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 |
|---|---|---|---|---|
23,029 | <p>I used computer program to generate numbers of this form, then verify if they are amicable. However, it took too long to finish, is there any special technique to verify if a pair is amicable? </p>
<p>One pair that I found is $m = 284 = 2^{2}.71, n = 220 = 2^{2}.5.11$ </p>
<p>Thanks,<br>
Chan</p>
| milcak | 5,105 | <p>If you are looking for a nice program for finding amicable pairs, there is a related question on <a href="https://stackoverflow.com/questions/4218056/how-to-optimize-code-for-finding-amicable-pairs">stackoverflow</a>. </p>
<p><a href="http://tutoree7.pastebin.com/wKvMAWpT" rel="nofollow noreferrer">Here</a> is a co... |
124 | <p>Harold Williams, Pablo Solis, and I were chatting and the following question came up.</p>
<p>In Lie group land (where you're doing differential geometry), given a finite-dimensional Lie algebra <em>g</em>, you can find a faithful representation <em>g</em> → End(V) by Ado's theorem. Then you can take the group ... | Theo Johnson-Freyd | 78 | <p>Over a field that's not C --- well, at least in non-zero characteristic --- , the problem is deeper than Ben suggests: the proof of Ado's theorem that I know requires characteristic zero. I think if the theorem were true in non-zero characteristic Mark Haiman would have said so --- he seemed to suggest in his class... |
124 | <p>Harold Williams, Pablo Solis, and I were chatting and the following question came up.</p>
<p>In Lie group land (where you're doing differential geometry), given a finite-dimensional Lie algebra <em>g</em>, you can find a faithful representation <em>g</em> → End(V) by Ado's theorem. Then you can take the group ... | Nicolás | 36,285 | <p>There is a very interesting way in which this statement can fail. If K is a number field and
G is an algebraic group over K having good reduction away from N,
then a Lie subalgebra h of Lie(G) is going to be algebraic if and only if the reduction mod P for primes P not dividing N is closed under p-th powers.</p>
<p... |
2,445,569 | <blockquote>
<p>Consider the limit
$$ \lim_{n \to \infty} \int_0^n e^{-px} \cos(ax)\, dx.$$</p>
<p>The exercise is asking about the values of $p$ and $a$ which the limit above exists.</p>
<p>a) exists for infinity values of $p$</p>
<p>b) exists for a finite values of $a$</p>
<p>c) exists for ev... | Harry Alli | 378,603 | <p>That symbol means: "is a member of...", or "belongs to". It is often used to state that a particular element is a member of a set.</p>
|
2,445,569 | <blockquote>
<p>Consider the limit
$$ \lim_{n \to \infty} \int_0^n e^{-px} \cos(ax)\, dx.$$</p>
<p>The exercise is asking about the values of $p$ and $a$ which the limit above exists.</p>
<p>a) exists for infinity values of $p$</p>
<p>b) exists for a finite values of $a$</p>
<p>c) exists for ev... | Miksu | 316,296 | <p>The $\in$ symbol means that the element before it belongs to some <strong>set</strong>. A better example would be $2\in \mathbb{N}$. This means that the number two is a natural number.</p>
<p>$(a,b)$ in the other hand means an open interval which includes all the real numbers that are greater than $a$ but less than... |
222,730 | <p>This is a qualifying exam problem from Indiana University.</p>
<p>Prove or provide a counterexample to the following statement: </p>
<p>If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function, then there exists a real number $L$ such that </p>
<p>$$\lim_{\epsilon \rightarrow 0} \int_{\epsilon \leq|x|\l... | Lukas Geyer | 43,179 | <p>This limit won't exist in general, a counterexample is $f(x) = \frac{1}{1+|\ln x|}$ for $x > 0$, $f(x)=0$ for $x\le0$, in which case the limit is $\infty$.</p>
|
3,496,882 | <p><span class="math-container">$ABC$</span> is a triangle with <span class="math-container">$AB < AC$</span>.
The tangent to the circumcircle at <span class="math-container">$A$</span> meets line <span class="math-container">$BC$</span> at <span class="math-container">$P$</span>. Points <span class="math-container... | MathematicsStudent1122 | 238,417 | <p>If we have a <span class="math-container">$C^1$</span> function <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> which is reasonably well-behaved as <span class="math-container">$x \to \infty$</span>, then <span class="math-container">$\sum_{k=1}^{n} f'(k) = O(f(n))$</span>. In your case, <span clas... |
2,167,441 | <p>i need some help here.</p>
<p>Let $A \in \mathbb{R}$ and $x$ be a limit point of the closure of $A$.</p>
<p>Let $(x_{n})$ be a sequence in closure of $A$ such that $x_{n}\neq x, \forall n\in \mathbb{N}$ and $x_{n}$ converges to $x$.</p>
<p>Define the sequence $(y_{n})$ as the following:</p>
<p>If $x_{n} \in A$, ... | projectilemotion | 323,432 | <p>Your solution is correct. Notice that your integral is of the form:
$$\int \frac{f'(x)}{f(x)}~dx=\ln|f(x)|+C$$
So you can easily obtain the solution if you know this fact.</p>
<hr>
<p>The online portal may have wanted you to omit the absolute value signs as @Ian suggested or it may be due to the brackets you've ad... |
3,878,905 | <p>Question: Given a complex number <em>z</em> such that <span class="math-container">$z^{13}=1$</span>, find the sum of all possible values of <span class="math-container">$z+z^3+z^4+z^9+z^{10}+z^{12}$</span>.</p>
<p>I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can... | Parcly Taxel | 357,390 | <p>As stated in comments, for a fixed <span class="math-container">$z=e^{i\theta}$</span> the given expression is equal to <span class="math-container">$2(\cos\theta+\cos3\theta+\cos4\theta)$</span>. As <span class="math-container">$z$</span> runs through its <span class="math-container">$13$</span> possible values, so... |
3,878,905 | <p>Question: Given a complex number <em>z</em> such that <span class="math-container">$z^{13}=1$</span>, find the sum of all possible values of <span class="math-container">$z+z^3+z^4+z^9+z^{10}+z^{12}$</span>.</p>
<p>I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can... | saulspatz | 235,128 | <p>Since <span class="math-container">$13$</span> is prime, the map <span class="math-container">$z\mapsto z^k$</span> is a permutation of the roots of <span class="math-container">$z^{13}-1$</span>, for <span class="math-container">$k=1,2,3,\dots,12$</span>. Thus, the answer is <span class="math-container">$6$</span>... |
5,381 | <p>How-to compute the <a href="http://en.wikipedia.org/wiki/Genus_%28mathematics%29#Orientable_surface">genus</a> of an algebraic <a href="http://en.wikipedia.org/wiki/Algebraic_curve">curve</a> in Mathematica ?</p>
<p>In my case the algebraic curve is explicitly defined by a polynomial.</p>
| madalina | 1,228 | <p>For computing the genus of a plane algebraic curve implicitly defined by a squarefree polynomial $f(x,y)$ there are different softwares available in the literature.</p>
<p><strong>Remark:</strong> I assume you are interested in computing the genus when the coefficients of the defining polynomial of the curve are ei... |
5,381 | <p>How-to compute the <a href="http://en.wikipedia.org/wiki/Genus_%28mathematics%29#Orientable_surface">genus</a> of an algebraic <a href="http://en.wikipedia.org/wiki/Algebraic_curve">curve</a> in Mathematica ?</p>
<p>In my case the algebraic curve is explicitly defined by a polynomial.</p>
| Daniel Lichtblau | 51 | <p>I will show a method that is conjectural, though i believe it is correct. It differs from the more common approach of using (quadratic) birational transformations to force singularities to be double points. More a detailed approach to that, see Madelina's response. Also I cover the exact case, although at least some... |
4,186,471 | <p>I'm given the series:
<span class="math-container">$$\sum_{n=1}^{+\infty}(n^2+n+1)x^{n-1}$$</span>
which is easy to see converges iff <span class="math-container">$|x|<1$</span>. What I'm missing is the idea behind the computation of the sum. I tried considering the series
<span class="math-container">$$\sum_{n=1... | David G. Stork | 210,401 | <p>Break it into terms:</p>
<p><span class="math-container">$$\sum\limits_{n=1}^\infty n^2 x^{n-1} = \frac{-x-1}{(x-1)^3}$$</span></p>
<p><span class="math-container">$$\sum\limits_{n=1}^\infty n x^{n-1} = \frac{1}{(x-1)^2}$$</span></p>
<p><span class="math-container">$$\sum\limits_{n=1}^\infty x^{n-1} = \frac{-1}{x-1}... |
4,071,323 | <p>In his paper, "Euler and the Partial Sums of the Prime Harmonic Series" (<a href="http://pollack.uga.edu/eulerprime.pdf" rel="nofollow noreferrer">pdf</a>), Paul Pollacks makes a small statement at the top of page 4:</p>
<p><span class="math-container">$$ \log \frac{3}{2} < (\frac{3}{2} - 1) = \frac12 $... | hunter | 108,129 | <p>For any <span class="math-container">$x > 1$</span> we have
<span class="math-container">$$
\log(x) < x - 1.
