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 |
|---|---|---|---|---|
2,509,095 | <p>Is there a very simple test to check if a line <em>segment</em> in $3D$ space cuts a plane?
It is assumed we have the coordinates of the endpoints of the line segment, so $p_1,p_2$ and that we have the equation of the plane: $z = d$ (so for simplicity we're assuming it's a plane orthogonal to the z-axis).</p>
| amd | 265,466 | <p>Represent the plane by the equation $ax+by+cz+d=0$ and plug the coordinates of the end points of the line segment into the left-hand side. If the resulting values have opposite signs, then the segment intersects the plane. If you get zero for either endpoint, then that point of course lies <em>on</em> the plane. </p>
<p>You can get the coefficients of the plane equation from a normal vector $\mathbf n$ and a point $\mathbf p$ on the plane via the point-normal form of the equation: $\mathbf n\cdot(\mathbf x-\mathbf p)=0$.</p>
|
438,336 | <p>This a two part question:</p>
<p>$1$: If three cards are selected at random without replacement. What is the probability that all three are Kings? In a deck of $52$ cards.</p>
<p>$2$: Can you please explain to me in lay man terms what is the difference between with and without replacement.</p>
<p>Thanks guys!</p>
| André Nicolas | 6,312 | <p><strong>Without Replacement:</strong> You shuffle the deck thoroughly, take out <strong>three</strong> cards. For this particular problem, the question is "What is the probability these cards are all Kings."</p>
<p><strong>With Replacement:</strong> Shuffle the deck, pick out <strong>one</strong> card, record what you got. Then put it <strong>back in the deck</strong>, shuffle, pick out one card, record what you got. Then put it back in the deck, pick out one card, record what you got. One might then ask for the probability that all three recorded cards were Kings. In the with replacement situation, it is possible, for example, to get the $\spadesuit$ King, or the $\diamondsuit$ Jack more than once. </p>
<p>For solving the "without replacement" problem, here are a couple of ways. There are $\binom{52}{3}$ equally likely ways to choose $3$ cards. There are $\binom{4}{3}$ ways to choose $3$ Kings. So our probability is $\binom{4}{3}/\binom{52}{3}$.</p>
<p>Or else imagine taking out the cards one at a time. The probability the first card taken out was a King is $\frac{4}{52}$. <strong>Given</strong> that the first card taken out was a King, the probability the second one was is $\frac{3}{51}$, since there are $51$ cards left of which $3$ are Kings. So the probability the first two cards were Kings is $\frac{4}{52}\cdot\frac{3}{51}$. **Given that the first two were Kings, the probability the third is is $\frac{2}{50}$. So the desired probability is $\frac{4}{52}\cdot\frac{3}{51}\cdot \frac{2}{50}$. </p>
<p><strong>Remark:</strong> We could solve the same three Kings problem under the "with replacement" condition. (You were not asked to do that,) The second approach we took above yields the answer $\left(\frac{4}{52}\right)^3$. Since we are replacing the card each time and shuffling, the probability of what the "next" card is is not changed by the knowledge that the first card was a King. </p>
|
918,788 | <p>How to do this integral</p>
<p>$$\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x$$</p>
<p>for any $k > 0$ ?.</p>
<p>I tried to use gamma function, but sometimes the series doesn't converge.</p>
| Mateus Sampaio | 101,351 | <p>Write
$$\cos kx = \frac{e^{ikx}+e^{-ikx}}{2}.$$
Using the identity that for $a>0$ and $b\in\mathbb{R}$,
$$\int_{-\infty}^{\infty}e^{-ax^2+ibx}dx=\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}},$$that can be obtained completing squares, we find for $a=1$ and $b=k$ that
$$\int_{-\infty}^{\infty}e^{-x^2}\cos kx dx=\int_{-\infty}^{\infty}e^{-x^2} \frac{e^{ikx}+e^{-ikx}}{2} dx=\sqrt{\pi}e^{-\frac{k^2}{4}}.$$</p>
|
240,700 | <p>How can I prove that every maximal ideal of $B= \mathbb{Z} [(1+\sqrt{5})/2] $ is a principal?</p>
<p>I know if I show that B has division with remainder, that means it is a Euclidean domain. It follows that B is PID, and then every maximal ideal is principal ideal in PID. </p>
<p>However, I haven't been able to show that $B$ has division with remainder.</p>
| kahen | 1,269 | <p>There are many different ways of proving that $\mathbb R^\infty$ is not a Banach space under <em>any</em> norm. The cleanest is probably to note that is has a countable basis, $(e_i)$, where $e_i(k) = \delta_{ik}$ (q.v. <a href="https://en.wikipedia.org/wiki/Kronecker_delta" rel="noreferrer">Kronecker delta</a>), but a simple application of the Baire Category Theorem gives that no countably-infinite-dimensional vector space can be a Banach space: each finite-dimensional subspace is closed and nowhere dense. </p>
|
3,511,118 | <p>I can't see how <span class="math-container">$$e^\left(2i\pi\right) = 1$$</span>
will result in:
<span class="math-container">$$e^\left(i\pi\right) +1 = 0$$</span>
thanks</p>
| José Carlos Santos | 446,262 | <p>My guess is that it's a typo and that they want you to compute <span class="math-container">$\displaystyle\frac{\mathrm d}{\mathrm dx}\int_3^xf(t)^2\,\mathrm dt$</span>.</p>
<p>Otherwise, you can say that <span class="math-container">$\displaystyle\int_3^xf(t)^2\,\mathrm dx=f(t)^2(x-3)$</span> and that therefore<span class="math-container">$$\frac{\mathrm d}{\mathrm dx}\int_3^xf(t)^2\,\mathrm dx=f(t)^2.$$</span></p>
|
3,405,914 | <blockquote>
<p>It's known that <span class="math-container">$\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$</span>.</p>
<p>Using the above statement, prove <span class="math-container">$\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$</span>.</p>
</blockquote>
<h2>My attempt</h2>
<p>Obviously, we want to reach a statement such as <span class="math-container">$$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$</span>
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:</p>
<p><span class="math-container">\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}</span></p>
<p>It seems quite obvious that <span class="math-container">$\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$</span>, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?</p>
| lhf | 589 | <p>Write
<span class="math-container">$$
\frac{3n-2}{3n+1} = 1-\frac{3}{3n+1}
$$</span>
Recall that
<span class="math-container">$$
\left(1-\frac{3}{3n+1}\right)^{3n+1} \to e^{-3}
$$</span>
Then
<span class="math-container">$$
\begin{align}
\left(1-\frac{3}{3n+1}\right)^{3n+1}
= &\left(1-\frac{3}{3n+1}\right)^{3n} \left(1-\frac{3}{3n+1}\right)
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{3n} \to e^{-3}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{n} \to e^{-1}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{2n} \to e^{-2}
\end{align}
$$</span></p>
|
3,405,914 | <blockquote>
<p>It's known that <span class="math-container">$\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$</span>.</p>
<p>Using the above statement, prove <span class="math-container">$\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$</span>.</p>
</blockquote>
<h2>My attempt</h2>
<p>Obviously, we want to reach a statement such as <span class="math-container">$$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$</span>
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:</p>
<p><span class="math-container">\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}</span></p>
<p>It seems quite obvious that <span class="math-container">$\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$</span>, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?</p>
| Danny Pak-Keung Chan | 374,270 | <p>Observe that <span class="math-container">\begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left(\frac{3n+1-3}{3n+1}\right)^{2n}\\
& = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}\left(1+\frac{(-3)}{3n+1}\right)^{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}</span>
We assume the fact without proof: For any <span class="math-container">$x\in\mathbb{R}$</span>, <span class="math-container">$\lim_{n\rightarrow\infty}(1+\frac{x}{n})^{n}=e^{x}$</span>.</p>
<p>Let <span class="math-container">$y_{n}=\left(1+\frac{(-3)}{n}\right)^{n}$</span>. Then,
<span class="math-container">\begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}</span>
Since <span class="math-container">$y_{n}\rightarrow e^{-3}$</span> and <span class="math-container">$(y_{3n+1})_{n}$</span> is a subsequence
of <span class="math-container">$(y_{n})_{n}$</span>, we have <span class="math-container">$y_{3n+1}\rightarrow e^{-3}$</span> too as <span class="math-container">$n\rightarrow\infty$</span>.
Note that <span class="math-container">$\left(1+\frac{(-3)}{3n+1}\right)^{-1}\rightarrow1$</span> as
<span class="math-container">$n\rightarrow\infty$</span>. Therefore
<span class="math-container">\begin{eqnarray*}
\lim_{n\rightarrow\infty}(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \lim_{n\rightarrow\infty}\left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}\\
& = & \left\{ e^{-3}\right\} ^{\frac{2}{3}}\\
& = & e^{-2}.
\end{eqnarray*}</span>
In the above, we have used the fact that <span class="math-container">$x\mapsto x^{\frac{2}{3}}$</span> is
continuous.</p>
|
3,405,914 | <blockquote>
<p>It's known that <span class="math-container">$\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$</span>.</p>
<p>Using the above statement, prove <span class="math-container">$\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$</span>.</p>
</blockquote>
<h2>My attempt</h2>
<p>Obviously, we want to reach a statement such as <span class="math-container">$$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$</span>
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:</p>
<p><span class="math-container">\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}</span></p>
<p>It seems quite obvious that <span class="math-container">$\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$</span>, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?</p>
| user | 505,767 | <p>We can use that</p>
<p><span class="math-container">$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac{3}{3n+1}\right)^{2n}=\left[\left(1-\frac{3}{3n+1}\right)^{-\frac{3n+1}3}\right]^{-\frac{6n}{3n+1}}\to e^{-2}$$</span></p>
|
2,882,985 | <p>Let $f_n(x)=\frac{1}{n}\boldsymbol 1_{[0,n]}(x)$. This sequence is bounded in $L^1(\mathbb R)$ since $\|f_n\|_{L^1}=1$. But why is there no subsequence that convergent weakly ? I know that if such subsequence exist (still denote $f_n$), then $\|f_n\|_{L^1}=1$ Let denote $f$ it's limit. Then, since $f_n\to 0$ pointwise then $f=0$ because $$\lim_{n\to \infty }\int_{\mathbb R}f_n\varphi=\int_{\mathbb R}\lim_{n\to \infty }f_n\varphi=0.$$</p>
<p>Therefore $\|f\|=0$. Now, for me there is no reason to have $$\lim_{n\to \infty }\|f_n\|=0.$$ Indeed, using semi lower continuity of the norm, we have that $$1=\liminf_{n\to \infty }\|f_n\|\geq \|f\|=0,$$
and thus, there is no contradiction. But I know that I should have a contradiction, but where ?</p>
| Lorenzo Quarisa | 402,046 | <p>As you noticed, if a subsequence $\left\{f_n\right\}$ of $\frac{1}{n}\mathbf{1}_{[0,n]}$ converges, it must converge to $0$. But for a sequence $\left\{f_n\right\}\subset L^1(\mathbb{R})$ we have $f_n\to 0$ weakly in $L^1(\mathbb{R})$ if and only if
$$\int_{-\infty}^{+\infty} f_n\varphi \to 0,\qquad \forall \varphi \in L^{\infty}(\mathbb{R}) $$
Which fails, for instance, with $\varphi=1$.</p>
|
2,050,760 | <p>The question:</p>
<blockquote>
<p>Find a recurrence for the number of n length ternary strings that contain "00", "11", or "22".</p>
</blockquote>
<p>My answer:</p>
<p>$3(a_{n-2}) + 3(a_{n-1} - 1)$</p>
<p>Proof:</p>
<p>Cases:</p>
<p>______________00 (a_(n-2))</p>
<p>______________11 (a_(n-2))</p>
<p>______________22 (a_(n-2))</p>
<p>______________0 (a_(n-1) - 1)</p>
<p>______________1 (a_(n-1) - 1)</p>
<p>______________2 (a_(n-1) - 1) (Subtract the case in which it ends with 00, 11, or 22)</p>
| Robert Israel | 8,508 | <p>The point is that
$$ \pmatrix{1 & a\cr 0 & 1\cr} \pmatrix{1 & b\cr 0 & 1\cr} = \pmatrix{1 & a+b\cr 0 & 1\cr}$$</p>
<p>The mapping takes $\pmatrix{1 & c\cr 0 & 1\cr}$ to $c$.</p>
|
3,572,842 | <p><strong>Context:</strong> 1st year BSc Mathematics, Vectors and Mechanics module, constant circular motion.</p>
<p>This may be trivial, but can someone tell me what's wrong with the following reasoning?</p>
<p><span class="math-container">$$\underline{e_r}=\underline{i}\cos\theta+\underline{j}\sin\theta=(1,\theta) \;\;(1);$$</span>
<span class="math-container">$$\underline{e_\theta}=\frac{d(\underline{e_r})}{d\theta}=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\;(2);$$</span>
so
<span class="math-container">$$(1),(2):\;\; \underline{e_\theta}=\frac{d}{d\theta}((1,\theta))=(1,\frac{d\theta}{d\theta})=(1,1) \;\;(3),$$</span>
so
<span class="math-container">$$(2),(3): \;\; (1,1)=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\; (4),$$</span>
an undesirable conclusion.</p>
| poetasis | 546,655 | <p>I belive the leg-sums are the same "set" as leg-differences except that the difference-set also contains <span class="math-container">$P^0=1$</span>.</p>
<p>What you have is the set of prime power products where
<span class="math-container">$\quad p_n\equiv\pm 1 \pmod 8$</span>.</p>
<p>The value <span class="math-container">$49=7^2$</span> and <span class="math-container">$\space119=7\cdot 17$</span></p>
<p>Under <span class="math-container">$100$</span>, <span class="math-container">$|A-B|\in \big\{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\big\}$</span>.</p>
<p>You can build a sum-set to confirm or disprove this.</p>
|
4,228,826 | <p>Consider the inequality
<span class="math-container">$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$</span>
for <span class="math-container">$0 < x < 1$</span>. The upper bound can be read off the Taylor expansion for <span class="math-container">$\sqrt{1-x}$</span> around <span class="math-container">$0$</span>,
<span class="math-container">$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$</span>
by noting that all the non linear terms are negative. Can the left side inequality be read-off the expansion by a similar reasoning? Please do not try to prove the left side inequality by other means (such as minimizing <span class="math-container">$\sqrt{1-x} - 1 + \frac{x}{2} + \frac{x^2}{2}$</span> using derivatives).</p>
| orangeskid | 168,051 | <p>We have a function
<span class="math-container">$$f(x) =\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128 }- \frac{7 x^5}{256}- \cdots = \\=1 - a_1 x - a_2 x^2 -a_3 x^3 \cdots $$</span>
The series converges for every <span class="math-container">$x\in [0,1]$</span>, so we have
<span class="math-container">$$a_1 + a_2 + \cdots = 1$$</span></p>
<p>Therefore we can write
<span class="math-container">$$1 - \frac{x}{2} - \frac{x^2}{2} = 1 - a_1 x - (a_2 + a_3 + \cdots ) x^2< \\ <1 - a_1 x - a_2 x^2 - a_3 x^3 -\cdots = f(x)$$</span> for <span class="math-container">$x\in (0,1)$</span>.</p>
|
193,001 | <p>This is from a GRE prep book, so I know the solution and process but I thought it was an interesting question: Explicitly evaluate $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right).$$</p>
| Juan S | 2,219 | <p>This telescopes, using the fact that $\text{arctan}(u)-\text{arctan}(v) = \text{arctan}(\frac{u-v}{1+uv})$</p>
<p>Specifically take $u=n+1$ and $v=n$. Then $$\text{arctan}\left(\frac1{n^2+n+1}\right) = \text{arctan}(n+1)-\text{arctan}(n)$$</p>
<p>This gives $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right) = \text{arctan}(m+1) - \pi/4$$</p>
|
193,001 | <p>This is from a GRE prep book, so I know the solution and process but I thought it was an interesting question: Explicitly evaluate $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right).$$</p>
| lab bhattacharjee | 33,337 | <p>By observation,
$\tan^{-1}\frac{1}{n^2+n+1}=\cot^{-1}(n^2+n+1)=\cot^{-1}\frac{n(n+1)+1}{n+1-n}$
$=\cot^{-1}(n)-\cot^{-1}(n+1)$</p>
<p>$\sum_{n=1}^{m}\tan^{-1}\left({\frac{1}{{n^2+n+1}}}\right)$
$=\frac{\pi}{4}-\cot^{-1}(m+1)$ using <a href="http://mathworld.wolfram.com/TelescopingSum.html" rel="noreferrer">this</a>.</p>
|
1,600,307 | <p>Let $n$ be an integer greater than 1, $\alpha$ be a real number, and consider the quadratic form $Q_{\alpha}$ given by: </p>
<p>for every $(x_1, ... , x_n) \in R^n$, </p>
<p>$$Q_{\alpha}(x_1,...,x_n)= \sum_{i=1}^n x_i^2 - \alpha(\sum_{i=1}^n x_i)^2$$</p>
<p>Find all the eigenvalues of $Q_{\alpha}$ in terms of $\alpha$ and $n$. What is the condition on $\alpha$ for $Q_{\alpha}$ to be positive-definite?</p>
<p>EDIT: I am currently expanding out every term, and then regrouping the common $x_ix_j$ terms to see what I can come up with and to see whether I can spot a matrix that I can put together...</p>
<p>Any ideas are welcome. </p>
<p>Thanks,</p>
| zyx | 14,120 | <p>For constant $S = \sum x_i$, the sum of squares is minimum when all $x_i$ are equal, at which point the form equals the rank 1 quadratic form $n(S/n)^2 - \alpha S^2 = S^2 (\frac{1}{n} - \alpha)$. </p>
<p>Hence definiteness requires $\alpha \leq \frac{1}{n}$. Subtracting the rank 1 form from $Q$ leaves the variance-like form $\sum (x_i - \bar{x})^2$ where $\bar{x}=S/n$. That is a decomposition of $Q$ as a sum of $n+1$ squares</p>
<p>$$ Q_{\alpha} = \sum_i (x_i - \bar{x})^2 + ( \frac{1}{n} - \alpha) (\sum_i x_i)^2 $$</p>
<p>The other part is essentially the calculation of eigenvalues of the all-1's matrix, one of the more frequently asked questions on the site.</p>
<p><a href="https://math.stackexchange.com/search?q=%22all+ones%22+matrix">https://math.stackexchange.com/search?q=%22all+ones%22+matrix</a> </p>
|
1,366,023 | <p>Here's a problem I was just working on:</p>
<blockquote>
<p>Let $f$ have an essential singularity at $0$. Show that there is a sequence of points $z_n \to 0$ such that $z_n^n f(z_n)$ tends to infinity.</p>
</blockquote>
<p>I know already that there exists a sequence $z_n \to 0$ such that $f(z_n)$ tends to any complex number I want, hence I can get a sequence that tends to infinity. The problem is that I need this sequence to tend to infinity really fast. </p>
<p>What I did so far was look at the function $g_n(z) := z^n f(z)$, $n \geq 1$. This obviously also has an essential singularity at $0$, so I can find a sequence $z_{n1}, z_{n2}, ...$ such that $\lim\limits_k g_n(z_{nk}) = \infty$. Do you think it's possible to extract from the array $z_{nk}, (n,k) \in \mathbb{N}^2$ a subsequence $z_l$ such that $g_l(w_l) \to \infty$ as $l \to \infty$? I tried for awhile but I'm just not very good at these kinds of arguments.</p>
| amirbd89 | 49,744 | <p>Yes. It is possible, but I think it will be easier to rebuild it from scratch than to extract such a sequence.</p>
<p>You can define $z_n$ in the following way:</p>
<p>For each $n$, there exists $z_n \in B(0,1/n)$ such that $|g_n(z_n)| > n$ (since, as you said yourself, there is a sequence that tends to zero and takes $g_n$ to $\infty$). This is your sequence. It should be easy to finish the proof from here.</p>
|
255,374 | <p>Does there exist any noncomputable set $A$ and probabilistic Turing machine $M$ such that $\forall n\in A$ $M(n)$ halts and outputs $1$ with probability at least $2/3$, and $\forall n\in\mathbb{N}\setminus A$ $M(n)$ halts and outputs $0$ with probability at least $2/3$? What if you only require that $M(n)$ is correct with probability greater than $1/2$?</p>
| Greg Kuperberg | 1,450 | <p>Every such decision problem is computable, even in the harder version of the problem, assuming that the transition probabilities are, say, fixed rational numbers. A deterministic algorithm can calculate the probability distribution on the set of states of this stochastic TM after each $t$ time steps, and then step through $t$ until the probability of halt at either yes or no exceeds $1/2$.</p>
|
1,220,923 | <p>Find the value of the integral
$$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx.$$
I tried the substitution $x=t^5$ to obtain
$$\int_0^\infty \frac{5t^6}{1+t^{10}}dt.$$
Now we can factor the denominator to polynomials of degree two (because we can easily find all roots of polynomial occured in the denominator of the former integral by using complex numbers) and then by using partial fraction decomposition method find the integral!</p>
<p>Is there any simple method to find the integral value??!!</p>
| Olivier Oloa | 118,798 | <p>One may recall the <a href="http://en.wikipedia.org/wiki/Beta_function">Euler beta function</a>
$$
B(a,b) =\int _0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},
$$ with a remarkable case
$$
\int _0^1 x^{a-1}(1-x)^{-a}dx=\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)}.
