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 |
|---|---|---|---|---|
2,161,294 | <p>I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:</p>
<p>Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infinite? Is it a dense set? Is in countable? Are their members irrational numbers??</p>
<p><strong>Many thanks in advance!!</strong></p>
| Joffan | 206,402 | <p>So we want values of $0<x<1$ such that $x+k= \frac{\large 1}{\large x}$ for positive integer $k$, meaning $x^2+kx-1 =0$. This has a positive solution in the range for every $k$.</p>
|
2,161,294 | <p>I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:</p>
<p>Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infinite? Is it a dense set? Is in countable? Are their members irrational numbers??</p>
<p><strong>Many thanks in advance!!</strong></p>
| Cye Waldman | 424,641 | <p>The golden ratio is one of only two numbers that share certain properties, such as that the golden ratio is one less than its square. I have a complete description of these <em>morphic</em> numbers in my response to another post here: <a href="https://math.stackexchange.com/questions/394111/what-real-number-is-exactly-one-less-than-its-cube/2195453#2195453">What real number is exactly one less than its cube?</a></p>
|
1,735,910 | <p>In <a href="https://www.youtube.com/watch?v=aHU-L3BLd_w">a recent video</a> the legendary Matt Parker claimed he kept flipping a two-sided (fair) coin untill he scored a sequence of ten consecutive 'switch flips', i.e. letting $T$ denote a tail and $H$ a head, then a sequence of ten switch flips is defined to be either $THTHTHTHTH$ or $HTHTHTHTHT$. He set up a contest and allowed each viewer to guess once at the exact amount of flips he needed to obtain such a sequence. The ten viewers with the ten closest answers would be awarded a prize.</p>
<p>The contest is over, so there is no incentive to keep a solution to the following problem to yourself. <em>What is the best number to bet?</em> Of course this somehow depends on how other viewers answer: you are more likely to win if your bet is not close to many other bets, so if a large number of viewers is mathematically inclined and bets the same number - say 1023 - then it no longer is the best (profit maximizing) bet. I've therefore simplified to the following question: let $X$ be the stochastic variable representing the number of flips needed untill a sequence of ten consecutive switch flips if obtained, then for which number $a \in \mathbb{N}$ does the expected value of the (absolute value of the) error
$$
\mathbb{E}[\vert X - a \vert]
$$
reach its minimal value? It is well-known that $a$ is the median of the (distribution of) $X$, but how can one compute it? Numerical approximations are welcome, theoretical (generalizable) results are preferred.</p>
<p>I found a way to compute the expected value of $X$ itself for general $n$ (i.e. the total number of coin flips needed to get a sequence of the form $THTHTHTH...$ or $HTHTHTHT...$ of length $n$). Let $\mathbb{E}_i$ denote the expected number of coin flips needed to get a desired sequence of length $n$, assuming we already have a sequence of length $i \in \mathbb{N}$. We immediately find
$$
\mathbb{E}_0 = 1 + \mathbb{E}_1
$$
since we are certain to have a sequence of length $1$ after one flip. Furthermore, for $1 \leq i \leq n-1$
$$
\mathbb{E}_i = \frac{1}{2}\left(\mathbb{E}_1 + 1\right) + \frac{1}{2}\left(\mathbb{E}_{i+1} + 1\right)
$$
since, given a sequence of $i$ flips which ends, say, on a tail, we have a $\frac{1}{2}$ chance to increase this to a sequence of $i+1$ flips (if we get, say, a head) and a $\frac{1}{2}$ chance to get back where we started, at $1$ flip. Using that $\mathbb{E}_n = 0$ the above gives us system of $n$ equations in the $n$ variables $\mathbb{E}_0, \ldots, \mathbb{E}_{n-1}$. One can easily check that the unique solution is given by
$$
\mathbb{E}_i = 2^{n} - 2^{i} \quad 0 \leq i \leq n
$$
Since Matt Parker started at $0$ and wanted to get $10$ flips, the expected value of the number of flips needed is $2^{10} - 1 = 1023$ and this should be a reasonable bet.</p>
<p>Does anyone know how to find the distribution of $X$ (or directly the median of $X$)? Like I said, analytical solutions are of course preferred, but any kind of method - even requiring numerical computations, but preferably not Monte Carlo simulations - would be interesting to me.</p>
<p><strong>EDIT:</strong> I found out that the problem can be reduced to a combinatorial problem. Indeed, we have that
$$
P(X \leq k) = \frac{\# \lbrace \text{sequences of length $k$ which contain a desired subsequence of length $n$}\rbrace}{2^k}
$$
where $2^k$ is the total number of sequences of length $k$, since every sequence of length $k$ is equally likely to occur. Let $S_k$ be the set of sequences of length $k$ of $0$'s and $1$'s (we identify tails with $0$ and heads with $1$). We have a map
$$
f: S_k \to S_{k-1}
$$
where, for any sequence $s \in S_k$, the $i$th element of $f(s)$ is $1$ if $s(i) \neq s(i+1)$ and $0$ is $s(i) = s(i+1)$. This map makes the desired sequences in $S_k$ correspond bijectively with the sequences in $S_{k-1}$ which contain $n-1$ zeroes in a row. Hence, it suffices to count the number of sequences of a given length $k-1$ which contain $n-1$ zeroes in a row. Any ideas on how to continue?</p>
| joriki | 6,622 | <p>The probability to get $10$ consecutive switch flips can be modelled as a Markov chain, with the states corresponding to the number of consecutive switch flips in the immediate past. The state in which $10$ consecutive switch flips have been encountered is absorbing. The stationary distribution with eigenvalue $1$ has probability $1$ at this absorbing state. There is a quasi-stationary distribution in which each state has half the probability of the one with one less switch flip and probability "leaks out" into the absorbing state at a rate of roughly $2^{-10}$. The median is the least integer $a$ for which $\textsf P(X\le a)\gt\frac12$. The time it takes for $\frac12$ to leak out is roughly given by $a\cdot2^{-10}=\log2$, or $a=2^{10}\log2\approx 710$.</p>
<p><a href="https://gist.github.com/joriki/5b531f630a907a74e9c723a6eb6896e6" rel="nofollow noreferrer">Here's Java code</a> that follows the evolution of the distribution of the process until the probability for $10$ switch flips exceeds $\frac12$. The median turns out to be $712$. The precision of the above estimate is in part gratuitous, since the process takes $10$ steps to equilibrate before probability starts leaking into the absorbing state. Nevertheless, the agreement shows that the quasi-stationary model is quite good, so $P(X\le x)\approx1-\left(1-2^{-10}\right)^t\approx1-\mathrm e^{-2^{-10}t}$ should be a good approximation.</p>
<p>Regarding the game-theoretic aspect of the problem, this is related to the <a href="https://math.stackexchange.com/questions/583157">ice-cream vendor problem</a>, where the flip numbers are mapped to the beach using the cumulative distribution function $P(X\le x)$. However, the conclusion in the continuous case that there is no equilibrium for $3$ players doesn't go through in the discrete case.</p>
|
499,476 | <p>Use mathematical induction to prove that the derivative of $f(x)=\sin(ax+b)$ is given by</p>
<p>$f^{(n)}(x)= (-1)^ka^n\sin(ax+b)$ if $n=2k$, and $(-1)^ka^n\cos(ax+b)$ if $n=2k+1$</p>
<p>for a number $k=0,1,2,3,...$</p>
<p>I have done som proofs by induction, but I seem to struggle as soon as trig functions appear.</p>
| AlexR | 86,940 | <p>Let $f(x) = \sin(ax+b)$ then we claim for $n\in\mathbb N_0$
$$\begin{align*}
f^{(2n)}(x) & = (-1)^n a^{2n} \sin(ax+b) \\
f^{(2n+1)}(x) & = (-1)^n a^{2n+1} \cos(ax+b)
\end{align*}$$
The induction start for $n=0$ is obvious. Then we show for $k\in\mathbb N_0$:
$$\begin{align*}
f^{(2k)}(x) & = f^{(2k-1)'}(x) \\
& \stackrel{\text{ind.}}{=} = ((-1)^{k-1} a^{2k-1} \cos(ax+b))'\\
& = (-1)^k a^{2k} \sin(ax+b) & \text{and}\\
f^{(2k+1)}(x) & = f^{(2k)'}(x) \\
& \stackrel{\text{ind.}}{=} ((-1)^k a^{2k} \sin(ax+b))' \\
& = (-1)^k a^{2k+1} \cos(ax+b)
\end{align*}$$
q.e.d.</p>
|
1,698,039 | <p>Alright, so let's say I have $$\frac{x^{-6}}{-x^{-4}}$$ The answer is $\dfrac{1}{x^2}$, but why isn't it $\dfrac{1}{-x^2}$?</p>
| frog1944 | 215,383 | <p>If the questions is asking; simplify $\frac {x^{-6}}{-x^{-4}}$, then the answer is $\frac {1}{-x^2}$. But if the answer is $\frac {1}{x^2}$, then I believe the question is $\frac {x^{-6}}{(-x)^{-4}}$.</p>
|
2,814,703 | <p>I am reading <a href="https://en.wikipedia.org/wiki/Lower_limit_topology" rel="nofollow noreferrer">lower limit topology</a> on Wikipedia, which states that the lower limit topology </p>
<blockquote>
<p>[...] is the topology generated by the basis of all half-open intervals $[a,b)$, where a and b are real numbers. [...] The lower limit topology is finer (has more open sets) than the standard topology on the real numbers (which is generated by the open intervals). The reason is that every open interval can be written as a (countably infinite) union of half-open intervals.</p>
</blockquote>
<p>I cannot see how to write $(a,b)$ as a countably infinite union of half-open intervals.</p>
| Henno Brandsma | 4,280 | <p>$$(a,b) = \bigcup \{[x,b): a < x < b \}$$</p>
<p>Every $[x,b) \subseteq (a,b)$ whenever $a < x < b$ for the right to left inclusion, and on the other hand, if $a < x < b$, $x \in [x,b)$, which shows the left to right inclusion. If you want a countable union at all cost (topologies are closed under all unions, but maybe you're doing measure theory?) then take $x$ to be all rationals in $(a,b)$ so </p>
<p>$$(a,b) = \bigcup \{[q,b): q \in \mathbb{Q} \text{ and } a < q < b\}$$ </p>
<p>but then the proof of equality is a little more involved: the right to left inclusion stays the same, but if $p \in (a,b)$ we first pick $q\ in \mathbb{Q}$ with $a < q < p$, and note that $p \in [q,b)$ which is a subset of the right hand side.</p>
|
2,406,043 | <p>Let the triangle $\triangle ABC$ have sides $a,b,c$ and be inscribed in a circle with radius $R$. If $p=\frac{a+b+c}{2}$ The radius of the circle can be expressed as</p>
<p>a) $$R=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{4abc}$$</p>
<p>b) $$R=\frac{4\sqrt{p(p-a)(p-b)(p-c)}}{abc}$$</p>
<p>c) $$R=\frac{abc}{4\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<p>d) $$R=\frac{4abc}{\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<hr>
<p>So clearly Heron's formula is being used here. I know that the radical is an expression of the area of the triangle. Let's denote $A_T=\sqrt{p(p-a)(p-b)(p-c)}$ and solve for it in all four cases:</p>
<p>a) $A_T=4Rabc$</p>
<p>b) $A_T=\frac{Rabc}{4}$</p>
<p>c) $A_T=\frac{abc}{4R}$</p>
<p>d) $A_T=\frac{4abc}{R}$</p>
<p>None of the RHS's even closely resembles to the area of the triangle of the form $A=\frac{base \cdot height}{2}.$ How should I do this? Keep in mind that this is from a test where one is to have about 3 minutes per question, so no complicated solution should be needed.</p>
| Michael Rozenberg | 190,319 | <p>The proof without AM-GM.</p>
<p>Let $x=a^3$, $y=b^3$ and $z=c^3$.</p>
<p>Hence, $abc=1$ and
$$x+y+z-3=a^3+b^3+c^3-3=a^3+b^3+c^3-3abc=$$
$$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=$$
$$=\frac{1}{2}(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)\geq0.$$
Done!</p>
|
2,489,498 | <p>A={a,b,c,d}</p>
<p>R={(a,b),(a,c),(c,b)}</p>
<p>According to the definition for transitive relation, if there is (a,b) and (b,c) there should be (a,c)</p>
<p>In the above relation there is (a,c),(c,b) as well as (a,b). Shouldn't it be transitive?</p>
| zwim | 399,263 | <p>To get rid of $\sqrt{x}$ you could substitute $u=\sqrt{x}$.</p>
<p>Then $\displaystyle I=\int_0^1 \sqrt{x}\ln(x)\,dx=\int_0^1 u\ln(u^2)2u\,du=4\int_0^1 u^2\ln(u)\,du$</p>
<p>Notice that $u\mapsto u^2\ln(u)$ is continuous on any $[\delta,1]$ with $0<\delta\ll 1$ so the part $\displaystyle \int_{\delta}^1 u^2\ln(u)\,du$ is finite.</p>
<p>Now in the interval $]0,\delta[$ let's study $f(u)=u\ln(u)$.</p>
<p>$f'(u)=\ln(u)+1\to-\infty$ so $f\searrow$ in a neighborhood of $0^+$ namely on $]0,\delta[$ for a small enough $\delta$.</p>
<p>We deduce that $f$ is bounded on this interval: $|u\ln(u)|<C$</p>
<p>[i.e. <em>in simple words you cannot decrease from $-\infty$, in math language $g(v)=f(\frac 1v)\nearrow$ and upper bounded $g(v)\le 0$ thus converges when $v\to+\infty$, and a convergent continuous function is bounded.</em> ].</p>
<p>So $\displaystyle \bigg|\int_0^\delta u^2\ln(u)\,du\bigg|\le \int_0^\delta u|u\ln(u)|\,du\le \int_0^\delta Cu\,du=\frac 12C\delta^2\ll 1$</p>
|
3,636,667 | <blockquote>
<p>Evaluate
<span class="math-container">$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$</span> </p>
</blockquote>
<p>Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good ideas yet, so I will appreciate any hints !</p>
| grand_chat | 215,011 | <p>The limit is <span class="math-container">$\frac1{p+1}$</span>. There is a nice closed form:
<span class="math-container">$$
\sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} = \frac1{p+1} \underbrace{(n+1)\cdots (n+p+1)}_{\text{$p+1$ factors}} -p!
$$</span>
(even nicer if you absorb <span class="math-container">$p!$</span> into the LHS as the <span class="math-container">$i=0$</span> term.)</p>
<hr>
<p><strong>Proof 1.</strong> Telescoping! Write
<span class="math-container">$$
\begin{align}
\frac{(p+i+1)!}{i!}-\frac{(p+i)!}{(i-1)!}=\frac{(p+i)!}{i!}[(p+i+1)-i]=\frac{(p+i)!}{i!}(p+1).
\end{align}
$$</span>
Sum from <span class="math-container">$i=1$</span> to <span class="math-container">$n$</span> to get
<span class="math-container">$$
\frac{(p+n+1)!}{n!}-(p+1)! = (p+1)\sum_{i=1}^n\frac{(p+i)!}{i!},
$$</span>
then divide through by <span class="math-container">$p+1$</span>. </p>
<hr>
<p><strong>Proof 2:</strong> Use the <a href="https://math.stackexchange.com/q/1490794/215011">Hockey Stick identity (H)</a>:
<span class="math-container">$$
\frac1{p!}\sum_{i=0}^n\frac{(p+i)!}{i!}=\sum_{i=0}^n {p+i\choose p}=\sum_{t=p}^{p+n}{t\choose p}\stackrel{H}={p+n+1\choose p+1}=\frac1{(p+1)!}\frac{(p+n+1)!}{n!}
$$</span>
Multiply through by <span class="math-container">$p!$</span>, and we're done.</p>
|
4,274,600 | <p><span class="math-container">$(x^3+x+1)^{-1} \mod (x^4+x+1)$</span> over <span class="math-container">$\text{GF}(2)$</span></p>
<p>I understand well how to solve the equation without inverse but don't know how to solve it with inverse.</p>
| Misha Lavrov | 383,078 | <p>For a problem with bigger polynomials, I'd want to try fancier tools. But in this problem, I'd be tempted to compute</p>
<p><span class="math-container">\begin{align}
A &= (x^3+x+1)(1) \bmod x^4+x+1 \\
B &= (x^3+x+1)(x) \bmod x^4+x+1 \\
C &= (x^3+x+1)(x^2) \bmod x^4+x+1 \\
D &= (x^3+x+1)(x^3) \bmod x^4+x+1
\end{align}</span></p>
<p>and then see what linear combination of <span class="math-container">$A, B, C, D$</span> adds up to <span class="math-container">$1$</span>. Then the inverse is the same linear combination of <span class="math-container">$1, x, x^2, x^3$</span>.</p>
<p>(Some of these, like <span class="math-container">$A$</span>, don't actually take work to compute. We can stop at <span class="math-container">$D$</span> because <span class="math-container">$x^4 \equiv x+1 \pmod{x^4+x+1}$</span>, so a hypothetical <span class="math-container">$E$</span> would just be the same as <span class="math-container">$A+D$</span>.)</p>
|
3,684,917 | <p>Let <span class="math-container">$C_{1}$</span> and <span class="math-container">$C_{2}$</span> be polytopes in <span class="math-container">$\mathbb{R}^{n}$</span> such that
<span class="math-container">$C_{1}=conv\left( V\right) $</span> with <span class="math-container">$V$</span> being a set of vertices. If
<span class="math-container">$V\subseteq C_{2}$</span>, my question is <span class="math-container">$C_{1}\subseteq C_{2}$</span>?</p>
| John Hughes | 114,036 | <p>I'm going to write a formula <span class="math-container">$H(a, b, n, p)$</span> for the number of items congruent to <span class="math-container">$n$</span>, modulo <span class="math-container">$p$</span>, in the interval <span class="math-container">$a \le k < b$</span>. If you want to apply it to get the answer to the question you've asked, you need to evaluate <span class="math-container">$H(a, b+1, n, p)$</span> to get the sum to be inclusive on both ends. I'm assuming here that <span class="math-container">$b \ge a$</span>. </p>
<p>Furthermore, I'm going to use the computer-scientist's convention that
<span class="math-container">$$
(x, y) \mapsto x \bmod y
$$</span>
is a <strong>function</strong> defined on pairs of integers, where <span class="math-container">$y$</span> must be positive, and that the value of this function is the number in the range <span class="math-container">$0, 1, \ldots, y-1$</span> that is congruent to <span class="math-container">$x$</span>, modulo <span class="math-container">$y$</span>.</p>
<p>Observe that for any <span class="math-container">$a, b, n, p$</span>, and <span class="math-container">$s$</span> we have
<span class="math-container">$$
H(a, b, n, p) = H(a-s, b-s, n-s, p),
$$</span>
so picking <span class="math-container">$s = a$</span>, we can simply compute our answer by computing
<span class="math-container">$$
H(a-a, b-a, n-a, p) = H(0, b-a, n-a, p).
$$</span>
Next observe that if we adjust <span class="math-container">$n-a$</span> by some multiple of <span class="math-container">$p$</span>, the answer remains the same, so if we say <span class="math-container">$n' = (n-a) \bmod p$</span>, then we only need to compute
<span class="math-container">$$
H(0, b-a, n', p)
$$</span>
and now <span class="math-container">$n'$</span> is a number between <span class="math-container">$0$</span> and <span class="math-container">$p-1$</span>. To simplify a little more, let's write <span class="math-container">$b' = b-a$</span>, so we seek to compute
<span class="math-container">$$
H(0, b', n', p).
$$</span>
In any span of <span class="math-container">$p$</span> sequential integers, there's ONE that's congruent to <span class="math-container">$n'$</span>, so let's look at how many such spans there are, starting at <span class="math-container">$0$</span>, and stopping while still less than <span class="math-container">$b'$</span>. That's exactly
<span class="math-container">$$
U(b', p) = \lfloor \frac{b'}{p} \rfloor.
$$</span>
What's left over is a sequence of fewer than <span class="math-container">$p$</span> numbers from <span class="math-container">$pU(b', p)$</span> to <span class="math-container">$b'$</span>, in which there might or might not be a number congruent to <span class="math-container">$n'$</span>. Taken <span class="math-container">$\bmod p$</span>, this sequence looks like
<span class="math-container">$$
0, 1, 2, \ldots, (b'-1) \bmod p
$$</span>
and we need to add one to our tally <strong>exactly</strong> if one of those numbers is <span class="math-container">$n'$</span>. In short, we get
<span class="math-container">$$
H(0, b', n', p) = U(b', p) + \begin{cases}
1 & n' < (b' \bmod p) \\
0 & n' \ge (b' \bmod p)
\end{cases}.
$$</span></p>
<p>Replacing this with the original values, we get
<span class="math-container">$$
H(a, b, n, p) = \lfloor \frac{b-a}{p} \rfloor +
\begin{cases}
1 & (n \bmod p) < ((b-a) \bmod p) \\
0 & (n \bmod p) \ge ((b-a) \bmod p)
\end{cases}.
$$</span></p>
<p>It's possible that there's some nice way to simplify this a little bit, but...I think I've said enough. </p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| Mhenni Benghorbal | 35,472 | <p>You can have instead the equivalent integral representation </p>
<blockquote>
<p>$$ I = \int_{0}^{1}\frac{\ln^2(u)\ln(1-u/2)}{u(u-2)}du \sim .5582373010. $$</p>
</blockquote>
<p>Try to evaluate the above integral. See my <a href="https://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk/276590#276590">answer</a>. See also <a href="https://math.stackexchange.com/questions/457371/alternating-harmonic-sum-sum-k-geq-1-frac-1kk3h-k">here</a>.</p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & =
\sum_{n = 1}^{\infty}{H_{n} \over 2^{n}}
\bracks{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}\,x^{n - 1}\,\dd x} =
{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}\sum_{n = 1}^{\infty}
\bracks{H_{n}\pars{x \over 2}^{n}}{\dd x \over x}
\\[5mm] &=
{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}
\bracks{-\,{\ln\pars{1 - x/2} \over 1 - x/2}}\,{\dd x \over x} =
-\,{1 \over 2}\int_{0}^{1/2}
{\ln^{2}\pars{2x}\ln\pars{1 - x} \over \pars{1 - x}x}\,\dd x
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1/2}{\ln^{2}\pars{2x}\ln\pars{1 - x} \over x}\,\dd x -
{1 \over 2}\int_{0}^{1/2}{\ln^{2}\pars{2x}\ln\pars{1 - x} \over 1 - x}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{2x}\,\dd x -
{1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{2\bracks{1 - x}}\ln\pars{x} \over x}
\,\dd x
\\[1cm] & =
-\int_{0}^{1/2}\mrm{Li}_{3}'\pars{x}\ln\pars{2x}\,\dd x
\\[5mm] & -
{1 \over 2}\,\ln^{2}\pars{2}\int_{1/2}^{1}{\ln\pars{x} \over x}\,\dd x -
\ln\pars{2}\int_{1/2}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x -
{1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\\[1cm] & =
\int_{0}^{1/2}\mrm{Li}_{4}'\pars{x}\dd x +
{1 \over 4}\,\ln^{4}\pars{2} +
\ln\pars{2}\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x -
{1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\\[1cm] & =
\mrm{Li}_{4}\pars{1 \over 2} + {1 \over 4}\,\ln^{4}\pars{2} +
\ln\pars{2}\bracks{%
\mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} -\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x}
\\[5mm] & -
{1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\\[1cm] & =
\mrm{Li}_{4}\pars{1 \over 2} + {1 \over 4}\,\ln^{4}\pars{2} +
\ln\pars{2}\bracks{%
\mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} - \mrm{Li}_{3}\pars{1} + \mrm{Li}_{3}\pars{1 \over 2}}
\\[5mm] & -
{1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\end{align}</p>
<blockquote>
<p>Since values of $\ds{\,\mrm{Li}_{2}\pars{1/2}}$ and
$\ds{\,\mrm{Li}_{3}\pars{1/2}}$ are
<a href="http://mathworld.wolfram.com/Polylogarithm.html" rel="noreferrer">well known</a> and
$\ds{\,\mrm{Li}_{3}\pars{1} = \zeta\pars{3}}$:</p>
</blockquote>
<p>\begin{align}
\sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & =
-\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} +
\,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2}\
\underbrace{\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x}
_{\ds{\equiv\ \mc{I}}}
\label{1}\tag{1}
\end{align}
<hr>
<strong>$\ds{\large\mc{I}:\ ?}$</strong>.
