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 |
|---|---|---|---|---|
1,285,213 | <p>Let $f\in P_2(\mathbb R)$, the space of second-order polynomials with real coefficients, and let the linear operator $T$ be defined as $T[f(x)] = f(0)+f(1)(x+x^2)$.</p>
<p>Is $T$ diagonalizable? If so, find a basis $\beta$ of $P_2(\mathbb R)$ in which $[T]_\beta$ is a diagonal matrix.</p>
| math.n00b | 135,233 | <p>You can write the operator in a matrix form by considering the action of this operator on $1$, $x$ and $x^2$ which is the standard basis for second-order polynomials.Then all you need to do is to deal with the eigenvectors of that matrix.</p>
|
1,907,743 | <p>I'm having trouble with a step in a paper which I believe boils down to the following inequality:
$$
\left\| \sum_{k\in\mathbb{Z}} f(\cdot+k) \right\|_{L^2(0,1)}
\leq c \|f\|_{L^2(\mathbb{R})}.
$$
I haven't come up with many ideas. Hitting the left-hand side with Minkowski, for example, produces something which... | Calvin Khor | 80,734 | <p>Inspiration from the Shannon sampling theorem: if you assume $\mathcal F_{\Bbb R} f$ is Schwartz with $\text{supp}\ f$ contained in $(-1/2,1/2)$ then $$ LHS^2 = \sum_k |\mathcal F_{\Bbb R}f(k)|^2 = \sum_k \int_{\Bbb R} f(y) e^{-2\pi i ky} \ \text dy = \sum_k ∫_{-1/2}^{1/2} f(y) e^{-2\pi i ky} \ \text dy = \sum_k |... |
1,701,935 | <p>I've been experimenting with recursive sequences lately and I've come up with this problem:</p>
<blockquote>
<p>Let $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$
<br><br>Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$ </p>
</blockquote>
<p><br><br> I know of $3$ ways... | Singh | 83,768 | <p>By continuity of cosine function, $a_n=\cos a_{n-1}$, which for any $a_0\in [-1,1]$ satisfies $a_n>0$ for all $n=1,2,...$, so can not intersect the open set $[-1,0)-\{a_0\}$. </p>
|
16,725 | <p>I was directed just now to a post with the following abbreviated time-line:</p>
<ul>
<li>Question was posted <strong>21</strong> hours ago</li>
<li>Question was closed as "unclear what you are asking" <strong>19</strong> hours ago</li>
<li>Question was deleted by the votes of three 10K users <strong>4</strong> hour... | stackErr | 59,787 | <p>Suggestion: Can we have a time limit(say 48-72 hours) and an edit limit (say 5 edits) on the question that is on hold such that if any of those limits are crossed then the question can be deleted. If none are crossed then the question cannot be deleted unless marked as spam/offensive/trolling/jokes (Thursdays sugges... |
1,905,898 | <p>The graph of quadratic function drawn on the interval $-1\leq x\leq 5$.</p>
<p><a href="https://i.stack.imgur.com/0eOzE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0eOzE.jpg" alt="enter image description here"></a></p>
<p>i.If the quadratic function is , $y=(x-2)^2-k$, find the value of $k$.... | Steve Suh | 362,486 | <p>If $\gamma<\alpha$, $\gamma$ is not an upper bound for L (since $\alpha$ was the <strong>least</strong> upper bound). It is written above in italics that every element in B must be an upper bound for L. Thus $\gamma$ is not in B.</p>
<p>We know that $\alpha \le x$ for every $x \in B$ because B is a set of upper ... |
1,905,898 | <p>The graph of quadratic function drawn on the interval $-1\leq x\leq 5$.</p>
<p><a href="https://i.stack.imgur.com/0eOzE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0eOzE.jpg" alt="enter image description here"></a></p>
<p>i.If the quadratic function is , $y=(x-2)^2-k$, find the value of $k$.... | fleablood | 280,126 | <p>L is the set of lower bounds of B.</p>
<p>Let $l\in L;b\in B $. As $l $ is a lower bound of $B $, $l\le b $ and therefore every element of $B $ is an upper bound of $L $.</p>
<p>So if $\gamma < \sup L $, $\gamma$ is not an upper bound of $L $ and thus not an element of $B $.</p>
<p>As for the second part. $\... |
384,700 | <p>This question addresses a hierarchy of linear recurrences
which arise from an attempt to generalize the Nekrasov-Okounkov
formula to the Young-Fibonacci setting.
A related posting</p>
<p><a href="https://mathoverflow.net/questions/384591/extensions-of-the-nekrasov-okounkov-formula">extensions of the Nekrasov-Okounko... | David E Speyer | 297 | <p>Let <span class="math-container">$G$</span> be any finite group. Then the group algebra <span class="math-container">$\mathbb{C}[G]$</span> is, as an algebra, isomorphic to <span class="math-container">$\bigoplus_V \mathrm{End}(V)$</span>, where the direct sum is over irreducible representations of <span class="math... |
3,742,189 | <p>in a game I play there's a chance to get a good item with 1/1000.
After 3200 runs I only got 1.</p>
<p>So how can I calculate how likely that is and I remember there are graphs which have 1 sigma and 2 sigma as vertical lines and you can tell what you can expect with 90% and 95% sureness.</p>
<p>Sorry if that's aske... | JMP | 210,189 | <p>You need to align the <strong>positive</strong> <span class="math-container">$x$</span>'s, so the second equation becomes <span class="math-container">$x\gt-1$</span>, and now you can see that your addition involves the two different inequality symbols.</p>
<p>Solve for the first# equation: <span class="math-contain... |
3,514,547 | <p>The problem is as follows:</p>
<p>The figure from below shows vectors <span class="math-container">$\vec{A}$</span> and <span class="math-container">$\vec{B}$</span>. It is known that <span class="math-container">$A=B=3$</span>. Find <span class="math-container">$\vec{E}=(\vec{A}+\vec{B})\times(\vec{A}-\vec{B})$</s... | José Carlos Santos | 446,262 | <p>The answer is <span class="math-container">$-9\vec k$</span>. In fact, the angle between <span class="math-container">$\vec A$</span> and <span class="math-container">$\vec B$</span> has <span class="math-container">$30^\circ$</span> degrees. On the other hand<span class="math-container">\begin{align}\left(\vec A+\v... |
1,006,562 | <p>So I am trying to figure out the limit</p>
<p>$$\lim_{x\to 0} \tan x \csc (2x)$$</p>
<p>I am not sure what action needs to be done to solve this and would appreciate any help to solving this. </p>
| JukesOnYou | 148,363 | <p>$$
\lim_{x\to 0} \tan x \csc (2x)= \lim_{x\to 0} \dfrac{\sin x}{2\cos^2x\sin x} = 1/2
$$</p>
|
2,428,243 | <p>How can I evalute this product??</p>
<p>$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$</p>
<p>Unfortunately, I have no idea.</p>
| Donald Splutterwit | 404,247 | <p>\begin{eqnarray*}
P=\prod_{i=1}^{\infty} (n^{-i})^{n^{-i}} = \prod_{i=1}^{\infty} n^{-in^{-i}} = n^{-\sum_{i=1}^{\infty}in^{-i}}
\end{eqnarray*}
$\sum_{i=1}^{\infty}ix^{i}= \frac{x}{(1-x)^2}$
\begin{eqnarray*}
P=n^{-\frac{n}{(n-1)^2}}
\end{eqnarray*}</p>
|
2,428,243 | <p>How can I evalute this product??</p>
<p>$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$</p>
<p>Unfortunately, I have no idea.</p>
| John Lou | 404,782 | <p>HINT:</p>
<p>$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}} = \prod_{i=1}^{\infty} \frac{1}{n^{\frac{i}{n^i}}} = \frac{1}{n^{\frac{1}{n}}} \cdot \frac{1}{n^{\frac{2}{n^2}}}\cdot\frac{1}{n^{\frac{3}{n^3}}}...$$</p>
<p>$$\prod_{i=1}^{k} {(n^{-i})}^{n^{-i}} = \frac{1}{n^{\frac{n^{k-1} + 2 \cdot n^{k-2}...(k-1) \cdot n + ... |
856,958 | <p>$$|x+y|=|x|+|y| \iff |xy|>0$$</p>
<p>I tried to prove the above inequality but i cant find a way. I tried assuming the first condition is true and tried to derive the second part of it but it seems i can't find a way to get through. I'm new to real analysis and it would be in great help if someone can provide me... | Community | -1 | <p>You want prove that
$$|x+y|=|x|+|y|\iff xy>0$$
which means that $x$ and $y$ have the same sign: to prove the necessary condition we do it by contrapositive so let $x$ and $y$ with opposite signs and prove that $|x+y|\ne|x|+|y|$. Can you take it from here?</p>
|
856,958 | <p>$$|x+y|=|x|+|y| \iff |xy|>0$$</p>
<p>I tried to prove the above inequality but i cant find a way. I tried assuming the first condition is true and tried to derive the second part of it but it seems i can't find a way to get through. I'm new to real analysis and it would be in great help if someone can provide me... | Martin R | 42,969 | <p>Actually $|x+y|=|x|+|y| \iff xy \ge 0$, i.e. if $x$ and $y$ have the same sign
or one of them is zero.</p>
<p>One way to see this is from</p>
<p>$$ |x + y|^2 = (x + y)^2 = x^2 + 2xy + y^2 $$</p>
<p>and</p>
<p>$$ (|x| + |y|)^2 = |x|^2 + 2|x||y| + |y|^2 = x^2 + 2|xy| + y^2$$</p>
<p>so that $|x+y|=|x|+|y| \iff xy... |
422,233 | <p>I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over <strong>Q</strong>.</p>
<p>I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (True... | alans | 80,264 | <p>$$\alpha-\alpha\sqrt{5}+\alpha\sqrt{7}=3\sqrt{5}-2\sqrt{7}+\sqrt{35},$$
$$\alpha-\sqrt{35}=(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7},$$
$$(\alpha-\sqrt{35})^2=[(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7}]^2,$$
$$\alpha^2+35-2\alpha\sqrt{35}=5(\alpha+3)^2+7(\alpha+2)^2-2\sqrt{35}(\alpha+2)(\alpha+3),$$
$$2\sqrt{35}(\alpha^2+4\... |
666,217 | <p>If $a^2+b^2 \le 2$ then show that $a+b \le2$</p>
<p>I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result</p>
| gt6989b | 16,192 | <p>Suffices to show if $a^2+b^2 = 2$ then $a+b \leq 2$. From the constraint consider
$$
f(a) = a + \sqrt{2-a^2}
$$
and we need to prove $f(a) \leq 2$ over $[0,\sqrt{2}]$.</p>
<p>$$
f'(a) = 1 - \frac{a}{\sqrt{2-a^2}} \Leftrightarrow a = 1
$$
which is a maximum by 1st derivative test.