$$</span>
For a geometric proof, note that both functions vanish at <span class="math-container">$x = 1$</span> and that the graph of <span class="math-container">$\log(x)$</span> is concave down (so is... |
2,123,914 | <blockquote>
<p>$$y''-2y'+y=\frac{e^x}{x}$$</p>
</blockquote>
<p>The solution is $$y=xe^xln|x|-xe^x+c_{1}e^x+c_{2}e^x$$</p>
<p>It is said that $y_{1}=e^x$ and $y_{2}=xe^x$ are the solution to the homogeneous equation.</p>
<p>How can I infer it? why for example $y_{1}=ln|x|$ and $y_{2}=xe^x$ are the solutions?</p>
... | S.C.B. | 310,930 | <p>$$y''-2y'+y=0 \iff y''-y'-y'+y=0 \iff (y'-y)'-(y'-y)=0$$</p>
<p>Note that $$(y'-y)'-(y'-y)=0 \iff y'-y=ce^{x} \iff (y'-y)e^{-x}=c$$
For some constant $c$. Now note $$(y'-y)e^{-x}=(ye^{-x})'$$
So we have that $$(ye^{-x})'=c \iff ye^{-x}=cx+b$$
For some constant $c,b$. So your answer is incorrect. The answer should b... |
1,929,319 | <p>We know that a polynomial in single variable/indeterminate is defined algebraically as-</p>
<blockquote>
<p>An expression of the form $a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{2}x^2+a_{1}x+a_{0}$, <p> where $a_n,\dots,a_0$ are constants/parameters and $x$ is the indeterminate/variable.</p>
</blockquote>
<p>But I was won... | jnyan | 365,230 | <p>A curve which after certain successive differentiations gives zero at all points. You will get a straight line parallel to x axis at the end at all points.</p>
|
3,444,112 | <p>I have to find the following limit:</p>
<p><span class="math-container">$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}}$$</span></p>
<p>This is what I tried:</p>
<p><span class="math-container">$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}... | Xiaohai Zhang | 628,555 | <p><span class="math-container">$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k+1}{k}\dfrac{k+1}{k+2}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \frac{ \dfrac{k+1}{k}}{\dfrac{k+2}{k+1}}}$$</span>... |
3,020,597 | <p>I am aware of the formula of this method. However, is it true that the method produces more accurate polynomial when the <span class="math-container">$x$</span> points are closer to each other? if so or not, why? </p>
<p>Moreover, If I am given a set of <span class="math-container">$n$</span> points i.e. <span clas... | Dante Grevino | 616,680 | <p>I suppose that the answer depends on what do mean with 'the "best" interpolating polynomial'. If you want to minimize the supremum norm of the polynomial <span class="math-container">$\prod_{j=0}^n(x-x_j)$</span> the answer is that you have to take the zeros of the Chebyschev's polynomial of degree <span class="math... |
1,540,667 | <blockquote>
<p>Consider a triangle having integer sides such that so side is greater than $4$ units.How many such triangles are possible?</p>
</blockquote>
<p>I could not solve it by trying to use combinatorics.</p>
<p>So, how to do it?</p>
<p>Thanks fro any help.</p>
<p>The original question came from here: <a ... | Ross Millikan | 1,827 | <p>With numbers this small, just list them by hand. Let the sides be $a,b,c$ with $a \le b \le c$ How many are there with $c=1?$ With $c=2?$ Etc. If the limit were higher, I would solve it by hand for small $c$, then look for a general expression. It should not be more than quadratic in $c$. Then summing over $c$... |
1,540,667 | <blockquote>
<p>Consider a triangle having integer sides such that so side is greater than $4$ units.How many such triangles are possible?</p>
</blockquote>
<p>I could not solve it by trying to use combinatorics.</p>
<p>So, how to do it?</p>
<p>Thanks fro any help.</p>
<p>The original question came from here: <a ... | Ben Longo | 137,131 | <p>More generally, the number of integer triangles with a greatest side length of <span class="math-container">$c$</span> is given by</p>
<p><span class="math-container">$$n=\left\lceil \frac{c+1}{2}\right\rceil \cdot \left\lfloor\frac{c+1}{2} \right\rfloor $$</span></p>
<h3>Proof</h3>
<p>Let <span class="math-containe... |
630,330 | <p>Please correct me if I am wrong. I believe that it is unknown whether the decimal representation of $\pi$ contains all strings, or lists on 10 digits. I would like to know if there is a number that is known that does in fact contain all of them, without the number being constructed explicitly to do so. Is there a nu... | Igor Rivin | 109,865 | <p>The are countably many (finite) strings of digits. Enumerate them as $s_1, s_2, \dotsc, s_n, \dotsc$ and then concatenate them into $x=0.s_1 s_2 \dots s_n \dots$ The number $x$ is the number you seek.</p>
|
2,600,692 | <p>How can I determine the length of the sides $b$, and the altitude $h$ of an <strong>isosceles triangle</strong> given its area $A= 24 \text{cm}^2$ and its perimeter $P= 16\text{cm}$?</p>
<p><img src="https://i.stack.imgur.com/FxGfi.gif"/></p>
| fourierwho | 81,375 | <p>I always see this result (or the idea of the proof) attributed to <a href="https://en.wikipedia.org/wiki/Krylov%E2%80%93Bogoliubov_averaging_method" rel="nofollow noreferrer">Krylov and Bogoliubov</a>. Let $\nu$ be any Radon probability measure on $X$. For each $n$, let $\varphi_{*}^{n}\nu$ be the measure given by... |
706,461 | <p>Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm,
$$\lim_{n\to\infty} \ln((n!)^{1/n}) = \lim_{n\to\infty} (1/n)\ln(n!) = \lim_{n\to\infty} (\ln(n!)/n)$$
Applying L'hopital's r... | Hanul Jeon | 53,976 | <p>$$
\begin{aligned}
\lim_{n\to\infty} (n!)^{1/n} &=\lim_{n\to\infty} \exp(\tfrac{1}{n} \ln n!)\\
&= \lim_{n\to\infty} \exp[\tfrac{1}{n} (\ln 1+\ln 2+\cdots + \ln n)]\\
&\ge \lim_{n\to\infty}\exp \left[ \frac{1}{n} \int_1 ^n \ln x dx\right]\\
&=\lim_{n\to\infty} \exp \frac{n\ln n -n+1}{n}
\end{aligned... |
2,544,418 | <p>Let $R>0$ be the radius of convergence of a power series $Σa_nx^n$. Is it not uniformly convergent in $(-R,R)$? My book goes out of its way to say that if $[a,b]⊂(-R,R)$, then the power series converges uniformly in $[a,b]$. Can't we just say that it is uniformly convergent in $(-R,R)$? </p>
| Fred | 380,717 | <p>No. Example: $ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$.</p>
<p>Suppose that $s_n(x):=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$ converges uniformly on $(-1,1)$ to $\frac{1}{1-x}$. </p>
<p>Then , to $ \epsilon=1$ we get $N \in \mathbb N$ such that</p>
<p>$\frac{|x|^{N+1}}{1-x}=|s_N(x)-\frac{1}{1-x}|<... |
1,392,273 | <p>I'm a math major at university and my tutor told me that for most people it's best to focus on either algebra or analysis, however I have trouble understanding the difference between them.</p>
<p>What are the actual differences between algebra and analysis, and how do I tell which is for me?</p>
| jeo15 | 252,647 | <p>Typically, a math major takes after calculus: linear algebra, differential equations, elementary real analysis, abstract algebra and possibly complex analysis. You really don't specialize in algebra or analysis until you get to grad school in mathematics.</p>
|
1,829,082 | <p>I was wondering how many base cases are needed and when to stop (in general).
