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 |
|---|---|---|---|---|
1,456,224 | <p>I've been asked to compute the Euler-Lagrange equation and second variation of the functional $$I[y]=\int_{a}^{b}(y'^2+y^4)dx$$
with boundary conditions $y(a)=\alpha$, $y(b)=\beta$. It's easy to see that $$I[y+\delta y]=I[y]+\int_{a}^{b}\delta y(4y^{3}-2y'') dx+\int_{a}^{b}(6y^{2}\delta y^{2}+\delta y'^{2})dx$$
So t... | Community | -1 | <p>The Euler-Lagrange equation </p>
<p>$$y^4 - (y')^2 = k$$</p>
<p>is actually not too hard to solve once you know $y(a) = y(b) = 0$. Note that $y(a) = 0$ imply $k \le 0$. If $k=0$, then $(y')^2 = y^4$ and you can check that the only solution (using $y(a) =0$ and $y'' - 4y^3 = 0$) is $y(t) = 0$. </p>
<p>If $k<0$,... |
2,503,306 | <p>Suppose $g{^n}$=e. Show the order of $g$ divides $n$.</p>
<p>Would I use Eulers Theorem???;</p>
<p>$a{^{\phi p}}$ $\equiv1 \pmod p$</p>
<p>$a{^{p-1}}\equiv1 \pmod p$</p>
<p>$a{^p}\equiv a\pmod p$</p>
<p>So then I would have </p>
<p>$g{^n}\equiv g\pmod n$</p>
<p>then I think you use the $\gcd$, which states $\... | lhf | 589 | <p>Consider $A = \{ k \in \mathbb Z : g^k =e \}$. Then $A$ is subgroup of $\mathbb Z$ and so $A=m\mathbb Z$, where $m$ is the smallest positive element of $A$. By definition, $m$ is the order of $g$. Thus, if $n\in A$, then $n$ is a multiple of $m$.</p>
<p>This approach may work for you if you see the subgroups of $\m... |
1,537,648 | <p>For example let's say we have a password combination of (a,b,c,d), if the password length was 1 then we'll have 4 possible passwords (a,b,c,d), now if the length was 2 then we'll have 20 possible passwords (a,b,...,dc,dd), I calculated this manually, I want the rule of calculating probability?</p>
| Iceman | 91,561 | <p>Suppose you want a password of length up to $n$. Suppose also that you have $m$ possible digits to pick from...</p>
<p>The total possible 1-passwords would just be $m$ (m letters)</p>
<p>The total possible 2-passwords would just be $m^2$ (m letters times m letters)</p>
<p>...</p>
<p>The total possible n-passwor... |
138,708 | <p>I'm dealing with derivatives of scalar functions of matrices and wondering if Mathematica can help me here.</p>
<p>The standard approach of expanding it in terms of components is cumbersome. As an motivating example, I want to minimize the following function, where $X$ is a matrix</p>
<p>$$f(X) = \text{tr}(X'X)$$... | Carl Woll | 45,431 | <p>In this post I discuss a function <code>MatrixD</code> which attempts to take a matrix derivative following the guidelines given in the <a href="http://www2.imm.dtu.dk/pubdb/views/edoc_download.php/3274/pdf/imm3274.pdf" rel="noreferrer">The Matrix Cookbook</a>.</p>
<p>I still want to take advantage of the normal pa... |
1,267,395 | <p>Julie is required to pay a 2 percent tax on all income over 3,000. She also has to pay 2.5 percent on all income over 20,000. She earned more than 20,000 and paid 992.50 what was her total income</p>
| user70160 | 228,922 | <p>She earns more than $20,000$</p>
<p>So she pays $2\%$ on her her income between $3,000$ and $20,000$, i.e. she pays $2\%$ on $17,000$ which gives:</p>
<p>$17,000\times2\% = 340$</p>
<p>Subtract that from the given total tax burden of $992.5$ to get the tax she paid for her income above $20,000$ as follows:</p>
<... |
1,457,063 | <p>I am utterly confused on how to solve this problem. I found a lemma that says $|A\cup B|=|A|+|B|$ is true if the two sets are disjoint which makes sense, but how do I prove the entire statement. </p>
| Sak | 5,209 | <p>Consider the function $f:A\cup B\to A\sqcup \{0\}$ and $g:A\cup B\to B\sqcup\{0\}$ given by $f(x)=x$ if $x\in A$ and $f(x)=0$ in other case; $g(y)=y$ if $y\in B$ and $g(0)=0$ in other case. </p>
<p>These functions are surjective. Then, $|A\cup B|\leq |A|$ and $|A\cup B|\leq |B|$, and therefore $|A\cup B|\leq |A|+|... |
1,174,359 | <p>I have the following problem where n is a positive integer $(n >= 1)$:</p>
<p>Prove that $\frac{1}{2n}\le\frac{1*3*5*...*(2n-1)}{2*4*...*2n}$</p>
<p>I know that I must start with the basic step showing that $P(1)$ is true as follows:
$1/(2*1) = 1/2$ so $P(1)$ is true.</p>
<p>Now follows the induction step wher... | Hippalectryon | 150,347 | <p>This is wrong. Because $f'(a)=b$ doesn't mean at all that $(f^{-1})'(a)=\frac{1}b$</p>
<p>(quick counterexample on $\Bbb{R}^+_*$ for $x=1$ : take $f(x)=e^x,f^{-1}(x)=\ln x$,$f'(1)=e,(f^{-1})'(1)=1$)</p>
<p>What you should do : find the inverse function of $f$, then compute its derivative the given point.</p>
<p>H... |
1,174,359 | <p>I have the following problem where n is a positive integer $(n >= 1)$:</p>
<p>Prove that $\frac{1}{2n}\le\frac{1*3*5*...*(2n-1)}{2*4*...*2n}$</p>
<p>I know that I must start with the basic step showing that $P(1)$ is true as follows:
$1/(2*1) = 1/2$ so $P(1)$ is true.</p>
<p>Now follows the induction step wher... | Mathemagician1234 | 7,012 | <p>y= f(x) = $\sqrt{5x}$ $\rightarrow$ $y^2$ = $5x$ $\rightarrow$ $\frac{1}{5}y^2$ = x </p>
<p>So $f^{-1}(x)$ = $\frac{1}{5}x^2$ , which is easily verified by taking $f^{-1}(f(x))$= x. </p>
<p>Now:
$(f^{-1})'(x)$ = $\frac{2}{5}x$. The slope of the tangent line of $f^{-1}$ at (4,$\frac{16}{5}$) is $(f^{-1})'(4)$ = $\... |
56,394 | <p>Hi!</p>
<p>While studying C*-algebras I found 2 different definitions for non degenerate representations (<em>-homomorphisms $\pi:\mathcal{A} \rightarrow B(\mathcal{h})$ where $\mathcal{A}$ is a C</em>-algebra and $B(\mathcal{h})$ is the space of bounded linear operators on some Hilbert space $\mathcal{h}$):</p>
<... | Stefan Waldmann | 12,482 | <p>In fact, for unital $C^*$-algebras non-degeneracy just means $\pi(1) = 1$. In the non-unital case there is even a sharper statement than your item (2): One can find for every $\phi$ and every $\epsilon > 0$ another vector $\psi$ and a <em>positive</em> algebra element $a \in \mathcal{A}^+$ with
\begin{equation}
... |
2,439,111 | <p>By definition, a function $ f: \mathbb{R} \rightarrow\mathbb{R}$ is linear iff</p>
<ol>
<li>$f(x+y)=f(x)+f(y)$ $ \forall x,y \in \mathbb{R}$ </li>
<li>$ f(bx) = bf(x)$ $ \forall b,x \in \mathbb{R}$</li>
</ol>
<p>I am trying to prove the following statement: </p>
<p>If $ f $ is a linear map defined above then $f$... | Community | -1 | <ul>
<li><p>if part: $a(x+y) = (ax) + (ay)$ and $a(bx) = b(ax)$.</p></li>
<li><p>only if part: $f(x)=f(x\cdot1)=xf(1)=ax$.</p></li>
</ul>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="mat... | Jack D'Aurizio | 44,121 | <p>$\sqrt{2017}\approx\sqrt{2000}=20\sqrt{5}\approx 20\cdot 2.236 \approx 45$ and
