Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
For what $n$ do the equations $ab = c, bc = a, ca = b$ have solutions mod $n$? In $\mathbb Z/(12)$ the elements $5, 7, 11$ have the property that the product of any two of them equals the third:
$$5 \times 7 = 11$$
$$7 \times 11 = 5$$
$$11 \times 5 = 7$$
I'm interested in generalizations of this. For what integers $n... | If we write your equations $\pmod 3$ and $\pmod 4$ we get
$$\pmod 3 \quad \pmod 4\\
-1 \times 1 = -1\quad 1 \times -1 = -1\\
1 \times -1=-1 \quad -1 \times -1=1\\
-1 \times -1=1 \quad -1 \times 1=-1$$
and you can apply the Chinese Remainder Theorem to get your values. One way to get other solutions is to replace $3,4$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trignometry problem: If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$ If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$
What I did:
$\sin^2\theta + 3\cos\theta = 2$
$3\cos\theta - 1 = 1 - \sin^2\theta$
$3\cos\theta - 1 = \cos^2\theta$
$\cos^3\... | From $3\cos\theta-1=\cos^2\theta$, let $u=\cos\theta$. Then you have the quadratic $u^2-3u+1=0$. Solving this quadratic gives $u_+=\frac{3+\sqrt{5}}2$ and $u_-=\frac{3-\sqrt{5}}2$. Note that $u_+>1$, and therefore it is not a valid solution to $u=\cos\theta$.
Now, you need to find $u^3+u^{-3}$. Plugging in $u_-$, y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$. The following question is taken from JEE practice set.
The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$.
My Attemp... | Another approach would be to consider that there are two cases to examine:
• $ \ x^2 - 3ax + b \ $ has a real "double zero" $ \ r \ \ , \ $ so $ \ 3a \ = \ 2r \ $ and $ \ b \ = \ r^2 \ \ ; \ $ or
• the polynomial has a "complex-conjugate pair" of zeroes $ \ \rho \ \pm \ i·\sigma \ \ , \ \ \rho \ , \ \sigma \ $ rea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^... | Too long for a comment
Make you life much easier using logarithmic differentiation
$$y=(x^4-1)\sqrt[3]{x^2-1} \implies \log(y)=\log(x^4-1)+\frac 13 \log(x^2-1)$$
$$\frac {y'}y=\frac {4x^3}{x^4-1}+\frac 13\frac {2x}{x^2-1}=\frac{2 x\left(7 x^2+1\right)}{3 \left(x^4-1\right)}$$
$$y'=\frac {y'}y \times y= ??? $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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evaluate $\sum_{i=0}^{n-1}\sum_{j=i+1}^{n+1} {n+1\choose j}{n\choose i}$
Evaluate $\sum_{i=0}^{n-1}\sum_{j=i+1}^{n+1} {n+1\choose j}{n\choose i}$.
Below is a summary of a solution based off of a problem in the summation chapter of the book Problem Solving Through Problems by Loren Larson.
Multiply both sides of the s... | We seek to evaluate
$$\sum_{p=0}^{n-1} \sum_{q=p+1}^{n+1}
{n+1\choose q} {n\choose p}.$$
This is
$$\sum_{p=0}^{n-1} {n\choose p}
\sum_{q=0}^{n-p} {n+1\choose q}
\\ = \sum_{p=0}^{n-1} {n\choose p}
[v^{n-p}] \frac{1}{1-v} \sum_{q\ge 0} {n+1\choose q} v^q
\\ = [v^n] \frac{1}{1-v}
\sum_{p=0}^{n-1} {n\choose p} v^p (1+v)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what c... | They simplified exponents when defining $\xi,$ but you could also write the equations
$$\begin{align*}a &= \xi(20\xi+13),\\b &= \xi^2(20\xi^2+13),\\c &= \xi^3(20\xi^3+13),\\d &= \xi^4(20\xi^4+13)\end{align*}$$
(actually, they swapped $c$ and $d$, but that doesn't change the problem). Consider the polynomial
$$f(x) = [x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Adding 1 to each entry of continued fractions Here we denote $[a_0,...,a_n]$ as the continued fraction of some rational number. If I take $p/q=[a_0,a_1,...,a_n]$ to $p'/q'=[a_0+1,a_1+1,...,a_n+1]$, are there any nice properties I can say about $p'/q'$?
| We have $p' = p + q$ and $q' = q$. This is because, in a continued fraction, the $n$-th term is $a_n = \lfloor \frac{p}{q} \rfloor$, the greatest integer less than or equal to $\frac{p}{q}$. But we have
$$
\frac{p}{q} = a_0 \frac{1}{1} + \frac{1}{a_1 \frac{1}{1} + \frac{1}{\ddots + \frac{1}{a_n}}}
$$
so
$$
\frac{p + 1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} ... | First, we start with the following.
$$I(m,1)=\int_0^1\frac{x^{m-1}}{1+x}dx$$
We can substitute $u=-\ln{x}$ to get the following integral.
$$I(m,1)=\int_0^\infty\frac{e^{-mu}}{1+e^{-u}}du$$
When we apply the geometric series to $\frac{1}{1-(-e^{-u})}$, switch the order of integration and summation, then integrate, we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\oint_C \frac{z}{(z+1)(z+3)} dz$ I want to evaluate $\oint \frac{z}{(z+1)(z+3)} dz$ where $C$ is the rectangle with the edges $2\pm i$, $-2\pm i$.
My attempt:
I used partial fraction expansion to express $\frac{z}{(z+1)(z+3)} = -\frac{1}{2(z+1)} +\frac{3}{2(z+3)}$.
Therefore,
\begin{align*}
\oint_C \frac{z}{... | Since $-3$ is not an interior point (as you wrote), the integral is $0$, by the Cauchy integral theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fr... | $$\frac{3}{10}\int\frac{10}{x^4+10x^2+25}dx$$
Dividing numerator and denominator by $x^2$
$$\frac{3}{10}\int\frac{\frac{10}{x^2}}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{\frac5{x^2}+1+\frac5{x^2}-1}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{1+\frac5{x^2}}{x^2+10+(\frac5x)^2}dx-\frac3{10}\int\frac{1-\frac5{x^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why are the derivatives of $\frac{x^2+1}{x^2+x+1}$ and $\frac{-x}{x^2+x+1}$ the same?
Why are the derivatives of these functions the same?
$$\frac{x^2+1}{x^2+x+1} \qquad\qquad \frac{-x}{x^2+x+1}$$
original exercise text (See part (e).)
I have tried to answer this question and consulted the answer booklet but this did... | $$
{d\over dx} (f(x)+1)={df\over dx}+{d1\over dx}={df\over dx}
$$
so we can add $1$ to a function and the derivative of the sum is the same as the derivative of the function. So
$$
\begin{align}
\frac{-x}{x^2+x+1}+1
&=\frac{-x}{x^2+x+1}+\frac{x^2+x+1}{x^2+x+1}\\
&={x^2+x+1-x\over x^2+x+1}\\
&={x^2+1\over x^2+x+1}\\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)... | This answer is just an elaboration on KStarGamer's comment. I myself went for the calculation idea.
Suppose the sum is converging. We have then
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
$$\implies S'(x)=\sum_{n=2}^{\infty}\left(-\frac{2x}{n^2}n^2\frac{1}{1-\frac{x^2}{n^2}}+2x\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\t... | Without using the derivatives, or Taylor or o(little) or O(big),$$\lim \limits_{x \to 0}\left( \cot^2 x-{1 \over x^2}\right) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} = $$$$\lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} $$$$\cdot\lim \limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there an identity for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$? Is there a simple relation for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$
like there is for $\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)$?
Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k... | A possible approach is to use (here $n$ is still odd!) $$\prod_{k=1}^{n-1}\left(1+x^2\tan^2\frac{k\pi}{n}\right)=\left(\frac{(1+x)^n+(1-x)^n}{2}\right)^2$$ obtained by factoring the RHS over $\mathbb{C}$, or as $P_n(1+x,x-1)/P_n(1,-1)$ where $$P_n(a,b)=\prod_{k=0}^{n-1}\left(a^2+b^2-2ab\cos\frac{2k\pi}{n}\right)=(a^n-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find all the real numbers an expression might take when $a, b, c$ are complex numbers of the same modulus.
Let $a, b, c$ be three complex numbers of the same modulus. Find all real numbers that might be equal to:
$$x = \frac{a^3+b^3+c^3}{abc}$$
It is obvious that when all three numbers are equal, we might write that:... | Let us implement Mark Bennet's idea.
Let $\frac{a^3}{abc}=e^{i\alpha}$ and $\frac{b^3}{abc}=e^{i\beta}$, where $\alpha,\beta\in[0,2\pi)$. Then $\frac{c^3}{abc}=\frac{abc}{a^3}\frac{abc}{b^3}=\frac1{e^{i\alpha}}\frac1{e^{i\beta}}=e^{i(-(\alpha+\beta))}.$
$$\begin{aligned}x &= e^{i\alpha} + e^{i\beta}+e^{i(-(\alpha+\beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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short weierstrass form for cubic I have some question about the derivation of the short Weierstrass form. In https://www.staff.uni-mainz.de/dfesti/EllipticCurvesNotes.pdf this note, I follow the derivation till the point $y^2 = β_0x^3 + β_1x^2 + β_2x + β_3$, but then it says that using the transformation $x' = x + β_1/... | You're right, there's a step missing here.
Starting from $$y^2 = \beta_0x^3 + \beta_1x^2 + \beta_2+ \beta_3$$ we can get rid of the (nonzero) $\beta_0$ as follows: send $y\mapsto \beta_0^2y$ and $x\mapsto \beta_0x$, which gives us $$\beta_0^4y^2 = \beta_0^4x^3 + \beta_0^2\beta_1x^2 + \beta_0\beta_2x+ \beta_3$$ and afte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What is the value of $\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$? $$\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$$
*
*When I solved this took the constant out of the integral and then multiplied $x^2$
by the bracket and evaluated the integral to get:
$$4 \, \left[ \frac{x^4}{4} +\frac{2x^3}{3}-\frac{x^5}{5} \right]^2_{-... | Let us do it step by step:
\begin{align*}
\int_{-1}^{2}4x^{2}(x + 2 - x^{2})\mathrm{d}x & = \int_{-1}^{2}(4x^{3} + 8x^{2} - 4x^{4})\mathrm{d}x\\\\
& = \left(x^{4} + \frac{8x^{3}}{3} - \frac{4x^{5}}{5}\right)\bigg\rvert_{-1}^{+2}\\\\
& = \left(16 + \frac{64}{3} - \frac{128}{5}\right) - \left(1 - \frac{8}{3} + \frac{4}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} d... | $$\ln(x-2)=\ln(((x-2)^2)^{1/2})=\frac{1}{2}\ln(x-2)^2$$
Can you continue?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Proving $\sum_{cyc}\frac{ab}{a+b+2c} \le \frac{1}{4}(a+b+c)$ for positive real $a$, $b$, $c$ Prove that $$\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \le \frac{1}{4}(a+b+c)$$
for positive real numbers $a$, $b$, and $c$.
| By AM-HM
$$\sum_{sym} \frac{bc}{2a+b+c} \leq \sum_{sym} \frac{bc}{4}(\frac{1}{a+b}+\frac{1}{a+c})=\frac{a+b+c}{4}.$$
Seems like you are new to inequalities. To study this topic, I recommend you this book.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this general nested radical for $\pi$ true? We have,
I. Liu Hui (c. 300 AD)
$$\pi \approx 3\cdot2^{\color{red}8}\times \underbrace{\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\sqrt{\color{blue}1}}}}}}}}}}}_{\color{red}{10}\text{ square roots}}$$
II. Viete (c. 1590 AD)
$$\p... | Firstly, we define the recursive sequence
\begin{equation}
\begin{split}
A_{k+1}&=\sqrt{2+A_k}\\
A_0&=\sqrt{\beta}=2\cos\left(\frac{\pi}{\alpha}\right)\\
\end{split}
\end{equation}
and let $B_k=\sqrt{2-A_k}$. Your claim is that
$$\pi=\lim_{k\rightarrow\infty}2^k\alpha B_k$$
To prove this claim, we shall first show by i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Extracting coefficients from a generating function Recently, I found that the generating function for a sequence I am interested in is
$$s(x) = -\frac{3 \, x^{3} + x^{2} + 2 \, x}{2 \, x^{3} + x - 1}.$$
Naturally, I am now keen on extracting the $n$th coefficient of the Taylor expansion of $s(x)$ without the help of a ... | We can derive the $n$-th coefficient by makeing a geometric series expansion of
\begin{align*}
\color{blue}{s(x)}&\color{blue}{=-\frac{3x^3+x^2+2x}{2x^3+x-1}}\\
&\color{blue}{=2x+3x^2+6x^3+10x^4+16x^5+28x^6+\cdots}
\end{align*}
We use the coefficient of operator $[x^n]$ to derive the coefficient of $x^n$ of a series.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Ste... | *
*Finds the roots of $x^2-x+1$; they are complex
*Pick one of them and prove that is a root of $x^{6n+2}-x^{6n+1}+1$; automatically the other one is also a root therefore the two polynomials have common roots, hence the one with lesser number of roots is a factor of the other polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proof Of A Gamma Function - Double Factorial Identity How does one prove the following identity?
$$\sqrt{(-1)^n\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}} = \frac{(2n-1)!!}{2^n}$$
I attempted to prove this using the definition of the double factorial, however I couldn't continue and feel like there is a better method. I would... | We use the identity
\begin{align*}
\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin (\pi z)}\qquad\qquad z\notin \mathbb{Z}
\end{align*}
evaluated at $z=n-\frac{1}{2}$ and obtain
\begin{align*}
\Gamma\left(1-\left(n+1/2\right)\right)\Gamma\left(n+1/2\right)&=\frac{\pi}{\sin\left(\pi\,\frac{2n+1}{2}\right)}\\
\color{blue}{\Gamma(1/... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = ... | The original parabola is
$ y = - x^2 + 4 x + 8 $
Shifting by $(h, k)$ gives the parabola,
$ y = - (x - h)^2 + 4 (x - h) + 8 + k $
Points $(1,1)$ and $(3,5)$ are on the new parabola, so
$ 1 = - (1 - h)^2 + 4 (1 - h) + 8 + k $
$ 5 = - (3 - h)^2 + 4 (3 - h) + 8 + k $
Substracting,
$ - 4 = - (1^2 - 3^2 - 2 h (1 - 3 )) + 4 ... | {
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Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2... | You made mistake in the second integral. It should be
$$
I_2 = \frac{\theta}{2} + C_2 = \frac{\arcsin(x)}{2} + C_2
$$
| {
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$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$ with one-variable calculus solution Prove that $$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$$
My textbook says that we need to note that the integral can be tran... | Clearly
\begin{eqnarray}
&&\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x\\
&=&\int_{0}^{\pi}\sqrt{x^2+\sin^2x}\mathrm{d}x+\int_{0}^{\pi}\sin x\ln(x+\sqrt{x^2+\sin^2x})\mathrm{d}\sin x\\
&=&\int_L\sqrt{x^2+y^2}\mathrm{d}x+y\ln(x+\sqrt{x^2+y^2})\mathrm{d}y\\
&=&\int_{L+l}\sqrt{x^2+y^... | {
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Real Analysis fundamental theorems of Calculus contradiction? Evaluate: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt.$
Solution: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt = \frac{7x^6}{4}.$
Proof:
Consider the function's $F:[a,b] \rightarrow \mathbb{R}$ and $f:[a,b] \rightarrow \mathbb{R}$, define: $F(t) = \frac{1}{4}x^3t^4 + c$ and ... | It should be noted that the Leibniz Integral Rule deals with integrals like the one in question.