For example, I have 4-cent and 11-cent stamps and I need to determine the amount of postage I can make, the cases I have that work are:
$4,8,11,12,15,16,17,19,20,22,23,24,26,27,28,30 \geq$</p>
<p>Now what I'm stuck on is: what stops me fr... | Joffan | 206,402 | <p>The strong induction you might be thinking of for this problem would start with the four cases:
\begin{align}
30&=11+11+4+4 \\
31&=11+4+4+4+4+4 \\
32&=4+4+4+4+4+4+4+4 \\
33&=11+11+11 \\
\end{align}</p>
<p>Then any subsequent case $k>33$ can run induction on the basis that there is a solution for ... |
25,527 | <p>I know how to find a point on a circle given a radius and an angle, but my knowledge of trigonometry doesn't extend much further than that. My question is probably best explained diagrammatically:</p>
<p><img src="https://i.stack.imgur.com/EsPnr.jpg" alt="enter image description here"></p>
<p>Given a centre point,... | Matt Calhoun | 1,417 | <p>Write $(x_1,y_1)=(Rcos\theta_1,Rsin\theta_1)$ and $(x_2,y_2)=(Rcos\theta_2,Rsin\theta_2)$, where $R = \sqrt{x_1^2+y_1^2}$ is the radius of the circle, and solve for $\theta_1$ and $\theta_2$; then clearly $(x_?,y_?)=(Rcos(\theta_1+\delta),Rsin(\theta_1+\delta)$, where $\delta = \frac{\theta_1-\theta_2}{2}$.</p>
<p>... |
399,865 | <p>Let $A$ be a linear map from the vector space $X$ to $Y$ and $T$ be a subspace of $X$ . I want to understand what is the meaning of saying that the restriction $A_{|T}: \to A(T)$ is invertible. Could anybody explain me what is this restriction map?</p>
<p>Thanks for the help.</p>
| Alex Provost | 59,556 | <p>This applies to a function between sets in general, not only vector spaces.</p>
<p>Given $f:X \rightarrow Y$ and a subset $T \subset X$, the restriction $f_{|T}:T \rightarrow Y$ is by definition $f_{|T}(x) = f(x)$. That is, the function is the same, except that its domain is restricted.</p>
<p>If $f$ is a linear m... |
1,640,993 | <p>Components $1$ and $2$ are connected in parallel, so that subsystem works iff either $1$ or $2$ works;since $3$ and $4$ are connected in series, that subsystem works
iff both 3 and 4 work. If components work independently of of one another and $P(\text{component works})=0.9$, calculate $P(\text{system works})$.</p>
... | Em. | 290,196 | <p>\begin{align*}
P(C|D) &= \frac{P(C \cap D )}{P(D)}\\
&=\frac{P(C \cap (B\cup C))}{P(D)}\\
&= \frac{P(CB\cup C)}{P(D)}\\
&=\frac{P(CB\cup C)}{P(D)}\\
&=\frac{P(CB)+P(C)-P(BC)}{P(D)}\\
&=\frac{P(\varnothing)+P(C)-P(\varnothing)}{P(D)}\\
&=\frac{P(C)}{P(D)}
\end{align*}
This is a long winded... |
3,146,277 | <blockquote>
<p>Let <span class="math-container">$\langle a_n\rangle$</span> be a sequence where <span class="math-container">$a_n = \sum\limits_{j=1}^{n}(-1)^j2/n$</span></p>
<ol>
<li><p>is oscillatory</p>
</li>
<li><p>diverges to <span class="math-container">$+\infty$</span></p>
</li>
<li><p>diverges to <span class="... | Fred | 380,717 | <p>We have <span class="math-container">$a_n=\frac{2}{n}\sum_{j=1}^n(-1)^j=\frac{1}{n} \cdot (1-(-1)^{n+1})$</span> and</p>
<p><span class="math-container">$|a_n| \le 2/n.$</span> for all <span class="math-container">$n$</span></p>
<p>Can you proceed ?</p>
|
2,705,733 | <p>I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$<Br></p>
<p>$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?<br>
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ <br>am I right?
<br>
$4 \sum_{i=0}^{2n-1-n} i+n $ <br>
$4 (\sum_{i=0}^{n... | fleablood | 280,126 | <p>That is how you do it, yes.</p>
<p>I suppose that as I am the type of person who <em>always</em> makes an indexing area to play it safe I'd to this</p>
<p>$\sum_{i=n}^{2n-1} 3+4i = $</p>
<p>$\sum_{j= 0}^{n-1} (3 + 4(j + n)) = $</p>
<p>$3\sum_{j= 0}^{n-1} 1 + 4\sum_{j= 0}^{n-1} j + n\sum_{j= 0}^{n-1} 1=$</p>
<p>... |
2,705,733 | <p>I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$<Br></p>
<p>$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?<br>
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ <br>am I right?
<br>
$4 \sum_{i=0}^{2n-1-n} i+n $ <br>
$4 (\sum_{i=0}^{n... | epi163sqrt | 132,007 | <blockquote>
<p><strong>Hint:</strong> Note the brackets $\sum_{i=n}^{2n-1}\color{blue}{(} 3+4i\color{blue}{)}$ @robjohn used in his answer. It is crucial to use them, since the sum <em>without brackets</em> is
\begin{align*}
\color{blue}{\sum_{i=n}^{2n-1} 3+4i}
=\left(\sum_{i=n}^{2n-1} 3\right)+4i
=3\left(\sum_{... |
491,851 | <p>Given that <span class="math-container">$719$</span> is prime, find the least positive residue of <span class="math-container">$11^{721} \pmod {719}$</span>, without using modular exponentiation.</p>
<p>So, I know how to use modular exponentiation and have done it to get the answer <span class="math-container">$612... | Kazi Ziaul Hassan | 657,123 | <p>This is a simple application of Fermat's Little theorem.</p>
<p>**
if M is prime, then to find a<sup>b</sup> % M,</p>
<p>r = b % (M-1)</p>
<p>c = a % M</p>
<p>a<sup>b</sup> % M = c<sup>r</sup> % M</p>
<p>**</p>
<p>721 % 718 = 3</p>
<p>11 % 719 = 11</p>
<p>Therefore, 11<sup>721</sup> % 719 = 11<sup>3</sup> %... |
327,239 | <p>I wanted to show my nephew(16) a simple approach to calculate the distance between two geo-locations. The mathematical knowledge of a 16-year old boy is limited to simple geometrical shapes like triangle, circles, etc.</p>
<p>My approach bases on the assumption that you can draw the two locations as two points (A a... | copper.hat | 27,978 | <p>Let $p(\theta, \phi) = r(\cos \phi \cos \theta, -\cos \phi \sin \theta,\sin \phi)$ (latitude is $\phi$ measured north, longitude is $\theta$ measured west, the point $0°$, $0 °$ corresponds to $(r,0,0)$, the north pole is $(0,0,r)$).</p>
<p>Then the Euclidean distance is $d=\|p(\theta_1,\phi_1)-p(\theta_2,\phi_2)\|... |
98,820 | <p>In the course of exploring <a href="https://mathematica.stackexchange.com/a/77812/531">@kglr's answer</a> to <a href="https://mathematica.stackexchange.com/q/77784/531">Edmund's question</a> gain some insight into how I might answer <a href="https://mathematica.stackexchange.com/questions/98817/how-do-i-overlay-thin... | Anti Earth | 32,591 | <p>This is what <a href="https://reference.wolfram.com/language/ref/AbsoluteOptions.html" rel="nofollow noreferrer"><code>AbsoluteOptions</code></a> is for, but it actually doesn't work for me (v11.2), even though the doc states:</p>
<blockquote>
<p>gives the absolute settings of options specified in an expression s... |
614,083 | <p>Remember that we've already proven the following, for any real symmetric <span class="math-container">$n\times n$</span> matrix <span class="math-container">$M$</span>:<br />
(i) Each eigenvalue of <span class="math-container">$M$</span> is real.<br />
(ii) Each eigenvector can be chosen to be real.<br />
(iii) Eig... | Squirtle | 29,507 | <p>If it were... we could not have transitivity, for then everything would be equal.</p>
|
614,083 | <p>Remember that we've already proven the following, for any real symmetric <span class="math-container">$n\times n$</span> matrix <span class="math-container">$M$</span>:<br />
(i) Each eigenvalue of <span class="math-container">$M$</span> is real.<br />
(ii) Each eigenvector can be chosen to be real.<br />
(iii) Eig... | D Left Adjoint to U | 26,327 | <p>Let $x, y$ be toleranced quantities. That is $x = x_0 \pm \sigma_x$, and similarly for $y$. Then $x + y = x_0 + y_0 \pm (\sigma_x + \sigma_y)$. So as you can see the tolerance gets propogated to the result of calculation and increased. This is refered to as <em>propogation of uncertainty</em> and has been studie... |
4,622,525 | <blockquote>
<p>The maximum area of a triangle whose sides <span class="math-container">$a,b,c$</span> satisfy <span class="math-container">$0\le a\le1,1\le b\le2,2\le c\le3$</span></p>
</blockquote>
<hr />
<p>We can clearly see that <span class="math-container">$1,2,3$</span> as sides does not make a triangle so we ca... | Hagen von Eitzen | 39,174 | <p>Suppose we found the optimal value of <span class="math-container">$b$</span>. Then the area is <span class="math-container">$\frac 12hb$</span> where <span class="math-container">$h$</span> is the height and clearly <span class="math-container">$h\le a$</span>. It follows that <span class="math-container">$a$</spa... |
3,148,095 | <blockquote>