\begin{align}
\mc{I} & \equiv
\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x
\\[5mm] & =
{1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{1 - x} \over x}\dd x -
{1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{x} \over x}\dd x -
{1 \over 3}\int_{1/2}^{1}\!\ln^{3}\pars{1 - x \over x}{\dd x \over x} +
\int_{1/2}^{1}\!{\ln\pars{1 - x}\ln^{2}\pars{x} \over x}\,\dd x
\\[5mm] & =
{1 \over 3}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\dd x +
{1 \over 12}\,\ln^{4}\pars{2} +
{1 \over 3}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x -
\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x
\\[1cm] & =
{1 \over 3}\bracks{-\ln^{4}\pars{2} - 3\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\dd x} +
{1 \over 12}\,\ln^{4}\pars{2} -
\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x
\\[5mm] &
-\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x
\\[1cm] & =
-\,{1 \over 4}\,\ln^{4}\pars{2} -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x -
\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x
\end{align}</p>
<blockquote>
<p>The remaining integrals can be straightforward evaluated by successive <em>integration by parts</em> and by using the $\ds{\,\mrm{Li}_{s}}$ <em>recursive property</em>. Namely,</p>
</blockquote>
<p>\begin{align}
&\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x =
\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} -
2\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x
\\[5mm] & =
\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} -
2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x}
+ 2\int\mrm{Li}_{4}'\pars{x}\,\dd x
\\[5mm] & =\
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x}
+ 2\,\mrm{Li}_{4}\pars{x}}}
\end{align}
such that
\begin{equation}
\mc{I} \equiv
\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x =\
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
-\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}}}\label{2}\tag{2}
\end{equation}
<hr>
With \eqref{1} and \eqref{2}:
\begin{align}
\sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & =
-\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} +
\,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2}
\bracks{-\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}}
\\[5mm] & =\ \bbox[25px,#ffe,border:1px dotted navy]{\ds{%
{1 \over 720}\,\pi^{4} + {1 \over 24}\,\ln^{4}\pars{2} -
{1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2}}}\
\approx 0.5582
\end{align}</p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| Dennis Orton | 761,314 | <p>By first finding the following integral by using the algebraic identity
<span class="math-container">$a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$</span> one can easily prove avoiding Euler sums that:
<span class="math-container">$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)$$</span>
Now:
<span class="math-container">$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=\frac{1}{2}\ln \left(2\right)\int _0^1\frac{\ln ^2\left(x\right)}{1-\frac{x}{2}}\:dx+\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-\frac{x}{2}\right)}{1-\frac{x}{2}}\:dx$$</span>
<span class="math-container">$$=2\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}$$</span>
<span class="math-container">$$=2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1}{2}\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$</span>
<span class="math-container">$$=\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{3}\ln ^4\left(2\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$</span>
By making use of the result we find:
<span class="math-container">$$\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}=\frac{1}{8}\zeta \left(4\right)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln \left(2\right)\zeta \left(3\right)+\frac{1}{24}\ln ^4\left(2\right)$$</span></p>
|
1,285,621 | <p>Below is the joint distribution of Boolean random variables X1, X2 and X3. How do I find variance and expectation of X2? I understand that variance is
"average of squares of difference from mean value". But here we have boolean values. I am confused. Can somebody suggest how to do this? I am a newbie to probability and random variables.</p>
<p><img src="https://i.stack.imgur.com/sOmBM.png" alt="" /></p>
| AlexR | 86,940 | <p>It's no different than an integer variable with values $\{0,1\}$. Naturally, both EV and variance will be in the real interval $[0,1]$.<br>
What you have to do is simply find the distribution of $X_2$ from the table - this will be parameterized by $p\in[0,1]$ with $P(X_2=1) = p$ and $P(X_2=0)=1-p$.<br>
The expected value will then be $\mathbb E(X_2) = p$ and the variance will be $\mathrm{Var}(X_2) = p(1-p)^2 + (1-p)p^2 = p(1-p)$</p>
|
1,285,621 | <p>Below is the joint distribution of Boolean random variables X1, X2 and X3. How do I find variance and expectation of X2? I understand that variance is
"average of squares of difference from mean value". But here we have boolean values. I am confused. Can somebody suggest how to do this? I am a newbie to probability and random variables.</p>
<p><img src="https://i.stack.imgur.com/sOmBM.png" alt="" /></p>
| Lawrence | 214,733 | <p>It looks like you're asking why the variance and expected value of a boolean variable isn't itself boolean. The problem goes away if you consider variance to be a measure of the variable rather than the variable itself, and consider 'expected value' to be a form of scaled probability rather than the domain value to 'expect'.</p>
<p>For example, given instances {0,1,1}, the mean is 2/3, which isn't binary; variance is analogous. Likewise, if you consider the same 3 instances to represent something you could earn in a single 3-outcome event, then assuming equal probability of each outcome, in 100 events, you would expect your earnings to be represented by 2/3 * 100, which isn't binary.</p>
|
1,285,621 | <p>Below is the joint distribution of Boolean random variables X1, X2 and X3. How do I find variance and expectation of X2? I understand that variance is
"average of squares of difference from mean value". But here we have boolean values. I am confused. Can somebody suggest how to do this? I am a newbie to probability and random variables.</p>
<p><img src="https://i.stack.imgur.com/sOmBM.png" alt="" /></p>
| BruceET | 221,800 | <p>For reference, I am repeating the PDF (or PMF) table of $X_2.$
Then $E(X_2) = \sum_{i=0}^6 v_i p_i,$ the terms and total of
which are in the third row. (My notation may different from
that of your book, so make sure you connect the table with
whatever formula your book has.)</p>
<pre><code> value 0 0 1 1 0 0 1 1 Total
prob. .04 .08 .08 .20 .06 .12 .12 .30 1.00
product 0 0 .08 .2 0 0 .12 .30 0.70
</code></pre>
<p>The result is that $E(X_2) = 0+0+.08+\cdots+.30 = .70.$
To find the variance $V(X_2),$ there are two possible paths.
The easiest is to use the general formula $$V(Y) = E(Y^2) - [E(Y)]^2.$$</p>
<p>For your 0-1 random variable, $E(X) = E(X^2)$ because $0^2 = 0$ and $1^2 = 1$ are the only two values involved. Applied to $X_2,$
the general formula gives $$V(X_2) = .7 - (.7)^2 = .7 - .49 = .21.$$
This is the point I was trying to make in my note.</p>
<p>A second way to find the variance of $X_2$ is to use
the definition of the variance of a discrete random variable:
$$ V(Y) = \sum p_i(v_i - \mu)^2,$$
where the sum is taken over all values for which a probability
has been defined, and $\mu = E(Y).$ In order to illustrate this, I will make a new
table with rows 'values' and 'prob' as before, but with new rows
'diff' to show $(v_i - \mu)$, 'sqr' to show $(v_i - \mu)^2$, and
'Prod2' to show $p_i(v_i - \mu)^2.$</p>
<pre><code> values 0 0 1 1 0 0 1 1 Total
prob .04 .08 .08 .20 .06 .12 .12 .30 1.0
dif -.7 -.7 .3 .3 -.7 -.7 .3 .3 ---
sqr .49 .49 .09 .09 .49 .49 .09 .09 ---
Prod2 .0196 .0392 .0072 .0180 .0294 .0588 .0108 .0270 0.21
</code></pre>
<p>This gives the same variance as before. But the previous method
is simpler in your case.</p>
<p>I hope these illustrations have helped you understand the basic
formulas and that you can find means and variances of the other
two random variables in your problem.</p>
|
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even number of 2s and the right hand side has an odd number of 2s? </p>
<p>I will be grateful for your help Best regards</p>
| Jack M | 30,481 | <p>The basic argument that $\sqrt 2$ is irrational is generalized using the fundamental theorem of arithmetic.</p>
<p>Supose $(\frac p q)^{12} = 2$. Then $p^{12} = 2 q^{12}$. The prime factorization on the left hand side contains a number of $2$-s which is a multiple of $12$, but the prime factorization on the right hand side contains a number of $2$-s which is one more than a multiple of $12$.</p>
|
105,857 | <p>Let $\mathcal{O}$ be the ring of integers in an algebraic number field. Is $\text{SL}_2(\mathcal{O})$ generated by elementary matrices? If it isn't, is there any other natural generating set for it?</p>
<p>The usual argument shows that this is true for $\mathcal{O} = \mathbb{Z}$ (or, more generally, a Euclidean domain). However, I haven't been able to generalize this to other rings of integers.</p>
| Alex | 57,771 | <p>If $\mathcal O$ be the ring of integers in an algebraic number field, whether $SL_2(\mathcal O)$ is generated by elementary matrices depends on the field $k$:</p>
<ul>
<li><p>If $k = \Bbb Q$, or $k = \Bbb Q(\sqrt{-D}$ for $D\in\{1,2,3,7,11\}$, then $SL_2(\mathcal O)$ is generated by elementary matrices.</p></li>
<li><p>If $k = \Bbb Q(\sqrt{-D})$ for $D$ any other squarefree integer, then $SL_2(\mathcal O)$ is not generated by the elementary matrices. However, the index of the subgroup generated by elementary matrices in $SL_2(\mathcal O)$ is finite, so we can add a finite number of non-elementary matrices to the generating set, to ensure $SL_2(\mathcal O)$ will be generated by this set.</p></li>
<li><p>For all other $k$, $SL_2(\mathcal O)$ is not only generated by elementary matrices, but a bounded number of matrices is enough to generate any matrix. The bound depends on the field $k$.</p></li>
</ul>
|
801,081 | <p>I was doing some school work and got bored so I started messing with k-gonal numbers. I started with the triangular numbers, square numbers and looked for patterns. I noticed something.</p>
<p>Let $n^{(k)}$ denote the $n$-th $k$-gonal number. For example, $3^{(3)}$ is the third triangular number, 6.</p>
<p>I found that there was an easy way to compute the formula for each $k$-gonal and noticed that the
$$n^{(k)}=n^{(k-1)}+(n-1)^{(3)}$$</p>
<p>So to find the formula for the $n$-th pentagonal number,
$$n^{(5)}=n^{(4)}+(n-1)^{(3)}$$
$$n^{(5)}=n^2+\frac{n(n-1)}{2}$$
$$n^{(5)}=\frac{n(3n-1)}{2}$$</p>
<p>So after doing this a bunch of times, I think I found the pattern...</p>
<p>Let $f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ such that
$$f(n,k)=\frac{n[(k-2)n-(k-4)]}{2}$$
Is this the formula for the $n$-th $k$-gonal number? Are there any other intersting formulas that come out of the 2 x 2 array of these numbers like the one I derived above?</p>
| Janaka Rodrigo | 1,043,137 | <p>Method (iii) <br/>
Let u(k,n) be the n th k - gonal number <br/>
Consider the regular polygon of k sides which represents u(k,n) <br/>
When straight lines drawn through a selected vertex joining other vertices there will be k-2 triangles and out of which any triangle can be used to represent the triangle number u(3,n) <br/>
Now by the each of remaining k-3 triangles we can represent the triangle number u(3,n-1). Here we have n-1 instead of n because these triangles have n-1 dots(points) on the base and sides. <br/>
Therefore, u(k,n) = u(3,n) + (k-3)u(3,n-1) <br/>
u(k,n) = n(n+1)/2 + (k-3)(n-1)n/2 <br/>
u(k,n) = (1/2)n{ (n-1)k-2n +4 }</p>
|
2,199,222 | <p>I have the feeling of being stuck or missing something trying to prove
$$ \lim_{N\to\infty}\sum_{k=1}^{N} \frac{1}{N+k} =\int_{1}^{2} \frac{1}{x} dx = ln(2)$$</p>
<p>Using Riemann-Sums I have shown that $$\int_{1}^{a} \frac{1}{x} dx=\lim_{N\to\infty}\sum_{k=1}^{N} (a^{1/N}-1)=\lim_{N\to\infty}N(a^{1/N}-1)=\lim_{h\to 0}\frac{a^h-1}{h}=ln(a)$$</p>
<p>So I would have to show that </p>
<p>$$\lim_{N\to\infty}\sum_{k=1}^{N} \frac{1}{N+k}=\lim_{N\to\infty}\sum_{k=1}^{N} (2^{1/N}-1)$$</p>
<p>However the summands are not equal. How does one prove this equality?</p>
| user209663 | 209,663 | <p>I think it's called Skew Hermitian matrix. </p>
<p>Here is the Wikipedia link <a href="https://en.wikipedia.org/wiki/Skew-Hermitian_matrix" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Skew-Hermitian_matrix</a><a href="https://en.wikipedia.org/wiki/Skew-Hermitian_matrix" rel="nofollow noreferrer">1</a></p>
|
1,749,730 | <p>What is the maximum number of faces of totally convex solid that one can "see" from a point? </p>
<p>...and, more importantly, how can I ask this question better? (I'm a college student with little experience in asking well formed questions, much less answering them.) </p>
<p>By "see" I mean something like this: you point a camera from a point at the solid, and look at the picture. How many of the faces of the solid look like faces and not just lines? Let's assume that the lens is just a point in space (no lenses wider than the solid itself) and that the camera is a finite distance from the solid. I know this is a crude definition... if you have any ideas for a more rigorous definition, this would be awesome, then maybe there's ways to prove the answer to my question mathematically. </p>
<p>For example, in the picture of <a href="https://i.stack.imgur.com/Ue0Fg.jpg" rel="noreferrer">this cube</a>, you can see 3 faces. This is the maximum you can see for a cube. How can that be proved? </p>
<p>What methods might you use to prove this for a convex solid of any size and shape? Are there ways to do so using only relatively basic math (Multivariable calc, linear algebra, high school geometry)? </p>
| Ethan Bolker | 72,858 | <p>You can always find a place from which to see at least half the faces.</p>
<p>To see why, start by considering a polyhedron with central symmetry. Imagine
a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane perpendicular to your line of sight. Then think about what you see from far enough away in the opposite direction. You can see all the faces from one side or the other and no face from both sides, so the symmetry says you see half each time.</p>
<p>Four of the five regular polyhedra have a center of symmetry. The tetrahedron does not: there's no place to put the origin that allows invariance under the map $x \to -x$.</p>
<p>Even without central symmetry, you see all the faces from one side or the other, so you see at least half from at least one side. Pyramids represent an extreme case. You can see all but one face from one direction and just one from the other, as @almagest points out in a comment.</p>
<p>Since the polyhedron has only finitely many faces, "far enough away" in the preceding proof does not have to be at infinity (though it may be pretty far). As @JohhHughes comments, if you put your camera close enough to any face that's the only face you'll see.</p>
<p>Note: the arguments work in all dimensions. They are particularly easy to visualize in the plane. (On the line they're trivial.)</p>
|
1,749,730 | <p>What is the maximum number of faces of totally convex solid that one can "see" from a point? </p>
<p>...and, more importantly, how can I ask this question better? (I'm a college student with little experience in asking well formed questions, much less answering them.) </p>
<p>By "see" I mean something like this: you point a camera from a point at the solid, and look at the picture. How many of the faces of the solid look like faces and not just lines? Let's assume that the lens is just a point in space (no lenses wider than the solid itself) and that the camera is a finite distance from the solid. I know this is a crude definition... if you have any ideas for a more rigorous definition, this would be awesome, then maybe there's ways to prove the answer to my question mathematically. </p>
<p>For example, in the picture of <a href="https://i.stack.imgur.com/Ue0Fg.jpg" rel="noreferrer">this cube</a>, you can see 3 faces. This is the maximum you can see for a cube. How can that be proved? </p>
<p>What methods might you use to prove this for a convex solid of any size and shape? Are there ways to do so using only relatively basic math (Multivariable calc, linear algebra, high school geometry)? </p>
| Joseph Malkevitch | 1,369 | <p>Perhaps this paper which deals with the situation in a rather abstract way, and for higher dimensional polytopes might interest some: <a href="http://www-math.mit.edu/~rstan/transparencies/vis.pdf" rel="nofollow">http://www-math.mit.edu/~rstan/transparencies/vis.pdf</a></p>
|
555,045 | <p>Let $K$ be a quadratic number field.
Let $R$ be an order of $K$, $D$ its discriminant.
By <a href="https://math.stackexchange.com/questions/546281/on-a-certain-basis-of-an-order-of-a-quadratic-number-field">this question</a>, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.</p>
<p>Let $x_1,\cdots, x_n$ be a sequence of elements of $R$.
We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.</p>
<p><strong>My Question</strong>
Is the following proposition correct?
If yes, how do you prove it?</p>
<p><strong>Proposition</strong></p>
<p><strong>Case 1</strong> $D$ is even.</p>
<p>$P = [2, \omega]$ is a prime ideal and $2R = P^2$.</p>
<p><strong>Case 2</strong> $D \equiv 1$ (mod $8$).</p>
<p>$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.
Moreover $P' = \sigma(P)$, where $\sigma$ is the unique non-identity automorphism of $K/\mathbb{Q}$.</p>
<p><strong>Case 3</strong> $D \equiv 5$ (mod $8$).</p>
<p>$2R$ is a prime ideal.</p>
| Makoto Kato | 28,422 | <p>I realized after I posted this question that the proposition is not correct.
I will prove a corrected version of the proposition.</p>
<p>We need some notation.
Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$.
We denote $\sigma(\alpha)$ by $\alpha'$ for $\alpha \in R$.
We denote $\sigma(I)$ by $I'$ for an ideal $I$ of $R$.</p>
<p><strong>Proposition</strong>
Let $K$ be a quadratic number field, $d$ its discriminant.
Let $R$ be an order of $K$, $D$ its discriminant.
By <a href="https://math.stackexchange.com/questions/191830/discriminant-of-a-binary-quadratic-form-and-an-order-of-a-quadratic-number-field">this question</a>, $D \equiv 0$ or $1$ (mod $4$).
By <a href="https://math.stackexchange.com/questions/546281/on-a-certain-basis-of-an-order-of-a-quadratic-number-field">this question</a>, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.
Let $f$ be the order of $\mathcal{O}_K/R$ as a $\mathbb{Z}$-module.
Then $D = f^2d$ by <a href="https://math.stackexchange.com/questions/528122/on-orders-of-a-quadratic-number-field">this question</a>.
<strong>We suppose gcd$(f, 2) = 1$</strong>.</p>
<p><strong>Case 1</strong> $D$ is even.</p>
<p>Since $D \equiv 0$ (mod $4$), $D \equiv 0, 4$ (mod $8$).
If $D \equiv 0$ (mod $8$), let $P = [2, \omega]$.
If $D \equiv 4$ (mod $8$), let $P = [2, 1 + \omega]$.
Then $P$ is a prime ideal and $2R = P^2$.
Moreover $P = P'$.</p>
<p><strong>Case 2</strong> $D \equiv 1$ (mod $8$).</p>
<p>$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.</p>
<p><strong>Case 3</strong> $D \equiv 5$ (mod $8$).</p>
<p>$2R$ is a prime ideal.</p>
<p>We need the following lemmas to prove the proposition.</p>
<p><strong>Lemma 1</strong>
Let $K, R, D, \omega$ be as in the proposition.
Let $P = [2, b + \omega]$, where $b$ is a rational integer.
Then $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$).
Moreover, if $P$ satisfies this condition, $P$ is a prime ideal.</p>
<p>Proof:
By <a href="https://math.stackexchange.com/questions/553942/condition-for-a-b-omega-to-be-the-canonical-basis-of-an-ideal-of-a-quadrat">this question</a>, $P = [2, b + \omega]$ is an ideal if and only if $N_{K/\mathbb{Q}}(b + \omega) \equiv 0$ (mod $2$).
$N_{K/\mathbb{Q}}(b + \omega) = (b + \omega)(b + \omega') = \frac{2b + D + \sqrt D}{2}\frac{2b + D - \sqrt D}{2}
= \frac{(2b + D)^2 - D}{4}$.
Hence $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$)
Since $N(P) = 2$, $P$ is a prime ideal.</p>
<p><strong>Lemma 2</strong>
Let $K, R, D, \omega$ be as in the proposition.
Suppose gcd$(f, 2) = 1$ and there exist no prime ideals of the form $P = [2, b + \omega]$,
where $b$ is an integer.
Then $2R$ is a prime ideal of $R$.</p>
<p>Proof:
Let $P$ be a prime ideal of $R$ lying over $2$.
Then $P \cap \mathbb{Z} = 2\mathbb{Z}$.
By <a href="https://math.stackexchange.com/questions/552516/canonical-basis-of-an-ideal-of-a-quadratic-order">this question</a>, there exist integers $b, c$ such that $P = [2, b + c\omega],
c \gt 0, 2 \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$).
Then $c = 1$ or $2$.
By the assumption, $c = 2$.
Hence $P = [2, 2\omega] = 2R$.</p>
<p><strong>Proof of the proposition</strong></p>
<p><strong>Case 1</strong> $D$ is even.</p>
<p>Let $P = [2, b + \omega]$, where $b$ is an integer.
We may assume that $b = 0$ or $1$.
By Lemma 1, $P$ is a prime ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$).</p>
<p>Suppose $D \equiv 0$ (mod $8$).
Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 0$.
Hence $P = [2, \omega]$ is an ideal of $R$.
$P' = [2, \omega'] = [2, D - \omega] = [2, -\omega] = [2, \omega] = P$.</p>
<p>Suppose $D \equiv 4$ (mod $8$).
Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 1$.
Hence $P = [2, 1 + \omega]$ is an ideal of $R$.
$P' = [2, 1 + \omega'] = [2, 1 + D - \omega] = [2, -1 - D + \omega] = [2, 1 + \omega] = P$.</p>
<p>Since gcd$(f, 2) = 1$, $P$ is regular by <a href="https://math.stackexchange.com/questions/556643/criterion-on-whether-a-given-ideal-of-a-quadratic-order-is-regular-or-not">this quuestion</a>.
Hence $PP' = 2R$ by <a href="https://math.stackexchange.com/questions/562238/norm-of-a-regular-ideal-of-an-order-of-an-algebraic-number-field-which-is-galois">this question</a>.</p>
<p><strong>Case 2</strong> $D \equiv 1$ (mod $8$).</p>
<p>If $b = 0, 1$, then
$(2b + D)^2 - D \equiv (2b + 1)^2 - 1 \equiv 0$ (mod $8$).
Hence, by Lemma 1, $P = [2, \omega]$ and $Q = [2, 1 + \omega]$ are prime ideals of $R$.