$$44^2 = 1936,\qquad 45^2=2025$$
hence $\sqrt{2017}\in\color{red}{\left(44,45\right)}$.</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="mat... | Bram28 | 256,001 | <p>For a rough estimate, I'd first divide by $100$ and think about $20.17$, for $\sqrt{2017} = 10 \sqrt{20.17}$. In fact, I would just consider $\sqrt{20}$: </p>
<p>You know $4^2=16$ and $5^2 =25$, so $\sqrt{20}$ is between $4$ and $5$, and is in fact close to the middle of them, i.e close to $4.5$. Hence, $\sqrt{201... |
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="mat... | Jam | 161,490 | <p>There are well enough good answers here, but no one's suggested this method yet, so I'll add it.</p>
<h3>Method <span class="math-container">$1$</span> - (Secant Approximation)</h3>
<p>We can take two simple perfect squares that straddle <span class="math-container">$2017$</span>. We're just getting a rough estimate... |
930,949 | <p>Given that the circle C has center $(a,b)$ where $a$ and $b$ are positive constants and that C touches the $x$-axis and that the line $y=x$ is a tangent to C show that $a = (1 + \sqrt{2})b$</p>
| Mauro ALLEGRANZA | 108,274 | <p>The language of <em>first-order</em> logic is made of :</p>
<ul>
<li><p>sentential <em>connectives</em></p></li>
<li><p><em>quantifiers</em></p></li>
<li><p>the <em>equality</em> symbol</p></li>
<li><p>countable many (individual) <em>variables</em> : $x_1, x_2, \ldots$</p></li>
<li><p>a set (possibly empty) of <em>... |
2,805,975 | <p>For matrices $A, B$, I would like to show and understand the intuition behind the following identity
$$
(A+B)^{-1} = A^{-1} - (A+B)^{-1} B A^{-1}
$$
assuming the inverses exist.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$(A+B)\big [ A^{-1} - (A+B)^{-1} B A^{-1}\big ] =I + BA^{-1}-BA^{-1}=I$$</p>
<p>Thus $$(A+B)^{-1} = A^{-1} - (A+B)^{-1} B A^{-1}$$</p>
<p>The intuition behind it is probably $$ \frac {1}{a+b} = \frac {1}{a} -\frac {1}{a+b}\frac {b}{a}$$</p>
<p>where a and b are real numbers. </p>
|
1,278,719 | <p>This is a problem from Artin's book "Algebra". In the fifth miscellaneous problem of the chapter "Vector spaces", he has asked to prove that:</p>
<p>If $\alpha$ is a cube root of $2$, then the real numbers $a+b\alpha +c\alpha ^2$ with $a,b,c \in \mathbb{Q}$ form a field.</p>
<p>I am stuck at proving this. For exam... | Trevor J Richards | 5,952 | <p>This series converges <strong>conditionally</strong> for every $x$. To see this, fix some $x\geq0$. Then once $n>x$, $\dfrac{1}{n+x}\geq\dfrac{1}{2n}$, so by the comparison test, the series $\displaystyle\sum_n\dfrac{1}{n+x}$ diverges. Similarly if $x<0$.</p>
|
4,572,505 | <p>There are many approximations of <span class="math-container">$\pi$</span> using trigonometric and rational numbers. But I created this one: <span class="math-container">$$\pi \approx \sqrt[11]{294204}$$</span> Which is correct to almost <span class="math-container">$8$</span> decimal places. Are there any other app... | Tito Piezas III | 4,781 | <p>The well-known <span class="math-container">$\pi \approx \frac{\ln\left(5280^3\,+\,744\right)}{\sqrt{67}}$</span> and <span class="math-container">$\pi \approx \frac{\ln\left(640320^3\,+\,744\right)}{\sqrt{163}}$</span> involves the <em>j-function</em> and integers. But we can also use the <em>Dedekind eta function<... |
1,949,966 | <h2>Q 1a</h2>
<p>Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?</p>
<p>I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.</p>
<p>To be more explicit, I want as many properties to hol... | Dirk | 3,148 | <p>The absolute value is quite a different thing than a square. A square simply comes from multiplication and nothing else. Especially, a square does not need an order on the underlying structure. However, the absolute value can only be defined after an order in defined by setting
$$
|x| = \begin{cases} x & x\geq 0... |
1,706,939 | <p>Can anyone share an easy way to approximate $\log_2(x)$, given $x$ is between $0$ and 1?</p>
<p>I'm trying to solve this using an old fashioned calculator (i.e. no logs)</p>
<p>Thanks!</p>
<p>EDIT: I realize that I stepped a bit ahead. The x comes in the form of a fraction, e.g. 3/8, which is indeed between 0 and... | Community | -1 | <p>First normalize the value to the range $[1,2)$, multiplying by $2$ as long as necessary (the number of multiplies will form the integer part of the logarithm).</p>
<p>Then use the formula</p>
<p>$$\log_2\left(\frac{1+t}{1-t}\right)=\frac2{\ln(2)}\left(t+\frac{t^3}3+\frac{t^5}5\cdots\right)$$ evaluating for </p>
<... |
288,340 | <p>I am having difficulty understanding the recursive definition of a language. The problem asked how to write this non recursively. But I want to understand just how a recursive definition of a language works.</p>
<p>Recursive definition of a subset of L of $\{a,b\}^*$.</p>
<p>Basis : $a\in L$ </p>
<p>Recursive D... | Community | -1 | <pre><code> a
/ \
aa ab
/ \ / \
aaa aab aab abb
/ \ / \ / \ / \
aaaa aaab aaab aabb aaab aabb aabb abbb
</code></pre>
|
2,041,251 | <p>Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.</p>
<p>I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help ... | Paramanand Singh | 72,031 | <p>Use Taylor's theorem to get $$f\left(\frac{1}{2}\right)=f(0)+\frac{1}{2}\cdot f'(0)+\frac{1}{8}\cdot f''(c)$$ for some $c\in(0,1/2)$ and your answer follows. </p>
<p>Note that the result can not be obtained by mean value theorem directly, but rather one should observe that proofs of mean value theorem and Taylor's ... |
1,079,262 | <p>How many permutations exist in the string $ABCDEFG$, starting from the smallest possible combination if the only direction allowed is forward? For example, B is the smallest possible combination in the string $BDEF$. The only direction being forward, $BD, BE, BF$ are larger permutations, etc.</p>
<p>NOTE: You can't... | Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcomma... |
1,079,262 | <p>How many permutations exist in the string $ABCDEFG$, starting from the smallest possible combination if the only direction allowed is forward? For example, B is the smallest possible combination in the string $BDEF$. The only direction being forward, $BD, BE, BF$ are larger permutations, etc.</p>
<p>NOTE: You can't... | robjohn | 13,854 | <p>Your approach is correct. Perhaps more direct is
$$
\begin{align}
\frac{e^{k\pi}(1+z+z^2/2+O(z^3))}{z^2(1-z^2/3+O(z^4))}
&=\frac{e^{k\pi}(1+z+5z^2/6+O(z^3))}{z^2}\\
