>Proposition</strong>: let <span class="math-container">$X$</span> be a topological space, then TFAE:</p>
<ol>
<li><span class="math-container">$X$</span> is connected.</li>
<li>If <span class="math-container">$O \subseteq X$</span> is clopen (closed and open), <span class="math-container">$O=\emptyset$</span> or <span class="math-container">$O=X$</span>.</li>
<li>Every continuous <span class="math-container">$f:X \to \{0,1\}$</span> (the two point set in the discrete topology) is constant.</li>
</ol>
<p><strong>Proof</strong>: <span class="math-container">$1 \implies 2$</span>: If <span class="math-container">$O$</span> were clopen and not empty and not equal to <span class="math-container">$X$</span>, <span class="math-container">$\{O, X\setminus O\}$</span> would be a disconnection of <span class="math-container">$X$</span> (both sets are open, non-empty and trivially disjoint), contradiction. </p>
<p><span class="math-container">$2\implies 3$</span>: If <span class="math-container">$f: X \to \{0,1\}$</span> is continuous, then <span class="math-container">$f^{-1}[\{0\}]$</span> is clopen as the inverse image of a clopen set, so is empty (and <span class="math-container">$f \equiv 1$</span>) or <span class="math-container">$X$</span> (and <span class="math-container">$f \equiv 0$</span>).</p>
<p><span class="math-container">$3 \implies 1$</span>: Suppose <span class="math-container">$X= U \cup V$</span> were a disconnection of <span class="math-container">$X$</span>. Then the function <span class="math-container">$f$</span> sending all points of <span class="math-container">$U$</span> to <span class="math-container">$0$</span> and all points of <span class="math-container">$V$</span> to <span class="math-container">$1$</span> is well-defined (<span class="math-container">$U$</span> and <span class="math-container">$V$</span> are disjoint) and continuous (pasting lemma using that <span class="math-container">$U,V$</span> are open and constant maps are continuous) and non-constant (both sets are non-empty). This contradiction shows <span class="math-container">$X$</span> is connected.</p>
<p>As <span class="math-container">$[0,1]$</span> is connected, no such map as you look for exists.</p>
|
306,178 | <p>Given
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.</p>
| Mhenni Benghorbal | 35,472 | <p>Use the first derivative test and prove the function</p>
<p>$$ f(x)=\left(1+\frac{1}{x}\right)^{x+1}\hspace{-6 mm},\qquad \quad x \geq 1, $$</p>
<p>is decreasing on $[1,\infty]$. That is prove $f'(x)<0$ on $[1,\infty]$. </p>
|
3,425,780 | <p>Let <span class="math-container">$H_i$</span> be a <span class="math-container">$\mathbb C$</span>-Hilbert space and <span class="math-container">$T$</span> be a densely-defined linear operator from <span class="math-container">$H_1$</span> to <span class="math-container">$H_2$</span>.</p>
<blockquote>
<p>How can we show that if <span class="math-container">$T$</span> is injective and <span class="math-container">$\operatorname{im}T$</span> is dense, then <span class="math-container">$T^\ast$</span> is injective as well? I've read that the reason is that <span class="math-container">$$\ker T^\ast=(\operatorname{im}T)^\perp=\{0\}\tag1,$$</span> but I don't get why <span class="math-container">$(1)$</span> holds.</p>
</blockquote>
<p>I know that for a general densely-defined <span class="math-container">$T$</span>, <span class="math-container">$\ker T^\ast=(\operatorname{im}T)^\perp$</span> and hence <span class="math-container">$(\ker T^\ast)^\perp=\overline{\operatorname{im}T}$</span>. On the other hand, the identity <span class="math-container">$\ker T=(\operatorname{im}T^\ast)^\perp$</span> can only be concluded, when <span class="math-container">$T$</span> is closable (since this is equivalent to <span class="math-container">$T^\ast$</span> being densely-defined).</p>
<p>So, assuming <span class="math-container">$\operatorname{im}T$</span> is dense, the only thing I was able to infer is that <span class="math-container">$$(\ker T^\ast)^\perp=\overline{\operatorname{im}T}=H_2\tag2.$$</span> Now if<span class="math-container">$T^\ast$</span> would be continuous, then <span class="math-container">$\ker T^\ast$</span> would be closed and hence <span class="math-container">$H_2=\ker T^\ast\oplus(\ker T^\ast)^\perp$</span>, which would immediately yield <span class="math-container">$\ker T^\ast=\{0\}$</span> and hence the claim.</p>
| IrbidMath | 255,977 | <p>We know that the vertex x coordinates is <span class="math-container">$\frac{-b}{2a} = \frac{1}{4}$</span> solve for <span class="math-container">$b = \frac{-a}{2}$</span> now the point <span class="math-container">$\frac{1}{4} , \frac{-9}{8}$</span> satisfies the parabola so </p>
<p><span class="math-container">$\frac{a}{16} - \frac{a}{8} + c = \frac{-9}{8}$</span> </p>
<p><span class="math-container">$a-2a + 16c = -18 $</span> </p>
<p><span class="math-container">$c = \frac{-18 +a}{16}$</span> finally </p>
<p><span class="math-container">$a+b+c = a + \frac{-a}{2} +\frac{-18+a}{16} \in \mathbb{Z}$</span> find the solution and pick the smallest one </p>
<p>we will have <span class="math-container">$16 \mid 9a-18 $</span> </p>
|
4,212,199 | <p><span class="math-container">$$\frac15\int_0^5(4+2^{0.1t^2})dt$$</span>
This question is from part (b) of question 1 in the AP Calculus BC 2019 FRQ. I've tried using u-substitution where <span class="math-container">$u = 0.1 t^2$</span>, but this won't work because of the <span class="math-container">$2t$</span> that will result when taking the derivative of <span class="math-container">$u$</span>. Integration by parts wouldn't work as the integral doesn't involve a product, and trigonometric substitution wouldn't work either. Is there a technique that would allow me to evaluate this?</p>
| John Douma | 69,810 | <p>Your integral is wrong. As you know, the total work done is given by the total force exerted over a distance. In this case the force on the cable is variable. We know that the starting weight is <span class="math-container">$200$</span> pounds and decreases by <span class="math-container">$2$</span> pounds for every foot that the cable is raised. Therefore, the force is given by <span class="math-container">$$F(x)=200-2x$$</span> where <span class="math-container">$x$</span> is the distance above the ground of the bottom of the cable. From this we get that the total work is</p>
<p><span class="math-container">$$\int_{0}^{80}(200-2x) dx=9600\text{ ft lb}$$</span></p>
|
226,488 | <p>I am currently working on a challenge problem where I need to show that there is a point $x \in \mathbb{R_+}$ such that $\cos(x) = 0$ using only a few properties of the cosine function. In particular, the only properties of the cosine function that I can use are:</p>
<ul>
<li><p>$\cos(x)$ is continuous</p></li>
<li><p>$\cos(x) = Re(\exp(z))$ for $z \in \mathbb{C}$</p></li>
<li>$\cos^2(x) + \sin^2(x)=1$</li>
<li>$\displaystyle \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n!)}x^{2n}$</li>
<li>$\displaystyle \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$</li>
</ul>
<p>My strategy is to use the intermediate value theorem on the interval $[0,a]$ since it's easy to show that $\cos(0) = 1$. If I could show that there is a point $a \in \mathbb{R}_+$ s.t. $\cos(a) < 0$, then the IVT and the continuity of the cosine function would allow me to conclude that there has to be some $x\in [0,a]$ such that $\cos(x) = 0$.</p>
| Eric Angle | 35,995 | <p>The circle is divided into $n$ wedges with angle
$$
\theta = \frac{2 \pi}{n}.
$$
Consider the triangle formed by the center of the circle (point $A$), one of the two adjacent points (point $B$), and the midpoint of the line joining the two adjacent points (point $C$). Now, $\angle BAC = \theta / 2 = \pi / n$, $AB = d/2$, and $BC = m / 2$, where $d$ is the diameter of the circle, so
$$
AB \sin \left(\angle BAC\right) = BC \Rightarrow \frac{d}{2} \sin\left(\frac{\pi}{n}\right) = \frac{m}{2} \Rightarrow d = \frac{m}{\sin \left(\pi / n\right)}.
$$</p>
|
2,673,078 | <p>Generalize the Monty Hall problem where there are $n \geq 3$ doors, of
which Monty opens $m$ goat doors, with $1 \leq m \leq n$.<br><strong>Original Monty Hall Problem:</strong> There are $3$ doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don’t want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is Door $1$. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening Door $2$and Door $3$, he chooses Door $2$ and Door $3$ with equal probability .<br>Find the probability that the strategy of always switching succeeds, given that Monty opens Door $2$.</p>
<p><strong>My approach:</strong><br>
Let $C_i$ be the event that car is behind the door $i$,
$O_i$ be the event that Monty opened door $i$ and $X_i$ be the event that intially I chose door $i$. Here $i=1,2,3,...,n$.<br>
Let's start with case where I chose $X_1$. Then:<br>
$P(O_{j_1, j_2, ..., j_m}|C_1, X_1) = {{n-1}\choose{m}}(\frac{1}{n-1})^m$, here $j \in$ {$m$ doors out of $n-1$, i.e., exclude Door$1$ }<br>
$P(O_{k_1, k_2, ..., k_m}|C_t, X_1) = {{n-2}\choose{m}}(\frac{1}{n-2})^m$,
here $k \in$ {$m$ doors out of $n-2$, i.e., exclude Door$1$ & Door$t$}, $t \in$ {$2,3, ..., n$}<br>
Also, $P(C_r|X_s) = \frac{1}{n}$, here $r,s \in$ {$1,2,...,n$}</p>
<p>Probability of winning by switching is,</p>
<p>$$P(C_3 | O_{k_1, k_2, ..., k_m}, X_1) = \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{m-doors}|X_1)}$$</p>
<p>$$= \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{j_1, j_2, ..., j_m}|C_1, X_1).P(C_1|X_1) + \sum_{t=2}^n(P(O_{k_1, k_2, ..., k_m}|C_t, X_1).P(C_t|X_1))}$$</p>
<p>$$ = \frac{{{n-2}\choose{m}}(\frac{1}{n-2})^m.\frac{1}{n}}{(\frac{1}{n}).({{n-1}\choose{m}}(\frac{1}{n-1})^m + {{n-2}\choose{m}}(\frac{1}{n-2})^m.(n-1))}$$</p>
<p>$$= \frac{(n-m-1)(n-1)^{m-1}}{(n-2)^m + (n-m-1)(n-1)^m}$$</p>
<p>However, the correct answer is $\frac{(n-1)}{(n-m-1).n}$. Any insights to what I have done wrong.</p>
| David K | 139,123 | <p>A correct answer to the problem has been posted.
But you asked where your approach went wrong.</p>
<p>Suppose the event $C_1 \cap X_1$ is given.
In that event, according to the rules of the game, Monty will open
$m$ doors.
Apparently you have no information that would make one set of doors more
likely to be opened than another, I assume every set of doors is equally likely.
(You mention $\frac12 \geq p \geq 1$ at one point, but it seems to be for the three-door problem, not the generalized problem, and you ignore it in your solution.)</p>
<p>Since there are $\binom{n-1}{m}$ distinct sets of doors that Monty could open, the probability to open the set
$J = \{j_1,j_2,\ldots,j_m\}$ (where $1\not\in J$), given $C_1 \cap X_1,$ is
$$P(O_{j_1, j_2,\ldots, j_m}\mid C_1\cap X_1) = \frac{1}{\binom{n-1}{m}}.$$</p>
<p>That's one error in your computation.</p>
<p>Notice that in the event $X_1,$ just like in the event $C_1\cap X_1,$
there are $\binom{n-1}{m}$ distinct sets of doors that Monty could open,
and every set of doors is equally as likely as any other.
(Every set is <em>not</em> equally likely when $C_2\cap X_1,$ but each of the events
$C_2,$ $C_3,$ $C_4,$ and so forth is as just as likely as $C_2$;
by symmetry, when all is added up, no subset of doors is more likely than another.) Therefore
$$P(O_{j_1, j_2,\ldots, j_m}\mid X_1) = \frac{1}{\binom{n-1}{m}}.$$</p>
<p>If $C_1 \cap X_t$ is given, where $t \neq 1,$
Monty must open $m$ of the remaining $n - 2$ doors.
Following reasoning similar to the above, but considering that there
are only $n-2$ available doors, the probability to open any particular set
$K = \{k_1, k_2, \ldots , k_m\},$ where $1 \not\in K$ and $t \not\in K,$ is
$$P(O_{k_1, k_2,\ldots, k_m}\mid C_1\cap X_t) = \frac{1}{\binom{n-2}{m}}.$$</p>
<p>That's another error in your computation.</p>
<p>Since Monty must open <em>some</em> set of $m$ doors no matter where the car is or which door you chose,
$P(O_\mathrm{m-doors} \mid X_1) = 1,$
not what you substituted. But in fact it was an error to put that
probability where you put it in the first place, because
$$
P(C_3 \mid O_{k_1, k_2,\ldots, k_m} \cap X_1)
\neq \frac{P(O_{k_1, k_2,\ldots, k_m} \mid C_3\cap X_1) P(C_3 \mid X_1)}
{P(O_\mathrm{m-doors} \mid X_1)}.
$$</p>
<p>A formula that you could have used is one that considers the <em>same</em> set of $m$ doors in all three places where $m$ doors occur in the equation,
\begin{align}
P(C_3 \mid O_{k_1, k_2,\ldots, k_m} \cap X_1)
&= \frac{P(O_{k_1, k_2,\ldots, k_m} \mid C_3\cap X_1) P(C_3 \mid X_1)}
{P(O_{k_1, k_2,\ldots, k_m} \mid X_1)} \\
&= \frac{\left(1 / \binom{n-2}{m}\right) (1/n)}
{\left(1 / \binom{n-1}{m}\right)} \\
&= \frac{n - 1}{n(n - m - 1)}.
