Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Properties of Matrix Product I am studying for my exam next week and the teacher has posted previous exams online.
I have the following question, given:
$A = \begin{bmatrix}4&-2&2\\2&4&-4\\1&1&0\end{bmatrix}$
$u = \begin{bmatrix}1\\3\\2\end{bmatrix}$
Find $A^5*u$ without any calculations...
HINT: Properties of the Matr... | Note that
$$A u = \begin{pmatrix}4 & -2 & 2\\ 2 & 4 & -4\\1 & 1 & 0\end{pmatrix}\begin{pmatrix}1\\3\\2\end{pmatrix} = \begin{pmatrix}2 \\ 6 \\ 4\end{pmatrix} = 2 \begin{pmatrix}1 \\ 3 \\ 2\end{pmatrix} = 2 u$$
Then for any $n>0$
$$A^n u = A^{n-1} (A u)= 2^n u$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\... | Your algebra is cooky!
Note that
$$ \left( \frac{1}{2} \right)^2 \sum_{n=2}^\infty \left( \frac{1}{2} \right)^n = \sum_{n=2}^\infty \left( \frac{1}{2} \right)^{n+2} = \sum_{n=0}^\infty \left( \frac{1}{2} \right)^{n + 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
... | Both your solutions are correct. To see your first and second solution align note that one solution of $x = (-1)^{1/3}$ is $$x = -1$$
So one possible solution of your original problem is $$-16*(-1)^{1/3} = 16$$
This can be seen by observing that $$(-1)^3 = -1$$
In the complex plane $x= (-8)^{4/3}$ has multiple solution... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more si... | Hint:
Since $$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)$$
Solve $z^2+z+1=0$ and $z^2-z+1=0$ by using the quadratic formula:
$$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
$$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction proof I'm having trouble with: $1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}$ So I'm being asked to use induction to prove that for every $x\in\{a\ |\ a\in R, a\neq 1\}$ and for every $n\in N$
$$
1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}
$$
I have no trouble proving it for $n=1$ :
$$
1+x = \frac{1-x^2}{1-x}
... | $$\frac{1-x^{k+1}}{1-x} +x^{k+1}= \frac{1-x^{k+1}}{1-x} + \frac{1-x}{1-x}x^{k+1} = \frac{1-x^{k+1} + (1-x)x^{k+1}}{1-x} = \dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Sum of roots of equation system It is necessary to find the number of $x+y+z$ (sum of roots of equation) in the range $\left(\frac{-5\pi}{4}; \frac{13\pi}{2} \right)$:
$$
\left\{
\begin{array}{c}
\sin x+\sin y=2\cos z\\
\sin y + \sin z= 2\cos x \\
\sin z + \sin x = 2 \cos y
\end{array}
\right.
$$
I tried different wa... | If you are interested in only one solution, choose $x=y=z$, simplifying to only one equation:
$\sin(x)=\cos(x)$
which is e.g. satisfied for $x=\frac{\pi}{4}$. Your soultion would then be $x+y+z=3x=\frac{3\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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evaluating a cosine function using sine function
If $\sin^2(\frac\pi9)+\sin^2(\frac{2\pi}9)+\sin^2(\frac{3\pi}9)+\sin^2(\frac{4\pi}9)=\frac94$, evaluate $\cos^2(\frac\pi9)+\cos^2(\frac{2\pi}9)+{}$ $\cos^2(\frac{3\pi}9)+\cos^2(\frac{4\pi}9)$.
I know the identity $\sin^2(x)+\cos^2(x)=1$
I am thinking of replacing $\sin... | No, it's not right. There is an error in the last step, you should subtract $\frac{9}{4}$ as follows
$$\cos^2\left(\frac{\pi}{9}\right)+\cos^2\left(\frac{2\pi}{9}\right)+\cos^2\left(\frac{3\pi}{9}\right)+\cos^2\left(\frac{4\pi}{9}\right)=1+1+1+1-\frac{9}{4}=\frac{16-9}{4}=\color{blue}{\frac{7}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding Laurent Series With Square Denominator I'm lost. We very, very briefly went over Laurent series in our last class and I have no idea how to deal with them. How might one find:
$f(z)=\frac{1}{(z^2-1)^2}$
Find the Laurent series in the annulus $0 < |z-1| < 2$.