Since $f(0), f(1), f(\sqrt{2})$ are... |
555,446 | <p>Given this shape: <img src="https://i.stack.imgur.com/1rRsC.png" alt="diagram showing a 4000 unit wide cyan square with a 400 unit wide red square in the middle"></p>
<h1>Is it possible to divide the cyan area into 5 equal area shapes</h1>
<p>such that:</p>
<ol>
<li>Each shape is the same</li>
<li>Each shape has ... | Kent Fredric | 106,097 | <p>I don't think it possible to have all shapes identical, because the criteria of "touching the outside" and "touching the inner square" basically means there has to be at least 5 edges running from center to outside.</p>
<p>And because it is impossible to draw 5 edges from a square in such a way that all edges leave... |
4,310,003 | <p>Suppose you have a non empty set <span class="math-container">$X$</span>, and suppose that for every function <span class="math-container">$f : X \rightarrow X$</span>, if <span class="math-container">$f$</span> is surjective, then it is also injective. Does it necessarily follow that <span class="math-container">$... | Laxmi Narayan Bhandari | 931,957 | <p>We start with the substitution <span class="math-container">$e^x=t$</span>. This yields</p>
<p><span class="math-container">$$I = \int\limits_0^\infty\frac{\mathrm dt}{1+t^4} $$</span></p>
<p>Now using <span class="math-container">$t^4=y$</span>,</p>
<p><span class="math-container">$$\begin{align}I &= \frac14\in... |
325,186 | <p>If <span class="math-container">$p$</span> is a prime then the zeta function for an algebraic curve <span class="math-container">$V$</span> over <span class="math-container">$\mathbb{F}_p$</span> is defined to be
<span class="math-container">$$\zeta_{V,p}(s) := \exp\left(\sum_{m\geq 1} \frac{N_m}{m}(p^{-s})^m\right)... | Wojowu | 30,186 | <p>The definition using exponential of such an ad hoc looking series is admittedly not too illuminating. You mention that the series looks vaguely logarithmic, and that's true because of denominator <span class="math-container">$m$</span>. But then we can ask, why include <span class="math-container">$m$</span> in the ... |
3,673,613 | <p>I have to find out if <span class="math-container">$\displaystyle\sum_{n=2}^{\infty}$$\dfrac{\cos(\frac{\pi n}{2}) }{\sqrt n \log(n) }$</span> is absolute convergent, conditional convergent or divergent. I think it's divergent while the value for <span class="math-container">$\cos\left(\dfrac{\pi n}{2}\right)$</span... | user8675309 | 735,806 | <p>I take OP's statement<br>
<em>We can easily prove this inequality</em><br>
<span class="math-container">$\dim(\ker((A-\lambda I)(A-\psi I))) \geq \dim(\ker(A-\lambda I)) + \dim(\ker(A-\psi I))$</span><br>
as a given </p>
<p>we also know that<br>
<span class="math-container">$\dim(\ker((A-\lambda I)(A-\psi I))) \leq... |
172,080 | <p>Here is a fun integral I am trying to evaluate:</p>
<p>$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$</p>
<p>I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial se... | Random Variable | 16,033 | <p>There is a <a href="https://math.stackexchange.com/questions/776903/lobachevskys-formula-for-integrals">theorem</a> that states if <span class="math-container">$f(x)$</span> is continuous and <span class="math-container">$\pi$</span>-periodic on <span class="math-container">$\mathbb{R}$</span>, then <span class="mat... |
172,080 | <p>Here is a fun integral I am trying to evaluate:</p>
<p>$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$</p>
<p>I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial se... | user149844 | 149,844 | <p>I am just adding the proof of the identity for those who have interest:
$$ \sin^{2n+1} x = \frac{1}{4^n}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$
Using the complex representation and the Binomial Theorem, we have
$$\begin{aligned}
\sin^{2n+1}x&=\left(\frac{\mathrm{e}^{ix}... |
2,637,914 | <p>I would like to teach students about the pertinence of the Axiom of Infinity. Are there any high school-level theorems of arithmetic, algebra, or calculus, whose proof depends on the Axiom of Infinity? If there are no such examples, what would be the simplest theorem which demands the Axiom of Infinity?</p>
<p>It... | Akababa | 87,988 | <p>I think you're doing induction backwards; if you assume $n+1$ is true you're already done the inductive step. Assume $n$ is true, that is:
$$\sum_{i=1}^n\frac{i}{i+1}\leq \frac{n^2}{n+1}$$
and try to prove that $$\sum_{i=1}^{n+1}\frac{i}{i+1}\leq \frac{(n+1)^2}{n+2}$$</p>
|
4,076,324 | <p>Let <span class="math-container">$ax+b$</span> be the group of affine transformations <span class="math-container">$x\mapsto ax+b$</span> with <span class="math-container">$a>0$</span> and <span class="math-container">$b\in \mathbb{R}$</span>. How do you topologize this group? As a group it is isomorphic to the... | Igor Rivin | 109,865 | <p><span class="math-container">$a>0, b\in \mathbb{R}$</span> gives you a pretty obvious homeomorphism with the (open) upper halfspace. Is that locally compact?</p>
|
4,076,324 | <p>Let <span class="math-container">$ax+b$</span> be the group of affine transformations <span class="math-container">$x\mapsto ax+b$</span> with <span class="math-container">$a>0$</span> and <span class="math-container">$b\in \mathbb{R}$</span>. How do you topologize this group? As a group it is isomorphic to the... | nullUser | 17,459 | <p>The topology on this group is inherited from <span class="math-container">$\mathbb{R}^2$</span>, and when identified in <span class="math-container">$\mathbb{R}^2$</span> it is an open half-plane. Definitely locally-compact.</p>
|
4,076,324 | <p>Let <span class="math-container">$ax+b$</span> be the group of affine transformations <span class="math-container">$x\mapsto ax+b$</span> with <span class="math-container">$a>0$</span> and <span class="math-container">$b\in \mathbb{R}$</span>. How do you topologize this group? As a group it is isomorphic to the... | José Carlos Santos | 446,262 | <p>Let <span class="math-container">$G$</span> be your group and consider the bijection<span class="math-container">$$\begin{array}{rccc}\psi\colon&(0,\infty)\times\Bbb R&\longrightarrow&G\\&(a,b)&\mapsto&\begin{bmatrix}a&b\\0&1\end{bmatrix}.\end{array}$$</span>Then, consider the distanc... |
1,219,129 | <p>For any vector space $V$ over $\mathbb{C}$, let $X$ be a set whose cardinality is the dimension of $V$. Then $V \cong \bigoplus\limits_{i \in X} \mathbb{C}$ as vector spaces.</p>
<p>Is there a similar description of arbitrary Hilbert spaces? Is there something they all "look" like?</p>
| Alex Zorn | 73,104 | <p>Every Hilbert space is isomorphic to $L^2(X)$ for some measure space $X$. This is a generalization of Tomek's answer.</p>
<p>Now this is redundant, since $X$ is not determined up to isomorphism by $H$. However, this formulation sheds light onto the spectral theorem:</p>
<p>If $A$ is a self-adjoint (possibly unboun... |
4,612 | <p>I would like to make a slope field. Here is the code</p>
<pre><code>slopefield =