$P' = [2, \omega'] = [2, D - \omega] = [2, - D + \omega] = [2, 1 + \omega] = Q$.
Since gcd$(f, 2) = 1$, $P$ is regular by <a href="https://math.stackexchange.com/questions/556643/criterion-on-whether-a-given-ideal-of-a-quadratic-order-is-regular-or-not">this question</a>.
Hence $PP' = 2R$ by <a href="https://math.stackexchange.com/questions/562238/norm-of-a-regular-ideal-of-an-order-of-an-algebraic-number-field-which-is-galois">this question</a>.</p>
<p><strong>Case 3</strong> $D \equiv 5$ (mod $8$).</p>
<p>Consider the following congruence equation.
$(2b + D)^2 - D \equiv (2b + 5)^2 - 5 \equiv 0$ (mod $8$).
Since $b$ does not satisfy this congruence equation when $b = 0$ or $1$,
there exist no ideals of the form $[2, b + \omega]$.
Hence $2R$ is a prime ideal by Lemma 2.</p>
|
266,832 | <p>Let $p(x) = \sum_{k \geq 0} a_k x^k$ where the $a_k$'s are IID random variables taken from a mean-zero random variable taking finitely many values in $\mathbb{R}$; it clearly converges for $-1<x<1$. Is it a.s. true that the sign of $p(x)$ oscillates infinitely often as $x \rightarrow 1^-$? That is, is it the case (with probability 1) that there exist $x_1 < x_2 < x_3 < \dots$ in (0,1) such that $p(x_k)$ has the same sign as $(-1)^k$?</p>
<p>I imagine that this is well-known for $P(1) = P(-1) = 1/2$ (where $P(r)$ denotes the probability that $a_k = r$); the case that most interests me is $P(1) = P(-1) = 1/4$ and $P(0) = 1/2$, but I'm sure that the same technique settles both cases.</p>
<p>I am also interested in knowing about the magnitude of the oscillations.</p>
| Robert Israel | 13,650 | <p>We can construct inductively a sequence $t_n \to 1-$ and an increasing sequence $K_n$ of positive integers
such that with probability $> 1 - 1/n^2$, $f(t_n)$ has the same sign as
$V_n = \sum_{k=K_{n-1}+1}^{K_n} a_k t_n^k$. Since $\sum 1/n^2 < \infty$,
almost surely $f(t_n)$ has the same sign as $V_n$ for all but finitely many $n$. Almost surely the independent random variables $V_n$ change sign infinitely often, so a.s. $f(t_n)$ changes sign infinitely often.</p>
|
4,510,795 | <p>I need to find the % difference between two numbers. One person told me to use <span class="math-container">$\frac{x-y} {x} $</span>, another told me to use <span class="math-container">$\frac x y$</span> <span class="math-container">$- 1$</span> . Who is right?</p>
<p>Example:
Today's price: <span class="math-container">$23892$</span>
Yesterday's price: <span class="math-container">$23941$</span></p>
<p>Using <span class="math-container">$\frac{x-y} {x} $</span>: <span class="math-container">$-0.21%.$</span>
Using <span class="math-container">$\frac x y$</span> <span class="math-container">$- 1$</span> : <span class="math-container">$-0.20%$</span></p>
<p>It's almost the same result, but not quite. How to intuitively understand why?</p>
| Samuel Lelièvre | 162,322 | <p>Sage's <code>solve</code> has an optional parameter <code>to_poly_solve</code>.</p>
<p>It helps in this case:</p>
<pre><code>sage: x = var('x')
sage: rad_pol = sqrt(x) + sqrt(1 - x) - 1
sage: solve(rad_pol, x)
[sqrt(x) == -sqrt(-x + 1) + 1]
sage: solve(rad_pol, x, to_poly_solve=True)
[x == 0, x == 1]
</code></pre>
<p>The documentation of <code>solve</code> can be accessed</p>
<ul>
<li>from Sage
<ul>
<li>in text mode:
<pre><code>sage: solve?
</code></pre>
</li>
<li>rendered in html:
<pre><code>sage: browse_sage_doc(solve)
</code></pre>
</li>
</ul>
</li>
<li>in the built documentation
<ul>
<li>online:
<a href="https://doc.sagemath.org/" rel="nofollow noreferrer">https://doc.sagemath.org/</a><br />
in particular:
<a href="https://doc.sagemath.org/html/en/reference/calculus/sage/symbolic/expression.html?#sage.symbolic.expression.Expression.solve" rel="nofollow noreferrer">solve</a></li>
<li>offline:
<code>file:///path_to_sage_doc/index.html</code><br />
in particular
<code>file:///path_to_sage_doc/sage/html/en/reference/calculus/sage/symbolic/expression.html?#sage.symbolic.expression.Expression.solve</code></li>
</ul>
</li>
</ul>
<p>Depending on how you installed Sage, typing <code>sage.env.SAGE_DOC</code>
in a Sage session might give you a path. If so, you can replace
<code>/path_to_sage_doc</code> with that path in the "offline" urls above.</p>
|
1,500,827 | <p>A composite number $n$ is a Fermat-pseudoprime to base $a$, if</p>
<p>$$a^{n-1}\equiv\ 1\ (\ mod\ n)$$</p>
<p>If $n-1=2^s\times t$ , $t$ odd , $n$ is a strong a-PRP, if either
$2^t\equiv 1\ (\ mod\ n)$ or there is a number $u$ with $0\le u<s$ and
$\large 2^{2^u\times t}\equiv -1\ (\ mod\ n\ )$</p>
<p>I want to find the composite numbers, which are strong PRP to the bases
$2,3,5$ upto , lets say , $10^9$ in an efficient way (faster than just check all the composite numbers).</p>
<p>Is this possible ? If yes, how ?</p>
| Brian M. Scott | 12,042 | <p>Evaluating a few of the differences by hand is a good idea, but in the end you have to do it symbolically rather than just numerically:</p>
<p>$$\begin{align*}
a_{n+1}-a_n&=\left(1+\frac1{2^n}\right)-\left(1+\frac1{2^{n-1}}\right)\\
&=\frac1{2^n}-\frac1{2^{n-1}}\\
&=\frac1{2^n}-\frac2{2^n}\\
&=-\frac1{2^n}\;,
\end{align*}$$</p>
<p>so</p>
<p>$$a_{n+1}=a_n-\frac1{2^n}\;.$$</p>
<p>That’s probably the most straightforward approach. Alternatively, you might notice that since $a_n-1=\frac1{2^{n-1}}$ and $a_{n+1}-1=\frac1{2^n}=\frac12\cdot\frac2{2^{n-1}}$, we always have </p>
<p>$$a_{n+1}-1=\frac12(a_n-1)=\frac12a_n-\frac12\;,$$</p>
<p>and hence </p>
<p>$$a_{n+1}=\frac12a_n+\frac12\;.$$</p>
|
2,475,757 | <p>I want to determine if the following integrals converge or diverge.</p>
<ol>
<li>$\int_{0}^\infty \frac{\sqrt{x}}{\sqrt[3]{x^5+1}}dx.$</li>
<li>$\int_{0}^\infty \sin\frac{1}{x^2+1}dx$.</li>
<li>$\int_{\sqrt{2}}^2 \frac{dx}{\sqrt{x^2-2}}dx.$</li>
<li>$\int_{0}^1 \frac{\ln{x}}{x}dx.$</li>
</ol>
<hr>
<p><strong>(1):</strong> Here I can raise the integrand to the third power and use that $$\int_{0}^\infty\frac{\sqrt{x}}{\sqrt[3]{x^5+1}}\leq\int_{0}^\infty\frac{x^{3/2}}{x^5+1},$$</p>
<p>Clearly the RHS is convergent since the denominator grow faster than the numerator. Is this correct reasoning?</p>
<p><strong>(2):</strong> Just by looking at it I can say that as $x\rightarrow \infty$ then the argument for $\sin$ approaches $0$, so the entire function approaches 0, thus the integral is convergent. Same question as the above, is this reasoning correct? And how can one show this analytically?</p>
<p><strong>(3):</strong> Having trouble with this one. Clearly the function is not defined at $x=\sqrt{2}$, should I instead then be looking at $$-\lim_{n\rightarrow \sqrt{2}}\int_{2}^{n}\frac{dx}{\sqrt{x^2-2}}?$$</p>
<p><strong>(4):</strong> This one seemed simple at first glance. I used the function $\ln(x)$ for comparison. $$\int_{0}^1\frac{\ln{x}}{x}dx\le \int_{0}^1 \ln{x}dx,$$</p>
<p>I know that the right integral is convergent since its value is $-1$, but why is the left integral divergent?</p>
| user247327 | 247,327 | <p>Oh, bother! Amanda R just pointed out the balls are numbered 1 to 3, not 0 to 2! The computation is basically the same.</p>
<p>The largest of the two numbers will be 2 for (1, 2), (2 ,1), or (2, 2). That is three out of the total 3*3= 16 possible results drawing two balls. P(2)= 3/9.</p>
<p>(Although you didn't ask, 1 is the "larger" only for (1,1). The probability is P(1)= 1/9.
1 is the larger only for (1, 1). The probability is P(1)= 1/9. </p>
<p>Three is larger for (1, 3), (2, 3), (3, 1), (3, 2) and (3, 3). The probability P(3)= 5/9.</p>
<p>Of course, 3/9+ 1/9+ 5/9= 9/9= 1.)</p>
<p>Thanks Amanda R! </p>
|
4,302,213 | <blockquote>
<p>Let <span class="math-container">$R,S$</span> be rings and <span class="math-container">$\varphi : R\to S$</span> be a ring homomorphism. Verify that</p>
<ol>
<li><span class="math-container">$\varphi(na) = n\varphi(a)$</span> for all <span class="math-container">$n\in\mathbb Z$</span> and <span class="math-container">$a\in R$</span>.</li>
<li><span class="math-container">$\varphi(a^n) = (\varphi(a))^n$</span> for all <span class="math-container">$n\in\mathbb Z^+$</span> and all <span class="math-container">$a\in R$</span>.</li>
<li>If <span class="math-container">$A$</span> is a subring of <span class="math-container">$R$</span>, then <span class="math-container">$\varphi(A) = \{\varphi(a):a\in A\}$</span> is a subring of <span class="math-container">$S$</span>.</li>
</ol>
</blockquote>
<p>For (1) I have the following:
<span class="math-container">$$
\begin{split}
\varphi(na) &= \varphi((n-1)a+a) \\
&= \varphi((n-1)a)+\varphi(a) \\
&= (n-1)\varphi(a)+\varphi(a)\\
&= n\varphi(a).
\end{split}
$$</span></p>
<p>For (2): For <span class="math-container">$n=1$</span>, we have <span class="math-container">$\varphi(a) = \varphi(a)$</span>. Suppose for some <span class="math-container">$n\in\mathbb Z^+$</span> that <span class="math-container">$\varphi(a^n)=(\varphi(a))^n$</span>. Observe:
<span class="math-container">$$
\begin{split}
\varphi(a^n) & =(\varphi(a))^n \\
\varphi(a)\cdot \varphi(a^n) & =\varphi(a)\cdot (\varphi(a))^n \\
\varphi(a \cdot a^n) & =(\varphi(a))^{n+1} \\
\varphi( a^{n+1}) & =(\varphi(a))^{n+1} \\
\end{split}
$$</span></p>
<p>For (3): I verify for <span class="math-container">$a,b\in\varphi(A)$</span> then <span class="math-container">$a-b\in\varphi(A)$</span> and <span class="math-container">$a\cdot b\in \varphi(A)$</span> (for brevity).</p>
<ul>
<li>Well, <span class="math-container">$0\in A \implies 0 \in \varphi(A) $</span></li>
<li>For <span class="math-container">$a,b \in \varphi(A)$</span> we have <span class="math-container">$\varphi(a) - \varphi(b) = \varphi(a-b)$</span>, thus, verified, because <span class="math-container">$a-b\in A$</span>.</li>
<li>For <span class="math-container">$a,b \in \varphi(A)$</span>, we have <span class="math-container">$\varphi(a) \cdot \varphi(b) = \varphi (a\cdot b)$</span>, thus, verified, because <span class="math-container">$a\cdot b \in A$</span>.</li>
</ul>
<p>Did I do these right? Feedback appreciated!</p>
| Wuestenfux | 417,848 | <p>For the first property,
use induction to show that <span class="math-container">$\phi(na) = n\phi(a)$</span> for all <span class="math-container">$a$</span> and <span class="math-container">$n\geq 0$</span>.
This is actually what you have done.</p>
<p>For negative <span class="math-container">$n$</span> one must be careful.
For <span class="math-container">$n>0$</span> put
<span class="math-container">$(-n)a = -(na)$</span>, which is the additive inverse of <span class="math-container">$na$</span>.
This part is then true simply by definition.</p>
|
3,347,342 | <blockquote>
<p><span class="math-container">$$\frac{2}{5}^{\frac{6-5x}{2+5x}}<\frac{25}{4}$$</span></p>
</blockquote>
<p>I can write this as
<span class="math-container">$$\frac25 ^{\frac{6-5x}{2+5x}} <\frac25 ^{-2}$$</span>
Therefore <span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span>
Solving it , we get <span class="math-container">$x\in (-2, -\frac 25)$</span></p>
<p>The correct answer is <span class="math-container">$x\in (-\infty , -2)\cup (-\frac 25 , \infty)$</span></p>
<p>I feel it’s got something to do with signs. Maybe in the part where I wrote <span class="math-container">$\frac 25 ^{-2}$</span> </p>
<p>If I change the inequality there, I arrive at the answer. But I disagree. I haven’t changed the number at all, so the sign should not change. What’s the right answer?</p>
| Mostafa Ayaz | 518,023 | <p>Note that <span class="math-container">$$a^x>a^y\implies \begin{cases}x>y&,\quad a>1\\x<y&,\quad 0<a<1\end{cases}$$</span></p>
|
2,347,820 | <p>What is the solution to $\log_{10} x -x=2?$</p>
<p>I have tried to solve it but I couldn't. I've got to $x^x =200$.</p>
| Robert Israel | 8,508 | <p>Assuming you're using the principal branch of log,
$$x = -\frac{W(-100 \ln(10))}{\ln(10)} $$
where $W$ is the principal branch of the Lambert W function. Other branches of log would correspond to other branches of Lambert W. Since $-100 \ln(10) < -1/e$, no solutions are real.</p>
|
2,347,820 | <p>What is the solution to $\log_{10} x -x=2?$</p>
<p>I have tried to solve it but I couldn't. I've got to $x^x =200$.</p>
| Jack D'Aurizio | 44,121 | <p>If $x=10^{x+2}$ has a real solution such solution has to be $\geq 0$, since $10^{x+2}\geq 0$ for any $x\in\mathbb{R}$.<br>
But if $x\geq 0$ and $x=10^{x+2}$ then $x\geq 100$ since $10^{x+2}$ is an increasing function.<br>
If $x\geq 100$ and $x=10^{x+2}$ then $x\geq 10^{102}\geq 10^{10^2}$ for the same reason.<br>
At the next step we get $x\geq 10^{10^{10^2}}$ and by iterating this argument it should be clear that there are no real solutions.</p>
|
2,811,870 | <p>This is a question from Brilliant.org</p>
<blockquote>
<p>The triangle $ABC$ has $AB = 9$ and $AC:BC = 40:41$. What is the maximum possible area of $ABC$?</p>
</blockquote>
<p>For this question, I considered the equation $A=\frac 12ab\sin\theta$.</p>
<p>Since $\sin\theta\le 1$, then $A$ is maximised when $\sin\theta = 1$.</p>
<p>This meant $ABC$ was a right-angled triangle, after some working I got the answer as 180.
However, it was wrong, Brilliant said it is not necessarily a right-angled triangle and then used Heron's formula to find the maximised area, 820.</p>
<p>I checked other posts which were similar to my question such as, <a href="https://math.stackexchange.com/q/2096023">How to maximize the area of a triangle, given two sides?</a>. However, they followed the same method I did</p>
<p>I am interested in why a right-angled triangle would <em>not</em> maximise the area in this case and what is wrong with my logic?</p>
| dan_fulea | 550,003 | <p>As mentioned in a comment, the best way to realize where is the error is to consider the similar problem where $AB=9$ is given and fixed, and where $AC:BC=1$. Then in this case the triangle is isosceles, and we have the option to choose its height as big as we want. The bigger this heigth $h$, (the smaller the angle in $C$, and) the bigger the area. There is no limit for the area, we can let $h$ tend to infinity, the area goes to infinity!</p>
<p>As seen also in this example, in the given situation the sides $a=BC$, $b=AC$ and the angle in $C$ variate together in a "complicated way", and contribute more or less to the area. A maximum for $\sin \hat C$ does not lead to a maximal area, since the sides $a,b$, and the height $h$ are dramatically constrained to be "small".</p>
<hr>
<p>Now let us also give a simple solution to the "brilliant" problem.</p>
<p>First of all, let us consider the following example of a triangle satisfying the given conditions, the one with sides $9,40,41$.
Yes, it has a right angle, but not there where i supposed first that the question supposed it to be... In a (bad) picture:</p>
<p><a href="https://i.stack.imgur.com/aqiBd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aqiBd.png" alt="Triangle ABC''"></a></p>
<p>(Sorry, i cannot get it smaller. But maybe it is psychologically good so.)</p>
<p>A possible choice of $C$ is as above, in the position which makes the angle in $B$ a right angle. An other position is the symmetrical position, which makes the angle in $A$ a right angle. With the one or the other choice we get a first weak attempt to maximize the area, it is $\frac 12\cdot 9\cdot 40=180$.</p>
<p>Now let us try to get a better guess. Consider a point on $AB$ which divides $AB$ in two parts, $s$ and $9-s$.</p>
<p>In fact we allow
$$s\in\Bbb R\ ,$$
for a negative value of $s$ we consider the point on $AB$ to be on the half-line $(AB$ "after" $B$.</p>
<p>We draw a long perpendicular in this point to $AB$, choose on it that one point (in the "upper" half-plane) $C''$, so that the sides
$AC$ and $BC$ are in the given proportion:
$$
\frac{AC''}{BC''}=\frac{41}{40}\ .
$$
This is maybe possible, maybe impossible.
(For instance, for $s=9/2$ this is impossible.) If it is possible, we want to get the height $h=h(s)$ as a function of $s$, that we will maximize.</p>
<p>The above figure is maybe wrong, maybe $C''$ is so that the corresponding height is $\le 40$, and maybe it is there where $?C''$ is placed, well, we have to compute. (Yes, it is wrong.) </p>
<p><strong>Note:</strong> This is somehow related to the posted question. It turns out, that the maximal height is obtained for a negative $s$, (which is relatively big in absolute value,) the point on the line $AB$ which realizes the maximal height is not on the segment $AB$.</p>
<p><strong>Computations.</strong> Note that from
$$
\frac{41}{40}
=\frac{AC''}{BC''}=
\frac
{\sin \hat B}
{\sin \hat A}
=
\frac
{h/\sqrt{h^2+s^2}}
{h/\sqrt{h^2+(9-s)^2}}
=
\frac
{\sqrt{h^2+(9-s)^2}}
{\sqrt{h^2+s^2}}
$$
we get a simple relation determining $h$,
$$
\frac{41^2}{40^2}
=
\frac
{h^2+(9-s)^2}
{h^2+s^2}
\ ,
$$
so
$$
\begin{aligned}
\underbrace{(41^2-40^2)}_{=9^2}h^2
&=
40^2(9-s)^2-41^2s^2
\\
&=
9(40-9s)(s+360)\ .
\end{aligned}
$$
The roots of the last expression are $40/9$ and $-360$.
To have a positive quantity, we need $40-9s\ge 0$, so $s\le 40/9$.</p>
<p>The maximal value of the expression is taken exactly in the middle of the segment determined by the roots, which is $s^*=-1600/9$. For this value of the $s$-variable we get a corresponding maximal $h^*$, which is
$$
h^*
=\frac 19\sqrt{9\cdot (40+1600)\left(-\frac {1600}9+360\right)}
=
\frac {1640}9\ .
$$
So the maximal area is $\frac 12 \cdot AB\cdot h^*=\frac{1640}2=820$.</p>
|
997,587 | <p>The first sequence given is 3, 7, 16, 41, 77,....
I really am quite stuck on this because I can't seem to find any relationship between one term and the terms prior to it. I first noticed that it seemed like we were adding a perfect square to each one, since 3+4=7, 7+9=16, etc. But we skipped over adding the perfect square of 16 to anything so that must not be a useful idea.</p>
<p>The next problem is a recursive sequence given by $3s_{n-1}+2$ and I need to write it as a closed-form formula. I believe I should use something involving $3^n$ and I have tried subtracting $2^n$ but it only works for the first two terms.</p>
| Brian M. Scott | 12,042 | <p>The first one is a mistake: either it was intended to be $3,7,16,32,57,\ldots$, corresponding to the description $s_1=3$, $s_n=s_{n-1}+n^2$ for $n>1$, or there is simply not enough information to allow one to guess how it is intended to differ from that sequence.</p>
<p>For the second one you can try ‘unwinding’ it:</p>
<p>$$\begin{align*}
s_n&=3s_{n-1}+2\\
&=3(3s_{n-2}+2)+2\\
&=3^2s_{n-2}+3\cdot2+2\\
&=3^2(3s_{n-3}+2)+3\cdot2+2\\
&=3^3s_{n-3}+3^2\cdot2+3\cdot2+2\\
&\;\vdots\\
&=3^ks_{n-k}+3^{k-1}\cdot2+3^{k-2}\cdot2+\ldots+3\cdot2+2\\
&=3^ks_{n-k}+2\sum_{\ell=0}^{k-1}3^\ell\;.