$$</p>
<p>Then, by the change of variable $\displaystyle x=\frac{1}{1+t^{10}}$, giving $\displaystyle t=x^{-1/10}(1-x)^{1/10}$, we get
$$
\int_0^\infty \frac{5t^6}{1+t^{10}}dt=\frac12\int _0^1 x^{-7/10}(1-x)^{7/10-1}dx=\frac{\pi}{2\sin(7\pi/10)}=\frac{\sqrt{5}-1}{2}\pi.
$$</p>
|
2,962,203 | <p>I got stuck at : <span class="math-container">$a^2/b^2 = 12+2 \sqrt 35$</span></p>
<p>I understand that <span class="math-container">$12$</span> is rational and now I need to prove that <span class="math-container">$\sqrt{35}$</span> is irrational.</p>
<p>so I defined <span class="math-container">$∀c,d∈R$</span> while <span class="math-container">$d$</span> isn't <span class="math-container">$0$</span> that:
<span class="math-container">$c^2/d^2 = \sqrt 35$</span>
so <span class="math-container">$- c^2=(d^2)\sqrt{35}$</span>
It means that <span class="math-container">$c$</span> divide with <span class="math-container">$5$</span> and <span class="math-container">$7$</span>?
Also, how do I prove that if for example <span class="math-container">$X^2/4$</span> then <span class="math-container">$X/4$</span>? </p>
| lhf | 589 | <p>You're on the right track.</p>
<p>Consider the powers of <span class="math-container">$5$</span> that divide both sides of <span class="math-container">$c^2=35d^2$</span>. You have an even number for the RHS but an odd number for the LHS.</p>
<p>Indeed, if <span class="math-container">$5^m$</span> is the largest power of <span class="math-container">$5$</span> that divides <span class="math-container">$c$</span> and <span class="math-container">$5^n$</span> is the largest power of <span class="math-container">$5$</span> that divides <span class="math-container">$d$</span>, then we get <span class="math-container">$2m=2n+1$</span>, a contradiction.</p>
|
2,962,203 | <p>I got stuck at : <span class="math-container">$a^2/b^2 = 12+2 \sqrt 35$</span></p>
<p>I understand that <span class="math-container">$12$</span> is rational and now I need to prove that <span class="math-container">$\sqrt{35}$</span> is irrational.</p>
<p>so I defined <span class="math-container">$∀c,d∈R$</span> while <span class="math-container">$d$</span> isn't <span class="math-container">$0$</span> that:
<span class="math-container">$c^2/d^2 = \sqrt 35$</span>
so <span class="math-container">$- c^2=(d^2)\sqrt{35}$</span>
It means that <span class="math-container">$c$</span> divide with <span class="math-container">$5$</span> and <span class="math-container">$7$</span>?
Also, how do I prove that if for example <span class="math-container">$X^2/4$</span> then <span class="math-container">$X/4$</span>? </p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> <span class="math-container">$ $</span> Subtract <span class="math-container">$12$</span> then square it again to show that it's a root of a polynomial with integer coef's that's monic (lead coef <span class="math-container">$= 1).\,$</span> Now apply the <a href="https://math.stackexchange.com/a/658058/242">Rational Root Test</a> (or equivalent).</p>
|
2,962,203 | <p>I got stuck at : <span class="math-container">$a^2/b^2 = 12+2 \sqrt 35$</span></p>
<p>I understand that <span class="math-container">$12$</span> is rational and now I need to prove that <span class="math-container">$\sqrt{35}$</span> is irrational.</p>
<p>so I defined <span class="math-container">$∀c,d∈R$</span> while <span class="math-container">$d$</span> isn't <span class="math-container">$0$</span> that:
<span class="math-container">$c^2/d^2 = \sqrt 35$</span>
so <span class="math-container">$- c^2=(d^2)\sqrt{35}$</span>
It means that <span class="math-container">$c$</span> divide with <span class="math-container">$5$</span> and <span class="math-container">$7$</span>?
Also, how do I prove that if for example <span class="math-container">$X^2/4$</span> then <span class="math-container">$X/4$</span>? </p>
| Rishabh Khandelwal | 606,037 | <p>Assuming √5 + √7 is rational,
√5 + √7 = a ∕ b
Squaring both sides,
5 + 7 + 2√35 = a²/b²
⇒2√35 = a²/b² - 12
⇒2√35 = (a²-12b²)/b²
⇒√35 = (a²-12b²)/2b²</p>
<p>According to our assumption,
(a²-12b²)/2b² should have been rational
but actually it is irrational.
Hence our assumption is wrong.
Therefore, √5 + √7 is irrational</p>
|
42,787 | <p>I am using <code>ListPlot</code> to display from 5 to 12 lines of busy data. The individual time series in my data are not easy to distinguish visually, as may be evident below, because the colors are not sufficiently different.</p>
<p><img src="https://i.stack.imgur.com/PiMMh.png" alt="enter image description here"></p>
<p>I have been trying to use <code>PlotStyle</code>, <code>ColorData</code>and related functions to get better colors. I would rather not have to specify a specific list of colors because the number of plot items varies from test to test. I created a toy plot to experiment with - the problem is illustrated by lines "F" and "G", which seem to be almost the same color. <code>PlotStyle</code> -> <code>ColorData</code> doesn't seem to work. Is there a simple way to do this?</p>
<pre><code>ListPlot[Table[i*Range[0, 10], {i, 1, 5, 0.5}]
, Frame -> True, Joined -> True, PlotRange -> All
, PlotLegends -> SwatchLegend[Characters["ABCDEFGHIGKLMNOP"]]
, PlotStyle -> ColorData["TemperatureMap"] ]
</code></pre>
<p><img src="https://i.stack.imgur.com/Wdi0O.png" alt="enter image description here"></p>
<p>It looks like </p>
<pre><code>ListLinePlot[Table[data2*i, {i, k}], PlotStyle -> Thick,
ColorFunction -> Function[{x1, x2}, ColorData[c1][x2]]]
</code></pre>
<p>from another <a href="https://mathematica.stackexchange.com/questions/27131/plotstyle-in-listplot-change-color-scheme-manually-choose-color-of-first-plot?rq=1">question</a> may be the answer. I didn't see that before. I'll try it out. I don't think I really understand <code>ColorData</code>. Meanwhile, if anyone has generally enlightening comments, I would appreciate them.</p>
| Sjoerd C. de Vries | 57 | <p>The key is to use one of the <em>indexed</em> color sets. You can find them in the color schemes palette or generate them using <code>ColorData[c]</code>. The j-th color in scheme i can be obtained using <code>ColorData[i][j]</code>.</p>
<p>To generate a set of colors for use in <code>PlotStyle</code> you can use</p>
<pre><code>PlotStyle->ColorData[2]/@Range[8].
</code></pre>
<p>Most color schemes only have a finite set of colors to choose from, except scheme 1 and 63. Scheme 1 happens to be the default color scheme.</p>
<p>The following code shows the color schemes as colored spheres in RGB space.</p>
<pre><code>Partition[
Table[
Graphics3D[{#,
Sphere[{##}, 1/10]& @@ #}& /@ (ColorData[c]/@Range[50]/. GrayLevel[g_]->RGBColor[g, g, g]),
Lighting -> "Neutral",
PlotRange -> {{0, 1}, {0, 1}, {0, 1}}, PlotRangePadding -> 1/10,
PlotLabel -> c, ViewPoint -> {10, 5, 7}
],
{c, 1, 93}
],
6, 6, {1, 1}, {}] // GraphicsGrid
</code></pre>
<p><img src="https://i.stack.imgur.com/337Cf.png" alt="Mathematica graphics"></p>
<p>You can choose a standard scheme or use your own colors, specifying the RGB values manually using <code>RGBColor</code> or <code>Hue</code>.</p>
|
779,095 | <p>Let
$$f(x,y)=\left\{ \begin{matrix} \frac{x^2y}{x^4+y^2} & (x,y)\neq(0,0) \\0 & (x,y)=(0,0)\end{matrix}\right.$$</p>
<p>It is easy to prove that the $f$ is not continuous at $(0,0)$ (doing the limit along the curve $y=x^2$).</p>
<p>I want to know whether it is possible to define the partial derivatives of $f$ at $(0,0)$ and find the directions $\vec v$ such that $D_vf(0,0)$ is defined.</p>
<p>I've calculated the partial derivatives of $f$ for $(x,y)\neq(0,0)$:
$$\frac{\partial f}{\partial x}=\frac{2xy(x^4+y^2)-4yx^5}{(x^4+y^2)^2}$$
$$\frac{\partial f}{\partial y}=\frac{x^2(x^4+y^2)-2x^2y^2}{(x^4+y^2)^2}$$</p>
<p>Neither of them is continuous at $(0,0)$. However, if we take for example the line $y=x$, then $$\lim_{(x,y)\to (0,0)}_{x=y}\frac{\partial f}{\partial x}=-2$$
$$\lim_{(x,y)\to (0,0)}_{x=y}\frac{\partial f}{\partial y}=-1$$</p>
<p>So I'm tempted to say that even though the partial derivatives do not exist at the origin, $D_{(1/\sqrt2),1/\sqrt2)}f(0,0)=\frac{1}{\sqrt2}(-2,-1)$.</p>
<p>Is this correct? If so, how could I find all the vectors $\vec v$ such that $D_\vec v f(0,0)$ is defined</p>
| Wlod AA | 490,755 | <p>Each finite pos (partially ordered set) has to have at least one minimal element.</p>
<p>Consider 3-set <span class="math-container">$\ X:=\{a\ b\ c\}.\ $</span> First, let <span class="math-container">$a$</span> be the <strong>unique</strong> minimal element. There are exactly three different partial orders in <span class="math-container">$\{b\ c\}.\ $</span> The same when <span class="math-container">$\ b\ $</span> or <span class="math-container">$\ c\ $</span> is the only minimal element in <span class="math-container">$\ X.\ $</span> Each of those partial orders in a 2-element set extends uniquely over the third element that is minimal. Every two of these partial orders in <span class="math-container">$\ X\ $</span> are different. Thus, there are exactly <span class="math-container">$9=3\cdot 3\ $</span>
partial orders in <span class="math-container">$\ X\ $</span> that have exactly one minimal element.</p>
<p>Now, let <span class="math-container">$\ a\ $</span> and <span class="math-container">$\ b\ $</span> be the only minimal elements in <span class="math-container">$\ X\ $</span> (so that <span class="math-container">$\ c\ $</span> is not minimal). We have exactly three partial orders like this in <span class="math-container">$\ X.\ $</span> If the lonely non-minimal element is <span class="math-container">$\ b\ $</span> or <span class="math-container">$\ a\ $</span>, each time we get 3 different partial orders, and all of such <span class="math-container">$\ 3\cdot 3=9\ $</span> partial orders are different. Thus, so far, we have 9+9=18 different partial orders.</p>
<p>Finally, there is exactly one partial order such that all elements of <span class="math-container">$\ X\ $</span> are minimal. This gives us finally the answer 1+18=19.</p>
<p><strong>THEOREM</strong> There are exactly <span class="math-container">$\ 19\ $</span> different partial orders in any fixed 3-set.</p>
|
2,642,144 | <p>How would I prove or disprove the following statement?
$ \forall a \in \mathbb{Z} \forall b \in \mathbb{N}$ , if $a < b$ then $a^2 < b^2$</p>
| Cameron Buie | 28,900 | <p>To disprove it, one simply provides a counterexample, as the other answers have addressed.</p>
<p>If I were to attempt to prove it, here's how I might begin--and how one might discover that it is false, and find a key to generating counterexamples.</p>
<blockquote>
<p>Since <span class="math-container">$x<y$</span> is equivalent to <span class="math-container">$y-x$</span> being positive, then we want to show that <span class="math-container">$b^2-a^2$</span> is positive whenever <span class="math-container">$a\in\Bbb Z$</span> and <span class="math-container">$b\in\Bbb N$</span> are such that <span class="math-container">$b-a$</span> is positive.</p>
<p>Now, <span class="math-container">$b^2-a^2=(b-a)(b+a),$</span> and so since <span class="math-container">$b-a$</span> is positive, then for <span class="math-container">$b^2-a^2$</span> to be positive, we must show that <span class="math-container">$b+a$</span> is positive.</p>
</blockquote>
<p>Here, though, we'd be stymied. We'd be okay so long as <span class="math-container">$|a|<|b|,$</span> but if <span class="math-container">$|a|\ge|b|,$</span> then we'd have to have <span class="math-container">$a\le -b,$</span> so that <span class="math-container">$b+a\le b+-b=0.$</span></p>
<p>Now, if we'd instead had <span class="math-container">$a\in\Bbb N$</span> and <span class="math-container">$b\in\Bbb Z,$</span> then we'd be just fine. In that case, we'd simply note that <span class="math-container">$-a\le a,$</span> so that <span class="math-container">$0<b-a\le b+a,$</span> as desired.</p>
|
8,312 | <p>Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? </p>
| Qiaochu Yuan | 232 | <p>The two are the same if the roots of <span class="math-container">$f$</span> form an integral basis of the ring of integers of <span class="math-container">$\mathbb{Q}[x]/f(x)$</span> (e.g. if <span class="math-container">$f$</span> is a cyclotomic polynomial) because then, well, they're defined by the same determinant (see <a href="http://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field#Examples" rel="nofollow noreferrer">Wikipedia</a>), but in general they don't. In the general case <span class="math-container">$\mathbb{Z}[\alpha_1, ... \alpha_n]$</span> is an order in <span class="math-container">$\mathcal{O}_K$</span> so one can write the <span class="math-container">$\alpha_i$</span> as an integer linear combination of an integral basis, so the matrices whose determinants define the two discriminants should be related by the square of a matrix with integral entries, hence integral determinant. </p>
<p>In fact if I'm not totally mistaken, the quotient of the two discriminants should be precisely the index of <span class="math-container">$\mathbb{Z}[\alpha_1, ... \alpha_n]$</span> in <span class="math-container">$\mathcal{O}_K$</span> as lattices, or maybe its square...?</p>
<p>In any case, since the discriminant of the field is defined in terms of <span class="math-container">$\mathcal{O}_K$</span> it is the "right" choice for carrying information about, for example, ramification. One can see this even in the quadratic case: if <span class="math-container">$d \equiv 1 \bmod 4$</span> then the discriminant of <span class="math-container">$x^2 - d$</span> is <span class="math-container">$4d$</span> but the discriminant of <span class="math-container">$\mathbb{Q}(\sqrt{d})$</span> is <span class="math-container">$d$</span>, and the latter is the "right" choice because <span class="math-container">$2$</span> doesn't ramify in <span class="math-container">$\mathbb{Z} \left[ \frac{1 + \sqrt{d}}{2} \right]$</span>. </p>
|
8,312 | <p>Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? </p>
| Community | -1 | <p>In response to Qiaochu,</p>
<p>$Disc(f)/Disc(\mathcal{O}_K)$ is the square of the index of $\mathbb{Z}[ \alpha _1, \ldots , \alpha _n ]$ in $\mathcal{O}_K$. The index itself is the determinant of the change of basis matrix from $(\alpha _1, \ldots , \alpha _n )$ to an integral basis for $\mathcal{O}_K$. This matrix is squared when taking the discriminant.</p>
|
8,312 | <p>Let $f\in\mathbb{Z}[x]$ be monic and irreducible, let $K=$ splitting field of $f$ over $\mathbb{Q}$. What can we say about the relationship between $disc(f)$ and $\Delta_K$? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? </p>
| Jiangwei Xue | 9,430 | <p>I think there was some confusion about the splitting field and the field $\mathbb{Q}[x]/(f(x))$, which is isomorphic to the field generated by one root of $f(x)$. (We always assume that $f(x)$ is monic irreducible.)</p>
<p>Let $\alpha$ be a root of $f(x)$, and $L=\mathbb{Q}(\alpha)$ be the field generated by $\alpha$, $\mathbb{Z}[\alpha]$ be the subring of $\mathcal{O}_L$ generated by $\alpha$. The the discriminant of $f(x)$ is the discriminant of the lattice $\mathbb{Z}[\alpha]$. So $\mathrm{Disc}(f)/\mathrm{Disc}(\mathcal{O}_L)$ is the square of $[\mathcal{O}_L: \mathbb{Z}[\alpha]]$. (See III.3 of Lang's "Algebraic number theory"). </p>
<p>However, for splitting fields, these things hardly compares. For example, take $f(x)=x^4-x+1$, then the discriminant of $f(x)$ is 229 (a prime, which coincides with the discriminant of the field $L$ in this case) , but the discriminant of the splitting field of $f(x)$ is $229^{12}$ (calculated using Pari), which has 28-digits. (Well, it is not hard to show the discriminant of the splitting field of $f(x)$ share the same prime divisors as the field $L$.)</p>
<p>Sorry about bring up a really old question. It is just I asked myself the same thing today. </p>
|
1,716,656 | <p>I am having trouble solving this problem</p>
<blockquote>
<p>Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?</p>
</blockquote>
<p>My attempt:</p>
<p>I first want to find the deposit per month.</p>
<p>I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,</p>
<p>$D*100(Ia_{30|0.08}) = 100,000$</p>
<p>However, the $D$ I got was 8.12, which is clearly not right.</p>
<p>Can someone help?</p>
| browngreen | 321,445 | <p>The probability of rolling a 1 and 3 is 1/18. Same for the probability of 2&4, 3&5, and 4&6.</p>
<p>So the overall probability of the dice being two apart equals 4/18 = 2/9.</p>
|
1,411,305 | <p>I have been trying to solve the following problem:</p>
<blockquote>
<p>What is the probability that among 3 random digits, there appear
exactly 2 different ones?</p>
</blockquote>
<p>The formula for no repititions is:</p>
<pre><code>(n*(n-1)...(n-r+1))/n^r
</code></pre>
<p>So, for the first digit there are 10 possibilities, for the second 9.