I'm not even really sure how to begin. Is there anyon... | This illustrates the standard technique in pendantic detail:
$$
\begin{align}
\frac{1}{(z^{2}-1)^{2}}
& =\frac{1}{(z-1)^{2}}\frac{1}{(z+1)^{2}}\\
& =\frac{1}{(z-1)^{2}}\frac{1}{((z-1)+2)^{2}} \\
& =\frac{1}{(z-1)^{2}}\frac{1}{4(1+\frac{z-1}{2})^{2}} \\
& =-\frac{1}{(z-1)^{2}}\frac{1}{2}\frac{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derive a ϴ(1) formula for a Recurrence relation I'm given a piece wise function with sequence $a_0$ $a_1$ etc
$$a_n = \begin{cases}8 & n=0\\-7 & n=1\\25 & n=2\\7a_{(n-2)}+6a_{n-3} & otherwise\end{cases}$$
I'm asked to derive a ϴ(1) formula for $a_n$, by solving the recurrence relation.
I'm still learning about recurr... | A generatingfunctionological solution is to define $A(z) = \sum_{n \ge 0} a_n z^n$, shift the recurrence by 3, multiply by $z^n$ and recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 3} z^n
&= 7 \sum_{n \ge 0} a_{n + 1} z^n + 6 \sum_{n \ge 0} a_n z^n \\
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
&= 7 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find conditional distribution $ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right) $ $f(x,y) =x+y$, $0<x<1$ ,$0<y<1$
find
$ \mathbb{P}\left(\frac{1}{4} \le Y \le \frac{3}{4} \bigm| X = \frac{1}{3}\right)$
For now,
$$fx(x) = x+\frac{1}{2} $$, $$fy(y) = y+\frac{1}{2} $$
and $$(y \mid x)... | You are on your way. But please note that the conditional density of $Y$ given that $X=1/3$ is $\frac{1/3+y}{1/3+1/2}$, not $\frac{x+y}{1/3+1/2}$. Now integrate from $1/4$ to $3/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the complex number $z$ such that it satisfies: 1.$|z+\frac{1}{z}|=\frac{\sqrt{13}}{2} $, 2.$[Im (z)]^2+ [Re(z)]^2=2$ Find the complex number $z$ such that it satisfies: $$\left|z+\frac{1}{z}\right|=\frac{\sqrt{13}}{2} $$$$\Im (z)^2+ \Re(z)^2=2$$$$\frac{\pi}{2}<\arg(z)<\pi$$ then find $$z^{1991}$$
Know I was thinki... | Assuming $z=-a+bi$ with $a,b\in\mathbb{R}^+$ so $a> 0$ and $b> 0$:
*
*The Absolute value:
$$\left|(-a+bi)+\frac{1}{(-a+bi)}\right|=\left|\frac{1+(-a+bi)^2}{(-a+bi)}\right|=\frac{|1+(-a+bi)^2|}{|(-a+bi)|}=\sqrt{\frac{(1+(-a)^2-b^2)^2+(2(-a)b)^2}{(-a)^2+b^2}}=\sqrt{\frac{(1+a^2-b^2)^2+(2ab)^2}{a^2+b^2}}$$
*The rea... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b$
Prove that for all positive integers $b$ that $$\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b.$$
My idea is induction, but I cannot figure stuff out on the inductive step.
| This can be reduced to a telescoping sum.
$$
\begin{align}
\sum_{a=1}^b\frac{a\cdot a!\binom{b}{a}}{b^a}
&=\sum_{a=1}^b\frac{[b-(b-a)]\frac{b!}{(b-a)!}}{b^a}\tag{1}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-\sum_{a=1}^{b-1}\frac{\frac{b!}{(b-a-1)!}}{b^a}\tag{2}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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prove trig equivalence $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
I want to use the the $e^{ix}$ identities but I'm not sure if it can be done that way, let alone how to do it.