In the current example the integral takes the following form (with a slight generalization).
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= (x \, x)^n \, \frac{d}{dx}(x) - (x \cdot 0)^n \, \frac{d}{dx}(0) + ... | {
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Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition Prove continuity of $f: [\frac{1}{2},\infty ) \rightarrow \mathbb{R}: x \mapsto \sqrt{2x-1}$ for $x_0>\frac{1}{2}$ with Epsilon-delta definition of continuity
show:
$\forall ... | For $x_0=\frac{1}{2}$:
$|f(x)-f(\frac{1}{2})|=|f(x)|=\left|\sqrt{2x-1}\right|<\epsilon$
$\forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| \stackrel{!}{<} \epsilon$
choose: $\delta:= \frac{\epsilon^{2}}{2} \Rightarrow \forall x \in ( \frac{1}{2},\frac{1}{2}+\delta): \left|\sqrt{2x-1}\right| < \e... | {
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Can we evaluate $\int_0^\infty x^k\frac{ae^{bx}}{\left(1+ae^{bx}\right)^2}e^{-\frac{x^2}{2}}dx$ This sequence of integrals come from the following expectations $$E\left(X^kf(X)\right),\quad f(x;a,b)=\frac{ae^{bx}}{\left(1+e^{bx}\right)^2},\quad a>0,~b>0,~k=0,1,2,\cdots,$$
where $X\sim N(0,1)$. So we can express them as... | The farthest I could go
$$x^k\frac{ae^{bx}}{1+ae^{bx}}e^{-\frac{x^2}{2}}=\sum_{n=0}^\infty (-1)^n\, x^k\,e^{-\frac{x^2}{2}}\left(\frac{e^{-b x}}{a}\right)^n$$
Defining
$$J_{n,k}=\int_0^\infty x^k\,e^{-\frac{x^2}{2}}e^{-nb x}\,dx$$ Using Kummer confluent hypergeometric functions, they write (with $t=bn$)
$$J_{n,k}=2^{\f... | {
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Prove that for every positive integer $n$ there do not exist four positive integers $a,b,c,d$ with $ad = bc$ and $n^2 < a < b < c < d < (n+1)^2$. Here is the problem Timothy Gowers was trying to solve in this YouTube video.
Prove that for every positive integer $n$ there do not exist four positive integers $a,b,c,d$ w... | All variables are positive integers.
It is enough to prove the following claim.
Claim: Suppose $n^2\le a<b\le c<d\le (n+1)^2$ and $ad=bc$. Then $a=n^2$, $b=c=n(n+1)$, $d=(n+1)^2$.
Proof:
$ad=bc\implies\frac dc=\frac ba.$ Hence $$d-c=c(\frac dc-1)> a(\frac ba-1)=b-a.$$
Let $k=b-a$. Since $d-c$ and $b-a$ are integers, $d... | {
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Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than... | Both conditions and the hypothesis are homogenous, so we can scale the variables until $a=1$. Then the problem is equivalent to the implication:
$$b<\sqrt{c}\land c<\frac{2b}{1+b}\Rightarrow b^2+c^2\le1+bc.$$
New conditions are more convenient and can be rewritten to:
$$b^2<c<\frac{2b}{1+b}.$$
Let's devide the conditio... | {
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Probability on Number Theory Problem: Suppose that $a,b,c \in \{1,2,3,\cdots,1000\}$ are randomly selected with replacement. Find the probability that $abc+ab+2a$ is divisible by $5$.
Answer given from the worksheet: $33/125$
My answer: $\frac{641}{3125}$
Attempt: Since $abc+ab+2a = a(bc+b+2)$, either $a \equiv 0 \pmo... | Brute forcing this in Python, I get the same answer of $41/125 = 0.328$ previously given more thoroughly by John Omielan, which differs from the worksheet's answer of $33/125$:
import itertools as it
N = 100
combos = list(it.product(range(1,N+1), repeat=3))
exprs = [a*b*c + a*b + 2*a for (a,b,c) in combos]
print(len([x... | {
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Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | Here is an alternate solution using residues. Starting from where you left off, we can use the complex definitions of $\sin(x)$ and $\cos(x)$ to transform the integral into
$$\int_{0}^{2\pi}\frac{4e^{2ix}\left(1+e^{4ix}\right)}{6e^{4ix}+e^{8ix}+1}dx.$$
Since the integrand is periodic on $\pi$, let $z=e^{2ix}$. The cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$ Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$
substituting $y=ux$ so that $y' = u + x\frac{du}{dx}$"
$u + x\frac{du}{dx} = \frac{2x^2u}{ux^2-x^2} = \frac{2u}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{2u}{u^2-1} - u = \frac{2u}{u^2... | $$y' = \frac{2xy}{y^2-x^2}$$
Since $y'=\dfrac 1 {x'}$:
$$\dfrac 1 {x'} = \frac{2xy}{y^2-x^2}$$
$${y^2-x^2} = {2xx'y}$$
Note that $2xx'=(x^2)'$:
$$y^2 = {(x^2)'y}+x^2$$
$$y^2 = {(x^2y)'}$$
Integrate:
$$\dfrac {y^3}3=x^2y+C$$
| {
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Evaluating $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ in the most efficient way I have come across a tricky question while studying logarithms.
$$\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$$
While plugging it into a calculator brings a seemingly simple answer, I cannot find a way to star... | Rewrite $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ as $\log_3(1+(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. You can use the Difference of Squares formula to find $\log_3(1+(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. If you use the Difference of Squares formula 5 times, you get $\log_... | {
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"source": "stackexchange",
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Dodecahedron, angle between edge and face. In an effort to build a dodecahedron frame in Fusion360 I need to know some of the angles. Looking around I found out that the angle between an edge and a face on a regular dodecahedron is $121.7^\circ$ but I couldn't find the mathematical formula nor the way to calculate this... | This angle can be derived from the dihedral angle $\delta = \arccos (-1/\sqrt{5})$ of the dodecahedron using a vector method. (The dihedral angle $\delta$ is derived during construction of the dodecahedron - eg see this answer to How does this proof of the regular dodecahedron's existence fail?).