<p>Let <span class="math-container">$ f:\{z|\; |z| \lt 1\} \rightarrow \mathbb C $</span> be a non constant analytic function. Which of the following conditions can possibly be satisfied by <span class="math-container">$f$</span> ?</p>
<ol>
<li><span class="math-container">$f(\frac{1}{n})=f(\frac{... | hmakholm left over Monica | 14,366 | <p>You don't need anything as fancy as the identify theorem to reject possibility 2 -- not even anything specific to <em>complex</em> analysis.</p>
<p>First, the only way for <span class="math-container">$f$</span> to be even <em>continuous</em> is if <span class="math-container">$f(0)=0$</span>.</p>
<p>And then, if ... |
185,591 | <p>Let $b \in (0,1)$, $m\in \mathbb{N}$ and $a>0$. I want to bound
$$\sum_{k=m+1}^\infty b^{k^a} \leq c \; b^{m^a}, $$
where $c>0$ is independent from $m$.</p>
<p>Is there a simple way of proving this inequality with $c$ not beeing too large?</p>
<p>Edit: $c$ can only be independent from $m$ if $a>1$. Otherw... | Christian Remling | 48,839 | <p>This cannot work for $a<1$: we then have that $(m+m^{1-a})^a = m^a (1+m^{-a})^a = m^a +O(1)$, so the terms of your series show no decay (except for a multiplicative constant) over an interval of length $\gtrsim m^{1-a}$. Thus
$\sum_{k>m} b^{k^a} \gtrsim m^{1-a} b^{m^a}$ and
$$
b^{-m^a} \sum_{k>m} b^{k^a} \t... |
185,591 | <p>Let $b \in (0,1)$, $m\in \mathbb{N}$ and $a>0$. I want to bound
$$\sum_{k=m+1}^\infty b^{k^a} \leq c \; b^{m^a}, $$
where $c>0$ is independent from $m$.</p>
<p>Is there a simple way of proving this inequality with $c$ not beeing too large?</p>
<p>Edit: $c$ can only be independent from $m$ if $a>1$. Otherw... | Pietro Majer | 6,101 | <p>For $a\ge 1$, the quantity $(m+k+1)^a-m^a$ is increasing w.r.to $m\ge0$ (because $x\mapsto x^a$ is convex). So for $0<b<1$, the term $b^{(m+k+1)^a-m^a} $ is decreasing w.r.to $m\ge0$, and so is $\sum_{k\ge0}b^{(m+k+1)^a-m^a}=b^{-m^a} \sum_{k> m}b^{k^a}$. That is, we have
the required bound with $$c=c(b):... |
1,777,540 | <p>Well-orders on $\mathbb{R}$ are interesting because they witness the "alephness" of the cardinality of the continuum, and they're therefore connected to the continuum problem. It seems reasonable, therefore, to ask: "How well-behaved can well-orders on $\mathbb{R}$ be?"</p>
<p>This is too vague for the website, so ... | Eric Wofsey | 86,856 | <p>Yes, such a well-order exists (in ZF) as long as there exists any well-ordering of $\mathbb{R}$. Indeed, let $\kappa$ be the least ordinal which is in bijection with $\mathbb{R}$ (also known as the cardinal of $\mathbb{R}$), let $f:\mathbb{R}\to\kappa$ be a bijection, and define $x\unlhd y$ to mean $f(x)\leq f(y)$.... |
93,096 | <p>Hi, what are the "best" values for lenght of needle (l) and distance between paralles (d) for an accurate approximation of pi? Does it have to be l-d-1.0 or ld? Thanx</p>
| Allen Knutson | 391 | <p>I'm guessing that the longer the needle the better (meaning, highest $\ell/d$ ratio). It seems like you'll get significant experimental issues from the cases where the needle just barely crosses-or-doesn't-cross a line, and if your needle is long enough to cross lots of lines then this becomes less relevant. Probabl... |
782,812 | <p>I'm trying to solve this recurrence relation: </p>
<p>$$a_m = 8 \cdot a_{m-1} + 10^{m-1}, a_1=1$$</p>
<p>By the change of variable $\displaystyle b_m = \frac{a_m}{10^m}$ I obtained this linear non-homogenous relation:</p>
<p>$$b_m = \frac{8}{10}b_{m-1} + \frac{1}{10}, b_1 = \frac{1}{10}$$</p>
<p>Now I'm trying t... | berkeleychocolate | 145,540 | <p>I agree with your first equation with A but not the equation with A^m. Your approach would work if your first equation related (b_m, b_m-1) with (b_m-1, b_m-2) but it doesn't.</p>
<p>A simple way to solve this is to let b_m = c_m + k and substitute in into the recursion for b_m to get= .</p>
<p>c_m = . 8 c_m-1 + ... |
265,101 | <p>Let</p>
<pre><code>L1={1,2,4,6,8,9,10,14,15,17,21,22,72,76,80,96,106,116,117}
</code></pre>
<p><strong>Question1)</strong> Group from 10-th to 14-th elements in <code>L1</code>.</p>
<p>So that the output becomes</p>
<pre><code>L2={1,2,4,6,8,9,10,14,15,{17,21,22,72,76},80,96,106,116,117}
</code></pre>
<p>A possible s... | Syed | 81,355 | <p><strong>2nd attempt</strong></p>
<pre><code>L1 = {1, 2, 4, 6, 8, 9, 10, 14, 15, 17, 21, 22, 72, 76, 80, 96, 106,
116, 117}
</code></pre>
<p>Here I specify ranges (not in order but non-overlapping)</p>
<pre><code>pos = List /@ # & /@ {Range[10, 14], Range[5, 7] , Range[16, 16]}
</code></pre>
<p>and I insert <c... |
1,174,076 | <p>So I have this basic absolute value problem: $|x-5|<|x+1$|. From what I understand, I need to consider what happens in every case. There are four cases, right? One where both sides are positive, one where both sides are negative and two where one side is negative and the other is positive.</p>
<p>The case were b... | Asinomás | 33,907 | <p>The left hand side is the distance from $x$ to $5$ and the right hand side is the distance from $x$ to $-1$. When is $x$ closer to $5$? The midpoint is $2$. So when $x>2$</p>
|
2,626,630 | <p>$$|\{n\in N:n\mid p^2q\}|$$
$p$ is an odd prime number, $q$ is a prime number and $p$ does not equal $q$.</p>
<p>I dont understand what they're asking me to find, is it to find a value of $n$ that divides into $p^2q$?</p>
<p>I need all the help i can get here, very new to this<br>
thanks</p>
| quasi | 400,434 | <p>Since $101 \equiv 1\;(\text{mod}\;10)$, it follows that $101^{6} \equiv 1\;(\text{mod}\;10)$.
<p>
Since $101$ is not a multiple of $7$, it follows, by Fermat's Little Theorem, that $101^6\equiv 1\;(\text{mod}\;7)$.
<p>
Then $101^6 - 1$ is a multiple of both $10$ and $7$, hence it's a multiple of $70$.
<p>
Since $101... |
3,686,313 | <p>I'm currently learning linear algebra and I have stumbled across the following example in my book without a solution.</p>
<p>A matrix <span class="math-container">$B \in \mathbb {R}^{n x n} $</span> is called idempotent if <span class="math-container">$BB = B$</span>. Prove that if A is idempotent and fullfills <sp... | Minus One-Twelfth | 643,882 | <p><strong>Hint:</strong> <span class="math-container">$AA^{-1}=I_n$</span> by definition. </p>
|
48,366 | <p>Does the Laplace equation on a rectangle with Dirichlet boundary conditions at two opposing sides and Neumann boundary conditions at the other two, always have a solution? If it does, is it unique? Is the same true for the discrete Laplace equation (with the standard five-point laplacian)?</p>
| dranxo | 6,360 | <p>The solution to the Dirichlet problem is unique by the maximum principle. For the same reason, the solution to the Neumann problem is unique up to a constant.</p>
<p>At worst your problem would fail uniqueness by a constant, but those Dirichlet conditions you've got will prevent this.</p>
<p>This carries over to t... |
1,369,805 | <p>How do I calculate the follwing Limit:</p>
<p><span class="math-container">$$ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)dt}{{\int_{0}^{x^2} \sin(t)dt}} $$</span></p>
<p>I have been solving an exam from my University's collection of previous exams, and in one of them, there wasn't an official solution, b... | Bort | 256,093 | <p>neither the formula in the post nor in the picture makes sense. I assume you are looking for</p>
<p>$$ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)\mathrm{d}t}{{\int_{0}^{x^2} \sin(t)\mathrm{d}t}} $$</p>
<p>It's obvious that both terms individually go to zero, thus, you can apply L'Hospital's rule (from... |
1,369,805 | <p>How do I calculate the follwing Limit:</p>
<p><span class="math-container">$$ \lim \limits_{x \to 0^+} \frac{\int_{0}^{x} (e^{t^2}-1)dt}{{\int_{0}^{x^2} \sin(t)dt}} $$</span></p>
<p>I have been solving an exam from my University's collection of previous exams, and in one of them, there wasn't an official solution, b... | Claude Leibovici | 82,404 | <p>Starting from user1337's answer, consider $$A=\frac{e^{x^2}-1}{2x\sin(x^2)}$$ when $x$ is small.</p>
<p>Now use Taylor expansions $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ and replace $y$ by $x^2$ so $$A=\frac{x^2+\frac{x^4}{2}+\cdots}{2x\Big(x^2-\frac{x^6}{6} +\cdots... |
2,792,485 | <p>I need to prove that for all $n\in\mathbb{N}$ the next equality holds: $$\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=0$$First, I thought in induction.