Any tips would be appreciated.
| From $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$, get:
$$\require{cancel}\begin{align}
\sin x+\sin y&=\sin\left(\frac{x+y}{2}+\frac{x-y}{2}\right)+\sin\left(\frac{x+y}{2}-\frac{x-y}{2}\right) \\[2ex]
&=\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)+\cancel{\cos\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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suppose p is fermat prime and p>3, show that $\left(\frac{7}{p}\right) = -1$ suppose p is fermat prime, show that $\left(\frac{7}{p}\right) = -1$
We wish to solve $\left(\frac{7}{p}\right) = -1$ for $p$. By quadratic reciprocity and the fact that $7 \equiv 3 \pmod{4}$ we know $\left(\frac{7}{p}\right) = - \left(\frac{p... | Fermat primes are of the form $p=2^{2^{k}}+1$ for some $k\in\Bbb Z_{\ge 0}$. If $k=0$, your claim is wrong; but it's true if $k\ge 1$. I'll assume $k\ge 1$.
We know $p\equiv 1\pmod{4}$, so by Quadratic Reciprocity $\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)$.
We know $2^{2^k}+1\equiv \{5,3\}\pmod{7}$, none of wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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taylor expansion of $\sinh(x)$
I would like to find taylor expansion of $sh(x)$
My thoughts
indeed,
note that : $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ then
\begin{align}
\sinh(x)&=\frac{e^x-e^{-x}}{2} \\
&=\frac{1}{2}\left( e^x-e^{-x} \right)\\
&\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^n\frac{x^k}{k!}+o(x^n)-\sum_... | How about a rather simple derivation like the one below:
$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$
and
$$\exp(-x)= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
So when you subtract the two equations:
$$\exp(x) - \exp(-x) = 2x + 2\frac{x^3}{3!} + 2\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating the integral $\int (x^2-1)^{\frac{-3}{2}}dx$ I need to solve the differential equation:
$$\displaystyle f'(x) = (x^2-1)^{-\frac{3}{2}}, f(2)=\frac{3-2\sqrt 3}{3}$$
Which basically amounts to solving the integral $\int (x^2-1)^{-\frac{3}{2}} \mathrm{d} x$. I was thinkng of using $\int (x^2-1)^{-\frac{1}{2}} \... | I thought it might be instructive to present an approach that uses integration by parts, rather than hyperbolic trigonometric substitution. To that end, we begin by writing
$$\begin{align}
\frac{1}{(x^2-1)^{3/2}}&=\frac{(1-x^2)+x^2}{(x^2-1)^{3/2}}\\\\
&=\frac{x^2}{(x^2-1)^{3/2}}-\frac{1}{\sqrt{x^2-1}}
\end{align} \tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find a polynomial with integral coefficients whose zeros include $\sqrt{2} + \sqrt{5}$. Find a polynomial with integral coefficients whose zeros include $\sqrt{2} + \sqrt{5}$.
I think I can use $-3= (\sqrt{2} + \sqrt{5})(\sqrt{2} - \sqrt{5})$ and a certain telescopic factorisation. The problem is that I don't know how ... | You may observe that
$$
(\sqrt{2} + \sqrt{5})^2=7+2\sqrt{10}
$$ giving
$$\left((\sqrt{2} + \sqrt{5})^2-7\right)^2= (2\sqrt{10})^2=40.$$ Thus $\sqrt{2} + \sqrt{5}$ is a root of
$$P(X)= (X^2-7)^2-40=X^4-14X^2+9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think i... | Hint: $$\cos\left(\dfrac{21}{4}\pi\right)=\cos\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\dfrac{21}{4}\pi\right)=\sin\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\\ \cos\left(\dfrac{11}{4}\pi\right)=\cos\left(\left(\dfrac{10}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solving a partial sum.... Hey can anyone help with this? This is the classic NPV equation:
$$\texttt{NPV = -CapEx} + \sum_{i=0}^n \frac{\texttt{Revenue − Costs}}{(1+\texttt{Discount})^i}$$
For my purposes all the elements are know except costs.
I need to isolate costs in this equation. Is this possible?
Thanks,
Mike
| Your equation is
$NPV = -CapEx + \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad |+CapEX$
$NPV + CapEx = \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad $
Splitting the fraction
$NPV + CapEx = \sum_{i=0}^n \left( \frac{R }{(1+d)^i} -\frac{C }{(1+d)^i} \right) \quad $
Factoring out $\frac{1}{(1+d)^i}$
$NPV + CapEx = \sum_{i=0}^n \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
| Your long division is wrong and you can easily check it by doing the computation:
$$
x^3+\frac{4x^3}{x^2-4}=\frac{x^5}{x^2-4}
$$
Moreover, the degree of the remainder should be less than the degree of the denominator.