\end{align}</p>
|
3,129,192 | <p>I've been trying to integrate
<span class="math-container">$$\int\frac{dx}{(x-a)(x-b)}$$</span></p>
<p>By using the substitution
<span class="math-container">$$x=a \cos^2 \theta + b \sin^2 \theta$$</span></p>
<p>The only problem here is I arrived at the result
<span class="math-container">$$\frac{2}{a-b} \ln |\csc 2\theta - \cot 2\theta|+c$$</span></p>
<p>and I having trouble on how to substitute from <span class="math-container">$\theta$</span> to <span class="math-container">$x$</span>.</p>
| tomasz | 30,222 | <p><strong>Hint</strong>: Since both spaces are clearly compact Hausdorff, it is enough to find a continuous bijection. Parametrise the annulus by polar coordinates, and then do some scaling.</p>
|
3,712,188 | <blockquote>
<p>Consider the vector field <span class="math-container">$\vec{u}=(xy^2,x^2y,xyz^2)$</span></p>
</blockquote>
<p>The curl of the vector field is <span class="math-container">$$\nabla \times\vec{u}=(xz^2,-yz^2,0)$$</span>
Consider the line integral of <span class="math-container">$\vec{u}$</span> around the ellipse <span class="math-container">$C$</span> <span class="math-container">$x^2+4y^2=1, z=-1$</span>.</p>
<p>With <span class="math-container">$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a=1, b=\frac{1}{2}$</span>, the parameterisation gives <span class="math-container">$$\vec{r}=(x,y,z)=(cos\theta,\frac{1}{2}sin\theta,-1)$$</span>
<span class="math-container">$$d\vec{r}=(-sin\theta,\frac{1}{2}cos\theta,0))$$</span></p>
<p><span class="math-container">$$\vec{u}=(\frac{1}{4}cos\theta sin^2\theta,\frac{1}{2}cos^2\theta sin\theta,\frac{1}{2}sin\theta cos\theta) $$</span>
<span class="math-container">$$\oint_{C} \vec{u} \cdot d\vec{r}=\frac{1}{4}\int^{2\pi}_0sin\theta cos\theta(cos^2\theta -sin^2\theta)d\theta=0$$</span>
(I got the same result by working in cartesian cooridnates without parameterizing the curve)</p>
<p>But this does not make sense because
<span class="math-container">$$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r}$$</span>
so if a vector field is conservative, its curl should be zero.</p>
<p>Can someone please explain where my conceptual errors lie? </p>
| hdighfan | 796,243 | <p>The reason why your approach seems to fail is because this expression is trivially just <span class="math-container">$0$</span>. I'm assuming there's been a transcription error, but if not, then we have <span class="math-container">$$\begin{align}&\lim_{n\to\infty}\frac{1}{n^{n+1}}\sum_{k=1}^nk^p\\
=&\lim_{n\to\infty}\frac1{n^{n-p}}\left(\frac 1n\sum_{k=1}^n\left(\frac kn\right)^p\right)\end{align}$$</span></p>
<p>where the right bracket is the integral expression you're looking for (which is bounded by <span class="math-container">$1$</span>), and the left goes to <span class="math-container">$0$</span>.</p>
<p>Perhaps your sum should be <span class="math-container">$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^nk^p$$</span> in which case you just want the right bracket, which is evaluatable as a Riemann Sum.</p>
|
3,712,188 | <blockquote>
<p>Consider the vector field <span class="math-container">$\vec{u}=(xy^2,x^2y,xyz^2)$</span></p>
</blockquote>
<p>The curl of the vector field is <span class="math-container">$$\nabla \times\vec{u}=(xz^2,-yz^2,0)$$</span>
Consider the line integral of <span class="math-container">$\vec{u}$</span> around the ellipse <span class="math-container">$C$</span> <span class="math-container">$x^2+4y^2=1, z=-1$</span>.</p>
<p>With <span class="math-container">$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a=1, b=\frac{1}{2}$</span>, the parameterisation gives <span class="math-container">$$\vec{r}=(x,y,z)=(cos\theta,\frac{1}{2}sin\theta,-1)$$</span>
<span class="math-container">$$d\vec{r}=(-sin\theta,\frac{1}{2}cos\theta,0))$$</span></p>
<p><span class="math-container">$$\vec{u}=(\frac{1}{4}cos\theta sin^2\theta,\frac{1}{2}cos^2\theta sin\theta,\frac{1}{2}sin\theta cos\theta) $$</span>
<span class="math-container">$$\oint_{C} \vec{u} \cdot d\vec{r}=\frac{1}{4}\int^{2\pi}_0sin\theta cos\theta(cos^2\theta -sin^2\theta)d\theta=0$$</span>
(I got the same result by working in cartesian cooridnates without parameterizing the curve)</p>
<p>But this does not make sense because
<span class="math-container">$$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r}$$</span>
so if a vector field is conservative, its curl should be zero.</p>
<p>Can someone please explain where my conceptual errors lie? </p>
| Kavi Rama Murthy | 142,385 | <p>Hint: <span class="math-container">$\int_1^{n} x^{p}dx= \sum\limits_{k=2}^{n}\int_{k-1}^{k}x^{p}dx$</span>. Note that <span class="math-container">$\int_{k-1}^{k}x^{p}dx$</span> lies between <span class="math-container">$k^{p}$</span> and <span class="math-container">$(k-1)^{p}$</span>. Using this, conclude that <span class="math-container">$\sum\limits_{k=1}^{n} k^{p}$</span> lies between <span class="math-container">$\int_1^{n} x^{p}dx+n^{p}$</span> and <span class="math-container">$\int_1^{n} x^{p}dx+1 -(n+1)^{p}$</span>. Of course <span class="math-container">$\int_1^{n} x^{p}dx=\frac {n^{p+1}-1} {p+1}$</span>. This should make it easy for you to find the limit for different values of <span class="math-container">$p$</span>. </p>
|
3,664,272 | <h2><strong>MOTIVATION</strong></h2>
<p>I am considering investing a significant amount of money into a raffle. In order to decide the number of entries I purchase, I would like to find probability distributions for the number of prizes I will win with respect to the number of entries I purchase.</p>
<h2><strong>HOW THE RAFFLE WORKS</strong></h2>
<p>Total entries: 1000</p>
<p>Winning entries (# of prizes): 20</p>
<p>How it actually works is in 20 rounds of 50 entries. </p>
<ul>
<li>Entries 1-50 have a 1/50 chance to win prize 1</li>
<li>Entries 51-100 have a 1/50 chance to win prize 2</li>
</ul>
<p>... </p>
<ul>
<li>Entries 951-1000 have a 1/50 chance to win prize 20</li>
</ul>
<p>The entry numbers are purchased in order, so technically if I can get entries 1-50 then I have a 100% chance to win prize 1. However, I don't expect I will be able to do this since many people will be trying to buy entries at the same time. For simplicity, perhaps we can just assume that my entries will be evenly distributed across all 20 rounds (see <strong>BONUS</strong> below for my thoughts on how this change impacts the solution and please correct me if I am wrong).</p>
<h2><strong>INITIAL THOUGHTS</strong></h2>
<p>From some quick research I think the estimate for my odds of winning ONE prize is approximately like this:</p>
<p>1 - [ (1000-n) / 1000 ]^20</p>
<p>where n = number of entries I purchase</p>
<h2><strong>WHAT I WANT TO KNOW</strong></h2>
<p>What I actually want is how to calculate the probability distribution of the number of prizes I win. So not just whether I win 1 prize or not.</p>
<p>Given n where n is the number of entries I purchase, I want to know the average (mean) number of prizes I should expect to win and the surrounding distribution. This way I can decide my risk tolerance and choose how many entries (n) it is worth it for me to buy.</p>
<h2><strong>BONUS</strong></h2>
<p>I mentioned we can simplify the problem to assume my entries will be even distributed across all 20 rounds, but I am curious what the optimal strategy would be if I could choose my entry numbers.</p>
<p>For example, if n = 100 entries, is it best to buy entries 1-100 and have a 100% chance to win 2 prizes? Or would having a more even distribution be better. For example, having 5 entries in each of the 20 rounds ?</p>
<p>In other words, I could have:</p>
<ul>
<li>100% chance to win in 2 rounds (win 2 prizes) and 0% chance to win in
the other 18 rounds</li>
<li>10% chance to win in all 20 rounds</li>
</ul>
<p>My understanding is that in both cases my expected number of wins is 2. The difference is that in the first case it is guaranteed whereas in the second place I could get lucky and win more or unlucky and win less. Correct?</p>
<p>Extrapolating from that, it seems like the more evenly distributed the entry numbers are across rounds, the more uncertainty in the number of prizes I will actually win. However, the expected number (mean) of the distribution should always be the same. Is this true?</p>
| Sirswagger21 | 785,596 | <p>Generally, you are correct in that the expected number of the distribution would more or less be the same. Obviously, going for a split in each is a high risk, high return probability.</p>
<p>The thing is, as you stated earlier, there is no way you will get a sure 100% for both raffles 1 and 2. Therefore, I estimate the highest probability you will get for 1 individual raffle is about 50%, although this could widely vary.</p>
<h1>5 Tickets in 20 Raffles</h1>
<p>Now, for some math. Let's calculate the probability you get <strong>less than 2 wins</strong> when investing 5 in each raffle.</p>
<p>For 1 win, it's <span class="math-container">$\binom{20}{1} \cdot (\frac{1}{10})^1 \cdot (\frac{9}{10})^{19} =$</span> 27.017%.</p>
<p>And for 0, it's 12.158%.</p>
<p>Adding them up, we get the total probability as 39.175%.</p>
<p>The probability of you <strong>getting 2 when investing 5 in each is 28.518%.</strong>, through a similar concept.</p>
<p>Now, to calculate the probability of getting <strong>more than 2</strong>, we just add the probabilities from 0 to 2 and subtract that sum from 1.</p>
<p>The probability is 1 - 0.67333 = 32.667%.</p>
<p>Summing everything up,</p>
<p>The probability of getting <strong>less than 2 wins</strong> is <strong>39.175%</strong>.</p>
<p>The probability of getting <strong>exactly 2 wins</strong> is <strong>28.518%</strong>.</p>
<p>The probability of getting <strong>more than 2 wins</strong> is <strong>32.667%</strong>.</p>
<p>As you can see, it's actually a larger chance of getting under 2 wins than above.</p>
<h1>10 Tickets in 10 Raffles</h1>
<p>Now, we calculate the probabilities for when you enter 10 raffles with 10 tickets each.</p>
<p>Similar reasoning, but just change up the numbers a bit.</p>
<p>For 1 win, it's <span class="math-container">$\binom{10}{1} \cdot (\frac{1}{5})^1 \cdot (\frac{4}{5})^9 =$</span> 26.844%.</p>
<p>And for 0, it's 10.737%.</p>
<p>Therefore, <strong>getting under 2 wins is 37.581%.</strong></p>
<p>Getting <strong>exactly 2 wins is 30.199%.</strong></p>
<p>And getting <strong>more than 2 wins is 1 - 0.67780 = 32.22%.</strong></p>
<p>Summing everything up,</p>
<p>The probability of getting <strong>less than 2 wins</strong> is <strong>37.581%</strong>.</p>
<p>The probability of getting <strong>exactly 2 wins</strong> is <strong>30.199%</strong>.</p>
<p>The probability of getting <strong>more than 2 wins</strong> is <strong>32.22%</strong>.</p>
<p>As you can see, investing <em>5 tickets in 20 raffles</em> gives you a <strong>higher chance of getting less than 2 wins</strong>, but also gives you a <strong>higher chance of getting more than 2 wins.</strong> However, the <strong>difference</strong> between the less than 2 wins percentage is much larger than the difference between the more than 2 wins percentage.</p>
<p>Using this data, make your own decision! Hope you win more than 2, at least :D</p>
<p>-FruDe</p>
<p>P.S. This was my first ever math answer on StackExchange, tell me what you think!</p>
|
2,904,359 | <p><strong>Definition.</strong> A formula <span class="math-container">$\phi(x,a)$</span> <em>divides</em> over a set <span class="math-container">$B$</span> if there are <span class="math-container">$k<\mathbb{N}$</span> and a sequence <span class="math-container">$(a_i)_{i<\omega}$</span> such that</p>
<p>(1) <span class="math-container">$\text{tp}(a/B)=\text{tp}(a_i/B)$</span>, for all <span class="math-container">$i<\omega$</span>;</p>
<p>(2) <span class="math-container">$\{\phi(x,a_i)\}_{i<\omega}$</span> is <span class="math-container">$k$</span>-inconsistent.</p>
<hr />
<p><strong>Definition.</strong> A formula <span class="math-container">$\phi(x,a)$</span> <em>forks</em> over a set <span class="math-container">$B$</span> if there are <span class="math-container">$n\in\mathbb{N}$</span> and formulas <span class="math-container">$\psi(x,b_1),\dots, \psi(x,b_n)$</span> such that</p>
<p>(1) for each <span class="math-container">$i=1,\dots, n$</span>, the formula <span class="math-container">$\psi_i(x,b_i)$</span> divides over <span class="math-container">$B$</span>;</p>
<p>(2) <span class="math-container">$\phi(x,a)\models \bigvee_{i=1}^{n} \psi_i(x,b_i)$</span>.</p>
<hr />
<p>It is clear that dividing implies forking.</p>
<p><strong>Question.</strong> Does forking always imply dividing? If no, is there a formula that forks but does not divide?</p>
| Mostafa Mirabi | 589,696 | <p>No, forking does not always imply dividing. In simple theories dividing and forking are the same, but they are not the same in general. Look at the following example. </p>
<p><strong>Example.</strong> Let $\mathcal{L}=\{ R^{(3)} \}$ be a language which consists of a ternary relation. Consider the $\mathcal{L}$-structure $\mathcal{M}=\big(\mathbb{S}^1, R \big)$ where $\mathbb{S}^1$ is the unit circle around the origin on the plane and $R(x,y,z)$ holds if and only if $y$ lies on the shorter arc between $x$ and $z$, ordered clock-wise, including the endpoints. Now, let $a,b$ and $c$ be three equidistant points on $\mathbb{S}^1$. Then</p>
<p>$$\mathcal{M}\models \forall x\bigg( R(a,x,b)\vee R(b,x,c) \vee R(c,x,a) \bigg)$$</p>
<p><strong>Claim(1).</strong> Each of $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$.</p>
<p>Proof of Claim(1).</p>
<p>We will show, for instance, the formula $R(a,x,b)$ 2-divides over $\emptyset$. Let $a, a_0,b_0, a_1,b_1,\dots, b$ be a sequence of consecutive points on $\mathbb{S}^1$.
Then $(a_ib_i)_{i<\omega}$ is a sequence such that </p>
<p>i) $\text{tp}(a_i,b_i)=\text{tp}(a,b)$ for each $i<\omega$;</p>
<p>ii) $\big\{ R(a_i,x,b_i) \big\}_{i<\omega}$ is 2-inconsistent.</p>
<p>Therefore $R(a,x,b)$ 2-divides over $\emptyset$. Similarly we can prove that $R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$ as well.</p>
<p><strong>Claim(2).</strong> The formula $x=x$ forks over $\emptyset$ but does not divide over $\emptyset$.</p>
<p>Proof of Claim(2). </p>
<p>Since $x=x\models R(a,x,b)\vee R(b,x,c)\vee R(c,x,a)$ and by Claim(1) each of $R(a,x,b), R(b,x,c)$ and $R(c,x,a)$ 2-divides over $\emptyset$, $x=x$ forks over $\emptyset$. But $x=x$ does not divide over $\emptyset$, because if $x=x$ divides over $\emptyset$, then there are $k<\omega$ and an $\emptyset$-indiscernible sequence $(a_i)_{i<\omega}$ such that $\big\{ x=x \big\}_{i<\omega}$ is $k$-inconsistent which is a contradiction. </p>
<p>Therefore the formula $x=x$ forks but does not divide.</p>
|
284,995 | <p>There is a theorem that if a Hopf algebra $H$ is commutative or cocommutative, then $S^2=id_H$, where $S$ denotes the antipode.</p>
<p>May I know if the converse is true?</p>
<p>(i.e. if $S^2=id_H$, does it follow that $H$ is commutative or cocommutative?)</p>
<p>If no, what would be the simplest counter-example?</p>
<p>Sincere thanks for any help!</p>
| JP McCarthy | 19,352 | <p>The Eight Dimensional Kac-Paljutkin Hopf Algebra has an involutive antipode but is neither commutative nor cocommutative.</p>
|
61,174 | <p>I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers. First, take the numbers between zero and one, exclusive. The following sequence of real numbers is suggested so that we have bijection. </p>
<p>0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.</p>
<p>Obviously, this includes repeats, but this set is countable. Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable. Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable. </p>
<p>Please help me with this. I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.</p>
<p>Thanks.</p>
| hmakholm left over Monica | 14,366 | <p>Your function ignores all the real numbers whose decimal representations are not finite, such as </p>
<p>$\dfrac13=0.3333\ldots$</p>
<p>The subset of real numbers that do have finite decimal representations is indeed countable (also because they are all rational and $\mathbb Q$ is countable).</p>
|
61,174 | <p>I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers. First, take the numbers between zero and one, exclusive. The following sequence of real numbers is suggested so that we have bijection. </p>
<p>0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.</p>
<p>Obviously, this includes repeats, but this set is countable. Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable. Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable. </p>
<p>Please help me with this. I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.</p>
<p>Thanks.</p>
| Liam Gavin Murray | 217,943 | <p>The set you have shown is a list of all rationals between $0$ and $1$ that can be written in the form $x /10^n$ with $x\in\mathbb{Z}$, which is countable. But the full set of reals between $0$ and $1$ is bigger.</p>
<p>All reals are the limit of some sub-sequence of this sequence, but not all are in this sequence, e.g. $\sqrt{2}=1.14142\ldots$ or $\frac{1}{3}=0.33333\ldots$.</p>
|
2,939,605 | <p>The given task goes as follows:</p>
<blockquote>
<p>Show that <span class="math-container">$ f: \mathbb{R} \longrightarrow \mathbb{R}$</span> defined by <span class="math-container">$f(x) = \sqrt{1 + x^2} $</span> is not a polynomial function.</p>
</blockquote>
<p>I tried this approach - if <span class="math-container">$f(x)$</span> is a <span class="math-container">$n$</span>-degree polynomial function, then the <span class="math-container">$(n+1)$</span>-st derivative equals to 0 and I was trying to determine the <span class="math-container">$k$</span>-th derivative of <span class="math-container">$f(x)$</span> (and show it differs from 0 for any <span class="math-container">$k$</span>) but without success. Since <span class="math-container">$f(x)$</span> is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas? </p>
| dxiv | 291,201 | <p>Alt. hint: following up on OP's idea to use derivatives, note that <span class="math-container">$\,f(x) \cdot f'(x) = x\,$</span>, but the only polynomials which satisfy that identity are <span class="math-container">$\,f(x)=\pm x\,$</span> (<em>why?</em>).</p>
<p>Alternatively, show that there exists no polynomial <span class="math-container">$\,f\,$</span> such that <span class="math-container">$\,f^2(x)=x^2+1\,$</span>.</p>
|
2,846,700 | <p>Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for $f(x) + g(x)$ to be </p>
<ol>
<li>even;</li>
<li>odd?</li>
</ol>
<p>For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.</p>
<p>It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.</p>
<p>I've tried to do it the following way:
Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:</p>
<p>$$
h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\
g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function}
$$</p>
<p>I'm stuck at that point.</p>
<blockquote>
<p>How can I prove/disprove that $f(x) + g(x)$ may be even?</p>
</blockquote>
| Florian | 185,854 | <p>Well, by definition $f(x)+g(x)$ is even if and only if $f(x)+g(x)=f(-x)+g(-x)$ for all $x$. If $g(x)$ is even, it satisfies $g(x)=g(-x)$ for all $x$. Inserting this into the previous equation gives $f(x)=f(-x)$ for all $x$, hence $f$ must be even.</p>
<p>Why did it work for odd? Here the definition requires $f(x)+g(x)=-f(-x)-g(-x)$. Since $g(x)$ is even, this transforms to $f(x)+g(x)=-f(-x)-g(x)$ i.e. $g(x) = -\frac 12 (f(x)+f(-x))$, which your example satisfies. In other words: if $f$ is not even, <em>subtracting</em> its even <em>part</em> [*] makes it odd.</p>
<p>[*] This refers to the fact that you can decompose every function $f(x)$ into its even part $f_e(x)$ and its odd part $f_o(x)$ such that $f(x) = f_e(x)+f_o(x)$, where $f_e(x) =\frac 12 (f(x)+f(-x))$ and $f_o(x) =\frac 12 (f(x)-f(-x))$.</p>
|
72,098 | <p>This is my first time using Stack Exchange, but it looks like a good resource. I am here to ask a couple questions about my homework. We're working from the latest edition of <em>Abstract Algebra</em> by Herstein. This is problem #3 from p. 73 of that book, and it reads as follows:</p>
<blockquote>
<p>Let $G$ be any group and $A(G)$ the set of all 1-1 mappings of $G$, as a set, onto itself. Define $L_{a} \colon G \to G$ by $L_{a}(x)=xa^{-1}$. Prove that:<br>
(a) $L_{a} = A(G)$.<br>
(b) $L_{a}L_{b} = L_{ab}$.<br>
(c) The mapping $\psi:G \rightarrow A(G)$ defined by $\psi(a) = L_{a}$ is a monomorphism of G into A(G).</p>
</blockquote>
<p>I believe that I have answered parts b & c correctly, but have some concerns about the rigorousness of my approaches to all three problems. I will start out here by including my proposed solution to part a. Any advice or comments would be greatly appreciated.</p>
<p>(a) $A(G)$, as the set of all 1-1 mappings of $G$ onto itself, can be represented as the set of all operations on an element $x \in G$ such that the result is also in $G$. Since $G$ is a group, we know that this set can be represented as the set of all group multiplications $yx \mid x \in G, y \in G$ for a given element $x$. This is because any $x$ element can be fixed as the first parameter, while the $y$ elements are taken over every element of G. We have then that $yx \in G$, due to G's closure under group multiplication. Furthermore, any element $e \in G$ has an inverse element $e^{-1} \in G$, since $G$ is a group. This means that we can consider any element in $G$ as the inverse of its inverse: $e = (e^{-1})^{-1}$. This, when plugged in for our variable $x \in G$ above, gives us that $\forall x \in G, \forall y \in G, yx^{-1} \in G$. This is exactly our given mapping $L_{a}$ above, with the labels rearranged. This shows that $L_{a}$ is a mapping from $G \rightarrow G$, as required. In order to show why this set contains every such possible mapping, we will assume that there is a mapping $M_{a}(x) : G \rightarrow G$ such that $M_{a}(x) \notin L_{a}$. This mapping, as a mapping from G to G, must take the form of a group multiplication: $M_{a}(x) = x*a, x\in G$. However, since G is a group, we have again that $a = (a^{-1})^{-1}$, or, if we let $m = a^{-1}$, that $a = m^{-1}$, and so our mapping can be rewritten as $M_{a}(x) = x*m^{-1} m \in G$. However, this is exactly the same mapping as our above $L_{a}(x)$, showing that every mapping in $A(G)$ can indeed be written as a multiplication between some element $x \in G$ and another element's inverse $a^{-1} \in G$, and thus that $L_{a} \in A(G)$. $\blacksquare$</p>
<p>Thanks for taking the time to check this out, even if you don't feel you can offer any help.