\end{align*}$$</p>
<p>Depending on whether the first term that you’re given is $s_0$ or $s_1$, you can unwind it to $k=n$ or $k=n-1$ to get an expression for $s_n$ in terms of $n$. That expression will still contain a summation, but it’s the sum of a geometric series, so you can get it in closed form.</p>
|
3,611,072 | <p>Show that if a prime <span class="math-container">$p ≠ 3$</span> is such that <span class="math-container">$p≡1$</span> (mod 3) then p can be written as <span class="math-container">$a^2-ab+b^2$</span> where a and b are integers. </p>
<p>I have no idea how to approach this question, so any help much appreciated! This is in the context of algebraic number theory, so I'm not sure if it's helpful to consider rings of integers or anything like that. </p>
| Tob Ernack | 275,602 | <p>Given that <span class="math-container">$p = a^2 - ab + b^2$</span>, we can factor over <span class="math-container">$\mathbb{Z}[\zeta_3]$</span> (ring of Eisenstein integers) to obtain <span class="math-container">$p = \left(a + \zeta_3b\right)\left(a + \zeta_3^2b\right)$</span>.</p>
<p>Therefore the ideal <span class="math-container">$(p)$</span> is not inert over <span class="math-container">$\mathbb{Z}[\zeta_3]$</span>.</p>
<p>Then by the Dedekind factorization theorem we must have that <span class="math-container">$X^2 + X + 1$</span> splits into linear factors in <span class="math-container">$\mathbb{F}_p[X]$</span>, i.e. there are roots in <span class="math-container">$\mathbb{F}_p$</span>.</p>
<p>But the roots are formally <span class="math-container">$\frac{-1 \pm \sqrt{-3}}{2}$</span>, so <span class="math-container">$-3$</span> must be a perfect square mod <span class="math-container">$p$</span>. By quadratic reciprocity, this means that <span class="math-container">$p \equiv 1 \pmod 3$</span> as desired.</p>
<hr>
<p>The converse is also true: given that <span class="math-container">$p \equiv 1 \pmod 3$</span> we use quadratic reciprocity to show that <span class="math-container">$-3$</span> is a perfect square mod <span class="math-container">$p$</span>. Then the roots <span class="math-container">$\frac{-1 \pm \sqrt{-3}}{2}$</span> of the minimal polynomial <span class="math-container">$X^2 + X + 1$</span> are in <span class="math-container">$\mathbb{F}_p$</span>, so <span class="math-container">$X^2 + X + 1$</span> splits in <span class="math-container">$\mathbb{F}_p[X]$</span>.</p>
<p>By the Dedekind factorization theorem, the ideal <span class="math-container">$(p)$</span> factors into a product of two prime ideals in <span class="math-container">$\mathbb{Z}[\zeta_3]$</span>. Since <span class="math-container">$\mathbb{Z}[\zeta_3]$</span> is a PID, we can write <span class="math-container">$(p) = (\alpha)(\beta) = (\alpha\beta)$</span>, then it follows that <span class="math-container">$p = u\alpha\beta$</span> for some <span class="math-container">$\alpha, \beta \in \mathbb{Z}[\zeta_3]$</span> nonunits and <span class="math-container">$u$</span> some unit.</p>
<p>Finally taking norms we must have <span class="math-container">$N_{\mathbb{Q}(\zeta_3)/\mathbb{Q}}(\alpha) = p$</span>, so <span class="math-container">$p$</span> is the norm of <span class="math-container">$\alpha = a + b\zeta_3$</span>, which is <span class="math-container">$a^2 - ab + b^2$</span>.</p>
|
3,611,072 | <p>Show that if a prime <span class="math-container">$p ≠ 3$</span> is such that <span class="math-container">$p≡1$</span> (mod 3) then p can be written as <span class="math-container">$a^2-ab+b^2$</span> where a and b are integers. </p>
<p>I have no idea how to approach this question, so any help much appreciated! This is in the context of algebraic number theory, so I'm not sure if it's helpful to consider rings of integers or anything like that. </p>
| nguyen quang do | 300,700 | <p>This is just a matter of quadratic reciprocity law.</p>
<p>Let <span class="math-container">$\omega$</span> be a primitive cubic root of unity and <span class="math-container">$K=\mathbf Q(\omega)= \mathbf Q(\sqrt {-3})$</span> . It is known that the ring of integers of <span class="math-container">$K$</span> is <span class="math-container">$A=\mathbf Z[\omega]=\mathbf Z[\sqrt {-3}]$</span>, which is a PID. Knowing that <span class="math-container">$N(a+b\omega)=a^2 - ab +b^2$</span> (where <span class="math-container">$N$</span> is the norm map defined by <span class="math-container">$N(z)=z.\bar z$</span>), your question amounts to the characterisation of a rational prime <span class="math-container">$p$</span> which is a norm from <span class="math-container">$A$</span>, i.e. <span class="math-container">$p$</span> splits completely in <span class="math-container">$A$</span>, or equivalently, <span class="math-container">$p$</span> is odd and <span class="math-container">$-3$</span> is a quadratic residue mod <span class="math-container">$p$</span> (note that the case <span class="math-container">$p=2$</span> is excluded because <span class="math-container">$-3\neq 1$</span> mod <span class="math-container">$8$</span>), i.e. <span class="math-container">$(\frac {-3}{p})=1$</span> . But <span class="math-container">$(\frac {-3}{p})=(\frac {3}{p})(\frac {-1}{p})=(\frac {3}{p})$</span> because <span class="math-container">$p$</span> is odd, and quadratic reciprocity asserts here that <span class="math-container">$(\frac {3}{p})=(\frac {p}{3})$</span> . Finally, the wanted <span class="math-container">$p$</span> is characterized by <span class="math-container">$(\frac {p}{3})=1=(\frac {-3}{p})$</span>. The first equality implies that <span class="math-container">$p\equiv 1$</span> mod <span class="math-container">$3$</span>, and the second that <span class="math-container">$p$</span> splits, as we have seen above.</p>
|
672,744 | <p>Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.</p>
<p>Having trouble evaluating the integral: </p>
<p>Solved for $x$:</p>
<ul>
<li>$x=0, y=1$</li>
<li>$x=2, y=13$</li>
</ul>
<p>$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\sqrt\frac{y-1}3'}^2\,dy$$</p>
<p>I got stuck at 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+(1/12)+(1/(y-1))}</p>
<p>any help would be great thanks</p>
| Semsem | 117,040 | <p>Note that $((\sqrt{\frac{y-1}{3}})')^2=(\frac{1}{2\sqrt{\frac{y-1}{3}}} (1/3))^2=\frac{1}{4(y-1)}$
$$=\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\frac{1}{4(y-1)}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y-1}3+\frac{1}{12}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y}3-\frac{1}{4}}dy
\\= 6\pi\{(\frac{y}3-\frac{1}{4})^\frac{3}{2}\}_1^{13}$$</p>
|
1,896,008 | <p>Is the following statement correct: </p>
<blockquote>
<p>If $A$ and $B$ are closed subsets of $[0,\infty)$, then $A+B=\{x+y:x \in A,y \in B\}$ is closed in $[0,\infty)$.</p>
</blockquote>
| Yaddle | 333,729 | <p>Let $(c_n)_{n \in \mathbb N}$ a convergent sequence in $A + B$ with
$$c := \lim_{n \to \infty} c_n.$$
We need to show, that $c \in A+B$. There exist sequences $(a_n)_{n \in \mathbb N}$ in $A$ and $(b_n)_{n \in \mathbb N}$ in $B$ with $c_n = a_n + b_n$. Since $A,B \subseteq [0, \infty)$, we have that $a_n \leq c_n$ for all $n \in \mathbb N$ and since $(c_n)_{n \in \mathbb N}$ converges it is bounded. Hence $(a_n)_{n \in \mathbb N}$ and $(b_n)_{n \in \mathbb N}$ are bounded. With the Bolzano Weierstrass theorem we get that there exist convergent subsequences of $(a_n)_{n \in \mathbb N}$ and $(b_n)_{n \in \mathbb N}$. W.l.o.g. we can choose $(a_n)_{n \in \mathbb N}$ and $(b_n)_{n \in \mathbb N}$. Hence there exist
$$a := \lim_{n \to \infty} a_n \quad \text{and} \quad b := \lim_{n \to \infty} b_n.$$
Since $A$ and $B$ are closed, we have $a \in A$ and $b \in B$. Thus it follows that $$c = \lim_{n \to \infty} c_n = \lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = a + b \in A + B.$$
I hope that it helps you :)</p>
|
2,153,340 | <p>Let $G$ be a group, and $C$ a set of proper subgroups of $G$.</p>
<p>Each subgroup in $C$ is normal subgroup of $G$.</p>
<p>For $G_1 , G_2\in C$, if $G_1 \ne G_2$ then $G_1\cap G_2=\{e_G\}$</p>
<p>$\bigcup\limits_{H\in C}H= G$.</p>
<p>Need to prove that G is Abelian group, hint someone?</p>
| user300 | 226,798 | <p>Hint: Let $x,y\in G$. Then there exist $G_1,G_2\in C$ such that $x\in G_1,y\in G_2$. Then show $xyx^{-1}y^{-1}\in G_1\cap G_2$</p>
|
3,670,240 | <p>It' not a physics question, just ..coincidence ;) (i'm concerned about mathematical rightness of it)</p>
<p>Let's consider <span class="math-container">$U,T,S,P,V\in\mathbb{R_{>0}}$</span> such that
<span class="math-container">$$dU=TdS-PdV$$</span></p>
<ul>
<li>Based on this, how we can rigorously proof that <span class="math-container">$U=U(S,V)$</span>?</li>
</ul>
<hr>
<p>Attempt 1: (probably inconclusive, see 'Attempt 2')</p>
<p>Let us consider
<span class="math-container">$$A, X, Y \in \mathbb{R}\;\;\mid\;\; A=A(X,Y)\;\;\;\wedge\;\;\; dA=dU$$</span></p>
<p>Then
<span class="math-container">$$dA=\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY$$</span>
Requirement <span class="math-container">$dA=dU$</span> implies
<span class="math-container">$$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY=TdS-PdV$$</span>
or
<span class="math-container">$$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY-TdS+PdV=0$$</span>
Now, since <span class="math-container">$dX, dY, dS$</span> and <span class="math-container">$dV$</span> are arbitrary, to make the sum null, what they multiply must be zero, and since <span class="math-container">$T,P$</span> are not null by definition, only possibilities are that
<span class="math-container">$$X=S\;\wedge\;Y=V \qquad\text{or}\qquad Y=S\;\wedge\;X=V$$</span>
in either case, we obtain
<span class="math-container">$$\frac{\partial A}{\partial S}\bigg|_V=T,\qquad\frac{\partial A}{\partial V}\bigg|_S=-P$$</span>
(I've considered <span class="math-container">$A$</span> being just function of two variables <span class="math-container">$X,Y$</span>, but this is not restrictive since if more than two variables were present in <span class="math-container">$A$</span> dependencies, the result woudn't change, as the additional partial derivatives appearing in <span class="math-container">$dA$</span> expansion would have been necessarily set to <span class="math-container">$0$</span>, eliminating thus their dependency in <span class="math-container">$A$</span>)</p>
<p>Also follows that </p>
<p><span class="math-container">$$A=A(S, V)$$</span></p>
<p>Then, being <span class="math-container">$dA=dU\,[..]\Rightarrow\,U=U(S,V)$</span></p>
<p>Some question about this attempt</p>
<ol>
<li>How to properly carry on last step, if all was correct so far? (simply saying that <span class="math-container">$A$</span> and <span class="math-container">$U$</span> differ by a constant as a consequence to mean value theorem? but how we can say this if still we don't know <span class="math-container">$U$</span> dependencies..?)</li>
<li>Has sense to look for <span class="math-container">$A$</span> such that <span class="math-container">$dA=dU$</span> if <span class="math-container">$A$</span> initially is not function of the same variables as <span class="math-container">$U$</span>?</li>
<li>Seems that to make the above reasoning work, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have to be independent one with respect to the other, but what if we cannot require this for <span class="math-container">$S$</span> and <span class="math-container">$V$</span>?</li>
</ol>
<hr>
<p>Attempt 2: (als inconclusive see 'Attempt 3')</p>
<p>From <span class="math-container">$dU=TdS-PdV$</span>, we have
<span class="math-container">$$\frac{dU}{dS}=T-P\,\frac{dV}{dS}\qquad\text{and}\qquad\frac{dU}{dV}=T\,\frac{dS}{dV}-P$$</span>
Then
<span class="math-container">$$\frac{dU}{dS}\bigg|_{V}=\Bigg(T-P\,\frac{dV}{dS}\bigg)\Bigg|_{V}=T\qquad\text{and}\qquad\frac{dU}{dV}\bigg|_{S}=\Bigg(T\,\frac{dS}{dV}-P\bigg)\Bigg|_{S}=P$$</span>
Eventually
<span class="math-container">$$dU=\frac{dU}{dS}\bigg|_{V}\,dS+\frac{dU}{dV}\bigg|_{S}\,dV$$</span></p>
<p>But here arises the problem, if i were sure that <span class="math-container">$U$</span> would just depend on <span class="math-container">$S,\,V$</span>, we could have written (you can check <a href="https://en.wikipedia.org/wiki/Partial_derivative#Basic_definition" rel="nofollow noreferrer">wikipedia page on this</a>)
<span class="math-container">$$dU=\frac{\partial U}{\partial S}\,dS+\frac{\partial U}{\partial V}\,dV$$</span>
and maybe arrive to the conclusion <span class="math-container">$U=U(S,V)$</span> in some way, but being the reasoning 'circular' we cannot do so..</p>
<p>So also this way seems inconclusive.. i wrote it in the hope of maybe clicking some ideas in the answerer, thanks!</p>
<hr>
<p>Attempt 3: posted in answer</p>
| Giorgio Pastasciutta | 660,461 | <p>Thinking back, i couldn't accept that such a simple question had such a long and complicated proof, so eventually i come up with:</p>
<p>Assume <span class="math-container">$U=U(S,V)$</span>, this implies
<span class="math-container">$$\frac{dU}{dA}\bigg|_{S,\,V}=0\qquad\forall\, A\;\text{not dependent on S,V,U}$$</span>
Substituting, we have
<span class="math-container">$$\frac{dU}{dA}\bigg|_{S,\,V}=\bigg(T\;\frac{dS}{dA}-P\;\frac{dV}{dA}\bigg)\bigg|_{S,\,V}=0-0=0$$</span>
and since this holds for any A not dependent on S,V,U, assumption is proved.</p>
<p>Note: it does not matter if <span class="math-container">$S$</span> and/or <span class="math-container">$V$</span> depend on A, their variation must be 0 anyways.</p>
|
2,066,455 | <p>I ask this question mainly to resolve (hopefully) and error with the following problem. </p>
<p>The United States Court consists of $3$ women and $6$ men. In how many ways can a $3$-member committee be formed if each committee must have at least one woman?</p>
<p>My approach:
Since each group needs at least one woman, there are $\binom31$ ways to pick the first one. After that, both men and women are allowed. There are $\binom82$ ways of doing this. </p>
<p>This amounts to $\binom31\times\binom82 = 84$ ways, but the answer is $64$. I don't see how this is the answer for two reasons. The first is that there should be at least somewhere a $3$ multiplied but $64$ has no powers of $3$. The second is that I wrote a counter in Java to look at all strings from the list $a,b,c,d,e,f,g,h,$ and $i$ choosen $3$ at a time and count how many words contained $a,b,$ or $c$. </p>
<p>What seems to be the problem?</p>
| AlgorithmsX | 355,874 | <p>You have $8$ choices for the person sitting on one end (because there are only $8$ people that can be on either end), you have $7$ choices for the person sitting on the other end (because you already put a person on the end and you still can't put two people on the end), and then you have $8!$ for the rest of the people. Your total is therefore $8*7*8!$.</p>
|
4,269,898 | <p>I've a question concerning inverse limits, since I don't usually work with them this extensively.</p>
<p>I'm considering the inverse limit of the following "bi-inverse system" of <span class="math-container">$R$</span>-modules and black arrows <span class="math-container">$f_{\bullet,\bullet}$</span>, and <span class="math-container">$g_{\bullet,\bullet}$</span>. Since inverse limits commutes among themselves we can define <span class="math-container">$$A := \varprojlim_{i,j}A_{i,j}$$</span> regardless of the order we take them. Let <span class="math-container">$$\widetilde{A}:=\varprojlim_k A_{k,k}$$</span> be the limit of the diagonal of this inverse system.</p>
<p><a href="https://i.stack.imgur.com/UUsPK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UUsPK.png" alt="enter image description here" /></a></p>
<p>It's easy to see we have a unique map <span class="math-container">$$\widetilde{A} \to A$$</span> that makes everything commute (induced by the obvious maps in the system).</p>
<blockquote>
<p>I was wondering if it's "easy to see" if there's a map going in the other direction and then prove that <span class="math-container">$A\cong \widetilde{A}$</span> maybe.</p>
</blockquote>
| Ittay Weiss | 30,953 | <p>A functor <span class="math-container">$j\colon C\to D$</span> between small categories is initial if pulling back limits along it does not alter limits (in the precise sense that the evident induced morphism is an isomorphism). So, you are asking whether the diagonal inclusion in your case is initial.</p>
<p>If we write <span class="math-container">$N$</span> for the category whose objects are the naturals, and with <span class="math-container">$x\to y$</span> precisely when <span class="math-container">$x\ge y$</span>, then your diagram inclusion is the diagonal <span class="math-container">$\Delta \colon N\to N\times N$</span>. So, the question is whether <span class="math-container">$\Delta$</span> is initial. This is the case, and can be verified either directly from the definition: <span class="math-container">$\Delta$</span> is initial if for any object <span class="math-container">$(m,n)\in N\times N$</span> the slice category <span class="math-container">$\Delta/(m,n)$</span> is connected. Alternatively, every left adjoint functor is initial, and <span class="math-container">$\Delta$</span> is a left adjoint, its right adjoint is given by the join function <span class="math-container">$(m,n)\mapsto m\vee n$</span> (the max function).</p>
|
88,565 | <p>Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:</p>
<p><span class="math-container">$$x^2\ne x\implies x\ne 1$$</span></p>
<p>I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when <span class="math-container">$x^2$</span> is not equal to <span class="math-container">$x$</span>, <span class="math-container">$x$</span> also can't be <span class="math-container">$0$</span> and because <span class="math-container">$0$</span> isn't excluded as a possible value of <span class="math-container">$x$</span>, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.<br><br>
<strong>Why I think the logical sentence above is true:</strong><br>
My understanding of the implication symbol <span class="math-container">$\implies$</span> is the following:
If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether <span class="math-container">$x$</span> can be <span class="math-container">$0$</span>. Maybe <span class="math-container">$x$</span> can't be <span class="math-container">$-\pi i$</span> too, but as I see it, it doesn't really matter, as long as <span class="math-container">$x \ne 1$</span> holds. And it always holds when <span class="math-container">$x^2 \ne x$</span>, therefore the sentence is true.</p>
<h3>TL;DR:</h3>
<p><strong><span class="math-container">$x^2 \ne x \implies x \ne 1$</span>: Is this sentence true or false, and why?</strong></p>
<p>Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.</p>
| P.K. | 34,397 | <p><em>Thing to note.</em> This is called <a href="http://en.wikipedia.org/wiki/Truth_tables#Logical_implication" rel="noreferrer">logical implication</a>.</p>
<blockquote>
<p>$x^2≠x⟹x≠1$: Is this sentence true or false, and why?</p>
</blockquote>
<p>We can always check that using an example. Let us look at this implication as $\rm P\implies Q$. Now we shall consider cases:</p>
<ul>
<li><strong>Case 1:</strong> If we consider $x = 0$, then $\rm P$ is false, and $\rm Q$ is true. </li>
<li><strong>Case 2:</strong> If we consider $x = 1$, then $\rm P$ is false, and $\rm Q$ is false as well.</li>
<li><strong>Case 3:</strong> If we consider each value except $x=0$ and $x = 1$, then both $\rm P$ and $\rm Q$ will be true since $x^2 = x \iff x^2 - x = 0 \iff x(x - 1) = 0$ which means that $x=0$ and $x = 1$ are the only possibilities. </li>
</ul>
<p>Fortunately, our truth tables tell us that logical implication will hold true as far as we are not having $\rm P$ true and $\rm Q$ false. Look at the cases above; none of them has $\rm P$ true and $Q$ false. Thus <strong>Case 1, Case 2 and Case 3 are all true</strong> according to mathematical logic, so $\rm P\implies Q$ is true, or in other words: <strong>$x^2 \ne x \implies x \ne 1$ is true</strong>.</p>
<hr>
<p>I apologize for being late, but I always have my two cents to offer... thank you.</p>
|
2,757,870 | <p>I've come to this:
$$f: \mathbb{N} \to \{\ldots, -6,-4,-2,0,2,4,6, \ldots\},\qquad f(n) =
\begin{cases}
2n & \text{ if } n \text{ is odd} \\
-n & \text{ if } n \text{ is even}
\end{cases}$$</p>
<p>I don't know what to do with this though. I never know how to format a proof correctly.</p>
| fleablood | 280,126 | <p>Well, you've found your bijection. So the next step is to prove it is a bijection..... which you can't because it is not.</p>
<p>First prove it is surjective: that for any $e\in E$ that there is an $n\in \mathbb N$ so that $f(n) = e$. </p>
<p>Failure of pf: let $e=2k $. In $k\le 0$ then $f(-e) = e$.