&=\color{#C00000}{e^{k\pi}}\left(\frac1{z^2}+\color{#C00000}{\frac1z}+\frac56+O(z)\right)
\end{align}
$$
Thus, the residue at $z=k\in\mathbb{Z}$... |
106,031 | <p>I need to "monochromize" a large amount of plots (mostly coming from <code>ListPlot</code>) and export them to PDF. The problem is that I no longer have the data used to generate the plots, I only have notebooks that contain the plots. I attempted copy-pasting one plot and then something along the lines of <code>Sho... | Sjoerd C. de Vries | 57 | <p>Example plot:</p>
<pre><code>Plot[{Cos[x], Sin[x], Tan[x]}, {x, 0, 2 π}]
</code></pre>
<p><img src="https://i.stack.imgur.com/DBYMS.png" alt="Mathematica graphics"></p>
<p>Now, just copy this plot from the notebook (using ctrl-C) and paste it in the front of following expression:</p>
<pre><code>/. a : RGBColor[_... |
1,595,118 | <p>What is value of $a+b+c+d+e$? If given :</p>
<p>$$abcde=45$$</p>
<p>And $a,b, c, d, e$ all are distinct integer.</p>
<hr>
<p>My attempt :</p>
<p>I calculated, $45 = 3^2 \times 5$.</p>
<blockquote>
<p>Can you explain, how do I find the distinct values of $a,b, c, d, e$ ?</p>
</blockquote>
| Panglossian Oporopolist | 218,322 | <p>Indeed... the values would be $5,3,-3,-1,1$. So adding them should give you $5$.</p>
|
185,061 | <p>Let $p$ be a prime number. For which finite $p$-groups $H$ is there a finite $p$-group $G$ such that $[G,G] \cong H$?</p>
| Dietrich Burde | 32,332 | <p>There are some results for special cases. Burnside has proved in $1912$ that,
if $G$ is a non-metabelian $p$-group, then the centre of the derived group of $G$ cannot be
cyclic. In particular, a non-Abelian group of order $p^3$ cannot be the derived group of
a $p$-group. Blackburn later described the 2-generator gr... |
2,065,254 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is twice differentiable.</p>
<p>We know that:
$$\lim_{x\to-\infty}\ f(x) = 1$$</p>
<p>$$\lim_{x\to\infty}\ f(x) = 0$$</p>
<p>$$f(0) = \pi$$</p>
<p>We have to prove that there exist at least two points of the function in which $f''(x) = 0$.</p>
<p>How could w... | Barry Cipra | 86,747 | <p>For the heck of it, let's do a proof that invokes only the Intermediate and Mean Value Theorems. In particular, let's eschew any explicit use of the Extreme Value Theorem.</p>
<p>If $f$ ever takes the same value infinitely often, the Mean Value (or Rolle's) Theorem applied twice shows there are two values of $x$ f... |
3,043,296 | <p>Prop: For sets A and B, say A ~ B iff there exists a bijection from A to B. Then ~ is an equivalence relation on sets.</p>
<p>I understand that an equivalence relation holds the properties of reflexive, symmetric, and transitive. I am also aware of their definitions, however, I am struggling to write a proof for th... | Community | -1 | <p>We only need to show three properties: </p>
<p><strong>Reflexiveness:</strong></p>
<p>Does there exist for ANY set <span class="math-container">$A$</span> a bijective function between the set and itself?
If so, we can say that <span class="math-container">$A \sim A$</span>.</p>
<p>Hint, what if there were a func... |
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<spa... | Robert Z | 299,698 | <p>Combinatorial proof of
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
that is, after dividing by $n!$,
$$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n}=\binom{2n+1}{n+1}.$$
The RHS counts the number of $\{0,1\}$-strings of length $2n+1$ with $(n+1)$ $1$s. The LHS ... |
4,280,894 | <blockquote>
<p>Given
<span class="math-container">$$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$</span>
calculate
<span class="math-container">$$
\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}
$$</span>
where <span class="math-container">$f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \te... | River Li | 584,414 | <p><em>Alternative proof</em>:</p>
<p>Let
<span class="math-container">\begin{align*}
J_n &= \int_0^1 (1 - x^2 + x^3)^n \,\mathrm{d} x, \\
K_n &= \int_0^1 (1 - x^2 + x^3)^n \ln (x + 2)\,\mathrm{d} x.
\end{align*}</span></p>
<p>First, we have
<span class="math-container">$$K_n \ge \int_0^1 (1 - x^2 + x^3)^n \l... |
4,280,894 | <blockquote>
<p>Given
<span class="math-container">$$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$</span>
calculate
<span class="math-container">$$
\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}
$$</span>
where <span class="math-container">$f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \te... | Piquancy | 979,182 | <p>Fisrtly,we prove that:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span>
Perform piecewise calculation:<span class="math-container">$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\i... |
4,180,750 | <p>Let's consider inner product space with vectors <span class="math-container">$x, y, z$</span> which satisfies:</p>
<p><span class="math-container">$$\|x+y+z\|^2 = 14$$</span></p>
<p><span class="math-container">$$\|x+y-z\|^2 = 2$$</span></p>
<p><span class="math-container">$$\|x-y+z\|^2 = 6$$</span></p>
<p><span cla... | José Carlos Santos | 446,262 | <p>Assuming that you are working over <span class="math-container">$\Bbb R$</span>, let:</p>
<ul>
<li><span class="math-container">$s=\|x\|^2+\|y\|^2+\|z\|^2$</span>;</li>
<li><span class="math-container">$a=2\langle x,y\rangle$</span>;</li>
<li><span class="math-container">$b=2\langle x,z\rangle$</span>;</li>
<li><spa... |
1,705,081 | <p>It's a matrix solved with least squares equations (probaly). I used some calculator but can't get his outcome. If you have a way how to get to this please explain how.</p>
<p>[The math on that image is:
$$A = \left[\matrix{4&3&1&0&1\cr 5&2&1&0&1\cr 4&2&1&1&1\cr 3&... | iiivooo | 218,986 | <p>Your answer is correct. To compare the logarithm you should notice that
$$\lim_{x->\infty}\frac{(\log(x))^p}{x}=0$$
for any positive constant $p$ and from this you immediately see that $(\log n)^3$ has smaller rate that $\sqrt n\log n$ and both have smaller order than all polynomials. </p>
|
21,752 | <blockquote>
<p>"Let $P$ be the change-of-basis matrix
from a basis $S$ to a basis $S'$ in a
vector space $V$. Then, for any vector
$v \in V$, we have $$P[v]_{S'}=[v]_{S}
\text{ and hence, } P^{-1}[v]_{S} =
[v]_{S'}$$</p>
<p>Namely, if we multiply the coordinates
of $v$ in the original b... | Agustí Roig | 664 | <p>Everybody studying the change of basis affair should work out some simple examples like the following. Consider this basis in $\mathbb{R}^2$:</p>
<p>$$
v_1 = (1,1) \qquad \text{and} \qquad v_2 = (1,-1) \ .