<strong>EDIT</strong>: Thanks to those who pointed out that I had used =, not $\in$, above by mistake. Also appreciated is the edit to italicize my variables. Now I know how to as well. :) </p>
<p><strong>EDIT</strong>: The responses so far have been so helpful, I'd like to put the other two parts of my solution up to solicit feedback on them as well. Hopefully they aren't as muddled as the first part's was.<br>
<strong>changed a bit in response to feedback</strong><br>
(b)
$L_{a}(x) = xa^{-1}$<br>
$L_{b}(x) = xb^{-1}$<br>
$(L_{a}L_{b})(x) = L_{a}(L_{b}(x))$<br>
$L_{a}(L_{b}(x)) = L_{a}(xb^-1)$<br>
$L_{a}(xb^-1) = xb^{-1}a^{-1}$<br>
$xb^{-1}a^{-1} = x(ab)^{-1} = L_{ab}(x)$<br>
$L_{a}(x)L_{b}(x) = L_{ab}(x) \quad\blacksquare$ </p>
<p>(c)
$\psi(a) = L_{a}(x) = xa^{-1}$. To show that this mapping is a monomorphism of $G$ into $A(G)$, we will first rely on part (a) to state that $\psi(a) = L_{a}$ is indeed a mapping from $G$ to $A(G)$. Now we must show that $\psi$ is a monomorphism of $G$ into $A(G)$. First we will show that $\psi$ is a homomorphism of $G$ into $A(G)$. To this end, we will appeal to the results of our calculations in part (b) to state that $L_{a}L_{b} = L_{ab}$, which of course implies that $\psi(a)\psi(b) = L_{a}L_{b} = L_{ab} = \psi(ab)$, which proves that $\psi$ is a homomorphism. In order to continue and show that $\psi$ is a monomorphism, we must show that it is an injective (1-1) mapping. Let $\psi(a) = Z = \psi(b)$. This can be written as:
$\psi(a) = L_{a}(e) = ea^{-1} = Z$<br>
$\psi(b) = L_{b}(e) = eb^{-1} = Z$<br>
$Z = ea^{-1} = eb^{-1}$<br>
$a^{-1} = b^{-1} \Rightarrow a = b$<br>
The last line of the above follows from the uniqueness of inverse elements in $G$. This shows that the only way for two output values of $\psi$ to be equal is for their inputs to be equal as well, and thus $\psi$ is an injective homomorphism, or a monomorphism from $G$ to $A(G)$. $\blacksquare$</p>
| Zev Chonoles | 264 | <p>Take a look at the problem again:</p>
<p><img src="https://i.stack.imgur.com/CRArC.png" alt="enter image description here"></p>
<p>Part a) says that
$$L_a\in A(G),$$
where the symbol in the middle denotes "is an element of" (<a href="http://en.wikipedia.org/wiki/Element_(mathematics)#Notation_and_terminology" rel="nofollow noreferrer">see here</a>), not
$$L_a=A(G).$$
The latter expression doesn't make sense; $L_a$ is a function from $G$ to $G$, while $A(G)$ is a collection of functions. We want to prove that $L_a$ is a member of this collection.</p>
|
2,735,854 | <p>How many permutations of the word $STRESSLESSNESS$ begin OR end with an $E$? </p>
<p>Correct me if I'm wrong, but you would have to subtract the permutations where $E$ begins AND ends the permutation?</p>
| Dan | 508,768 | <p><em>STRESSLESSNESS</em> is 14 letters long.</p>
<p>If we take away an <strong>e</strong>, it becomes 13 letters long.</p>
<p>The permutations of this 13 letter string is:</p>
<p>$$\frac{13!}{7!2!}$$</p>
<p>Since the <strong>E</strong> can occur at the beginning or the end, we multiply this by 2 to get:
$$\frac{13!}{7!}$$</p>
<p>We then take into account a possibility of an E at the beginning and at the end:</p>
<p>$$\frac{13!}{7!}-\frac{12!}{7!}=1140480$$</p>
|
186,310 | <p>I am trying to evaluate: </p>
<pre><code>For[i = 1, i < 100000, i++, h[0] = hs;
h[i]=hi /. Check[FindRoot[Xfnew[a, w, hi, mxs + i*step]*92*10^23) == 1198/10000, {hi, h[i - 1]}]];
If [h[i] == 25, Break[], AppendTo[hin, {mxs + i*step, h[i]}]]];
</code></pre>
<p>where mxs , hs, and step has some values, and hin is initially empty list.</p>
<p>The reason I am using Findroot in a loop, is that my function of Xfnew is seriously complicated, and Findroot can only find solutions very close to the actual solution. So I cannot evaluate Findroot over a wide range. So I am using small steps, and using the previous step's result as the starting point for FindRoot.</p>
<p>If FindRoot cannot evaluate this, then it returns some error message, and the Check option sets the solution to 25, in which case I break the for loop, otherwise it keeps adding to the list.</p>
<p>I want to know if there is a faster way to do the same thing in Mathematica, (like using Table, or other functions which generally people recommend using, over for loop).</p>
<p>Thank you for your help. </p>
| Titus | 30,703 | <p>An example with Table. <code>f[x_]</code> is a quadratic and every step the constant term changes. The idea is to store the last solution every iteration and use it in next i's solution, THEN store the new solution. I have tried it in various setups (including this) and it works, but feel free to check in your application.</p>
<pre><code>actualroot:=k (* set initially e.g. = 0*)
f[x]:= x^2 - 3 x - a
Table[actualroot=FindRoot[f[x]==0, {x, actualroot}][[1, 2]], {a, 1, 25}]
</code></pre>
<p>Output is</p>
<blockquote>
<p>{-0.302776, -0.561553, -0.791288, -1., -1.19258, -1.37228, -1.54138,
-1.70156, -1.8541, -2., -2.14005, -2.27492, -2.40512, -2.53113,
-2.65331, -2.772, -2.88748, -3., -3.10977, -3.21699, -3.32183,
-3.42443, -3.52494, -3.62348, -3.72015}</p>
</blockquote>
<p>A note: I do not know your original function, but for simplicity my step is to increase a by 1 every iteration, which is a term in f(x). It is straightforward to apply it to your case, I think.</p>
|
2,698,284 | <p>Here is the problem: </p>
<p>Let A be a set with n elements. Find an expression for S(n, 2), the number of partitions of A into exactly two subsets. You can either start with the general recurrence for S(n, k), or count S(n, 2) directly. </p>
<p>I'm having trouble understanding what exactly it wants me to do. So far I know that it wants me to find an expression that gives the number of combinations of A into sets like (x, y). I got the equation: (n - 1) * 2 for number of partitions of A into two parts, but I don't know understand what the problem means by creating two subsets. What am I doing wrong?</p>
| Ross Millikan | 1,827 | <p>Choose one subset of $A$. The other subset in the partition is everything else. You then need to divide by $2$ because you could have chosen the other subset first and gotten the original subset as the everything else.</p>
|
2,698,284 | <p>Here is the problem: </p>
<p>Let A be a set with n elements. Find an expression for S(n, 2), the number of partitions of A into exactly two subsets. You can either start with the general recurrence for S(n, k), or count S(n, 2) directly. </p>
<p>I'm having trouble understanding what exactly it wants me to do. So far I know that it wants me to find an expression that gives the number of combinations of A into sets like (x, y). I got the equation: (n - 1) * 2 for number of partitions of A into two parts, but I don't know understand what the problem means by creating two subsets. What am I doing wrong?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Suppose your set is $\{ 1,3,8\}$
Our partitions are $$\{\{1\},\{3,8\}\}\\ \{\{3\},\{1,8\}\}\\ \{\{8\},\{1,3\}\} $$</p>
<p>Each partition has 2 disjoint sets whose union is the original set.</p>
|
2,655,178 | <p>I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$. </p>
<p>I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as. </p>
<p>If anyone could provide their working out for this problem it would be extremely enlightening. </p>
| amd | 265,466 | <p>An equation of the cubic that passes through the four points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and $(x_4,y_4)$ is $$\begin{vmatrix} x^3 & x^2 & x & y & 1 \\ x_1^3 & x_1^2 & x_1 & y_1 & 1 \\ x_2^3 & x_2^2 & x_2 & y_2 & 1 \\ x_3^3 & x_3^2 & x_3 & y_3 & 1 \\ x_4^3 & x_4^2 & x_4 & y_4 & 1 \end{vmatrix} = 0.$$ Plug in the coordinates of your points and simplify.</p>
|
212,466 | <p>Consider the list <code>list={1,2,3,4,5,6,7,8}</code> and imagine I wanted to change it to <code>{{1,2,3},{4,5,6},{7,8}}</code>. How can I do this?</p>
<p>Using <code>ArrayReshape</code> in the usual manner</p>
<pre><code>list = {1, 2, 3, 4, 5, 6, 7, 8};
ArrayReshape[list, {3, 3}]
</code></pre>
<p>yields <code>{{1, 2, 3}, {4, 5, 6}, {7, 8, 0}}</code>. Is it possible to use <code>ArrayReshape</code> and avoid that extra 0 at the end? Essentially, I want to reshape my list according to its size.</p>
| MarcoB | 27,951 | <p>Using <code>Partition</code> with <code>UpTo</code> is clearly the most expedient answer here, as shown above. However, this got me wondering how one could force <code>ArrayReshape</code> into service anyway, just for fun. For instance, we could twist the use of the padding parameter in <code>ArrayReshape</code> to force it to pad with <code>Nothing</code> (or the <a href="https://mathematica.stackexchange.com/a/3705/27951">vanishing function</a> <code>##&[]</code>), as in either of the following, which is then removed when evaluated:</p>
<pre><code>ArrayReshape[Range[8], {3, 3}, Hold@Nothing] // ReleaseHold
ArrayReshape[Range[8], {3, 3}, Inactive@Nothing] // Activate
</code></pre>
<blockquote>
<p><code>{{1, 2, 3}, {4, 5, 6}, {7, 8}}</code> </p>
</blockquote>
<hr>
<p>After posting the answer, I refreshed the page to see that this was <a href="https://mathematica.stackexchange.com/questions/212466/avoiding-zeros-in-arrayreshape/212476#comment544159_212466">brought up in comments as well</a>.</p>
|
1,390,593 | <p>"Since the operation of left multiplication is faithful, $G$ is isomorphic to its image in $\operatorname{Perm}(G)$. If $G$ has order $n$, $\operatorname{Perm}(G)$ is isomorphic to $S_n$." - Artin's Algebra</p>
<p>So left multiplication as an action means $\phi_g(x)=gx$</p>
<p>So proving isomorphism, means bijective homomorphism, but I don't understand:</p>
<p>Injective: $ga_1=b$ and $ga_2=b$ $\implies a_1=g^{-1}b=a_2,\square$</p>
<p>Surjective: $\forall b\in G, ga=b$ where you can let $a=g^{-1}b$ where clearly $g^{-1}b\in G$ whenever $g,b\in G$. $\square$</p>
<p>But then homomorphism fails since $\phi_g(ab)=gab\ne\phi_g(a)\phi_g(b)=gagb$</p>
| Bernard | 202,857 | <p>You're confusing $G$ as a group and $G$ as left-acting on itself. The homomorphism is this:
\begin{align*}
\varphi\colon G&\to \operatorname{Perm}G\\
g&\mapsto (\varphi_g\colon a\mapsto ga)
\end{align*}</p>
<p>It is a homomorphism because $\;\varphi_{gg'}\colon a\mapsto(gg')a=g(g'a)=\varphi_g\bigl(\varphi_{g'}(a)\bigr)$, hence $\;\varphi_{gg'}=\varphi_g\circ\varphi_{g'}$.</p>
<p>It is injective because $\;\varphi_g=\operatorname{id}_G\;$ means that, if $ga=a$ for all $a\in G$ (actually one $a$ is enough), then $g=e$.</p>
|
2,054,338 | <p>I had this question on my exam and I thought I could solve it using the row echelon method. Well, I couldn't, and I still don't know how to solve it. We were asked to "determine $a$ such the equation system has a unique solution" and to determine $a$ itself. The system had three equations: </p>
<p>$2x - y + az = 3$ </p>
<p>$3x - 4y + 2az = 1$</p>
<p>$x + y - z = 2$.</p>
<p>Could you provide me with a step-by-step of how to solve this?</p>
<p>Thank you</p>
| Robert Israel | 8,508 | <p>Hint: divide by $a$ and integrate.</p>
|
2,411,811 | <p>Determine the limit:</p>
<p>$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}$$</p>
<p>I know that the answer is $0$, I am just unsure of how to solve this. Thanks.</p>
| imranfat | 64,546 | <p>Let's go completely NUTS on this one. Let $2h=x$ so that the limit becomes$$2\lim_{x\to 0}\frac{1-\cos(x)}{x}$$ Now let $x=\pi/2-t$ so that the limit becomes$$2\lim_{t\to \pi/2}\frac{1-\cos(\pi/2-t)}{\pi/2-t}=2\lim_{t\to \pi/2}\frac{sint-1}{t-\pi/2}$$
Now apply the limit definition of the derivative for $sine$ at $t=\pi/2$ to arrive at $2\cos\pi/2=0$</p>
|
2,607,733 | <p>Prove the inlcusion-exclusion principle using the fact that for three pairwise disjoint sets $X$, $Y$, $Z$:</p>
<p>$$|X\cup{Y}\cup{Z|}=|X|+|Y|+|Z|$$</p>
<p>I tried setting $X\cup{Y}=A$, but arrive at $|X\cup{Y|}=|X|+|Y|$ due to the sets being disjoint.</p>
| Community | -1 | <p>Other answers are pointing to the formal reasons why $d(x,y)=\|x+y\|$ is not a metric. However, I would also suggest that you build your geometrical intuition here.</p>
<p>In geometry, when you have points $A$ and $B$ and the origin $O$ in, say, a plane, you would calculate the vector $\vec{AB}=\vec{OB}-\vec{OA}$ and then you would use the ordinary vector norm to calculate the distance of $A$ to $B$:</p>
<p>$$d(A, B)=\|\vec{AB}\|=\|\vec{OB}-\vec{OA}\|=\|\vec{B}-\vec{A}\|$$</p>
<p>(if we agree to not write $O$ and just identify any point $X$ with the vector $\vec{X}=\vec{OX}$).</p>
<p><strong>That</strong> is where the minus sign comes from. It would not make much sense to use the plus sign ($\|\vec{OB}+\vec{OA}\|$). You can calculate with it, but you certainly don't have a clear geometrical intuition behind it.</p>
|
1,098,587 | <blockquote>
<p>Prove that for a bipartite graph $G$ on $n$ vertices the number of edges in $G$ is at most $\frac{n^2}{4}$. </p>
</blockquote>
<p>I used induction on $n$. </p>
<p>Induction hypothesis: Suppose for a bipartite graph with less than $n$ vertices the result holds true. </p>
<p>Now take a bipartite graph on $n$ vertices.Let $x,y$ be two vertices in $G$ where an edge exist between $x$ and $y$. Now remove these two vertices from $G$ and consider this graph $G'$. $G'$ has at most ${(n-2)^2}\over4$. Add these two vertices back. Then the number of edges $G$ can have is at most </p>
<p>$$|E(G')|+d(x)+d(y)-1$$ </p>
<p>My question is in my proof I took $d(x) + d(y) \le n$, where $d(x)$ denotes the degree of vertex $x$. Can I consider $d(x)+d(y) \leq n$? I thought the maximum number of edges is obtained at the situation $K_{\frac n 2,\frac n 2}$</p>
| abhishek kumar | 589,020 | <p>suppose the bipartite graph has m vertices in set1 and n vertices in set2 . the sum of degrees of all vertices would be 2 * e= 2 * m * n
thus, e = m * n
and m+n = v
so
n*(v-n)= e
solve this quadratic equation and discriminant . </p>
|
3,166,201 | <blockquote>
<p>If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are prime numbers which satisfy <span class="math-container">$x^2-2y^2=1$</span>, solve for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
</blockquote>
<p><strong>My attempt</strong>: </p>
<p><span class="math-container">$x^2-2y^2=1$</span></p>
<p><span class="math-container">$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$</span></p>
<p><span class="math-container">$\implies (x+\sqrt{2}y)=1$</span> and <span class="math-container">$(x-\sqrt{2}y)=1$</span></p>
<p><span class="math-container">$\implies x=1$</span> and <span class="math-container">$y=0$</span></p>
<p>Clearly <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?</p>
| Dr. Mathva | 588,272 | <p>What about </p>
<blockquote>
<p><span class="math-container">\begin{align*}&x^2-2y^2=1\tag{1}\\\iff & x^2-1=(x+1)(x-1)=2y^2\end{align*}</span></p>
</blockquote>
<p>Since <span class="math-container">$2\mid (x+1)(x-1)$</span>, we conclude that both <span class="math-container">$(x+1)$</span> and <span class="math-container">$(x-1)$</span> have to be even, and hence <span class="math-container">$$4\mid 2y^2\implies 2\mid y^2\implies 2\mid y$$</span> and since <span class="math-container">$y$</span> is prime, <span class="math-container">$\color{red}{y=2}$</span>. Can you end it now?</p>
<p>From (1), it follows immediately that <span class="math-container">$x^2-2y^2=x^2-8=1$</span>. Thus,
the only solution is <span class="math-container">$\color{blue}{(3, 2)}$</span>.</p>
<hr>
<p><strong><em>Addendum</em></strong></p>
<p>The problem with your method is that for <span class="math-container">$a,b\in\mathbb R$</span></p>
<p><span class="math-container">$$a·b=1\not\Rightarrow a=1\;\text{ and }\;b=1$$</span></p>
<p>In fact, this only works if <span class="math-container">$$a·b=0\implies a=0\;\text{ or }\;b=0$$</span></p>
|
356,279 | <p><strong>Definitions</strong> (<a href="https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467" rel="nofollow noreferrer">https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467</a>):</p>
<p>Given a planar convex region <span class="math-container">$C$</span> (could be smooth or polygonal), an <strong>area bisector</strong> of <span class="math-container">$C$</span> is any line that partitions <span class="math-container">$C$</span> into 2 pieces of equal area. A <strong>'fair bisector'</strong> is a line that partitions <span class="math-container">$C$</span> into 2 pieces of <em>equal area and equal perimeter</em>.</p>
<p>Thru every point on the boundary of <span class="math-container">$C$</span>, an area bisector can be drawn (for a description of their properties, please see 'Mathematical Omnibus' by Fuchs and Tabachnikov, Lecture 11). But it can be seen that a convex planar region can have just a single fair bisector (eg. for a thin isosceles triangle, the only fair bisector is the bisector of its apex angle) or a finite number of them (in which case, their number is necessarily odd as can be seen from simple continuity arguments; see reference at the top) or infinitely many.</p>
<p><strong>Observations:</strong> For regions with a center of symmetry such as a circular disk or ellipse or regular polygon with even number of sides, all fair bisectors are concurrent. But, numerically, we see that for a general convex region <span class="math-container">$C$</span> with finitely many fair bisectors, the fair bisectors are not necessarily concurrent but usually very close to being so. Clearly,for a general <span class="math-container">$C$</span> with exactly 3 fair bisectors, they determine a small triangular region deep in the interior of <span class="math-container">$C$</span>. For <span class="math-container">$C$</span>s with more fair bisectors, their many possible intersections will divide the interior of <span class="math-container">$C$</span> into many regions. Let us refer to the union of those regions which do not share the outer boundary of <span class="math-container">$C$</span> as the 'core' of <span class="math-container">$C$</span>.