If $k>0$ and $k $ is odd then $f (k)=2k=e $. If $k>0$ and $k $ is even... well, you're hosed. $f (even)\le 0$. And $f (odd)=2*odd) $ but nothing gives you a positive $2*even $. It can't be done.</p>
<p>Next prove it is injective. That if $f(n) = f(m)$ then $n = m$. Or if $n \ne m$ then $f(n) \ne f(m)$.</p>
<p>Failure of a pf: Let $m \ne n$. Case 1: $n, m$ are both odd. Then $f(n)=2n$ and $f(m) = 2m$ but $2m\ne 2n$. so far so good. Case 2: $n, m$ are both odd. Then $f(n) = -n$ and $f(m) -m$ but $-n \ne -m$. So far so good.</p>
<p>Case 3: $n = 2k$ is even and $m = 2j + 1$ is odd. Then $f(n) = -2k$. And $f(m) = 4j + 2$. Now if $-2k = 4j + 2$ then $k = -2j - 1$. That is certainly possible. $f(-2j - 1) = -4j -2 = f(4j+2)$ but $-2j-1 \ne 4j+2$. So it is not injective.</p>
<p>===</p>
<p>My advice would be when it comes to guessing the bijection $f: \mathbb N \to \{...-6, -4, -2, 0 , 2, 4, 6,....\}$ it be easier to try to find</p>
<p>$f^{-1}= g: \{...-6, -4, -2, 0 , 2, 4, 6,....\}\to \mathbb N$ to get a better idea of a strategy. </p>
<p>Here we have to figure out what to do with the negative and the positive integers. We need to send the negatives to one half of the integers and to send the positives (and 0) to the other. </p>
<p>So we have to send the negative values to the odd or evens and the positive (and 0) to the evens or odds.</p>
<p>Since $0$ isn't natural (unless it is) and we don't want zero we should send $0$ and then negatives to the odds.</p>
<p>If $2k \le 0; g(2k) = - (2k + 1)$</p>
<p>And we can send the positive evens to ... the positive evens.</p>
<p>If $2k > 0$ then $g(2k) = 2k$.</p>
<p>Now if $g: \{...-6, -4, -2, 0 , 2, 4, 6,....\}\to \mathbb N$ is a bijection then $f = g^{-1}: \mathbb N\to \{...-6, -4, -2, 0 , 2, 4, 6,....\}$ will be too.</p>
<p>We don't actually <em>have</em> to express $f(x)$ but... well,....</p>
<p>$f:\mathbb N \to \{...-6, -4, -2, 0 , 2, 4, 6,....\}\to \mathbb N$. If $n= 2k+1$ is odd, $f(n) = -2k$. If $n=2k$ is even, $f(n) = n=2k$.</p>
<p>Now we must prove $f$ or $g$ is a bijection. I'll do $f$ because... well, it's harder.</p>
<p>1) Show $f$ is surjective. If $e = 2k \in E$ show there is an $n \in \mathbb N$ so that $f(n) = e$.</p>
<p>If $e > 0$ then $f(e) = e$. If $e =2k \le 0$ then $f(-2k + 1) = 2k = e$. </p>
<p>So $f$ is surjective.</p>
<p>2) Show $f$ is injective. If $f(n) = f(m) = 2k$ prove that if $n\ne m$ then $f(n) \ne f(m)$.</p>
<p>Case 1: $n,m$ both odd. $n=2k + 1; m = 2j + 1; k\ne j$. Then $f(2k+1) = -2k$ and $f(2j+1) = -2j$ and $-2k \ne -2j$.</p>
<p>Case 2: $n = 2k ; m=2j$ are both even $k \ne j$. Then $f(2k) =2k$ and $f(2j) = 2j$ and $2k\ne 2j$.</p>
<p>Case 3: $n = 2k +1$ is odd and $m = 2j$ is even. Then $f(n)= -2k \le 0 < 2j = f(m)$. So $f(n) \ne f(m)$.</p>
<p>Case 4: $n$ even and $m$ odd would be exactly the same as case 3:</p>
|
2,670,082 | <p>I have quite an interesting infinite totient sum. My task is to evaluate</p>
<p>$\sum_{n=1}^{\infty} \frac{\phi(n)}{5^n +1}.$</p>
<p>The problem is that I have no idea how to go from here as I have never seen such a problem before. The usual techique of writing $n$ and $\phi(n)$ in terms of the prime factorization of $n$ doesn't work. I get $\frac{\phi(n)}{5^n+1}=\frac{\prod_i p_i^{e_i-1} (p_i-1)}{5^{\prod p_i^{e_i}}+1}$, which turns into a complete crocodile when plugged back into the summation.</p>
<p>I know that the problem has an elegant, closed form solution because it appeared on a problem set that is expected to be solved by mere high school students. Furthermore, the only way to simplify such an infinite sum is to turn it into a numerical answer. Wolfram Alpha is unable to give a closed form solution, so I know that there is some clever trick to demolish this problem that I have not seen before. Does anyone have any hints or ideas on how do deal with this?</p>
<p>My only guess is that 5 is quite a peculiar number, and so it seems reasonable that the sum $\sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$ should also have a closed form solution for any integer $a$.</p>
<p>Update: Summing up to $n=1000$ using Wolfram-Alpha (it seems that my mistake was asking it to sum infinitely many terms) gives $0.22569444444444...$ with no end to the relentless onslaught of the 4's. Therefore I believe that the sum is $\frac{65}{288} = \frac{5 \cdot 13}{2^5 \cdot 3^2}.$ I have split up the possible answer into prime factors so that the clever trick I had earlier alluded to may be hopefully found. However, I do not have the luxury of pulling out a computational device on a math competition, and so I hope to find the actual solution, i.e. the method through one may derive this answer.</p>
<p>Update: We define $f(a) = \sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$. Here are some reasonable values for prime $a$ in case someone may try to use these to find the solution. At this point, I believe the only way to find a rigorous solution is to work backwards from the likely solution given by WA.</p>
<p>$f(2)=22569444444444...=\frac{5 \cdot 13}{2^5 3^2}$</p>
<p>$f(3)=0.$</p>
<p>$f(5)=0.$</p>
<p>$f(7)=0.$</p>
<p>$f(11)=0.09319444444...=\frac{11 \cdot 61}{2^5 3^2 5^2}$</p>
<p>Additional Update: With the help of achille hui, I have found that $$\sum_{n=1}^{\infty} \frac{\phi(n)}{1-(-1)^n q^n} = \frac{q}{(q+1)^2}.$$</p>
<p>This is almost the right sum.</p>
| David | 119,775 | <p>Basic facts -</p>
<ol>
<li>Using the idea behind Achille Hui's hint and noting that odd numbers can only have odd factors,</li>
</ol>
<blockquote>
<p><span class="math-container">$$\color{red}{\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
\frac{\phi(n)q^n}{1-q^{2n}}}
=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
\sum_{\textstyle{m=1\atop m\ \rm odd}}^\infty\phi(n)q^{mn}
=\sum_{\textstyle{d=1\atop d\ \rm odd}}^\infty\sum_{n\mid d}\phi(n)q^d
=\sum_{\textstyle{d=1\atop d\ \rm odd}}^\infty dq^d
\color{red}{=\frac{q+q^3}{(1-q^2)^2}}\ .$$</span>
2. The extraordinary telescoping sum
<span class="math-container">$$\color{red}{\sum_{k=0}^\infty\frac{2^k}{a^{2^k}+1}}
=\sum_{k=0}^\infty\left(\frac{2^k}{a^{2^k}-1}
-\frac{2^{k+1}}{a^{2^{k+1}}-1}\right)
\color{red}{=\frac1{a-1}}\ .$$</span>
3. <span class="math-container">${\Bbb Z}^+$</span> is the disjoint union of the sets
<span class="math-container">$$\{\,2^km\mid m\ \hbox{is odd}\,\}$$</span>
for <span class="math-container">$k\ge0$</span>.
4. If <span class="math-container">$n$</span> is odd and <span class="math-container">$k\ge1$</span> then
<span class="math-container">$$\phi(2^kn)=2^{k-1}\phi(n)\ .$$</span></p>
</blockquote>
<p>And now for <span class="math-container">$0<q<1$</span> we have
<span class="math-container">$$\eqalign{
\sum_{n=1}^\infty \frac{\phi(n)}{q^{-n}+1}
&=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty\frac{\phi(n)}{q^{-n}+1}
+\frac{\phi(n)}{q^{-2n}+1}+\frac{2\phi(n)}{q^{-4n}+1}
+\frac{4\phi(n)}{q^{-8n}+1}+\cdots\cr
&=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty\phi(n)
\left(\frac1{q^{-n}+1}+\frac1{q^{-2n}-1}\right)
\qquad\qquad\hbox{using (2) twice}\cr
&=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
\frac{\phi(n)q^n}{1-q^{2n}}\cr
&=\frac{q+q^3}{(1-q^2)^2}\ .\cr}$$</span>
Finally, we get your sum by taking <span class="math-container">$q=\frac15$</span>:
<span class="math-container">$$\sum_{n=1}^\infty\frac{\phi(n)}{5^n+1}=\frac{130}{24^2}=\frac{65}{288}\ .$$</span></p>
|
84,138 | <p>I recently have a paper rejected from a very good (but not the top) journal. The referee report said the result was good and certainly belong there, but he did not think I did enough to back up my claims (it was a rather long and harsh criticism at the exposition). Now I know for sure that my result is good and my proofs are correct, should I resubmit to the same journal after rewriting it.</p>
| rview | 20,137 | <p>yes you should resubmit it</p>
|
3,935,494 | <p>Usually the inverse of a square <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> is defined as a matrix <span class="math-container">$A'$</span> such that:</p>
<p><span class="math-container">$A \cdot A' = A' \cdot A = E$</span></p>
<p>where <span class="math-container">$E$</span> is the identity matrix.</p>
<p>From this definition they prove uniqueness but using significantly the fact that <span class="math-container">$A'$</span> is both right and left inverse.</p>
<p>But what if... we define right and left inverse matrices separately. Can we then prove that:</p>
<p>(1) the right inverse is unique (when it exists)<br />
(2) the left inverse is unique (when it exists)<br />
(3) the right inverse equals the left one</p>
<p>I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?</p>
| Math Lover | 801,574 | <p>In the second problem, please note your set excludes the case where none of the fetus is a baby boy as you know at least one of them is.</p>
<p>So the probability in the second case that all three fetuses are baby boys <span class="math-container">$ \displaystyle = \frac{0.485^3}{1-0.515^3}$</span></p>
|
3,935,494 | <p>Usually the inverse of a square <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> is defined as a matrix <span class="math-container">$A'$</span> such that:</p>
<p><span class="math-container">$A \cdot A' = A' \cdot A = E$</span></p>
<p>where <span class="math-container">$E$</span> is the identity matrix.</p>
<p>From this definition they prove uniqueness but using significantly the fact that <span class="math-container">$A'$</span> is both right and left inverse.</p>
<p>But what if... we define right and left inverse matrices separately. Can we then prove that:</p>
<p>(1) the right inverse is unique (when it exists)<br />
(2) the left inverse is unique (when it exists)<br />
(3) the right inverse equals the left one</p>
<p>I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?</p>
| Mehul | 857,496 | <p>now, I am not sure about an "official" sol but here is mine.</p>
<p>now in these cases, I would use sets and a Venn diagram. let's name the persons 1,2,3
now let A = {events|1 has a boy} and similarly for B and C so the Venn diagram would look like-<a href="https://i.stack.imgur.com/HDVit.jpg" rel="nofollow noreferrer">Venn diagram</a></p>
<p>so now the if at least one of them has a boy then we are in A<span class="math-container">$\cup$</span>B<span class="math-container">$\cup$</span>C and so the required probability is n(A<span class="math-container">$\cap$</span>B<span class="math-container">$\cap$</span>C)/n(A<span class="math-container">$\cup$</span>B<span class="math-container">$\cup$</span>C) = <span class="math-container">$0.485^3/(1-0.515^3)$</span></p>
|
80,918 | <p>Could anyone please tell me what could be the math function to get the number of zeros in given decimal representation of numbers? I scratched my head on Combination and Permutation but couldn't come up with generic answer. The number length can be up to 1000 digits, so you can represent a number as a String.</p>
<p>For example, if numbers range is $1-100$, the answer should be $11$, for $1-200$, it's $20$ and so on! Now, how would you find total number of zeros between $1-19447494833737292827272\cdots 444$ ( or any big number)?</p>
<p>Thanks.</p>
| Arturo Magidin | 742 | <p>Say you are writing all $d$ digit numbers, writing leading zeros. Then you would write out $d\times 10^d$ digits total, with each digit occurring exactly $\frac{1}{10}$th of the time, so you would write a total of $d\times 10^{d-1}$ zeros.</p>
<p>So, for example, to write all numbers between $0$ and $99$, writing the numbers less than $10$ as $0d$, you would write $2\times 10^2 = 200$ digits, of which $2\times 10=20$ are zeros.</p>
<p>Of course, of these you don't want to count the leading $0$s of $00-09$, so you subtract $10$; and you don't want to count the second $0$ from $00$, so you subtract another one, giving you a total of $9$ zeros. So you need to adjust the count above.</p>
<p>If you write all numbers of up to $d$ digits, then the number of $0$s, including leading $0$s, is $d\times 10^{d-1}$. Then you subtract the number of leading $0$s that appear in the left-most digit (there are $10^{d-1}$ of them), those that appear in the second-left-most digit when the left-most is $0$ (there are $10^{d-2}$ of them); those that appear in the third left-most digit when the first two are $0$ (there are $10^{d-3}$ of them) and so on. So you get
$$d\times 10^{d-1} - (1+10+10^2+\cdots + 10^{d-1}) = \frac{(9(d-1)-1)10^{d-1} + 1}{9}.$$
So, for example, for $d=2$ (from $1$ through $99$), you get
$$\frac{(9-1)10^1 + 1}{9} = \frac{81}{9} =9,$$
same as above. For $d=3$, (from $1$ through $999$) you get
$$\frac{(9(2) - 1)10^2 + 1}{9} = \frac{1701}{9} = 189.$$</p>
<p>What if you are doing something slightly different, as you write, say, only the numbers between $1$ and $751$?</p>
<p>You can count the zeros from $1$ to $99$ as above. </p>
<p>Then count the number of zeros in numbers of the form $7bx$ with $1\leq b\leq 5$. There's one for each value of $b$, for a total of $5$.</p>
<p>Then count the number of zeros in numbers of the form $70x$; there's $11$ of them.</p>
<p>Then count the number of zeros in numbers of the form $axy$ with $1\leq a\leq 6$; there's $10^{2-1}=10$ of them (you are just counting <em>all</em> zeros in numbers up to 2 digits, counting leading $0$s).</p>
<p>So after counting all the way to $99$, you then add:</p>
<ol>
<li>One zero for each number $7bx$, $1\leq b\leq 5$: total, $5$ (the middle digit of $751$).</li>
<li>Zeros for each number $70x$; total, $11$.</li>
<li>Zeros for each number $axy$ with $1\leq a\leq 6$; total, $6\times 10^{2-1} = 60$. </li>
</ol>
<p>There are 9 zeros from $1$ through $99$; and then there are $60+11+5=76$ zeros from $100$ through $751$, for a total of $85$ zeros from $1$ through $751$.</p>
|
2,280,666 | <p>Let AD be the altitude corresponding to the hypotenuse BC of the right triangle ABC. The circle of diameter AD intersects AB at M and AC at N shown. Prove $\frac{BM}{CN}$= $\bigg(\frac{AB}{AC}\bigg)^{3}$.</p>
<p>So far I have...</p>
<p>The power of B is $BD^{2}=(BM)(BA)$</p>
<p>The power of C is $CD^{2}=(CN)(CA)$</p>
<p>I am stuck after this. Any help would be appreciated!!</p>
| CY Aries | 268,334 | <p>As $\triangle ABD\sim\triangle CAD$,</p>
<p>$$\frac{BD}{AD}=\frac{AD}{CD}=\frac{AB}{CA}$$</p>
<p>Therefore,</p>
<p>$$\frac{BD}{CD}=\frac{BD}{AD}\cdot\frac{AD}{CD}=\left(\frac{AB}{CA}\right)^2$$</p>
<p>and thus</p>
<p>$$\frac{BD^2}{CD^2}=\left(\frac{AB}{CA}\right)^4$$</p>
<p>Using power of a point,</p>
<p>$$\frac{BM\cdot AB}{CN\cdot CA}=\left(\frac{AB}{CA}\right)^4$$</p>
<p>$$\frac{BM}{CN}=\left(\frac{AB}{AC}\right)^3$$</p>
|
2,280,666 | <p>Let AD be the altitude corresponding to the hypotenuse BC of the right triangle ABC. The circle of diameter AD intersects AB at M and AC at N shown. Prove $\frac{BM}{CN}$= $\bigg(\frac{AB}{AC}\bigg)^{3}$.</p>
<p>So far I have...</p>
<p>The power of B is $BD^{2}=(BM)(BA)$</p>
<p>The power of C is $CD^{2}=(CN)(CA)$</p>
<p>I am stuck after this. Any help would be appreciated!!</p>
| Peter Szilas | 408,605 | <p>$\angle AMD = 90 °$, so $MD$ || $AC$.</p>
<p>(Thales circle over $AD$.)</p>
<p>Intercept Theorem:</p>
<p>1) $BM/BA$ = $BD/BC$ .</p>
<p>$\angle AND = 90°$, so $ND$ || $AB$.</p>
<p>(Thales circle over $AD$.)</p>
<p>Intercept Theorem:</p>
<p>2) $CN/CA$ = $CD/CB$.</p>
<p>Dividing1) by 2):</p>
<p>$BM/CN$ × $CA/BA$ = $BD/CD$ ;</p>
<p>$BM/CN$ = $BA/CA$ × $BD/CD$.</p>
<p>To express the ratio $BD/CD$ we look at similar triangles.</p>
<p>$\triangle ADC$ is similar to $\triangle CAD$ is similar to $\triangle CAB$:</p>
<p>Let $\angle DAB$ = $\angle BCA$ = $\beta$.</p>
<p>1) $AB/AC$ = $tan(\beta)$.</p>
<p>2) $AD/DC$ = $tan(\beta)$.</p>
<p>3) $BD/AD$ = $tan(\beta)$.</p>
<p>Multiplying: 2) × 3) :</p>
<p>$BD/CD$ = $[tan(\beta)]^2$.</p>
<p>Hence using 1):</p>
<p>$BD/CD$ = $(AB/AC)^2$.</p>
<p>Altogether: </p>
<p>$BM/CN$ = $(AB/AC)^3$.</p>
|
2,251,998 | <p>Here is a question that I am working on:</p>
<blockquote>
<p>If $G$ is a group such that <em>every</em> non-identity element has order $2$, show that $G$ is abelian (commutative).</p>
</blockquote>
<p><strong>My attempt</strong></p>
<p>Suppose that for all $a \in G$, we have $$a^2 = Id$$</p>
<p>My goal is to show that $ab = ba$</p>
<p>Since $(ab)^2 = abab$ = Id, we multiply both left sides by $ba$ and we have</p>
<p>$(ba)abab = baId$</p>
<p>$\Rightarrow ba^2bab = ba$</p>
<p>$\Rightarrow bIdbab = ba$</p>
<p>$\Rightarrow b^2ab = ba$</p>
<p>$\Rightarrow Idab = ba$</p>
<p>$\Rightarrow ab = ba$</p>
<p>This condition will hold if I multiply $(ba)$ on the right side as well.</p>
<p>How does this look? </p>
| TMM | 11,176 | <p>See <a href="https://isc.carma.newcastle.edu.au/advanced" rel="nofollow noreferrer">inverse symbolic calculator</a>, which I think was originally coined Plouffe's inverter. It does exactly what you want. If you enter your decimals there, it will give your candidate solution as one match, other close matches for instance being $(16+9^{2/3} \cdot 11^{2/3}) / 100 \approx 0.37400477$.</p>
|
137,136 | <p><a href="https://i.stack.imgur.com/tk4kk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tk4kk.png" alt="Mathcad example of solving coil impedance"></a></p>
<p>I am new to Mathematica.</p>
<p>I am trying to figure out how to do the same thing in Mathematica as the image depicts done in Mathcad.</p>
<p>Declaring the variables with units I figured out.</p>
<pre><code>U = Quantity[100, "Volts"];
f = Quantity[50, "Hz"];
R1 = Quantity[20, "Ohms"];
L1 = Quantity[0.005, "Henries"];
XL1 = UnitSimplify[2 π f L1]
</code></pre>
<p>I am stuck on how define Z1 with the angle.</p>
| yode | 21,532 | <p>If we change the last <code>Transpose</code> into a <code>f</code>,it will return a uncalculate structure</p>
<pre><code>Dataset[{{0, 10}, {2, 11}, {3, 12}}][Transpose /* Map[MinMax] /* f]
</code></pre>
<blockquote>
<p>f[{{0, 3}, {10, 12}}]</p>
</blockquote>
<p>But the <code>Transpose</code> doesn't work here,so I tend to consider it as a bug about <code>Dataset</code> operate</p>
<hr>
<p>And the following operation is right in current rule</p>
<pre><code>Dataset[{{0, 10}, {2, 11}, {3, 12}}][Transpose /* Map[MinMax], All]
</code></pre>
<p><a href="http://o8aucf9ny.bkt.clouddn.com/2017-03-06-17-22-30.png" rel="nofollow noreferrer">http://o8aucf9ny.bkt.clouddn.com/2017-03-06-17-22-30.png</a> </p>
|
1,550,603 | <p>I got confirmed from a graduate school starting from next year and I will major algebraic geometry.</p>
<p>Until now, I have never thought that I study little things than others with my age. However, I heard that <strong>some</strong> of my colleagues already studied Hartshorne at least once and quite a few of them have read Rudin's RCA when they were undergaduates. It's kinda unbelievable to me, but it seems like if they really <strong>did</strong> study and understood, then they will write absolutely a better Ph.D thesis than mine.</p>
<p>So I'm now very worrying myself. I want to know whether this situation is general. Is it recommenable to study graduate subjects as early as possible? Or are there people here who experienced the same thing too? Was that beneficial?</p>
<p>Between "studying each thing deep and slow" and "skimming many subjects as fast as possible", which one is better?</p>
| Noah Schweber | 28,111 | <p><em>Your question is very broad, and I'm not sure this fully addresses it; but this is too long for a comment, and hopefully you find it useful nonetheless.</em></p>
<hr>
<p>I think there's a couple false assumptions here.</p>
<p>First, that there is a "better" way to approach studying mathematics. People vary wildly in how they learn, and ultimately I think it's best to find an approach to math that works well for you. Certainly you shouldn't discount the value of mastering difficult material early (and although rare, it is definitely believable for an undergrad to master such material), but at the same time, just knowing a bunch of stuff doesn't make you a mathematician. Grad school is much more about <em>learning to be a mathematician</em> than it is about learning material.</p>
<p>Second, and far more importantly, your line</p>
<blockquote>
<p>It seems like if they really did study and understood, then they will write absolutely a better Ph.D thesis than mine.</p>
</blockquote>
<p>This is something I'm still struggling to understand intuitively, so saying this in answer to your question also helps me internalize it: <strong>mathematics is not linear</strong>. Even ignoring the fact mentioned above that mathematics $\not=$ a bunch of facts, and even ignoring the fact that one's rate of learning changes over time, it is impossible to guess ahead of time how your thesis will compare with someone else's simply because there are so many different facets of mathematics. As you go through grad school, regardless of where you start relative to your peers, you will eventually find yourself an expert in some small area, just as they will in their own small areas. What contributions you make to this small area will surprise you, and it's pointless to try to guess ahead of time whether they will "match up" (however you might measure that) with someone else's. </p>
<p>Your thesis is not predetermined; it will be the product a number of things, including the growth you experience as a mathematician as you go through grad school (as well as a fair amount of chance, let's be honest). Certainly knowing more things at the beginning is an advantage, but it's by no means a dispositive one.</p>
|
211,705 | <p>I am given a table of possible <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> values that can be generated in a casino. In the game, both are generated with each turn.</p>
<p><img src="https://i.stack.imgur.com/G0nLn.jpg" alt="enter image description here" /></p>
<blockquote>
<p>The questions asks me to determine the minimum fee that should be charged per turn so that that casino doesn't lose money, if the payouts are:</p>
<p><span class="math-container">$a) \ 8X_1$</span></p>
<p><span class="math-container">$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128}$</span></p>
<p><span class="math-container">$c) \ 8X_1X_2$</span></p>
</blockquote>
<p>For a), I added up the total possible odds for each value <span class="math-container">$X_1$</span> could take, <span class="math-container">$$ 0\cdot\frac{2}{16} + 1\cdot\frac{4}{16}+2\cdot\frac{6}{16} + 3\cdot\frac{4}{16} = \frac{7}{4}$$</span>
and determined that the casino would need to charge at least <span class="math-container">$\$1.75$</span> per turn.</p>
<p>For b) I tried the pretty much the same thing, only breaking it down into individual situations, for example; <span class="math-container">$$P\{X_1=0, X_2 =1\} = \frac{1}{16}$$</span> <span class="math-container">$$P\{X_1=0, X_2 =2\} = \frac{1}{16}$$</span> <span class="math-container">$$etc...$$</span></p>
<blockquote>
<p>My question is, is there a faster, more efficient way to solve these types of problems? What if there were many more possible outcomes?</p>
</blockquote>
| John Gowers | 26,267 | <p>What you are trying to do is to find the value of </p>
<p>$$\mathbb{E}(Y)=\sum_yy\mathbb{P}(Y=y)$$</p>
<p>for various random variables $Y$ depending on $X_1,X_2$. Notice that this is exactly what you did in the first question: you worked out the possible values for $Y=8X_1$, and then multiplied each value by the probability of it occurring, and then added up the results. </p>
<p>If $Y=4X_1+8(X_2)^2-\frac{51}{128}$, you <em>could</em> work out the possible values of that rather nasty expression, find the probability of each one occurring, and then do the same process to find $\mathbb{E}(Y)$, but it is actually easier to use the following identity: </p>
<p>$$\mathbb{E}(aU+bV)=\sum_ux\mathbb{P}(aU+bV=x)$$
$$=a\sum_uu\mathbb{P}(U=u)+b\sum_v\mathbb{P}(V=v)=a\mathbb{E}(U)+b\mathbb{E}(V)$$</p>
<p>Getting from the first line to the second is non-trivial, but I'll leave that to you as an exercise, if you're not allowed just to quote it. Using that identity should make the problem a lot easier. </p>
|
1,146,802 | <p>Often seen similar systems of equations. Usually consider such systems in which decisions no. Such as there. <a href="https://math.stackexchange.com/questions/1146460/is-there-a-b-c-d-in-mathbb-n-so-that-a2b2-c2-b2c2-d2">Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$?</a></p>
<p>I think it would be more interesting to solve the system in which there are solutions. For example to find out whether such a system solution?</p>
<p>$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=k^2\end{aligned}\right.$$</p>
<p>What is the right approach? And how to solve it?</p>
| Kieren MacMillan | 93,271 | <p>The <span class="math-container">$2.n.m$</span> Diophantine equations have well-known complete solutions (a.k.a. parameterizations) for all <span class="math-container">$n,m \ge 1$</span> (for example, see <a href="https://www.jstor.org/stable/3620159" rel="nofollow noreferrer">Bradley's paper</a> or <a href="https://www.jstor.org/stable/2302941" rel="nofollow noreferrer">Barnett's paper</a>).</p>
<p>In particular, the solutions for the <span class="math-container">$2.1.2$</span> and <span class="math-container">$2.1.3$</span> equations you are asking about have been known for centuries. If <span class="math-container">$q$</span> is odd (which is essentially equivalent to “all primitive integer solutions”), you’re looking for
<span class="math-container">\begin{align}
k &= r^2+s^2 \\
c &= 2rs \\
q &= r^2-s^2
\end{align}</span>
where
<span class="math-container">$$q = t^2+u^2+v^2+z^2$$</span>
and either
<span class="math-container">$$c = t^2+u^2-v^2-z^2 \tag{$\star$}$$</span>
or
<span class="math-container">$$c = 2(tz-uv)\tag{$\star\star$}$$</span>
With respect to <span class="math-container">$q$</span>, you are are ultimately finding <span class="math-container">$r,s,t,u,v,z$</span> such that
<span class="math-container">\begin{align}
r^2 &= s^2+t^2+u^2+v^2+z^2;
\end{align}</span>
this is the <span class="math-container">$2.1.5$</span> Diophantine equation, for which the complete solution (a.k.a. parameterization) is known. You then simply need to intersect that result with the solutions for (<span class="math-container">$\star$</span>) and (<span class="math-container">$\star\star$</span>) to obtain the complete solution to your original system of equations.</p>
|
1,535,731 | <p>I am currently self-learning about matrix exponential and found that determining the matrix of a diagonalizable matrix is pretty straight forward :).</p>
<p>I do not, however, know how to find the exponential matrix of a non-diagonalizable matrix.</p>
<p>For example, consider the matrix
$$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$</p>
<p>Is there any process of finding the exponential matrix of a non-diagonalizable matrix? If so, can someone please show me an example of the process? :). I am not looking for an answer of the above mentioned matrix (since I just made it up), but rather I'm interested in the actual method of finding the matrix exponential to apply to other examples :)</p>
| Ben Grossmann | 81,360 | <p>There are two facts that are usually used for this computation:</p>
<blockquote>
<p><strong>Theorem:</strong> Suppose that $A$ and $B$ commute (i.e. $AB = BA$). Then $\exp(A + B) = \exp(A)\exp(B)$</p>
<p><strong>Theorem:</strong> Any (square) matrix $A$ can be written as $A = D + N$ where $D$ and $N$ are such that $D$ is diagonalizable, $N$ is nilpotent, and $ND = DN$</p>
</blockquote>
<p>With that, we have enough information to compute the exponential of every matrix.</p>
<p>For your example, we have
$$
D = \pmatrix{1&0\\0&1} = I, \quad N = \pmatrix{0&0\\1&0}
$$
we find that
$$
\exp(D) = eI\\
\exp(N) = I + N + \frac 12 N^2 + \cdots = I + N + 0 = I + N
$$
So, we have
$$
\exp(D + N) = \exp(D) \exp(N) = (eI)(I+N) = e(I+N) = \\
\pmatrix{e&0\\e&e}
$$</p>
|
3,668,101 | <p>I know that if <span class="math-container">$n \bmod k \le k-1$</span> then this sum is converge then it has finite sum, I just guess it's <span class="math-container">$\ln(k)$</span> because when <span class="math-container">$k=1$</span> sum is <span class="math-container">$0=ln(1)$</span>. I really don't know how to find it. Please help me.</p>
| Tony419 | 786,419 | <p>Also, @xpaul 's computation shows that the initial sum is asymptotically equal to <span class="math-container">$\log k$</span>. Indeed, @xpaul derived
<span class="math-container">\begin{eqnarray}
&&\sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{(mk+r)\mod(k)}{(mk+r)(mk+r+1)}\\
&=&\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{r}{(mk+r)(mk+r+1)}=\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)}
\end{eqnarray}</span>
and we can estimate the inner sum as follows
<span class="math-container">$$
\frac{1}{r(r+1)}<\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)}< \frac{1}{r(r+1)} +
\sum_{m=1}^{\infty }\frac{1}{(mk)^2}< \frac{1}{r(r+1)}+\frac{10}{k^2}.