If the long division is performed correctly, you find
$$
\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{x^2-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$
How to compute
$$\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$$
I'm interested in more ways of computing this integral.
There is always the straight forward method to decompose into simple elements
$\dfrac{1}{(1 + t^{4})}$ it w... | Here is another method using purely clever substitutions. Note that
$$
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
=
\int_{0}^\infty \frac{w^2}{1+w^4} \mathrm{d}w
$$
Where the last integral comes from the substitution $w \mapsto 1/x$. Addition now gives that
$$
\begin{align*}
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
& = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How do I complete a proof for an inequality by using the triangle inequality theorem? Prove that the inequality will hold for every real number, $x$
$$\left| 2+x \right| \le \left| 2x+1 \right| +\left| 1-x \right| $$
Proof: This proof is by case analysis.
Case 1:
1) Let $a=2x+1$ and let $b=1-x$ and $a,b\in\mathbb{R}$... | By the triangle inequality $|a| + |b| \ge |a + b|$
Let $a = 2x + 1$. Let $b = 1 - x.$
Then $|2x + 1| + |1 - x| \ge |2x + 1 + 1 - x| = |x + 2|$
| {
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"url": "https://math.stackexchange.com/questions/1511920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
| $4^n+10\cdot 9^{2n-2}=4\cdot 4^{n-1}+(7+3)(9^2)^{n-1}=4\cdot 4^{n-1}+(7+3)(81)^{n-1}$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81)^{n-1}=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1}+4^{n-1})$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})+3\cdot 4^{n-1}=7\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n... | {
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"url": "https://math.stackexchange.com/questions/1512443",
"timestamp": "2023-03-29T00:00:00",
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} |
Sum of the following series upto n terms:$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$ Find the sum of the following series up to n terms:
$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$
My attempt:
First,I tried to apply binomial series,but i cannot understand how to apply it.
Then ,I tried as follows:
$\sum_{k=1}^n \frac {k}{(... | It is a telescopic series. We have:
$$ \frac{k}{(k+1)(k+2)} = \frac{2}{k+2}-\frac{1}{k+1}$$
hence:
$$ \frac{k}{(k+1)(k+2)}2^k = \frac{2^{k+1}}{k+2}-\frac{2^k}{k+1} $$
and:
$$ \sum_{k=1}^{n}\frac{k}{(k+1)(k+2)}2^k = -1+\frac{2^{n+1}}{n+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expression for recurrence relation $a_n$ using exponential generating functions $a_0 = 2$, $a_n = na_{n - 1} - n!$ for $n \geq 1$.
Let $$f(x) = \sum_{n \geq 0}a_n\frac{x^n}{n!}.$$
Multiplying each term in the relation by $\frac{x^n}{n!}$ and summing over values for which the relation is defined and we get:
$$
\begin{... | I'll take off where you end up. You want:
$\begin{align}
n! [x^n] \frac{2 - 3 x}{(1 - x)^2}
&= n! [x^n] (2 - 3 x) (1 - x)^{-2} \\
&= n! \left(
2 [x^n] (1- x)^{-2} - 3 [x^{n - 1}] (1 - x)^{-2}
\right) \\
&= n! \left(
2 (-1)^n \binom{-2}{n} - 3 (-1)^{n - 1} \binom{-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve using the limit's definition The question is :
I am stuck here :
$$| \sqrt{x+1} - \sqrt{x} -1 | / \sqrt{x} + 1$$
i know that the numerator is negetive so i must change it in order to delete the absolute value yet i still don't know how to proceed.
Thanks in advance !
| Let $\epsilon >0$, we are seeking for $\delta >0$, such that , for all $x\geq 0$ with $x>\delta$, we have $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| <
\epsilon$$
Indeed, $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| = \Bigg| \frac{\sqrt{x+1} -\sqrt{x}-1}{\sqrt{x}+1} \Bigg| $$
Note that , for all $x\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~... | Here is a solution using vieta's formula.