The core must lie deep inside <span class="math-container">$C$</span>.</p>
<p><strong>Questions:</strong></p>
<ol>
<li><p>For which convex shape of <span class="math-container">$C$</span> is the area of the 'core' of <span class="math-container">$C$</span> the largest as a fraction of the area of <span class="math-container">$C$</span>? Intuitively, a relatively large core is a measure of the asymmetry of <span class="math-container">$C$</span>. Can one say (say) that such a shape is always one with exactly 3 fair bisectors?</p>
</li>
<li><p>Generalizing a bit, what about lines that break off the same fraction <span class="math-container">$t$</span> of the area and outer boundary length of <span class="math-container">$C$</span>? For a circular disk, it appears that only for <span class="math-container">$t=1/2$</span>, we have such lines (any diameter). Are there <span class="math-container">$C$</span>'s for which such lines exist for several (maybe even arbitrarily many) different values of <span class="math-container">$t$</span>?</p>
<p>Guess: All centrally symmetric convex regions (rectangles, ellipses,...) appear to give such only one single partitioning line that divides both area and outer perimeter in same ration - only for <span class="math-container">$t=1/2$</span>. But general convex regions with no symmetry might give infinitely many such lines - one such partitioning line for each orientation - and a different value <span class="math-container">$t$</span> for each orientation. And the set of these lines might even have interesting envelopes.</p>
</li>
</ol>
<p>These questions have obvious higher dimensional analogs.</p>
| Joseph O'Rourke | 6,094 | <p>This is not an answer, and not even that helpful, but
I wanted to see the central pattern formed by the collection of
perimeter bisectors.
<hr />
<img src="https://i.stack.imgur.com/W8N0K.jpg" width="250" />
<br />
<img src="https://i.stack.imgur.com/d0u0l.jpg" width="250" /></p>
<hr />
|
1,612,859 | <p>Take $i(G)$ to be the independence number of $G$, i.e. the maximum
number of
pairwise nonadjacent vertices in $G.$ I want to show that if $G$ has
$n$ vertices and $\frac{nk}{2}$ edges where $k \geq 1,$
$i(G) \geq \frac{n}{k+1}.$ I'm having difficulties making the
combinatorial argument on this graph. I get that if the graph has
$\frac{nk}{2}$ edges, $2|n \vee 2|k.$ However, beyond
this, I am having few insights. I am wondering, how might I best
represent pairwise nonadjacent vertices using this knowledge of
the number of edges in this graph?</p>
| Leen Droogendijk | 95,972 | <p>If you are allowed to use Turan's theorem, a simpler solution is possible.</p>
<p>$\overline G$ has $e=\binom n2-\frac{nk}2=\frac n2(n-1-k)$ edges.
Now $i(G)\geq\frac n{k+1}$ $\iff$ $\overline G$ has a clique of size at least $\frac n{k+1}$.
From Turan we know that avoiding a clique of size $r+1$ requires $e\leq(1-\frac1r)\frac{n^2}2$.</p>
<p>So we solve $\frac n2(n-1-k)\leq(1-\frac1r)\frac{n^2}2$, which is equivalent to $r\geq\frac n{k+1}$.</p>
<p>So (with $r=\lceil \frac n{k+1}\rceil-1$), avoiding a clique of size $\lceil \frac n{k+1}\rceil$ requires $r\geq\frac n{k+1}$.</p>
<p>Since this is impossible a clique of size $\lceil \frac n{k+1}\rceil$ must exist in $\overline G$, so an independent set of that size must exist in $G$.</p>
|
549,411 | <p>How would I go about generalizing the product rule to the product of $n$ functions $\psi_1(x), \ \psi_2(x), ..., \ \psi_n(x)$? That is, I'm hoping to obtain an expression for</p>
<p>$$
\frac{d}{dx} \prod_{j = 1}^n \psi_j(x)
$$</p>
| Marius Damian | 118,834 | <p>Suppose that exists $m,n\in\mathbb{N^*}$ such that $[n\sqrt{2}]=[(2+\sqrt{2})m]=t\in\mathbb{N^*}.$
Then $t<n\sqrt{2}<n+1,\quad t<(2+\sqrt{2})m<t+1\implies \dfrac{t}{\sqrt{2}}<n<\dfrac{t+1}{\sqrt{2}},\quad \dfrac{t}{2+\sqrt{2}}<m<\dfrac{t+1}{2+\sqrt{2}}\stackrel{(+)}{\implies} t<n+m<t+1,\;\text{false}\implies A\cap B=\emptyset.$</p>
|
631,348 | <p>Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?</p>
<p>P.S. using Kummer's test doesn't count</p>
| ziang chen | 38,195 | <p>If $L>1$, choose $\epsilon\gt 0$, $L-\epsilon>1$ then </p>
<p>$$1-\frac{L-\epsilon}{n} > \frac{a_{n+1}}{a_n}$$</p>
<p>Choose $p$ such that $1\lt p\lt L-\epsilon$, $\sum\frac1{n^p}$ converges. $b_n=\frac1{n^p}$, if $n$ big enough, then</p>
<p>$$\frac{b_{n+1}}{b_n}=(1-\frac{1}{n+1})^p=1-\frac{p}{n}+O(\frac1{n^2})\gt 1-\frac{L-\epsilon}{n}> \frac{a_{n+1}}{a_n}$$ </p>
<p>so $\sum a_n$ converges</p>
|
512,960 | <p>Let's say I have a statement: if p then q.<br>
The converse would be: if q then p.<br>
The inverse would be: if not p then not q.<br>
The contraposition would be: if not q then not p.<br>
What would you call the following? if not p then q.<br>
Thanks.</p>
| Rustyn | 53,783 | <p>$$
p \Rightarrow q \equiv \lnot p \lor q $$</p>
<p>$$
\lnot p \Rightarrow q \equiv p\lor q
$$
It is logically equivalent to "$p$ or $q$"</p>
|
2,202,048 | <p>There's a box with $N$ balls in it. One of them is red and the others white. What's the probability of getting the red ball at the $k$th try ( if you're not putting them back in?) where $k = 1,2,3,\ldots,N$</p>
| Sum | 369,413 | <p>The answer will be straightforward,</p>
<p>If the red ball has to come at the $k^{th}$ instance, we must draw white balls till the $(k-1)^{th}$ instance. Probability of getting white balls till $(k-1)^{th}$ instance will be:</p>
<p>$$\frac{N-1}{N} \frac{N-2}{N-1} \cdots \frac{N-k}{N-k+1} = \frac{N-k}{N}$$</p>
<p>Probability of getting red ball at the $k^{th}$ instance will be, $\frac{1}{N-k}$</p>
<p>Probability of getting consecutive white balls will be $1$.</p>
<p>Overall probability,</p>
<p>$$\frac{N-k}{N} \cdot \frac{1}{N-k} \cdot 1 = \frac{1}{N}$$</p>
|
947 | <p>I'm looking for the algorithm that efficiently locates the "loneliest person on the planet", where "loneliest" is defined as:</p>
<p>Maximum minimum distance to another person — that is, the person for whom the closest other person is farthest away.</p>
<p>Assume a (admittedly miraculous) input of the list of the exact latitude/longitude of every person on Earth at a particular time.</p>
<p>Also take as provided a function <span class="math-container">$d(p_1, p_2)$</span> that returns the distance on the surface of the earth between <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> - I know this is not trivial, but it's "just spherical geometry" and not the important (to me) part of the question.</p>
<p>What's the most efficient way to find the loneliest person?</p>
<p>Certainly one solution is to calculate <span class="math-container">$d(\ldots)$</span> for every pair of people on the globe, then sort every person's list of distances in ascending order, take the first item from every list and sort those in descending order and take the largest. But that involves <span class="math-container">$n(n-1)$</span> invocations of <span class="math-container">$d(\ldots)$</span>, <span class="math-container">$n$</span> sorts of <span class="math-container">$n-1$</span> items and one last sort of <span class="math-container">$n$</span> items. Last I checked, <span class="math-container">$n$</span> in this case is somewhere north of six billion, right? So we can do better?</p>
| Scott Morrison | 3 | <p>Let's iteratively come up with a candidate, "the loneliest person so far", and at each step weed out lots of people who we can see aren't as lonely.</p>
<p>Pick a person at random, and calculate their "loneliness". Since they're the first, they are by default the loneliest person so far.</p>
<p>Now let's start searching for a new candidate. Pick a person at random, find everyone within the our current record loneliness. Either this set is empty, or not. If it's empty, we've found a new loneliest person, calculate their loneliness, and continue. Otherwise, mark everyone in that ball as happy, which just means we remove them from the pool of people we sample randomly (but not from any future calculations of loneliness).</p>
<p>This eventually finds the loneliest person.</p>
<p>Before we think about how efficient this is, let's make an optimisation. Optionally, pretend the world is a torus (following a fine tradition; Arnol'd's Classical Mechanics does this in a footnote while talking about weather forecasting). As usual, it's not essential.</p>
<p>Now, before we begin, sort everyone into two lists, one by latitude and one by longitude. We can now use this to efficiently find everyone within some radius, without having to evaluate every pair-wise distance. We can also improve the other step, calculating the loneliness of a new candidate.</p>
<p>Finally, on second thoughts I've decided that actually working out the complexity of this algorithm sounds like too much work right now. :-)</p>
|
2,566,546 | <p>Show that for any $x_1 < x_2$ and $y_1 < y_2$ one has
$P(x_1 < X ≤ x_2, y_1 < Y ≤ y_2) = F(x_2, y_2) + F(x_1, y_1) − F(x_1, y_2) − F(x_2, y_1)$.</p>
<p>Would I just need to split the LHS to something that gives me the right?</p>
| ajotatxe | 132,456 | <p>Consider
$$Q=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$$</p>
<p>Let $P=Q^2$. Since $P=I$, $P$ is a projection, but $Q$ is not.</p>
|
1,160,699 | <p>In my discrete math book, I was tasked with finding a counterexample for this:</p>
<blockquote>
<p>If $n$ is prime, then $2^n-1$ is prime.</p>
</blockquote>
<p>Does there exist a counterexample for such a statement? Also, am I wrong in thinking that when something asks for a counterexample, is it looking for some logic that proves the original statement to be false?</p>
<p>Any help is appreciated, as I've got a test on subjects like this tomorrow.</p>
| mathamphetamines | 126,882 | <p>A counter-example is one instance in which the statement does not hold to be true. Let n=11. $2^{11} - 1$ is a composite number.</p>
<p>$2^{11}-1=2047 = 23 \cdot 89$</p>
|
1,160,699 | <p>In my discrete math book, I was tasked with finding a counterexample for this:</p>
<blockquote>
<p>If $n$ is prime, then $2^n-1$ is prime.</p>
</blockquote>
<p>Does there exist a counterexample for such a statement? Also, am I wrong in thinking that when something asks for a counterexample, is it looking for some logic that proves the original statement to be false?</p>
<p>Any help is appreciated, as I've got a test on subjects like this tomorrow.</p>
| Robert Soupe | 149,436 | <p>No, it's not. Among the first twenty thousand primes $p$, only thirty-one of them give primes when plugged into the formula $2^p - 1$. In fact, you could have your computer give you a pseudorandom prime number $p < 10^7$ and I would wager you \$100 that $2^p - 1$ is composite.</p>
<p>For a counterexample, at least the ones that are easy to find, you don't really need logic, you just need a single example to counter the assertion. Try $$\frac{2^{23} - 1}{47}.$$ But like I suggested, there are plenty more where those came from, see <a href="http://oeis.org/A054723" rel="nofollow">Sloane's A054723</a>.</p>
<p>The primes $p$ that do give prime $2^p - 1$ are sometimes called "Mersenne primes" (though some use the term for the prime $2^p - 1$ rather than for $p$, calling that a "Mersenne exponent") after a medieval French math amateur who correctly identified some and incorrectly identified others.</p>
<p>Without computers, it was very tough going finding Mersenne primes. Barely a dozen of them were known prior to World War II. But even with our fancy computers, we know less than fifty of them, see <a href="http://www.mersenne.org/primes/" rel="nofollow">http://www.mersenne.org/primes/</a> for a lot more info than I can give here.</p>
<p>Here's a related question for which you do need logic: prove that if $n$ is composite, then $2^n - 1$ is also composite.</p>
|
3,122,612 | <p>I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:</p>
<p><span class="math-container">$\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n$</span></p>
<p>I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).</p>
<p><a href="http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf" rel="nofollow noreferrer">http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf</a></p>
<p>EDIT: The full equation is</p>
<p><span class="math-container">$\frac{\partial A}{\partial z}+\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n \frac{\partial^{n} A}{\partial t^{n}}=-i \frac{\chi^{(2)}\omega}{2nc}AA^{*}e^{-i\Delta \textbf{k}\cdot \textbf{z}}$</span></p>
<p>so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..</p>
| Community | -1 | <p><span class="math-container">$$\sum_{n=1}^\infty\frac{(-ik)^n}{(-i)\,n!}=i(e^{-ik}-1)=i(\cos k-1-i\sin k).$$</span></p>
<hr>
<p>This can be rewritten as</p>
<p><span class="math-container">$$2i\cos\frac k2\left(\cos\frac k2-i\sin\frac k2\right)=2i\cos\frac k2e^{-ik/2}$$</span> but there is little benefit.</p>
|
1,025,588 | <p>I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem:</p>
<p>If $a, b, c \gt 0$ prove that $$ a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.$$</p>
<p>I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you!</p>
| Adriano | 76,987 | <p>One trick that shows up a lot with AM-GM inequalities is to apply AM-GM several times, then combine them all together somehow. Let's try that. AM-GM tells us that:
$$
x + y + z \geq 3 \sqrt[3]{xyz} \tag{$\star$}
$$
Now for what values of $x,y,z$ would the LHS and RHS of $(\star)$ match up with some of the terms in the LHS and RHS of the desired inequality? In particular, suppose that we want the cube root to be $a^2b$. Then $xyz = a^6b^3$, which suggests that we take $x = y = a^3$ and $z = b^3$. Repeating this, we obtain:
\begin{align*}
a^3 + a^3 + b^3 &\geq 3 \sqrt[3]{a^3a^3b^3} = 3a^2b \\
b^3 + b^3 + c^3 &\geq 3 \sqrt[3]{b^3b^3c^3} = 3b^2c \\
c^3 + c^3 + a^3 &\geq 3 \sqrt[3]{c^3c^3a^3} = 3c^2a \\
\end{align*}
Adding up the above three inequalities and dividing through by $3$ yields the desired inequality. $~~\blacksquare$</p>
|
1,429,563 | <blockquote>
<p>Let <span class="math-container">$\{F_n\}_{n\geq 0}$</span> be the Fibonacci sequence.</p>
<p>Prove that the number of primes <span class="math-container">$p$</span> so that <span class="math-container">$p\mid F_{p-1}$</span> is infinite.</p>
</blockquote>
<hr />
<p>I tried to use induction, to no avail.</p>
| N. S. | 9,176 | <p><strong>Hint 1:</strong> Prove that if the equation
$$x^2-x-1=0 \pmod{p}$$
has two solutions $x_1 \neq x_2$ then there exists constants $C_1, C_2$ such that</p>
<p>$$F_n \equiv C_1x_1^n+ C_2x_2^n \pmod{p}$$</p>
<p><strong>Hint 2:</strong> In this case what can you say about $F_{p-1}$ and $F_0$?</p>
<p><strong>Hint 3:</strong> For $p >5$ the $$x^2-x-1=0 \pmod{p}$$
has two solutions $x_1 \neq x_2$ if and only if $5$ is a quadratic residue modulo $p$. </p>
<p>Quadratic reciprocity tells you that this happens exactly when $p$ has certain remainders modulo $5$, and Dirichclet Theorem telss you that there are infinitely many such primes..</p>
|
484,550 | <p>The problem is as follows: let $n_1, n_2,..., n_t$ be positive integers. Prove that if $n_1+n_2+...+n_t-t+1$ objects are placed into $t$ boxes, then for some $i, i=1, 2, ..., t$, the $i$th box contains at least $n_i$ objects. </p>
<p>I'm having difficulty getting started in developing a proof because I have no intuition as to why this should be true or whether or not it actually is. Could someone help get me started?</p>
| Michael Joyce | 17,673 | <p>Suppose, for the sake of contradiction, that the $i$-th box always contains $< n_i$ objects. Then what is the largest the total number of objects could be? (This occurs when the $i$-th box has $n_i - 1$ objects.) But how many objects are there in total?</p>
|
987,802 | <p>I've solved forming $8$ digit number using $1,1,2,2,2,3,3,3$. We have 8 digits: two 1, three 2 and three 3. First I put three 2's in 8 possible places. Number of putting 2's is $\frac{8!}{3!5!}$ . After putting 2's we have 5 possible empty places left. We put three 3's: $\frac{5!}{3!2!}$ . Then we have 2 empty places and two 1's: $= \frac{2!}{0!2!} $. The answer will be multiplication of all these.</p>
| paw88789 | 147,810 | <p>Line up the violinists say in alphabetical order. </p>
<p>Now pick a violist, a cellist, and a bass for the first violinist ($10^3$ ways).</p>
<p>Now pick a violist, a cellist, and a bass for the second violinist ($9^3$ ways).</p>
<p>And so on.</p>
<p>Do you see what happens from here?</p>
|
987,802 | <p>I've solved forming $8$ digit number using $1,1,2,2,2,3,3,3$. We have 8 digits: two 1, three 2 and three 3. First I put three 2's in 8 possible places. Number of putting 2's is $\frac{8!}{3!5!}$ . After putting 2's we have 5 possible empty places left. We put three 3's: $\frac{5!}{3!2!}$ . Then we have 2 empty places and two 1's: $= \frac{2!}{0!2!} $. The answer will be multiplication of all these.</p>
| Brian M. Scott | 12,042 | <p>Alternatively, after lining up the violinists as in paw88789’s answer, you can choose all of the violists at once by lining them up in front of the violinists; how many ways are there to line up $10$ people? Then do the same with the cellists, and again with the bassists. It’s the same idea, but with the choices made one instrument at a time instead of one quartet at a time.</p>
<p>The key to both answers is fixing the order of the violinists: that’s what ensures that you don’t count any division into quartets twice.</p>
<p><strong>Added:</strong> Your $\frac{40!}{10!4!}$ can be interpreted in the following way. There are $40!$ ways to line up all $40$ students. Say that we don’t care in which order the ten cellists appear in the lineup: keeping their overall positions in the lineup, they can be permuted in $10!$ different ways, so we divide by $10!$. The fraction $\frac{40!}{10!}$ then counts the number of ways to line up the $40$ students if we treat the cellists as indistinguishable, i.e., we care only where they are in the lineup, not which one is in which of those $10$ positions. Finally, we decide that we’ll also treat the four tallest violists as indistinguishable: we care which $4$ positions they occupy as a group, but we don’t care which of them is in which of those positions. That gives us $\frac{40!}{10!4!}$ lineups that differ in ways that we care about. I think that you can probably see that this bears little resemblance to what we want to count.</p>