$$</span>
Plugging this to the previous expression we get
<span class="math-container">\begin{align*}
\sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{r=1}^{k-1}\frac{1}{r+1}+\mathcal{O}(k^{-1})=\log k +\mathcal{O}(1)
\end{align*}</span></p>
|
255,811 | <p>I'm recalling this question from memory, so I may be messing it up a bit.</p>
<p>Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.</p>
<p>Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$. Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.</p>
<p>I'm not sure if this is right or where to go from here.</p>
| user1551 | 1,551 | <p>Apply MVT to $g(x) = \int (ax^2+bx+c) dx$.</p>
|
1,373,170 | <p>How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,</p>
<p>where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.</p>
<p>I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular case.</p>
<p>As a side comment: the values of $N$ I am working with are $N\approx 10^6$. </p>
| johannesvalks | 155,865 | <p>Write an algorithm to solve the problem...</p>
<p>Let</p>
<blockquote>
<p>$$
Q = \left( \sum_{m=1}^{n} \exp(k_m y) - 1 \right)^2.
$$</p>
</blockquote>
<p>In case we have the solution, we have
$$
Q = 0
$$.</p>
<p>For changes in $Q$ we have</p>
<blockquote>
<p>$$
\delta Q = 2 Q \left( \sum_{m=1}^{n} k_m \exp(k_m y) \right) \delta y.
$$</p>
</blockquote>
<p>For each step use</p>
<blockquote>
<p>$$
\delta y = - 2 Q \left( \sum_{m=1}^{n} k_m \exp(k_m y) \right),
$$</p>
</blockquote>
<p>whence</p>
<blockquote>
<p>$$
\delta Q \le 0,
$$</p>
</blockquote>
<p>so we get closer to the solution.</p>
<p>Once you have found $y$, the solution for $x$ is given by</p>
<blockquote>
<p>$$
x = \frac{1}{y}.
$$</p>
</blockquote>
<p>This would do the trick.</p>
|
1,373,170 | <p>How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,</p>
<p>where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.</p>
<p>I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular case.</p>
<p>As a side comment: the values of $N$ I am working with are $N\approx 10^6$. </p>
| marty cohen | 13,079 | <p>Let
$f(x)
= e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}-1
=\sum_{i=1}^N e^{k_i/x}-1
$,
and let
$K = \sum_{i=1}^N k_i
$.</p>
<p>The restrictions that
$x > 0$
and
$k_i < 0$
are important in what follows.</p>
<p>$f'(x)
=\sum -\frac{k_i}{x^2}e^{k_i/x}
=-\frac1{x^2}\sum k_ie^{k_i/x}
=\frac1{x^2}\sum |k_i|e^{k_i/x}
$,
so
$f'(x) > 0$.
This means that
your function has
at most one root.
Since $f(x) < 0$
for small $x$
and
$f(x) > 0$
for large $x$,
$f$ has exactly one positive root.</p>
<p>To get simple bounds
on the root of
$f(x) = 0$,
let
$k_{min} = \min(k_i)$
and
$k_{max} = \max(k_i)$.
Note that,
since the $k_i < 0$,
$|k_{min}|
>|k_{max}|
$.</p>
<p>Then
$e^{k_{min}/x}
\le e^{k_i/x}
\le e^{k_{max}/x}
$
so
$e^{k_{min}/x}
\le \dfrac{\sum_{i=1}^N e^{k_i/x}}{N}
\le e^{k_{max}/x}
$.
If
$f(x) = 0
$, then
$e^{k_{min}/x}
\le \dfrac{1}{N}
\le e^{k_{max}/x}
$
or
$k_{min}/x
\le -\ln(N)
\le k_{max}/x
$
or,
since $k_i < 0$,
$|k_{min}|/x
\ge \ln(N)
\ge |k_{max}|/x
$
or
$\frac{|k_{max}|}{\ln(N)}
\le x
\le \frac{|k_{min}|}{\ln(N)}
$.</p>
<p>Another bound can be gotten using
$e^z \ge 1+z$.
$f(x)
\ge \sum (1+k_i/x) - 1
= N+\frac1{x}\sum k_i - 1
= N+\frac{K}{x} - 1
$.
If
$f(x) = 0$,
then
$1-N
\ge \frac{K}{x}
= -\frac{|K|}{x}
$
or
$N-1
\le \frac{|K|}{x}
$
or
$x \le \frac{|K|}{N-1}
$.</p>
<p>This can be improved
by using the inequality
between the arithmetic and geometric means.</p>
<p>$\begin{array}\\
\dfrac{\sum_{i=1}^N e^{k_i/x}}{N}
&\ge \left(\prod_{i=1}^N e^{k_i/x}\right)^{1/N}\\
&= e^{\sum_{i=1}^N k_i/(Nx)}\\
&= e^{K/(Nx)}\\
\end{array}
$</p>
<p>Therefore,
if $f(x) = 0$,
$\dfrac1{N}
\ge e^{K/(Nx)}
$
or
$-\ln(N)
\ge K/(Nx)
$
or
$\ln(N) \le |K|/(Nx)
$
or
$x \le \dfrac{|K|}{N\ln(N)}
$.</p>
<p>So an initial $x$
could be
$x_0
=\frac{|K|}{N\ln(N)}
$.</p>
<p>We can then apply
Newton's iteration,
$x_{n+1}
=x_n - \frac{f(x_n)}{f'(x_n)}
$
a few times
and see what happens.
It should be OK.</p>
|
1,951 | <p>In <a href="https://matheducators.stackexchange.com/a/1949/704">this answer</a>, user <a href="https://matheducators.stackexchange.com/users/942/robert-talbert">Robert Talbert</a> stated that</p>
<blockquote>
<p>There are some amazing things you can do pedagogically with clickers.</p>
</blockquote>
<p>I'd like to see some examples. (Not that I'm eager to try that myself, but I'm just curious.) As I stated in one of the comments to the linked question,</p>
<blockquote>
<p>I can't see much point in using such devices, at least in the context of teaching.</p>
</blockquote>
<p>– but I'm probably mistaken. Please enlighten me, and at the same time give MESE users some neat ideas! :)</p>
| Robert Talbert | 942 | <p>First of all I would highly recommend <a href="http://amzn.com/B002UHTTYO">Derek Bruff's definitive book on this subject</a>. There are more good ideas in that book than any one faculty member can expect to implement. If there's anything I do with clickers in the classroom that works, it's probably appropriated from Derek in some way. </p>
<p>That said, here are three ways I use clickers on a regular basis: </p>
<ul>
<li><em>Warmup/entry quizzes</em>. Students are given 2-3 questions at the beginning of class to activate knowledge that they will need from the day, usually some combination of review from the previous class or some major point from the pre-class reading and exercises that they were supposed to finish. Since I usually teach with a flipped classroom format, I grade these (well, the clicker software grades them and I just download the data) for a light preparation grade. </li>
<li><em>Peer instruction</em>. This is a teaching style that Eric Mazur of Harvard University Physics pioneered. The class is organized around a series of small lectures that set up conceptual questions that are targeted at common or important misconceptions. Students use the clickers to vote on the questions, then if there's widespread disagreement on the answer (which there usually is if the question is good) the students are put into pairs and tasked to defend their choice to the other person for two minutes. Then students vote again on the same question after they've had a chance to discuss. Quite often, the second round of voting converges either on near-unanimity on the right answer or else it's 50/50. There's an enormous literature on the effectiveness of peer instruction if you want to Google it and learn more. </li>
<li><em>Formative assessment during a lecture</em>. If your instructional tastes are more aimed at traditional lecture, you can insert "concept check" questions at significant moments in the lecture to check for student understanding. That way you have real-time data -- collected without having the personal or social biases inherent in hand-raising and other methods of getting student feedback -- that show whether your lecture is hitting home or not. </li>
</ul>
<p>Again, though, Derek's book is awesome and encyclopedic, so check that out. </p>
|
1,672,847 | <p>A stick of total length $1$ is split at a randomly selected point $X$, i.e. $X$ is uniformly distributed in the interval $[0, 1]$.</p>
<p>Determine the expected length of the piece that contains the point $1/3$.</p>
<p>I've figured out so far that I need to determine a function $f(x)$ so that the length of the piece will be equal to $L=f(x)$, but I don't know how to work from here?</p>
| André Nicolas | 6,312 | <p>Given that the length $X$ is $\le 1/3$, the expectation of $X$ is $1/6$, and the expected length of the piece that contains $1/3$ is $5/6$.</p>
<p>Given that $X\gt 1/3$, the expectation of $X$ is $2/3$, so the expected length of the piece that contains $1/3$ is $1/3$.</p>
<p>Thus by the Law of Total Expectation, the expected length of the piece that contains $1/3$ is $(1/3)(5/6)+(2/3)(1/3)$.</p>
<p><em>Remark</em>: The result is $1/2$. I am feeling foolish, since the same argument gives $1/2$ if $1/3$ is replaced by any number $a$. The answer $1/2$ should be "obvious." Maybe tomorrow.</p>
|
2,440,802 | <p>The number of positive integers that $n$ can take in between the range $100$ to $200$.</p>
<p>I tried a lot using the prime factorization method but no use. </p>
| Piquito | 219,998 | <p>What I want is just to find out a solution. I feel the one I have found is the only but I have not stopped to prove this.</p>
<p>The identities $$(8t+2)^2-(8t+2)-2=8(8t^2+3t)\\(27s+24)^2+2(27s+24)-3=27(27s^2+50s+23)$$ establish two parametrizations, respectively for the diophantine equations $$n^2-n-2=8y\\n^2+2n-3=27z$$</p>
<p>We have$$100\lt8t+2\lt 200\iff 13\le t\le 24$$ which corresponds with
$$n\in\{106,114,122,130,138,146,154,162,170,178,\color{red}{186},194\}$$</p>
<p>We have also$$100\lt 27s+24\lt 200\iff 3\le s\le 6$$ which corresponds with
$$n\in\{105,132,159,\color{red}{186}\}$$
In fact one has
$$(186)^2-186-2=\color{red}{8}\cdot4301\\(186)^2+2(186)-3=\color{red}{27}\cdot1295$$</p>
|
2,136,411 | <p>According to the Theorem 12.7 of the book Analytic Nymber Theory by Apostol, $$\zeta(1-s) = 2(2\pi)^{-s} \Gamma(s) \cos \big(\frac{\pi s}{2}) \zeta(s)$$ which results in (as the book also says) that $\zeta(-2n) =0$ for $n=1,2,3, \dots$, the so-called trival zeros of $\zeta(s)$. </p>
<p>But how on earth $\zeta(-2n) = \sum_{i=1}^{\infty} \frac{1}{i^{-2n}}= \sum_{i=1}^{\infty} i^{2n}=\infty=0$? </p>
| Jan Eerland | 226,665 | <p>Notice, that the Riemann zeta function is only defined when the $\Re\left(\text{s}\right)>1$, so:</p>
<p>$$\zeta\left(\text{s}\right)=\sum_{\text{n}\in\mathbb{N}^+}\frac{1}{\text{n}^\text{s}}\space\space\space\to\space\space\space\zeta\left(-2\text{s}\right)\ne\sum_{\text{n}\in\mathbb{N}^+}\frac{1}{\text{n}^{-2\text{s}}}=\sum_{\text{n}\in\mathbb{N}^+}\text{n}^{2\text{s}}\tag1$$</p>
|
2,558,870 | <p>Suppose $f:[0,1]\to \mathbb{R}$ is uniformly continuous, and $(p_n)_{n\in\mathbb{N}}$ is a sequence of polynomial functions converging uniformly to $f$.</p>
<p>Does it follow that $\mathcal{F}=\{p_n\mid n\in\mathbb{N}\}\cup \{f\}$ is equicontinuous?</p>
<p>Also, if $C_n$ are the Lipschitz constants of the polynomials $p_n$, does it follow that $C_n<\infty$ for all $n$, and $\lim_{n\to\infty} C_n=\infty?$</p>
<p>I'm preparing for a test, but I'm not sure how to go about answering these two question. Any hints or tips as to what to look for would be appreciated. </p>
| user | 505,767 | <p>$$(A-pI)v_i=q_iv_i-pv_i\implies (A-pI)^{-1}(A-pI)v_i=(A-pI)^{-1}(q_i-p)v_i\implies (A-pI)^{-1}v_i=(q_i-p)^{-1}v_i \quad \square$$</p>
|
2,970,370 | <p>For <span class="math-container">$f \in C^0([0,1])$</span>, I have the following partial differential equation:</p>
<p><span class="math-container">$$u''(x) = f(x)$$</span> in <span class="math-container">$\Omega = (0,1)$</span>
<span class="math-container">$$u'(0) = u'(1) = 0$$</span></p>
<p>Why is this equation not well posed in <span class="math-container">$H^1(\Omega)$</span>?</p>
<p>--> Is it because of the missing boundary conditions? </p>
| Lutz Lehmann | 115,115 | <p>Set <span class="math-container">$v=u'$</span> then you are trying to solve the problem
<span class="math-container">$$
v'=f(x)\\
v(0)=v(1)=0
$$</span>
But <span class="math-container">$$v(1)-v(0)=\int_0^1v'(s)\,ds=\int_0^1f(s)\,ds$$</span>
so the equality of the boundary values is only possible if <span class="math-container">$f$</span> has mean zero.</p>
<p>In general you only get one boundary condition for a first-order ODE. As solutions, if any exist at all, are in <span class="math-container">$C^2$</span>,
<span class="math-container">$$
u(t)=u(0)+\int_0^t(t-s)f(s)\,ds,
$$</span>
the discussion of <span class="math-container">$H^1$</span> properties is a side-issue.</p>
|
4,381,145 | <blockquote>
<p>Show that the three vector fields <span class="math-container">$X = y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}, Y = z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}$</span> and <span class="math-container">$Z=x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$</span> on <span class="math-container">$\Bbb R^3$</span> are tangent to the <span class="math-container">$2$</span>-sphere <span class="math-container">$\Bbb S^2$</span>.</p>
</blockquote>
<p>I have a definition that a vector field is tangent to submanifold <span class="math-container">$S \subseteq M$</span> if for all <span class="math-container">$p \in S$</span> the tangent vector <span class="math-container">$X_p$</span> is in <span class="math-container">$T_pS \subseteq T_pM$</span>.</p>
<p>I don't really know how to approach the problem. Why is <span class="math-container">$X_p$</span> neccessarily a tangent vector? Using coordinate charts it seems that <span class="math-container">$X_p$</span> is of form <span class="math-container">$$X_p = \sum_{i=1}^n X^i(p) \frac{\partial}{\partial x^i} \bigg|_p$$</span> but what is this <span class="math-container">$X^i(p)$</span> here?</p>
<p>John Lee's book also suggests that <span class="math-container">$X$</span> is tangent to a submanifold <span class="math-container">$S$</span> if and only if <span class="math-container">$(Xf) \mid_S = 0$</span> for every <span class="math-container">$f \in C^\infty(M)$</span> such that <span class="math-container">$f\mid_S \equiv 0$</span>.</p>
<p>Can I use either one of these definitions here?</p>
| Clara B. | 572,257 | <p>By Proposition <span class="math-container">$5.38$</span> in John Lee's book "Smooth manifolds" (2. Edition), <span class="math-container">$T_p\mathbb S^2=\ker df_{|p}$</span>, where <span class="math-container">$f:\mathbb R^3\to\mathbb R$</span>, <span class="math-container">$(x,y,z)\mapsto x^2+y^2+z^2-1$</span>. Now <span class="math-container">$\frac 12\mathrm{grad}f(x,y,z)=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}$</span>. So for <span class="math-container">$p=(x,y,z)\in\mathbb S^2$</span>, a tangent vector <span class="math-container">$v=\alpha\frac{\partial}{\partial x}+\beta\frac{\partial}{\partial y}+\gamma\frac{\partial}{\partial z}\in T_p\mathbb R^3$</span> is tangent to <span class="math-container">$\mathbb S^2$</span> if and only if <span class="math-container">$\alpha x+\beta y+\gamma z=0$</span>.</p>
|
391,333 | <p>It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$. My question is very simple.</p>
<blockquote>
<p>There are $3$-tuples $(p, q, \alpha) \in
\mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
such that $\alpha\geq 2$ and $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$ </p>
</blockquote>
| Calvin Lin | 54,563 | <p>[The following claims should be obvious, and will not be proven.]</p>
<p>Claim: The degree of $ \sum_{i=1}^n i^r$ is $r+1$.</p>
<p>Claim: The leading coefficient of $ \sum_{i=1}^n i^r$ is $\frac{1}{r+1} $.</p>
<p>The first claim gives us $p+1 = \alpha (q+1) $. The second claim gives us $\frac{1}{p+1} = \left( \frac{ 1}{q+1} \right)^\alpha$.</p>
<p>Multiplying these two equations, we get $(q+1) = \alpha^\frac{1}{\alpha-1}$</p>
<p>Claim: $1 \leq \alpha^{\frac{1}{\alpha-1}} \leq 2$, with equality if and only if $\alpha = 1, 2$. </p>
<p>With this claim, the RHS is never an integer except for $\alpha = 1, 2 $.</p>
<p>If $\alpha = 1$, then $p+1 = q+1, p=q$.<br>
If $\alpha = 2$, this forces $q+1=2$ or $q=1$, and that $p+1 = 2(q+1) $ so $p=3$.</p>
<p>Finally, verify that $(p,p,1) $ and $(3,1,2)$ are solutions.</p>
|
2,669,292 | <p>$g_n(x) = \frac{\ln(1+x/n)}{n}$ on $\mathbb{R}$.