Next, the third digit should be the same as previous two: it is binomial(2,1)
And there are total 10^3 possibilities of ordering with replacement:
My formula is:</p>
<pre><code>(10*9*binomial(2,1))/10^3
</code></pre>
<p>But it leads to the wrong answer.
Could you please help me understand where is my mistake and the logic of solving such problems?
Thank you </p>
| robjohn | 13,854 | <p>There are $10$ choices for the digit that appears twice and then $9$ choices for the digit that appears once. There are $\binom{3}{1}=3$ ways to arrange the digits. This gives $10\cdot9\cdot3=270$ choices out of $1000$. That is, a $0.27$ probability.</p>
|
2,243,542 | <p>In a previous question here <a href="https://math.stackexchange.com/q/2240195/369757">Can we define the Cantor Set in this way?</a></p>
<p>we defined a family of sets $ \left\{ C_0,C_1,C_2,C_3,\dots \right\}$</p>
<p>We can call this set $S_1$ , where the values of these elements is</p>
<p>$C_0 = \left\{ 0.0 \right\}$</p>
<p>$C_1 = \left\{ 0.0 , 0.2 \right\}$</p>
<p>$C_2 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 \right\}$</p>
<p>$C_3 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 ,0.002 , 0.202 ,0.022 , 0.222 \right\}$</p>
<p>and so on...</p>
<p>We can then take the union of $\bigcup S_1$ and get some countable set $X$</p>
<p>We will define a set $S_2$ to be all the members of $S_1$ that are not redundant.</p>
<p>$C_0$ is redundant because it can be removed from $S_1$ and the union will still be the same.</p>
<p>In fact all the elements of $S_1$ are redundant so our set</p>
<p>$\bigcup S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots \right\} = \varnothing \neq X \to$ Contradiction</p>
<p>We removed all the sets from the union that did not contribute any information, and all the information disappeared.</p>
<p>What went wrong here?</p>
| Noah Schweber | 28,111 | <p>First, note that this isn't really a question about the Cantor set, just about the combinatorics of infinite sets in general.</p>
<p>E.g. we can recast it in terms of sets of natural numbers: let $A_n=\{1, 2, 3, ..., n\}$, and think about $\bigcup A_n$. Or, even easier, we could just take $B_n=\{0\}$ (that's not a typo) and think about $\bigcup B_n$.</p>
<hr>
<p>Now as to the answer to your question, <em>nothing went wrong</em> except your intuition for how infinite sets behave. Even though removing <em>any finite number</em> of the $A_n$s leaves the union of the rest unchanged, that's no reason to believe that removing <em>infinitely many</em> of the $A_n$s should do the same. There are lots of times we see a "phase transition" between finite behavior and infinite behavior - e.g. changing finitely many terms of a divergent series to $0$ leaves the result divergent, but if you change infinitely many terms to $0$ it might converge.</p>
|
3,430,812 | <p>Consider the set of integers, <span class="math-container">$\Bbb{Z}$</span>. Now consider the sequence of sets which we get as we divide each of the integers by <span class="math-container">$2, 3, 4, \ldots$</span>.</p>
<p>Obviously, as we increase the divisor, the elements of the resulting sets will get closer and closer.</p>
<p><strong>Question:</strong> In the limit as <span class="math-container">$\text{divisor}\to\infty$</span>, what will the "limiting" set be?
(I don't think it could be <span class="math-container">$\Bbb{R}$</span>.)</p>
| Ali Ashja' | 437,913 | <p><b>1)</b> Let <span class="math-container">$S_n = \{ \frac{z}{n} \ | \ z \in \mathbb{Z} \}$</span> and <span class="math-container">$p_i$</span> be <span class="math-container">$i$</span>-th prime integer.</p>
<p><b>2)</b> It has no limit! Because since <span class="math-container">$(n,n+1)=1$</span> we have <span class="math-container">$S_n \cap S_{n+1} = \mathbb{Z}$</span>, so always new set miss any non integer rationals included in previous one and get some new ones.</p>
<p><b>3)</b> But <span class="math-container">$\limsup$</span>, exists. If you consider <span class="math-container">$a_n=\prod_{i=1}^n p_i^n$</span>, then set sequence <span class="math-container">$S_{a_n}$</span> is an strict increasing sequence, with respect to inclusion order, that for every <span class="math-container">$m$</span> there is a <span class="math-container">$k$</span> that <span class="math-container">$m | a_k$</span>, so <span class="math-container">$S_m \subseteq S_{a_k}$</span>, Therefore it tends to <span class="math-container">$\mathbb{Q}$</span>.</p>
<p><b>4)</b> Also <span class="math-container">$\liminf$</span>, exists. As we see <span class="math-container">$S_n \cap S_{n+1} = \mathbb{Z}$</span>, so it tends to <span class="math-container">$\mathbb{Z}$</span>.</p>
|
3,430,812 | <p>Consider the set of integers, <span class="math-container">$\Bbb{Z}$</span>. Now consider the sequence of sets which we get as we divide each of the integers by <span class="math-container">$2, 3, 4, \ldots$</span>.</p>
<p>Obviously, as we increase the divisor, the elements of the resulting sets will get closer and closer.</p>
<p><strong>Question:</strong> In the limit as <span class="math-container">$\text{divisor}\to\infty$</span>, what will the "limiting" set be?
(I don't think it could be <span class="math-container">$\Bbb{R}$</span>.)</p>
| Paul Sinclair | 258,282 | <p>The three answers thus far assume by limits of the sets you mean the common value of the set-theoretic <span class="math-container">$\liminf$</span> and <span class="math-container">$\limsup$</span> (where convergence means they agree). This is a highly reasonable assumption, given that you did not specify a meaning for the limit of sets yourself.</p>
<p>However, I want to point out that there are other possibilities for defining a limit of sets. For example, given a sequence of sets <span class="math-container">$(S_n)$</span> with <span class="math-container">$\forall n, S_n \subseteq X$</span> for some topological space <span class="math-container">$X$</span>, you could define</p>
<p><span class="math-container">$$\lim_n S_n = \{x\in X\mid \exists (s_n) \subset X, s_n \to x \wedge \forall n, s_n \in S_n\}$$</span></p>
<p>By this definition with <span class="math-container">$X = \Bbb R$</span>, the limit of your sets is indeed <span class="math-container">$\Bbb R$</span>.</p>
|
316,055 | <p>I have no idea of how to solve the following: </p>
<p>$$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{3x}$$</p>
<p>I know about the notable special limit $$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$$, and I know that I have to do some algebraic manipulation and change what I have above to the notable one, but I can't quite see how.</p>
| muzzlator | 60,855 | <p>A handy thing to know is that $e^x = 1 + x + O(x^2) \approx 1 + x $ for $x$ near $0$.</p>
<p>Using this, you will see immediately that the limit is $\frac{1}{3}$.</p>
<p>Being a bit more rigorous, you may notice that $$\lim_{x\rightarrow 0} \frac{e^x - 1}{3x} = \frac{1}{3} \lim_{x\rightarrow 0} \frac{e^x - e^0}{x} = \frac{1}{3}\frac{d}{dx} e^x \big|_{x=0} $$</p>
|
316,055 | <p>I have no idea of how to solve the following: </p>
<p>$$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{3x}$$</p>
<p>I know about the notable special limit $$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$$, and I know that I have to do some algebraic manipulation and change what I have above to the notable one, but I can't quite see how.</p>
| Joe | 24,942 | <p>Using the fact that
$$\lim_{x\to 0} \frac{e^x-1}{x} = 1$$</p>
<p>We can simply rewrite your limit as </p>
<p>$$\frac13\lim_{x\to 0} \frac{e^x-1}{x}$$</p>
<p>Which is $\frac13 \cdot1 = \frac13$.</p>
<p><strong>Alternative</strong>: note that this is of form $\frac00$, and both functions are continuous in the neighborhood of $0$, so we can use <a href="http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="nofollow">L'Hopital's Rule</a>, which allows us to rewrite our limit as</p>
<p>$$\lim_{x\to 0} \frac{e^x}{3} = \frac13 $$</p>
<p><strong>Alternative 2</strong>: use the Taylor Series for $e^x$ around $0$.</p>
<p>Recall that $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ Writing out the first few terms yields $1+x+O(x^2)$ where $O(x^2)$ represents the higher order terms on or past the order of $x^2$. Since we are in a neighborhood around $0$, note that these higher order terms all go to $0$ since $x$ is going to $0$.</p>
<p>Thus our limit now is
$$\lim_{x\to 0} \frac{(1+x)-1}{3x} = \lim_{x\to 0} \frac{x}{3x} = \frac13$$</p>
|
3,165,781 | <p>Ok, I am a bit confused.</p>
<p>Does first countability mean that for every element <span class="math-container">$x$</span> in the space, there is a collection of open sets and each of those open sets are countable containing <span class="math-container">$x$</span> and for every open neighborhood of <span class="math-container">$x$</span>, one of those countable open set is contained in. </p>
<p>OR</p>
<p>For every element <span class="math-container">$x$</span> in the space, there is a collection of open sets, where the number of sets in this collection is countably many, that contains <span class="math-container">$x$</span> and for every neighborhood of <span class="math-container">$x$</span>, one of those open sets (not necessarily countable) is contained in?</p>
<p>Thank you in advanced. </p>
| Henno Brandsma | 4,280 | <p>It's more the second one, put formally: for each <span class="math-container">$x \in X$</span> there is an at most countable collection <span class="math-container">$U_n(x), n \in \mathbb{N}$</span> of open neighbourhoods of <span class="math-container">$x$</span> such that for each open set <span class="math-container">$O$</span> that contains <span class="math-container">$x$</span> there is an <span class="math-container">$n$</span> such that <span class="math-container">$U_n(x) \subseteq O$</span>.</p>
<p>FYI as a curiosity, if the <span class="math-container">$U_n(x)$</span> are just required to be open (but not always a neighbourhood of <span class="math-container">$x$</span>) but do satify the final condition, these are not called local bases, but local pseudo-bases, and <span class="math-container">$X$</span> is not called first countable but of countable pseudocharacter.</p>
|
202,034 | <p>Is finding the largest prime factor of a number computationally easier than factoring the number into powers of primes? </p>
| binn | 39,264 | <p>no, see</p>
<p><a href="https://mathoverflow.net/questions/104043/saying-things-rapidly-about-integer-factorisations">https://mathoverflow.net/questions/104043/saying-things-rapidly-about-integer-factorisations</a></p>
|
177,774 | <blockquote>
<p>Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio).</p>
</blockquote>
<p>The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work.</p>
| user758556 | 31,023 | <p>The trick you mentioned $\frac{d}{dx}[x^{x}] = \frac{d}{dx} e^{x \ln{x}}$ still works. :)</p>
<p>Apply the chain rule:
$e^{x \ln{x}}\frac{d}{dx}[x \ln{x}]$</p>
<p>And then the product rule:
$e^{x \ln{x}}(\ln{x}+x\frac{1}{x})$</p>
<p>Simplify:
$x^x(1+\ln{x})$</p>
<p>Edit: You wanted the value of the derivative evaluated at $x = 1$, so just substitute in and you get 1.</p>
|
177,774 | <blockquote>
<p>Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio).</p>
</blockquote>
<p>The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work.</p>
| Vincenzo Tibullo | 6,266 | <p>Using the definition:
$$
\begin{align}
f'(1)&=\lim_{x\rightarrow1}\frac{x^x-1}{x-1}\\
&=\lim_{x\rightarrow1}\frac{e^{x\log{x}}-1}{x-1}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{y}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{(1+y)\log(1+y)}\frac{(1+y)\log(1+y)}{y}\\
&=\lim_{t\rightarrow0}\frac{e^{t}-1}{t}\lim_{y\rightarrow0}(1+y)\frac{\log(1+y)}{y}=1
\end{align}
$$
where $y=x-1$, and $t=(1+y)\log(1+y).$ </p>
|
347,385 | <p>Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that:
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$</p>
<p>and how to find the smallest constant $C$ which satisfies
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$</p>
| Georges Henry | 77,984 | <p>write $g=f'$ and observe that $f(0)=-\int_0^{1/2}(1-2t)g(t)dt$ from $\int_0^{1/2}f(x)dx=0.$ Therefore $$(\int_0^{1}f(x)dx)^2=(\int_0^1g(t)\min(t,1-t)dt)^2\leq \int_0^{1}g(t)^2dt\times \int_0^1(\min(t,1-t)^2dt$$ from Schwarz. </p>
|
2,153,743 | <p>I have the relation $R = \{(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)\}$ on the set $\{1,2,3,4\}$. I have to find [1] and [4] but I don't really understand what that means. </p>
<p>I get that $[a]$ is the set of all elements of $A$ related (by $R$) to $a$ so $[a]=\{x\in A : x$ $R$ $a\}$ right? But I don't get the significance of the number 1 or 4 inside of the brackets ([1] and [4]).</p>
| Vlad Z | 287,055 | <p>Your R relationship tells you which two elements are "equal" to each other, for instance 1=1, 1=2, 2=1, and so on...</p>
<p>[1] is a set of all the elements that are equal to 1 (that is the first element of all the pairs in R that have the second element 1), so [1] = {1, 2}.
Also [4]={4}</p>
|
1,812,956 | <blockquote>
<p>Find the equation of the normal to the curve with equation $4x^2+xy^2-3y^3=56$ at the point $(-5,2)$.</p>
</blockquote>
<p>I know that the normal to a curve is $$-\frac{1}{f'(x)}$$
And when I differentiate the curve implicitly I get $$-\frac{8x-y^2}{6y^2}$$</p>
<p>Substituting that into the equation for a normal you get a positive reciprocal $6y^2/(8x-y^2)$
But apparently this is wrong, I'm given the points $(-5,2)$ how are these useful?</p>
| Annalise | 332,290 | <p>The derivative is $8x+y^2+2yx\dfrac{dy}{dx}-9y^2\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=\dfrac{-y^2-8x}{2yx-9y^2}$</p>
<p>Substituting the point $(-5,2)$ we have $\dfrac{dy}{dx}=(-4+40)/(-20-36)=-\frac{9}{14}$</p>
<p>Since we want the normal line, we take the inverse reciprocal of the slope: $\frac{14}{9}$.</p>
<p>We can use the point slope form to find the line: $y-2=\frac{14}{9}(x+5)$</p>
|
3,387,138 | <p>First Definition.
A modular form of level n and dimension -k is an analytic function <span class="math-container">$F$</span> of <span class="math-container">$\omega_1 $</span> and <span class="math-container">$\omega_2$</span> satisfying the following properties :</p>
<ol>
<li><span class="math-container">$F(\omega_1,\omega_2)$</span> is holomorphic and unique for all <span class="math-container">$\omega_1,\omega_2$</span> , where <span class="math-container">$Im(\omega_1/\omega_2)>0$</span> .</li>
</ol>
<p>2.<span class="math-container">$F(\lambda\omega_1,\lambda\omega_2)=\lambda^{-k}F(\omega_1,\omega_2)$</span> for all <span class="math-container">$\lambda\neq0$</span> .</p>
<p>3.<span class="math-container">$F(a\omega_1+b\omega_2,c\omega_1+d\omega_2)=F(\omega_1,\omega_2)$</span> , if <span class="math-container">$a,b,c,d$</span> are rational integers with <span class="math-container">$$ad-bc=1 $$</span> <span class="math-container">$$a\equiv d \equiv1 ,b\equiv c \equiv0 (mod \ n)$$</span></p>
<ol start="4">
<li>The function <span class="math-container">$$F(\tau)=\omega_2^kF(\omega_1,\omega_2)$$</span> has a power series in <span class="math-container">$\tau=\infty$</span> <span class="math-container">$$F(\tau)=\sum_{m=0}^{\infty} c_m e^{2\pi im\tau/n} \ ,Im(\tau)>0$$</span> </li>
</ol>
<p>Second Definition .</p>
<p>Let <span class="math-container">$k \in \mathbb{Z}$</span> . A function <span class="math-container">$f:\mathbb{H} \rightarrow\mathbb{C}$</span> is is a modular form of weight k for <span class="math-container">$SL_2(\mathbb{Z})$</span> if it satisfies the following properties :</p>
<p>1) <span class="math-container">$f$</span> is holomorphic on <span class="math-container">$\mathbb{H} .$</span></p>
<p>2)<span class="math-container">$f|_k M=f$</span> for all <span class="math-container">$M \in SL_2(\mathbb{Z})$</span> .</p>
<p>3) <span class="math-container">$f$</span> has a Fourier expansion <span class="math-container">$$f(\tau)=\sum_{n=0}^{\infty} a_n e^{2\pi i n\tau}$$</span></p>
<p>I do not understand why these two definitions are equivalent .</p>
<p>Thanks for the help .</p>
| hunter | 108,129 | <p>The recipe to convert:</p>
<p>Given <span class="math-container">$f$</span>, set <span class="math-container">$F(\omega_1, \omega_2) = f(\omega_1/\omega_2)$</span>.</p>
<p>Given <span class="math-container">$F$</span>, set <span class="math-container">$f(\tau) = F(\tau, 1)$</span>. </p>
<p>(You'll have to check that if <span class="math-container">$f$</span> has all the properties it's supposed to then the corresponding <span class="math-container">$F$</span> does and vice-versa.)</p>
<p>The "big picture" is that it is a "function" on the space of isomorphism classes of lattice (up to homothety-- this isn't literally true unless <span class="math-container">$k=0$</span>). You can either specify the lattice by giving two generators, or by scaling such that one of the generators is <span class="math-container">$1$</span> and giving the other generator.</p>
|
152,880 | <p>I know that for every $n\in\mathbb{N}$, $n\ge 1$, there exists $p(x)\in\mathbb{F}_p[x]$ s.t. $\deg p(x)=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.</p>
<blockquote>
<p>I am interested in counting how many such $p(x)$ there exist (that is, given $n\in\mathbb{N}$, $n\ge 1$, how many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$).</p>
</blockquote>
<p>I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist $X$ irreducible polynomials of degree $n$ where $X$ is the number of...".</p>
<p>What are your thoughts ?</p>
| Martin Brandenburg | 1,650 | <p>The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$ equals</p>
<p>$$\frac{1}{n} \cdot \sum_{d|n} p^d \mu\left(\frac{n}{d}\right)$$</p>
<p>where $\mu$ is the Möbius function. This follows rather easily from the Möbius inversion formula. You can find details <a href="http://people.virginia.edu/~mve2x/7751_Fall2009/irreducible.pdf">here</a>. Note that this in particular implies the existence of one irreducible polynomial and therefore of the field with $p^n$ elements.</p>
|
311,677 | <p>The problem from the book. </p>
<blockquote>
<p>$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 -y$ </p>
</blockquote>
<p>I understand the solution till this part. </p>
<p>$\ln \vert 6 - y \vert = x + C$ </p>
<p>The solution in the book is $6 - Ce^{-x}$ </p>
<p>My issue this that this solution, from the book, doesn't seem to resolve the issue of the abs value of $\vert 6 - y\vert$ </p>
| ryang | 21,813 | <ol>
<li>Here's a rigorous solution:
<span class="math-container">$$\begin{align}
&\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 -y \\
\frac1{6-y}\dfrac{\mathrm{d}y}{\mathrm{d}x} &= 1 \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ \ \ y=\bbox[pink]{6} \\
\int\frac{\mathrm{d}y}{6-y} &= \int1{\mathrm{d}x} \\
-\ln \lvert 6-y\rvert &= x +C' \\
\lvert 6-y\rvert &= e^{-x-C'} \\
6-y= e^{-x-C'}\ \ \ &\text{or}\ \ \ y-6= e^{-x-C'}\\
y&=\bbox[pink]{6\pm e^{-C'}e^{-x}}.