<p>An alternative correct argument based on dividing out unwanted duplicates could go like this. Say we have quartets $1$ through $10$. There are $10!$ ways to assign the violinists to them, $10!$ ways to assign the violists to them, and so on, for a total of $10!^4$. However, the numbering of the quartets was arbitrary: if I permuted the numbers, which I can do in $10!$ ways, I’d still have the same quartets. To correct for that, I divide by $10!$, getting $10!^3$.</p>
|
3,672,814 | <p><span class="math-container">$g_n(y)=\sum_{k=1}^{n} \frac{\sin ky}{k},x \in [0,\pi]$</span>. So i try to find the pointwise limit function first, following this approach in this post <a href="https://math.stackexchange.com/questions/13490/proving-that-the-sequence-f-nx-sum-limits-k-1n-frac-sinkxk-is">Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$</a>. I find it to be <span class="math-container">$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=g(x) ,\, x\in(0,2\pi).$</span>
I tried to prove it by Dirichlet test:<br />
take <span class="math-container">$h_n(x)=\frac{1}{n}\quad \text{and} \quad u_n(x)=\sin(nx) ,
\text{then} |\sum_{k=1}^n\sin(kx)| \leq \frac{1}{\sin(\frac{x}{2})} \leq \frac{1}{\sin(\frac{\delta}{2})}$</span>.<br />
I want to prove or disprove that <span class="math-container">$(g_n)$</span> converge uniformly on <span class="math-container">$[0,\pi]$</span>.</p>
| MathematicsStudent1122 | 238,417 | <p><span class="math-container">$\sum_{n \geq 1} \frac{\sin nx}{n}$</span> is exactly the Fourier series of <span class="math-container">$f(x)=\frac{\pi-x}{2}$</span> (<span class="math-container">$x \in [0, 2\pi)$</span>) extended <span class="math-container">$2\pi$</span> periodically to <span class="math-container">$\mathbb{R}$</span>. Were the convergence of the Fourier series uniform, then the Fourier series would in fact converge pointwise to <span class="math-container">$f$</span> (see <a href="https://math.stackexchange.com/questions/2707753/prove-uniform-convergence-of-a-fourier-series">this</a>) but notice that at <span class="math-container">$x=0$</span> <span class="math-container">$f$</span> fails to coincide with its Fourier series. Hence we deduce that convergence is not uniform. </p>
|
183,058 | <p>I'm having difficulty producing a proper animated gif (for inclusion in a PowerPoint presentation). The code I have is this:</p>
<pre><code> HSBhsl[h_, s_, l_] :=
Module[{H, S, B}, H = h; B = l + s (1 - Abs[2 l - 1])/2;
S = (2 B - l)/B; {H, S, B}]
TestCube[\[Theta]_] :=
Graphics3D[Rotate[{EdgeForm[], Hue[HSBhsl[32/250, 228/250, 105/256]],
Cuboid[{0, 0, 0}, {1, 1, 1}]}, -\[Theta] Degree, {0, 0, 1}, {1, 1,
1}], Lighting -> {{"Ambient",
RGBColor[0.7, 0.7, 0.7]}, {"Directional",
RGBColor[0.35, 0.35, 0.35], ImageScaled[{0, 5, 0}]}},
Boxed -> False, Background -> None,
PlotRange -> 4 {{-1, 1}, {-1, 1}, {-1, 1}}, ViewAngle -> Pi/10000]
</code></pre>
<p>The animation is then produced like so:</p>
<pre><code>list = Table[
Show[TestCube[t], ViewPoint -> {1000, 1000, 1000}], {t, 0, 360, 1}];
SetDirectory@NotebookDirectory[];
Export["CubesAnimation.gif", list, "TransparentColor" -> White]
</code></pre>
<p>Unfortunately, this produces an amimation that seems to stack images on top of each other, thus giving the appearance of a "trace" left by the previous frame. I tried adding <code>"TransitionEffect" -> Background</code>, but this had no effect.</p>
<p>How do I get a proper GIF animation, with a transparent background, of the rotating cube?</p>
| Piotr Wendykier | 9,239 | <p>To export to an animated PNG with transparency you should convert <code>Graphics</code> to <code>Image</code> first: </p>
<pre><code>list = Table[Image[Show[TestCube[t], ViewPoint -> {1000, 1000, 1000}]], {t,
0, 360, 1}];
</code></pre>
<p>Then, you call <code>Export</code>: </p>
<pre><code>In[9]:= Export["CubesAnimation.png", list]
Out[9]= "CubesAnimation.png"
In[10]:= ImageMeasurements[Import["CubesAnimation.png", {"ImageList", 1}], "Transparency"]
Out[10]= True
</code></pre>
|
3,080,294 | <blockquote>
<p>A cone has its guiding curve to the circle <span class="math-container">$x^2+y^2+2ax+2by=0$</span> and passes through a fixed point <span class="math-container">$(0,0,c)$</span>. If the section of the cone by plane <span class="math-container">$y=0$</span> is a rectangular hyperbola. Prove that the vertex lies on fixed circle <span class="math-container">$x^2+y^2+z^2+2ax+2by=0$</span> and <span class="math-container">$2ax+2by+cz=0$</span>.</p>
</blockquote>
<p>Attempt:</p>
<p>Using equation of circle and fixed point, equation of cone can be found out, it comes:</p>
<p><span class="math-container">$$cx^2+cy^2-2axz-2byz+2acx+2bcy=0$$</span></p>
<p>Its section by <span class="math-container">$y=0$</span> plane comes out to be
<span class="math-container">$$cx^2-2axz+2ax=0$$</span></p>
<p>I am unable to proceed further. Please help</p>
| Blue | 409 | <p>That the guiding curve is a circle implies that the axis of the cone is perpendicular to the plane of the circle (that is, the <span class="math-container">$xy$</span>-plane, which we'll call "the horizontal") and passes through the circle's center, <span class="math-container">$(-a,-b,0)$</span>. </p>
<p>Also, it is "known" (and shown in <a href="https://math.stackexchange.com/a/137462/409">this answer</a>) that the eccentricity of a conic is given by
<span class="math-container">$$e = \frac{\sin \angle P}{\sin \angle C} \tag{1}$$</span>
where <span class="math-container">$\angle C$</span> and <span class="math-container">$\angle P$</span> are the "cone angle" and "plane angle", the angles with the horizontal made by the cone's generator-lines and the cutting plane, respectively. Now, a rectangular hyperbola has eccentricity <span class="math-container">$\sqrt{2}$</span>, and the plane <span class="math-container">$y=0$</span> makes <span class="math-container">$\angle P = 90^\circ$</span> with the horizontal; we find, then, that <span class="math-container">$\angle C = 45^\circ$</span>. Consequently, the height of the cone's vertex above (or below) the <span class="math-container">$xy$</span>-plane must be equal to the guiding circle's radius, <span class="math-container">$\sqrt{a^2+b^2}$</span>.</p>
<p>The vertex of the cone is thus
<span class="math-container">$$V = (-a,-b,\pm\sqrt{a^2+b^2}) \tag{2}$$</span>
which clearly satisfies the target equation
<span class="math-container">$$x^2 + y^2 + z^2 + 2 a x + 2 b y = 0 \tag{3}$$</span>
for the sphere that has the guiding circle as a great circle.</p>
<p>Note that we have not yet introduced the "fixed point" <span class="math-container">$C := (0,0,c)$</span>. Since we have deduced the cone's vertex without this point, we find that <span class="math-container">$c$</span> is not actually a free parameter in this problem. Indeed, one readily sees that, because <span class="math-container">$\overline{OV}$</span> and <span class="math-container">$\overline{VC}$</span> are generators making <span class="math-container">$45^\circ$</span> angles with the horizontal, <span class="math-container">$\triangle OVC$</span> is an isosceles right triangle with apex <span class="math-container">$V$</span>; necessarily, the <span class="math-container">$z$</span>-coordinate of <span class="math-container">$C$</span> must be twice the <span class="math-container">$z$</span>-coordinate of <span class="math-container">$V$</span>: that is, <span class="math-container">$c = \pm 2 \sqrt{a^2+b^2}$</span>. Thus, the target plane
<span class="math-container">$$2 a x + 2 b y + c = 0 \qquad\to\qquad a x + b y \pm z \sqrt{a^2+b^2} = 0 \tag{4}$$</span>
is easily seen to be satisfied by <span class="math-container">$V$</span>. <span class="math-container">$\square$</span></p>
|
4,322,897 | <p>Let <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> be a sequence in <span class="math-container">$\Bbb{R}^n$</span> and <span class="math-container">$z\in\Bbb{R}^n$</span> be given. If <span class="math-container">$\{\|z-y_n\|\}_{n=1}^\infty$</span> is a convergent sequence, what can we say about <span class="math-container">$\{y_n\}_{n=1}^\infty$</span>? More precisely, can we conclude that <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> is bounded?</p>
<p>For some reason, I'd like to extract a convergent subsequence from <span class="math-container">$\{y_n\}_{n=1}^\infty$</span>, and this can be done by showing that <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> is bounded. Here is my attempt. Since <span class="math-container">$\{\|z-y_n\|\}_{n=1}^\infty$</span> converges, it must be bounded. Then <span class="math-container">$\exists r>0$</span> s.t. <span class="math-container">$\forall n\in\Bbb N$</span>, <span class="math-container">$\left|\|z-y_n\|\right|<r$</span>, but this amounts to saying that each term of <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> falls into an open ball centered at <span class="math-container">$z$</span>. Thus, <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> must be a bounded sequence. Is my attempt correct, please? Thank you.</p>
| Literally an Orange | 314,721 | <p>If <span class="math-container">$x_n$</span> is a real-valued sequence then <span class="math-container">$x_n \to x$</span> implies that for some large <span class="math-container">$N$</span>, <span class="math-container">$|x_n-x|<1$</span> for <span class="math-container">$N \leq n$</span>. Now let <span class="math-container">$M=max_{n<N}\{|x_n|\}+1+2|x|$</span> and notice that <span class="math-container">$|x_n| \leq |x_n-x|+|x|$</span></p>
|
2,772,190 | <p>The question:</p>
<blockquote>
<p>Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$</p>
</blockquote>
<p>So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution set, $|x^2+2x+2|\ge-1$ as indicated by the roots of the RHS equation, but this is where I get stuck. </p>
<p>Where am I going wrong?</p>
| Servaes | 30,382 | <p>Obviously $|x^2-3x-4|$ is greater than or equal to $0$, because you take the absolute value. This observation tells you nothing. In stead, distinguish cases where the quadratics are positive or negative.</p>
|
2,772,190 | <p>The question:</p>
<blockquote>
<p>Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$</p>
</blockquote>
<p>So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution set, $|x^2+2x+2|\ge-1$ as indicated by the roots of the RHS equation, but this is where I get stuck. </p>
<p>Where am I going wrong?</p>
| Cesareo | 397,348 | <p>Knowing that $x^2+2x+2 > 0$</p>
<p>The equation is equivalent to</p>
<p>$$
x^2+2x+2 = \vert x^2-3x-4\vert
$$</p>
<p>which is equivalent to</p>
<p>$$
x^2+2x+2 =
\left\{\begin{array}{lcl}-x^2+3x+4 & \rightarrow & x = \{\frac{1\pm\sqrt{17}}{4}\}\\
x^2-3x-4 & \rightarrow & x = -\frac{6}{5}\end{array}\right.
$$</p>
|
1,012,985 | <p>Let $A = \left[ \begin{matrix} 3 & 2 & 1\\ 5 & 0 &1\end{matrix}\right]$, </p>
<p>how can I know if there is a matrix $N$ , st. $AN=0$ (N is not a zero matrix) </p>
| klimenkov | 161,422 | <p>Let $N = \pmatrix{x_1 & x_2 \\ x_3 & x_4 \\ x_5 & x_6}$.</p>
<p>If you mutiply $A$ and $N$ you will get 4 equations with 6 variables. If you let any 2 of these variables be non-zero, you will get the system of 4 equations with 4 variables. When you solve it, you will get the elements of your matrix and some of them will be non-zero.</p>
|
4,363,384 | <p>Let <span class="math-container">$A_i$</span>, <span class="math-container">$i \in I$</span> subsets of a space X given topology <span class="math-container">$\tau$</span>. Show that <span class="math-container">$\overline{\bigcup_i A_i }\subset \bigcup_i\overline{ A_i }$</span> does not hold necessarily :</p>
<p>My counterexample is in <span class="math-container">$\Bbb N$</span> with the co-finite topology and <span class="math-container">$A_k=\{2k\}=\overline{ A_k }$</span> because singletons are closed. Hence, <span class="math-container">$\bigcup_i\overline{ A_i } = 2\Bbb N$</span>. However, <span class="math-container">$\overline{\bigcup_i A_i }$</span> is the smallest closed set containing <span class="math-container">$2\Bbb N$</span>, since closed sets are finite or equal to <span class="math-container">$\Bbb N$</span>, then <span class="math-container">$\overline{\bigcup_i A_i }=\Bbb N$</span> so <span class="math-container">$\overline{\bigcup_i A_i }\nsubseteq \bigcup_i\overline{ A_i }$</span>.</p>
<p>But here, since <span class="math-container">$I$</span> is countable, it seems to disagree with the fact that <span class="math-container">$\overline{A \cup B}=\overline{A} \cup \overline{B}$</span>. Is my counterexample wrong ? Should I find a case with <span class="math-container">$I$</span> uncountable ?</p>
| tomasz | 30,222 | <p>Your example is fine.</p>
<p>The equality <span class="math-container">$\overline{A\cup B}=\bar A\cup \bar B$</span> only tells you about <em>binary</em> unions (and, by induction, finite unions), not arbitrary ones. This is related to the fact that a <em>finite</em> union of closed sets is closed (but not an arbitrary one).</p>
|
1,675,411 | <p>So far I have this:</p>
<p>First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.</p>
<p>Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.</p>
<p>So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?</p>
| MathematicsStudent1122 | 238,417 | <p>The base case is clear. The inductive step is done as follows. Suppose it holds for some $k\geq5$. That is, $k^2<2^k$</p>
<p>We know $(k+1)^2<2k^2 \Longleftrightarrow \frac 12 (k+1)^2 - k^2 < 0$ for $k\geq 5$ (this can be shown rigorously determining the roots of the polynomial and noting that it changes sign only at those roots). </p>
<p>Then we have inequalities $(k+1)^2 <2k^2 < 2 \cdot 2^k = 2^{k+1}$</p>
|
915,542 | <p>After working on an ODE I find I am needing to solve the integral </p>
<p>$$\int \frac{u}{b - au - u^2}\mathrm{d}u$$</p>
<p>Trig subs, banging heads against walls, and sobbing have not yielded a solution. Yet. </p>
<p>Could use a hand, thanks.</p>
| Jean-Claude Arbaut | 43,608 | <p>You can write $b-au-u^2=\frac{a^2+4b}{4}-(u+\frac a2)^2$</p>
<p>There are thus three cases to consider</p>
<p><strong>A. $a^2+4b=0$</strong></p>
<p>Then your integral is</p>
<p>$$\int\frac{\mathrm{d}u}{b-au-u^2}=-\int\frac{\mathrm{d}u}{(u+\frac a2)^2}=\frac{1}{u+\frac a2}+C$$</p>
<p><strong>B. $a^2+4b>0$</strong></p>
<p>The trinomial $b-au-u^2$ has two real roots $\alpha,\beta$</p>
<p>$$\alpha=\frac{-a+\sqrt{a^2+4b}}{2}$$
$$\beta=\frac{-a-\sqrt{a^2+4b}}{2}$$</p>
<p>Partial fraction decomposition yields</p>
<p>$$\frac{1}{b-au-u^2}=\frac{1}{\beta-\alpha}\frac{1}{u-\alpha}-\frac{1}{\beta-\alpha}\frac{1}{u-\beta}$$</p>
<p>Hence</p>
<p>$$\int\frac{\mathrm{d}u}{b-au-u^2}=\frac{1}{\beta-\alpha}\int \left(\frac{1}{u-\alpha}-\frac{1}{u-\beta}\right)\,\mathrm{d}u=\frac{1}{\beta-\alpha}\ln \left|\frac{u-\alpha}{u-\beta}\right|+C$$</p>
<p><strong>C. $a^2+4b<0$</strong></p>
<p>The trinomial $b-au-u^2$ has two complex roots. Let's apply the change of variable $u=\lambda t$ with $\lambda=\frac12\sqrt{|a^2+4b|}$:</p>
<p>$$\int\frac{\mathrm{d}u}{b-au-u^2}=\int\frac{\mathrm{d}u}{\frac{a^2+4b}{4}-(u+\frac a2)^2}=\int \frac{\lambda\,\mathrm{d}t}{-\lambda^2-(\lambda t+\frac a2)^2}\\=-\frac1\lambda\int \frac{\mathrm{d}t}{1+(t+\frac a{2\lambda})^2}=-\frac1\lambda\mathrm{Arctan}\left(t+\frac{a}{2\lambda}\right)+C\\=-\frac{2}{\sqrt{|a^2+4b|}}\mathrm{Arctan}\left(\frac{2u+a}{\sqrt{|a^2+4b|}}\right)+C$$</p>
<hr>
<p>There is another solution in the case $a^2+4b>0$</p>
<p>Let's apply the change of variable $u=\lambda t$ with $\lambda=\frac12\sqrt{a^2+4b}$:</p>
<p>$$\int\frac{\mathrm{d}u}{b-au-u^2}=\int\frac{\mathrm{d}u}{\frac{a^2+4b}{4}-(u+\frac a2)^2}=\int \frac{\lambda\,\mathrm{d}t}{\lambda^2-(\lambda t+\frac a2)^2}\\=\frac1\lambda\int \frac{\mathrm{d}t}{1-(t+\frac a{2\lambda})^2}=\frac1\lambda\mathrm{Argth}\left(t+\frac{a}{2\lambda}\right)+C\\=\frac{2}{\sqrt{a^2+4b}}\mathrm{Argth}\left(\frac{2u+a}{\sqrt{a^2+4b}}\right)+C$$</p>
<p>It's valid if $\left|\frac{2u+a}{\sqrt{a^2+4b}}\right|<1$. Outside of this interval, the integral is instead</p>
<p>$$\frac{2}{\sqrt{a^2+4b}}\mathrm{Argcoth}\left(\frac{2u+a}{\sqrt{a^2+4b}}\right)+C$$</p>
<p>Notice that $\mathrm{Argth} x=\frac12\ln\frac{1+x}{1-x}$, defined for $|x|<1$ whereas $\mathrm{Argcoth} x=\frac12\ln\frac{x+1}{x-1}$, defined for $|x|>1$. Their derivative is the same, $\frac{1}{1-x^2}$.</p>
|
2,509,230 | <p>I was studying Chevalley-Serre relations, which can be summed up to these</p>
<p><span class="math-container">$$\tag{S1}\left[h_{i},\,h_{j}\right]=0$$</span></p>
<p><span class="math-container">$$\tag{S2}\left[e_{i},\,f_{i}\right]=h_{i} \quad \left[e_{i},\,f_{j}\right]=0 \quad\text{for } i\neq j$$</span></p>
<p><span class="math-container">$$\tag{S3} \left[h_{i},\,e_{j}\right]=A_{ij}e_{j}
\quad \left[h_{i},\,f_{j}\right]=-A_{ij}f_{j}$$</span></p>
<p><span class="math-container">$$\tag{S4} \text{ad}\left(e_{i}\right)^{1-A_{ij}}\left(e_{j}\right)=0
\quad\;\; \text{ad}\left(f_{i}\right)^{1-A_{ij}}\left(f_{j}\right)=0 \quad\text{for } i\neq j$$</span> </p>
<p>where <span class="math-container">$A_{ij}$</span> are the coefficients of the Cartan matrix. Now it seems to me that relations (S1),(S2), and (S3) are really quite natural, but I don't fully understand relations in (S4). Does anybody has an insight on what does those relations mean?</p>
| Tobias Kildetoft | 2,538 | <p>It might help if we label the generators by the corresponding roots instead (since this is where they come from). When we do this, we get that if $[e_{\alpha_i},e_{\alpha_j}]$ is in the root space corresponding to $\alpha_i + \alpha_j$, so since there are only finitely many roots, at some point, if we keep taking brackets with the same $e_{\alpha_i}$, we need to get $0$.</p>
<p>That was the reason coming from actually looking at the Lie algebra we know we should get. But there is another way to look at this: What would happen if we left these out?</p>
<p>As it happens, if we leave out the relation, we no longer get a finite dimensional Lie algebra, so what the relation really does is make sure the Lie algebra stays small.</p>
|
2,056,979 | <p>I am attempting:
<a href="https://i.stack.imgur.com/FM80Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FM80Z.png" alt="enter image description here"></a></p>
<p>My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.