VectorPlot[{1, .005 * p*(10 - p) }, {t, -1.5, 20}, {p, -10, 16},
Ticks -> None, AxesLabel -> {t, p}, Axes -> True,
VectorScale -> {Tiny, Automatic, None}, VectorPoints -> 15]
</code></pre>
<p>I solved the diffe... | Jens | 245 | <p>To plot the vector field and the streamlines (curves) together, there are two other plot functions that are specialized for this purpose: </p>
<ul>
<li><a href="http://reference.wolfram.com/mathematica/ref/StreamPlot.html" rel="nofollow noreferrer"><code>StreamPlot</code></a></li>
<li><a href="http://reference.wolf... |
3,189,173 | <p>What will be the remainder when <span class="math-container">$2^{87} -1$</span> is divided by <span class="math-container">$89$</span>?</p>
<p>I tried it solving by Euler's remainder theorem by separating terms:</p>
<p><span class="math-container">$$ \frac {2^{87}}{89} - \frac{1}{89}$$</span></p>
<p><span class="... | Phicar | 78,870 | <p>So <span class="math-container">$$2^{88}\equiv 1\pmod {89}$$</span> and <span class="math-container">$(2,89)=1$</span> so
<span class="math-container">$$2(2^{87}-1)=2^{88}-2\equiv 1-2=-1 \pmod {89}$$</span> so <span class="math-container">$$2^{87}-1\equiv -2^{-1}\pmod {89},$$</span>
but <span class="math-container">... |
2,477,676 | <p>I'm supposed to prove that for any Random Variable X, </p>
<p>$E[X^4] \ge \frac 14 P(X^2\ge \frac 12)$</p>
<p>I tried substituting the definitions of expected value and of the probability into the inequality, but that gets me no where. </p>
<p>Any tips on where to go with this proof? Would a moment generating fun... | Abhiram Natarajan | 481,835 | <p>Use <a href="https://en.wikipedia.org/wiki/Markov%27s_inequality" rel="nofollow noreferrer">Markov's inequality</a>. For a random variable $$X \ge 0, P[X \ge a] \le \frac{E[X]}{a}.$$</p>
<p>We have</p>
<p>\begin{align}
P[X^2 \ge \frac{1}{2}] &= P[X^4 \ge \frac{1}{4}] \qquad \textit{[$X^2 \ge 0$, and $(\cdot)^2... |
3,073,832 | <p>I need to understand the meaning of this mathematical concept: "undecided/undecidable". </p>
<p>I know what it means in the English dictionary. But, I don't know what it means mathematically.</p>
<p>If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.<... | J.G. | 56,861 | <p>If a proposition <span class="math-container">$p$</span> can be stated in the language of a theory <span class="math-container">$T$</span>, we say <span class="math-container">$p$</span> is undecidable in <span class="math-container">$T$</span> if <span class="math-container">$T$</span> contains neither a proof nor ... |
3,464,615 | <p>A novel process of manufacturing laptop screens is under test. In recent tests, it is found that 75% of the screens are acceptable. What is the most probable number of acceptable screens in the next batch of 10 screens and what is the probability?</p>
<p>Does that mean 7 screens out of 10 will pass with a probabili... | Community | -1 | <p>Tabulating (3/4)^x * (1/4)^(10-x) * (10 C x) for x<11, we get that 8 is the most likely number with probability about 28.2%.</p>
|
2,115,532 | <blockquote>
<p>Let $\mu$ be a $\sigma$-finite measure on $(A,\mathcal{A})$. Then
there are finite measures $(\mu_n)_{n \in \mathbb{N}}$ on
$(X,\mathcal{A})$ such that $$\mu = \sum_{n \in \mathbb{N}}\mu_n$$</p>
</blockquote>
<p>So if $\mu$ is $\sigma$-finite, we have that $$X = \bigcup_{n \in \mathbb{N}}X_n$$ fo... | operatorerror | 210,391 | <p>Try evaluating
$$
\int_{\gamma}\frac{e^z}{z}dz
$$
around the unit circle. This will be easy to tackle using the integral formula you mentioned. </p>
<p>Let's just make sure it's the right integral. Parametrize the path as
$z=e^{it}\implies dz=ie^{it}dt$ and
$$
\int_{\gamma}\frac{e^z}{z}dz=
-i\int_{0}^{2\pi} e^{e... |
3,042,802 | <p>For each <span class="math-container">$n ≥ 1$</span>, let <span class="math-container">$T_n = \{x ∈ l_2(N) : ||x||_1 ≤ n \}$</span>.</p>
<p>For <span class="math-container">$n ≥ 1$</span>, is <span class="math-container">$T_n$</span> an absorbing subset of <span class="math-container">$l_2(N) $</span>, but why?
I w... | user289143 | 289,143 | <p>Since <span class="math-container">$l_1(\mathbb{N})$</span> is a proper subset of <span class="math-container">$l_2(\mathbb{N})$</span> we can take <span class="math-container">$x \in l_2(\mathbb{N})- l_1(\mathbb{N})$</span>, i.e. <span class="math-container">$||x||_2< \infty$</span> and <span class="math-contain... |
358,102 | <p>How would I go about doing this?</p>
<p>I assume it is some integral I have to solve, but I have no idea what.</p>
<p>(Note:Not a physicist so please excuse incompetence with regard standard notation.)</p>
<p>Context is I want to estime the energy of N point particles spread over the unit sphere. This is an equat... | Arthur | 15,500 | <p>Let a vertex $v$ be in two different strongly connected components $G_1$ and $G_2$. Then there is one vertex $v_1 \in G_1\setminus G_2$ and one vertex $v_2 \in G_2\setminus G_1$ so that $v$ is strongly connected to both of them. Therefore $v_1$ and $v_2$ are also strongly connected to eachother via $v$, and thus hav... |
54,506 | <p><a href="http://www.hardocp.com/news/2011/07/29/batman_equation/" rel="noreferrer">HardOCP</a> has an image with an equation which apparently draws the Batman logo. Is this for real?</p>
<p><img src="https://i.stack.imgur.com/VYKfg.jpg" alt="Batman logo"></p>
<p><strong>Batman Equation in text form:</strong>
\beg... | Willie Wong | 1,543 | <p>Looking at the equation, it looks like it contains terms of the form
$$ \sqrt{\frac{| |x| - 1 |}{|x| - 1}} $$
which evaluates to
$$\begin{cases} 1 & |x| > 1\\ i & |x| < 1\end{cases} $$</p>
<p>Since any non-zero real number $y$ cannot be equal to a purely imaginary non-zero number, the presence of tha... |
54,506 | <p><a href="http://www.hardocp.com/news/2011/07/29/batman_equation/" rel="noreferrer">HardOCP</a> has an image with an equation which apparently draws the Batman logo. Is this for real?</p>
<p><img src="https://i.stack.imgur.com/VYKfg.jpg" alt="Batman logo"></p>
<p><strong>Batman Equation in text form:</strong>
\beg... | ShreevatsaR | 205 | <p>As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)</p>
<hr>
<p>The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \fr... |
54,506 | <p><a href="http://www.hardocp.com/news/2011/07/29/batman_equation/" rel="noreferrer">HardOCP</a> has an image with an equation which apparently draws the Batman logo. Is this for real?</p>
<p><img src="https://i.stack.imgur.com/VYKfg.jpg" alt="Batman logo"></p>
<p><strong>Batman Equation in text form:</strong>
\beg... | J. M. ain't a mathematician | 498 | <p>Since people (not from this site, but still...) keep bugging me, and I am unable to edit my previous answer, here's <em>Mathematica</em> code for plotting this monster:</p>
<pre><code>Plot[{With[{w = 3 Sqrt[1 - (x/7)^2],
l = 6/7 Sqrt[10] + (3 + x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x + 1)^2],
h = ... |
2,461,615 | <p>I am still at college. I need to solve this problem.</p>
<p>The total amount to receive in 1 year is 17500 CAD.