Consider a group of th... | {
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"source": "stackexchange",
"question_score": "6",
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\l... | We could also take a "factorization" approach. Since
$$ \log_{4n} 40\sqrt{3} \ \ = \ \ \log_{3n} 45 \ \ = \ \ \alpha \ \ , $$
we can write
$$ (4n)^\alpha \ \ = \ \ 2^3·3^{1/2}·5 \ \ \ , \ \ \ (3n)^\alpha \ \ = \ \ 3^2·5 $$ $$ \Rightarrow \ \ n^\alpha \ \ = \ \ 2^{3 \ - \ 2·\alpha}·3^{1/2}·5 \ \ = \ \ (2^0)·3^{2 \ - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
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"source": "stackexchange",
"question_score": "5",
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Combination with repeated elements If I have a set of $4$ numbers, for example $(1, 2, 3, 3)$, how do I calculate the $C(4, 3)$? The order doesn't matter. So, there are three different combinations that I can get - $(1, 2, 3), (1, 3, 3)$ and $(2, 3, 3)$.
Another example, if I have the word $BABY$, how do I calculate $C... | With a little algebra, we can calculate the desired numbers. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. In this way we can write, for example
\begin{align*}
C(4,2)=\binom{4}{2}&=[x^2](1+x)^4\\
&=\color{blue}{[x^2]}\left(1+4x+\color{blue}{6}x^2+4x^3+1\right)\color{blue}{=... | {
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Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| Using Henry's approach, but with different estimates. If you know that the first three digits of $\ \sqrt{2}\ $ is $\ 1.41,\ $ then you have: $\ 1.40 < \sqrt{2} < 1.42857\ldots,\ $ i.e.,
$$\ \frac{7}{5} < \sqrt{2} < \frac{10}{7}.$$
So,
$$ \left( 9^{\sqrt{2}} \right)^7 < \left( 9^{ \frac{10}{7}} \right)^7 = 9^{10} = {81... | {
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Making integration problems such that $\int{}f^{-1}(x)\text{ d}x$ is difficult and $\int{}f(x)\text{ d}x$ is easy. There is an amazing formula to integrate the inverse of a function:
$$\int{f^{-1}(x)\text{ d}x}=x f^{-1}(x)-F\bigg(f^{-1}(x)\bigg)+c, \text{where }F(x)=\int{f(x)\text{ d}x}$$
I know how to derive this form... | Another example:
Take $$f(x) = -\frac{1}{x \sqrt{1 - x^2}} , \qquad x \in \left(0, \frac{1}{\sqrt{2}}\right)$$
(we've restricted the domain of $f$ so that it has an inverse). Then,
$$f^{-1}(x) = \frac{1}{\sqrt{2}} \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x} .$$ Then, we can transform the integral
$$\int f^{-1}(x)\,dx = \fr... | {
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Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve f... | Since the question asked for $\sin 2x$, after failures using other methods, I found this one:
$$\sin x + \cos x + \tan x + \cot x + \csc x+ \sec x = 7$$
$$(\sin x + \cos x) + \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} + \frac{1}{\sin x}+\frac{1}{\cos x}=7$$
$$(\sin x + \cos x) + \frac{\sin^2x+\cos^2x}{\sin x\cos x} + ... | {
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How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
| $$x^6 + 3x^3+2x^2+x+1= (x + 1)^2 \left( {\left( {x - \tfrac{1}{2}} \right)^4 + \tfrac{1}{2}\left( {x - \tfrac{1}{2}} \right)^2 + x^2 + \tfrac{{13}}{{16}}} \right) \ge 0$$
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Primitive Polynomials over $\mathbb{Z}_3$ Which of the following polynomials are primitive over $\mathbb{Z}_3$?
*
*$x^3 + x^2 + x + 1$
*$x^3+x^2+x+2$
*$x^3+2x+1$
So I now in order to be primitive the polynomials have to be irreducible, which is only true for $x^3+x^2+x+2$ and $x^3+2x+1$.
But how do I show if they a... | Hint
There are $4=\dfrac {\varphi (3^3-1)}3$ primitive third degree polynomials over $\Bbb F_3.$
Let $t$ be a root of $x^3+2x+1.$ Take $\alpha=t+(x^3+2x+1),$ a primitive $26$th root of unity in $\Bbb F_{27}$.
We know $\alpha $ is primitive because $t^2\neq1$ and by the Frobenius automorphism, $$t^9=(t^3)^3=(t-1)^3=t... | {
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Inequality with increasing variables
If $n,k\ge2$, and $0\le a_0\le a_1\le\cdots$, prove that
\[\left(\frac{1}{k n} \sum_{l=0}^{k n-1} a_{l}\right)^{k} \geq \frac{1}{n} \sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}.\]
This inequality is an improvement of AM-GM inequality. For $n=3, k=2$, $0\le a_0\le a_1\le\cdots\le a... | (With the homogenization of $\sum a_i = kn$.)
First, we consider the $ k = 2, n = 2$ case.
Suppose $ a + b + c + d = 4, a \leq b \leq c \leq d$, we want to show that $ ac + bd \leq 2$.
Intuitvely, if $a \leq d$ are fixed, we'd want to increase $b$ and decrease $c$ subject to $ b \leq c $.
Hence, we replace $(a, b, c, d... | {
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$\epsilon$–$\delta$ proof that $\lim_{x\to x_0}\frac{1}{x} = \frac{1}{x_0}$ for all $x_0\neq 0$; how to identify $\delta$? I would like to use the $\epsilon$–$\delta$ definition of the limit of a function to show that
$$\lim_{x\to x_0} \frac{1}{x} = \frac{1}{x_0}$$
But I'm having trouble identifying a $\delta>0$ for ar... | Given $\ \varepsilon > 0.$
If $\varepsilon < \frac{1}{\vert x_0 \vert },$ then $\vert \varepsilon x_0 \vert<1,\ $ and so $$ x \in \left( \underbrace{\ \frac{x_0}{1 + \varepsilon x_0}\ }_{a}, \underbrace{\ \frac{x_0}{1- \varepsilon x_0}\ }_{b} \right) \implies \frac{1}{x} \in \left( \frac{1 - \varepsilon x_0}{x_0}, \fra... | {
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Find $a+b +c$, if $\sin{x}+\sin^2{x}=1$ and $\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$ There is my problem :
Find $a+b +c$,
if $$\sin{x}+\sin^2{x}=1$$ and $$\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$$
I'm sorry, I can't solve this problem but I really want to know the solution.
I know that $\cos^2{x}... | $\sin^2 x+\sin x-1=0$ gives $\sin x = \frac{-1\pm\sqrt{5}}{2}$, so that
$$\cos^2 x = 1-\sin^2 x = 1-\frac{3\pm\sqrt{5}}{2} = \frac{-1\pm\sqrt{5}}{2}.$$
Then you want to solve
$$(\cos^2 x)^6+a(\cos^2 x)^5 + b(\cos^2 x)^4 + c(\cos^2 x)^3 = 1.$$
After some algebra, this reduces to (for the positive sign on $\sqrt{5}$)
\be... | {
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Where've I gone wrong on this Maclaurin series expansion I keep getting the same answer, and I know it's wrong, but I'm not seeing where I've gone wrong.
I'm supposed to find the first 4 terms of $e^{-x} \cos(x)$ using Taylor series expansion and multiplying the terms.