<a href="https://i.stack.imgur.com/9onWu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9onWu.jpg" alt="enter image description here"></a></p>
| paw88789 | 147,810 | <p>Hint: $4^{n+2}+5^{2n+1}=4(4^{n+1})+25(5^{2n-1})=4(4^{n+1}+5^{2n-1})+21(5^{2n-1})$.</p>
|
2,056,979 | <p>I am attempting:
<a href="https://i.stack.imgur.com/FM80Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FM80Z.png" alt="enter image description here"></a></p>
<p>My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.
<a href="https://i.stack.imgur.com/9onWu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9onWu.jpg" alt="enter image description here"></a></p>
| Bill Dubuque | 242 | <p>Before considering your proof, let's gather some insight from a simpler proof using congruences. Below the inductive step follows very simply by using $\,\rm{\color{#C00}{CPR}} = $ <a href="https://math.stackexchange.com/a/879262/242">Congruence Product Rule</a> to multiply the first two congruences </p>
<p>$$\begin{align}
{\rm mod}\,\ 7\!:\qquad\ 4\,\ &\equiv\,\ 5^{\large 2}\\[0.3em]
4^{\large K+1}&\equiv -5^{\large 2K-1}\ \ \ {\rm i.e.}\ \ P(K)\\
\overset{\rm{\color{#C00}{CPR}}}\Longrightarrow\ \ \ 4^{\large K+2}&\equiv -5^{\large 2K+1}\ \ \ {\rm i.e.}\ \ P(K\!+\!1)
\end{align}$$</p>
<hr>
<p>The common inductive proofs using divisibility in other answers effectively do the same thing, i.e. they repeat the <em>proof</em> of the Congruence Product Rule in this special case, but expressed in divisibility vs. congruence language (<a href="https://math.stackexchange.com/a/1239206/242">e.g. see here</a>). But the product rule is <em>much</em> less arithmetically intuitive when expressed as unstructured divisibilities, which greatly complicates the discovery of the inductive step. I explain this at length in other answers, e.g. <a href="https://math.stackexchange.com/a/1286562/242">see here.</a></p>
<p>If congruences are unfamiliar then you can instead use the rule in divisibility form as below. This will allow you to structure the induction in the above intuitive arithmetical Product Rule form. </p>
<p>$$\begin{align} {\rm mod}\,\ m\!:\, A\equiv a,\, B\equiv b&\ \ \,\Longrightarrow\,\ \ AB\equiv ab\qquad\text{Congruence Product Rule}\\[3pt]
m\mid A-a,\ B-b&\,\Rightarrow\, m\mid AB-ab\qquad\text{Divisibility Product Rule}\\[4pt]
{\bf Proof}\quad (A-a)B+a(B&-b)\, = AB-ab\end{align}$$</p>
<hr>
<p>To finish the proof that you started we can proceed as follows</p>
<p>$$\begin{align} f(k\!+\!1) - f(k) &=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}{24}\cdot 5^{2k-1}\\
&=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}3 \cdot 5^{2k-1} + \color{#0a0}{21}\cdot 5^{2k-1}\\
&=\, 3\, f(k) + 7n\\
\Rightarrow\qquad f(k\!+\!1)\, &=\, 4\, f(k) + 7n\\[0,3em]
\Rightarrow\ \ 7\mid f(k\!+\!1)\,\ &{\rm if}\,\ 7\mid f(k), \ \ {\rm i.e.}\ \ P(k\!+\!1)\ \ {\rm if}\ \ P(k)
\end{align}$$</p>
<p>Note that the above says that $\ f(k\!+\!1)\equiv 4\,f(k)\ \pmod{7}\,$</p>
<p>so an easy induction shows that $\ f(k)\equiv 4^{\large k-1}\, f(1)\pmod 7,\ $ so $\ 7\mid f(k)\iff 7\mid f(1)$ </p>
<p>Now how using the Product Rule as above makes it much clearer that incrementing the index amounts simply to multiplication by $\,4,\,$ when viewed modulo $\,7.\,$ Once that innate arithmetical structure hs been revealed, the proof is easy.</p>
|
4,126,527 | <p>Suppose <span class="math-container">$w_1$</span>, <span class="math-container">$w_2$</span>, ..., <span class="math-container">$w_{n-1}$</span> are the complex roots not equal to <span class="math-container">$1$</span> of <span class="math-container">$z^n-1=0$</span>, where <span class="math-container">$n$</span> is odd.</p>
<p>Show: <span class="math-container">$\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}$</span>.</p>
<p>Hello, I am struggling with this complex number proof. Any help is welcomed; thank you in advance.</p>
<p>I have tried expanding the LHS and comparing it to the RHS but am not able to make them equal; I got this far:
<span class="math-container">$\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}$</span></p>
<p>Part 2 of the question:
Hence show: <span class="math-container">$$\sum_{k=1}^{n-1} \frac{1-\bar{w_k}}{1+w_k} = n$$</span></p>
<p>Any suggestions?</p>
| user | 293,846 | <p>Since <span class="math-container">$n$</span> is odd we know: <span class="math-container">$w\ne-1$</span>.
Therefore, using <span class="math-container">$(1+w)(1+\bar w)\ne0$</span> and <span class="math-container">$w\bar w=1$</span>:
<span class="math-container">$$\begin{align}
&\frac{1-\bar{w}}{1+w}+\frac{1-w}{1+\bar{w}}=2-w-\bar{w}\\
\iff&\frac{2-w^2-\bar{w}^2}{2+w+\bar{w}}=2-w-\bar{w}\\
\iff&2-w^2-\bar{w}^2=2-w^2-\bar{w}^2,
\end{align}$$</span>
the latter equation being obvious identity.</p>
|
2,062,671 | <p>How many sequences of positive integer numbers $\{a_n\}$ such that $$a_0 =1, a_1 = 2,|a_{n+2}a_n - a_{n+1}^2| = 1 ?$$</p>
| Will Jagy | 10,400 | <p>Ambiguous question. </p>
<p>Define the rational numbers
$$ b_n = \frac{a_{n+1} + a_{n-1}}{a_n}. $$
IF we always have the constant $C$ with
$$ a_n a_{n+2} - a_{n+1}^2 = C, $$
then a simple calculation shows that
$$ \frac{b_{n+1}}{b_n} = 1, $$ so that we have a constant $\tau$ with
$$ \frac{a_{n+1} + a_{n-1}}{a_n} = \tau,$$
or
$$ a_{n+2} = \tau a_{n+1} - a_n. $$
This is an ordinary linear recurrence, solutions easy enough.</p>
<p>HOWEVER,
if $ a_n a_{n+2} - a_{n+1}^2 = (-1)^{z_n}, $ for some integer sequence $z_n,$ there are infinitely many possibilities. </p>
<p>AT LEAST, IT appears that way at first glance. It is also possible that, whichever way we start, $1,2,3,4,5,$ or $1,2,5,13, 34,$ is the way we must continue forever, in which case there are exactly two solutions. More work required. The person who asked should pursue this. </p>
<p>IT NOW seems the other answer is exactly correct. Either consecutive numbers, or Fibonacci numbers, or alternate Fibonacci, or the $c_{n+2} = 2 c_{n+1 } + 2 c_n$ from the other answer. </p>
<p>The hint for beginning a proof is just that, with $a_{n-1} > 2,$ it is not possible for both
$$ \frac{a_n^2 - 1}{a_{n-1}} $$ and
$$ \frac{a_n^2 + 1}{a_{n-1}} $$
to be integers. That is, once $a_3$ and $a_4$ are chosen and legal, the rest of the sequence is forced by the condition of remaining integers. The four sequences begin
$$ 1, 2, 3, 4, $$
$$ 1, 2, 3, 5, $$
$$ 1, 2, 5, 12, $$
$$ 1, 2, 5, 13, $$</p>
|
4,264,808 | <p>So we know that <span class="math-container">$\frac{n^2}{2} \geq \frac{n}{2}$</span>, but I'm stuck proving that <span class="math-container">$\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$</span>. Am I missing something?</p>
| eyeballfrog | 395,748 | <p>For all <span class="math-container">$t > 0$</span>, we obviously have <span class="math-container">$t > \sqrt{t} - 1$</span>. Therefore we also have
<span class="math-container">$$
e^t > 2^t > 2^{\sqrt{t} - 1},
$$</span>
or equivalently,
<span class="math-container">$$
\frac{e^{2t}}{2^\sqrt{t}} > \frac{e^t}{2}.
$$</span>
Now just let <span class="math-container">$t = \ln x$</span>.</p>
|
4,264,808 | <p>So we know that <span class="math-container">$\frac{n^2}{2} \geq \frac{n}{2}$</span>, but I'm stuck proving that <span class="math-container">$\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$</span>. Am I missing something?</p>
| DanielWainfleet | 254,665 | <p>For <span class="math-container">$n\ge 1$</span> we have</p>
<p><span class="math-container">$n^2/2^{\sqrt {\log n}}\ge n/2\iff$</span></p>
<p><span class="math-container">$ n/2^{\sqrt {\log n}}\ge 1/2 \iff $</span></p>
<p><span class="math-container">$ 2n\ge 2^{\sqrt {\log n}}\iff$</span></p>
<p><span class="math-container">$\log 2 +\log n\ge (\sqrt {\log n}\,)\log 2\iff $</span></p>
<p><span class="math-container">$\log 2 +\log n - (\sqrt {\log n})\log 2\ge 0 \iff$</span></p>
<p><span class="math-container">$(\sqrt {\log n}-(\log 2)/2))^2+\log 2 -(\log 2)/2)^2\ge 0.$</span></p>
|
3,441,647 | <p>I was asked to find the domain of <span class="math-container">$$\arcsin[\frac{x^2+1}{2x}]$$</span>
My first step was <span class="math-container">$$-1\leq\frac{x^2+1}{2x} \leq1$$</span></p>
<p>What I don't understand is why I cannot cross multiply to get <span class="math-container">$$-2x\le {1+x}^{2} \le 2x$$</span> and then solve the inequality? I tried doing this and got the wrong answer.</p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$y=\arcsin\dfrac{x^2+1}{2x}$</span></p>
<p><span class="math-container">$x^2-2x\sin y+1=0$</span></p>
<p><span class="math-container">$x=\sin y\pm i\cos y$</span></p>
<p>As <span class="math-container">$x$</span> is real, <span class="math-container">$\cos y$</span> must be <span class="math-container">$0$</span></p>
<p><span class="math-container">$\implies\sin y=\pm1$</span></p>
|
2,418,274 | <p>AP=DS=CR=BQ=2a, ABCD and PQRS are squares with a side a
find the angle between AR and BC..
i got $$AR=\sqrt{6}a$$can i use ARQ triangle in order to find the corresponding angle..</p>
<p><a href="https://i.stack.imgur.com/QtRaT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QtRaT.jpg" alt="enter image description here"></a></p>
| orangeskid | 168,051 | <p>The triangle $\Delta ACR$ has a right angle at $C$. The side $AC = \sqrt{2} CR$, so $\tan \alpha = \frac{1}{\sqrt{2}}$, and you get your $\alpha = \arctan \frac{1}{\sqrt{2}}= \ldots$. </p>
<p>Btw, the large diagonal is $\sqrt{3} \cdot \textrm{edge} = 2\sqrt{3} a$, if you want to use $\arcsin$. </p>
|
1,348,763 | <p>Using the comparison test I am supposed to figure out whether this integral converges or diverges. what other function should I use? Also, the inequality stating that $1/\sqrt{e^x+1}$ is larger or smaller than another function must be proven.</p>
| Marco Cantarini | 171,547 | <p>Note that we can calculate the precise value of the integral, not only the convergence. We have $$\int_{1}^{\infty}\frac{1}{\sqrt{1+e^{x}}}dx\overset{{\scriptstyle e^{x}=u}}{=}\int_{e}^{\infty}\frac{1}{\sqrt{u+1}u}du\overset{{\scriptstyle u+1=v}}{=}\int_{1+e}^{\infty}\frac{1}{\left(v-1\right)\sqrt{v}}dv\overset{{\scriptstyle t=\sqrt{v}}}{=}2\int_{\sqrt{1+e}}^{\infty}\frac{1}{t^{2}-1}dt
$$ and now the integral is easy to solve $$=\lim_{a\rightarrow\infty}\left.\left(-2\tanh^{-1}\left(t\right)\right)\right|_{\sqrt{1+e}}^{a}=2\coth^{-1}\left(\sqrt{1+e}\right).