\end{align}$$</span>
Observe that <span class="math-container">$\pm e^{-C'}$</span> is a nonzero arbitrary constant.<br />
Combining the two sub-answers (in pink) gives the general solution
<span class="math-container">$\bbox[yellow]{y=6+Ce^{-x}}$</span>.</li>
</ol>
<p><br />
2. Alternatively, this solution avoids dealing with the modulus function, and is more compact to boot:
<span class="math-container">$$\begin{align}
\dfrac{\mathrm{d}y}{\mathrm{d}x} &= 6 -y\\
\dfrac{\mathrm{d}y}{\mathrm{d}x} +y &= 6\\
\dfrac{\mathrm{d}y}{\mathrm{d}x}e^x +ye^x &= 6e^x \\
\dfrac{\mathrm{d}}{\mathrm{d}x} (ye^x) &= 6e^x\\
ye^x &= \int6e^x{\mathrm{d}x}\\
&=6e^x+C\\
\bbox[yellow]{y}&\bbox[yellow]{=6+Ce^{-x}}.
\end{align}$$</span></p>
|
3,148,076 | <p>CONTEXT: Challenge question set by uni lecturer for discrete mathematics course</p>
<p>Question: Prove the following statement is true using proof by contradiction: </p>
<p>For all positive integers <span class="math-container">$x$</span>, if <span class="math-container">$x$</span>, <span class="math-container">$x+2$</span> and <span class="math-container">$x+4$</span> are all prime, then <span class="math-container">$x=3$</span>.</p>
<p>I know I'd do this by trying to prove the negation of the statement, but then failing to do so and hence 'contradicting' myself. </p>
<p>I've also found the negation to be that there exists a positive prime integer <span class="math-container">$x$</span> such that +2 and +4 are prime, but <span class="math-container">$x≠3$</span></p>
<p>I'm stuck at the part of the proof where you show that the negation is false.</p>
| NazimJ | 533,809 | <p>Assume not, so <span class="math-container">$\exists x$</span> positive integer s.t. <span class="math-container">$x, x+2, x+4$</span> prime but <span class="math-container">$x\neq3$</span>.</p>
<p>But we can look at the numbers modulo 3, and realize that there are 3 possiblilities:</p>
<p>1) <span class="math-container">$x\equiv 0 (mod 3)$</span>. So <span class="math-container">$x+2\equiv 2 (mod 3)$</span> and <span class="math-container">$x+4\equiv 1 (mod 3)$</span></p>
<p>2) <span class="math-container">$x\equiv 1 (mod 3)$</span>. So <span class="math-container">$x+2\equiv 0 (mod 3)$</span> and <span class="math-container">$x+4\equiv 2 (mod 3)$</span></p>
<p>3) <span class="math-container">$x\equiv 2 (mod 3)$</span>. So <span class="math-container">$x+2\equiv 1 (mod 3)$</span> and <span class="math-container">$x+4\equiv 0 (mod 3)$</span></p>
<p>In any event, one of the 3 numbers must be divisble by 3, which means it is not prime (unless it is equal to 3). Which is only possible if <span class="math-container">$x=3$</span></p>
|
1,436,867 | <p>I don´t know an example wich $ \rho (Ax,Ay)< \rho (x,y) $ $ \forall x\neq y $ is not sufficient for the existence of a fixed point .
can anybody help me? please</p>
| Brian M. Scott | 12,042 | <p>HINT: Take your space to be $\Bbb R$, and try $A(x)=x-f(x)$, where $f$ is positive and increasing and satisfies </p>
<p>$$\frac{f(y)-f(x)}{y-x}<1$$</p>
<p>whenever $x<y$. You can get such an $f$ by tinkering with the arctangent function.</p>
|
2,038,189 | <p>(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)</p>
<p>I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.</p>
<p>However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?</p>
<p>I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".</p>
<p>I have a theory:</p>
<ul>
<li>Invert both sides and then multiply both sides by 2;</li>
</ul>
<p>Is this correct?</p>
| Kaynex | 296,320 | <p>You are allowed to invert both sides, given you invert the entire side, like such:
$$2 + x = \frac1y \rightarrow \frac{1}{2 + x} = y$$
A common mistake is to invert only one term. Note that "inverting" happens because we can multiply both sides of the equation by the product of both sides. Take for example:
$$\frac{1}{2 + x} = \frac{1}{y} \rightarrow \frac{y(2 + x)}{2 + x} = \frac{y(2 + x)}{y}\rightarrow y = 2 + x$$
We did the above by multiplying both sides by y(2 + x), which is allowed, as long as it is done to both sides. You can solve your example in the same way.</p>
|
2,038,189 | <p>(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)</p>
<p>I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.</p>
<p>However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?</p>
<p>I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".</p>
<p>I have a theory:</p>
<ul>
<li>Invert both sides and then multiply both sides by 2;</li>
</ul>
<p>Is this correct?</p>
| q.Then | 222,237 | <p>Well...
$$\begin{equation}\begin{aligned}5 &= \frac{2}{x} &&\text{From question}\\5x&=2 &&\text{Multiply by x on each side}\\ x&=\frac{2}{5} &&\text{Divide by 5 on each side}
\end{aligned}\end{equation}$$
Or
$$\begin{equation}\begin{aligned}\frac{2}{5} &= x &&\text{From question}\\2&=5x &&\text{Multiply by 5 on each side}\\ \frac{2}{x}&=5 &&\text{Divide by x on each side}
\end{aligned}\end{equation}$$</p>
|
229,966 | <p>I want to put a title to the plotlegends I am using. I get a solution <a href="https://mathematica.stackexchange.com/questions/201353/title-for-plotlegends">here</a> which says to use <code>PlotLegends -> SwatchLegend[{0, 3.3, 6.7, 10, 13, 17, 20}, LegendLabel -> "mu"]</code>. But I also want to place the legends where I want like using <code>PlotLegends -> Placed[Range[1, 6, 1], {0.2, 0.3}]</code>.</p>
<p>How can I do both?</p>
<p>Edit:
As the answer was given, the wrapping works, but there is still a problem, i.e my plot looks like this: <a href="https://i.stack.imgur.com/Mad6w.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Mad6w.png" alt="enter image description here" /></a></p>
<p>Where I want to name the legends as "H".
But when I do that Swatchlegend thing it becomes like this :
<a href="https://i.stack.imgur.com/In3YR.png" rel="noreferrer"><img src="https://i.stack.imgur.com/In3YR.png" alt="enter image description here" /></a></p>
<p>I want to keep the markers and colors same.
What should I do?</p>
| Tim Laska | 61,809 | <p>You are probably looking for <a href="https://reference.wolfram.com/language/ref/PointLegend.html" rel="noreferrer"><code>PointLegend</code></a>, but you should provide more details so others can reproduce your results.</p>
<pre><code>mu = {0, 3.3, 6.7, 10, 13, 17, 20};
pl = PointLegend[mu, LegendLabel -> "mu", LegendFunction -> "Frame",
LegendLayout -> "Row", LegendMarkers -> Automatic];
fns = Table[n^(1/p), {p, 7}, {n, 10}];
ListLinePlot[fns, PlotMarkers -> Automatic,
PlotLegends -> Placed[pl, Top]]
</code></pre>
<p><a href="https://i.stack.imgur.com/WE2WK.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WE2WK.png" alt="enter image description here" /></a></p>
|
3,630,421 | <p>If <span class="math-container">$x+y = 5$</span>, <span class="math-container">$xy = 1$</span> and <span class="math-container">$x > y$</span>, then <span class="math-container">$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= ?$</span> The answer key gives for the asnwer <span class="math-container">$\frac{\sqrt{21}}{3}$</span>, <span class="math-container">$\frac{7}{\sqrt{21}}$</span>, <span class="math-container">$\frac{\sqrt{7}}{\sqrt{3}}$</span>, <span class="math-container">$\frac{7}{3}$</span>.
<span class="math-container">$$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} * \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}= \frac{x+2\sqrt{x}\sqrt{y}+y}{x-y}= \frac{5+2\sqrt{1}}{x-y}= \frac{7}{x-y}$$</span>
Once <span class="math-container">$$x=5-y $$</span>
<span class="math-container">$$(5-y)y=1 $$</span>
<span class="math-container">$$y^2-5y+1 $$</span>
<span class="math-container">$$y=\frac{5\pm \:\sqrt{21}}{2} $$</span>
Considering <span class="math-container">$x=5-y$</span>, in order to <span class="math-container">$x>y$</span>, the value of y should be: <span class="math-container">$y=\frac{5-\sqrt{21}}{2} $</span>. Therefore, x will be <span class="math-container">$y=\frac{5+\sqrt{21}}{2}.$</span> Now solving <span class="math-container">$x-y:$</span></p>
<p><span class="math-container">$$x-y\Rightarrow \frac{5+\sqrt{21}}{2}- \frac{5-\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y\Rightarrow \frac{5-5+\sqrt{21}+\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y\Rightarrow \frac{2\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y= \sqrt{21}$$</span>
Back to <span class="math-container">$\frac{7}{x-y}$</span>
<span class="math-container">$$\frac{7}{x-y}= \frac{7}{\sqrt{21}} $$</span>
<span class="math-container">$$∴\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= \frac{7}{\sqrt{21}} $$</span></p>
<p>I did solve it but it wasn't a nice solution. Can anyone give a cleaver and more interesting solution? I am studying for a test and wish there were a quick way to solve this kind of question, I would appreciate some interesting ideas.</p>
| user170231 | 170,231 | <p>If you're amenable to a multivariate approach, you can place the base of the pyramid in the <span class="math-container">$x,y$</span> plane so that the vertices of the base are located at <span class="math-container">$\left(\pm\frac\ell{\sqrt2},0,0\right)$</span> and <span class="math-container">$\left(0,\pm\frac\ell{\sqrt2},0\right)$</span>, and the apex is at <span class="math-container">$\left(0,0,h\right)$</span>, where <span class="math-container">$\ell$</span> is the length of one side of the base and <span class="math-container">$h$</span> is the height of the pyramid.</p>
<p>Then the volume of the pyramid is <span class="math-container">$4$</span> times the volume of the tetrahedral chunks in each of the four octants above the <span class="math-container">$x,y$</span> plane. The chunk in the first octant is bounded by the planes <span class="math-container">$x=0$</span>, <span class="math-container">$y=0$</span>, <span class="math-container">$z=0$</span>, and <span class="math-container">$\frac{\sqrt2}\ell x+\frac{\sqrt2}\ell y+\frac1hz=1$</span>, and the volume of this chunk is</p>
<p><span class="math-container">$$T=\int_0^{\frac\ell{\sqrt2}}\int_0^{\frac\ell{\sqrt2}-x}\int_0^{h-\frac{\sqrt2h}\ell x-\frac{\sqrt2h}\ell y}\mathrm dz\,\mathrm dy\,\mathrm dx=\frac{h\ell^2}{12}$$</span></p>
<p>and so the total volume would be <span class="math-container">$P=4T=\dfrac{h\ell^2}3$</span>.</p>
|
1,363,860 | <p>This problem is for my own exploration, not for class. The problem goes as follows:</p>
<blockquote>
<p>There are <span class="math-container">$n$</span> pairs of people with restraining orders against one another. However, all <span class="math-container">$2n$</span> people are friends with the other <span class="math-container">$2n-2$</span> people (does not include the person they have the restraining order against). So, all <span class="math-container">$2n$</span> people are at a party. Each person wants to stand closer than <span class="math-container">$1$</span> to every person he is friends with. However, he must stand further than <span class="math-container">$1$</span> from the person who has a restraining order against him. We assume that the party is held in <span class="math-container">$\mathbb{R}^2$</span>.</p>
</blockquote>
<blockquote>
<p>How large can <span class="math-container">$n$</span> be before this becomes impossible?</p>
</blockquote>
<p>I conjecture than <span class="math-container">$n$</span> must be finite. However, it would also be useful if <span class="math-container">$n$</span> could be proven to be strictly countable.</p>
<p><strong>EDIT:</strong> This can be done for all positive integers <span class="math-container">$n$</span>. Can this be done for uncountably many pairs?</p>
| Batominovski | 72,152 | <p>I shall prove that this task is not possible if there are uncountably many persons. Let $J$ be an (uncountable) index set, and for $j \in J$, $x_j$ and $y_j$ are two people with a restraining order against one another. Suppose that it is possible to put these people on $\mathbb{R}^d$ for some $d\in\mathbb{N}$ so that the required distance conditions are met. </p>
<p>Let $S$ be the set of points where these people are situated. By abuse of notation, the points of $S$ are also called $x_j$ and $y_j$, where $j\in J$. Clearly, no two persons can be placed at the same location.</p>
<p>We first claim that there exists a point $p$ in $S$ such that, for any positive real number $\epsilon$, the open ball $B(p,\epsilon):=\left\{q\in\mathbb{R}^d\,\Big|\,\|p-q\|_2<\epsilon\right\}$ contains a point in $S$ other than $p$. We shall prove this claim by contradiction. If the opposite of the claim above is true, then for every $p\in S$, there exists $\epsilon_p>0$ such that $B\left(p,\epsilon_p\right)\cap S=\{p\}$. Clearly, for each $p\in S$, there exists a point $q_p\in B\left(p,\epsilon_p\right)$ with rational coordinates. Hence, the map $S\to \mathbb{Q}^d$ sending $p\in S$ to $q_p$ is an injective function from an uncountable set to a countable set, which is a contradiction. Therefore, there exists a point $p \in S$, which we shall assume without loss of generality that $p=y_\alpha$ for some $\alpha \in J$, such that any ball $B(p,\epsilon)$, where $\epsilon>0$, contains a point in $S$ other than $p$.</p>
<p>Now, we look at $x_\alpha$. Let $r:=\left\|x_\alpha-y_\alpha\right\|_2$. By the assumption, $r>1$. Take $\varepsilon:=\frac{r-1}{2}$ so that $\varepsilon>0$. The ball $B\left(y_\alpha,\varepsilon\right)$ contains a point $q \in S$ with $q\neq p$, which must be friends with $x_\alpha$. However, by the Triangle Inequality,
$$\left\|x_\alpha-q\right\|_2\geq \left\|x_\alpha -y_\alpha\right\|_2 - \left\|y_\alpha-q\right\|_2>r-\epsilon=\frac{r+1}{2}>1\,.$$
This is a contradiction. Ergo, the task is impossible.</p>
<p>Now, what is left is when there are infinitely but countably many persons. I conjecture that the task is still impossible.</p>
<p><strong>P.S.</strong> In my proof, I used the $2$-norm $\|\bullet\|_2$ (also known as the Euclidean norm). However, you can use any metric on $\mathbb{R}^d$ that is equivalent to the metric associated to the Euclidean norm.</p>
<hr>
<p><strong>EDIT:</strong> This is my attempt to prove that the task is impossible when $J$ is countable, say $J:=\mathbb{N}$. Suppose contrary that it is possible, and we use the same settings as before. </p>
<p>We now observe that $S$ is a bounded set, whence, due to the Heine-Borel Theorem, its topological closure $\bar{S}$ is compact, making it also sequentially compact. Therefore, the sequence $\left\{x_i\right\}_{i\in\mathbb{N}}$ has a subsequence $\left\{x_{i_j}\right\}_{j\in\mathbb{N}}$ converging to a point $u$ in $\bar{S}$. Now, the sequence $\left\{y_{i_j}\right\}_{j\in\mathbb{N}}$ has a subsequence $\left\{y_{i_{j_k}}\right\}_{k\in\mathbb{N}}$ converging to a point $v\in\bar{S}$. We denote $u_k$ for $x_{i_{j_k}}$ and $v_k$ for $y_{i_{j_k}}$ for all $k\in\mathbb{N}$. Hence, $\displaystyle\lim_{k\to\infty}\,u_k=u$ and $\displaystyle\lim_{k\to\infty}\,v_k=v$. </p>
<p>Note that $\left\|u_i-v_i\right\|_2 >1$ and $\left\|u_i-v_j\right\|_2<1$ for all $i,j\in\mathbb{N}$ such that $i \neq j$. Thus, $$\left\|u-v_j\right\|_2=\displaystyle \lim_{k\to\infty}\,\left\|u_k-v_j\right\|_2\leq 1 \text{ and }\left\|u_i-v\right\|=\displaystyle\lim_{k\to\infty}\,\left\|u_i-v_k\right\|_2\leq 1$$ for all $i,j \in \mathbb{N}$. Ergo, $\|u-v\|_2=\displaystyle\lim_{j\to\infty}\,\left\|u-v_j\right\|_2\leq 1$. However, $$\|u-v\|_2=\displaystyle\lim_{i\to\infty}\,\left\|u_i-v_i\right\|_2\geq 1\,.$$ That is, $\|u-v\|_2=1$.</p>
<p>I shall leave the rest to smart individuals.</p>
|
19,495 | <p>I was told that one of the most efficient tools (e.g. in terms of computations relevant to physics, but also in terms of guessing heuristically mathematical facts) that physicists use is the so called "Feynman path integral", which, as far as I understand, means "integrating" a functional (action) on some infinite-dimentional space of configurations (fields) of a system.</p>
<p>Unfortunately, it seems that, except for some few instances like Gaussian-type integrals, the quotation marks cannot be eliminated in the term "integration", cause a mathematically sound integration theory on infinite-dimensional spaces — I was told — has not been invented yet.</p>
<p>I would like to know the state of the art of the attempts to make this "path integral" into a well-defined mathematical entity.</p>
<p>Difficulties of analytical nature are certainly present, but I read somewhere that perhaps the true nature of path integral would be hidden in some combinatorial or higher-categorical structures which are not yet understood...</p>
<p>Edit: I should be more precise about the kind of answer that I expected to this question. I was not asking about reference for books/articles in which the path integral is treated at length and in detail. I'd have just liked to have some "fresh", (relatively) concise and not too-specialistic account of the situation; something like: "Essentially the problems are due to this and this, and there have been approaches X, Y, Z that focus on A, B, C; some progress have been made in ... but problems remain in ...".</p>
| Theo Johnson-Freyd | 78 | <p>There is a relatively large literature on path integrals. The best book that I know of is Johnson and Lapidus, <em>The Feynman Integral and Feynman's Operational Calculus</em>, 2000. See also the books and papers by Cecile DeWitt-Morette.</p>
|
4,487,654 | <blockquote>
<p>Demonstrate recursively that</p>
<p><span class="math-container">$$\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$$</span></p>
</blockquote>
<p><strong>My work:</strong></p>
<p>Define</p>
<p><span class="math-container">$$a_n = \prod_{k = 0}^n (1 + x^{2^k}) = (1 + x^{2^n})a_{n - 1} \iff a_n - (1 + x^{2^n})a_{n - 1} = 0$$</span>
<span class="math-container">$$A(x) = \sum_{n = 0}^\infty a_nx^n = a_0 + \sum_{n = 1}^\infty a_{n}x^n = 1 + x + \sum_{n = 1}^\infty a_{n}x^n$$</span>
<span class="math-container">$$\implies xA(x) = \sum_{n = 1}^\infty a_{n-1}x^n$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">\begin{align}