$$</p>
<p>Or, since we are going to stress the bases and coordinates thing, we could write it this way</p>
... |
1,651,991 | <p>Let $p(x)$ be an odd degree polynomial and let $q(x)=(p(x))^2+ 2p(x)-2$ </p>
<p>a) The equation $q(x)=p(x)$ admits atleast two distinct real solutions.</p>
<p>b) The equation $q(x)=0$ admits atleast two distinct real solutions.</p>
<p>c) The equation $p(x)q(x)=4$ admits atleast two distinct real solutions.</p>
<... | GoodDeeds | 307,825 | <p>The straight line joining $(-2,-0.5)$ and $(0.25,0.5)$ is given by
$$y-0.5=\frac{0.5-(-0.5)}{0.25-(-2)}(x-0.25)$$
$$y-0.5=\frac{1}{2.25}(x-0.25)$$
$$2.25y-1.125=x-0.25$$
$$2.25y-x=0.875$$
$$18y-8x=7$$</p>
<p>Now, substituting $x=-1$, we get,
$$18y+8=7$$
$$y=-\frac{1}{18}$$</p>
|
1,634,807 | <p>The full question:</p>
<p>Having an equivalence relation $\sim$ on $\Bbb N$ defined by: $a \sim b$ meaning $a,b\in\Bbb N$ such that $a=b*10^k,$ for some $k\in\Bbb Z$, give a complete set of equivalence class representatives.</p>
<p>I am having trouble visualising these. I'm thinking you would need the whole set $\... | Asinomás | 33,907 | <p>The numbers that are not multiples of $10$ give us a complete set of representatives, and in fact we have exactly one representative for each class. Let $X$ be the set of all positive numbers not divisible by $10$</p>
<p>To see why it is a complete set of representatives take $n\in\mathbb N$, suppose $10^k$ is the ... |
4,398,864 | <p>How many ways are there to select a three digit number <span class="math-container">$\underline{A}\ \underline{B}\ \underline{C}$</span> so that <span class="math-container">$A \neq B$</span>, <span class="math-container">$B \leq C$</span>, and <span class="math-container">$A < C$</span>?</p>
<hr />
<p>I found th... | lulu | 252,071 | <p>If the three digits are distinct and non-zero, then there are two ways to arrange any selection of three digits from <span class="math-container">$\{1,\cdots, 9\}$</span>. Thus this gives us <span class="math-container">$2\times \binom 93=168$</span> cases.</p>
<p>If <span class="math-container">$B=0$</span>, there... |
2,418,440 | <p><strong>Defn:</strong> Let $f$ be a function from $\mathbb{R}$ into a set $X$. We say that $f$ is <em>periodic</em> if there exists $p>0$ such that for all $x\in \mathbb{R}$, we have $f(x+p)=f(x)$. </p>
<p><strong>Prove</strong>: If $f$ is a continuous periodic function from $\mathbb{R}$ into a metric space $M$,... | DanielWainfleet | 254,665 | <p>For $r>0$ let $s_r\in (0,p)$ such that $\forall x,y\in [0,2p]\;(|x-y|<s_r\implies |f(x)-f(y)|<r). $</p>
<p>For any $x,y\in \mathbb R$ with $|x-y|<s_r$ there exists $n\in \mathbb Z$ such that $\{x,y\}\subset [np, (n+2)p].$ Because if $n=\max \{m\in \mathbb Z: mp\leq \min (x,y)\}$ then $\min (x,y)<(... |
2,523,660 | <p>I'm trying to solve the next question:</p>
<p>For all $m\in I=(0,1)$ there is a subset $A_m \subseteq \mathbb{R}$ that $A_{m} = \{ a\in \mathbb{R} : a-\lfloor a \rfloor < m \} $. Find $\bigcap\limits_{m\in I} A_{m}$</p>
<p>So I think that the solution is $\bigcap\limits_{m\in I} A_m=\mathbb{Z}$, and I tried to ... | William Elliot | 426,203 | <p>$\pi$ is continuous. Thus its inverse image of an open set is open and a closed set closed. No, the inverse is not an open nor a closed map because it is not a map from X to X×Y. </p>
<p>It is a function from the power set of X to the power set of X×Y and those two power sets do not have a topology. Thus it is ... |
3,328,737 | <p>For any rational number, <span class="math-container">$\frac{p}{q}$</span> , <span class="math-container">$p$</span> and <span class="math-container">$q$</span> should be integers, <span class="math-container">$q\neq0$</span> and <span class="math-container">$p,q$</span> should not have any common factors.
Now, if w... | Ahmed Hossam | 430,756 | <p><a href="https://artofproblemsolving.com/wiki/index.php/Proof_by_contradiction" rel="nofollow noreferrer">https://artofproblemsolving.com/wiki/index.php/Proof_by_contradiction</a></p>
<p>Proof (by contradiction) that <span class="math-container">$\sqrt{2}$</span> is irrational.</p>
<p><strong>Assume for a minute, ... |
1,646,135 | <p>I have the subset $\left[0,1\right] \backslash \mathbb{Q}$ in $\mathbb{R} \backslash \mathbb{Q}$. </p>
<p>Am I right in thinking that this set is open and not closed in the space given?</p>
<p>Also, how do I go about finding the interior, closure and boundary?</p>
| lulu | 252,071 | <p>No such function exists. </p>
<p>To see that, suppose you had such an $f$. Let's compute a few examples.</p>
<p>Case I: $$A=\{1,2,3\},\,B=\{3,4,5\},C=\{1,4,6\}$$</p>
<p>Then it is easy to compute everything in your expression. We get $$0=f(3,3,3,1,1,1)$$</p>
<p>Case II: $$A=\{1,2,3\},\,B=\{1,4,5\},C=\{1,6,7\... |
3,446,084 | <p>Suppose we have coprime integers <span class="math-container">$(a,b)$</span> and let <span class="math-container">$\ell \in \mathbb{Z}$</span> be arbitrary. The general solution to the linear Diophantine equation <span class="math-container">$ax+by=\ell$</span> is given by <span class="math-container">$x=\ell x' + b... | A.J. | 654,406 | <p>Suppose <span class="math-container">$\,(x',y')\,$</span> is the minimal solution to the <span class="math-container">$\,ax+by=1\,$</span> generated by the extended Euclidean algorithm. Then as you pointed out, <span class="math-container">$\,(nx',ny')\,$</span> will be a solution to <span class="math-container">$\,... |
2,922,881 | <p>I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$</p>
<p>The point $z_0=0$ seems to be a singularity. I'm not sure but I think it's also a non-removable one. I tried using the Taylor expansion of $e^x$ and $\cos{x}$ as that usually hel... | eloiprime | 180,579 | <p>Assuming that $|z|=1$ is positively oriented (which is the convention), then your working looks fine to me. The residue theorem does not care about the type of singularities within the simply-connected domain around which you're integrating, just that there are finitely many such singularities. Furthermore, the meth... |
1,926,839 | <p>I have to find a function $f\in C^\infty(\mathbb{R})$ which, for fixed $a,b\in \mathbb{R}$, $a<b$, is identically 1 for $x\le a$, identically $0$ for $x\ge b$ and decreases in $a\le x\le b$. I've tried many times to write $f$ in $a\le x\le b$ using an exponential, but it didn't work. Can you help me?</p>
| Ian | 83,396 | <p>The function</p>
<p>$$f (x)=\begin {cases} e^{-1/x^2} & x>0 \\ 0 & otherwise \end {cases} $$
is smooth, increases to 1 from 0. Subtracting it from 1 gives a function that decreases from 1 to 0. Pick your favorite smooth increasing homeomorphism $h $ from $(a,b) $ to $(0,\infty)$. Then define $g(x)=1-f (h... |
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and func... | Alex Becker | 8,173 | <p>I'm afraid the person who told you that was wrong. There is no difference between a mapping and a function, they are just different terms used for the same mathematical object. Generally, I say "mapping" when I want to emphasize that what I am talking about pairing elements in one set with elements in another set, a... |
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and func... | Will Jagy | 10,400 | <p>Not that much difference in the long run. When I use the word function I generally mean that a point maps to a single point. So, if a point might map to several points, I am not going to use that word, more likely mapping or transformation. In a recent article I had one of these, each point went to several points, a... |
20,224 | <p>I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof? </p>
| David R. MacIver | 4,959 | <p>Should have searched a bit harder before asking this one. This is an immediate consequence of the Hewitt-Marczewski-Pondiczery theorem:</p>