And the university pays its students each 2 weeks (26 payments per year). </p>
<p>How much does a student have to receive for 4 months?
I have calculated this in 2 ways (both seem ok) but results are di... | Ross Millikan | 1,827 | <p>There are more than $8$ two week periods in four months. If you are paid by the month, the first calculation is correct and the total should be $5833$. On average there are $17$ weeks (and a little) in four months so you would get eight full two week paychecks and one more smaller one. The checks will be smaller ... |
128,708 | <p>My lecture notes say that for every bilinear form there exists a linear operator such that $$\tau (v,w) = v.(Tw)$$ and that there must exist some other linear operator $S$ such that $$(Sv).w = v.(Tw).$$ I understand everything up to there but then it says that it's easy to see that in an orthonormal basis, the matri... | Robert Israel | 8,508 | <p>The matrices that preserve the set $P$ of probability vectors are those whose columns are members of $P$. This is obvious since if $x \in P$, $M x$ is a convex combination of the columns of $M$ with coefficients given by the entries of $x$. Each column of $M$ must be in $P$ (take $x$ to be a vector with a single $... |
128,708 | <p>My lecture notes say that for every bilinear form there exists a linear operator such that $$\tau (v,w) = v.(Tw)$$ and that there must exist some other linear operator $S$ such that $$(Sv).w = v.(Tw).$$ I understand everything up to there but then it says that it's easy to see that in an orthonormal basis, the matri... | Hugo Nava Kopp | 130,222 | <p>Since you originally asked about $L^1$ spaces I dared to add this comment. </p>
<p>If one wants to preserve <strong>the integral</strong> in (finite-dimensional and with finite measure ) $L^1$ spaces rather than <strong>the norm</strong> of $\ell^p$, the matrices $M$ that do this <strong><em>are more general than ... |
2,046,521 | <p>Of course, faster calculations help solve problems quickly. But does that also mean that faster calculations open more opportunities for a career in mathematics (like a researcher)? I like mathematics and can spend weeks trying to solve any problem or understanding any concept. But nowadays, there are many contests ... | Vidyanshu Mishra | 363,566 | <p>After reading you question I remembered an interview given by Scott flansburg who is worldwide accepted as the fastest calculating hymen in the world ( I can't provide that video to you as I found that one luckily). He said in the video that calculation is purely a strong consequence of strong logic and ability to f... |
1,756,685 | <p>For natural numbers—that is, integers greater than or equal to 1—prove that: <br/>
$n^{2n+1}\ge(n+1)^{n+1}(n-1)^{n}$ <br/></p>
<p>Equivalently, show that $(1-1/n)^n$ is strictly increasing.</p>
| Rene Schipperus | 149,912 | <p>One can check that it can be rearranged to</p>
<p>$$1-\frac{1}{n+1}\leq \left(1-\frac{1}{(n+1)^2}\right)^{n+1}$$</p>
<p>And this is an application of the Bernoulli inequality $1+kx\leq (1+x)^k$ </p>
|
1,756,685 | <p>For natural numbers—that is, integers greater than or equal to 1—prove that: <br/>
$n^{2n+1}\ge(n+1)^{n+1}(n-1)^{n}$ <br/></p>
<p>Equivalently, show that $(1-1/n)^n$ is strictly increasing.</p>
| TOM | 118,685 | <p>For $n \ge 2$, the first inequality is equal to </p>
<p>$(1 + \frac{1}{n^2 - 1})^n \ge 1 + \frac{1}{n}$.</p>
<p>This is obvious by the following.</p>
<p>$(1 + \frac{1}{n^2 - 1})^n \ge 1 + \frac{n}{n^2 - 1} \ge 1 + \frac{1}{n}$.</p>
|
1,355,509 | <p>In my mathematical travels, I've stumbled upon the implicit formula $y^2+x^2+\frac{y}{x}=1$ and found that every graphing program I've plugged it in to seems to believe that there is large set of points which satisfy the equation $(y^2+x^2+\frac{y}{x})^{-1}=1$ which do not satisfy the original equation and this has ... | Will Jagy | 10,400 | <p>rotationally symmetric about the origin; your version does not allow $x=0$ but the curve becomes a smooth variety if $(0,0)$ is included</p>
<p>The smooth implicit function is
$$ x^3 + x y^2 + y - x = 0 $$
with gradient
$$ \left\langle 3 x^2 + y^2 - 1, 2xy + 1 \right\rangle $$</p>
<p>For large $|y|,$ solving the... |
18,511 | <p>I have a notebook written in Mathematica 8 in which I imported Tiff images and everything worked fine. Since I installed Mathematica 9, I get the error:</p>
<pre><code>In[14]:= Files[[1]][[1]]
Import[Files[[1]][[1]],"TIFF"]
Out[14]= Growth_1_130124_1353/Growth_1_130124_1353_T0001.tif
During evaluation of In[14]:= I... | cormullion | 61 | <p>Looks like it might be a problem introduced with the colorprofile additions. Perhaps Mathematica's getting confused with the grayscale model and RGB profile? Opens fine in Preview.</p>
<p><img src="https://i.stack.imgur.com/p5diK.png" alt="image details"> </p>
|
737,689 | <p>I have to prove that in a partially ordered set, only one of </p>
<blockquote>
<p>$$x<y,x=y,x>y$$ </p>
</blockquote>
<p>can hold. </p>
<p>My book says if both $x<y$ and $x=y$ hold, then this will imply $x<x$, which is a contradiction (contradicting irreflexivity). </p>
<p>I don't understand how thi... | Laz | 139,851 | <p>Let us prove the statement not only for <span class="math-container">$p$</span> prime, but for all natural numbers <span class="math-container">$n$</span>.<br />
Indeed, <span class="math-container">$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1})$</span>, <span class="math-container">$\forall a,b,n\in \mathb... |
1,136,278 | <p>Prove that $n(n-1)<3^n$ for all $n≥2$. By induction.
What I did: </p>
<p>Step 1- Base case:
Keep n=2</p>
<p>$2(2-1)<3^2$</p>
<p>$2<9$ Thus it holds.</p>
<p>Step 2- Hypothesis: </p>
<p>Assume: $k(k-1)<3^k$</p>
<p>Step 3- Induction:
We wish to prove that:</p>
<p>$(k+1)(k)$<$3^k.3^1$</p>
<p>We ... | Community | -1 | <p>Other approach (not induction): by the binomial theorem,
$$(1+2)^n=1+n.2+\frac12n(n-1).2^2+\frac1{3!}n(n-1)(n-2).2^3\cdots>n(n-1).$$</p>
|
579,907 | <p>Let $G$ be a group and let $H$ be a normal subgroup.</p>
<p>Prove that if $S\subseteq G$ generates $G$,
then the set $\{sH\mid s∈S\} ⊆ G/H$ generates $G/H$.</p>
<p>I have no idea how to deal with the question above.