A(x) - (1 + x^{2^n})xA(x) &= 1 + x + \sum_{n = 1}^\infty a_{n}x^n - \sum_{n = 1}^\infty (1 + x^{2^n})a_{n-1}x^n\\
A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x + \sum_{n = 1}^{\infty} (a_n - (1 + x^{2^k})a_{n - 1})x^n\\
A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x\\
A(x) &= \frac{1 + x}{1 - x(1 + x^{2^n})}\\
&= \frac{1 + x}{1 - x - x^{2^n + 1}}
\end{align}</span></p>
<p>To find <span class="math-container">$a_n$</span>, I now want to transform <span class="math-container">$A(x)$</span> in the form <span class="math-container">$\sum_{k = 0}^\infty a_nx^n$</span> and then take the limit as <span class="math-container">$n \to \infty$</span>. However, I’m unsure how to do this since I don’t think the expression can be decomposed into partial fractions.</p>
<p><strong>My question:</strong></p>
<ul>
<li>Is my approach correct?</li>
<li>How do I continue?</li>
</ul>
<p><strong>Note:</strong> I know that there are <a href="https://math.stackexchange.com/a/4487579/1072645">other solutions</a>, but I specifically want to see if defining a recurrence can yield a solution.</p>
| Joshua Tilley | 389,601 | <p>Let <span class="math-container">$x_{mn}$</span> be the <span class="math-container">$m^{th}$</span> root of <span class="math-container">$\sin\pi x=x/2n$</span> counting from zero, so <span class="math-container">$m$</span> ranges from <span class="math-container">$0,...,2n-1$</span> as noted. Then <span class="math-container">$a_n=\sum_{m=0}^{2n-1}x_{mn}$</span>. As you said <span class="math-container">$\lim_{n\rightarrow \infty}x_{mn}=m$</span>.</p>
<p>I will make a squeeze argument using some bounds for <span class="math-container">$x_{mn}$</span>. Evaluate <span class="math-container">$\sin\pi x-\frac{x}{2n}$</span> at <span class="math-container">$m+\tfrac{1}{2}$</span> and we get alternating positive and negative answers, showing that there is a root in each interval <span class="math-container">$(m-\tfrac{1}{2},m+\tfrac{1}{2})$</span> for all <span class="math-container">$m=0,...,2n-1$</span>. To get the bound, we also need to show there is at most one root in each interval, which I haven't figured out yet, but shouldn't be too hard. This shows that</p>
<p><span class="math-container">$$m-\tfrac{1}{2}\le x_{mn}\le m+\tfrac{1}{2}$$</span>
and hence</p>
<p><span class="math-container">$$\frac{1}{n^2}\sum_{m=0}^{2n-1}(m-\tfrac{1}{2})\le\frac{1}{n^2}\sum_{m=0}^{2n-1}x_{mn}\le\frac{1}{n^2}\sum_{m=0}^{2n-1}(m+\tfrac{1}{2})$$</span>
But the LHS and RHS both tend to <span class="math-container">$2$</span> as <span class="math-container">$n\rightarrow \infty$</span>.</p>
|
576,379 | <p>I know how to show that $f(x)=x^2$ is uniformly continuous, but I am confused when it is $x^2 +x$</p>
| John Hughes | 114,036 | <p>Quick wise-guy answer: every continuous function on a compact set is uniformly continuous. Heine-Borel tells you that $[0, 1]$ is compact. So your function is unif. continuous on $[0, 1]$, hence on the subset $(0, 1)$. </p>
<p>Alternative: In general, doing an epsilon-delta proof for a differentiable function $f$ at a point $x$, you want to choose something like $\delta = \frac{\epsilon}{f'(x)}$, assuming that $f'$ isn't changing too fast. A good bet is to say $\delta = \frac{\epsilon}{2f'(x)}$, and then say "but also, never pick $\delta > 1$ (or some other amount $A$ with the property that between $x-A$ and $x+A$, $f'(x)$ doesn't change by more than a factor of $2$". The case where $f'(x) = 0$ can be messy, but if you've seen the proof of continuity for $x^2$, you should be OK. </p>
<p>So what's that tell you? Well, your function has derivative $f'(x) = 2x + 1$ on the interval, i.e., the derivative is between $1$ and $3$. So you can safely choose $\delta = \epsilon/6$ to prove continuity at any point $x$ of the interval. </p>
|
1,874,634 | <blockquote>
<p>Corollary (of Schur's Lemma): Every irreducible complex representation of a finite abelian group G is one-dimensional.</p>
</blockquote>
<p>My question is now, why has the group to be abelian? As far as I know, we want the representation $\rho(g)$ to be a $Hom_G(V,V)$, where $V$ is the representation space. Isn't this always the case (i.e. even if the $\rho(g)$ is not abelian) as it is by definition a function $G \rightarrow GL(V)$?</p>
| dasWesen | 306,898 | <p>I had the same question, for maybe longer than a year, but because of a stupid mistake in understanding Schur's. Here it is, in case ( hoping ;) ) that someone else might do the same mistake:</p>
<p>Schur's lemma says something about >any< linear $\psi$ so that </p>
<p>(*) $ \;\;\;\; \psi \rho (g) = \rho (g) \psi$ $\;$ $\forall g$</p>
<p>for some irreducible representation $\rho$. (Correct me if there are more assumptions.) $\psi$ is not assumed to be irreducible. If you would assume it is irreducible, it would follow from Schur's lemma that it's one dimensional, already at this stage. But you don't know that.</p>
<p>(In the corollary, you have $\psi = \rho $, so $\psi$ is irreducible now. But you still need (*), which you get from G's abelianness, as the others have pointed out.)</p>
|
1,491,484 | <p>Let $a,b,x \in Z^+$. Prove that $\operatorname{lcm}(ax,bx) = \operatorname{lcm}(a,b)\cdot x$.</p>
<p>Here are my thoughts: </p>
<p>Let $d = \operatorname{lcm}(ax, bx)$. By definition $ax|d$ and $bx|d$. Now it can be seen that $a|d$ and $b|d$. So, let e = lcm(a,b). e is merely the lcm(ax, bx) (which equals d) multiplied by x. So, ex = d, which means that x $\cdot$ lcm(a,b) = d." </p>
<p>I am not certain of my proof's validity and I feel it is too informal to be considered valid. Any thoughts?</p>
| Oiler | 270,500 | <p>A nice fact that you can use is that for any $a,b\in \mathbb{Z}$, $$\text{lcm}(a,b) = \frac{ab}{\gcd(a,b)}.$$ So then you have that $$\text{lcm}(ax,bx)=\frac{x^2ab}{\gcd(ax,bx)}= \frac{x^2ab}{x \cdot \gcd(a,b)} = x \cdot \left( \frac{ab}{\gcd(a,b)} \right) = x \cdot \text{lcm}(a,b).$$</p>
|
207,778 | <p>I want to save expressions as well as their names in a file.</p>
<pre><code> func[i_] := i;
Do[func[i] >>> out.m,{i,1,3}];
</code></pre>
<p>The output is </p>
<pre><code> cat out.m
1
2
3
</code></pre>
<p>However the desired output is</p>
<pre><code> cat out.m
func[1] = 1;
func[2] = 2;
func[3] = 3;
</code></pre>
<p><code>Save</code> does not save here.</p>
| Fraccalo | 40,354 | <pre><code>list = ToString[#] <> "=" <> ToString[ReleaseHold@#] <> ";" &@
HoldForm@func[#] & /@ Range[3]
(# >>> out.m) & /@ list
</code></pre>
|
1,660,289 | <p>I want to find the line that passes through $(3,1,-2)$ and intersects at a right angle the line $x=-1+t, y=-2+t, z=-1+t$. </p>
<p>The line that passes through $(3,1,-2)$ is of the form $l(t)=(3,1,-2)+ \lambda u, \lambda \in \mathbb{R}$ where $u$ is a parallel vector to the line. </p>
<p>There will be a $\lambda \in \mathbb{R}$ such that $3+ \lambda u_1=-1+t, 1+ \lambda u_2=-2+t, -2+ \lambda u_3=-1+t$. </p>
<p>Is it right so far? How can we continue?</p>
| Julián Aguirre | 4,791 | <p>This is an answer to your last demand:</p>
<pre><code>Do[
q = Prime[n] Times@@@Rest[Subsets[Table[Prime[k], {k, 2, n - 1}]]];
twin = Intersection[Select[q - 4, PrimeQ] + 2, Select[q - 2, PrimeQ]];
Print["n = ", n, " - ", twin, " - ", Length[twin]],
{n, 3, 20}]
</code></pre>
<p>The pattern breaks at $n=9$, since there are $8$ and not $7$ pairs of twin primes. For $n=20$ there are $2674$ pairs.</p>
<p>I have included the possibility of $m=2$, since you use it (although in the question you say $2<m<n$.)</p>
|
363,166 | <p>For valuation rings I know examples which are Noetherian. </p>
<blockquote>
<p>I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? </p>
</blockquote>
<p>I am very eager to know. Thanks.</p>
| Hagen Knaf | 2,479 | <p>Consider the tower of domains</p>
<p>$$
K[x]\subset K[x^{1/2}]\subset \cdots \subset K[x^{1/2^k}]\subset\cdots
$$</p>
<p>where $K$ is a field and $x$ is transcendental over $K$. Every ring in the chain is a polynomial ring in one variable over $K$. Thus the localizations $O_k:=K[x^{1/2^k}]_{P_k}$, where $P_k$ is the prime ideal generated by $x^{1/2^k}$ are discrete valuation rings. Since $P_{k+1}\cap K[x^{1/2^k}]=P_k$ one has $O_k\subset O_{k+1}$ and $M_{k+1}\cap O_k =M_k$ for the maximal ideals $M_k$ of the rings $O_k$.</p>
<p>Now $O:=\bigcup\limits_k O_k$ is a non-noetherian valuation ring of the field $K(x^{1/2^k} : k\in\mathbb{N})$. The value group of an associated valuation is order-isomorphic to the subgroup $\{z/2^k : z\in\mathbb{Z}, k\in\mathbb{N}\}\subset\mathbb{Q}$. Hence this example yields a non-noetherian valuation ring of Krull dimension $1$.</p>
|
363,166 | <p>For valuation rings I know examples which are Noetherian. </p>
<blockquote>
<p>I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? </p>
</blockquote>
<p>I am very eager to know. Thanks.</p>
| mr.bigproblem | 63,621 | <p>Valuation rings that have dimension $\geq 2$ are not Noetherian. The dimension of a valuation ring is equal to the rank of its value group. </p>
<p>To get a simple example of a valuation ring that has dimension $2$, take $R = k[x,y]$, where $k$ is a field. Define the standard valuation $v: k(x,y) \rightarrow \mathbb{Z}^2$ with $v(x) = (1,0) \leq v(y) = (0,1)$, and take the value of a polynomial as the minimal values among those of its monomials. The value group is $\mathbb{Z}^2$, which has rank $2$. So the valuation ring is not Noetherian. This example is "standard" in the sense that it is encountered more often. However, Hagen's example is more interesting.</p>
|
363,166 | <p>For valuation rings I know examples which are Noetherian. </p>
<blockquote>
<p>I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind? </p>
</blockquote>
<p>I am very eager to know. Thanks.</p>
| Uri Brezner | 91,827 | <p>In order to obtain a non Noetherian valuation ring, take $\mathbb{Z}^2$ with the lexicographic order.
Define the valuation $v:k(x,y)^* \to \mathbb{Z}^2$ as follows: for any $a \in k^*$ and $0 \le n,m \in \mathbb{Z}$ set $v(ax^ny^m)=(n,m)$.
For a polynomial $\: f=\sum f_i \in k[x,y]^*$ set $v(f)= \inf \{v(f_0),...,v(f_d)\} $ where the $f_i$ are distinct monomials.
Finally for a rational function $f \in k(x,y)^*$ there are $ g,h \in k[x,y]$ such that $f= \frac{g}{h}$ set $v(f)= v(g)-v(h)$.
The corresponding valuation ring $R_v= \{f \:|\: v(f) \ge 0\}\cup \{0\}$ contains $k[x,y]$, but it also contains $xy^{-1}$ since $(0,0) < (1,-1)$. In fact $xy^n \in R_v$ for any $n \in \mathbb{Z}$.
It follows that $R_v=k[x,y,x/y,x/y^2,x/y^3...]_{(y)}$.</p>
|
70,946 | <p>I'm an REU student who has just recently been thrown into a dynamical system problem without basically any background in the subject. My project advisor has told me that I should represent regions of my dynamical system by letters and look at the sequence of letters formed by the trajectory of a point under the iteration of my map.</p>
<p>He claims that it's a common result that if two points share the same sequence, then this sequence of letters is periodic. I've asked around among some of the other students, and they said that this is sometimes called symbolic dynamics, but none of them remembers this sort of result. I've also searched the internet, but it's possible that my google-fu is weak, since I didn't find any answers that way.</p>
<p>To go one step further, there are obvious cases where it is false- take $S^1\times I$, and encode the regions as $A$ corresponds to $[0,\pi)\times I$ and $B$ corresponds to $[\pi,2\pi)\times I$ with map $f(x,y)=(x+1\mod{2\pi},y)$. Obviously any two points $(x,y)$ and $(x,z)$ with $y\neq z$ will have the same sequence, but since 1 is an irrational multiple of $2\pi$, the trajectory will never be periodic.</p>
<p>I'm interested in the general theory and common techniques applied to the question:</p>
<blockquote>
<p>Represent a dynamical system by associating symbols with regions of the space. When is it true that if two distinct points's trajectories have the same sequence of symbols, then the sequence of symbols is periodic?</p>
</blockquote>
<p>Any answers, examples, or specific references would be greatly appreciated.</p>
| Gjergji Zaimi | 2,384 | <p>Intuitively this should happen for a large class of dynamical systems, but I don't know the right necessary and sufficient conditions. </p>
<p>A class of examples satisfying this is given by polyhedral billiards, where you assign a symbol to each face and correspond orbits to sequences in the obvious manner. It is a result of G Galperin, T Krüger and S Troubetzkoy in their paper "Local instability of orbits in polygonal and polyhedral billiards" (1995), that if a trajectory has a periodic sequence then it must be a periodic trajectory and that if there are two different trajectories with the same symbolic sequences then the symbolic sequences are periodic.</p>
|
1,942,364 | <p>How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.</p>
<p>How can I go about counting the number of squares of each size?</p>
| fleablood | 280,126 | <p>Let the vertices of the $n \times n$ grid by $\{(x,y)| 0\le x \le n; 0 \le y \le n\}$. </p>
<p>(Is that what an $n \times n$ grid is? A $1 \times 1$ has <strong>$4$</strong> vertices and an $n \times n$ grid has $(n+1)^2$ vertices? Or is a $1 \times 1$ grid a single point? I'm assuming the former.)</p>
<p>A $k \times k$ square will have an lower left hand vertex as $(x,y)$ and a upper right hand vertex at $(x+k, y+k)$ with the stipulation $0 \le x; 0 \le y; x+k \le n; y+k \le n$ or in other words: $0 \le x \le n-k; 0 \le y \le n-k$.</p>
<p>There are $n-k+1$ possible options for $x$ and $n-k+1$ options for $y$ so there $(n-k+1)^2$ $k\times k$ squares.</p>
<p>So the total number of squares is $\sum{k=1}^n(n-k+1)^2$. Let $j = n-k+1$ and we have #squares = $\sum_{j = n;j--}^1j^2 = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}6$</p>
<p>Or to put it more simply.... A $k \times k$ square has a side of length $k$. There are $n-(k-1)$ ways to choose this side from the grid that is $n$ long so for squares of lengths $1,2, ...., n$ there are $n, n-1....., 1$ way to choose the horizontal side and $n, n-1,...., 1$ ways to choose the vertical sides. So there are $n^2, (n-1)^2,....., 1^2$ possible $1\times 1, 2\times 2, ...,n \times n$ squares. So there are $\sum k^2 = \frac{n(n+1)(2n+1)}6$ total squares.</p>
|
2,979,103 | <blockquote>
<p>Let <span class="math-container">$S$</span> be the region <span class="math-container">$\{z:0<|z|<\sqrt{2}, \ 0 < \text{arg}(z) <
\pi/4\}$</span>. Determine the image of <span class="math-container">$S$</span> under the transformation</p>
<p><span class="math-container">$$f(z)=\frac{z^2+2}{z^2+1}.$$</span></p>
</blockquote>
<p>I'm facing some difficulties on problems of this nature that includes Möbius transformations. First I can see that under <span class="math-container">$z\mapsto z^2$</span> doubles the argument and squares the modulus. So now we have that our <span class="math-container">$S$</span> is transformed to
<span class="math-container">$$S_1=\{z:0<|z|<2, \ 0 < \text{arg}(z) <
\pi/2\}$$</span></p>
<p>We can now let <span class="math-container">$M(z)=\frac{z+2}{z+1}.$</span> But here we need to parametrize the boundary of <span class="math-container">$S_1$</span>. Thus we have that</p>
<p><span class="math-container">$$\gamma_1=2it, \quad t\in[0,1]$$</span>
<span class="math-container">$$\gamma_2=2t, \quad t\in[0,1]$$</span>
<span class="math-container">$$\gamma_3=2e^{it}, \quad t\in[0,\pi/2]$$</span></p>
<p>So, </p>
<p><span class="math-container">$$M(\gamma_1)=\frac{2it+2}{2it+1}=\frac{4t^2+2}{4t^2+1}-i\frac{2t}{4t^2+1},$$</span></p>
<p><span class="math-container">$$M(\gamma_2)=\frac{2t+2}{2t+1}$$</span></p>
<p><span class="math-container">$$M(\gamma_3)=\frac{2e^{it}+2}{2e^{it}+1}$$</span></p>
<p>substituting <span class="math-container">$t=0$</span> and <span class="math-container">$t=1$</span> in <span class="math-container">$M(\gamma_1)$</span> we get the two points <span class="math-container">$2$</span> and <span class="math-container">$\frac{6}{5}-i\frac{2}{5}$</span>. This means that <span class="math-container">$\gamma_1$</span> is mapped to a circle or line going through those points.</p>
<p>Doing the same for <span class="math-container">$M(\gamma_2)$</span> we get that the two points are <span class="math-container">$2$</span> and <span class="math-container">$\frac{4}{3}$</span>.</p>
<p>However, plotting these points dont really make sense. Also, I'm not sure how to handle <span class="math-container">$M(\gamma_3)$</span>, it gets very messy. This leads me to assume that the method I'm using is very inefficient.</p>
<p>Any way around this?</p>
| TurlocTheRed | 397,318 | <p>I think this might help:</p>
<p><span class="math-container">$$f(z)=\frac{2+z^2}{1+z^2}=1+\frac{1}{1+z^2}$$</span></p>
<p>If <span class="math-container">$z=x+iy$</span>:</p>
<p><span class="math-container">$$\frac{1}{1+z^2}=\frac{1}{(x^2-y^2+1)+2ixy}=\frac{(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$</span></p>
<p>So:</p>
<p><span class="math-container">$$f(z)=\frac{(x^2-y^2+1)^2+4x^2y^2+(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$</span></p>
|
2,817,507 | <p>I came across the following argument in my discrete maths textbook:</p>
<p>Since $n=O(n), 2n=O(n)$ etc., we have:
$$
S(n)=\sum_{k=1}^nkn=\sum_{k=1}^nO(n)=O(n^2)
$$</p>
<p>The accompanying question in the book is: <strong>What is wrong with the above argument?</strong></p>
<p><strong>Attempt:</strong> Performing the summation for $n=2,3$ gives:
$$
\sum_{k=1}^2k(2)\leq\sum_{k=1}^22(2)=2(2^2)=O(2^2)
$$
$$
\sum_{k=1}^3k(3)\leq\sum_{k=1}^33(3)=3(3^2)=O(3^2)
$$
Thus, it seems like the argument is correct. However, I know that I am not correct.</p>
| R.Jackson | 552,485 | <p>$k$ is a variable that runs from $0$ to $n$, therefore in the small cases your textbook showed you, it appears to be $O(n)$, however, the ending terms of the sum are $..., (n-2)n, (n-1)n, n^2$ which are clearly not $O(n)$. Therefore, the argument of the sum is not $O(n)$ but rather ${O(n)}^2$. This means that the sum in question, once evaluated, is $O(n)^3$. </p>
|
2,132,936 | <p>How do you simplify this problem?