If $n=0$ then $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=\lim\limits_{x\to 0}{e^{-1/|x|}}=0$. If we suppose that $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=0$ the... | farruhota | 425,072 | <p>Make the change: $y=\frac 1x$. Evaluate the left hand and right hand limits:
$$\lim\limits_{x\to 0+}\dfrac{e^{-1/|x|}}{x^n}=\lim\limits_{x\to 0+}\dfrac{e^{-1/x}}{x^n}=\lim\limits_{y\to +\infty}\dfrac{e^{-y}}{\frac{1}{y^n}}=\lim\limits_{y\to +\infty}\dfrac{y^n}{e^{y}}=0,$$
because the exponential function grows faste... |
2,357,297 | <p>Suppose I have lost my keys.
There is:</p>
<ol>
<li>Probability 0.7 i lost it in the main room. </li>
<li>Probability 0.2 i lost it in the bedroom. </li>
<li>Probability 0.1 I lost it in the bathroom.</li>
</ol>
<p>So I look for the keys in the main room first but don't find it.
There is a probability 0.35 I mis... | B. Goddard | 362,009 | <p>If you do the partial fraction expansion, the summand becomes</p>
<p>$$\frac{1}{2^{r+1}}\left( \frac{2}{r} - \frac{1}{r+1}\right) = \frac{1}{2^r r} - \frac{1}{2^{r+1}(r+1)}, $$</p>
<p>So the sequence is telescoping. All terms cancel except the first and so sum equals $\frac12$.</p>
|
1,698,470 | <p>I need to find a numerical solution for $T(x)$ to </p>
<p>\begin{equation}
aT'' - bT^4 + c = 0, \ \ \ T(0) = T_0, \ \ \ T'(L) = 0
\end{equation}
where $a$, $b$, $c$, $T_0$ and $L$ are positive constants. This is a heat conduction and emission problem with boundary conditions, and FYI we know that $T'(0)<0$. </p... | Community | -1 | <p>If this helps,</p>
<p>$$aT'T'' +(c - bT^4)T' = 0$$</p>
<p>then</p>
<p>$$\frac a2T'^2+(cT-\frac b5T^5)=C.$$</p>
|
1,887,807 | <p>I would like to evaluate the following indefinite integral: </p>
<blockquote>
<p>$$\int\frac{1}{\sin^4\theta+\cos^4\theta}\, d\theta$$</p>
</blockquote>
| florence | 343,842 | <p>$$\int \frac{1}{\sin^4t + \cos^4t} dt = \int \frac{1}{\sin^4(t)+\cos^4(t)}\frac{\sec^4(t)}{\sec^4(t)} dt = \int \frac{\sec^2(t)(1+\tan^2(t))}{\tan^4(t)+1} dt = $$
(letting $u = \tan(t), du = \sec^2(t)$)
$$\int \frac{1+u^2}{1+u^4} du = \int \frac{1+u^2}{(u^2-\sqrt{2}u+1)(u^2+\sqrt{2}u+1)} du = $$$$\frac{1}{2}\int \fr... |
1,384,426 | <p>Solve the given equation. Let k be any integer. </p>
<p>$$\csc^2 θ = 5 \cot θ + 7$$</p>
<p>I just need the first step or two please. I tried converting it:</p>
<p>$$\frac{1}{\sin^2 θ} = \frac {5\cosθ}{\sinθ} + 7$$</p>
<p>Then I tried a number of different ways to simplify it but it didn't work out</p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice, $$csc^2\theta=5\cot \theta+7$$ </p>
<p>$$\cot^2\theta+1=5\cot \theta+7$$ $$\cot^2\theta-5\cot \theta-6=0$$
$$(\cot\theta -6)(\cot\theta+1)=0$$
$$\implies \cot\theta-6=0 \iff \tan \theta=\frac{1}{6}\iff \color{blue}{\theta=n\pi+\tan^{-1}\left(\frac{1}{6}\right)}$$</p>
<p>$$\implies \cot\theta+1=0 \iff \tan ... |
2,983,345 | <p><span class="math-container">$3^{10^{100}} \pmod {13}$</span></p>
<p>It was originally <span class="math-container">$(10^{100})^{(10^{100})}$</span> (i.e. googol^googol), but I simplified it down to what I have there. However, I can't seem to reduce the exponent without changing the value of the entire expression.<... | B. Goddard | 362,009 | <p>Note that <span class="math-container">$3^3 \equiv 1 \pmod{13}$</span>, so you need only to reduce the exponent modulo <span class="math-container">$3$</span>. </p>
<p><span class="math-container">$$10^{100} \equiv 1^{100} \equiv 1 \pmod{3},$$</span></p>
<p>s0</p>
<p><span class="math-container">$$3^{10^{100}} \... |
1,549,904 | <p>If you have an investment fund which has four options $x_1,x_2,x_3,x_4$</p>
<p>More should be invested in the combination of funds $x_2$ and $x_3$ than funds $x_1$ and $x_4$ by a ratio of at least $1.5:1$.</p>
<p>The constraint I have formed is as follows:</p>
<p>$$1.5(x_2 + x_3) - (x_1 + x_4) \geq 0.$$</p>
<p>W... | Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>There is a shortcut which, may be, you could like. Consider $$y=\log(1+e^x) \implies y'=\frac{e^x}{e^x+1}=\frac{1}{1+e^{-x}}$$ </p>
<ul>
<li>replace $e^{-x}$ by its Taylor series at $x=0$</li>
<li>perform the long division</li>
<li>integrate each term without forgetting the integration... |
12 | <p>All introductory calculus books that I have seen spend most of their chapters on differential calculus talking about derivatives, with at most a short section defining differentials as <span class="math-container">$dy = f'(x) \, dx$</span>. However, differentials are useful for understanding a lot of things, like l... | Michael Bächtold | 590 | <p>Was Silvanus Thompsons lovely "<a href="http://en.wikipedia.org/wiki/Calculus_Made_Easy" rel="noreferrer">Calculus made easy</a>" mentioned already? It's a classic (100 years old) freely available on <a href="http://www.gutenberg.org/ebooks/33283" rel="noreferrer">gutenberg.com</a>. Some opinions of it can be found ... |
40,763 | <p>Suppose that $G$ is a finite group, and that $H$ and $K$ are normal subgroups of $G$ with trivial intersection, and suppose that $H$ and $K$ are isomorphic. Is it true that the quotient groups $G/H$ and $G/K$ are isomorphic? </p>
| Pete L. Clark | 299 | <p>As Mariano has shown, the answer is a clear <strong>no</strong>.</p>
<p>The best repair I can think of is the following: suppose that $H$ and $K$ are normal subgroups of a group $G$ such that there exists an automorphism $\varphi: G \rightarrow G$ with $\varphi(H) = K$. Then $G/H \cong G/K$: indeed, the isomorphis... |
3,851,476 | <p>I've been given the question</p>
<blockquote>
<p>Let <span class="math-container">$A= \begin{pmatrix}\lambda &-2 \\ -3 & \lambda-1 \end{pmatrix}$</span></p>
<blockquote>
<p>a) for what value/s of <span class="math-container">$\lambda$</span> is <span class="math-container">$A$</span> invertible?</p>
</block... | Nevzat Eren Akkaya | 611,386 | <p>You need to find Det (A) first.A is inverteble if and only if
Det (A) is not equal to 0.Then you can use gauss elimination and elementary row operations or cramers rule
Det(A)=(x-3)(x+2) so it is inverteble for all x which is not equal to 3 or -2</p>
|
3,339,064 | <p>Let <span class="math-container">$R$</span> be a commutative ring with identity and let <span class="math-container">$f\colon R \to R[X]$</span> be the natural ring homomorphism. Let <span class="math-container">$I$</span> be an ideal of <span class="math-container">$R$</span>. What I have shown is that <span class=... | E.R | 325,912 | <p>Fact 1: Consider the ring epimomoephis <span class="math-container">$\psi:R[x]\rightarrow (R/I)[x]$</span> by <span class="math-container">$\psi(a_0+a_1x+...+a_nx^n)=(a_0+I)+(a_1+I)x+...+(a_n+I)x^n$</span>. Clearly, <span class="math-container">$ker(\psi)=I^e=I[x]$</span>. Thus, <span class="math-container">$R[x]/I[... |