Can somebody please give me some help?</p>
| Ittay Weiss | 30,953 | <p>You can prove in general that if $\psi:G_1\to G_2$ is a surjective group homomorphism, then if $S\subseteq G_1$ generates $G_1$, then $\psi(S)=\{\psi(s)\mid s\in S\}$ generates $G_2$. The proof is quite straightforward, just follows the meaning of being a generating set. </p>
<p>Now, to conclude what you need to sh... |
4,128,050 | <p><a href="https://i.stack.imgur.com/ybNTh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ybNTh.jpg" alt="enter image description here" /></a></p>
<p>I think we should use the corollary to solve this problem.</p>
<p>For the first question, I think we can use the property that <span class="math-con... | Mod.esty | 766,784 | <p>I think the hardest part of the question is how to prove <span class="math-container">$H_1(G ; \mathbb{Z}^{'}) = \mathbb{Z}$</span>.</p>
<p>We just need to prove that the kernel of <span class="math-container">$J\otimes \mathbb{Z^{'}} \longrightarrow \mathbb{Z}[G]\otimes \mathbb{Z^{'}}$</span> is <span class="math-... |
23,485 | <p>Forgive me if this has been brought up either here or meta.se before, but I could not find it in either.</p>
<p>On a user's activity page, beneath their reputation, there is a label that says "top x% overall". Clicking this brings you to the user reputation leagues. However, there appears to be a flaw here.</p>
<p... | quid | 85,306 | <p>There is a problem, yet it is a bit different from the one claimed. There is no problem with the computation of the percentile. </p>
<p>Let us look at a site with a much smaller number of users to simplify manual checks. </p>
<p>The site Retro Computing has 29 pages of users (a page holds 36 names). This gives a ... |
23,485 | <p>Forgive me if this has been brought up either here or meta.se before, but I could not find it in either.</p>
<p>On a user's activity page, beneath their reputation, there is a label that says "top x% overall". Clicking this brings you to the user reputation leagues. However, there appears to be a flaw here.</p>
<p... | Sklivvz | 272 | <p>This has been fixed. The issue was in the way the "1+" group was calculated. </p>
<p>All the "n+" groups were exactly what they should have been, the number users with rep of at least "n", so basically "200+" includes the "500+" users.</p>
<p>The "1+" was instead calculated assuming the list was of non overlapping... |
2,147,458 | <p>Solve the following integral:
$$
\frac{2}{\pi}\int_{-\pi}^\pi\frac{\sin\frac{9x}{2}}{\sin\frac{x}{2}}dx
$$</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo... |
2,936,028 | <p>The question is:</p>
<p>Prove that If the sum of the elements of each row of a square matrix is k, then the sum of the elements in each row of the inverse matrix is 1/k ?</p>
<p>In the text book the answer is:</p>
<p>Let A be <span class="math-container">${m\times m}$</span>, non-singular, with the stated propert... | Stefan Lafon | 582,769 | <p>Let $M$ be the matrix and $u$ be the vector with 1 for all its elements.
Then saying that the sum of all the elements in the rows of $M$ is $k$ is equivalent to saying that $$Mu= ku$$
Now multiply that equation by $M^{-1}$ to the left:
$$u=kM^{-1}u$$ or $$\frac 1k u=M^{-1}u$$
Which means that the sum of the element... |
206,305 | <p>Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$ by the definition of the limit. $s \geq 0$ and $s_n$ is a sequence of non-negative real numbers.</p>
<p>This is my preliminary computation:</p>
<p>$|\sqrt{s_n} - \sqrt{s}| < \epsilon$</p>
<p>multiply by the conjugate:</p>
<p>$|\dfrac{s_n - s}{\sqrt{s_n}+\sqr... | robjohn | 13,854 | <p>Since $s_n\to s$, for any $\epsilon$, we can find an $N$ so that for all $n\ge N$ we have $|s_n-s|<\epsilon\sqrt{s}$. Then
$$
|\sqrt{s_n}-\sqrt{s}|=\left|\frac{s_n-s}{\sqrt{s_n}+\sqrt{s}}\right|\le\left|\frac{s_n-s}{\sqrt{s}}\right|
$$
to get that
$$
|\sqrt{s_n}-\sqrt{s}|\le\epsilon
$$</p>
|
2,993,979 | <p>I tried to determine if <span class="math-container">$n\cdot \arctan (\frac 1n)$</span> is divergent or convergent. </p>
<p>My solution is in the two pictures. I really have no clue as how to solve it, so I tried something, but it cannot be right. At least that's what I think.</p>
<p>I am sorry in advance for my ... | Claude Leibovici | 82,404 | <p>If we speak about the sequence
<span class="math-container">$$a_n=n \tan ^{-1}\left(\frac{1}{n}\right)$$</span> let <span class="math-container">$x=\frac{1}{n}$</span> and consider
<span class="math-container">$$y=\frac{\tan ^{-1}\left({x}\right)}x$$</span> and use Taylor series of <span class="math-container">$\ta... |
1,617,462 | <p>Is this a line or a plane, I thought it would be a plane where z=0 always so it will be the xy plane.</p>
<p>Also: what will be the normal vector for this if it is a plane?</p>
| Sri-Amirthan Theivendran | 302,692 | <p>It is the collection of vectors that are orthogonal to $(2,-1,0)$ and hence a plane. </p>
|
1,617,462 | <p>Is this a line or a plane, I thought it would be a plane where z=0 always so it will be the xy plane.</p>
<p>Also: what will be the normal vector for this if it is a plane?</p>
| Eli Rose | 123,848 | <p>It's true that $z = 0$ in the equation, but don't think of the equation as <em>requiring</em> $z = 0$ -- instead think of it as <em>putting no conditions</em> on $z$. $z$ doesn't appear in the equation, hence it can be anything.</p>
<p>So this is not the $xy$-plane, but a different plane: the set $\{(x, y, z) \in \... |
784,032 | <p>Find the remainder when $6!$ is divided by 7.</p>
<p>I know that you can answer this question by computing $6! = 720$ and then using short division, but is there a way to find the remainder without using short division?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ In analogy with <strong>Gauss's trick</strong> (see below), to simplify the product we pair up each number with its (multiplicative) inverse mod $7.\,$ Thus $$ 6! = 1\cdot (\overbrace{2\cdot 4}^{\equiv \,1})(\overbrace{3\cdot5}^{\equiv\, 1})\cdot 6 \equiv 1\cdot 1\cdot 1\cdot 6\equiv6\pmod... |
884,362 | <blockquote>
<p>Compute the integral
$$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$</p>
</blockquote>
<p>My Try: I substitute $$\cos(x)=u$$</p>
<p>but it did not help. Please help me to solve this.Thanks </p>
| David | 119,775 | <p>As an <em>indefinite</em> integral this would be hard, maybe impossible, but there is a clever trick for the definite integral. Let
$$I=\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx\ .$$
Substituting $x=2\pi-t$ gives
$$I=\int_0^{2\pi}\frac{(2\pi-t)\cos(t)}{5+2\cos^2(t)}\,dt
=\int_0^{2\pi}\frac{(2\pi-x)\cos(x)}{5... |
3,897,689 | <p>i have the equation:
<span class="math-container">$$y'+2y\:=1$$</span></p>
<p>and i solve it the regular way for first order differential equation:
<span class="math-container">$$y'\:=1-2y$$</span>
<span class="math-container">$$\frac{dy}{dx}=1-2y$$</span>
<span class="math-container">$$\int \:\frac{1}{1-2y}dy=\int ... | user577215664 | 475,762 | <p><span class="math-container">$$y'+2y\:=1$$</span>
With integrating factor method:
<span class="math-container">$$(ye^{2x})'=e^{2x}$$</span>
<span class="math-container">$$ye^{2x}=\dfrac 12 e^{2x}+K$$</span>
<span class="math-container">$$\boxed {y(x)=\dfrac 12 +Ke^{-2x}}$$</span>
Then we can rewrite this as:
<span c... |
3,814,195 | <p>As an applied science student, I've been taught math as a tool. And although I've been studying <strong>a lot</strong> throughout the years, I always felt like I am missing depth. Then I read geodude's answer on this <a href="https://math.stackexchange.com/questions/721364/why-dont-taylor-series-represent-the-enti... | awkward | 76,172 | <p>Assuming you are interested in applications (given your background), my favorite book for applications of complex analysis is <em>Fundamentals of Complex Analysis: with Applications to Engineering and Science</em> by E.B. Saff and A.D. Snider. Their coverage of residue theory, in particular, is more extensive than ... |
63,633 | <p>(This question came up in a conversation with my professor last week.)</p>
<p>Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$.
<br>
Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ?