<p>Let $m \geq \aleph_0$. If $\{X_s : s \in S\}$ are topological spaces with $d(X_s) \leq m$ and $|S| \leq 2^m$ then $d(\prod_s X_s) \leq m$.</p>
|
20,224 | <p>I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof? </p>
| Henno Brandsma | 2,060 | <p>This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2010;task=show_msg;msg=0487.0001" rel="nofollow">here</a>. It's in fact not that hard, the fact for a product of copies of 2 point discrete spaces already implies... |
3,299,469 | <p>Consider the set <span class="math-container">$A=\{n\ a \}$</span> where <span class="math-container">$a>0$</span> is a constant and <span class="math-container">$n \in \mathbb{N}$</span></p>
<p><strong>How shall we write this set <span class="math-container">$A$</span> in set theory?</strong></p>
<p>If we writ... | J.G. | 56,861 | <p>You can also write it as <span class="math-container">$a\Bbb N$</span>.</p>
|
4,049,293 | <p>I am learning about the cross entropy, defined by Wikipedia as
<span class="math-container">$$H(P,Q)=-\text{E}_P[\log Q]$$</span>
for distributions <span class="math-container">$P,Q$</span>.</p>
<p>I'm not happy with that notation, because it implies symmetry, <span class="math-container">$H(X,Y)$</span> is often us... | Hagen von Eitzen | 39,174 | <p><span class="math-container">$$ \begin{align}f(n+1)-f(n)&=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac7{15}
\\&=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac7{15}\\&=\text{integer}+\frac15+\frac13+\frac 7{15}\end{align}$$</span>
is an integer for all integers <span class="math-container">$n$</sp... |
194,671 | <p>I'm searching for two symbols - considering they exist - (1) unknown value; (2) unknown probability.</p>
<p><strong>Note</strong>: I thought that $x$ was used in a temporary context, whenever I see it, it remains unknown until an evaluation is made. I was thinking in a "unknown and impossible to be known" context. ... | Aang | 33,989 | <p>Generally used Symbol for unknown value is $x$ and other letters.</p>
<p>and</p>
<p>unknown probability of some event $A$ is denoted as $P(A)$ or $P_A$.</p>
|
194,671 | <p>I'm searching for two symbols - considering they exist - (1) unknown value; (2) unknown probability.</p>
<p><strong>Note</strong>: I thought that $x$ was used in a temporary context, whenever I see it, it remains unknown until an evaluation is made. I was thinking in a "unknown and impossible to be known" context. ... | Trevor Wilson | 39,378 | <p>One convention of notation is to use symbols from near the end of the alphabet (e.g., $x$, $y$, $z$, $t$ as others have answered) for "unknown" values and symbols from near the beginning of the alphabet (e.g., $a$, $b$, $c$) for "known" or "fixed" values. Of course conventions vary between fields of mathematics.</p... |
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| Clive Newstead | 19,542 | <p><strong>Hint:</strong> If <span class="math-container">$f(x) = x^3 + x^2 - 2x - 1$</span>, then what is <span class="math-container">$f(x+2)$</span>?</p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| José Carlos Santos | 446,262 | <p>Apply Eisenstein's criterion to <span class="math-container">$(x+2)^3+(x+2)^2-2(x+2)-1$</span>, with <span class="math-container">$p=7$</span>.</p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| Bill Dubuque | 242 | <p>We don't need to pull the Eisenstein shift out of a hat like magic. I explain how to find it below.</p>
<p><strong>Hint</strong> <span class="math-container">$\, \bmod\color{#c00}7\!:\,\ f(x) \equiv (x\!-\!2)^{3} \ $</span> is a <em>prime power</em>. </p>
<p>So Eisenstein works on <span class="math-container">$\,... |
14,583 | <p>Given a <code>Graph</code> with an automatically computed layout (i.e. not explicitly given <code>VertexCoordinates</code>, but using a <code>GraphLayout</code> method), how can we extract the coordinates of the vertices?</p>
<pre><code>In[]:= g = RandomGraph[{10, 20}, GraphLayout -> "SpringEmbedding"]
Out[]= &l... | ssch | 1,517 | <p>In version 8, you can use:</p>
<pre><code>VertexCoordinates /. AbsoluteOptions[g, VertexCoordinates]
</code></pre>
<p><code>AbsoluteOptions</code> is usually a good bet when other things just return <code>Automatic</code></p>
<p>In version 9, there's the <a href="http://reference.wolfram.com/mathematica/ref/Graph... |
1,631,505 | <p>Give the precise meaning of the limit- $$\lim_{x\to-\infty} f(x) = +\infty$$ (x is going to negative infinity, the symbol is hard to see)</p>
<p>I know that as $x$ gets smaller and smaller, $f(x)$ gets larger and larger, but how do I put that in terms of a precise definition?</p>
| Zhanxiong | 192,408 | <p>The definition is standard:</p>
<blockquote>
<p>For any $M > 0$, there exists $K < 0$ (usually sufficiently small) such that for all $x < K$, it holds $f(x) > M$.</p>
</blockquote>
<p>To make analogy to the normal function limit, you might also treat $-\infty$ as a special point, whose neighbors are ... |
3,637,785 | <p>Prove that
<span class="math-container">$${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n},$$</span></p>
<p><span class="math-container">$$\frac{1}n {2n -2 \choose n-1} 2^{-2n +1} = (-1)^{n-1} {\frac12 \choose n}.$$</span></p>
<p>The second part can be proved by replacing <span class="math-container">$n$</spa... | Olivier Roche | 649,615 | <p>Assume for contradiction that there are non zero natural numbers <span class="math-container">$a, b$</span> such that <span class="math-container">$a^2 = 2 b^2$</span>. Consider <span class="math-container">$a \neq 0 \in \mathbb{N}$</span> minimal such that there is <span class="math-container">$b \in \mathbb{N}$</s... |
2,573,492 | <p>In my course "Introduction To Algebraic Topology" I had following test problem:</p>
<blockquote>
<p>Exemplify a topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$.</p>
</blockquote>
<p>I was supposed to use this theorem:</p>
<blockquote>
<p>Let $Y$ be a simply connected topological space. If a ... | Andres Mejia | 297,998 | <p>There is an explicit construction:</p>
<p>Consider the CW complex given by three $0$ cells, three $1$ cells and a two cell pasted in (this is a filled in triangle.)</p>
<p>Now, considering the circle with a base point, and pasting in this $CW$-complex by identifying all three vertices with the base point (in a thr... |
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https... | Misha | 21,684 | <p>Igor, the following is not an answer but, I think, is as close to an answer as one can get at this time. First of all, Bill Thurston is making two good points:</p>
<p>(a) There is no algorithm in the context of hyperbolic groups, as it follows from the Rips' 1982 construction. </p>
<p>(b) At this stage, there is n... |
912,426 | <p>A bag contains six chips, numbered 1 through 6. If two chips are chosen at random without replacement and the values on those two chips are multiplied, what is the probability that this product will be greater than 20?</p>
<p>I tried to solve by counting the total possibilities (36) and solving for 6 choices that w... | amcalde | 168,694 | <p>$$\prod_{n=1}^N \frac{2n-1}{2n} = \exp\left\{ \log \left( \prod_{n=1}^N \frac{2n-1}{2n} \right)\right\} = \exp\left\{ \sum_{n=1}^N \log \left(\frac{2n-1}{2n} \right)\right\}$$</p>
<p>$$ = \exp\left\{ \sum_{n=1}^N \log(2n-1) - \log(2n)\right\} = \exp\left\{ \sum_{n=1}^N \log(2n)+\log(1-\frac{1}{2n}) - \log(2n)\righ... |
478,566 | <p>I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.</p>
<p>Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?</p>
<p>My proof... | Community | -1 | <p>Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?</p>
|
38,193 | <p>For simplicity, let me pick a particular instance of Gödel's Second Incompleteness
Theorem:</p>