2,740,207 | <p>I am trying to evaluate the following integral:
\begin{align}
\int_0^{2\pi}\exp(-C)\cdot\exp\left(a\sin\phi+b\cos\phi\right)\,d\phi\,,
\end{align}
where $C,a,b$ are large constants. I know that this eventually reduces to the modified Bessel function of the first kind. However, since $C, a, b$ are very large (the ter... | TomTom314 | 553,125 | <p>The group homomorphism provides an isomorphism $S_n/\text{ker}(f)\cong \text{im}(f)$ whereby $\text{ker}(f)$ is a normal subgroup. In this case you have $\ker(f)\in\{1,A_n,S_n\}$. G abelian implies $\ker(f)\neq 1$ and therefore $S_n/\text{ker}(f)\cong 1$ or $S_n/\text{ker}(f)\cong\mathbb{Z}/ 2\mathbb{Z}$</p>
<p>Fo... |
2,011,284 | <p>I know that a singular matrix cannot have an inverse, but I was curious as to whether or not a matrix with a non-zero determinant can have an inverse. And from what I understand the requirement for a matrix to be non-singular is that it has a determinant that is not zero.</p>
| John Wayland Bales | 246,513 | <p>Every square matrix $\mathbf{A}$ has an adjugate matrix $\textbf{adj}(\mathbf{A})$ formed from the cofactors of $\mathbf{A}$, even if $\mathbf{A}$ is singular. The inverse of a square matrix is the matrix </p>
<p>\begin{equation}
\mathbf{A}^{-1}=\dfrac{\textbf{adj}\mathbf{(A)}}{\textbf{det}\mathbf{(A)}} \tag{1}
\en... |
633,724 | <p>Prove $e^x, e^{2x}..., e^{nx}$ is linear independent on the vector space of $\mathbb{R} \to \mathbb{R}$</p>
<p>isn't it suffice to say that $e^y$ for any $y \in \mathbb{R}$ is in $\mathbb{R}^+$<br>
Therefore, there aren't $\gamma_1, ...\gamma_n$ such that $\gamma_1e^x+\gamma_2e^{2x}...+\gamma_ne^{nx}=0$.<br>
Theref... | Davide Giraudo | 9,849 | <p>Positiveness of the exponential is not enough as pointed out in the comments and anorton's answer. </p>
<p>Start from the equation $$\forall x\in\mathbb R, \quad \sum_{j=1}^n\gamma_je^{jx}=0.$$
Multiply this equation by $e^{-nx}$. We get for any $x$,
$$\gamma_n+\sum_{j=1}^{n-1}\gamma_je^{(j-n)x}=0.$$
Now, letting $... |
633,724 | <p>Prove $e^x, e^{2x}..., e^{nx}$ is linear independent on the vector space of $\mathbb{R} \to \mathbb{R}$</p>
<p>isn't it suffice to say that $e^y$ for any $y \in \mathbb{R}$ is in $\mathbb{R}^+$<br>
Therefore, there aren't $\gamma_1, ...\gamma_n$ such that $\gamma_1e^x+\gamma_2e^{2x}...+\gamma_ne^{nx}=0$.<br>
Theref... | Robert Lewis | 67,071 | <p>To show that the functions $e^{\alpha_i x}$ are linearly independent over $\Bbb R$ for distinct, <em>real</em> but otherwise arbitrary $\alpha_i$, then the arguments presented by Davide Giraudo and julien in their answers seem to be the way to go. So, +1 for each of their answers and lauds for their fine work!</p>
... |
633,724 | <p>Prove $e^x, e^{2x}..., e^{nx}$ is linear independent on the vector space of $\mathbb{R} \to \mathbb{R}$</p>
<p>isn't it suffice to say that $e^y$ for any $y \in \mathbb{R}$ is in $\mathbb{R}^+$<br>
Therefore, there aren't $\gamma_1, ...\gamma_n$ such that $\gamma_1e^x+\gamma_2e^{2x}...+\gamma_ne^{nx}=0$.<br>
Theref... | LeeNeverGup | 104,910 | <p>You can show it using the <a href="http://en.wikipedia.org/wiki/Wronskian" rel="nofollow noreferrer">wronskian</a>:</p>
<p>Let $f_i=e^{ix}$</p>
<p>Then,
$$W(f_1,f_2,...,f_n)=
\left| \begin{array}{ccc}
e^x & e^{2x} & ... & e^{nx}\\
e^x & 2e^{2x} & ...& ne^{nx} \\
e^x & 4e^{2x} &... &... |
1,132,024 | <p>Let $a,b,c \in \Bbb Z$. Show that the Diophantine equation $ax+by = b+c$ is solvable in integer $x , y$ iff $ax+by =c$ is solvable.</p>
<p>We know that a Diophantine equation $ax+by =c$ is solvable iff $d|c$ where $d=gcd(a,b)$.
From here how can we do the above proof??</p>
| user26486 | 107,671 | <p>$(x,y)=(m,n)$ satisfies $ax+by=c$ iff $(x,y)=(m,n+1)$ satisfies $ax+by=b+c$.<br>
Thus $\exists (x,y)(ax+by=c)\iff\exists (x,y)(ax+by=b+c)$.</p>
|
42,964 | <p>I have an array $v_{ijk}$ which is effectively a rank-$3$ tensor with dimensions $3\times3\times3$, and I need to contract it with $e^{ijk}$, i.e. evaluate $v_{ijk} e^{ijk}$ (see <a href="http://en.wikipedia.org/wiki/Einstein_notation" rel="noreferrer">Einstein convention</a>).<br>
Can anyone tell me how that could... | Никита Кузнецов | 63,518 | <p>@Artes is right, but I have to mention also that if contracting with the Levi-Civitta tensor is something you are going to do many times, it is way more computationally efficient to use the direct summation over the coordinates. Also, it is usefull to calculate the tensor once and then use it without calling the Lev... |
3,820,373 | <p>I'm reading about The Bayes Problem in textbook <em>A Probabilistic Theory of Pattern Recognition</em> by Devroye et al.</p>
<p><a href="https://i.stack.imgur.com/UELeT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UELeT.png" alt="enter image description here" /></a>
<a href="https://i.stack.img... | dmh | 352,034 | <p>Im not sure I understand your question so please let me know if i havent answered it: I believe you have a misunderstanding about <span class="math-container">$\eta$</span>. It is the probability that <span class="math-container">$Y=1$</span> given the value of <span class="math-container">$X$</span>, so it is in ge... |
3,681,096 | <p>This is probably a relatively simple question and one that betrays my lack of knowledge of basic mathematics but it seems to be something I can't quite grasp even though I suspect the answer is quite intuitive. </p>
<p>Suppose we have a number line from 1 to 10. Were I to count the distance from 1 to 10, I would ge... | fleablood | 280,126 | <p>Get in the habit of <em>always</em> calling the starting point <span class="math-container">$0$</span> because the starting point is <span class="math-container">$0$</span> distance from itself. So if you start at <span class="math-container">$1$</span> and end at <span class="math-container">$10$</span> that in t... |
2,346,841 | <p>So I have been trying to solve this question, </p>
<p>$$
\int_{0}^{\pi/2}\left\{\,\cot\left(x\right)\,\right\}\,\mathrm{d}x
$$
where $\left\{\right\}$ means the fractional part.
$\left(\,x - \left\lfloor\,x\,\right\rfloor\,\right)$.</p>
<p>Progress so far, </p>
<ul>
<li><code>MATLAB</code> gives out the following... | Thomas Andrews | 7,933 | <p>Just a start:</p>
<p>The typical way of doing this sort of problem is to let $a_n=\cot^{-1} n=\arctan\frac{1}{n}$. Then the integral can be written:</p>
<p>$$\sum_{n=0}^{\infty} \int_{a_{n+1}}^{a_{n}} ((\cot x)-n)\,dx=\sum_{n=0}^{\infty}\left(n(a_{n+1}-a_{n})+ \int_{a_{n+1}}^{a_n}\cot x\,dx\right)$$</p>
<p>You ge... |
2,346,841 | <p>So I have been trying to solve this question, </p>
<p>$$
\int_{0}^{\pi/2}\left\{\,\cot\left(x\right)\,\right\}\,\mathrm{d}x
$$
where $\left\{\right\}$ means the fractional part.