<br>
What if $G$ is finite?</p>
| Qiaochu Yuan | 290 | <p>Here's a comment which might as well be written down. If $f$ is required to be an inner automorphism, then for $G$ finite this question can be understood using the character table of $G$:</p>
<blockquote>
<p>$x$ is conjugate to its inverse if and only if $\chi(x)$ is real for all characters $\chi$.</p>
</blockquo... |
2,512,736 | <p>I do not understand how this result is a special case of theorem 9.1, could anyone explain this for me please?</p>
<p><a href="https://i.stack.imgur.com/hsgYr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hsgYr.png" alt="enter image description here"></a></p>
<p>This is theorem 9.1:</p>
<p><a... | Community | -1 | <p>$F'(X):H\in M_n\rightarrow A^TH+HA-HUX-XUH$ where $U=BR^{-1}B^T$.</p>
<p>Let $K=F'(X_i)^{-1}F(X_i)$.</p>
<p>At each step, you must solve in $K$ this linear equation:</p>
<p>$A^TK+KA-KUX_i-X_iUK=A^TX_i+X_iA-X_iUX_i+Q$, or</p>
<p>$(A^T-X_iU)K+K(A-UX_i)=A^TX_i+X_iA-X_iUX_i+Q$, which is a Sylvester equation.</p>
<p... |
94,440 | <p>In Sean Carroll's <em>Spacetime and Geometry</em>, a formula is given as
$${\nabla _\mu }{\nabla _\sigma }{K^\rho } = {R^\rho }_{\sigma \mu \nu }{K^\nu },$$</p>
<p>where $K^\mu$ is a Killing vector satisfying Killing's equation ${\nabla _\mu }{K_\nu } +{\nabla _\nu }{K_\mu }=0$ and the convention of Riemann curvatu... | Gravity_CK | 762,708 | <p>We could try to solve it the other way that if a vector obeys the first condition then it must be a Killing vector.</p>
<p>We assume,
<span class="math-container">$$
\nabla_{\mu} \nabla_{\sigma} K^{\rho} = R_{\sigma \mu \nu}^{\rho} k^{\nu}-\text { (i) }
$$</span>
<span class="math-container">$$ where
[\nabla_{\mu},... |
2,991,825 | <p>I'm trying to find the general solution to this matrix
<span class="math-container">\begin{bmatrix}1&-2&1&3&0\\2&-4&4&6&4\\ -2&4&-1&-6&2\\1&-2&-3&3&-8\end{bmatrix}</span></p>
<p>Ax=<span class="math-container">$\begin{bmatrix}1&6&0&-7&\... | Angina Seng | 436,618 | <p>Use <span class="math-container">$\sin x=x+O(x^3)$</span> as <span class="math-container">$x\to0$</span>. Then
<span class="math-container">$$\sin\frac\pi{2^{n+1}}=\frac{\pi}{2^{n+1}}+O(2^{-3n})$$</span>
and
<span class="math-container">$$2^n\sin\frac\pi{2^{n+1}}=\frac{2^n\pi}{2^{n+1}}+O(2^{-2n})$$</span>
etc.</p>
|
365,483 | <p>Let <span class="math-container">$f\colon X\to \mathbb{A}^n_{\mathbb{C}}$</span> be a morphism of <span class="math-container">$\mathbb{C}$</span>-schemes. Suppose <span class="math-container">$f$</span> is (a) separated, (b) flat, (c) locally of finite type, (d) all fibers are quasi-compact, is <span class="math-co... | Angelo | 4,790 | <p>Let <span class="math-container">$X$</span> be the scheme obtained by gluing the generic points of all <span class="math-container">$\operatorname{Spec}\mathcal{O}_p$</span> for all closed points <span class="math-container">$p$</span> of <span class="math-container">$\mathbb{A}^1_{\mathbb C}$</span>. The obvious mo... |
4,228,512 | <p>The question is: does the sequence of characteristic functions <span class="math-container">$f_k(x) := \chi_{[-\frac{1}{k}, \frac{1}{k}]}(x)$</span> converge in distributional sense to the Dirac delta?</p>
<p>In order to answer I followed this approach, but I fear I'm neglecting something important in my lines:</p>
... | Kavi Rama Murthy | 142,385 | <p>You are making things too commplicated. <span class="math-container">$\psi$</span> is a bounded function and if <span class="math-container">$|\psi| \leq M$</span> we get <span class="math-container">$|\int_{-1/k}^{1/k} \psi (x)dx|\leq \frac M {2k} \to 0$</span>.</p>
|
338,535 | <p>Suppose that $f$ is a function defined on the set of natural numbers such that $$f(1)+ 2^2f(2)+ 3^2f(3)+...+n^2f(n) = n^3f(n)$$ for all positive integers $n$.
Given that $f(1)= 2013$, find the value of $f(2013)$.</p>
| masmoudihoussem | 64,548 | <p>This is an easy question.</p>
<p>Let's prove first that for every non negative integer, the following holds:</p>
<p>$$n^2 f(n)=f(1)$$</p>
<p>For $n=2$:</p>
<p>$$f(1)+2^{2} f(2)=2^{3} f(2)$$
$$2^2 f(2)=f(1)$$</p>
<p>Suppose that for every $p$ less than $n$:</p>
<p>$$p^2 f(p)=f(1)$$</p>
<p>Then by hypothesis</p... |
338,535 | <p>Suppose that $f$ is a function defined on the set of natural numbers such that $$f(1)+ 2^2f(2)+ 3^2f(3)+...+n^2f(n) = n^3f(n)$$ for all positive integers $n$.
Given that $f(1)= 2013$, find the value of $f(2013)$.</p>
| lab bhattacharjee | 33,337 | <p>We have $$\sum_{1\le r\le n} r^2f(r)=n^3f(n)$$</p>
<p>Putting $n=m, \sum_{1\le r\le m} r^2f(r)=m^3f(m)$</p>
<p>Putting $n=m+1, \sum_{1\le r\le m+1} r^2f(r)=(m+1)^3f(m+1)$</p>
<p>On subtraction, $$m^3f(m)=f(m+1)\{(m+1)^3-(m+1)^2\}$$</p>
<p>$$f(m+1)=f(m)\cdot\left(\frac m{m+1}\right)^2=f(m-1)\cdot\left(\frac {m(m-... |
2,208,943 | <p>I am about to finish my first year of studying mathematics at university and have completed the basic linear algebra/calculus sequence. I have started to look at some real analysis and have really enjoyed it so far.</p>
<p>One thing I feel I am lacking in is motivation. That is, the difference in rigour between the... | polfosol | 301,977 | <p>Some other answers have already provided excellent insights. But let's look at the problem this way: <em>Where does the need for rigor originates</em>? I think the answer lies behind one word: counter-intuition.</p>
<p>When someone is developing or creating mathematics, they mostly need to have an intuition about w... |
2,208,943 | <p>I am about to finish my first year of studying mathematics at university and have completed the basic linear algebra/calculus sequence. I have started to look at some real analysis and have really enjoyed it so far.</p>
<p>One thing I feel I am lacking in is motivation. That is, the difference in rigour between the... | user64742 | 289,789 | <p>The purpose of "rigor" is to prove that when you claim something in mathematics it actually is legitimately true. If you wish to ask "why" then it is a fairly simple answer:</p>
<p>When we use calculus in machinery, programming, and to solve problems in science at a much larger scale than just a handful of expert s... |
3,489,212 | <p>Playing around I found a series which looks to converge to the square root function.</p>
<p><span class="math-container">$$\sqrt{p^2+q}\overset{?}{=}p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)$$</span></p>
<p>Is it correct?</p>
| Community | -1 | <p>No.</p>
<p><span class="math-container">$$p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)=p\left(1+\frac q{2p^2}\frac1{1+\dfrac q{2p^2}}\right)=p+\frac{pq}{2p^2+q}\ne\sqrt{p^2+q}.$$</span></p>
|
1,393,265 | <p>How to prove that$(n!)^{1/n}$ tends to infinity as limit tends to infinity?
I tried to do this by expanding $n!$ as $n\times (n-1)\times (n-2)\cdots 4\times3\times2\times 1$ and taking out n common from each factor so that I can have $n$ outside the radical sign, But then the last terms would be $(4/n)\times(3/n)\ti... | Zhanxiong | 192,408 | <p>Denote $(n!)^{1/n}$ by $a_n$, then
$$\log a_n = \frac{1}{n}\log n! = \frac{\log 1 + \log 2 + \cdots + \log n}{n}.$$
By the celebrated <a href="https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation#Definition" rel="nofollow">Cesaro's theorem</a> (note the result also holds if the general term tends to $\infty$), since ... |
1,393,265 | <p>How to prove that$(n!)^{1/n}$ tends to infinity as limit tends to infinity?