<p>ZFC (Zermelo-Fraenkel Set Theory plus the Axiom of Choice, the usual foundation of mathematics) does not prove Con(ZFC), where Con(ZFC) is a formula that expresses that
ZFC is consistent.</p>
<p>(Here ZFC can be replac... | Nik Weaver | 23,141 | <p>Yudkowsky and Herreshoff have a (messy but) <a href="https://intelligence.org/files/TilingAgentsDraft.pdf" rel="nofollow noreferrer">great paper</a> which relates the second incompleteness theorem to issues in theoretical artificial intelligence. (<a href="https://arxiv.org/pdf/1312.3626.pdf" rel="nofollow noreferre... |
233,367 | <p>A set of $m$ non-zero <strong><em>rationals</em></strong> {$a_1, a_2, ... , a_m$} is called a <em><a href="https://web.math.pmf.unizg.hr/~duje/intro.html" rel="nofollow">rational Diophantine $m$-tuple</a></em> if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties... | Tito Piezas III | 12,905 | <p>(<em>Too long for a comment, but may help in a generalization.</em>)</p>
<p>After some sleuthing around, it turns out $(1),(2),(3)$ can be encapsulated in the single equation,</p>
<p>$$(a b c d e + 2a b c + a + b + c - d - e)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d e + 1)\tag1$$</p>
<p>which I think is by Dujella. For... |
3,977,531 | <p>The Cantor-Bendixson theorem implies that any closed subset of the Cantor set <span class="math-container">$\mathcal{C}$</span> can be described as a disjoint union of a set <span class="math-container">$\mathcal{C}_c$</span> that is homeomorphic to the original Cantor set, and a countable open set <span class="math... | Mees de Vries | 75,429 | <p>The "countable open set <span class="math-container">$\mathcal C_o$</span>" is not open as a subset of the cantor set <span class="math-container">$\mathcal C$</span>, it is open as a subset of the set that can be written as <span class="math-container">$\mathcal C_c \sqcup \mathcal C_o$</span>. Non-empty ... |
3,977,531 | <p>The Cantor-Bendixson theorem implies that any closed subset of the Cantor set <span class="math-container">$\mathcal{C}$</span> can be described as a disjoint union of a set <span class="math-container">$\mathcal{C}_c$</span> that is homeomorphic to the original Cantor set, and a countable open set <span class="math... | Henno Brandsma | 4,280 | <p>The Cantor-Bendixsson theorem says that <span class="math-container">$C$</span> is a union of it's scattered part (open) and its perfect part (closed). The scattered part here is empty (as <span class="math-container">$C$</span> has no isolated points, and <span class="math-container">$C'=\emptyset$</span>, and ther... |
1,884,958 | <p>How do I show that <span class="math-container">$$\text{Var}(aX+b)=a^2\text{Var}(X).$$</span>
Since I am reading statistics for the first time, I don't have any idea how to start.</p>
<p>Thanks for helping me.</p>
| Sachchidanand Prasad | 249,258 | <p>See the solution is easy but at least you have to try once. Just applying the definition of variance you will get the desired result. Although I am writing the solution but please try by yourself.</p>
<blockquote class="spoiler">
<p> $$\begin{align*}\mathsf{Var}(X)& =\mathsf E[X^2]-\mathsf E[X]^2\\[2ex]\impli... |
1,736,376 | <p>Let $R$ be a ring and $I$ the set of non-invertible elements of $R$. </p>
<p>If $(I,+)$ is an additive subgroup of $(R,+)$, then show that $I$ is an ideal of $R$ and so $R$ is local. </p>
<p>$$$$ </p>
<p>I have done the following: </p>
<p>Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $\forall a,... | user1043065 | 34,162 | <blockquote>
<p>Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $∀a,b∈I
> : ab∈I.$</p>
</blockquote>
<p>This reasoning is wrong. Moreover, you don't need to prove this statement. The product of a non-invertible element and any other element is non-invertible. Therefore it is immediate that $IR\subs... |
1,237,528 | <p>$$ \displaystyle {\int_{0}^{z}} \sqrt {1 + \tan^2(\dfrac{\pi}{4} \dfrac{z}{H} )} dz $$</p>
<p>_</p>
<p>$$ gives $$ </p>
<p>_</p>
<p>$$ \dfrac{4H}{\pi} {\sinh^{-1}} ( {\tan \dfrac{\pi}{4} \dfrac{z}{H} } ) $$</p>
<p>Please advise solution</p>
<p>edit:- </p>
<p>I can get to </p>
<p>$$\dfrac{4H}{\pi} \displaysty... | GFauxPas | 173,170 | <p>Use the identity:</p>
<p>$$\tan^2 \theta + 1 = \sec^2 \theta$$</p>
<p>Though the answer won't necessarily look like the answer given. Does it have to be in terms of $\sinh^{-1}$?</p>
|
118,873 | <p>I understand that the Mellin transform of a modular form is expected to satisfy RH when it is an eigenform of all Hecke operators, in which case it has an Euler product. Now about when the form is not an eigenform: Is it known a case where the zeros are all in the critical strip?</p>
| Marc Palm | 10,400 | <p>Perhaps I don't understand the question, but in its current form the answer is no.</p>
<p>A general modular form of fixed weight will be a linear combination of Hecke eigenforms of that weight. The Gamma factors will imply that there are trivial zeros outside the critical stripe.</p>
<p>But I guess you have an L-f... |
2,916,099 | <p>Find a Mobius transformation $T$ from the unit disk to the right half plane with condition $T(0)=3$.</p>
<p>First, the transformation from the unit circle to the upper half plane is $T_1(z)=(1-i)\frac{z-i}{z-1}$.</p>
<p>So from the unit circle to the right half plane, $T_2(z)=-i(1-i)\frac{z-i}{z-1}$</p>
<p>How ca... | Donald Splutterwit | 404,247 | <p>It is well worth knowing that the conformal map $z \rightarrow \frac{1+z}{1-z}$ permutes the regions shown below $(1234)$ and act similarly on the lower half plane.</p>
<p><a href="https://i.stack.imgur.com/zfRg2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zfRg2.png" alt="enter image descript... |
2,178,318 | <p>On a test I wrote an implication arrow "$\implies$" to show that I deduced one statement from the previous one, but I didn't get full score since it was more accurate to use an equivalence arrow "$\iff$".
For example:
$$ 2x = 4 \implies x = 2 $$
but it's also true the other way around:
$$ 2x = 4 \impliedby x = 2$$
s... | Bram28 | 256,001 | <p>As your own very example shows: just because the implication goes one way doesn't mean that it doesn't go the other way as well. In your case, it goes from left to right <em>and</em> from right to left, so we can write $P \Leftrightarrow Q$. But this does <em>not</em> mean that one of $P \Rightarrow Q$ or $Q \Right... |
1,321,544 | <p>How do you evaluate the following?</p>
<p>$$\cos\left[\cos^{-1}\left(\frac{3}{4}\right)\right]$$</p>
<p>To me the cosine of an arc cosine is just the value, which would be $3/4$.</p>
| frosh | 211,697 | <p>It is a fact that $f \circ f^{-1}(x)=x$ which is <a href="http://en.wikipedia.org/wiki/Identity_function" rel="nofollow">Identity Function</a>.</p>
|
3,203,282 | <p>Given that <span class="math-container">$C[-\pi,\pi]$</span> is complete:
How can we prove, by using the supremum norm, that the space:</p>
<p><span class="math-container">$$C_p[-\pi,\pi]=\{f\in C[-\pi,\pi]\mid f(-\pi)=f(\pi)\}$$</span></p>
<p>is also complete? thank you!</p>
| Siong Thye Goh | 306,553 | <p>We know that <span class="math-container">$$\cos (y) =\sum_{n=0}^\infty \frac{(-1)^n y^{2n}}{(2n)!}= 1-\frac{y^2}{2!}+\frac{y^4}{4!}+\ldots$$</span></p>
<p>from Taylor series of cosine. </p>
<p>We just replace <span class="math-container">$y=2x$</span>.</p>
|
2,358,838 | <p>I can see the answer to this in my textbook; however, I am not quite sure how to solve this for myself . . . the book has the following:</p>
<blockquote>
<p>To take advantage of the inductive hypothesis, we use these steps:</p>
<p>$ 7^{(k+1)+2} + 8^{2(k+1)+1} = 7^{k+3} + 8^{2k+3} $</p>
<p>$$
= 7\cdot7^... | Paolo Leonetti | 45,736 | <p>\begin{align}
7^{n+2}+8^{2n+1}&=49\cdot 7^n+8\cdot 64^n \\
&\equiv -8\cdot 7^n+8\cdot 7^n\equiv 0\bmod{57}.