$\left(\,x - \left\lfloor\,x\,\right\rfloor\,\right)$.</p>
<p>Progress so far, </p>
<ul>
<li><code>MATLAB</code> gives out the following... | Jack D'Aurizio | 44,121 | <p>As a further addendum,
$$ \gamma = \lim_{N\to +\infty}\sum_{n=1}^{+\infty}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right] \tag{1}$$
hence the whole problem boils down to evaluating
$$ A = \sum_{n\geq 1}\left[\frac{1}{n}-\arctan\frac{1}{n}\right]\stackrel{\mathcal{L}^{-1}}{=}\int_{0}^{+\infty}\frac{1-\frac{\s... |
2,022,062 | <p>Does anybody know a function $f$ <strong>that is continuous on $[0, \infty)$</strong>, so that $\int_{0}^{\infty} f(x)dx$ converges but $\int_{0}^{\infty} f(x)^2dx$ diverges? Thanks</p>
<p>EDIT: Sorry! I meant that $f$ is also continuous there.</p>
| Mark McClure | 21,361 | <p>A simple function with this property is $f(x)=\sin(x)/\sqrt{x}$, whose graph looks like so:</p>
<p><a href="https://i.stack.imgur.com/iYEEk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYEEk.png" alt="enter image description here"></a></p>
<p>A simple appeal to the alternating series test sho... |
2,460,255 | <p>How does $F=\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n c_{ij}\dot q_i\dot q_j\implies\frac{\partial{\ F}}{\partial\dot{q_r}}=\sum_{j=1}^nc_{q_rj}\dot q_j$?</p>
<p>I'm not getting how $c_{ij} $ gets transformed to $c_{q_rj}$ and how does the factor $\frac{1}{2}$ vanish...</p>
<p>The original source is <a href="https://www... | jinawee | 73,833 | <p>$c_{q_r j}$ makes no sense to me, the correct answer would be $c_{rj}$ since $\frac{\partial \dot q_i}{\partial \dot q_r}=\delta_{ir}$.</p>
<p>For the $1/2$ factor, try to apply the product rule of derivatives.</p>
|
2,460,255 | <p>How does $F=\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n c_{ij}\dot q_i\dot q_j\implies\frac{\partial{\ F}}{\partial\dot{q_r}}=\sum_{j=1}^nc_{q_rj}\dot q_j$?</p>
<p>I'm not getting how $c_{ij} $ gets transformed to $c_{q_rj}$ and how does the factor $\frac{1}{2}$ vanish...</p>
<p>The original source is <a href="https://www... | alexjo | 103,399 | <p>I think that it is just an abuse of notation to highlight the variable $q_r$ and it's equivalent to</p>
<p>$$
\frac{\partial F}{\partial \dot q_r}=\sum_{j=1}^{n}c_{rj}\dot q_j=-Q^N_r
$$</p>
|
3,379,470 | <p>If there is a number <span class="math-container">$x = 0.333333333$</span> where 3 is repeated infinitely how to convert it to fraction ?
It is said that <span class="math-container">$-> 0.(3) = \frac{3}{10} - 1$</span> and <span class="math-container">$0.33 = \frac{33}{100} - 1$</span>, But I can't understand W... | Quanto | 686,284 | <p>Observe that the solid is symmetric with respect to the y-axis. So, it is efficient to integrate using cylindrical coordinates along the y-direction. Rewrite the original integral as</p>
<p><span class="math-container">$$\iiint_E\sqrt{3x^2+3z^2}\,dV =\int_0^8 dy \int_0^{2\pi}d\theta\int_0^{\sqrt{y/2}} \sqrt{3}r\>... |
28,602 | <p>I am <a href="http://rads.stackoverflow.com/amzn/click/0470073330" rel="nofollow">reading</a>:</p>
<blockquote>
A right cylinder with cross section $\Omega$ is a solid that is formed by translating $\Omega$ along a line, or _axis_, that is perpendicular to it.
</blockquote>
<p>So my understanding now is a cylinder... | Scott Carter | 722 | <p>The authors of that calculus book are giving you <em>their</em> definition of a cylinder for the purposes of working an integral. Certainly, an extrusion would be an effective word here, but not a word that would be widely recognized. An alternative term might be <em>prism</em>, but prism evokes the idea of $\Omega$... |
3,329,650 | <p>Let <span class="math-container">$f$</span> be a complex function <span class="math-container">$f(z) = \frac{z^2 +1}{z(z-i)^3}$</span>. It needs to be put jnto Laurent's series, so that it will converge in a set <span class="math-container">$D = \{ g(z) \mid Re(z) > 0 \}$</span>, where <span class="math-containe... | Martin R | 42,969 | <p>One option is to write <span class="math-container">$g$</span> as a composition of “simple” transformations:
<span class="math-container">$$
g(z) = 2i \left( 1 + \frac{1}{z-1}\right)
$$</span>
and then successively determine the image of the right half-plane.</p>
<p>Another option is to use that fact that <span c... |
4,072,810 | <p>Currently trying to prove that <span class="math-container">$\log(n+2)/\log(n+1)$</span> converges to <span class="math-container">$1$</span> with analysis tools.</p>
<p>if I can prove that <span class="math-container">$\forall \varepsilon >0$</span>, <span class="math-container">$\exists N$</span> such that
<sp... | Moishe Kohan | 84,907 | <p>The first thing to resolve such a confusion is to know the correct definitions (which can be found for instance <a href="https://en.wikipedia.org/wiki/Simple_Lie_group" rel="nofollow noreferrer">here</a>).</p>
<p><strong>Definition.</strong> A connected Lie group <span class="math-container">$G$</span> is said to be... |
1,286,720 | <blockquote>
<p>to get a license, one's need to pass a 3 stages exam.
<br>
If one's fail in one stage he can not continue to the next stage.
<br> The probability he will pass the first stage is $0.9$, if he passed the first stage, the probability he will pass the second test is $0.8$.
if he passed the first... | Asinomás | 33,907 | <p>You made a small slip up when working $\bmod 2$, this is because $0^n$ is always congruent to $0$ no matter what congruence you are working with . I repeat this step once again.</p>
<p>$8^{38}\equiv 3^{38}\equiv3^{9\cdot4}3^{2}\equiv(3^{4})^{9}3^2\equiv1^9\cdot3^2\equiv9\equiv 4\bmod 5$</p>
<p>$8^{38}\equiv0 \bmod... |
1,286,720 | <blockquote>
<p>to get a license, one's need to pass a 3 stages exam.
<br>
If one's fail in one stage he can not continue to the next stage.
<br> The probability he will pass the first stage is $0.9$, if he passed the first stage, the probability he will pass the second test is $0.8$.
if he passed the first... | barak manos | 131,263 | <p>Here is a solution that does not make use of <em>Fermat's Little Theorem</em>.</p>
<hr>
<p><strong>The Algorithm:</strong></p>
<ul>
<li><p>Input: $x=8,e=38,n=210$</p></li>
<li><p>Output: $y=1$</p></li>
<li><p>Repeat until $e=0$:</p>
<ul>
<li><p>If $e\equiv1\pmod2$ then set $y=yx\bmod{n}$</p></li>
<li><p>Set $x=x... |
978,124 | <p>My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.</p>
<p><em>Page 40. Exercise 10.</em> Prove by induction, that for $n\ge1$ we have $$\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{m=1}^{2n}\frac{(-1)^{m+1}}{m}.$$ <em>The attempt at a solution</em>: To prove by induction,... | mvw | 86,776 | <p>The second sum has two summands.</p>
|
3,551,494 | <p>Can someone help me out with <span class="math-container">$$\lim\limits_{x\rightarrow 0^+}x\log(1+x^{-1})?$$</span> I tried Taylor's expansion to no avail.</p>
| K.K.McDonald | 302,349 | <p><span class="math-container">$$\lim_{x\rightarrow 0}x\log(1+x^{-1})=\lim_{x\rightarrow 0}x\log(x^{-1}(x+1))=\lim_{x\rightarrow 0}x\log(x+1)-\lim_{x\rightarrow 0}x\log x = 0-0=0$$</span></p>
<p>We know that <span class="math-container">$\lim_{x\rightarrow 0}x\log x=\lim_{x\rightarrow 0}\frac{\log x}{\frac{1}{x}} $</... |
26,511 | <p>I have spent quite some time trying to figure out what the fastest way is to get a list lists of all subexpressions and their positions. I have tried things with MapIndexed, which seems ideal in cases where we have an expression in which no functions with Hold-Attributes are present that we can let evaluate freely. ... | Mr.Wizard | 121 | <p>I'm not certain I understand your goals, though I too wish there were a cleaner way to do this.<br>
I presume that you are dissatisfied with the performance of this fairly direct solution:</p>
<pre><code>index[expr_] := {Extract[expr, #, HoldComplete], #} & /@ Position[expr, _]
</code></pre>
<p>Your own method... |
31,473 | <p>In a recent post <a href="https://math.meta.stackexchange.com/questions/31425/autofilters-for-hot-network-questions">Autofilters for Hot Network Questions</a> the possibility was raised to created a list of words which when contained in a title of a question would prevent the question from being included in the HNQ ... | quid | 85,306 | <p>I propose the following word</p>
<p><strong>trick</strong></p>
<p>for inclusion in the HNQ block-list.</p>
<p>Please up-vote if you agree with the inclusion. Down-vote if you disagree with the inclusion.</p>
|
374,761 | <p>It is probably an easy question, but somehow I am stuck.</p>
<p><strong>Question</strong> Is the following statement true? If yes, how to prove it?</p>
<blockquote>
<p>Suppose that <span class="math-container">$f\in C^1(\mathbb{R}^n)$</span> is convex and
<span class="math-container">$$
\langle\nabla f(x)-\nabla f(y... | Borbei | 127,661 | <p>A partial answer: if you're willing to strengthen your assumption and suppose <span class="math-container">$f\in C^2(\mathbb{R}^n)$</span>, then yes. Otherwise, I'm not sure.</p>
<hr />
<p>Functions that satisfy -
<span class="math-container">$$\langle\nabla f(x)-\nabla f(y),x-y\rangle \leq L\|x-y\|^2$$</span>
are c... |
374,761 | <p>It is probably an easy question, but somehow I am stuck.</p>
<p><strong>Question</strong> Is the following statement true? If yes, how to prove it?</p>
<blockquote>
<p>Suppose that <span class="math-container">$f\in C^1(\mathbb{R}^n)$</span> is convex and
<span class="math-container">$$
\langle\nabla f(x)-\nabla f(y... | Denis Serre | 8,799 | <p><strong>Yes</strong></p>
<p>Consider first the case where <span class="math-container">$f\in{\cal C}^2$</span>. Then
<span class="math-container">$$\nabla f(y)-\nabla f(x)=\int_0^1{\rm D}^2f(x+t(y-x))\cdot(y-x)\,dt.$$</span>
There follows
<span class="math-container">$$\|\nabla f(y)-\nabla f(x)\|\le\|y-x\|\int_0^1\|... |
2,890,207 | <p>Consider the following system of equations where $x,y,z$ are variables and for a constant $\mathrm C$, $y \times z = \mathrm C \neq 0$</p>
<p>\begin{equation}
\left\{
\begin{array}{lcl}
a_1x + b_1y + c_1z &= 0\\
a_2x + b_2y + c_2z &= 0\\
\end{array}
\right.