I tried to do this by expanding $n!$ as $n\times (n-1)\times (n-2)\cdots 4\times3\times2\times 1$ and taking out n common from each factor so that I can have $n$ outside the radical sign, But then the last terms would be $(4/n)\times(3/n)\ti... | David Holden | 79,543 | <p>$$
\lim_{n \to \infty} (n!)^{\frac1n} = \lim_{n \to \infty} \exp\left({\frac1n}\sum_{k=1}^n \log k\right)
$$
for any $n \gt 1$ we have
$$
n\log n -n+1 = \int_1^n \log x dx \lt \sum_{k=1}^n \log k\ \lt \int_1^n \log (x+1) dx \\= (n+1)\log(n+1) -(n+1) -2\log 2 +2
$$
i.e.
$$
\log \frac{n}{e} +\frac1{n} \lt \frac1{n} \... |
3,453,408 | <p>I'm reading through some lecture notes and see this in the context of solving ODEs:
<span class="math-container">$$\int\frac{dy}{y}=\int\frac{dx}{x} \rightarrow \ln{|y|}=\ln{|x|}+\ln{|C|}$$</span> why is the constant of integration natural logged here?</p>
| Quanto | 686,284 | <p>Normally, you need the boundary values to solve the ODE. Assume <span class="math-container">$y(x_0)=y_0$</span>, then the solution is,</p>
<p><span class="math-container">$$\ln |y| - \ln|y_0| = \ln |x| - \ln|x_0| $$</span></p>
<p>Thus, <span class="math-container">$\ln|C|$</span> is necessary and is to be determi... |
45,570 | <p>I'm writing a little package in Mathematica for geology where a particular stone may be approximated as an hemisphere. Anyway this is a rough estimation because a real hemisphere has its height as loong as its radius. Instead, a reservoir stone (for an hydrocarbon) has often a form of a section of an hemisphere, its... | m_goldberg | 3,066 | <h3>Edit</h3>
<p>I now have a better understanding of what you are looking for.</p>
<p>To get plot centered at the origin defined in terms of the radius and height, then you can use <a href="http://reference.wolfram.com/mathematica/ref/SphericalPlot3D.html" rel="nofollow noreferrer"><code>SphericalPlot3D</code></a> as ... |
4,021,994 | <p>I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered <a href="https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_3" rel="nofollow noreferrer">this</a> problem I tried to reason out an algebra formula for it</p>
<blockquote>
<p>For ... | Alan | 175,602 | <p>From the step <span class="math-container">$x-.75x=12.50$</span> simplify to <span class="math-container">$.25x=12.5$</span>, divide by .25 to immediately get <span class="math-container">$x=50$</span>. No looping/brute force needed.</p>
|
319,262 | <p>If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17? </p>
<p>This was from a textbook called "Discrete math and its application", however it does not provide solution for this question... | joriki | 6,622 | <p>Gerry's answer shows that the average sum of the triples is $16.5$. If there's no sum above $17$, then at least five sums have to be $17$ for the average to be $16.5$. Since two successive sums can't be equal, at most five sums are $17$, and thus exactly five sums are $17$, and thus the other five sums are $16$ and ... |
3,386,530 | <p>Let <span class="math-container">$(\Omega,\mathcal{F},\mathbb{P})$</span> be a probability space and <span class="math-container">$(\mathcal{X},d)$</span> be a complete, separable, locally compact metric space. Suppose that <span class="math-container">$X,X_1,X_2,X_3,... : \Omega\to\mathcal{X}$</span> are <span clas... | Kavi Rama Murthy | 142,385 | <p><span class="math-container">$\sqrt {(z_1-x_1)^{2}+(z_2-x_2)^{2}} =\frac 1 2 \min\{r,s\} <s$</span> so <span class="math-container">$(z_1,z_2) \in E$</span>. There is no mistake in the manual. </p>
<p><span class="math-container">$s$</span> is chosen in a particular way and that condition is not met in your ex... |
98,798 | <p>I used of this command to draw a sphere </p>
<pre><code>Graphics3D[{Specularity[White, 50], ColorData["Atoms", "Ag"],
Sphere[{0, 0, 0}, .7]}, Lighting -> "Neutral", Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/TL15F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TL... | Bob Hanlon | 9,362 | <p><a href="http://reference.wolfram.com/language/ref/RegionPlot3D.html" rel="nofollow noreferrer"><code>RegionPlot3D</code></a> has a <a href="http://reference.wolfram.com/language/ref/Mesh.html" rel="nofollow noreferrer"><code>Mesh</code></a> option.</p>
<pre><code>RegionQ[Sphere[{0, 0, 0}, .7]]
(* True *)
Regio... |
98,798 | <p>I used of this command to draw a sphere </p>
<pre><code>Graphics3D[{Specularity[White, 50], ColorData["Atoms", "Ag"],
Sphere[{0, 0, 0}, .7]}, Lighting -> "Neutral", Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/TL15F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TL... | Zviovich | 1,096 | <pre><code>latitude[r_, a_] :=
Line[Table[{r Cos[a] Sin[b], r Sin[a] Sin[b], r Cos[b]}, {b, 0,
2 Pi, .1}]]
longitude[r_, b_] :=
Line[Table[{r Cos[a] Sin[b], r Sin[a] Sin[b], r Cos[b]}, {a, 0,
2 Pi, .1}]]
orbit[r_, a_, incline_] := Rotate[latitude[r, a], incline, {1, 0, 0}]
Graphics3D[{Specularity[White, 0... |
395,791 | <p>I am searching for examples of manifolds which are not symmetric spaces but where Jacobi fields can be computed in closed form. For now, I am aware of</p>
<ul>
<li>Gaussian distribution with the Wasserstein metric: <a href="https://arxiv.org/pdf/2012.07106.pdf" rel="noreferrer">https://arxiv.org/pdf/2012.07106.pdf</... | Robert Bryant | 13,972 | <p>A particularly simple non-homogeneous example in which one can explicitly integrate the Jacobi equations is the complete metric on <span class="math-container">$\mathbb{R}^2$</span> given by
<span class="math-container">$$
g = (x^2{+}y^2{+}2)\bigl(\mathrm{d}x^2+\mathrm{d}y^2\bigr).
$$</span>
It has Gauss curvature <... |
2,259,840 | <blockquote>
<p>Points $P$, $Q$, and $R$ lie on the same line. Three semi-circles with the diameters $PQ$, $QR$, and $PR$ are drawn on the same side of the line segment $PR$. (That is, suppose we have an <a href="https://en.wikipedia.org/wiki/Arbelos" rel="nofollow noreferrer">arbelos</a>.) The centers of the semi-ci... | Jack D'Aurizio | 44,121 | <p>I will outline an approach that can be ultimately used to prove Descartes' (kissing circles) theorem too. The key idea is to perform a circle inversion and to keep track of some distances.</p>
<p><a href="https://i.stack.imgur.com/P6Pm1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P6Pm1.png" a... |
3,999,488 | <p><strong>Question:</strong> How is the differentiation of <span class="math-container">$xy=constant$</span> equal to <span class="math-container">$x\text{d}y+y\text{d}x$</span>?</p>
<p><strong>My Approach:</strong> I first tried using partial differentiation, which I know very little of. Basically, it's the different... | Abhinav Tahlani | 739,290 | <p>Consider a function in a single variable only, say for example <span class="math-container">$f(x)=xsinx$</span>. How would you go about finding <span class="math-container">$f'(x)?$</span> Certainly, you would have the privilege of using <em>chain rule</em> to differentiate <span class="math-container">$f(x)$</span>... |
3,476,022 | <p>I was watching this Mathologer video (<a href="https://youtu.be/YuIIjLr6vUA?t=1652" rel="noreferrer">https://youtu.be/YuIIjLr6vUA?t=1652</a>) and he says at 27:32</p>
<blockquote>
<p>First, suppose that our initial <em>chunk</em> is part of a parabola, or if you like a cubic, or any polynomial. If I then tell you... | Peter LeFanu Lumsdaine | 2,439 | <p>If <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are polynomials agreeing on infinitely many points, then <span class="math-container">$p-q$</span> is a polynomial that’s 0 on infinitely many points.</p>
<p>But if a polynomial <span class="math-container">$f$</span> of degree <... |
627,871 | <p>Let $\mathbf{A}$ be an algebra (in the sense of universal algebra) of some signature $\Sigma$. By <em>quasi-identity</em> I mean the formula of the form</p>