Let $s=x+y$ and $p=xy$. Then $s+p=44$ and $sp=448$. Therefore $s$ and $p$ are the roots of the quadratic $z^2-44z+448=0$. Thus it follows that $s=16$ and $p=28$. Hence $$x^2+y^2=(x+y)^2-2xy=s^2-2p=256-56=\boxed{200}.$$
But note that if $s=28$ and $p=16$, then
$$x^2+y^2=(x+y)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to solve this limit without using L'Hospital's Rule? $$
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan ... | Let $u := \frac{1}{x}$, then $$\lim_{u\to 0}\frac{\arctan 3u}{\arctan 2u} = \frac{3}{2}$$ because $\arctan u\sim u$, when $u\to 0$. As $x\to\infty$, then $|\arctan x|\to \frac{\pi}{2}$: absolute value signs unnecessary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
solve $x^4 + y^4 = x^3 + y^3 + 10$ for $x,y \in \mathbb{Z}$ solve $x^4 + y^4 = x^3 + y^3 + 10$ for $x,y \in \mathbb{Z}$.
I tried solving this by trying to find upper bounds for $|x|$ and $|y|$, therefor it is quite useful to write:
$x^4 + y^4 - x^3 - y^3 = .....$
where ... is in the form of a square.
I tried to write... | $x^4-x^3$, and $y^4-y^3$, are never negative and rarely less than $10$. There must be very few possibilities to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of the integral: $\int _{10}^{19} \frac{\sin x}{1+x^a}dx$
Find the minimum odd value of $a$, where$a>1,~ a \in \mathbb{N}$ such that $$\int_{10}^{19} \frac{\sin x}{1+x^a}dx<\frac{1}{9}$$
ATTEMPT:- Let $I(a)=\int _{10}^{19} \frac{\sin x}{1+x^a}dx$ then from Leibnitz's rule, $$I'(a)=-\int _{10}^{19} \f... | As the question is written, if you can show $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \frac 19$ you are done. But $\int _{10}^{19} \frac{\sin x}{1+x^3}dx \lt \int_{10}^{19}\frac 1{1001}dx=\frac {9}{1001}\lt \frac 19$ No need to evaluate the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
upper limit of $\cos (x^2)-\cos (x+1)^2$ is $2$ How to prove that upper limit of $\cos (x^2)-\cos (x+1)^2$ is $2$.
Can anyone show proof of this?
| Let's use the sum to product formula:
$$ \cos a + \cos b = \sin \tfrac{a+b}{2}\cos \tfrac{a-b}{2} $$
Hmm is that right? I never remember the correct combinations of + and - symbols. Let's see:
$$ \frac{1}{2} \left( e^{2\pi i \, \frac{a+b}{2}} - e^{-2\pi i \, \frac{a+b}{2}}\right) \times \frac{1}{2} \left( e^{2\pi i \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is the numerator of $\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}$ a power of $2$? I stumbled on something numerically, and was just starting to work on it, but it seemed fun enough to share.
Let $$f(n)=\sum_{k=0}^{n} \frac{(-1)^{k}}{2k+1}\binom{n}{k}$$
It appears, from the first few values, that $f(n)$ always has num... | Posting my own answer to my question after a few months because:
*
*This answer is directly related to the source of the question.
*This answer shows the numerator is a power of 2 without finding the explicit closed formula for the value.
This question originally came from taking this question and asking myself ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
Find the limit $\lim_{x \to 0} (2^x + \sin (3x)) ^{\cot(3x)}$ Please help, I have already tried every thing I can, but nothing works. I have no I idea what to do.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$
| While evaluating limit of expressions of type $\{f(x)\}^{g(x)}$ it is best to take logarithms. Let $L$ be the desired limit so that
\begin{align}
\log L &= \log\left(\lim_{x \to 0}(2^{x} + \sin 3x)^{\cot 3x}\right)\notag\\
&= \lim_{x \to 0}\log(2^{x} + \sin 3x)^{\cot 3x}\text{ (via continuity of log)}\notag\\
&= \lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to evaluate $\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$ $$\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$$
to evaluate the above equation, I used the formula: $$\sin^{-1} (x) + \sin^{-1} (y) = \sin^{-1}[x(1-y^2) + y(1-x^2)]$$
therefore I have got,
$$\sin^{-1} (\sqrt{2} ... | Let $\sin^{-1}(\sqrt2\sin\theta)=x\implies\sqrt2\sin\theta=\sin x$ and $-\dfrac\pi2\le x\le\dfrac\pi2$
Now $\sqrt{\cos2\theta}=\sqrt{1-\sin^2x}=|\cos x|$
As $\cos x\ge0,|\cos x|=+\cos x\implies\sin^{-1}\sqrt{\cos2\theta}=\sin^{-1}(\cos x)=\dfrac\pi2-\cos^{-1}(\cos x)$
Now $\cos^{-1}(\cos x)=\begin{cases} x &\mbox{if } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let z be a complex number such that $z^2 +z + 1/z^2 + 1/z + 1=0 $ If n is a natural number then find the value of $ z^{2012n} + z^{1006n} + 1/z^{2012n} +1/z^{1006n} $ is equal to.