$$ \frac {\mathrm d}{\mathrm dx}\left[(3x+1)^3\sqrt{x}\right] $$
$$= \frac {(3x+1)^3}{2\sqrt {x}} + 9\sqrt{x} (3x+1)^2 $$
$$\frac{(3x+1)^2(21x+1)}{2\sqrt x} $$</p>
| Community | -1 | <p>Use the product rule of differentiation, that is $\mathrm {d}(uv) = u\mathrm {dv} + v\mathrm {du} $. We thus get, $$\frac {d}{dx}[(3x+1)^3\sqrt{x}] $$ $$= (3x+1)^3\frac {d}{dx}[\sqrt {x}] + \sqrt {x}\frac {d}{dx}[(3x+1)^3] $$ $$= \frac {(3x+1)^3}{2\sqrt {x}} + 3 (3x+1)^2 (3)\sqrt{x} $$ $$= (3x+1)^2 [\frac {3x+1}{2\sqrt {x}} + 9\sqrt {x}] $$ $$= (3x+1)^2 [\frac {(3x+1) + 2\sqrt {x}[9\sqrt {x}]}{2\sqrt {x}}] $$ $$= \frac{(3x+1)^2 (21x+1)}{2\sqrt {x}} $$ Hope it helps. </p>
|
1,455,348 | <p>Recently, having realized I did not properly internalize it (shame on me!), I went back to the definition of continuity in metric spaces and I found a proposition for which I was looking for a proof.</p>
<p>Here there is the result and my "proof" (in the hope to get rid of the quotation marks).</p>
<p><em>[In general, I use <span class="math-container">$N_{\varepsilon, X} (x)$</span> to denote an open <span class="math-container">$\varepsilon$</span>-neighbourhood of <span class="math-container">$x \in X$</span>.]</em></p>
<blockquote>
<p><strong>Proposition:</strong> Let <span class="math-container">$\phi \in \mathbb{R}^X$</span> be a continuous function, with <span class="math-container">$X$</span> an arbitrary metric space. Then, the set <span class="math-container">$\{ x \mid \phi(x) \geq \alpha \}$</span> is closed.</p>
<p><em>Attempted proof:</em><br />
Let <span class="math-container">$\alpha$</span> be an arbitrary real number. We establish the result by showing that <span class="math-container">$\{ x \mid \phi(x) < \alpha \}$</span> is an open set in <span class="math-container">$X$</span>. Notice that for every <span class="math-container">$x \in X$</span>, if <span class="math-container">$\phi (x) < \alpha$</span>, then there is a <span class="math-container">$\varepsilon > 0$</span> such that there is an open neighbourhood <span class="math-container">$N_{\varepsilon, \mathbb{R}} (\phi (x)) < \alpha$</span>. Let <span class="math-container">$z \in X$</span> be arbitrary and such that <span class="math-container">$\phi (z) < \alpha$</span>. Hence, by the definition of continuity and the fact that <span class="math-container">$\phi$</span> is continuous, there is a <span class="math-container">$\delta (z, \varepsilon) > 0$</span> such that</p>
<p><span class="math-container">$$N_{\delta, X} (z) \subseteq \phi^{-1} ( N_{\varepsilon, \mathbb{R}} ( \phi (z)). $$</span></p>
<p>Hence, being <span class="math-container">$z \in X$</span> arbitrary, the proposition follows.</p>
</blockquote>
<p><em>Is this proof correct?</em></p>
<p>As always, any feedback is more than welcome.<br />
Thank you for your time.</p>
<hr />
<p><strong>Edit:</strong><br />
I know it is possible to proceed, as hinted by air in a comment below, through the fact that the if a function is continuous, then the preimage of a closed set is closed. However, I find this solution a bit too <em>topological</em>, in the sense that I really would like to know about this <span class="math-container">$\varepsilon - \delta$</span> proof, which – to me – has a stronger metric flavour.</p>
| layman | 131,740 | <p>I'm not sure why you need an $\epsilon$-$\delta$ proof when we aren't trying to prove continuity. We are just trying to prove $\{x \mid \phi(x) \geq a \}$ is closed if $\phi$ is continuous.</p>
<p>It suffices to show the complement is open. But since $\{x \mid \phi(x) \geq a \} = \phi^{-1}( [a, \infty) )$, and $f^{-1}(B^{c}) = (f^{-1}(B))^{c}$ for every function $f$ and every set $B$ in the codomain of $f$, we have:</p>
<p>$[a,\infty)^{c} = (-\infty, a)$ which is open. Since $\phi$ is continuous, $\phi^{-1}( (-\infty, a))$ is open. But $\phi^{-1}( (-\infty, a)) = \phi^{-1}( [a, \infty)^{c}) = (\phi^{-1}([a, \infty))^{c}$.</p>
<p>Thus, $(\phi^{-1}([a, \infty))^{c}$ is open, and so its complement, $\phi^{-1}([a,\infty))$ is closed. But $\phi^{-1}([a,\infty))$ is just another notation for $\{x \mid \phi(x) \geq a \}$, so our set is closed, as desired.</p>
|
652,660 | <p>Show $\lnot(p\land q) \equiv \lnot p \lor \lnot q$</p>
<p>this is my solution . Check it please </p>
<p><img src="https://i.stack.imgur.com/1y7DB.jpg" alt="enter image description here"></p>
| Juan | 124,049 | <p>Your solution is right but needs a few adjustments.
In the second line of $(\Rightarrow)$ it should be "is" not "are" in both cases.
Showing that "$p$ is true exactly when $q$ is true" is not enough to guarantee that $p$ and $q$ are equivalent, do you know why ?</p>
|
661,026 | <p>prove or disprove this
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$</p>
<p>this problem is from when Find this limit
$$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}^3}{\displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}^3}=\dfrac{1}{8}?$$</p>
<p>first,follow I can't
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$
.Thank you for you help</p>
| Pifagor | 125,067 | <p>Find the derivative ov the left side of your first equation and use it's zeros to find the minima of that function. They will, hopefully, all be positive :)</p>
|
661,026 | <p>prove or disprove this
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$</p>
<p>this problem is from when Find this limit
$$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}^3}{\displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}^3}=\dfrac{1}{8}?$$</p>
<p>first,follow I can't
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$
.Thank you for you help</p>
| Hagen von Eitzen | 39,174 | <p>For negative $x$, all summands are $\ge 0$ so $f(x)\ge 100$.
For $0\le x \le 4$, $f(x)\ge 100-x^3-x\ge 100-64-4>0$.
Note that
$$ f(x+1)=\frac12x^4+x^3-2x+\frac{197}2\ge (x^2-2)x+\frac{197}2>0$$
for $x>\sqrt 2$, so $f(x)>0$ for $x>\sqrt 2-1$. This covers all of $\mathbb R$.</p>
|
2,113 | <p>With respect to the stated reason for closure, I'd like to get some clarification as to what, precisely, "too localized" encompasses (at least a definition, or explanation, that is more specific and objective than the current "definition"). It just strikes me that some questions which might appear to some as being "too localized" are, in fact, of greater interest to the "world-wide" audience than some of the questions that are accepted and answered (and hence deemed not too localized), which in fact may only be of interest to PhD candidates, if not PhD researchers. While I'm not objecting at all to questions of the latter sort, you must admit that questions of the latter sort such are likely not of any interest whatsoever to the vast majority of those who participate in the world wide internet? (some questions, perhaps, only of interest to only a small community of a significant minority of mathematicians?) I'm not meaning to be argumentative, or to suggest that questions of the latter sort be "closed." I am simply a bit confused regarding some questions that <em>are</em> judged to be "too localized" and I'm attempting to reconcile this action with the given definition, as it stands.</p>
| Qiaochu Yuan | 232 | <blockquote>
<p>I think you're right - used for lack of "better" options. But if it doesn't satisfy "too localized", and there are no other options or available justifications for closing, then a question shouldn't be closed. It's unfair to the OP to justify closing by giving an arbitrary and uninformative reason for doing so.</p>
</blockquote>
<p>I disagree. Not every close reason is available on every SE site, I don't know who decided which reasons should be available where, and I don't see why we shouldn't close an inappropriate question just because the software prevents us from saying why. For example, on other SE sites there is a "noise or pointless" close reason, which I would love to have here, but which we don't have, so I have to make do with whatever I think the closest approximation is at any given time. </p>
|
3,043,780 | <p><a href="https://i.stack.imgur.com/h1M7D.png" rel="nofollow noreferrer">the image shows right-angled triangles in semi-circle</a></p>
<p>In Definite Integration, we know that area can be found by adding up the total area of each small divided parts.</p>
<p>So, base on the Definite Integration, we may say the area of circle is equal to
<span class="math-container">$\sum^{n}_{k=1}(h_k)$</span></p>
<p>And we also know that <span class="math-container">$A_k=2r(h_k)(1/2)=rh_k$</span> while <span class="math-container">$A$</span> is refer to the right-angled triangle's area.</p>
<p>so that, <span class="math-container">$A_k(1/r)=h_k$</span></p>
<p>As a result this equation comes out with fixed position of diameter:</p>
<p><span class="math-container">$\sum^{n}_{k=1}(h_k)=\pi r^2$</span></p>
<p><span class="math-container">$\sum^{n}_{k=1}(A_k)(1/r)=\pi r^2$</span></p>
<p><span class="math-container">$\sum^{n}_{k=1}(A_k)=\pi r^3$</span></p>
<p>I wish to know whether I am correct or not, thanks.</p>
| David Holden | 79,543 | <p>you might like to try an alternative method that may be more transparent and robust. </p>
<p>supposing Q lies within ABC. a test probe P moving from A towards B should initially observe a decrease in the distance <span class="math-container">$PQ = r_{\lambda}$</span>. if we parameterize the position of the probe as
<span class="math-container">$$
P_{\lambda} = \lambda A + (1-\lambda) B
$$</span></p>
<p>then for an internal point <span class="math-container">$Q$</span> we should have:</p>
<p><span class="math-container">$$
\frac{d|P_\lambda Q|^2}{d\lambda}\bigg|_{\lambda= 0} \lt 0
$$</span></p>
<p>if <span class="math-container">$A = (x_1,y_1), B= (x_2, y_2), C=(x_3,y_3)$</span>, and <span class="math-container">$Q=(x_0, y_0)$</span> then this test computes as:</p>
<p><span class="math-container">$$
(a_2-a_1)(a_1-a_0) +(b_2-b_1)(b_1-b_0) \lt 0
$$</span></p>
<p>with similar tests for the other two sides, BC and CA, obtained by cyclic permutation.</p>
|
787,358 | <p>Consider the equation: ay'' +by'+cy=0</p>
<p>If the roots of the corresponding characteristic equation are real, show that a solution to the differential equation either is everywhere zero or else can take on the value zero at most once.</p>
<p>hmm I have no idea how to do this one, I think it might have to do something with repeated roots?, but im now sure, any tips/solutions to this one? thanks! :D</p>
| RRL | 148,510 | <p>The general solution is of the form $C_1\, \exp(r_1 x) + C_2 \, \exp(r_2x)$ where $r_1$ and $r_2$ are, in this case, the real roots of $ar^2 + br +c = 0$.</p>
<p>The roots are $ r_1 = -\alpha + \beta$ and $ r_2 = -\alpha - \beta $ where $\alpha = \frac{b}{2a}$ and $\beta = \sqrt{\alpha^2 -\frac{c}{a}}$.</p>
<p>So the general solution is</p>
<p>$$C_1\exp(- \alpha x)\exp(\beta x) + C_2 \exp(- \alpha x)\exp(-\beta x)$$</p>
<p>If $C_1 = C_2 = 0$ then the solution is everywhere zero. If $C_1 = 0$ and $C_2 \neq 0$ or if $C_1 \neq 0$ and $C_2 = 0$ then the solution is never zero because the exponential function is positive. </p>
<p>Now suppose $C_1$ and $C_2$ are both non-zero. Setting the general solution to $0$ we see</p>
<p>$$\exp(-\alpha x)\left[C_1 \exp(\beta x) + C_2 exp(-\beta x)\right] = 0$$</p>
<p>Dividing both sides by $\exp(-\alpha x)$ and rearranging we get</p>
<p>$$\exp(2\beta x) = -\frac{C_2}{C_1}$$</p>
<p>which has one and only one solution for $x$ if the constants have opposite signs. In this case the solution of the differential equation is $0$ at only this single value of x.</p>
|
2,466,556 | <p><strong>Solution:</strong> </p>
<p>The vectors $\vec{AB}=(3,2,1)-(0,1,2)=3,1,-1$ and $\vec{AC}=(4,-1,0)-(0,1,2)=(4,-2,-2),$ are two direction vectors of the plane. A normal vector $\vec{n}$ to the plane is then given by $$\vec{n}=\vec{AB}\times\vec{AC}=(-4,2,-10).$$</p>
<p>Since $A$ is a point on the plane, we get</p>
<p>$$\vec{n}\cdot (x-0,y-1,z-2)=-4x+2(y-1)-10(z-2)=-4x+2y-10z+18=0.$$</p>
<hr>
<p>I dont understand the first part of the last equation. </p>
<ol>
<li>What vector is $(x-0,y-1,z-2)$?</li>
<li>Why do they take the dot-product of the above with the normal vector? (How does it give the equation of the plane?)</li>
</ol>
| Anurag A | 68,092 | <p>The idea of finding the equation of a plane is to have an equation that can help to locate any point $P(x,y,z)$ on the plane. In order to do that consider the vector $\vec{AP}=(x,y,z)-(0,1,2)$. This vector lies in the given plane. Now that we have a normal vector $\hat{n}$ (which by definition is orthogonal to any vector in the plane) we know that $\hat{n} \cdot \vec{AP}=0$. This gives the equation we are looking for.</p>
|
2,466,556 | <p><strong>Solution:</strong> </p>
<p>The vectors $\vec{AB}=(3,2,1)-(0,1,2)=3,1,-1$ and $\vec{AC}=(4,-1,0)-(0,1,2)=(4,-2,-2),$ are two direction vectors of the plane. A normal vector $\vec{n}$ to the plane is then given by $$\vec{n}=\vec{AB}\times\vec{AC}=(-4,2,-10).$$</p>
<p>Since $A$ is a point on the plane, we get</p>
<p>$$\vec{n}\cdot (x-0,y-1,z-2)=-4x+2(y-1)-10(z-2)=-4x+2y-10z+18=0.$$</p>
<hr>
<p>I dont understand the first part of the last equation. </p>
<ol>
<li>What vector is $(x-0,y-1,z-2)$?</li>
<li>Why do they take the dot-product of the above with the normal vector? (How does it give the equation of the plane?)</li>
</ol>
| zipirovich | 127,842 | <ol>
<li><p>The vector $\langle x-0,y-1,z-2\rangle$ is a vector connecting a generic point $P(x,y,z)$ lying in the desired plane with the point $A(0,1,2)$, which is also in this plane. Since both of them are in this plane, the vector connecting them $\overrightarrow{AP}=\langle x-0,y-1,z-2\rangle$ is in this plane.</p></li>
<li><p>The word "normal" means "perpendicular" in this context. So saying "$\overrightarrow{n}$ is normal to the plane" means that $n$ is perpendicular to the plane, and therefore it's perpendicular to any vector that lies in this plane. So we know that $\overrightarrow{n}\perp\overrightarrow{AP}$ must be true. And then there's the property of dot products: two vectors are perpendicular if and only if their dot product is zero.</p></li>
</ol>
|
881,831 | <p>It is trivial that a group $G$ is abelian if and only if every subgroup of $G$ with two generators is abelian (i.e., any two elements commute).</p>
<p>If $G$ is a nilpotent group, every subgroup with two generators must be nilpotent. Is the reciprocal true? More precisely:</p>
<blockquote>
<p>Let $G$ be a group and suppose that every subgroup of $G$ generated by
two elements is nilpotent (with uniformly bounded class if needed). Is
$G$ necessarily nilpotent?</p>
</blockquote>
| Mikko Korhonen | 17,384 | <p>If we do not require that the class of a subgroup generated by two elements is bounded by some fixed constant, here is one example.</p>
<p>Consider the infinite direct sum $G = G_1 \oplus G_2 \oplus G_3 \oplus \cdots$ where $G_i$ is nilpotent of class $i$. </p>
<p>Then $G$ is not nilpotent, since it has nilpotent subgroups of arbitrarily large class. But any finitely generated subgroup of $G$ is contained in $G_1 \oplus G_2 \oplus \cdots \oplus G_n$ for some $n$, and this subgroup is nilpotent of class $n$. In particular any subgroup generated by two elements is nilpotent.</p>
|
1,903,235 | <p>According to Wikipedia, </p>
<blockquote>
<p>Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions</p>
</blockquote>
<p>However, the article on Euclidean space states already refers to </p>
<blockquote>
<p>the n-dimensional Euclidean space.</p>
</blockquote>
<p>This would imply that Hilbert space and Euclidean space are synonymous, which seems silly.</p>
<p>What exactly is the difference between Hilbert space and Euclidean space? What would be an example of a non-Euclidean Hilbert space?</p>
| Community | -1 | <p>Hilbert space: a vector space together with an inner product, which is a Banach space with respect to the norm induced by the inner product </p>
<p>Euclidean space: a subset of $\mathbb R^n$ for some whole number $n$</p>
<p>A non-euclidean Hilbert space: $\ell_2(\mathbb R)$, the space of square summable real sequences, with the inner product $((x_n),(y_n)) = \sum_{n=1}^{\infty}x_n y_n$</p>
|
1,903,235 | <p>According to Wikipedia, </p>
<blockquote>
<p>Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions</p>
</blockquote>
<p>However, the article on Euclidean space states already refers to </p>
<blockquote>
<p>the n-dimensional Euclidean space.</p>
</blockquote>
<p>This would imply that Hilbert space and Euclidean space are synonymous, which seems silly.</p>
<p>What exactly is the difference between Hilbert space and Euclidean space? What would be an example of a non-Euclidean Hilbert space?</p>
| Luthier415Hz | 924,287 | <p>An Euclidean space is a normed linear space, that is, it has a norm and its elements are linear functions.</p>
<p>An Euclidean space has an inner product (scalar product):</p>
<p><span class="math-container">$$(x_\alpha,x_\beta)=0$$</span> for orthogonal elements</p>
<p>and</p>
<p><span class="math-container">$$(x_\alpha,x_\alpha)=1$$</span></p>
<p>This scalar product must satisfy the following:</p>
<p><span class="math-container">$$
\begin{array}
f\bullet\ being\ not\ negative (x,x>0)\\
\bullet\ being\ symmetric (x,y)=(y,x)\\
\bullet\ being\ linear: \lambda x, y)=\lambda(x,y)\\
\bullet\ (x,y+z)=(x,y)+(x,z)
\end{array}
$$</span></p>
<p>We say therefore that <em>any</em> space equipped with an inner product satisfying these properties is called Euclidean space.</p>
<p>Euclidean space is therefore a general term for a normed linear space with an inner product, where the norm can be</p>