The answer should be:
$$1-x+\frac{1}{3}x^3-\frac{1... | You are missing the term $\frac{x^4}{4!} \cdot 1$ from $(\cos x)(e^{-x})$. Everything else is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of $\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$
Find the sum of $$\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$$
The solution in the book is a lot different than what I tried
what they did in the book is say $k-1=t$ then they expanded from this point $\sum_{t=0}^{n-1} (t+1)(n) {{n-1} \choose {t-1}}$ and the fin... | Since $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$, differentiate twice to get
$$n(n-1)(1+x)^{n-2} = \sum_{k=2}^n k(k-1)\binom{n}{k}x^{k-2}$$
Substituting $n+1$ for $n$ we also get
$$(n+1)n(1+x)^{n-1} = \sum_{k=2}^{n+1} k(k-1)\binom{n+1}{k}x^{k-2}.$$
Substitute $x=1$ in each of these, giving
\begin{align*}
n(n-1)2^{n-2} &=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I change the order of integration $\int_{\pi/2}^{5\pi/2} \int_{\sin x}^{1} f(x,y)dydx$ How do I change the order of integration in $$\int_{\pi/2}^{5\pi/2} \int_{\sin x}^{1} f(x,y)dydx\;?$$
$y=\sin x$; $y=1$, $x=\pi/2$; $x= 5\pi/2$.
I can guess from here that $y$ is from $-1$ to $1$.
Then $x=\sin^{-1}(y)$ and $\s... | You need to find the equation of the form $x=a+b \cdot \text{sin}^{-1}(y)$ passing through $\left( \frac{3\pi}{2},-1\right)$ and $\left( \frac{\pi}{2},1\right).$ So $\frac{3\pi}{2}=a + b \cdot \text{sin}^{-1}(-1)$ and $\frac{\pi}{2}=a + b \cdot \text{sin}^{-1}(1)$. Now you have a system of 2 linear equations in $a$ an... | {
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Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$ In one of my analysis course, we considered the function
$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$$
Then my teacher told us that $f$ had the following behavior
$$
f(x)\sim\begin{cases}
\frac{-1}{4}... | First note that $$ x \int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy=o(x\log{x})$$
as $x\to 0^{+}$. Indeed, the integral $\int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy$ is convergent and $\lim_{x\to 0^{+}}\frac{x}{x\log{x}}=0$.
Second, using l'Hopital's rule one can compute
$$\lim_{x\to 0^{+}}
\frac{x \i... | {
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Power series of inverse function Let, $f$ be a bijective function on set of Real numbers. Let, $f(x) =\sum_{n=1}^{\infty}a_{n}x^n$ such that $a_{1}=2,a_{2}=4$ let, $f^{-1}(x) =\sum_{n=1}^{\infty} b_nx^n$
Then find value of $b_1$.
My approach:
we know,
$$\frac{1}{1-2x}=1+2x+4x^2+8x^3+\ldots $$
$$\frac{2x}{1-2x} = \sum_{... | You cannot assume anything about $f$ other than the facts that $a_1=2$ and $a_2=4$.
If $f^{-1}(x)=\sum_{n=1}^\infty b_nx^n$, then $f^{-1}\bigl(f(x)\bigr)=x$ means that$$b_1(a_1x+a_2x^2+a_3x^3+\cdots)+b_2(a_1x+a_2x^2+a_3x^3+\cdots)^2+\cdots=x.\label{a}\tag1$$But the coefficient of $x$ on the LHS of \eqref{a} is $b_1a_1$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Computing $\int_0^{\infty} x^n\left\lfloor\frac{1}{x}\right\rfloor d x=\frac{\zeta(n+1)}{n+1}$, where $n\in \mathbb N$. In the post, it was found that
$$
\int_0^{\infty} x\left\lfloor\frac{1}{x}\right\rfloor d x=\frac{\pi^2}{12}
$$
I want to generalise the integral as
$$
\int_0^{\infty} x^n\left\lfloor\frac{1}{x}\righ... | Here is an alternative method. Similarly to the comment by reuns in the linked post, we can do the following:
\begin{align*}
\int_0^\infty x^n\left\lfloor\frac 1x\right\rfloor dx
&=\int_0^\infty y^{-n}\lfloor y\rfloor\frac{dy}{y^2}\\
&=\int_0^\infty y^{-n-2}\sum_{\substack{m\in\mathbb N\\ 1\leq m\leq y}}1 dy\\
&=\sum_{... | {
"language": "en",
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"source": "stackexchange",
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Foci of ellipse My question is given a ellipse of the equation : $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ where $a>b$ then how we can find the coordinates of the foci. I want to find those coordinates without the presuming that the foci exists because most proofs I found online assume the properties of foci to be true an... | The sum of lengths of $(-f, 0)$ and $(f, 0)$ from point P is
$$S = \sqrt{(f+x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}+\sqrt{(f-x)^2+b^2\left(1-\frac{x^2}{a^2}\right)}$$
Your approach was correct upto this step. However, you need to realize what happens when you differentiate S with respect to f.
The condition $\frac{{\pa... | {
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How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(... | In terms of the Gaussian hypergeometric function
$$\int \frac{dx}{\left(x+\frac{1}{x}\right)^{2n}}=\frac{x^{1-2 n}}{2 n+1}\, _2F_1\left(2 n,\frac{2n+1}{2};\frac{2n+3}{2};-x^2\right)$$ which can also write
$$\int \frac{dx}{\left(x+\frac{1}{x}\right)^{2n}}=(-1)^{n+1}\frac{i}{2} B_{-x^2}\left(n+\frac{1}{2},1-2 n\right)$$... | {
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Proving $\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\frac12\sqrt{\frac{13+3\sqrt{13}}2}$
Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$
My Attempt
Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}... | with $ \; w = \cos \frac{2 \pi}{13} + i \sin \frac{2 \pi}{13} \; \; \; $ in mind, let
$$ x = -i \left( w - w^{25} + w^3 - w^{23} + w^9 - w^{17} \right) $$
we may calculate polynomials in $x.$ We may then apply the relation $w^{26 } = 1 $ repeatedly and express the outcome as sums of $w^{25}, w^{24}, ..., w^... | {
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisi... | Presumably you're only looking for a way to understand the induction problem, but you can note that $n^3+2n = n^3 - n + 3n = (n-1)(n)(n+1) + 3n$. Since any three consecutive integers has a multiple of three, we're adding two multiples of three and so get another multiple of 3.
| {
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How can I understand and prove the "sum and difference formulas" in trigonometry? The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \... | Consider a unit circle with $O$ as the centre. Let $P_{1}$, $P_2$ and $P_{3}$ be points on the circle making angles $A$, $B$ and $A−B$, respectively, with the positive direction of the X-axis.
We know that if two chords subtend equal angle at the centre, then the chords are equal and chords $P_{3}P_{0}$ and $P_1P_2$ s... | {
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | I have another method as well. From skimming the previous solutions, I don't think it is a duplicate of any of them
In Complex analysis, we learn that $\sin(\pi z) = \pi z\Pi_{n=1}^{\infty}\Big(1 - \frac{z^2}{n^2}\Big)$ which is an entire function with simple zer0s at the integers. We can differentiate term wise by un... | {
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Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}... | This question is now old enough for some more complete answers.
For number 1:
$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} =
\sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$
where $d$ is the common difference,
$$ = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right)
= \frac{a_n - a_1}{d(\sqrt{a_... | {
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
| Let $a\cos\theta+b\cot\theta=c$
$\implies a\cos\theta=c-b\cot\theta=\frac{c\sin\theta-b\cos\theta}{\sin\theta}$
$\implies a\cos\theta\sin\theta=c\sin\theta-b\cos\theta$
Putting $c=r\cos\alpha,b=r\sin\alpha$ where $r>0$
Squaring & adding we get $r^2=c^2+b^2\implies r=+\sqrt{b^2+c^2}$ and $\frac{\sin\alpha}b=\frac{\cos\a... | {
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Trigonometry Expression Is $(\sin \phi)^2$ is equal to $\sin^2\phi$?