\end{equation}</p>
<p>The most straightforward so... | Community | -1 | <p>I just loved Penner and Best's <em>Calculus</em>. I've in some sense been a P hd student for $28$ some odd years, (whether a good thing or a bad thing); but I just remember those $2$ volumes very fondly (used them in high school)...</p>
|
2,890,207 | <p>Consider the following system of equations where $x,y,z$ are variables and for a constant $\mathrm C$, $y \times z = \mathrm C \neq 0$</p>
<p>\begin{equation}
\left\{
\begin{array}{lcl}
a_1x + b_1y + c_1z &= 0\\
a_2x + b_2y + c_2z &= 0\\
\end{array}
\right.
\end{equation}</p>
<p>The most straightforward so... | David C. Ullrich | 248,223 | <p>Rudin "Principles of Mathematical Analysis". Also known as "baby Rudin"...</p>
|
1,786,403 | <p>Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP_i^2}$$</p>
<p>I took $P_1$ as $(3,0)$. Let other points be $P_i(3\... | David C. Ullrich | 248,223 | <p>If you multiply both sides by $x^2+1$ you get $$x^{17}-x=7(x^2+1)$$or $$x^{16}=1+7\frac{x^2+1}{x}.$$A little calculus shows that the right side is negative for all $x<0$. So $x>0$, and now again a little calculus shows that $$x^{16}\ge1+7(2).$$Show that $x\ne1$ and it follows that $x^{16}>15.$</p>
|
2,867,087 | <p>First of all i hope I'm at the right place to ask for help with this problem!</p>
<p>Let's assume a simple 2x2 grid in where all x and y values are not known. The total values are known though.</p>
<pre><code>x1 | x2 | xTotal
y1 | y2 | yTotal
____________________
xy1Total | xy2Total
</code></... | Empy2 | 81,790 | <p>The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions. </p>
<p>Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You c... |
3,688,628 | <p>If <span class="math-container">$a,b\in \mathbb{N}$</span> find all primes <span class="math-container">$p$</span> such that <span class="math-container">$a^3+b^3=p^2$</span></p>
<p>My approach-</p>
<p><span class="math-container">$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$</span> suppose <span class="math-container">$a+b=x$<... | Rushabh Mehta | 537,349 | <p>In the case that <span class="math-container">$a+b=a^2-ab+b^2$</span>, we have <span class="math-container">$a^2+b^2=ab+a+b$</span>, so <span class="math-container">$a^2+b^2=(a+1)b+a$</span>. Without loss of generality (since the formula is symmetric in <span class="math-container">$a,b$</span>) that <span class="ma... |
439,820 | <blockquote>
<p>Is $\large i^{i^i}$ real ? How to find it?</p>
</blockquote>
<p>Thank You!</p>
| Aang | 33,989 | <p>$i^i=e^{i\log i}$</p>
<p>Now on principal branch,using $i=e^{i\pi/2}\implies \log i=i\pi/2$ gives $i^i=e^{-\pi/2}$</p>
<p>Therefore, $i^{i^i}=i^{e^{-\pi/2}}=e^{e^{-\pi/2}\log i}=e^{i(\pi e^{-\pi/2})/2}=\cos\left(\pi \frac{e^{-\pi/2}}{2}\right)+i\sin\left(\pi \frac{e^{-\pi/2}}{2}\right)$</p>
<p>and hence its imagi... |
2,658,754 | <p>This is what I have to show:</p>
<blockquote>
<p>Let $f: G\to H$ be a homomorphism and let $p$ be prime. If range of $f$ has an element
of order $p$, then $G$ has an element of order $p$.</p>
</blockquote>
<p>I start off by proving the lemma:</p>
<blockquote>
<p><strong>Lemma</strong>. For each element $a\i... | Andres Mejia | 297,998 | <p><strong>hint:</strong> since $p \mid o(a)$, we have that</p>
<p>$a^{(mp)}=e$, so $(a^m)^p=e$. </p>
|
3,537,983 | <p>Okay so I have come across an exercise that wants me to compare some outputs of functions. However, I have to do this without a calculator <a href="https://i.stack.imgur.com/rNz5u.png" rel="nofollow noreferrer">Here it is</a></p>
<p>How can I calculate sine/cosine of a real number without using calculator? I notice... | Quanto | 686,284 | <p>Express <span class="math-container">$m$</span> as a function of <span class="math-container">$x\ge 0$</span>,</p>
<p><span class="math-container">$$m(x) = - \frac{x(x^2-x+4)}{(x^2+4)^2}\le0\tag 1$$</span></p>
<p>It is equivalent to find the lower bound of <span class="math-container">$m(x)$</span> for <span class... |
3,364,191 | <p>I've tried to write it in complex exponential form to see if I could spot any geometric series, which led me nowhere near a solution. Mathematica evaluates it to <span class="math-container">$\frac{3}{2}$</span>, any idea on how to arrive to this result?</p>
| Angina Seng | 436,618 | <p>Write <span class="math-container">$\zeta=\exp(2\pi i/N)$</span>. Then
<span class="math-container">$$\cos\left(\frac{2\pi k}N\right)=\frac{\zeta^k+\zeta^{-k}}{2}$$</span>
and then
<span class="math-container">$$\left|1+\cos\left(\frac{2\pi k}N\right)\right|^2
=\left(1+\cos\left(\frac{2\pi k}N\right)\right)^2
=\frac... |
1,145,378 | <p>I believe a function has to be bijective to be invertible, in fact Wikipedia says that an invertible function is another name for a bijection.
<a href="https://en.wikipedia.org/wiki/Bijection" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Bijection</a></p>
<p>But in answers to the question <a href="https:/... | mandez | 214,993 | <p>An invertible function shall be both injective and surjective, i.e Bijective! where every elemenet in the final set shall have one and only one anticident in the initial set so that the inverse function can exist!</p>
|
2,230,226 | <p>This is a contest math problem which I was not able to solve. A hint toward the solution would be helpful as well. </p>
<p>Problem: Show that there are infinitely many powers of 2 starting with the digit 7.</p>
<p>Thanks in advance. </p>
| P Vanchinathan | 28,915 | <p>I was intrigued. Not being smart enough to get a theoretic proof I used brute force. Definitely ugly. Only redeeming aspect is it is constructive.</p>
<p>Here is how I proceeded.</p>
<p>Remark: If two numbers are powers of 2 then so is their product.</p>
<p>IDEA: I will get one power of 2 starting with 7; then I'... |
10,864 | <p>I need to insert a very large expression in a <em>Mathematica</em> program. So I first
Put[large expression,"file"]
and then use </p>
<pre><code>Block[{}, Get["file"];...]
</code></pre>
<p>to implement the function. But the variables here is not local variables and may interact with other programs. I have tried ... | Dr. belisarius | 193 | <p>From the help:</p>
<pre><code>Block[{a = 1, b = 2, c = 3, d = 4}, Get["ExampleData/language"]]
</code></pre>
<p>or</p>
<pre><code>Module[{a = 1, b = 2, c = 3, d = 4}, Evaluate@Get["ExampleData/language"]]
</code></pre>
<p>If you want your variables to be global, simply</p>
<pre><code>a = 1; b = 2; c = 3; d = 4;... |
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