\end{align}</p>
|
65,304 | <p>I have a plane curve $C$ described by parametric equations $x(t)$ and $y(t)$ and a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. The line integral of $f$ along $C$ is the area of the "fence" whose path is governed by $C$ and height is governed by $f$.</p>
<p><img src="https://i.stack.imgur.com/4rmZy.png" alt="... | Verbeia | 8 | <p>This is just a small tweak of Belisarius' answer, using <code>MeshFunctions</code> to get the vertical lines and <code>BoundaryStyle</code> to get the "fence".</p>
<pre><code>ParametricPlot3D[{Cos[t], Sin[t], z (2 + Sin[t] Cos[t]^2)}, {t,
0, \[Pi]/2}, {z, 0, 1}, MeshFunctions -> {#2 &}, MeshStyle -> {R... |
3,452,707 | <p>It is well known that <span class="math-container">$\sum_{k=0}^n{n\choose k} =2^n$</span>.</p>
<p><strong>My question:</strong> If <span class="math-container">$z$</span> is the limit point of an infinite sequence of real numbers <span class="math-container">$\{ a_n \}$</span>, then does <span class="math-container"... | Calvin Khor | 80,734 | <p>Something far more general is true. Define the doubly half-infinite matrix <span class="math-container">$ c_{mn}$</span>,
<span class="math-container">$$ c_{mn} = \frac1{2^m}\binom{m}{n}\mathbb 1_{n\le m} $$</span>
then you're asking if
<span class="math-container">$$ t_m := \sum_{n=0}^\infty c_{mn} a_n \to z?$$</s... |
3,451,629 | <p>Let <span class="math-container">$f : \mathbb{R} \to \mathbb{R}$</span> be a continous map then which of the following cannot be the image of </p>
<p><span class="math-container">$[0,1)$</span> under <span class="math-container">$f$</span> ?</p>
<p>(a) <span class="math-container">$0$</span></p>
<p>(b) <span clas... | WoolierThanThou | 686,397 | <p>I think the only good way to go about this really is to apply the property that <span class="math-container">$f$</span> maps compacts to compacts. So we see that <span class="math-container">$f([0,1])$</span> is compact, and <span class="math-container">$f([0,1])=f([0,1))\cup f(\{1\})$</span>, but the latter is a si... |
1,453,067 | <p>My friend's professor raised this question in a coaching and he and I tried everything we could think of. But later I thought that since $\sin (2x) $ can have values only between -1 & +1 and anything but +1 makes the equation complex ( keeping in mind that the integral is meant to be non-complex), there is no so... | Adesh Tamrakar | 266,411 | <p>yes. its solvable
let $$\sin2x-1=t$$
then $$\cos2x= {\sqrt{t^2+2t}}$$
then after all simplification you'll get $$ \int \frac{2}{\sqrt{2-t}}.dt$$
now you can integrate it easily.</p>
|
1,478,142 | <p>Evaluate these limits by relating them to a derivative. </p>
<p>$\lim\limits_{x \to 0} \frac{\sqrt{\cos{x}}-1}{x}$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: rewrite it in the form
$$\frac{(\sqrt{\cos x}-1)(\sqrt{\cos x}+1)}{x(\sqrt{\cos x}+1)(\sqrt{\cos x}+1)}$$</p>
|
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to pro... | Indrayudh Roy | 70,140 | <p>The right hand side is Binomial theorem for we have $$(1+\frac{2}{n})^{n}=1+n.\frac{2}{n}+{n \choose 2}(\frac{2}{n})^{2}+\cdots \geq1+2=3$$ and hence $3^{\frac{1}{n}}\leq (1+\frac{2}{n})$, with equality holding only when $n=1$.(Since the $n$-th root fuction is increasing).</p>
|
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to pro... | Sandeep Thilakan | 124,957 | <p>The left inequality (rather both) is straightforward using calculus. Consider $f:(\frac{2}{3}, \infty) \to \mathbb{R}$ $$f(x)=3-(1+ \frac{2}{3-2x})^x$$</p>
<p>Clearly $f(1) = 0$. Differentiating,
$$f'(x)=6 \left( 1+\frac{3}{3x-2}\right)^x \log \left( 1+\frac{2}{3x-2}\right)\frac{1}{(3x-2)^2}$$</p>
<p>which is posi... |
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to pro... | achille hui | 59,379 | <p>For the LHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $\frac13$, we get</p>
<p>$$\begin{align}
&\left(1 + \frac{2}{3n-2}\right)^{-1}= 1 - \frac{2}{3n} = \frac{1}{n}\left( (n-1)\times 1 + \frac{1}{3}\right) \ge \frac{1}{\sqrt[n]{3}}\\
\implies &
1 + \frac{2}{3n-... |
1,784,912 | <p>In this question, I know that $\text{C},\text{R},\text{T},\text{A}\in\mathbb{R}^+$</p>
<p>I've this circuit (the bottom of the resitor is connected to earth ($0$)):
<a href="https://i.stack.imgur.com/hfKGJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hfKGJ.jpg" alt="enter image description her... | alexjo | 103,399 | <p>The transfer function is
$$
H(s)=\frac{V_{\text{out}}(s)}{V_{\text{in}}(s)}=\frac{s}{s+\frac{1}{\tau}}
$$
where $\tau=RC$. The input voltage is
$$
V_{\text{in}}(s)=\frac{A}{s}\tanh\left(\frac{sT}{4}\right)=\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\sum_{k=0}^{\infty} \mathrm e^{-kTs}
$$
and then
$$
V_{\t... |
71,117 | <p>I have this assertion: if $p$ is a prime such that $p\equiv 11 \pmod{56}$, then $p$ splits in $\mathbb{Z}[\sqrt{14}]$ (the discriminant of $\mathbb{Z}[\sqrt{14}]$ is $56$.)</p>
<p>Why? Does $p\equiv 11\pmod{56}$ imply $14$ is a quadratic residue mod $p$?</p>
| Eric Naslund | 6,075 | <p>$11$ is not a square modulo $56$. Suppose $11\equiv a^2 \pmod{56}$. Then reducing modulo $8$ we see that $$3\equiv a^2 \pmod{8}$$ which is impossible, since $1$ and $4$ are the only squares modulo $8$.</p>
<p>Here is one way to prove that $p\equiv 11 \pmod{56}$ does not factor. Since the norm is multiplicative, ... |
3,791,936 | <p>An advanced sum <a href="https://www.facebook.com/photo.php?fbid=3190290677734375&set=a.222846247812181&type=3&theater" rel="noreferrer">proposed</a> by Cornel Valean:</p>
<blockquote>
<p><span class="math-container">$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$$</span>
<span class="ma... | Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\exp... |
3,891,551 | <p>Let <span class="math-container">$S$</span> be the unit sphere <span class="math-container">$x^2+y^2+z^2=1$</span> with the outward pointing normal vector <strong>n</strong>. Calculate the flux for the vector field <span class="math-container">$\mathbf{f}(\mathbf{r})=4\mathbf{r}$</span> through <span class="math-con... | Math Lover | 801,574 | <p>In spherical coordinates,</p>
<p><span class="math-container">$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r \cos \phi$</span></p>
<p>Surface area element <span class="math-container">$dS = r^2 \sin \phi \ d \theta d \phi = \sin \phi \ d \theta d \phi \, $</span> (as <span class="math-container">$r... |
3,891,551 | <p>Let <span class="math-container">$S$</span> be the unit sphere <span class="math-container">$x^2+y^2+z^2=1$</span> with the outward pointing normal vector <strong>n</strong>. Calculate the flux for the vector field <span class="math-container">$\mathbf{f}(\mathbf{r})=4\mathbf{r}$</span> through <span class="math-con... | Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\exp... |