Can any one tell what is the ans for the below expression
$\sin^260$ + $\cos^260$ + $\tan^245$ + $\sec^260$ - $\csc^260$
| ($sin \phi$)^2 is equal to $sin^2\phi$.
$sin^260$ + $cos^260$ + $tan^245$ + $sec^260$ - $cosec^260$
The trick to that is to use a few trigonometric identities.
$\sin^2\theta + \cos^2\theta = 1$
$\tan 45 = 1$
The value of $\cos 60$ is $\frac{1}{2}$, so $\sec^260$ will evaluate to 4.
The value of $\sin 60$ is $\sqrt{\f... | {
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Completing the square problem The problem i am supposed to solve for $x$ by completing the square:
$3x^2+9x+5 = 0$
step 1. move constant to right: $3x^2+9x\quad\quad+5 = 0$
step 2. divide by $3$: $x^2+3\quad\quad+\frac{5}{3}$
step 3. $(\frac{1}{2}b)^2$: $(\frac{1}{2}\cdot 3)^2 = \frac{9}{4}$
step 4. add and subtract ... | As Theo Buehler said in a comment, your answer and technique are correct, just yielding a different form of the same answer (rationalize the denominator in your answer, then combine the fractions by finding a common denominator).
As an aside, while I'm sure that you're applying the technique as you were taught (those s... | {
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
| Problems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter wha... | {
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How to reduce congruence power modulo prime? If I have a congruence equation, says
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be
$$x^{14}x... | Yes, if you're looking for solutions of the equation mod $7$ then, since $\rm\:x=0\:$ is not a solution, you can in fact deduce that $\rm\:x^6 = 1\:$. If you couldn't exclude $\rm\:x=0\:$ then you'd instead need $\rm\:x^7 = x\:.$
| {
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Name of Formula $x^3+y^3=z^3+1$ I encountered the formula
$$x^3+y^3=z^3+1$$
with the condition, that
$$x<y<z$$
and wonder, whether it has got a specific name or whether it can be easily transformed into another well-known (family of) formula(s).
| $$X^3+ Y^3+ Z^3=1$$ is the formula which is known as harder factor and yours is a distorted and conditional form of harder factor
If $X+Y+Z=0$ then $X^3+ Y^3+ Z^3=1$.
In your question $X$ is less than $Y$ and $Y$ is less than $Z$ means the minimum possible difference between $X$ and $Y$, $Y$ and $Z$ is $1$.
At the same... | {
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Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$ Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and tran... | If you know how to solve linear recurences, this would simplify your computations:
\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}
\begin{eqnarray}
da_{n-1} &=& dba_{n-2}+cd^{n-1}
\end{eqnarray}
and subtract....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factor quadratic $ax^2+bx+c$? How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$
The answer is
$$(x+2)(x-1)$$
I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I h... | As Ross pointed out, and as was previously discussed, we know that
$$
(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab.
$$
Therefore, to factor a quadratic expression $x^2 + cx + d$, all one has to do is find two numbers that multiply to $d$ and add to $c$. Let $m$ and $n$ be those two numbers tha... | {
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Calculating arc length of a curve, stuck on dy/dx part (algebra mostly) The equation is:
$$x=\frac{1}{8}y^4 + \frac{1}{4}y^{-2},\qquad 1\leq y\leq 2.$$
I have the formula. I'm not sure how to write it out but this is what it says:
Length is equal to the integral (with $b$ and $a$ for limits) of the square root of $1+(d... | You squared incorrectly. The square of $a-b$ is not $a^2-b^2$. (Also, you should be using $\frac{dx}{dy}$, not $\frac{dy}{dx}$, because here your independent variable is $y$ and your dependent variable is $x$; your integral will be with respect to $y$, not with respect to $x$).
The square of $\displaystyle\frac{1}{2}y^... | {
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Complex number to polar form I need to take a raincheck with this problem. I want to make sure I haven't messed up some fundamental idea.
Convert the complex number $$-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i$$ to polar form.
I took the modulus as below,
$$\lvert-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i \rvert = \sqrt{(\dfrac{-1}{... | The problem is you forgot to add $\pi$ in calculating the argument. See the wiki page which shows all the formulas for computing the argument depending on $x$ and $y$.
Taking $z=x+yi$ to be your complex number, here $x$ is negative and $y$ is positive, so
$$
\text{arg}(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1... | {
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What are the fields with 4 elements? What are the fields with 4 elements?
| Let $K$ be your field.
The additive group of $K$ is an abelian group with four elements. The order of $1$ in this group divides $4$, so it is either $2$ or $4$. Were it $4$, we would have $1+1\neq0$ and $(1+1)\cdot(1+1)=0$, which is absurd in a field. It follows that $1+1=0$ in $K$. But then for all $x\in K$ we have $... | {
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Generating sequences using the linear congruential generator I came across the linear congruential generator on Wikipedia:
http://en.wikipedia.org/wiki/Linear_congruential_generator
I gather that for a particular choice of the modulus, multiplier and increment, the generator would generate a unique sequence. However, i... | Not every finite sequence can be obtained by a linear congruence generator. In fact, the one you request cannot.
Note that we must have $m\geq 23$ in order to "get" $22$ as an answer. We have that you are requiring:
$$\begin{align*}
2a + c &\equiv 11 \pmod{m}\\
11a+c &\equiv 5\pmod{m}\\
5a+c &\equiv 9\pmod{m}\\
9a+... | {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accord... |
I was wondering if there are alternative approaches to solve this problem
Yes, there is.
and is my proof entirely correct?
I think so. I couldn't find anything wrong with it.
Now let's start with another proof.
Let
$$
\{a,b,c,x,y,z \in \mathbb R_{\geq0} : a=x^{\frac{1}{3}},b=y^{\frac{1}{3}},c=z^{\frac{1}{3}}\}
$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 7
} |
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an e... | As is often the case, we can use Cauchy-Schwarz.
Since the $L^3$ norm is larger then the $L^4$ norm, we have that $$\left(a^{3}+b^{3}+c^{3}\right)^{\frac{1}{3}}\geq\left(a^{4}+b^{4}+c^{4}\right)^{\frac{1}{4}}$$ with equality if and only if $a=b=c.$ Using Cauchy Schwarz, we have that $$\sqrt{3\left(a^{4}+b^{4}+c^{4}\... | {
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"source": "stackexchange",
"question_score": "9",
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Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and... | First write $A=2b$. Then make appropriate use of $\tan b=\sin b/\cos b$, $\sin2b=2\sin b\cos b$, and $\cos2b=2\cos^2b-1=1-2\sin^2b$.
| {
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"source": "stackexchange",
"question_score": "13",
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How to solve these inequalities? How to solve these inequalities?
*
*If $a,b,c,d \gt 1$, prove that $8(abcd + 1) \gt
(a+1)(b+1)(c+1)(d+1)$.
*Prove that $ \cfrac{(a+b)xy}{ay+bx} \lt \cfrac{ax+by}{a+b}$
*Find the greatest value of $x^3y^5z^7$ when $2x^2+2y^2+2x^2=15$
Any hints/solution are welcome.
| Solution:
*
*Since $a,b,c,d>1$, then the following inequalities are true based on Rearrangement inequalities: if $x>1$ and $y>1$ then $(x-1)(y-1) > 0$, ie $xy+1 >x+y$.
$$
\begin{aligned}
abcd + 1 &> abc + d
\\
abcd + 1 &> abd + c
\\
abcd + 1 &> acd + b
\\
abcd + 1 &> bcd + a
\\
abcd + 1 &> ab + cd
\\
abcd + 1 &> ad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/56719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Difficulty in integrating I tried to integrate this by parts but it didn't work out. Any simple means of doing it.$$\int\sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx$$
| Put $2x+2 = 3 \tan(\theta)$ and see what happens.