<p>$$(\forall x_1) (\forall x_2) \dots (\forall x_n) \left(\left[\bigwedge_{i=1}^{k}t_i(x_1, \dots, x_n)=s_i(x_1, \dots, x_n)\right]\rightarrow t(x_1, \dots, x... | Bartek | 23,371 | <p>Another way of seeing this about cancellation is to notice that free semigroups/monoids are cancellative, and every semigroup/monoid is a quotient of one.</p>
|
1,688,762 | <p>$$\int \sqrt{\frac{x}{2-x}}dx$$</p>
<p>can be written as:</p>
<p>$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$</p>
<p>there is a formula that says that if we have the integral of the following type:</p>
<p>$$\int x^m(a+bx^n)^p dx,$$ </p>
<p>then:</p>
<ul>
<li>If $p \in \mathbb{Z}$ we simply use binomial expa... | MickG | 135,592 | <p>Let me try do derive that antiderivative. You computed:</p>
<p>$$f(x)=\underbrace{-2\arcsin\sqrt{\frac{2-x}{2}}}_{f_1(x)}\underbrace{-\sqrt{2x-x^2}}_{f_2(x)}.$$</p>
<p>The easiest term is clearly $f_2$:</p>
<p>$$f_2'(x)=-\frac{1}{2\sqrt{2x-x^2}}\frac{d}{dx}(2x-x^2)=\frac{x-1}{\sqrt{2x-x^2}}.$$</p>
<p>Now the mes... |
1,688,762 | <p>$$\int \sqrt{\frac{x}{2-x}}dx$$</p>
<p>can be written as:</p>
<p>$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$</p>
<p>there is a formula that says that if we have the integral of the following type:</p>
<p>$$\int x^m(a+bx^n)^p dx,$$ </p>
<p>then:</p>
<ul>
<li>If $p \in \mathbb{Z}$ we simply use binomial expa... | Machinato | 240,067 | <p>Alternative solution - let $x=2t^2$, then</p>
<p>$$I=\int\sqrt{\frac{x}{2-x}}\mathrm{d}x=4\int\frac{t^2}{\sqrt{1-t^2}}\mathrm{d}t=4J$$</p>
<p>By parts we have</p>
<p>$$J=-t\sqrt{1-t^2}+\int\sqrt{1-t^2}\;\mathrm{d}t = -t\sqrt{1-t^2}+\int\frac{1-t^2}{\sqrt{1-t^2}}\;\mathrm{d}t\!=\!-t\sqrt{1-t^2}+\arcsin t-J $$</p>
... |
1,688,762 | <p>$$\int \sqrt{\frac{x}{2-x}}dx$$</p>
<p>can be written as:</p>
<p>$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$</p>
<p>there is a formula that says that if we have the integral of the following type:</p>
<p>$$\int x^m(a+bx^n)^p dx,$$ </p>
<p>then:</p>
<ul>
<li>If $p \in \mathbb{Z}$ we simply use binomial expa... | notuserealname | 568,250 | <p>Let $u=\sqrt{2-x}$ then we simply want</p>
<p>$-2\int \sqrt{2-u^2}du$ which is simple after $u=\sqrt{2}\sin{v}$</p>
|
3,978,303 | <p><strong>Background</strong></p>
<p>The following Euler product for the Riemann zeta function is well known.</p>
<p><span class="math-container">$$ \sum_n \frac{1}{n^s} = \prod_p (1-\frac{1}{p^s})^{-1} $$</span></p>
<p>Here <span class="math-container">$n$</span> ranges over all integers, <span class="math-container"... | Thomas Andrews | 7,933 | <p>The way to prove this rigorously is to show that:</p>
<p><span class="math-container">$$\sum_{n\leq N}\frac1{n^s}\leq \prod_{p\leq N}\left(1-1/p^s\right)^{-1}\leq\sum_{n=1}^{\infty}\frac1{n^s}$$</span> This you can get because the product in the middle is finite, so you can use the argument without worrying about in... |
4,602,683 | <p>Let <span class="math-container">$\mathbb{F}$</span> be a field, and consider <span class="math-container">$\mathbb{F}^\mathbb{F}$</span> as an algebra over <span class="math-container">$\mathbb{F}$</span> with the standard function multiplication. Let <span class="math-container">$D$</span> be a linear transformati... | Eric Wofsey | 86,856 | <p>Let <span class="math-container">$\mathbb{F}=\mathbb{F}_2$</span> and consider <span class="math-container">$D:\mathbb{F}_2^{\mathbb{F}_2}\to\mathbb{F}_2^{\mathbb{F}_2}$</span> which sends the constant functions to <span class="math-container">$0$</span> and the nonconstant functions to <span class="math-container">... |
20,726 | <p>The following situation is ubiquitous in mathematical physics. Let
$\Lambda_N$
be a finite-size lattice with linear size
$N$. An typical example would be the subset of
$\mathbb{Z}\times\mathbb{Z}$
given by those pairs of integers
$(j,k)$ such that
$j,k \in$ {
$0,\ldots,N-1$}. On each vertex
$j$ of the latt... | jjcale | 17,261 | <p>For VBS quantum antiferromagnets in one dimension see also :</p>
<p>Ian Affleck, Tom Kennedy, Elliott H. Lieb and Hal Tasaki,
Valence bond ground states in isotropic quantum antiferromagnets.
Comm. Math. Phys., Volume 115, Number 3 (1988) </p>
<p>and</p>
<p>Stefan Knabe, Energy gaps and elementary excitations for... |
2,129,830 | <p>I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them. </p>
| MPW | 113,214 | <p>A punctured disk and a slit disk are easy examples in the plane.</p>
|
2,129,830 | <p>I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them. </p>
| Ilmari Karonen | 9,602 | <p>Since the complement of an open set is closed (and vice versa), and since the complement of the interior is the closure of the complement, we can rephrase your question equivalently as:</p>
<blockquote>
<p>Is every closed set the closure of some open set?</p>
</blockquote>
<p>This immediately suggests a countere... |
50,002 | <p>a general version: connected sums of closed manifold is orientable iff both are orientable.
I think this can be prove by using homology theory, but I don't know how.Thanks.</p>
| PseudoNeo | 7,085 | <p>You can also have a differential eye on that matter. I will use a less precise vocabulary than in the other answers.</p>
<p>A manifold is orientable if and only if, when you follow a (smooth) path, you never come back to the starting point with the orientation reversed (as happens for example in the Möbius band). T... |
942,470 | <p>I am trying to count how many functions there are from a set $A$ to a set $B$. The answer to this (and many textbook explanations) are readily available and accessible; I am <strong>not</strong> looking for the answer to that question and <strong>please do not post it</strong>. Instead I want to know what fundamen... | Andrew | 154,986 | <p>Technically, what you've done in your example is defined all possible functions $f:A \to \mathcal{P}(B)$. That is, you're sending elements of $A$ to elements of $\mathcal{P}(B)$. If you want to count functions $f:A \to B$, then the outputs must be <em>elements</em> of $B$, not <em>subsets</em> of $B$.</p>
<p>Anothe... |
1,480,331 | <blockquote>
<p>Let $A$ be an $m \times n$ matrix with $m < n$ and $\operatorname{rank}(A) = m$. Prove that there exist infinitely many matrices $B$ such that $AB = I$.</p>
</blockquote>
<p>Stumped. How do I begin to prove this?</p>
| lulu | 252,071 | <p>Think of it this way: Throwing a $2$ or a $4$ is meaningless, so ignore those cases. Without them we have only $4$ equally likely events: $\{6,Odd,Odd,Odd\}$. The probability of getting the $6$ first is then seen to be just $\frac 14$</p>
|
288,499 | <p>Simply stated, I've been trying for a long time to either find in the literature, or derive myself, a notion of path in Cech closure spaces, that specialises to paths in a topological space, and to graph-like paths in so-called "quasi-discrete closure spaces". </p>
<p>Let me recall the definitions:</p>
<p>A closur... | user2554 | 118,562 | <p>Gauss's procedure leads to Bolyai's result on the volume of orthoscheme tetrahedron, as I'll show here. However, Gauss's result is a little bit more limited than Bolyai, since Gauss refers to an orthoscheme tetrahedron of which 4 of the 12 face angles of the tetrahedron are right (each face is an hyperbolic right tr... |
4,368,464 | <p>How to solve <span class="math-container">$\sum_{i=1}^{n} \frac{P_i}{1+(d_i-d_1)x/365} = 0$</span> in spreadsheet?</p>
<p>We have already known that in Excel,</p>
<p>XIRR() find the root of the equation:
<span class="math-container">$\sum_{i=1}^{n} \frac{P_i}{(1+x)^{(d_i-d_1)/365}} = 0$</span>, which is the IRR (Int... | Arthur | 15,500 | <p>You're asking whether
<span class="math-container">$$
1+2+3+\cdots+n
$$</span>
has the same value as
<span class="math-container">$$
0+1+2+3+\cdots+n
$$</span>
And the answer is that of course those are the same.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.