263,650 | <p>As proposed by Quillen, Drinfeld, and Deligne and other important mathematicians, there is supposed to be a philosophy that, at least over a field of characteristic zero, assigns to every "deformation problem" a differential graded Lie algebra or $L_{\infty}$-algebra that controls it. </p>
<p>I've seen this idea re... | Igor Khavkine | 2,622 | <p>With Domenico's clear explanation, I can actually write down more or less explicitly the DGLA describing the deformations of Einstein metrics.</p>
<p>First, some notation. Let $\bar{g}_{ab}$ denote a given (background) Einstein metric, with corresponding Levi-Civita connection $\bar{\nabla}_a$, Riemann tensor $\bar... |
4,122,425 | <p>Let’s say a corona test is correct with <code>p=0.8</code>. If I now take two tests. What’s the probability that I get a correct result?</p>
<p>I think thought of <code>0.8*0.8</code>, but that makes now sense, since it should not decrease and <code>0.8+0.8</code> gives a probability over 1, which makes no sense eit... | Célio Augusto | 600,009 | <p>The probability that both tests are wrong is <span class="math-container">$(0.2)(0.2)=0.04$</span>. Therefore, the probability that <em>at least</em> one of them is right is <span class="math-container">$1-0.04=0.96$</span>.</p>
<p>This, if course, assumes that both tests are independent.</p>
|
2,420,727 | <p>I'm trying to evaluate </p>
<blockquote>
<p>$$\lim _{ x\to -\infty } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 } } $$</p>
</blockquote>
<p>by rationalizing the denominator, but I am not getting anywhere. Can someone please help me with this?</p>
<p>Thanks</p>
| haqnatural | 247,767 | <p>$$\lim _{ x\to -\infty } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 } } =\lim _{ x\to -\infty } \frac { x\left( 2-\frac { 3 }{ x } \right) }{ \left| x \right| \sqrt { \left( 1+\frac { 7 }{ x } -\frac { 2 }{ { x }^{ 2 } } \right) } } =\\ =\lim _{ x\to -\infty } \frac { x\left( 2-\frac { 3 }{ x } \right) }{ -x\sq... |
2,807,611 | <p>I know the answer is $n=6$, but can't figure out how to solve.
I tried dividing by $n!$, but didn't work because there isn't one in RHS to simplify... also tried using Gamma function properties, but didn't work either... </p>
<p>Any help would be appreciated.</p>
<p>Thanks.</p>
| David Taylor | 538,979 | <p>Because of $a_n = (5!)^{-1}((n+2)! - n!)$ in $\mathbb{N}$, so $n \geq 5$, and $a_n \equiv 0\pmod{n}$ for $n \geq 6$. However, $330 = 6 \cdot 55 \not \equiv 0 \pmod{7}$ therefore $a_n = 330$ for some of $n \leq 6$. Therefore $a_n = 330$ at $n = 6$. </p>
|
3,954,865 | <p>I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the cl... | tommik | 791,458 | <p>As population variance is unknown you have a t-score, not z-score.</p>
<p>I did not check your calculation but if the score is so high (in Absolute value) you will reject the null hypothesis for any significance level.</p>
<p>Thus you reject <span class="math-container">$H_0$</span> at <span class="math-container">$... |
3,954,865 | <p>I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the cl... | Botnakov N. | 452,350 | <p>I will give some formulas, which may be proved by standard methods.</p>
<p>We have <span class="math-container">$H_0: X_i \sim N(a, \sigma^2)$</span>,</p>
<p><span class="math-container">$a= 500$</span>, <span class="math-container">$n=9$</span>, <span class="math-container">$\overline{X} = \frac{\sum_{i=1}^n X_i}{n... |
21,238 | <p>Would someone be able to point me to a good resource explaining step by step the process for solving inhomogenous recurrence relations? (ie something of the form $ a_n = \sum{{b_i}{a_{n-i}}} + f(n)$ )</p>
| Heinrich Hartmann | 5,714 | <p>A topological remark:</p>
<p>If <span class="math-container">$E \subset Y$</span> is a closed analytic subspace of a smooth space <span class="math-container">$X$</span>, then the boundary
of a tubular neighbourhood is an (odd dimensional, real) sphere bundle over <span class="math-container">$E$</span>.</p>
<p>Thus... |
2,604,844 | <p>I have the folowing induction :</p>
<p>"Every graph with n vertices and zero edges is connected"</p>
<ul>
<li><p>Base:</p>
<p>For $n=1$ graph with one vertice is connected, hence a graph with $1$ vertex and zero edge.</p></li>
<li><p>Assumpution:</p>
<p>Every graph with $n-1$ vertices and zero edges is connected... | ArsenBerk | 505,611 | <p>The problem is here:</p>
<blockquote>
<p>For every graph with $n$ vertices and zero edges lets remove one vertice hence we get a graph with $n−1$ vertices and zero edges, by the assumpution the graph is connected, therefore the original graph is connected. </p>
</blockquote>
<p>Because first you supposed argumen... |
4,194,511 | <p><span class="math-container">$$ \frac {d^2y}{dx^2} + 5\frac{dy}{dx} = 15x^2 $$</span></p>
<p>It's solution is</p>
<p><span class="math-container">$$y = y_{h} + y_{p}$$</span></p>
<p>where <span class="math-container">$y_{h} $</span> is the solution for homogenous equation and <span class="math-container">$y_{p} $</s... | jjagmath | 571,433 | <p>You are right. It's a common mistake that high school teachers make when they teach to solve even the most simple equations.</p>
<p>As José Carlos Santos says in his answer, you can track which of the steps of a solution are only implications or are actually equivalences.</p>
<p>Here's an easy example where not all ... |
863,167 | <p>How can the signed area be 0? For example if you have 3 on positive x side and 3 on the negative x side then you get the signed area of 0? How can area be 0?</p>
| Ant | 66,711 | <p>You can interpret the integral of a function as "area" only if the function is positive.</p>
<p>Otherwise, you may interpret the result as the area under the positive part of the function minus the area above the negative part of the function. </p>
<p>(N.B. I'm guessing you are talking about integrals, but your p... |
1,498,048 | <p>I can´t prove this problem. Can you help me? The problem says:</p>
<p><em>If $\{X_n\}$ is a sequence of identically distributed random variables with finite mean, then
$$lim_{n\to\infty}\frac{1}{n}\mathbb{E}\Big[\max_{1\leq j\leq n} |Xj|\Big] = 0$$
[HINT: Use Exercise 17 to express the mean of the maximum.]</em></p... | Kavi Rama Murthy | 142,385 | <p>Use Dominated convergence theorem after applying Exercise 17. The dominating function is P(|X_1|>x). </p>
|
1,242,570 | <p>I want to use the standard definition $x_n \rightarrow x$ if for all $\epsilon>0$ there exists $N$ such that if $n>N$ then $|x_N-x|<\epsilon$. </p>
<p>So my claim is $x_n\rightarrow 0$ If I set $N=\epsilon^2,$ then the following expression $|\sqrt{n^2+1}-n-0|<\epsilon$ will hold true. I solved for $N$... | Adhvaitha | 228,265 | <p><strong>HINT</strong> We have
$$x_n = \sqrt{n^2+1} - n \implies x_n = \dfrac1{\sqrt{n^2+1}+n} < \dfrac1{2n}$$</p>
|
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