$\textbf{Added.}$ First observe that $$4x^{2}+8x+13= (2x+2)^{2} + 3^{2}.$$ So I hope you are aware of the fact that $\text{if you have an integral of the form}$, $1+x^{2}$, then one generally substitutes, $x= \tan(\theta)$. That's the case here as well. By doing that we... | {
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Show that $PSL(3,4)$ has no element of order $15$.
$PSL(3,4)$ has no element of order $15$. Thus it is no isomorphic to $A_8$.
Here, $PSL(3,4)$ denotes the $3 \times 3$ projective special linear group on the field with $4$ elements.
As listing all the elements takes too much work, is there any better way to prove the... | Here is a proof without use of Maschke's theorem.
D.J.S. Robinson, A Course in the Theory of Groups, 2d edition, exerc. 3.2.6, p. 79 (before stating Maschke's theorem, which only happens in chapter 8), asks for a proof that $PSL(3, 4)$ has no element of order 15. The author gives the following hint : "Suppose there is ... | {
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"source": "stackexchange",
"question_score": "10",
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What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
| Another method is to rewrite $\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2\theta$ as
$$\frac{1 + (\sin\theta + 3\cos\theta)^2}{2}.$$
The minimum value of this expression is $\frac12$, so the maximum value of the original expression is 2.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Find the square root of a matrix Let $A$ be the matrix
$$A = \left(\begin{array}{cc}
41 & 12\\
12 & 34
\end{array}\right).$$
I want to decompose it into the form of $B^2$.
I tried diagonalization , but can not move one step further.
Any thought on this? Thanks a lot!
ONE STEP FURTHER:
How to find a upper triangular ... | This is an expansion of Arturo's comment.
The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$
This is of the form $A=Q\Lambda Q^{-1}$... | {
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If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?
The answer says $1$, but I am not sure how to derive it.
| I'm not sure how different this is, but here is my version
$a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}
\quad$ (Note this implies $abc \ne 0$)
$a^2bc + ac = ab^2c + ab = abc^2 + bc$
$a(abc) + ac = b(abc) + ab = c(abc) + bc$
$\quad a(abc) + ac = b(abc) + ab \implies (a-b)abc=a(b-c)$
$$ a=b=c \ne 0 \; \text{ or }... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 5
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System of $\sqrt{7x+y}+\sqrt{x+y}=6$ and $\sqrt{x+y}-y+x=2$ $$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$
I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as:
$$(y-x)(y+x+4)+4-x-y=0$$
or
$$x_{1,2}=\fra... | This is the way I would solve the given system in my school days. It is very similar to the hint provided by Ross Millikan. It turns out that one of the equations becomes linear. I omit some intermediate steps. The system
$$\left\{
\begin{array}{c}
\sqrt{7x+y}+\sqrt{x+y}=6 \\
\sqrt{x+y}-y+x=2
\end{array}
\right.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying a simple fraction with exponent I am trying to simplify this fraction :
$
\dfrac{(3^2)(5^4)}{15^3}
$
The answer is : $ \dfrac{5}{3} $
I am trying to do the following: $ \dfrac{3^2}{15^3} \cdot \dfrac{5^4}{15^3} $ so ... $ \dfrac{1^{-3}}{3} \cdot \dfrac{1^1}{3} $
But that's not giving me the right answer, ... | For starters, $\displaystyle\frac{3^2}{15^3} \cdot \frac{5^4}{15^3} = \frac{(3^2)(5^4)}{(15^3)(15^3)} = \frac{(3^2)(5^4)}{15^6}$, not $\displaystyle\frac{(3^2)(5^4)}{15^3}$, so your first step isn’t right. Next, $\displaystyle\frac{3^2}{15^3}$ isn’t $\displaystyle\frac{1^{-3}}{3}$, and $\displaystyle\frac{5^4}{15^3}$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| Since $\frac{d}{dt}\sqrt{1+t^2} = \frac{t}{\sqrt{1+t^2}}$, we can integrate by parts to get
$$
\int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt = \int t\cdot \frac{t}{\sqrt{1+t^2}}\mathrm dt
= t\sqrt{1+t^2} - \int \sqrt{1+t^2}\mathrm dt.
$$
Cheating a little bit by looking at a table of integrals, we get that since
$$
\frac... | {
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"source": "stackexchange",
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Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+1... | $x^2-14x+49=x-5$
$x^2-15x+54=0$
$x^2-9x-6x+54=0$
$x(x-9)-6(x-9)=0$
$(x-6)(x-9)=0$
$x=6$, or $x=9$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A Particular Two-Variable System in a Group Suppose $a$ and $b$ are elements of a group $G$.
If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.
I've been trying to prove but still inconclusive. Please prove to me.
Thanks very much for proof.
| In this answer I give credit to Jyrki Lahtonen for the answer he posted.
There are holes in his post, so I sensed the need for a step by step answer (firstly to convince myself, but also other people in doubt), and so here it is.
$\bbox[5px,border:2px solid]{\begin{array}{cc}a^3=b^{-1}a^2b&(\alpha)\\b^3=a^{-1}b^2a &(\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/66573",
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"source": "stackexchange",
"question_score": "22",
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If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.
| To begin with
$$ 2x=(2x)^3 =8 x^3=8x \ . $$
Therefore $6x=0 \ \ \forall x$.
Also
$$ (x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3 $$
and
$$ (x-y)=(x-y)^3=x^3-x^2 y - xyx -y x^2 + x y^2 +yxy+ y^2 x -y^3
$$
Subtracting we get
$$ 2(x^2 y +xyx+yx^2)=0 $$
Multiply the last relation by $x$ on the left and... | {
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"answer_id": 1
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The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4... | Back when I was in academia, I taught the "how to prove stuff" course, and one of the first problems that I'd give (which, I admit, I borrowed from my graduate adviser) was along the same vein, namely: criticize the "proof" of the following "theorem" or rethink your life!
"Theorem": You have all the money you need.
"... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int^{\frac{\pi}{4}}_{0} \tan^3{x}$ given $2\tan^3x = \frac{d}{dx}\left(\tan^2x+2\ln \cos x\right)$ Find $\int\nolimits^{\frac{\pi}{4}}_{0} ( \tan^3{x} ) \space dx$ given $2\tan^3x = \frac{d}{dx}( \tan^2x+2\ln \cos x )$
$$\int\nolimits^{\frac{\pi}{4}}_{0} \tan^3{x} \space dx = \frac{1}{2}\left[\tan^2{x} + 2 \ln{\... | You should know that $\cos\frac\pi4=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. So $$\ln \cos\frac\pi4=\ln\frac{1}{\sqrt{2}} = -\ln\sqrt{2} = -\ln\left(2^{1/2}\right)= -\frac12\ln 2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Next number in series What are the basic/advanced strategies used to find the next number in series.