I tried rewriting it as $ t^2+t-1=0 $ where $ t=z+1/z $ and then find roots but I don't know how to use it to get required value.
| HINT:
As $z\ne0,$ multiply throughout by $z^2$
$$z^4+z^3+z^2+z+1=0\implies z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0\implies z=e^{2\pi m i/5}$$ where $m\equiv1,2,3,4\pmod5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)... | Some obfuscation using linear algebra:
Write $x + \frac{1}{x} = a$ and let
$$ p(\lambda) = \left( \lambda - x \right) \left( \lambda - \frac{1}{x} \right) = \lambda^2 - a\lambda + 1$$ be a polynomial whose roots are $x$ and $\frac{1}{x}$ and consider the companion matrix
$$ A = \left( \begin{matrix} 0 & -1 \\ 1 & a \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Convert integral to a series I have to find an infinitite series expansion for the integral:
$$\int \frac{x}{8+x^3} \, dx$$
First, I started by determining the Taylor series of the integrand
$$\frac{x}{8+x^3}=\frac{x}{8} \cdot \frac{1}{1-(-(x/2)^3)} = \frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3... | $$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$
$$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Mathematics Olympiad Question $a+b+c=7$, ... Given $a+b+c=7$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = 0.7$, need to find $\frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{a+c}$.
I have noted that these two differ by a factor of $10$. So I divided the first equation by $10$ and equated the two. But that did not lead... | The trick it's just to add and subtract $1$ from each fraction and then to factorize:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{b+c}-1+\frac{a+b+c}{c+a}-1+\frac{a+b+c}{a+b}-1=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})-3=7\cdot 0.7 -3=1.9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$ ... | First, transform your equation:
$$b^2x^2+a^2y^2=a^2b^2$$
$$a^2y^2=a^2b^2-b^2x^2$$
$$y^2=b^2-\frac{b^2}{a^2}x^2$$
$$xy^2=b^2x-\frac{b^2}{a^2}x^3$$
Next, differentiate and solve to find extrema:
$$b^2-3\frac{b^2}{a^2}x^2=0$$
$$b^2=3\frac{b^2}{a^2}x^2$$
$$a^2=3x^2$$
$$x=\pm\sqrt{\frac{1}{3}}a$$
Finally, evaluate $xy^2$ at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz$, wrong book solution?
Evaluate $$\iiint_\Omega z\,\mathrm dx\mathrm dy\mathrm dz,$$ where
$$\Omega = \{(x, y, z) \in \mathbb R^3 \mid y \geq 0,\ z \geq 0,\ x^2 + y^2 \leq 4,\ (y - 1)^2 + z^2 \leq 1\}.$$
From the second and last conditions we get a semi-cy... | I don't think you interpreted the integral correctly. These are the limits of integration I got.
\begin{align}
-2\leq &x\leq 2\\
0\leq &y\leq \sqrt{ 4-x^2}\\
0\leq &z\leq \sqrt{2y-y^2}
\end{align}
Then the integral is
\begin{gather}
\int_{-2}^2 \int_0^{\sqrt{ 4-x^2}}\int_0^{\sqrt{2y-y^2}}z\,dz\,dy\,dx\\
\int_{-2}^2 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
To find the greatest and least distances of a point on the ellipse $x^2+4y^2=4$ from the straight line $x+y=4$ How to use the method of Lagrange multipliers to find the greatest and least distances of a point on the ellipse $x^2+4y^2=4$ from the straight line $x+y=4$ ?
| $$x+y=k$$
$$x^2+4y^2=(k-y)^2+4y^2=5y^2-2ky+k^2=4$$
$$5y^2-2ky+k^2-4=0$$
$$D=k^2-5(k^2-4)=20-4k^2\geq0$$
$$k^2\leq5$$
$$-\sqrt5\leq k\leq\sqrt5$$
$$x+y=4 \quad vs.\quad x+y=\sqrt5$$
$$\text{Distance between above two lines }=\frac{4-\sqrt5}{\sqrt2}=2\sqrt2-\frac{\sqrt10}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin polynomial of tan(x) The method used to find the Maclaurin polynomial of sin(x), cos(x), and $e^x$ requires finding several derivatives of the function. However, you can only take a couple derivatives of tan(x) before it becomes unbearable to calculate.