Don't they all converge to 0?</p>
| José Carlos Santos | 446,262 | <p>Please note that$$g_n(ne^n)=\frac{\ln(1+e^n)}n>\frac{\ln(e^n)}n=1.$$Therefore, your sequence cannot converge uniformly to $0$. This, in spite of your (correct) statement that for each <em>individual</em> $x$, $\lim_{n\to\infty}g_n(x)=0$. But uniform convergence is not about that; that's the realm of <em>pointwise</em> convvergence.</p>
|
3,995,492 | <p>I have no clue how to do this, I manage to get I get that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$</span> but I can't get anywhere from there.</p>
| sirous | 346,566 | <p>Use this relation:</p>
<p><span class="math-container">$6\times 11-5\times 13=1$</span></p>
<p>So we can write:</p>
<p><span class="math-container">$$11^{36}\equiv (6\times 11-5\times 13=1) \ mod (13) \equiv 6 \times 11\ mod (13)$$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$11^{36}\equiv 6\times 11 \ mod(13)$</span></p>
<p>Dividing both sides by 11 we get:</p>
<p><span class="math-container">$11^{35} \equiv 6 \ mod(13)$</span></p>
|
3,078,097 | <blockquote>
<p>Why it is impossible to split the natural numbers into two sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> such that for distinct elements <span class="math-container">$m, n \in A$</span> we have <span class="math-container">$m + n \in B$</span> and vice-versa?</p>
</blockquote>
<p>Also, does vice-versa means that there are distinct elements such that <span class="math-container">$x + y \in A$</span>? </p>
<p>How do I show the proof?</p>
| bof | 111,012 | <p>It is a famous fact of <a href="https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Example:_R(3,_3)_=_6" rel="nofollow noreferrer">Ramsey theory</a> that, if each edge of <span class="math-container">$K_6$</span> (a complete graph on <span class="math-container">$6$</span> vertices) is colored red or blue, there will be a triangle whose edges are all the same color.</p>
<p>Suppose the set <span class="math-container">$S=\{1,2,3,4,5,6,7,8,9,10\}$</span> is partitioned into two sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. I claim that there are distinct numbers <span class="math-container">$m,n$</span> in one of those two sets such that <span class="math-container">$m+n$</span> is in the same set.</p>
<p>Consider the complete graph with vertex set <span class="math-container">$V=\{0,1,3,4,9,10\}$</span>. For <span class="math-container">$x,y\in V$</span>, color the edge <span class="math-container">$xy$</span> red if <span class="math-container">$|x-y|\in A$</span>, blue if <span class="math-container">$|x-y|\in B$</span>. Then there are three vertices <span class="math-container">$x\lt y\lt z$</span> in <span class="math-container">$V$</span> such that the edges <span class="math-container">$xy$</span>, <span class="math-container">$yz$</span>, <span class="math-container">$xz$</span> all have the same color. Let <span class="math-container">$m=y-x$</span> and <span class="math-container">$n=z-y$</span>, so that <span class="math-container">$m+n=z-x$</span>. Then <span class="math-container">$m,n\in S$</span>, and <span class="math-container">$m\ne n$</span> since the set <span class="math-container">$V$</span> does not contain <span class="math-container">$3$</span> numbers in arithmetic progression. If <span class="math-container">$xy$</span>, <span class="math-container">$yz$</span>, <span class="math-container">$xz$</span> are all red, then <span class="math-container">$m,n,m+n\in A$</span>; if they are all blue, then <span class="math-container">$m,n,m+n\in B$</span>.</p>
|
366,249 | <p>$3xy^2dx+2x^3dy$
where is the boundary of the region between the circles $x^2+y^2=25$ and $x^2+y^2=64$ having positive orientation.</p>
<p>Not quite sure how to evaluate this...</p>
| doraemonpaul | 30,938 | <p>Let $y=\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}$ ,</p>
<p>Then $y'=\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}$</p>
<p>$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}$</p>
<p>$\therefore(x+1)^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}+(x+1)\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$</p>
<p>$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r}+\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$</p>
<p>$\sum\limits_{n=0}^\infty((n+r)^2-1)a_n(x+1)^{n+r}=0$</p>
<p>Since there are not any indical equations present, that means $r$ can be chosen as any complex number.</p>
<p>However, in fact, take $r=1$ or $r=-1$ will bring the above equation most simplified.</p>
<p>Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $r=-1$ :</p>
<p>$\sum\limits_{n=0}^\infty((n-1)^2-1)a_n(x+1)^{n-1}=0$</p>
<p>$\sum\limits_{n=0}^\infty n(n-2)a_n(x+1)^{n-1}=0$</p>
<p>$\therefore n(n-2)a_n=0$</p>
<p>$\therefore\begin{cases}a_0=a_0\\a_2=a_2\\a_n=0~\forall n\in\mathbb{N}\setminus\{2\}\end{cases}$</p>
<p>Hence $y=\dfrac{C_1}{x+1}+C_2(x+1)$</p>
|
134,987 | <blockquote>
<p>$$3x^2 + 2y^4 = z^4$$</p>
</blockquote>
<p><em>How do I solve this??</em> I would like to use so-called "elementary number theory", not abstract algebra (e.g. $\mathbb{Z} ( \sqrt d)$) or elliptic curves.</p>
<p>Note: I'm not asking <em>what</em> the solutions are, but rather <em>how</em> to find them.</p>
<p>My instincts are: </p>
<ul>
<li>search the internet (I compared this equation with the ~280 here on MSE, and tried a variety of similar searches on uniquation.com ...)</li>
<li>search the 3 number theory books that I have</li>
<li>try to find solutions "by inspection" (possibly after reducing the order of the variables)</li>
<li>do some magic with modular arithmetic </li>
<li>use <a href="http://www.alpertron.com.ar/QUAD.HTM" rel="nofollow">Alpern's solver</a> - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak) </li>
</ul>
<p>I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!</p>
<blockquote>
<p>What is the number-theoretic approach to such problems? Is there a general method?</p>
</blockquote>
| Will Jagy | 10,400 | <p>EDIT: it would appear that what you want to know a procedure. So, put all the degree four terms together on one side of the equals sign and re-write that as a quadratic form in substitute variables, as below.</p>
<p>The reason you use 3 is this: write $$ 3 x^2 = u^2 - 2 v^2, $$ where we will eventually put back $u = z^2, v = y^2.$</p>
<p>Now see my answer at <a href="https://math.stackexchange.com/questions/134937/solving-a-diophantine-equation/134957#134957">Solving a Diophantine Equation</a> </p>
<p>The discriminant of the binary quadratic form $u^2 - 2 v^2$ is $8.$ As $3$ is odd we can calculate $(8|3) = -1.$ From
$$ u^2 - 2 v^2 \equiv 0 \pmod 3 $$ we find that
$$ u,v \equiv 0 \pmod 3. $$ </p>
<p>That is the quadratic forms part. When you finally put back $u = z^2, v = y^2$ you find that $$ 3 x^2 \equiv 0 \pmod {81}, \; \; x^2 \equiv 0 \pmod {27}, \; \; x \equiv 0 \pmod 9.$$ And so on.</p>
<p><strong><em>EXTRA CREDIT:</em></strong> Try $$ 5 x^2 = 2 y^4 + y^2 z^2 + 3 z^4. $$ First, what is the discriminant of $2 u^2 + u v + 3 v^2?$</p>
|
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)y$$
$$\implies\color{blue}{(D+1)(D+4)y=0}\tag{2}$$
$$\implies (D+1)y=0 \space\space\text{or}\space\space (D+4)y=0$$ which has solutions $$y=Ae^{-x}\space\space\text{or}\space\space y=Be^{-4x}\tag{3}$$ respectively, where $A$ and $B$ are both constants. </p>
<p>Now if $(D+4)y=0$, then $$(D+1)(D+4)y=(D+1)\cdot 0=0$$
so any solution of $(D + 4)y = 0$ is a solution of the differential equation $(1)$ or $(2)$. Similarly, any solution of $(D + 1)y = 0$ is a solution of $(1)$ or $(2)$. $\color{red}{\text{Since the two solutions (3) are linearly independent, a linear combination}}$ $\color{red}{\text{of them contains two arbitrary constants and so is the general solution.}}$ Thus $$y=Ae^{-x}+Be^{-4x}$$ is the general solution of $(1)$ or $(2)$.</p>
</blockquote>
<p>The part I don't understand in this extract is marked in $\color{red}{\mathrm{red}}$.</p>
<ol>
<li>Firstly; <em>How</em> do we know that the two solutions: $y=Ae^{-x}\space\text{and}\space y=Be^{-4x}$ are linearly independent?</li>
<li>Secondly; <em>Why</em> does a linear combination of linearly independent solutions give the general solution. Or, put in another way, I know that $y=Ae^{-x}\space\text{or}\space y=Be^{-4x}$ are both solutions. But <em>why</em> is their <strong>sum</strong> a solution: $y=Ae^{-x}+Be^{-4x}$? </li>
</ol>
| Justpassingby | 293,332 | <ol>
<li><p>Assume there are constants $A$ and $B$ such that the function $A\exp(-x)+B\exp(-4x)$, is identically zero. The constants $A$ and $B$ are (over)determined by filling in (for example) $x=\ln1=0,\ln2,\ln3$ so that the only possibility becomes $A=B=0.$</p></li>
<li><p>The solutions of a homogeneous equation or system of equations form a vector space. This can be verified directly with abstract solutions in the original equation, or it could even be a <em>definition</em> of homogeneity. What remains to be shown is that, in the case of a second-order ODE, the dimension of the solution space is 2. This can be concluded from the decomposed form (2) of the differential operator: there are only 2 places where you can supply arbitrary constants.</p></li>
</ol>
<p>More explicitly, if $y$ is a solution then $(D+1)(D+4)y=0$ so $(D+4)y$ is determined up to a constant. But for any specific $C$ the equation $(D+4)y$ determines $y$ up to another constant.</p>
|
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)y$$
$$\implies\color{blue}{(D+1)(D+4)y=0}\tag{2}$$
$$\implies (D+1)y=0 \space\space\text{or}\space\space (D+4)y=0$$ which has solutions $$y=Ae^{-x}\space\space\text{or}\space\space y=Be^{-4x}\tag{3}$$ respectively, where $A$ and $B$ are both constants. </p>
<p>Now if $(D+4)y=0$, then $$(D+1)(D+4)y=(D+1)\cdot 0=0$$
so any solution of $(D + 4)y = 0$ is a solution of the differential equation $(1)$ or $(2)$. Similarly, any solution of $(D + 1)y = 0$ is a solution of $(1)$ or $(2)$. $\color{red}{\text{Since the two solutions (3) are linearly independent, a linear combination}}$ $\color{red}{\text{of them contains two arbitrary constants and so is the general solution.}}$ Thus $$y=Ae^{-x}+Be^{-4x}$$ is the general solution of $(1)$ or $(2)$.</p>
</blockquote>
<p>The part I don't understand in this extract is marked in $\color{red}{\mathrm{red}}$.</p>
<ol>
<li>Firstly; <em>How</em> do we know that the two solutions: $y=Ae^{-x}\space\text{and}\space y=Be^{-4x}$ are linearly independent?</li>
<li>Secondly; <em>Why</em> does a linear combination of linearly independent solutions give the general solution. Or, put in another way, I know that $y=Ae^{-x}\space\text{or}\space y=Be^{-4x}$ are both solutions. But <em>why</em> is their <strong>sum</strong> a solution: $y=Ae^{-x}+Be^{-4x}$? </li>
</ol>
| BLAZE | 144,533 | <p>Given the general form of a second order ODE is
$$a\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0\tag{1}$$</p>
<p>where $a,b,c$ are known real coefficients. The most important property of $(1)$ is <em>linearity</em>. This means that if we have two solutions $y_1$ and $y_2$, then any linear combination of $y_1$ and $y_2$ is also a solution of $(1)$ i.e.
$$y(x)=Ay_1(x) + By_2(x)$$ is a solution of $(1)$ for any choice of constants $A$ and $B$. This is a direct consequence of the linearity of derivatives, which allows us to write
$$\begin{align}\\ & a\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy = a\frac{\mathrm{d}^2 }{\mathrm{d}x^2}\left(Ay_1 + By_2\right)+b\frac{\mathrm{d} }{\mathrm{d}x}\left(Ay_1 + By_2\right)+c\left(Ay_1 + By_2\right) \\ & = A\left(a\frac{\mathrm{d}^2 y_1}{\mathrm{d}x^2}+b\frac{\mathrm{d}y_1}{\mathrm{d}x}+cy_1\right)+B\left(a\frac{\mathrm{d}^2 y_2}{\mathrm{d}x^2}+b\frac{\mathrm{d}y_2}{\mathrm{d}x}+cy_2\right) \\ & =0 \end{align}$$</p>
<p>So the coefficients of $A$ and $B$ are zero.</p>
<blockquote>
<p>Linearity is satisfied if for some function $f$:<br>
$$f(x+y)=f(x)+f(y)$$ and $$f(ax)=af(x)$$ where $a$ is a constant.</p>
</blockquote>
|
3,167,261 | <blockquote>
<p>Let <span class="math-container">$\mathcal{O}$</span> be an open subset of the plane <span class="math-container">$\mathbb{R}^{2}$</span> and
let the mapping <span class="math-container">$F : \mathcal{O} \rightarrow \mathbb{R}^{2}$</span> be
represented by <span class="math-container">$F(x, y) = (u(x, y), v(x, y))$</span> for <span class="math-container">$(x, y)$</span> in
<span class="math-container">$\mathcal{O}$</span>. Then, we say the mapping <span class="math-container">$F : \mathcal{O} \rightarrow
\mathbb{R}^{2}$</span> is called a <em>Cauchy-Riemann mapping</em> provided that
each of the functions <span class="math-container">$u : \mathcal{O} \rightarrow \mathbb{R}$</span> and <span class="math-container">$v
: \mathcal{O} \rightarrow \mathbb{R}$</span> has continuous second-order
partial derivatives and <span class="math-container">$$\frac{\partial u}{\partial x}(x, y) =
\frac{\partial v}{\partial y}(x, y) \hspace{1em} \text{ and }
\hspace{1em} \frac{\partial u}{\partial y} = -\frac{\partial
v}{\partial x}(x, y)$$</span> </p>
<p>for all <span class="math-container">$(x, y)$</span> in <span class="math-container">$\mathcal{O}$</span>.</p>
</blockquote>
<p>Is it necessarily true that <span class="math-container">$u(x, y)$</span> and <span class="math-container">$v(x, y)$</span> are harmonic?</p>
| avs | 353,141 | <p>The real and imaginary parts of a holomorphic function are harmonic. Source: <a href="https://books.google.com/books/about/Elementary_Theory_of_Analytic_Functions.html?id=KsGbqTBjyoUC" rel="nofollow noreferrer">Cartan's <em>Elementary Theory of Analytic Functions</em></a>.</p>
|
1,177,782 | <p>I tried to prove the following theorem and was wondering if someone could please tell me if my proof can be fixed somehow...</p>
<p>Theorem: Let $H$ be a Hilbert space and $x_n\in H$ a bounded sequence. Then $x_n$ has a weakly convergent subsequence.</p>
<p>My idea for a proof:</p>
<p>The map $\phi: H \to H^\ast$ in the Riesz representation theorem is an isometry therefore $\varphi_n := \phi(x_n)$ is also bounded and therefore $\varphi_1(x_n)$ is a bounded sequencein $\mathbb R$. By Bolzano Weierstras it ha a convergent subsequence $\varphi_1(x_{n_{k_1}})$. (Say, $\varphi_1(x_{n_{k_1}})\to \varphi_1(x)$ for some $x$) Let $x_{n_1}$ be the argument of the first element in this sequence (apologies for the notation; the first element is also called $x_{n_1}$...). </p>
<p>The sequence $\varphi_2(x_{n_1})$ has a convergent subsequence $x_{n_{k_2}}$. Let $x_{n_2}$ be the first element in that sequence.</p>
<p>And so on. Then the resulting sequence $x_{n_k}$ has the property that for all $j$:</p>
<p>$$ \varphi_j(x_{n_k}) \to \varphi_j(x)$$</p>
<p>My only problem is that I only showed this limit for $\varphi_n$ that is, not for all $\varphi \in H^\ast$.</p>
<blockquote>
<p>Can this argument be fixed somehow?</p>
</blockquote>
| Vincent Boelens | 94,696 | <p>I think this can be done without invoking Banach-Alaoglu or the Axiom of Choice. I will sketch the proof. By the Riesz representation theorem (which as far as I can tell can be proven without Choice), a Hilbert space is reflexive. Furthermore, it is separable iff its dual is.</p>
<p>To show the weak convergence of the bounded sequence <span class="math-container">$(x_n)$</span> assume first that <span class="math-container">$H$</span> is separable and let <span class="math-container">$\{x'_1,x'_2,\ldots\}$</span> be a dense set in the dual space. Use a diagonal argument to extract a subsequence <span class="math-container">$(x_{n_k})$</span> such that <span class="math-container">$x'_m(x_{n_k})$</span> converges for all <span class="math-container">$m$</span>. If <span class="math-container">$x'$</span> is any functional and for <span class="math-container">$\epsilon>0$</span>, there is <span class="math-container">$x'_m$</span> such that <span class="math-container">$\|x'-x'_m\|<\epsilon$</span>. Then,
<span class="math-container">\begin{align}\|x'(x_{n_k})-x'(x_{n_l})\|&\le \|x'(x_{n_k})-x'_m(x_{n_k})\|+\|x_m'(x_{n_k})-x'_m(x_{n_l})\|\\&+\|x'_m(x_{n_l})-x'(x_{n_l})\|<(2M+1)\epsilon,\end{align}</span>
if <span class="math-container">$k$</span> and <span class="math-container">$l$</span> are large enough (define <span class="math-container">$M=\sup_n \|x_n\|$</span>). Hence, <span class="math-container">$(x'(x_{n_k}))$</span>
is a Cauchy sequence. It remains to be shown that the weak limit exists. Consider the linear map <span class="math-container">$\ell(x'):= \lim_k x'(x_{n_k})$</span>. This is well-defined by the previous argument and bounded, since <span class="math-container">$\ell(x')\le \|x'\|M$</span>. By reflexivity of <span class="math-container">$H$</span>, there is <span class="math-container">$x\in H$</span> such that <span class="math-container">$\lim_kx'(x_{n_k})=\ell(x')=x'(x)$</span>, which means exactly that <span class="math-container">$x$</span> is the weak limit of <span class="math-container">$(x_{n_k})$</span>.</p>
<p>For the general case, let <span class="math-container">$Y$</span> be the closed linear span of <span class="math-container">$\{x_1,x_2,\ldots\}$</span>. This is then a separable Hilbert space and by the previous argument, there is a subsequence <span class="math-container">$(x_{n_k})$</span> and <span class="math-container">$y\in Y$</span> such that <span class="math-container">$(y'(x_{n_k}))$</span> converges to <span class="math-container">$y'(y)$</span> for all <span class="math-container">$y'\in Y'$</span>.
It remains to be shown that <span class="math-container">$(x'(x_{n_k}))$</span> converges to <span class="math-container">$x'(y)$</span> for all <span class="math-container">$x'\in H'$</span>. But this is obvious, since the restriction of <span class="math-container">$x'$</span> to <span class="math-container">$Y$</span> is a functional on <span class="math-container">$Y$</span>.</p>
|
2,669,278 | <p>I noticed a strange thing with my calculator.<br>
When I start with any number like 1,2,3 or 1.2, 1.34 .... or even 0.<br>
And repeatedly take the cosine function of this number.<br>
I get the same following number. I don't thing this is a coincidence since it's happening with any number I try. </p>
<pre>0.99984774153108811295981076866798</pre>
<p>It's pretty astonishing the accuracy this number has. I wouldn't have asked this question if only only few 4 or 5 decimals of every number matched but it's it's 32 decimal places I get for every number I try.<br>
You got to try it yourself to believe it.<br>
I want to know if there's a reason behind this? And why don't other functions like sine or tangent show similar properties?