<p><span class="math-container">$$d(x,y)=||x-y||=\sqrt{(x-y)^2}$$</span></p>
<p>The norm is a distance in the linear normed space, between two points.</p>
<p>Now, a Hilbert space is an example of an Euclidean space. It has an inner product, <em>however it is of infinite many dimensions</em>. For instance, considering a particle in quantum mechanics as a dimension, one can form a Hilbert space of 10 dimensions for a system of 10 particles, where the particle must satisfy the orthogonality principle of the inner product stated at the top. Therefore, all the wavefunctions for the particles given above give an inner product equal to zero, since they have the same angle between one another. Have a look at the atomic orbital chart and see how they have the same angle preserved between each wavefunction.</p>
<p>This wavefunction model is then said to satisfy the conditions for the inner product, and you can have infinitely many of them. If you used an Euclidean space to describe a system of particles, you were restricted to 3 particles, that is, the three dimensions Euclidean space is restricted to. Therefore, you will see that Hilbert space is commonly used in quantum mechanics, with all its properties of the Euclidean space, but with its infinitely many dimensions.</p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| Qiaochu Yuan | 232 | <p>One of the most important results you learn in a first course on abstract algebra is <a href="http://en.wikipedia.org/wiki/Burnside%27s_lemma">Burnside's lemma</a>, which has many applications in combinatorics and number theory. <a href="http://qchu.wordpress.com/2009/06/13/gila-i-group-actions-and-equivalence-relations/">Some time ago</a> I wrote a series of blog posts leading up to a powerful corollary of Burnside's lemma called the <a href="http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem">Polya enumeration theorem</a> (including several applications, so you should look in those posts for them), which can be used to count many things; it was originally used to count chemical compounds. </p>
<p>The Polya enumeration theorem in turn can be used to prove a powerful result in combinatorics called the <a href="http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/">exponential formula</a>, which gives you an enormous amount of information about permutations. For example, using the exponential formula you can prove results like </p>
<blockquote>
<p>The number of fixed points of a random permutation of $n$ elements is asymptotically <a href="http://en.wikipedia.org/wiki/Poisson_distribution">Poisson distributed</a> with parameter $\lambda = 1$ as $n \to \infty$</p>
</blockquote>
<p>relatively easily.</p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| Makoto Kato | 28,422 | <p>Diophantine equations of the type $m = ax^2 + bxy + c^2$ can be solved by using algebraic number theory on quadratic number fields.</p>
<p>As for connections between integral binary quadratic forms and quardratic number fields, here are some examples:</p>
<p><a href="https://math.stackexchange.com/questions/191830/discriminant-of-a-binary-quadratic-form-and-an-order-of-a-quadratic-number-field">Discriminant of a binary quadratic form and an order of a quadratic number field</a></p>
<p><a href="https://math.stackexchange.com/questions/191933/a-binary-quadratic-form-and-an-ideal-of-an-order-of-a-quadratic-number-field">A binary quadratic form and an ideal of an order of a quadratic number field</a></p>
<p><a href="https://math.stackexchange.com/questions/192217/bijection-between-an-ideal-class-group-and-a-set-of-classes-of-binary-quadratic">Bijection between an ideal class group and a set of classes of binary quadratic forms.</a></p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| Eric O. Korman | 9 | <p>The product of all the residue classe modulo a prime $p$ is to -1 (mod p).</p>
|
1,511,078 | <p><strong>Show that the product of two upper (lower) triangular matrices is again upper (lower) triangular.</strong></p>
<p>I have problems in formulating proofs - although I am not 100% sure if this text requires one, as it uses the verb "show" instead of "prove". However, I have found on the internet the proof below but my problem is not just that I can't do one by myself, but also that it happens that I don't understand a proof which is already written - thing which let me think that I should quit these studies.</p>
<p><strong>My first question is: how to choose the right strategy for proving something, in this case as in others?</strong>. I guess it is also matter of interest...Interest for numbers and their properties. I've never had such interest, frankly - although, perhaps, it is not nice to say this here. The reason for which I started studying math and some physics over an year and a half ago is because I dreamed about and still dreaming to do a kind of research for which I am interested and I need scientific background.</p>
<p>Sorry for the digression but I am so upset and I need some general advice too. Coming back to the topic of the present proof, <strong>my second question is: could you explain me in plain English the idea and logic behind the following proof, please? Is this proof general enough to cover also my case and what changes do I need to add for it?</strong> - I ask this because the text to which I refer speaks about upper and lower triagular matrices, while the text of the proof I report here speaks about upper triangular matrices only. I also notice that the following proof considers only square matrices but also rectangular matrices can be upper/lower triangular.</p>
<p>The only thing I know about all this is what upper/lower triangular matrices are and how to perform multiplication between matrices. This is the proof found on the internet:</p>
<p>"Suppose that U and V are two upper triangular <em>n × n</em> matrices. By the row-column rule for matrix multiplication we know that the <em>(i,j)-th</em> entry of the product UV is $ui1v1j + ui2v2j +···+ uinvnj$. We need to show that if i > j then this expression evaluates to 0. In fact, we will show that every term $uikvkj$ of this expression evaluates to 0. To prove this, we consider two cases: • If i > k then $uik = 0$ since U is upper triangular. Hence $uikvkj = 0$. • If k > j then $vkj = 0$ since V is upper triangular. Hence $uikvkj = 0$. Since i > j, for every k weeither have $i > k$ or $k > j$ (possibly both) so the set wo cases cover all possibilities for k."</p>
| darkmoor | 91,474 | <p>I was interested on the same question, so allow me to exploit my logic, hopping of course to get comments for possible flaws. Suppose you have two lower triangular matrices $\mathbf{L}_1$ and $\mathbf{L}_2$ illustrated bellow</p>
<p>$$\mathbf{L}_1 =
\begin{bmatrix}
l_{11}^{(1)} & l_{12}^{(1)} & \dots & \dots & l_{1n}^{(1)} &\\
& l_{22}^{(1)} & l_{23}^{(1)} & \dots & \vdots &\\
& & l_{33}^{(1)} & & \vdots &\\
& & & \ddots & \vdots &\\
& & & & l_{nn}^{(1)} &\\
\end{bmatrix}~~~~~\mathbf{L}_2 =
\begin{bmatrix}
l_{11}^{(2)} & l_{12}^{(2)} & \dots & \dots & l_{1n}^{(2)} &\\
& l_{22}^{(2)} & l_{23}^{(2)} & \dots & \vdots &\\
& & l_{33}^{(2)} & & \vdots &\\
& & & \ddots & \vdots &\\
& & & & l_{nn}^{(2)} &\\
\end{bmatrix}$$ </p>
<p>We want to prove that the following product is a lower triangular matrix,</p>
<p>$$\mathbf{L}_1 \mathbf{L}_2 = \mathbf{L}_1 \big[ \mathbf{l}_1, \mathbf{l}_2^{(2)}, \dots, \mathbf{l}_n^{(2)} \big] = \big[ \mathbf{L}_1 \mathbf{l}_1^{(2)}, \mathbf{L}_1 \mathbf{l}_2^{(2)}, \dots, \mathbf{L}_1 \mathbf{l}_n^{(2)} \big]$$</p>
<p>As we can see, the $k$-th column of product matrix $\mathbf{L}_1 \mathbf{L}_2$ is given by $\mathbf{L}_1 \mathbf{l}_k^{(2)}$ which is the linear combination of the $\mathbf{L_1}$ matrix columns with coefficients defined by the $k$-th column vecor $\mathbf{l}_k^{(2)}$. <strong>Each of the product matrix columns $\big(\mathbf{L}_1 \mathbf{l}_k^{(2)}\big)$ have possible non-zeros entries only above the $k$-th element.</strong> </p>
<p>This is because, the new columns are linear combinations of the first $k$ columns $\mathbf{l}_k^{(1)}$ which by their turn have possible non-zero values above their $k$-th entry. This property comes form the fact that columns $\mathbf{l}_k^{(2)}$ have zero entries after their $k$-th element. </p>
<p>$\mathcal{Thanks~for~reading}$. </p>
|
3,935,811 | <p>While solving a bigger problem, I stumbeled upon a system of parametric equations
<span class="math-container">$$
\left\{
\begin{array}{ll}
\dfrac{x-a}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{x-c}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0\\
\dfrac{y-b}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{y-d}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0
\end{array}
\right.
$$</span>
I need to solve it. I don't actually need all of the solutions to this system, one is enough.</p>
<p>Thanks, in advance.</p>
| Claude Leibovici | 82,404 | <p>This seems to be : find the extremum of function
<span class="math-container">$$f(x,y)=\sqrt{(a-x)^2+(b-y)^2}+\sqrt{(c-x)^2+(d-y)^2}$$</span> If you do not have any constraint, it could be any point along the line explained by @Quanto in his/her answer.</p>
|
762,651 | <p>I have to prove that "any straight line $\alpha$ contained on a surface $S$ is an asymptotic curve and geodesic (modulo parametrization) of that surface $S$". Can I have hints at tackling this problem? It seems so general that I am not sure even how to formulate it well, let alone prove it. Intuitively, I imagine that the normal $n_{\alpha}$ to the line/curve is perpendicular to the normal vector $N_{S}$ to the surface $S$, thus resulting in the asymptoticity; alternatively, a straight line has curvature $k = 0$ everywhere, and so the result follows. Is this reasoning adequate for a proof of the first part? I also realize that both geodesics and straight lines are the paths of shortest distance between two points on given surfaces (here, both $S$), thus the straight line must be a geodesic of any surface which contains it; should I quantify this statement, though?</p>
<p>Let $\mathbb{H}^2 = \{(x, y) \in \mathbb{R}^2: y>0 \}$ be the hyperbolic plane with the Riemannian metric $(\mathbb{d}s)^2 =\frac{(\mathbb{d}x)^2+(\mathbb{d}y)^2}{y^2}$. Consider a "square" $P = \{ (x, y) \in \mathbb{H}^2: 1 \leq x,y \leq 2 \}$. I need to calculate the geodesic curvature of the sides of $P$ and, for Gaussian curvature $K$ of $P$, I have to calculate $\int_{P} (K) \mathbb{d}\sigma$, where $\mathbb{d}\sigma$ is the area measure element of $\mathbb{H}^2$. Just hints as to how to start would be helpful. (I see that I have the first fundamental form, from which I can derive the coefficients $E$, $F$, and $G$ and thereby (hopefully easily) Christoffel symbols and an expression for area, but I do not see how any of this takes the actual square into account. Only the coordinates at which I evaluate these quantities seem to come come from the square! But I would still like detailed examples of even these things, please.)</p>
| evil999man | 102,285 | <p>Generally $$a>b \implies f(a)> f(b), \forall \text{ strictly increasing f(x)}$$</p>
<p>Your statement is just for $x^2$ which is true for $\Bbb R^+$. You have to state this fact while writing it down.</p>
<p>Your proof is also correct...like M.B. replied</p>
|
410,105 | <p>I have this recurrence relation:</p>
<p>$$
R(1)=1, RE(1)=0, EE(1)=0$$</p>
<p>$$a(n)=R(n) + RE(n)$$</p>
<p>$$R(n)=EE(n-1)+RE(n-1),$$$$ RE(n)=R(n-1),$$$$ EE(n)=RE(n-1)
$$</p>
<p>How do I get $a(15)$?
What kind of method do I use?</p>
| Boris Novikov | 62,565 | <p>Substitute $EE(n)$ from the last equation:</p>
<p>$$R(n)=RE(n-2)+RE(n-1).$$</p>
<p>Then substitute $RE(n)$:</p>
<p>$$R(n)=R(n-3)+RE(n-2) \ \ \ (1)$$</p>
<p>$$a(n)=R(n) + R(n-1) \ \ \ (2)$$</p>
<p>Now solve the recurrent equation ($1$) and then ($2$).</p>
|
4,402,262 | <p>A class of 24 pupils consists of 11 girls and 13 boys. To form the class committee, four of the pupils are chosen at random as "Chairperson", "Vice-Chairperson", "Treasurer", and "Secretary". Find the number of ways the committee can be formed if<br>
(i) the committee consists of at least one girl and at least one boy,<br>
(ii) the "Treasurer" and "Secretary" are both girls,<br>
(iii) The teacher requires a group of four students to represent the class in a student survey. Find the number of ways this group of students can be selected if there must be at least 1 girl and at most 2 boys.</p>
<p>My answers:</p>
<p>(i) GBBG, GBBB, GBGG</p>
<p>(11C2 x 13C2) + (11C1 x 13C3) + (11C3 x 13C1) = 9581</p>
<p>(ii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>(iii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>I have checked the correct answers, it shows that (i) 229944 (ii) 50820 (iii) 6765</p>
<p>I don't see my (i) is wrong, is it the correct answer for (i) of 229944 as incorrect?</p>
<p>For (ii) why it is using 11P2 x 22P2 = 50820 for the answer? Why this is a permutation question?</p>
| G Tony Jacobs | 92,129 | <p>As soon as you name the different positions in the committee, it's a permutation question, not a combination question. If Sally, Betty, Joe, and Mason are the four people on the committee, having Sally as Chair, Betty as Vicechair, Joe as Treasurer and Mason as Secretary is a different outcome from some other assignment of jobs to those same individuals. That's why the correct answer for part (i) is equal to your answer, times <span class="math-container">$4!$</span>.</p>
<p>The only part of this question that calls for combinations, not permutations, is part (iii), because it's just a group of <span class="math-container">$4$</span> students being chosen, as opposed to four distinct positions being filled.</p>
|
4,402,262 | <p>A class of 24 pupils consists of 11 girls and 13 boys. To form the class committee, four of the pupils are chosen at random as "Chairperson", "Vice-Chairperson", "Treasurer", and "Secretary". Find the number of ways the committee can be formed if<br>
(i) the committee consists of at least one girl and at least one boy,<br>
(ii) the "Treasurer" and "Secretary" are both girls,<br>
(iii) The teacher requires a group of four students to represent the class in a student survey. Find the number of ways this group of students can be selected if there must be at least 1 girl and at most 2 boys.</p>
<p>My answers:</p>
<p>(i) GBBG, GBBB, GBGG</p>
<p>(11C2 x 13C2) + (11C1 x 13C3) + (11C3 x 13C1) = 9581</p>
<p>(ii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>(iii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>I have checked the correct answers, it shows that (i) 229944 (ii) 50820 (iii) 6765</p>
<p>I don't see my (i) is wrong, is it the correct answer for (i) of 229944 as incorrect?</p>
<p>For (ii) why it is using 11P2 x 22P2 = 50820 for the answer? Why this is a permutation question?</p>
| SlipEternal | 156,808 | <p>i) you are close. In each case, once you have selected the boys and girls for the committee, you need to permute their assigned roles within the committee. Because individual boys and girls are still people, they are all distinct. So, you should multiply your answer by <span class="math-container">$4!$</span></p>
<p>ii) here, calculating the permutations of individuals to jobs is a bit tricky.</p>
<p>Case 1: <span class="math-container">$gggg$</span>: you can freely permute the four girls.</p>
<p>Case 2: <span class="math-container">$bggg$</span>: step 1: choose where the boy will go (chair or vice), then permute the three girls.</p>
<p>Case 3: <span class="math-container">$bbgg$</span>: permute the boys among chair and vice, then permute the girls among treasurer and secretary.</p>
<p>iii) you got it correct. There are no "positions" in the group for the student survey, so no need to permute the students chosen.</p>
|
567,683 | <p>Let $F:\mathbb R^2\to \mathbb R^2$ be the force field with </p>
<p>$$F(x,y) = -\frac{(x,y)}{\sqrt{x^2 + y^2}}$$</p>
<p>the unit vector in the direction from $(x,y)$ to the origin. Calculate the work done against the force field in moving a particle from $(2a,0)$ to the origin along the top half of the circle $(x−a)^2+y^2=a^2$.</p>
<p>Okay, I tried to use the line integral and I set
$x=a+a\cos(t)$, $y= a\sin(t)$ and $t\in [0,\pi]$. Then the work should be </p>
<p>$$\int_0^\pi F(r (t))(r)′dt$$</p>
<p>But I can't got the right answer!!</p>
| Felix Marin | 85,343 | <p>$\vec{F} = -\nabla r$ where $r \equiv \sqrt{x^{2} + y^{2}}$
$$
\color{#0000ff}{\large W}
= \int_{\left(2a,0\right)}^{\left(0,0\right)}\vec{F}\cdot{\rm d}\vec{r}
=
\int_{\left(2a,0\right)}^{\left(0,0\right)}\left(-\nabla r\right)\cdot{\rm d}\vec{r}
=
\sqrt{\left(2a\right)^{2} + 0^{2}} - \sqrt{0^{2} + 0^{2}}
=
\color{#0000ff}{\large 2\left\vert a\right\vert}
$$</p>
|
1,408,036 | <p>Five points are drawn on the surface of an orange. Prove that it is possible to cut the orange in half in such a way that at least four of the points are on the same hemisphere. (Any points lying along the cut count as being on both hemispheres.)</p>
| johannesvalks | 155,865 | <p>Given a sphere with radius $1$.</p>
<p>Use the points</p>
<blockquote>
<p>$$
A_+(1,0,0), \quad A_-(-1,0,0), \quad B_+(0,1,0), \quad B_-(0,-1,0), \quad C(0,0,1)
$$</p>
</blockquote>
<p>And you <strong>cannot</strong> make such a cut.</p>
<p>As for any great circle, $(A_+,A_-)$ and $(B_+,B_-)$ are never on the same hemisphere.</p>
<p>That is if we exclude points on both hemispheres as being cut through a point.</p>
|
2,007,176 | <p>Assume that $f:(0,∞)→ℝ$ is twice differentiable with $f(x)>0$ and $f'(x)<0$ for all $x \in (0, ∞)$. Prove that $f''(x)$ cannot always be negative. </p>