$$</p>
|
1,556,503 | <blockquote>
<p>Assume that $f$ is $a$-strongly convex and $g$ is $b$-strongly convex. Is the sum $f+g$ strongly convex, and with what constant? </p>
</blockquote>
<p><strong>Definition:</strong> $f$ is $a$-strongly convex if
$$ f(x)-f(y) \le \left<f '(x),x-y\right>-\frac{a}{2}\|x-y\|^2$$
for all $x,y$</p>
| Jing Lu | 295,496 | <p>c-strongly convex, c>=a+b</p>
<p>proof:</p>
<p>f(x+y)>=f(x)+f'(x)y+f''(x)/2 *y^2</p>
<p>g(x+y)>=g(x)+g'(x)y+g''(x)/2 *y^2</p>
<p>where f' is the first order gradient, g' is the first order gradient, f''(x) is some positive number at least a, g''(x) is some positive number at least b, both of them depend on x,</p>
<p>adding them together, we have F(x)=f(x)+g(x) </p>
<p>F(x+y)>=F(x)+F'(x)y+F''(x)/2 *y^2</p>
<p>where F'=f'+g' is the gradient of F
F''=f''+g''>=a+b
based on the strongly convex definition, F is c strongly convex</p>
<p>c=a+b if f'' and g'' have the same min location
c>a+b if f'' and g'' have different min location.</p>
<p>First time to use this website, can i use latex here?</p>
|
1,556,503 | <blockquote>
<p>Assume that $f$ is $a$-strongly convex and $g$ is $b$-strongly convex. Is the sum $f+g$ strongly convex, and with what constant? </p>
</blockquote>
<p><strong>Definition:</strong> $f$ is $a$-strongly convex if
$$ f(x)-f(y) \le \left<f '(x),x-y\right>-\frac{a}{2}\|x-y\|^2$$
for all $x,y$</p>
| Guy Fsone | 385,707 | <p>This is totally pretty easy from your definition we have </p>
<p>$$f(x)-f(y) \le \left<f '(x),x-y\right>-\frac{a}{2}\|x-y\|^2$$
and
$$g(x)-g(y) \le \left<g '(x),x-y\right>-\frac{b}{2}\|x-y\|^2$$</p>
<p>summing up both inequality we obtain:
$$f(x) +g(x)-f(y)+g(y) \le\left<f '(x),x-y\right> +\left<g '(x),x-y\right>\\-\frac{a}{2}\|x-y\|^2-\frac{b}{2}\|x-y\|^2$$</p>
<p>that is $$(f+g)(x)-(f+g)(y) \le\left<(f +g)'(x),x-y\right> +\left<g '(x),x-y\right>-\frac{a+b}{2}\|x-y\|^2.$$</p>
<p>Hence $f+g$ is $a+b$-strongly convex </p>
|
194,128 | <p>I am trying to solve this problem by induction. The sad part is that I don't have a very strong grasp on solving by inductive proving methods. I understand that there is a base case and that I need an inductive step that will set $k = n$ and then one more step that basically sets $k = n + 1$. </p>
<p>Here is the problem I am trying to solve: </p>
<blockquote>
<p>If $f(n) = \sum_{i = 0}^n X_{i}$, then show by induction that $f(n) = f(n - 1) + X_{n-1}$.</p>
</blockquote>
<p>Can I have someone please try to point me in the right direction? </p>
<p>*EDIT: I update the formula to the correct one. I wasn't sure how to typeset it correctly and left errors in my math. Thank you for those that helped. I'm still having the problem but now I have the proper formula posted. </p>
| Clive Newstead | 19,542 | <p>The way induction works is based on two steps:</p>
<ol>
<li>Check a base case.</li>
<li>Show that if it holds for one number, then it holds for the next.</li>
</ol>
<p>More formally, let $P$ be a property, and let $P(n)$ be the assertion that the property holds true for the natural number $n$; so for instance, $P(n)$ could be the assertion that $\sum_1^n i = \frac{n(n+1)}{2}$. Then to prove by induction that $P(n)$ holds for all $n \ge 1$, it suffices to</p>
<ol>
<li>Check that $P(1)$ holds;</li>
<li>Check that, for any given value of $n$, if $P(n)$ holds then $P(n+1)$ holds.</li>
</ol>
<p>Then we have a sort of domino effect. $P(1)$ holds as you showed for (1); and $P(2)$ holds because $2=1+1$ and $P(1)$ holds; and $P(3)$ holds because $3=2+1$ and $P(2)$ holds; and $P(4)$ holds because $4=3+1$ and $P(3)$ holds; and so on.</p>
<p><strong>But</strong> here, induction is not necessary, since you can prove it directly for any given value of $n$. For any function $g$, $\displaystyle \sum_{i=1}^n g(i) = \sum_{i=1}^{n-1} g(i) + g(n)$ just by the definition of the $\sum$ symbol (unless you have a weird definition), and the result follows immediately from this fact.</p>
|
12,047 | <p>How to prove the following trigonometric identities ?</p>
<p>1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$
</p>
<p>I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?</p>
<p>2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $
</p>
<p>For this one I tried substituting them with the sides of the triangle, but not successful to the final result.</p>
<p>These are not my homework, I am trying to learn maths almost on my own, so ...</p>
| user3180 | 67,758 | <p>1) tan(a)tan(b) = 1 is equivilant to sin(a)sin(b) = Cos(a)cos(b) **using the fact that tan is sine divided by cosine. Next use
cos(a+b) = cos(a)cos(b) -sin(a)sin(b)</p>
<p>which implies sin(a)sin(b) = cos(a)cos(b) -cos(a+b)</p>
<p>Use this on the left hand side of ** to get </p>
<pre><code> cos(a+b) = 0
</code></pre>
<p>which implies a+b = Pi/2 or -Pi/2</p>
|
12,047 | <p>How to prove the following trigonometric identities ?</p>
<p>1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$
</p>
<p>I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?</p>
<p>2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $
</p>
<p>For this one I tried substituting them with the sides of the triangle, but not successful to the final result.</p>
<p>These are not my homework, I am trying to learn maths almost on my own, so ...</p>
| Américo Tavares | 752 | <p>Added (and corrected twice):</p>
<p>$$\tan \alpha \tan \beta =1\Leftrightarrow \dfrac{\sin \alpha }{\cos \alpha }%
\dfrac{\sin \beta }{\cos \beta }=1$$
</p>
<p>Multiplying by $\cos\alpha\cos\beta\ne 0$, gives</p>
<p>$$\sin \alpha \sin \beta -\cos \alpha\cos \beta =0 \Leftrightarrow \cos (\alpha +\beta )=0$$
</p>
<p>This is equivalent to </p>
<p>$$\alpha +\beta =\dfrac{\pi }{2}+n\pi,\qquad (\ast)$$
</p>
<p>to which we still have to add the condition written above ($\cos\alpha\cos\beta\ne 0$), which means the constraint </p>
<p>$$\alpha,\beta\ne\dfrac{\pi}{2}+n\pi,\qquad (\ast\ast)$$
</p>
<p>where $n$ is an integer.</p>
<p>Note: In the original equation $\tan \alpha \tan \beta =1$, neither $\alpha$ nor $\beta$ can be zero. The combined conditions $(\ast)$ and $(\ast\ast)$ assures that.</p>
<hr>
<p>The identity $$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\qquad \theta \neq (2n+1)\dfrac{\pi}{2}$$
</p>
<p>is equivalent to $$\sin ^{2}\theta +\cos ^{2}\theta =1.$$ </p>
<p>Indeed, if </p>
<p>$$\theta \neq (2n+1)\dfrac{\pi }{2}\Leftrightarrow \sin \theta \neq \pm 1 \Leftrightarrow \dfrac{\pm1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }\neq 0\Leftrightarrow \pm\sec \theta -\tan \theta \neq 0,$$
</p>
<p>then</p>
<p>$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta
-\tan \theta \right) =1$$
</p>
<p>$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$</p>
|
2,254,694 | <p>I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:</p>
<p>Attempt 1:</p>
<p>Write out the terms of $b_n$:</p>
<p>$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$</p>
<p>Adding up the terms you get:</p>
<p>$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$</p>
<p>But a dead end here.</p>
<p>Attempt 2:</p>
<p>For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$</p>
<p>Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. </p>
<p>For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. </p>
| Sangchul Lee | 9,340 | <p>Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then</p>
<p>$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$</p>
<p>Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any <em>fixed</em> $m$ and for any $n > m$, we have</p>
<p>$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$</p>
<p>Note here that</p>
<ul>
<li><p>$\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.</p></li>
<li><p>$\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.</p></li>
</ul>
<p>So, taking limsup as $n\to\infty$ yields</p>
<p>$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$</p>
<p>Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim.</p>
|
1,248,668 | <p>I do have a system of n equations with m variables where m > n with integer coefficients. I wish to find a set of integer solutions to this system (In my case n = 2 and m = 4). Could somebody tell me how I can do it? I already solved this system with Mathematica but I would like to redo these calculations by hand to understand how their were obtained.</p>
<p>The system is:
$\left\{
\begin{array}{l l}
4u - 3v + 4w + 3z = 1\\
-4v - 3u - 4z + 3w = 0
\end{array} \right.$</p>
| marekszpak | 147,160 | <p>Well, finally I have used old good Hermite Normal Form. I knew this method before but I thought about something which could be easier for bigger systems to solve by hand.</p>
<p>If something check paper <a href="https://www.math.uwaterloo.ca/~wgilbert/Research/GilbertPathria.pdf" rel="nofollow">Linear Diophantine Equations, W. J. Gilbert, A. Pathria, 1990</a>. </p>
<p>What they used is just unimodular row operations to obtain HNF and matrix U at once. In my case I had to work on symbol coefficients and this approach led me to all what I needed. Well, if you know any better way I will be happy to give it a try. I know about Smith Normal Form but it sounds to complex - well, I have to say that I have never tried to us SNF. Maybe I should.</p>
|
2,765,570 | <p>Let $E = \{\frac{1}{n} | n \in \mathbb{N}\}$. Show that the function $f(x) = 1 \text{ if } x \in E \text{ and } f(x) = 0 \text{ if } x \notin E$ is integrable on $[0,1]$.</p>
<p>I know of the definition of integrable but I'm having a hard time applying it to this function.</p>
| Olecram | 558,042 | <p>You can note that, the inferior sum and superior sum be equal to 0, because for every partition of interval [0,1], $$P=\{0=x1<...<xn=1\}$$ $$\inf_{x \in [x_k,x_{k+1}]} f(x)=0$$ because you always can take x that is not of form $$\frac{1}{n}$$,the same for sup, thus, f is integrable and your integral is equal to 0. </p>
|
495,064 | <p>I'm currently tackling the following problems:</p>
<p>a) Let $a,b \in \mathbb{Z}$ and let $m$ be a nonnegative integer. Prove that $(a,b)=1$ if and only if $(a^m,b)=1$.</p>
<p>If $(a^m,b)=1$ then $pa^m+qb=1$ for some integers $p$ and $q$. It follows that $(pa^{m-1})a+qb=1$, which shows that $(a,b)=1$. This took care of the ''$\Leftarrow$'' direction. I can't think of a way for the ''$\Rightarrow$'' direction. Can you help me here?</p>
<p>b) In addition to the above, let now $n$ be a nonnegative integer as well. Prove that $(a,b)=1$ if and only if $(a^m,b^n)=1$.</p>
<p>The ''$\Leftarrow$'' direction is completely analogous. How about the ''$\Rightarrow$'' direction? </p>
<p><strong>Edit:</strong> Wow, this really wasn't that hard. Here is what I ended up with:</p>
<p>(a) We prove this by contradiction. First assume that $(a,b)=1$ and suppose that $(a^m,b)=d>1$. Then there is a prime $p$ in the prime factorization of $d$ such that $p \mid a^m$ and $p \mid b$. But if $p \mid a^m$, then $p \mid a$, which implies that $p \mid a$ and $p \mid b$, which contradicts the fact that $(a,b)=1$. Thus, $(a^m,b)=1$.</p>
<p>Now assume that $(a^m,b)=1$ and $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and therefore $p \mid (a^m,b)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
<p>(b) This proof is analogous to the one in (a). Assume that $(a,b)=1$ and suppose that $(a^m,b^n)=d>1$. Then we have a prime $p$ such that $p \mid b^n$ and $p \mid a^n$. But then $p \mid a$ and $p \mid b$, from which it follows that $p \mid (a,b)$, which contradicts our initial assumption. Thus, $(a^m,b^n)=1$.</p>
<p>Now assume that $(a^m,b^n)=1$ and that $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and $p \mid b^n$. It follows that $p \mid (a^m,b^n)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
| André Nicolas | 6,312 | <p>The Bezout Theorem is I think not the best approach. I prefer the approach of some1.new4U. </p>
<p>But Bezout will work. Since $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$.</p>
<p>Take the $m$-th power of $ax+by$, using the Binomial Theorem. The first term is $a^mx^m$. The remaining terms all have a positive power of $b$ in them, so they add up to $bt$ for some integer $t$. </p>
<p>Thus $a^ms+bt=1$, where $s=x^m$. </p>
|
2,341,552 | <p>I am pretty sure that
$$\bigg|\int_{A}e^{it}\,dt\bigg|\leq2$$
for every measurable set $A\subseteq[-\pi,\pi]$,
but I cannot prove this...</p>
| Richard | 191,878 | <p>We have
$$ I=\int_a^be^{it}dt=\frac{1}{i}(e^{ib}-e^{ia}) $$
Then
$$ |I|=|-i||e^{ib}-e^{ia}|=|e^{ib}||1-e^{i(b-a)}|\le |1|+|e^{i(b-a)}|\le 1+1=2$$</p>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| Cody | 13,295 | <p>SOS always has the most clever and ingenious solutions, but if I may contribute something I have found interesting. A fun method of evaluating a whole slew of Euler sums is by using the residues of digamma. </p>
<p>By noting the identity, $\displaystyle \sum_{n=1}^{\infty}\frac{(H_{n})^{2}}{n^{2}}=2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}+\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}......[1]$</p>
<p>one can evaluate each of the sums on the right side and thus arrive at the quadratic Euler sum in question.</p>
<p>For the first sum on the right, begin by considering $\displaystyle f(z)=\frac{\left(\gamma+\psi(-z)\right)^{2}}{z^{2}}$ and, due to the poles of digamma, compute the residue at n (the positive integers). </p>
<p>As $z\to n$, the series is $\displaystyle\frac{1}{(z-n)^{2}}+\frac{2H_{n}}{z-n}+\cdot\cdot\cdot $</p>
<p>Thus, the residue is $\displaystyle\lim_{z\to n}\left[Res\left(\frac{1}{(z-n)^{2}}\cdot \frac{1}{z^{3}}\right)+Res\left(\frac{2H_{n}}{z-n}\cdot \frac{1}{z^{3}}\right)\right]$</p>
<p>$\displaystyle=\frac{-3}{n^{4}}+\frac{2H_{n}}{n^{3}}$</p>
<p>Sum these residues: $\displaystyle-3\sum_{n=1}^{\infty}\frac{1}{n^{4}}+2\sum_{n=1}^{\infty}\frac{2H_{n}}{n^{2}}$</p>
<p>By taking the Laurent expansion of f(z), the residue at z=0 is the coefficient of the 1/z term. </p>
<p>$\displaystyle \psi(-z)+\gamma = \frac{1}{z}-\zeta(2)z+\zeta(3)z^{2}-\zeta(4)z^{3}+\cdot\cdot\cdot$ </p>
<p>$\displaystyle f(z)=\frac{1}{z^{5}}-\frac{\pi^{2}}{3}\cdot \frac{1}{z^{3}}-2\zeta(3)\cdot \frac{1}{z^{2}}+\frac{\pi^{4}}{180}\cdot \frac{1}{z}+\cdot\cdot\cdot $</p>
<p>As can be seen, the residue at 0 is $\frac{\pi^{4}}{180}$</p>
<p>Put them together, set to 0, and get </p>
<p>$\displaystyle2\sum_{n=1}^{\infty}\frac{2H_{n}}{n^{3}}-3\sum_{n=1}^{\infty}\frac{1}{n^{4}}+\frac{\pi^{4}}{180}=0$</p>
<p>$\displaystyle2H-\frac{\pi^{4}}{30}+\frac{\pi^{4}}{180}=0$</p>
<p>$\displaystyle \boxed{\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=\frac{\pi^{4}}{72}}.......[2]$</p>
<p>Now, for the other sum on the right of [1], where $\displaystyle H_{n}^{(2)}=\sum_{k=1}^{n}\frac{1}{k^{2}}$</p>
<p>$\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}$</p>
<p>Due to symmetry of Euler sums, if we have a sum $\displaystyle S_{p,q}=\sum_{n=1}^{\infty}\frac{H_{n}^{(p)}}{n^{q}}$, and $p=q$, then by symmetry $S_{p,q}+S_{q,p}=\zeta(p)\zeta(q)+\zeta(p+q)$</p>
<p>So, in this case with $p=q=2$, then </p>
<p>$\displaystyle2\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}=\frac{\pi^{2}}{36}+\frac{\pi^{4}}{90}=\frac{7\pi^{4}}{180}$</p>
<p>$\displaystyle \boxed{\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}=\frac{7\pi^{4}}{360}}$</p>
<p>Now, add this to the result of the other sum in [2]:</p>
<p>$\displaystyle\frac{7\pi^{4}}{360}+2\cdot \frac{\pi^{4}}{72}=\frac{17\pi^{4}}{360}$</p>
<p><strong>EDIT:</strong></p>
<p>If I may expand somewhat on this sum using the same technique but a different f(z). Of course, it requires a couple known Euler sums as lemmata. </p>
<p>By considering $\displaystyle f(z)=\frac{(\gamma+\psi(-z))^{3}}{z^{2}}$, one can use the residues at 0 and the positive integers to find the sum.</p>
<p>Using the series for $\displaystyle(\gamma+\psi(-z))^{3}$ at z=n:</p>
<p>$\displaystyle \frac{1}{(z-n)^{3}}+\frac{3H_{n}}{(z-n)^{2}}+\frac{3(H_{n})^{2}}{z-n}-\frac{3H_{n}^{(2)}}{z-n}-\frac{\pi^{2}}{2(z-n)}+\cdot\cdot\cdot $</p>
<p>Thus, the residues at z=n are:</p>
<p>$\displaystyle\lim_{z\to n}\left(Res\left[\frac{1}{(z-n)^{3}}\cdot \frac{1}{z^{2}}\right]+Res\left[\frac{3H_{n}}{(z-n)^{2}}\cdot \frac{1}{z^{2}}\right]+Res\left[\frac{3(H_{n})^{2}}{z-n}\cdot \frac{1}{z^{2}}\right]-Res\left[\frac{H_{n}^{(2)}}{z-n}\cdot \frac{1}{z^{2}}\right]-Res\left[\frac{\pi^{2}}{2(z-n)}\right]\right)$</p>
<p>The first two require derivatives due to the pole at n being of order 3. But, we ultimately obtain the sums:</p>
<p>$\displaystyle 3\sum_{n=1}^{\infty}\frac{1}{n^{4}}-6\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}+3\sum_{n=1}^{\infty}\frac{(H_{n})^{2}}{n^{2}}-3\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}-\frac{\pi^{2}}{2}\sum_{n=1}^{\infty}\frac{1}{n^{2}}+\frac{\pi^{4}}{20}=0$</p>
<p>Also, the residue at z=0 is $\displaystyle\frac{\pi^{4}}{20}$, which can be found by using its Laurent expansion: </p>
<p>$\displaystyle f(z)=\frac{1}{z^{5}}-\frac{3\zeta(2)}{z^{3}}-\frac{3\zeta(3)}{z^{2}}+\frac{\pi^{4}}{20z}+\cdot\cdot\cdot $ </p>
<p>Sum residues, evaluate known sums, call the quadratic sum being found H, set to 0 and solve for H. </p>
<p>$\displaystyle=\frac{\pi^{4}}{30}-\frac{\pi^{4}}{12}+3H-\frac{7\pi^{4}}{120}+\frac{\pi^{4}}{20}-\frac{\pi^{4}}{12}=0$</p>
<p>$\displaystyle \sum_{n=1}^{\infty}\frac{(H_{n})^{2}}{n^{2}}=\frac{17\pi^{4}}{360}$</p>
<p>Random Variable is an expert in this method and has refined it very well.</p>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| Ali Shadhar | 432,085 | <p>Using the generating function:</p>
<p><span class="math-container">\begin{gather}
\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}x^{n}=\operatorname{Li}_4(x)-2\operatorname{Li}_4(1-x)+2\ln(1-x)\operatorname{Li}_3(1-x)+\frac12\operatorname{Li}_2^2(x)\nonumber\\
\qquad-\ln^2(1-x)\operatorname{Li}_2(1-x)-\frac13\ln(x)\ln^3(1-x)+2\zeta(4),
\end{gather}</span></p>
<p>we have the following results:</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{H_n^{2}}{n^2}=\frac{17}{4}\zeta(4);$$</span></p>
<p><span class="math-container">$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)+\frac1{12}\ln^4(2);$$</span></p>
<p><span class="math-container">$$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\operatorname{Li}_4\left(\frac12\right)+\frac{37}{16}\zeta(4)-\frac74\ln(2)\zeta(3)+\frac14\ln^2(2)\zeta(2)-\frac1{24}\ln^4(2).$$</span></p>
|
167,228 | <p>I have shown several plots by Show function:</p>
<pre><code>Show[p1, p2, p3, p4, p5, p6, p7, p8, PlotRange -> All]
</code></pre>
<p>I need to save data of all plots in one <code>txt</code> format, so that they can be recognizable from each other. How can I do that?</p>
<p>Each plot is a 2D plot say, $f_i$ as a function of $x$.</p>
| Carl Woll | 45,431 | <p>Polynomials with real coefficients have either real roots, or complex roots that come in conjugate pairs. Suppose $r$ and $\bar{r}$ are complex conjugates, then:</p>
<pre><code>(x - r)(x - Conjugate[r]) == x^2 - 2 Re[r] x + Abs[r]^2 //FullSimplify
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<p>The function <code>realFactor</code> uses the above:</p>
<pre><code>realFactor[poly_, x_] := With[
{
real = Flatten @ Values @ Solve[poly == 0, x, Reals],
complex = Flatten @ Values @ Solve[poly == 0 && Im[x]>0, x]
},
Times @@ Join[
x - real,
x^2 - 2 Re[#]x + Abs[#]^2& /@ complex
] //RootReduce
]
</code></pre>
<p>Your example:</p>
<pre><code>realFactor[x^4 + 1, x]
</code></pre>
<blockquote>
<p>(1 - Sqrt[2] x + x^2) (1 + Sqrt[2] x + x^2)</p>
</blockquote>
<p>More complicated examples:</p>
<pre><code>realFactor[x^5 + 1, x]
realFactor[x^6 + x + 1, x]
</code></pre>
<blockquote>
<p>(1 + x) (1 + 1/2 (-1 - Sqrt[5]) x + x^2) (1 + 1/2 (-1 + Sqrt[5]) x + x^2)</p>
<p>(x^2 + x Root[1 - 27 #1^3 - 18 #1^4 - 12 #1^5 - 26 #1^9 + 10 #1^10 + #1^15 &,
3] + Root[-1 + #1^3 + #1^4 + #1^5 + 2 #1^6 + #1^7 - 2 #1^9 -
2 #1^10 - #1^12 + #1^15 &, 1]) (x^2 +
x Root[1 - 27 #1^3 - 18 #1^4 - 12 #1^5 - 26 #1^9 + 10 #1^10 + #1^15 &, 2] +
Root[-1 + #1^3 + #1^4 + #1^5 + 2 #1^6 + #1^7 - 2 #1^9 -
2 #1^10 - #1^12 + #1^15 &, 2]) (x^2 +
x Root[1 - 27 #1^3 - 18 #1^4 - 12 #1^5 - 26 #1^9 + 10 #1^10 + #1^15 &, 1] +
Root[-1 + #1^3 + #1^4 + #1^5 + 2 #1^6 + #1^7 - 2 #1^9 -
2 #1^10 - #1^12 + #1^15 &, 3])</p>
</blockquote>
|
3,081,552 | <p>Is this <code>∅ ⊈ { ∅, 1, 2 }</code> true or false ? </p>
<p>Also, I am confuse since this <code>{ ∅, 1, 2 }</code> has already contain a <code>∅</code>, does it still contain another <code>∅</code> meaning like : <code>{ ∅, ∅, 1, 2 }</code> ? </p>
<p>Is <code>∅ ∈ { ∅, 1, 2 }</code> true ? & <code>{∅} ∈ { ∅, 1, 2 }</code> false ?</p>
| Michael Rozenberg | 190,319 | <p><span class="math-container">$$\oslash\subset\{\oslash,1,2\}$$</span> and <span class="math-container">$$\oslash\in\{\oslash,1,2\}.$$</span>
Also, <span class="math-container">$$\{\oslash\}\subset\{\oslash,1,2\}.$$</span></p>
|
479,671 | <p>Let $p$ and $q$ be two distinct primes. Prove that $$p^{q-1} + q^{p-1} =1 \mod pq$$ I try to used Fermat little theorem and I obtain the congruence $p^q +q^p=0 \mod pq$. From this I don know how to relate to the congruence on the question.</p>
| pritam | 33,736 | <p>By Fermat's little theorem $p^{q-1}\equiv 1\pmod{q}$ and $q^{p-1}\equiv 0\pmod{q}$, now adding these two $$p^{q-1}+q^{p-1}\equiv1\pmod{q}$$ Similarly, the above also holds modulo $p$. Now since $\gcd (p,q)=1$, this also holds modulo $pq$.</p>
|
4,531,507 | <p><a href="https://i.stack.imgur.com/rJmjH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rJmjH.jpg" alt="enter image description here" /></a></p>
<blockquote>
<p>Find the area of <span class="math-container">$ABCD$</span> if <span class="math-container">$PSC$</span> is a semicircle.</p>
</blockquote>
<p>I have used the tangent properties of circles. I also assumed many variables to get a relationship between product of sides of the rectangle <span class="math-container">$ABCD$</span> but eventually and unfortunately, I proved <span class="math-container">$$1=1$$</span></p>
<p>Any help is greatly appreciated.</p>
| gus f | 1,025,519 | <p>We can see in triangles PQC and QBC that</p>
<p><span class="math-container">$∠QPC = ∠BQC,∠QCP$</span> is common.