I know the simple ones such as addition, multiplication etc. But recently I came into a question that goes on something like 812, 819, 823, 835, 834, 851(Don't try to solve this, I changed some numbers and there is no s... | About your series:
Let
$$
\begin{eqnarray*}
\begin{split}
P(x)&:= 812\frac{(x-2)(x-3)x-4)(x-5)(x-6)}{(1-2)(1-3)(1-4)(1-5)(1-6)}+819\frac{(x-1)(x-3)(x-4)(x-5)(x-6)}{(2-1)(2-3)(2-4)(2-5)(2-6)}\\
&+823\frac{(x-1)(x-2)(x-4)(x-5)(x-6)}{(3-1)(3-2)(3-4)(3-5)(3-6)} +835\frac{(x-1)(x-2)(x-3)(x-5)(x-6)}{(4-1)(4-2)(4-3)(4-5)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If $n$ is an odd pseudoprime , then $2^n-1$ is also odd pseudoprime I have some problems understanding the following proof:
Definition: A composite number $n \in \mathbb{N}$ is called pseudo prime if $n \mid 2^{n-1}-1$ holds.
Theorem: If n is a odd pseudo prime number, then $2^n-1$ is also an odd pseudo prime number, t... | This solution is essentially the same as the one quoted by OP, with some detail added and some removed. The most useful change is the introduction of $N$, which allows us to have one less level of exponentiation.
We have $2^{n-1}\equiv 1 \pmod{n}$. Let $N=2^n-1$. We first show that $N$ is composite. Since $n$ is compo... | {
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Finding $\lim\limits_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}$ $$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$
With a first look this must give $1$ as a result but have a problem to explain it.
How can I do it?
Edit
I noticed that it is $\frac{\infty}{\infty}$.
$$\lim_{n \to \infty}{n^{n}\frac{(\... | We can write $(1^1+2^2+\cdots+n^n)/n^n$ as $a_n + b_n + 1$, where
$$
a_n = \frac{1^1+2^2+\cdots+(n-2)^{n-2}}{n^n} \text{ and } b_n = \frac{(n-1)^{n-1}}{n^n}.
$$
Both $a_n$ and $b_n$ are positive, and also
$$
a_n < \frac{(n-2)(n-2)^{n-2}}{n^n} < b_n < \frac{n^{n-1}}{n^n} = \frac1n.
$$
The squeeze theorem should allo... | {
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How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how ... | $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$$
$\leftrightarrow (x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \geq 9$.
Using Cauchy Inequality, we have $x+y+z\geq 3\sqrt[3]{xyz}$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\geq 3\sqrt[3]{\frac{1}{xyz}}$
$$\Rightarrow (x+y+z)\left(\frac{1}{x} + ... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4}... | The definition of the derivative, $\lim\limits_{x\to 4} \dfrac{f(x)-f(4)}{x-4}$ will always give you the indeterminate form $0/0$ if you plug in the number that $x$ is approaching. I.e. you get $\dfrac{f(4)-f(4)}{4-4}$.
So when you see
$$\lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{... | {
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Proofs from the BOOK: Bertrand's postulate: $\binom{2m+1}{m}\leq 2^{2m}$ I have a very hard proof from "Proofs from the BOOK". It's the section about Bertrand's postulate, page 8:
It's about the part, where the author says:
$$\binom{2m+1}{m}\leq 2^{2m}$$
because $\binom{2m+1}{m}=\binom{2m+1}{m+1}$ are the same in $\sum... | Since $2m+1$ is odd, there will be an even number of coefficients (since you have everything from $\binom{2m+1}{0}$ to $\binom{2m+1}{2m+1}$). Because the coefficients are symmetric, you can write:
$\sum \limits_{k=0}^{m} \binom{2m+1}{k}=\sum \limits_{k=m+1}^{2m+1} \binom{2m+1}{k}$
But $\sum \limits_{k=0}^{2m+1} \binom{... | {
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Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$ Using the Taylor expansion for $${(1+x)}^{-1/2}$$ we have $${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$$
for $|x|<1$.
But if $|a| <1$, how can we use the above fact to find
$$\sum_{n=0}^\infty \binom{2n}{n} a^n... | Write out what $\binom{-1/2}{n}$ means; i.e.,
$$
\begin{align}
\binom{-1/2}{n} &= \frac{(-1/2)(-3/2) \cdots ((-2n+1)/2)}{n!} = \frac{(-1)^n}{2^n} \frac{(1)(3) \cdots (2n-1)}{n!} \\
&= \frac{(-1)^n}{2^n} \frac{(2n)!}{2(4) \cdots (2n)n!} = \left(\frac{-1}{4}\right)^n \frac{(2n)!}{n!n!} \\
&= \left(\frac{-1}{4}\right... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to find this generating function this is the power series:
$$\sum_{i=0}^\infty n(n-1)^2 (n-2) z^n.$$
how can I find a generating function from it? I could use the third derivative but the $n-1$ is squared so I don't know what to do..
| We can use
$$
\sum_{n=0}^\infty\binom{n}{k}z^n=\frac{z^k}{(1-z)^{k+1}}\tag{1}
$$
which follows from differentiating $\sum\limits_{n=0}^\infty z^n=\frac{1}{1-z}$ repeatedly $k$ times and multiplying by $\dfrac{z^k}{k!}$, and
$$
n(n-1)^2 (n-2)=n(n-1)(n-2)(n-3)+2n(n-1)(n-2)\tag{2}
$$
which is an example of the fact th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/84062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Puzzle: numerical pattern recognition IF
7 - 3 = 10124
6 + 3 = 3279
5 – 2 = 763
11 + 2 = 92613
Then,
15 - 3 =?
Any ideas ? I dont know how I am supposed to go about solving puzzles like this one ? is there any strategy ? any algorithm ?
| This is all I can come up with:
1: Do the operation, the answer to that will be placed on the far right:
$
\begin{align}
7-3 &= 4 \\
6 + 3 &=9 \\
5-2 &= 3 \\
11 + 2 &= 13
\end{align}$
2: Take the solution from above, multiply by the second digit in the original expression. Place it to the left of the answer we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Integral Inequality Question I am currently working on the problem below and I am in need of help.
Consider the definite integral $\int_{1}^{2}\frac{1}{t}dt$.
(a)By the dividing the interval $1\leq t\leq 2$ into $n$ equal parts and choosing appropriate sample points, show that
$$\sum_{j=1}^{n}\frac{1}{n+j}< \int_{1}^{... | For part (a), we divide the interval $[1,2]$ into $n$ intervals which only interests at boundary points, i.e. $[1+\frac{j-1}{n},1+\frac{j}{n}]$ where $1\leq j\leq n$. Note that $\frac{1}{t}$ is an decreasing function, which implies that for $1\leq j\leq n$
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$ I absolutely know I am not doing this right. :[
Could I get some input or point back in the right direction?
My work done so far is shown below.
Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the E... | $$
\operatorname{var}\left( \frac 1 a X - b\right) = \operatorname{var}\left( \frac 1 a X \right) = \frac 1 {a^2} \operatorname{var}(X).
$$
$$
\operatorname E\left( \frac 1 a X-b\right) = \frac 1 a \operatorname E(X) - b.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Last coupon collected in the coupon collectors problem Consider the classical coupon collectors problem. Given a particular coupon $i$ we can ask for the probability that $i$ is the last coupon collected. We asked this question on cstheory and got a wonderful answer from James Martin introducing us to the idea of Poiss... | I have no idea if this bound will be of any use to you or not (or even how it compares to the inclusion-exclusion bound already posted), but you can get a quick(er) upper bound as follows. Without loss of generality let's assume the coupon we care about is coupon $k$ and that $n_1 \leq n_2 \leq \dots \leq n_{k-1}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.