<p>I know that it intuitively makes sense because if $f$ is always decreasing with a positive derivative, then at some point, it must be concave up. However, I'm not sure how to prove this rigorously. </p>
| Arthur | 15,500 | <p>Let $a = f'(1)$ and $b = f(1)$. Then we have $a < 0 < b$. Now assume $f''(x) \leq 0$ for all $x$. That means that $f'(x) \leq a$ for all $x > 1$, which again means that $f(x) \leq ax + b - a$ for all $x \geq 1$. Inserting some $x > \frac{a-b}{a}$ contradicts $f(x) > 0$.</p>
|
2,007,176 | <p>Assume that $f:(0,∞)→ℝ$ is twice differentiable with $f(x)>0$ and $f'(x)<0$ for all $x \in (0, ∞)$. Prove that $f''(x)$ cannot always be negative. </p>
<p>I know that it intuitively makes sense because if $f$ is always decreasing with a positive derivative, then at some point, it must be concave up. However, I'm not sure how to prove this rigorously. </p>
| Stefano | 387,021 | <p>Another way to see this is the following. $f$ is a positive strictly decreasing function on $(0, +\infty)$. Therefore the limit of $f(x)$ for $x \to +\infty$ exists and it is finite (in particular non-negative).</p>
<p>Now, if $f''(x)$ was alays negative, then $f$ would be strictly concave on $(0,+\infty)$ and the limit of $f(x)$ for $x\to+\infty$ would be $-\infty$, which is a contraddiction.</p>
|
3,105,664 | <p><span class="math-container">$$
I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx
$$</span></p>
<blockquote>
<p>Prove that
<span class="math-container">$$
I_n=\frac{1}{(n+1)!}+I_{n+1}
$$</span></p>
</blockquote>
<p>I tried integration by parts and still can't prove it, I appreciate any hint/answer. </p>
| Peter Foreman | 631,494 | <p>Following the proof of the recurrence relation by Zacky, we have
<span class="math-container">$$I_{n+1}=I_n-\frac{1}{(n+1)!}$$</span>
<span class="math-container">$$=I_{n-1}-\frac{1}{n!}-\frac{1}{(n+1)!}$$</span>
<span class="math-container">$$=I_0 - \sum_{k=1}^{n+1} \frac{1}{k!}$$</span>
We can calculate <span class="math-container">$I_0$</span> as follows
<span class="math-container">$$I_0 = \int_0^1 e^x dx=[e^x]_0^1=e-1$$</span>
Therefore the final solution is:
<span class="math-container">$$I_{n+1}=e-1 - \sum_{k=1}^{n+1} \frac{1}{k!}$$</span>
<span class="math-container">$$=e-\sum_{k=0}^{n+1} \frac{1}{k!}$$</span>
or equivalently;
<span class="math-container">$$I_n=e-\sum_{k=0}^{n} \frac{1}{k!}$$</span></p>
|
108,331 | <p>I find the frequent emergence of logarithms and even nested logarithms in number theory, especially the <a href="http://en.wikipedia.org/wiki/Prime_gap#Lower_bounds">prime number counting business</a>, somewhat unsettling. What is the reason for them?</p>
<p>Has it maybe to do with the series expansion of the logarithm? Or is there something inherently exponential in any of the relevant number distributions, like in complexity theory or combinatorical problems? I think maybe in how you can construct bigger integers out of smaller ones.</p>
| Pedro | 23,350 | <p><strong>From A source book in Mathematics:</strong> Gauss (1971, at the age of fourteen) was the first one to suggest, <em>in a purely empiraical way</em>, the asymptotic formula $ \displaystyle \frac{x}{\log{x}}$ for $\phi(x)$.$^1$ Later on (1792-1793,1849) he suggested another formula $ \displaystyle \int_2^x \frac{dx}{\log{x}}$ of which $ \displaystyle \frac{x}{\log{x}}$ is the leading term$^2$. Legendre, being of course, unaware of Gauss' results, suggested another <em>empirical</em> formula $ \displaystyle \frac{x}{A \log x+B}$ $^3$ and specified the constants $A$ and $B$ as $A=1$, <a href="http://en.wikipedia.org/wiki/Legendre%27s_constant" rel="nofollow">$B=-1.08366$</a>$^4$ in the second edition of the Essai. Legendre's formula, which Abel quoted as "the most marvelous in mathemtics",$^5$ is correct up to the leading term only. This fact was recognized by Dirichlet$^6$. In this note Dirichlet suggested another formula $\displaystyle \sum^x \frac{1}{\log n}$.</p>
<p>$1$: (Werke, Vol.X$_1$, p.11, 1917).</p>
<p>$2$: (Gauss's letter to Encke, 1849, Werke, Vol II, pp.444-447, 1876)</p>
<p>$3$: (Essai sur ka théorie des nombres, 1st. ed., pp 18-19, 1789)</p>
<p>$4$: [Not from the book, but Wikipedia] He guessed $B$ to be about $1.08366$, but regardless of its exact value, the existence of $B$ implies the prime number theorem.</p>
<p>$5$: (letter to Holmboe, Abel Memorial, 1902, Correspondence, p.5)</p>
<p>$6$: ("Sur L'uasge des séries infinies dans la théorie des nombres." Crelle's Journal, Vol 18., p.272, 1838, in his remrk written on the copy presented to Gauss. Cf.Dirichlet, Werke, Vol. 1 , p372, 1889)</p>
<p><strong>From Wikipedia:</strong> Initially, it might seem that since our numbering system is base 10, this base would be more "natural" than base $e$. But mathematically, the number $10$ is not particularly significant. Its use culturally—as the basis for many societies’ numbering systems—likely arises from humans’ typical number of fingers. Other cultures have based their counting systems on such choices as 5, 8, 12, 20, and 60.</p>
<p>$\log_e$ is a "natural" $\log$ because it automatically springs from, and appears so often in, mathematics. </p>
<p>Further senses of this naturalness make no use of calculus. As an example, there are a number of simple series involving the natural logarithm. Pietro Mengoli and Nicholas Mercator called it logarithmus naturalis a few decades before Newton and Leibniz developed calculus.</p>
|
4,415,559 | <p>I am trying to find the root of <span class="math-container">$f(x)=ln(x)-cos(x)$</span> by writing an algorithm for bisection and fixed-point iteration method. I am currently using python but whenever I'm running it using either of the two methods, it prints out "math domain error". I guess this is due to ln(x) when x becomes 0 or negative.</p>
<p>So, I asked myself if this manipulation is valid:</p>
<p>If <span class="math-container">$f(x)=ln(x)-cos(x)=0$</span>, then <span class="math-container">$ln(x)=cos(x)$</span>. It also follows that <span class="math-container">$x=e^{cos(x)}$</span> so we have a function, say <span class="math-container">$h(x)=x-e^{cos(x)}$</span>, that has same root with <span class="math-container">$f(x)$</span>. So, I tried using <span class="math-container">$h$</span> to find the root of <span class="math-container">$f$</span> and I resolved the error prompt I am getting whenever I use <span class="math-container">$f$</span> in my code. This is for bisection method, and I got the root that I want to get.</p>
<p>I still don't know what is the appropriate <span class="math-container">$g(x)$</span> should I take for fixed-point iteration method such that if <span class="math-container">$f(x)=0$</span>, then <span class="math-container">$x=g(x)$</span> and <span class="math-container">$g'(x)<1$</span> for some open interval.</p>
<p>First question: Is using an alternative function <span class="math-container">$h$</span> to solve for the actual root of <span class="math-container">$f$</span> valid?</p>
<p>Last question: What could be a possible <span class="math-container">$g(x)$</span> to use to find the root using fixed-point iteration method?</p>
<p>Any help would be appreciated.</p>
| emacs drives me nuts | 746,312 | <blockquote>
<p><span class="math-container">$g′(x)<1$</span> for some open interval</p>
</blockquote>
<p>What you need is the stronger <span class="math-container">$|g′(x)|<1$</span> for some interval containing the zero <span class="math-container">$x_0$</span> of <span class="math-container">$f$</span>.</p>
<p>Your choice <span class="math-container">$g(x) = e^{\cos x}$</span> has derivative <span class="math-container">$g'(x) = -e^{\cos x}\sin x$</span>. But with <span class="math-container">$x_0\approx 1.303$</span> that evaluates to <span class="math-container">$g'(x_0)\approx -1.26$</span>. Hence <span class="math-container">$g$</span> will not work.</p>
<blockquote>
<p>First question: Is using an alternative function <span class="math-container">$h$</span> to solve for the actual root of <span class="math-container">$f$</span> valid?</p>
</blockquote>
<p>Yes, you can do that. The question is for which purpose you are doing it...</p>
<p>One annoyance with <span class="math-container">$f$</span> and fixed-point methods can be that you come to a halt without a result whenever <span class="math-container">$x<0$</span> because <span class="math-container">$\ln x$</span> is not defined there. However, using <span class="math-container">$h$</span> instead has the issue that whenever <span class="math-container">$x<0$</span>, the iteration will trickle to <span class="math-container">$-\infty$</span> an not produce a result, either.</p>
<blockquote>
<p>Last question: What could be a possible <span class="math-container">$g(x)$</span> to use to find the root using fixed-point iteration method?</p>
</blockquote>
<p>As <span class="math-container">$x=g(x)=e^{\cos x}$</span> does not work, you can try splits like
<span class="math-container">$$x=\alpha x + (1-\alpha)x = g(x)$$</span>
and from there
<span class="math-container">$$ x = \frac1\alpha \big(g(x)-(1-\alpha)x\big)=g_\alpha(x)$$</span>
for some constant <span class="math-container">$\alpha$</span>, and then try to find some <span class="math-container">$\alpha$</span> such that
<span class="math-container">$$|g'_\alpha(x_0)| < 1$$</span></p>
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1,880,090 | <p>The solution states that the ball of radius $\epsilon >0$ around a real number $x$ always contains the non-real number $x+i\epsilon/2$. </p>
<p>I don't understand the answer, for every number $x \in \mathbb{R}$ there is an open ball, right? For every $x \in \mathbb{R}$ there is an $r>0$ such that I can form an open ball $B_r(x)\subset \mathbb{R}$.</p>
| yago | 141,261 | <p>I think you're confusing open balls in $\mathbb{R}$ and $\mathbb{C}$. If you want to prove that $\mathbb{R}$ is open in $\mathbb{C}$, you have to prove that for any $x \in \mathbb{R}$, there exists $\epsilon > 0$ such that $B_x(\epsilon) \subseteq \mathbb{R}$, where $B_x(\epsilon)$ is a ball in $\mathbb{C}$, meaning that $B_x(\epsilon) = \{ z \mid z \in \mathbb{C}, |z-x| < \epsilon\}$.</p>
<p>Since for any $x \in \mathbb{R}$ and $\epsilon > 0$, $x + i\frac{\epsilon}{2} \in B_x(\epsilon)$ but $x + i\frac{\epsilon}{2} \notin \mathbb{R}$, you can't find any such ball.</p>
|
206,723 | <p>Can any one explain why the probability that an integer is divisible by a prime $p$ (or any integer) is $1/p$?</p>
| Jeff P. | 18,440 | <p>See <a href="http://en.wikipedia.org/wiki/Coprime_integers#Probabilities" rel="nofollow">http://en.wikipedia.org/wiki/Coprime_integers#Probabilities</a>.</p>
|
3,933,296 | <p>What I already have,</p>
<ol>
<li>Palindrome in form XYZYX, where X can’t be 0.</li>
<li>Divisibility rule of 9: sum of digits is divisible by 9. So, we have 2(X+Y)+Z = 9M.</li>
<li>The first part is divisible by 9 if and only if X+Y is divisible by 9. So, we have 10 pairs out of 90. And each such pair the total sum is divisible by 9 when Z is also divisible by 9. There are 2 such Zs: 0, 9. So, there are 20 divisible palindromes.</li>
<li>If (X+Y) mod 9 = 1, then 2(X+Y) mod 9 = 2; and in order for the total sum to be divisible by 8, Z must have the remainder of 1 when divided by 9. There is 1 such Z: 1. And again, we have 10 xy pairs with the given remainder. So, this case yields 10*1 = 30 more palindromes.</li>
<li>Same logic as on previous step applies to the case when 2(X+Y) mod 9 = 2.</li>
<li>So, total number of divisible palindromes = 80?</li>
</ol>
<p>When using this method, I only get 80 numbers of 5-digit palindromes that are divisible by 9(?) i dont think im doing this method correctly, can someone show me whats going on here</p>
| Piquito | 219,998 | <p>The numbers are <span class="math-container">$abcba$</span> with <span class="math-container">$1\le a\le9$</span> and <span class="math-container">$0\le b,c\le9$</span> and <span class="math-container">$2(a+b)+c=9,18,27,36,45$</span>.</p>
<p><span class="math-container">$c=0$</span> gives <span class="math-container">$2(a+b)=18,36\Rightarrow a+b=9,18$</span> which gives <span class="math-container">$ab=18,27,36,45,90,99$</span> then <span class="math-container">$10$</span> examples.</p>
<p><span class="math-container">$c=1$</span> gives <span class="math-container">$a+b=4,13$</span> which gives <span class="math-container">$ab=13,22,40,49,58,67$</span> then <span class="math-container">$10$</span> examples.</p>
<p><span class="math-container">$c=2$</span> gives <span class="math-container">$a+b=8,17$</span> which gives <span class="math-container">$ab=17,26,35,44,80,89$</span> then <span class="math-container">$10$</span> examples.</p>
<p>We can verify that the seven other central values <span class="math-container">$c$</span> give also <span class="math-container">$10$</span> examples each of them.</p>
<p>Consequently there are <strong>100 examples</strong>.</p>
|
1,024,068 | <p>I need to solve these two equations . </p>
<p>$ 2x + 4y + 3x^{2} + 4xy =0$</p>
<p>$ 4x + 8y + 2x^{2} + 4y^{3}$ = $0 $</p>
<p>I have added them , subtracted them . Nothing is helping here . Can anyone give hints ? Thanks</p>
| Claude Leibovici | 82,404 | <p>It is not so hard. Just eleiminate $y$ from the first equation; this gives $$y=\frac{-3 x^2-2 x}{4 (x+1)}$$ Plug it in the second equation and simplify as much as you can; you should arrive to $$(-27 x^3-22 x^2+28 x+24)\frac{x^3}{16 (x+1)^3}=0$$ So, beside the trivial solution $x=0$, remain the roots of the cubic $$-27 x^3-22 x^2+28 x+24=0$$ Use Cardano to show that there is only one real root.</p>
<p>By inspection of $$f(x)=-27 x^3-22 x^2+28 x+24$$ $f(0)=24$, $f(1)=3$, $f(2)=-224$; so the solution is very close to $x=1$. To approximate it, use Newton starting at $x=1$; the first iterates are $\frac{100}{97}$, $\frac{24311038}{23603689}$ which is the solution for more than six significant figures.</p>
<p>If you finish the steps for solving a cubic, you should find that the exact solution is $$x=\frac{1}{81} \left(\sqrt[3]{150704-1296 \sqrt{1113}}+2 \sqrt[3]{ 18838+162
\sqrt{1113}}-22\right)$$</p>
|
1,520,028 | <p>I'm struggling to figure out how to find a bound on my error for this problem:</p>
<p>Let T_{6}(x) be the Taylor polynomial of degree 6 based at a = 0 for the function f(x)=\cos(x). Suppose you approximate f(x) by T_{6}(x). If |x|\leq 1, find a bound on the error in your approximation by using the alternating series estimate. </p>
<p>So far, I have</p>
<p>f(x) = cos(x), f(0) = 1</p>
<p>f^(1)(x) = -sin(x), f^(1)(0) = 0</p>
<p>f^(2)(x) = -cos(x), f^(2)(0) = -1</p>
<p>f^(3)(x) = sin(x), f^(3)(0) = 0</p>
<p>f^(4)(x) = cos(x), f^(4)(0) = 1</p>
<p>and</p>
<p>T6(x) = 1 - x^2/4! + x^4/8! -x^6/12!</p>
<p>and that [b_n] = x^8/16!</p>
<p>I'll be honest, this section is way over my head and I'm struggling to make any sense out of it, so I'm not sure that I'm taking the correct approach here in general.</p>
<p>Any help is appreciated!</p>
| abel | 9,252 | <p>the mcclaurin series $\cos x$ is $$\cos x = 1 - \frac1{2!} x^2 + \frac 1{4!}x^4 - \frac 1{6!}x^6 + \frac 1{8!} x^8 + \cdots $$ now, if you truncate this alternating series by $T_6,$ then the error committed is of the same sign as the next term and smaller in absolute value. therefore $$0 \le \cos x - T_6(x)\le \frac1{8!} x^8 \le \frac 18 \text{ for } |x| \le 1.$$</p>
|
1,017,989 | <p>Consider three disjoint circles not necessarily of same radii. How do you draw the smallest circle enclosing all these three circles? Where is its centre, and what is its radius? </p>
| achille hui | 59,379 | <p>If your goal is to "draw" the smallest circle and you don't really care
about the numerical values of its center and radii too much, You can construct
it in a geometrical manner.</p>
<p>Let's say the three circles are centered at $A$, $B$ and $C$ respectively.</p>
<p>There are two possibilities. </p>
<ol>
<li>The smallest circle is touching two of the circles. In this case,
the center of the smallest circle will be collinear with the center
of the two circles it touches.</li>
<li>The smallest circle is touching all three circles.</li>
</ol>
<p>It is clear how to construct the smallest circle in $1^{st}$ case. For the $2^{nd}$ case,</p>
<ul>
<li>Construct a line passing through $A$ and $B$.</li>
<li>Let $G$ be the intersection of line $AB$ with circle $A$ on the opposite side of point $B$. </li>
<li>Let $H$ be the intersection of line $AB$ with circle $B$ on the opposite side of point $A$. </li>
<li>Let $I$ be the mid-point of $G$ and $H$. </li>
<li>Construct a hyperbola passing through $I$ having $A$ and $B$ as foci (the red curve
in the picture below).</li>
<li>Repeat the same procedure to the two other combination of pairs of circles to
obtain three hyperbolas (the red, green and blue curves in picture below).</li>
</ul>
<p>These 3 hyperbolas will intersect at two points $P$ and $R$.
The point $P$ is the one which lies on the branch that contains the mid point $I$.
It will be the center of another candidate of smallest circle (the orange circle) you want. </p>
<p>At the end, we will obtain four candidates of the smallest circle and it is easy to check which one is the smallest one.</p>
<p>In any event, in the $2^{nd}$ case, one can compute the coordinates of the center
$P$ by first figuring out the equations for the three hyperbolas and then determine
their intersection. The algebra will be a mess and I'll let you have the fun
(if you really want that).</p>
<p><img src="https://i.stack.imgur.com/hOHpL.jpg" alt="A tales of three circles"></p>
|
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