Note that the calculator is set to degrees.</p>
| Brian Tung | 224,454 | <p>Your calculator must be operating in degrees. Since $0.9998\ldots$ degrees is very close to $0$ (being less than $1/90$ of the way from $0$ radians to $\pi/2$ radians), its cosine must be very close to $1$. What you are finding is the fixed point of the function $\cos \theta$, where $\theta$ is expressed in degrees—that is, the number of degrees $\theta$ where</p>
<p>$$
\theta = \cos \theta
$$</p>
<hr>
<p>Here's a graphical depiction of $\theta$ and $\cos \theta$ (with $\theta$ expressed in degrees). The fixed point is the intersection of these two curves:</p>
<p><a href="https://i.stack.imgur.com/LDhzZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LDhzZ.png" alt="enter image description here"></a></p>
<p>Since it's difficult to see this intersection at the above scale, here it is zoomed in, and you'll see that the intersection occurs very close to $(1, 1)$; in fact, it is $(0.9998\ldots, 0.9998\ldots)$, as you discovered:</p>
<p><a href="https://i.stack.imgur.com/itJ3z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/itJ3z.png" alt="enter image description here"></a></p>
|
33,543 | <p>Let $M$ be a filtered module over a filtered algebra $A$, and suppose $gr(M)$ is flat over $gr(A)$, where $gr$ means the associated graded module and algebra, respectively.</p>
<p>What can one say in general about the flatness of $M$ over $A$, or with relevant assumptions (for instance in the above, we should assume both filtrations are complete to avoid dumb counterexamples)? Are there good references for this sort of question? I have played with the various definitions of flatness trying to find an obvious relationship, but I find flatness proofs confusing.</p>
<p>The particular examples I have in mind are comparing $U(\mathfrak{g})$-modules to $S(\mathfrak{g})$-modules, and $D(X)$-modules to $O(T^*X)$-modules for affine varieties $X$, if it helps. I suspect the answer doesn't depend on any of the details though.</p>
| Mariano Suárez-Álvarez | 1,409 | <p>Let me suppose, as in your examples, that we have a base field $k$.</p>
<p>It is well known that to check that a right $A$-module $M$ is flat it is enough to show that whenever $I\leq_\ell A$ is a left ideal, the map $M\otimes_AI\to M\otimes_A A$ induced by the inclusion $I\to A$ is injective. This condition can be rewritten: $M$ is flat iff for each left ideal $I\leq_\ell A$ we have $\mathrm{Tor}^A_1(M,A/I)=0$.</p>
<p>So now suppose $A$ and $M$ are (exhaustively, separatedly, increasingly from zero) filtered in such a way that $\mathrm{gr}M$ is a flat $\mathrm{gr}A$-module. </p>
<p>Pick a left ideal $I\leq_\ell A$; notice that the filtration on $A$ induces a filtration on the quotient $A/I$. We can compute $\mathrm{Tor}^A_\bullet(M,A/I)$ as the homology of the homologically graded complex $$\cdots\to M\otimes_kA^{\otimes_kp}\otimes_kA/I\to M\otimes_kA^{\otimes_k(p-1)}\otimes_kA/I\to\cdots$$ with certain differentials whose formula does not fit in this margin, coming from the bar resolution. Now the filtrations on $M$, on $A$ and on $A/I$ all collaborate to provide a filtration of our complex. We've gotten ourselves a positively homologicaly graded with a canonically bounded below, increasing, exhaustive and separated filtration. The corresponding spectral sequence then converges, and its limit is $\mathrm{Tor}^A_\bullet(M,A/I)$. Its $E^0$ term is the complex
$$\cdots\to\mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_kp}\otimes_k\mathrm{gr}(A/I)\to \mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_k(p-1)}\otimes_k\mathrm{gr}(A/I)\to\cdots$$
with, again, the bar differential, and its homology, which is the $E^1$ page of the spectral sequence, is then precisely $\mathrm{Tor}^{\mathrm{gr}A}_\bullet(\mathrm{gr}M,\mathrm{gr}(A/I))$. Since we are assuming that $\mathrm{gr}M$ is $\mathrm{gr}A$-flat, this last $\mathrm{Tor}$ vanishes in positive degrees, so the limit of the spectral sequence also vanishes in positive degrees. In particular, $\mathrm{Tor}^A_1(M,A/I)=0$.</p>
<p><strong>NB:</strong> As Victor observed above in a comment, Bjork's <em>Rings of differential operators</em> proves in its Proposition 3.12 that $\mathrm{w.dim}_AM\leq\mathrm{w.dim}_{\mathrm{gr}A}\mathrm{gr}M$ (here $\mathrm{w.dim}$ is the <em>flat dimension</em>) from which it follows at once that $M$ is flat as soon as $\mathrm{gr}M$ is; the argument given is essentialy the same one as mine. I am very suprised about not having found this result in McConnell and Robson's!</p>
|
320,348 | <p>I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$</p>
<p>I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively things like $${2n+2\choose n+1}={2n+2\over n+1}{2n+1\choose n}=2\cdot {2n+1\over n+1}{2n\choose n},$$</p>
<p>But I don't think it can lead me anywhere. I would like the proof to be as simple as possible.</p>
| Ishan Banerjee | 52,488 | <p>Split the <span class="math-container">$2n$</span> elements into two groups of size <span class="math-container">$n$</span>
Then the no. of ways of choosing <span class="math-container">$n$</span> from the <span class="math-container">$2n$</span> is the no. of ways of choosing <span class="math-container">$i$</span> from the <strong>1st</strong> and <span class="math-container">$n-i$</span> from the <strong>2nd</strong> and letting <span class="math-container">$i$</span> vary.</p>
|
114,122 | <p>I am trying to figure out the maximum possible combinations of a (HEX) string, with the following rules:</p>
<ul>
<li>All characters in uppercase hex (ABCDEF0123456789)</li>
<li>The output string must be exactly 10 characters long</li>
<li>The string must contain at least 1 letter</li>
<li>The string must contain at least 1 number</li>
<li>A number or letter can not be represented more than 2 times</li>
</ul>
<p>I am thinking the easy way to go here (I am most likely wrong, so feel free to correct me):</p>
<ol>
<li>Total possible combinations: $16^{10} = 1,099,511,627,776$</li>
<li>Minus all combinations with just numbers: $10^{10} = 10,000,000,000$</li>
<li>Minus all combinations with just letters: $6^{10} = 60,466,176$</li>
<li>etc...</li>
</ol>
<p>Can someone could tell me if this is the right way to go and if so, how to get the total amount of possible combinations where a letter or a number occur more than twice.</p>
<p>Any input or help would be highly appreciated!</p>
<p>Muchos thanks!</p>
<p>PS.</p>
<p>I don't know if I tagged this question right, sorry :(</p>
<p>DS.</p>
| hmakholm left over Monica | 14,366 | <p>I think the sanest way to split this is by how many <em>different</em> letters and numbers is contained in the string. Call the <em>total</em> number of different symbols $k$. The possibilities are then:</p>
<pre><code>k=5 1+4 2+3 3+2 4+1
k=6 1+5 2+4 3+3 4+2 5+1
k=7 1+6 2+5 3+4 4+3 5+2 6+1
k=8 1+7 2+6 3+5 4+4 5+3 6+2
k=9 1+8 2+7 3+6 4+5 5+4 6+3
k=10 1+9 2+8 3+7 4+6 5+5 6+4
</code></pre>
<p>For each of the 33 entries in the table, the number of ways to select which hex digit appear at all is a simple product of binomial coefficients.</p>
<p>From there each <em>row</em> of the table can be summed and treated uniformly. First select which $10-k$ of the $k$ symbols are going to appear twice, for a factor of $\binom{k}{10-k}$. Then multiply by $10!$ for the ways to arrange the symbols you have selected, and divide by $2^{10-k}$ to account for the fact that each pair of two equal symbols cannot be distinguished.</p>
|
3,045,677 | <p>I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.</p>
<p>Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?</p>
<p>Thank you.</p>
| user2661923 | 464,411 | <p>Given sides a,b,and c, use the law of cosines to calculate the angle <span class="math-container">$\theta$</span> between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle <span class="math-container">$\theta.$</span> </p>
|
217,436 | <p>I don't know where to start... It's a multiple-choice question: I can choose from $\sqrt{2}, 0, 2, 1$</p>
<p>Thank you!</p>
| lab bhattacharjee | 33,337 | <p>Using <a href="http://en.wikipedia.org/wiki/Euler%27s_formula" rel="nofollow">this</a> or <a href="http://mathworld.wolfram.com/EulerFormula.html" rel="nofollow">this</a>, $$e^{\frac{i\pi}4}=\cos \frac{\pi}4 +i\sin\frac{\pi}4=\frac{1+i}{\sqrt 2}$$</p>
<p>$$(1-i)\cdot e^{\frac{i\pi}4}=(1-i)\cdot \frac{(1+i)}{\sqrt 2}=\frac{1-i^2}{\sqrt 2}=\sqrt 2$$</p>
|
217,436 | <p>I don't know where to start... It's a multiple-choice question: I can choose from $\sqrt{2}, 0, 2, 1$</p>
<p>Thank you!</p>
| dot dot | 21,681 | <p>$$e^{i\pi/4}(1-i)=(1-i)(1+i)\frac{\sqrt{2}}{2}=2\frac{\sqrt{2}}{2}=\sqrt{2}$$</p>
|
2,206,247 | <p><strong>Question:</strong> Consider the following non linear recurrence relation defined for $n \in \mathbb{N}$:</p>
<p>$$a_1=1, \ \ \ a_{n}=na_0+(n-1)a_1+(n-2)a_2+\cdots+2a_{n-2}+a_{n-1}$$</p>
<p>a) Calculate $a_1,a_2,a_3,a_4.$</p>
<p>b) Use induction to prove for all positive integers that:</p>
<p>$$a_n=\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{3+\sqrt{5}}{2}\right)^n-\left(\dfrac{3-\sqrt{5}}{2}\right)^n\right]$$
Hi all! I'm having trouble solving this problem. I have no problem with part (a), but I'm having lots of troubles with part (b). I proved the base case (which is quite trivial), but I'm having trouble for the inductive step (proving k->k+1).</p>
<p><a href="https://i.stack.imgur.com/JgGfv.png" rel="nofollow noreferrer">Attempt</a></p>
<p>I don't know what to do from this point. Thank you! </p>
| Χpẘ | 309,642 | <p>Yes it is related to triangle inequality. If the two sides of right triangle have lengths of $\sqrt{a}$ and $\sqrt{b}$, then hypotenuse has length $\sqrt{a+b}$. Therefore $\sqrt{a}+\sqrt{b} > \sqrt{a+b}$ </p>
|
1,238,210 | <p>How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?</p>
<p>P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D</p>
| Alessio Bocci | 386,931 | <p>It is vary simple:
$$y(x)=\int_0^xe^{t^2}dt=\frac{1}{2} \sqrt{\pi } \text{erfi}(x)$$
So:
$$\lim_{x \rightarrow+\infty}y(x)=+\infty$$</p>
|
1,934,033 | <p>I'm new here. I wish to ask a question regarding predicate logic:</p>
<p>I was given three predicates:</p>
<p><strong>parent(p,q): p is the parent of q.</strong></p>
<p><strong>female(p): p is a female.</strong></p>
<p><strong>p = q: p and q are the same person.</strong></p>
<p>Now, I was tasked with translating this sentence: Alice has a daughter.</p>
<p>My answer was: <strong>There exists a q such that parent(Alice,female(q)).</strong></p>
<p>The answer given is: <strong>There exists a q such that female(q) AND parent(Alice,q).</strong></p>
<p>Is it correct to have a predicate (in this case, female) within another predicate (in this case, parent)?</p>
<p>Much appreciated.</p>
| Anonymous Coward | 251,953 | <p>No, it is not in this particular case. You conclude that by substitution.</p>
<p>female(q) can have two values : true or false.</p>
<ul>
<li>For female q, female(q)=true you have : parent(Alice,true).</li>
<li>For non-female q, female(q)=false you have : parent(Alice,false).</li>
</ul>
<p>Since false and true are not humans (they are truth values) the proposition <strong>parent(Alice,female(q))</strong> is false for all values of <strong>q</strong>. </p>
<p>So there does not exist a <strong>q</strong> such that <strong>parent(Alice,female(q))</strong>. </p>
<p>Your predicate within a predicate is a well-formed formula. True and false can be used as terms of a predicate. But the formula does not express the required concept.</p>
|
64,643 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number">$a^{1/2}$ is either an integer or an irrational number</a> </p>
</blockquote>
<p>I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?</p>
| Ross Millikan | 1,827 | <p>If you follow through the usual proof for $\sqrt{2}$ substituting $3$ for $2$, it goes through just fine. Let $\sqrt{3}=\frac{p}{q}, p,q $ relatively prime. $3=\frac{p^2}{q^2}$, so $3$ divides $p$ and so on.</p>
|
12,057 | <p>Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?</p>
| Bill Dubuque | 242 | <p>Simply repeatedly apply <span class="math-container">$\rm\color{#C00}E = $</span> Euclid's lemma:</p>
<p><span class="math-container">$\begin{align} 1 &= (pb+qa,qb)\\
\overset{\rm\color{#C00}E }\iff\ \ \ \ \ \ \ 1 &= (pb+qa,q) = (pb,q) \overset{\rm\color{#C00}E }= (b,q)\quad\ {\rm via}\quad\ (p,q) = 1\\
{\rm and}\ \ 1 &= (pb+qa,b) = (qa,b) \overset{\rm\color{#C00}E }= (q,b)\quad\ {\rm via}\quad\ (a,b) = 1\end{align}$</span></p>
<p>Therefore <span class="math-container">$\ \ 1 = (pb+qa,qb) \iff 1 = (q,b)$</span></p>
|
4,249,573 | <p>I'm studying Set Theory in my own, with Goldrei's textbook. The chapter I'm reading is on order-isomorphism and well-ordering. One exercise asks (i) to argue that, in general, a collection of well-ordered sets order-isomorphic to a given well-ordered set is a proper class (rather than a set). The proof, IMHO, is fairly straightforward and applicable to any isomorphism class. Then the same exercises asks (ii) if there is a well-ordered set, X say, s.t. the collection of all well-ordered set, order-isomorphic to X, is a set (rather than a proper class). I stumbled on that one.</p>
| Lazy | 958,820 | <p>Let’s assume <span class="math-container">$X\neq\emptyset$</span> (else the property is trivially true). First note that you are looking at classes of order-isomorphic ordered sets, so pairs <span class="math-container">$(A,\leq)$</span> so that all for all such pairs exists a bijection that leaves the order intact. Let’s call this class <span class="math-container">$Z$</span>. On the other hand if there is a bijection between two sets <span class="math-container">$A,B$</span> then this will transform any ordering on <span class="math-container">$A$</span> into an ordering on <span class="math-container">$B$</span>.</p>
<p>Thus by projecting on the first coordinate we get <span class="math-container">$|X|$</span> where <span class="math-container">$|X|$</span> denotes the class of cardinality.</p>
<p>So if <span class="math-container">$Z$</span> is a set, then also <span class="math-container">$|X|$</span> is a set.</p>
<p>Now, can this be? Let’s differentiate two cases: If <span class="math-container">$|X|=n>0\in\mathbb N$</span> then <span class="math-container">$B:=\{0,1,\ldots,n-2,|X|\}$</span> would be a set of that cardinality. On the other hand, if <span class="math-container">$|X|$</span> is infinite then <span class="math-container">$B=X\cup\{|A|\}$</span> has the same cardinality as <span class="math-container">$X$</span>.</p>
<p>So in both cases we’d find some <span class="math-container">$B\in|X|$</span> so that <span class="math-container">$|X|\in B$</span>. By ZF this is not possible.</p>
<p>Thus this is not in fact a set.</p>
<p>EDIT: On the ZF part. Set <span class="math-container">$C=\{B,|X|\}$</span>. Then <span class="math-container">$C,B$</span> are not disjoint, for both contain <span class="math-container">$|X|$</span>, and <span class="math-container">$C,|X|$</span> are not disjoint, for both contain <span class="math-container">$B$</span>. Thus this violates the axiom of regularity.</p>
|
28,877 | <p>Since I self-study mathematical analysis without <em>formal</em> teacher, I can only appeal to help from out site most of the time. It's obvious that to grasp the underlying concepts in mathematics, we must roll the sleeves and solve problems.</p>
<p>It's clear that there are actually mistakes and misunderstanding that are too subtle for me to recognize, so it's very natural for me to ask for proof verifications.</p>
<p>Even though I tried to write my proofs as detailed and clear as possible, they seem to attract little attention from other users. It seems to me that proof checking is a boring and tedious job, but it is essential for me (and possibly for all of us) to know whether and where I get wrong.</p>
<p>How can I make my post for proof verification more attractive and consequently attract more attention?</p>
<p>Below are questions that i have not received any answer. Most of them are related to Cantor-Bernstein-Schröder theorem. It would be great if someone can help me improve them so that they get an answer. Thank you so much!</p>
<p><a href="https://math.stackexchange.com/questions/2813526/is-this-a-mistake-in-the-proof-of-halls-marriage-theorem-from-https-proofwiki">Is this a mistake in the proof of Hall's Marriage Theorem from https://proofwiki.org?</a></p>
<p><a href="https://math.stackexchange.com/questions/2748266/top-down-and-bottom-up-proofs-of-a-lemma-used-to-prove-cantor-bernstein-schr%C3%B6der">Top-down and Bottom-up proofs of a lemma used to prove Cantor-Bernstein-Schröder theorem and their connection</a> (this is the question that i would like to receive answer most)</p>
<p><a href="https://math.stackexchange.com/questions/2759971/is-my-proof-of-cantor-bernstein-schr%C3%B6der-theorem-correct">Is my proof of Cantor-Bernstein-Schröder theorem correct?</a></p>
<p><a href="https://math.stackexchange.com/questions/2751389/bottom-up-proof-of-a-lemma-used-to-prove-bernstein-schr%C3%B6der-theorem">Bottom-up proof of a lemma used to prove Bernstein-Schröder theorem</a></p>
<p><a href="https://math.stackexchange.com/questions/2749527/julius-k%C3%B6nigs-proof-of-schr%C3%B6der-bernstein-theorem">Julius König's proof of Schröder–Bernstein theorem</a></p>
| user21820 | 21,820 | <p>Not yet mentioned...</p>
<ol>
<li><p>Learn at least one formal deductive system for first-order logic (you can see my profile for <a href="http://math.stackexchange.com/a/1684204/21820">one variant of Fitch-style natural deduction</a> and <a href="http://math.stackexchange.com/a/1788516/21820">some basic examples</a>. All mathematics can be carried out in a suitably user-friendly such system, and you can mechanically check your proof for correctness. Mistakes are still possible but only due to carelessness in checking or skipping steps in the proof. I have myself caught quite a few mistakes when writing in such formal style.</p></li>
<li><p>Try asking for help/feedback in a chat-room. The most active is of course the main chat-room <a href="https://chat.stackexchange.com/rooms/36/mathematics">Mathematics</a>. But for logic-related stuff you can also come to <a href="https://chat.stackexchange.com/rooms/info/44058/logic">the Logic chat-room</a>. Of course, it still depends on whether people are interested and have the time to look at your stuff. =)</p></li>
</ol>
|
966,278 | <p>Given a recursive relation
$$a_n = \begin{cases}
(1 - 2b_n)a_{n-1} + b_n, & n > 1 \\
\frac{1}{2}, & n =1
\end{cases}
$$, how can I expression $a_n$ in term of $b_i, i \in \{1, 2, \dots n\}$?</p>
| Peter Huxford | 152,620 | <p>Suppose that $a_n=\frac{1}{2}$ for some $n$. Then according to the recurrence:</p>
<p>\begin{align*}
a_{n+1}&=(1-2b_{n+1})a_n+b_{n+1} \\
&= \frac{1}{2}-b_{n+1}+b_{n+1} \\
&= \frac{1}{2}
\end{align*}</p>
<p>Since $a_1=\frac{1}{2}$, by induction this shows that $a_n=\frac{1}{2}$ for all $n$.</p>
|
3,425,369 | <p>I'm taking a linear algebra course and I'm having trouble proving linear (in)dependence of functions. I understand that I have to prove that the <span class="math-container">$a_1f(x) + a_2g(x) = 0$</span> but I don't know how to actually do that. For example given a pair of functions 1 and t, how do you prove linear independence?</p>
| Matthew Leingang | 2,785 | <p>With polynomials, or functions in general, you can evaluate them at certain points and explore the consequences.</p>
<p>For instance, suppose that there exist scalars <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span> such that <span class="math-container">$a_1 \cdot 1 + a_2 \cdot t = 0$</span> as functions. This means the equation holds for all <span class="math-container">$t$</span>. Evaluating at <span class="math-container">$t=0$</span> gives <span class="math-container">$a_1 = 0$</span>. Evaluating at <span class="math-container">$t=1$</span> gives <span class="math-container">$a_1 + a_2 =0$</span>, but since <span class="math-container">$a_1 =0$</span> is already known, <span class="math-container">$a_2 = 0$</span> as well. Therefore <span class="math-container">$1$</span> and <span class="math-container">$t$</span> are linearly independent.</p>
|
1,285,774 | <p>I have looked at similar questions under 'Questions that may already have your answer" and unless I have missed it, I cannot find a similar question.</p>
<p>I am trying to answer the following:</p>
<p>Let $A = \left(\begin{matrix}
a & b \\
b & d \\
\end{matrix}\right)$ be a symmetric 2 x 2 matrix. Prove that $A$ is positive definite if and only if $a > 0$ and $\det(A) > 0$. [Hint: $ax^2 + 2bxy + dy^2 = a\left(x+\frac{b}{a}y\right)^2 + \left(d-\frac{b^2}{a}\right)y^2$.]</p>
<p>I can see that in order for $\det(A)$ to be greater than $0$, $ad > b^2$.
I have also tried to find the eigenvalues corresponding to the standard matrix $A$ (of the quadratic form) and somehow, I get to $ad < b^2$, which is a contradiction. Could someone point me in the right direction? </p>
| user99163 | 99,163 | <p>Hint: $A$ is positive definite iff $\vec x^{T}A \vec x>0$ for all $\vec x = (x, y) \in\mathbb{R^{2}}\setminus \{(0,0\}$</p>
<p>iff $ax^{2}+2bxy+dy^{2}>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$</p>
<p>iff $a(x+\frac{by}{a})^{2}+(d-\frac{b^{2}}{a})y^{2}>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$</p>
<p>iff $a\left((x+\frac{by}{2})^{2}+det(A)y^{2}\right)>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$</p>
<p>It might be interesting for you to note that this is referred to as Sylvester's criterion.</p>
|
2,342,124 | <p><a href="https://i.stack.imgur.com/QdbFG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QdbFG.png" alt="enter image description here"></a></p>
<p>Well this seems like <span class="math-container">$1-|t|$</span> for <span class="math-container">$|t|<1$</span> and <span class="math-container">$0$</span> for <span class="math-container">$|t|>1$</span> . Taking the Fourier transform <span class="math-container">$$X(ω) = \int_{-\infty}^\infty(1-|t|)e^{-jωt}dt\\=\int_{-\infty}^\infty e^{-jωt}dt -\int_{-\infty}^\infty|t|e^{-jωt}dt\\=2\piδ(ω)-\int_{-1}^1|t|e^{-jωt}dt\\=2πδ(ω)-\int_{-1}^0-te^{-jωt}dt-\int_{0}^1te^{-jωt}dt$$</span></p>
<p>The infinity integral goes from -1 to 1 since the function is zero elsewhere. Now I'm having trouble continuing with this. It doesn't seem to give me the result of the problem's solution which is <span class="math-container">$sinc^2(\frac{ω}{2\pi})$</span> , the sampling function.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\iiint_{x^{2} + y^{2} + z^{2} < 1}\root{x^{2} + y^{2} + z^{2}}\,
\dd x\,\dd y\,\dd z =
\iiint_{r < 1}r\,\dd^{3}\vec{r} =
\iiint_{r < 1}{\nabla\cdot\pars{r\vec{r} \over 4}}\,\dd^{3}\vec{r}
\\[5mm] = &\
\iint_{r = 1}{r\ \vec{r} \over 4}\cdot\,\dd\vec{S}\qquad
\pars{Gauss\ Divergence\ Theorem}
\\[5mm] = &\
{1 \over 4}\iint_{r = 1}{\vec{r}\cdot\,\dd\vec{S} \over r^{2}} =
{1 \over 4}\int_{\Omega_{\vec{r}}}\dd\Omega_{\vec{r}}
= {1 \over 4}\,4\pi = \bbx{\pi}\qquad
\pars{~\Omega_{\vec{r}}:\ Solid\ Angle~}
\end{align}</p>
|
1,156,738 | <p>Is it fine to say "Groups $A$ and $B$ are isomorphic." or should one say "Groups $A$ and $B$ are isomorphic to each other."?</p>
| mweiss | 124,095 | <p>Both formulations are common. You could also say "$A$ is isomorphic to $B$."</p>
|
200,322 | <p>Is there a compact topological space $(X,\tau)$ such that for no cardinal $\kappa$ there is a surjective continuous map $e:\{0,1\}^\kappa \to X$? </p>
<p>(We assume that $\{0,1\}$ is endowed with the discrete topology, and $\{0,1\}^\kappa$ has the product topology.)</p>
| Tomasz Kania | 15,129 | <p>To elaborate on Joseph's answer, the class of continuous images of Cantor cubes has a fancy name, they are the so called <a href="http://en.wikipedia.org/wiki/Dyadic_space" rel="nofollow">dyadic spaces</a>. There is a nice result by Haydon: every Dugundji space is dyadic. (A space $X$ is Dugundji if the conclusion of the Borsuk--Dungundji theorem holds for $X$.)</p>
<blockquote>
<p>R. Haydon, <a href="http://matwbn.icm.edu.pl/ksiazki/sm/sm52/sm5213.pdf" rel="nofollow">On a problem of Pełczyński: Milutin spaces, Dugundji spaces and AE(0-dim)</a>,
<em>Studia Math.</em> <strong>52</strong> (1974), 23-31.</p>
</blockquote>
<p>It is easy to see that the conclusion of the Borsuk--Dugundji theorem fails for $\beta \mathbb{N}$ (it is actually a paradigm counter-example).</p>
|
232,436 | <p>How do I solve? I've tried to multiply and divide by the conjugate cannot advance.
$$\lim_{x\rightarrow +\infty} \sqrt{(x-a)(x-b)}-x$$</p>
| André Nicolas | 6,312 | <p>Multiplying and dividing by the conjugate works fine. Let $x$ be positive and larger than $a$ and $b$. We quickly obtain
$$\frac{-ax-bx+ab}{\sqrt{(x-a)(x-b)}+x}.$$
Divide top and bottom by $x$. (That is another commonly useful kind of move.) We get
$$\frac{-a-b+\frac{ab}{x}}{\sqrt{\left(1-\frac{a}{x}\right)\left(1-\frac{b}{x}\right)}+1}.$$
Now finding the limit is straightforward.</p>
|
232,436 | <p>How do I solve? I've tried to multiply and divide by the conjugate cannot advance.
$$\lim_{x\rightarrow +\infty} \sqrt{(x-a)(x-b)}-x$$</p>
| juantheron | 14,311 | <p>$\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\sqrt{(x-a)(x-b)}-x\;,$ Now Using $\bf{A.M\geq G.M}$</p>
<p>Now When $x\rightarrow \infty,$ Then $(x-a)\;,(x-b)\rightarrow \infty$</p>
<p>So $\displaystyle \frac{(x-a)+(x-b)}{2}\geq \sqrt{(x-a)(x-b)}\Rightarrow x-\left(\frac{a+b}{2}\right)\geq \sqrt{(x-a)(x-b)}$</p>
<p>And equality hold when $(x-a) = (x-b)\rightarrow \infty$</p>
<p>$\displaystyle \Rightarrow \lim_{x\rightarrow \infty}x-\left(\frac{a+b}{2}\right)-x =\lim_{x\rightarrow \infty} \sqrt{(x-a)(x-b)}-x$</p>
<p>So $\displaystyle \lim_{x\rightarrow \infty}\sqrt{(x-a)(x-b)}-x = -\left(\frac{a+b}{2}\right)$</p>
|
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