<span class="math-container">$∠PQC = ∠QBC = 90$</span>
Therefore both triangles are similar.</p>
<p><span class="math-container">$\frac{QC}{PC}=\frac{BC}{QC}$</span>,
<span class="math-container">$PC∗BC=10^{2}=100$</span></p>
<p>Now We can say that one side CD is equal to the radius and so diameter <span class="math-container">$PC=2∗radius = 2*CD$</span></p>
<p>Area=<span class="math-container">$BC*CD = \frac{BC∗PC}{2}=\frac{100}{2}=50$</span></p>
|
4,531,507 | <p><a href="https://i.stack.imgur.com/rJmjH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rJmjH.jpg" alt="enter image description here" /></a></p>
<blockquote>
<p>Find the area of <span class="math-container">$ABCD$</span> if <span class="math-container">$PSC$</span> is a semicircle.</p>
</blockquote>
<p>I have used the tangent properties of circles. I also assumed many variables to get a relationship between product of sides of the rectangle <span class="math-container">$ABCD$</span> but eventually and unfortunately, I proved <span class="math-container">$$1=1$$</span></p>
<p>Any help is greatly appreciated.</p>
| Doug M | 317,176 | <p>Lets call lengths:<br />
<span class="math-container">$\overline {BC} = x\\
\overline {PB} = y\\
\overline {BQ} = z$</span></p>
<p><span class="math-container">$\frac {x+y}2$</span> is the raduis of the circle.<br />
<span class="math-container">$\frac {x^2+xy}{2}$</span> is the area of the rectangle.</p>
<p><span class="math-container">$x^2+z^2 = 10^2$</span> by Pythagoras</p>
<p><span class="math-container">$xy = z^2$</span> I not sure the name of the theorem that provideds this result.</p>
<p><span class="math-container">$x^2 + xy = 100$</span></p>
<p>But <span class="math-container">$x^2 + xy$</span> is double the area of the rectanlge</p>
<p><span class="math-container">$A = \frac {x^2+xy}{2} = 50$</span></p>
|
2,390,804 | <p>How does one compute the following integral?
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$</p>
<hr>
<p>I have tried extending $x$ to the complex plane then evaluating the following contour integral
$$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$
with the contour $C$ running along the whole real axis and then upper semicircle. I obtain
$$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$
but not the original integral.</p>
| JJacquelin | 108,514 | <p>For information :</p>
<p>$\int_{0}^{\infty} \frac{\sqrt{x}\sin(y\:x)}{1+x^2} dx$ is the Fourier Sine Transform of $\frac{\sqrt{x}}{1+x^2}\quad $ In the Harry Bateman's Tables of Integral Transforms an even more general formula can be seen on page 71, Eq.28 , the Fourier Sine Transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$ :
$$\frac{1}{2}a^{2\nu-2\mu}\frac{\Gamma(1+\nu)\Gamma(\mu-\nu)}{\Gamma(\mu+1)}y \:_1\text{F}_2(\nu+1;\nu+1-\mu,3/2;a^2y^2/4)\:+\:4^{\nu-\mu-1}\sqrt{\pi}\frac{\Gamma(\nu-\mu)}{\Gamma(\mu-\nu+3/2)}y^{2\mu-2\nu+1}\:_1\text{F}_2(\mu+1;\mu-\nu+3/2,\mu-\nu+1;a^2y^2/4)
$$
With$\quad y=1\quad;\quad a=1\quad;\quad \nu=1/4\quad;\quad \mu=0\quad\to\quad \int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =$<br>
$$=\frac{1}{2}\Gamma(5/4)\Gamma(-1/4) \:_1\text{F}_2(5/4;5/4,3/2;1/4)\:+\:4^{-3/4}\sqrt{\pi}\frac{\Gamma(5/4)}{\Gamma(5/4)}\:_1\text{F}_2(1;5/4,3/4;1/4)
$$
The Generalized Hypergeometric $_1$F$_2$ function (don't confuse with the well-known 2F1) reduces to functions of lower level in the particular cases :</p>
<p>$\:_1\text{F}_2(5/4;5/4,3/2;1/4)=\sinh(1)$</p>
<p>$\:_1\text{F}_2(1;5/4,3/4;1/4)=\frac{\sqrt{\pi}}{4e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$</p>
<p>and after simplification :
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx = -\frac{\pi}{\sqrt{2}}\sinh(1)+\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) \right)$$
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(e^2\text{erfi}(1)+\text{erf}(1) +1-e^2\right)$$
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$
For the Hypergeometric $_1$F$_2$ function, see : <a href="http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/02/" rel="nofollow noreferrer">http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F2/02/</a></p>
<p>About the functions erf, erfi, erfc, see : <a href="http://mathworld.wolfram.com/Erf.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/Erf.html</a> , <a href="http://mathworld.wolfram.com/Erfi.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/Erfi.html</a> , <a href="http://mathworld.wolfram.com/Erfc.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/Erfc.html</a></p>
|
2,993,958 | <blockquote>
<p>Do there exist three non null vectors <span class="math-container">$a,b,c$</span> with <span class="math-container">$$a.b=a.c$$</span> such that <span class="math-container">$b\ne c$</span>?</p>
</blockquote>
<p>My attempt:</p>
<p>a.(b-c)=0</p>
<p>But is it possible to say b-c=0 since a is non null?</p>
| nonuser | 463,553 | <p>Take isosceles triangle <span class="math-container">$ABC$</span> such that <span class="math-container">$AB = AC$</span> and let <span class="math-container">$M$</span> be a midpoint of <span class="math-container">$BC$</span>. Then <span class="math-container">$$\vec{AM} \cdot \vec{AB} = \vec{AM} \cdot \vec{AC}$$</span></p>
|
805,902 | <p>I'm struggling to figure out what I'm exactly required to do. The problem states</p>
<p>"Compute which lines through the point $(1, 0)$ that are tangent to the parabola defined by $y = x^2$."</p>
<p>I believe it's a simple question however I've been going around this for quite a bit. </p>
<p>I'll appreciate any kind of help! </p>
<p>Thank you!</p>
| mfl | 148,513 | <p>The equation of a tangent line to the parabola at $(x_0,x_0^2)$ is given by $y-x_0^2=2x_0(x-x_0).$ If $(1,0)$ is a point of this line we have $-x_0^2=2x_0(1-x_0),$ or equivalently, $x_0^2=2x_0.$ So $x_0=0$ or $x_0= 2.$ That is, $(1,0)$ lies in the tangents to the parabola at points $(0,0)$ and $(2,4).$ Their equations are $y=0$ and $y-4=4(x-2)$ </p>
|
805,902 | <p>I'm struggling to figure out what I'm exactly required to do. The problem states</p>
<p>"Compute which lines through the point $(1, 0)$ that are tangent to the parabola defined by $y = x^2$."</p>
<p>I believe it's a simple question however I've been going around this for quite a bit. </p>
<p>I'll appreciate any kind of help! </p>
<p>Thank you!</p>
| Joel Hernandez | 120,476 | <p>Some comments were really helpful but I believe none gave me the intuition for this, which is what I really didn't understand , however I'm truly grateful for everyone that commented or attempted to help me.</p>
<p>Here's my approach after dreaming about it the entire night. (Not just the intuition)</p>
<p>We will first determine a general equation for all lines tangent to the function $f(x)$ at any given x , to do so let's create a generic point for our lines, let's make the point $(a,f(a))$ or equivalently $(a,a^2)$</p>
<p>We proceed to get the slope at any given value by getting the $d/dx$
of our function $f(x)$ wich results to be $f'(x) = 2x$</p>
<p>Next, let's evaluate the slope at our generalized point $(a,a^2)$</p>
<p>$$f'(a) = 2a$$</p>
<p>By doing this we can start to construct our line equation with $2a$ as the slope.</p>
<p>$$y=2ax+b$$</p>
<p>We're missing the b value, and to solve for it we will use the point that we are sure lie on our line equation, that is our point $(a,a^2)$ , so let's substitute it into our equation.</p>
<p>$$a^2=2a(a)+b$$</p>
<p>which is equal to </p>
<p>$$a^2=2a^2+b$$</p>
<p>Next, let's move the non b terms to the left , resulting
$$-a^2=b$$</p>
<p>With that we can substitute our b to our first equation, resulting</p>
<p>$$y=2ax-a^2$$</p>
<p>This is the tricky part, we need to find which values of a would contain the point (1,0) (On this specific example) so we can substitute x = 1 and y = 0 on our equation, resulting</p>
<p>$$0=2a(1)-a^2$$</p>
<p>equivalently </p>
<p>$$0=2a-a^2$$</p>
<p>We proceed to factor out the a as </p>
<p>$$0=a(2-a)$$</p>
<p>We know that if our outcome it's zero then one of the terms being multiplied is going to be equal to zero, so we can say that either </p>
<p>$$a=0$$</p>
<p>or the term</p>
<p>$$2-a=0$$</p>
<p>Which we can easily solve for a resulting in</p>
<p>$$-a=-2$$</p>
<p>or $$a=2$$</p>
<p>so either a is equal to zero or a is equal to 2 , so we can substitute our "a's" on our first defined function for all tangent lines $y=2ax-a^2$, resulting in the line equations</p>
<p>$$y=2(2)x-(2^2)$$</p>
<p>Which is equivalent to $$y=4x-4$$</p>
<p>and $$y=2(0)x-(0^2)$$</p>
<p>Which is equivalent to $$y=0$$</p>
<p>So by doing these things we managed to get the equation of the lines tangent to $f(x)=x^2$ that go through the point (1,0)</p>
<p>I hope that this helps someone! </p>
<p>Regards.</p>
|
805,902 | <p>I'm struggling to figure out what I'm exactly required to do. The problem states</p>
<p>"Compute which lines through the point $(1, 0)$ that are tangent to the parabola defined by $y = x^2$."</p>
<p>I believe it's a simple question however I've been going around this for quite a bit. </p>
<p>I'll appreciate any kind of help! </p>
<p>Thank you!</p>
| André Nicolas | 6,312 | <p>We can use basic "algebra." First draw a picture. That will pick up the fact that the $x$-axis is such a tangent line, and that there appears to be another one. </p>
<p>Now for the algebra. The tangent line "kisses" the parabola, so meets it at a "double-point." </p>
<p>The generic (non-vertical) line through $(1,0)$ has equation $y=m(x-1)$. We want the equation $m(x-1)=x^2$ to have a double root. The equation can be rewritten as $x^2-mx+m=0$. This has a double root if the discriminant $m^2-4m$ is equal to $0$, which happens at $m=0$ (which we knew) and $m=4$. </p>
<p><strong>Remark:</strong> The above method goes back to Fermat, and a little later, Descartes.</p>
|
692,383 | <p>Let $p, q$ be prime numbers which may or may not be distinct.
Let $\mathbb{Q}_p$ be the field of $p$-adic numbers.
Let $\mathbb{Z}_p$ be the ring of $p$-adic integers.
We define similarly $\mathbb{Q}_q$ and $\mathbb{Z}_q$.</p>
<p>Let $A = \mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$.
Let $\lambda\colon \mathbb{Z}_p \rightarrow A$
and $\mu\colon \mathbb{Z}_q \rightarrow A$ be the canonical ring homomorphisms.
Let $B = \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$.
Let $\psi\colon B \rightarrow A$ be the ring homomorphism induced by $\lambda$ and $\mu$.</p>
<p>Is $\psi$ injective?</p>
| Martin Brandenburg | 1,650 | <p>Since $A$ is the localization of $B$ at the elements $p$ and $q$ (by formal nonsense), the question is if $p$ and $q$ are regular elements of $B=\mathbb{Z}_p \otimes \mathbb{Z}_q$. But this is because $p$ is a regular element of $\mathbb{Z}_p$ and $\mathbb{Z}_q$ is torsion-free and hence flat over $\mathbb{Z}$; similarly for $q$.</p>
|
3,841 | <p>Taking tori in symmetric products and "miraculously" proving that the Floer homology is independent of choices always seemed, well, miraculous. Some time ago Max Lipyanski explained to me the origins of this construction from gauge theory on surfaces, a la Atiyah-Floer conjecture, which I have then forgotten. What is the origin of Heegard Floer?</p>
| Mike Usher | 424 | <p>From Szabó's delightfully understated <a href="http://www.ams.org/notices/200704/comm-veblen-web.pdf" rel="nofollow noreferrer">response</a> (pdf) to receiving the Veblen prize:</p>
<blockquote>
<p>The joint work with Peter Ozsváth
which is noted here grew out of our
attempts to understand Seiberg–Witten
moduli spaces over three-manifolds
where the metric degenerates along a
surface. This led to the construction of Heegaard Floer homology
that involved both
topological tools, such as Heegaard diagrams, and
tools from symplectic geometry, such as holomorphic
disks with Lagrangian boundary constraints.
The time spent on investigating Heegaard Floer
homology and its relationship with problems in
low-dimensional topology was rather interesting.</p>
</blockquote>
<p>Of course, if one believes that Heegaard Floer homology is somehow the limit of monopole Floer homology as one degenerates the metric in some way that depends on the Heegaard diagram, then the independence of Heegaard Floer homology from the Heegaard diagram would fall out from the metric-independence of monopole Floer homology. Unfortunately, I can't seem to find references that give any sort of precise picture of how Ozsváth and Szabó came to think that this should be the case (though it might have been a baby analogue of the picture in <a href="http://www.math.purdue.edu/%7Eyjlee/publ/mcmaster.pdf" rel="nofollow noreferrer">this paper</a><sup>link broken</sup> (pdf) by <a href="https://www.semanticscholar.org/author/Yi-Jen-Lee/113180079" rel="nofollow noreferrer">Yi-Jen Lee</a>, written a few years later).</p>
<p>It perhaps bears mentioning that Heegaard Floer homology wasn't the first invariant that Ozsváth and Szabó constructed based on thinking about the interaction of the Seiberg-Witten equations with a Heegaard diagram—see <em><a href="https://arxiv.org/abs/math/0006192" rel="nofollow noreferrer">The Theta Divisor and Three-Manifold Invariants</a></em> and <em><a href="https://arxiv.org/abs/math/0006194" rel="nofollow noreferrer">The theta divisor and the Casson–Walker invariant</a></em>, which extract an invariant from the theta-divisor of the Heegaard surface, appear to have been based on thinking about what happens to the Seiberg–Witten equations when one has a neck <span class="math-container">$S\times [-T,T]$</span> (<span class="math-container">$S$</span> is the Heegaard surface) with the metric on <span class="math-container">$S$</span> at <span class="math-container">$t=-T$</span> itself having long cylinders over the compressing circles for one handlebody, while the metric on <span class="math-container">$S$</span> at <span class="math-container">$t=T$</span> has long cylinders over the compressing circles for the other handlebody.</p>
|
47,728 | <p>Function <code>CreateTemporary</code> can generate a unique file name like</p>
<pre><code>file = CreateTemporary[]
(* "C:\Documents and Settings\kkouptsov\Local \Settings\Temp\m-263a0380-3a03-49ac-95cb-d21390b2d3fd" *)
</code></pre>
<p>How do we generate a unique file name such as above example? The reason I don't use <code>CreateTemporary</code> because I don't need to actually create the file, but only need the file path.</p>
<p>I plan to generate many of these names in parallel, and they should be unique to each other.</p>
| Kuba | 5,478 | <p>Rather ugly but should work</p>
<pre><code>Block[{m},
While[FileExistsQ[
m = FileNameJoin[{Directory[], FromCharacterCode[RandomInteger[25, 35] + 97]}]]];
m]
</code></pre>
|
1,284,086 | <p>Given a state $(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$, is it possible to factorise out the $|0\rangle$ in the middle of both of them?</p>
<p>Alternatively written as $|101\rangle + |001\rangle$ for convenience.</p>
<p>For a state such as the following it is possible to factorise out:</p>
<p>$|101\rangle + |100\rangle = |1\rangle (|01\rangle + |00\rangle)$</p>
<p>as shown above. Is there any way to factorise out the middle tensor in such a form?</p>
| Simon S | 21,495 | <p>$(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$ is clearly a product state. It's tricky to find good notation to denote this and I don't think there's an established standard way. Here's one option:</p>
<p>Let $\sigma_{ij}$ be an automorphism that flips the $i$th and $j$th position of a tensor product $\bigotimes_{k=1}^n V$. Then</p>
<p>$$(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle) = \sigma_{12} \Big( \ |0\rangle \ \otimes \ \big( |1\rangle \otimes |1\rangle + |0\rangle \otimes |1\rangle \big) \ \Big) $$</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.