Is there a relatively easy way to find the Maclaurin pol... | Long division of series.
$$ \matrix{ & x + \frac{x^3}{3} + \frac{2 x^5}{15} + \dots
\cr 1 - \frac{x^2}{2} + \frac{x^4}{24} + \ldots & ) \overline{x - \frac{x^3}{6} + \frac{x^5}{120} + \dots}\cr
& x - \frac{x^3}{2} + \frac{x^5}{24} + \dots\cr & --------\cr
&
\frac{x^3}{3} - \frac{x^5}{30} + \dots\cr
& \frac{x^3}{3} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to calculate a determinant of a 2x2 symmetry block matrix? I'd like to calculate the determinant of the matrix:
$$
\begin{pmatrix}
-A & B^\star \\
-B & A^\star \\
\end{pmatrix}
$$
$A$, $B$ are $L\times L$ complex matrix.
I know that if $A$ and $B$ are real matrix, the determinant c... | Hint.
If $\mathbf {A^*}$ is invertible you can use the general result:
$$
\det
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D
\end{bmatrix}= \det (\mathbf D) \det(\mathbf A-\mathbf B\mathbf D^{-1}\mathbf C)
$$
that is a consequence of the identity:
$$
\begin{bmatrix}
\mathbf A&\mathbf B\\
\mathbf C&\mathbf D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$ Find the value of
$$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$$
$$\frac{2n}{n^4-n^2+1}=\frac{2n}{1-n^2(1-n^2)}$$
I am not able to split it into sum or difference of two $\arctan$s.Please help me.
| Notice that we can write
$$ \frac{2n}{n^4 - n^2 + 1} = \frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$
In view of the addition formula for the tangent, we find that
$$ \arctan \bigg( \frac{x-y}{1+xy} \bigg) = \arctan x - \arctan y $$
for any $x > y > 0$. Thus we have
$$ \arctan\bigg( \frac{2n}{n^4 - n^2 + 1} \bigg) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$ Prove the following equation ($|x|<1$)
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
I made this question and I have the following answer ... | Here is a different/more streamlined/possibly less formal approach. To prove that $$\prod_{n=1}^{\infty}(1-x^n+x^{2n})(1+x^{2n-1}+x^{4n-2})=1$$
Separate the first type of factors into factors coming from even $n$ and odd $n$:
$$\begin{align}
&\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1-x^{2n-1}+x^{4n-2})(1+x^{2n-1}+x^{4n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Uniform Random Variable: Correlation and Independence Let X be a uniform random variable defined on the interval $(0,1)$. If $Y = 6X^2−6X+1$, compute the correlation of X and Y . Are X and Y independent? Are X and Y uncorrelated?
So my work is.
$F(X) = \int{1} dx = x$
$E(X) = \frac{a+b}{2} = \frac{1}{2}$
$F(Y) = \int ... | That's too much work, you would rather do
$$E[XY] = E[X(6X^2-6X+1)].$$
Then use the properties of expectation.
For the covariance, I would proceed as follows:
\begin{align*}
\text{Cov}(X,Y) &= \text{Cov}(X,6X^2-6X+1)\\
&=6\text{Cov}(X,X^2)-6\text{Cov}(X,X)+\text{Cov}(X,1)\\
&=6\left[E[X^3]-E[X]E[X^2]\right]-6\text{Var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the angles of $\triangle ABC$ In a $\triangle ABC$, from vertex $C$, the median to $AB$, the angle bisector of $\angle BCA$ and the perpendicular to $AB$ divides angle $\angle BCA$ into four equal parts.
The task is to compute angles in $\triangle ABC$.
Thanks for any help.
|
Applying sine law to triangle ACD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – 3x)}{\sin x} = \dfrac {\cos 3x}{\sin x}$
Applying sine law to triangle BCD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – x)}{\sin 3x} = \dfrac {\cos x}{\sin 3x}$
Then $\sin 3x \cos3x – \sin x \cos x = 0$
By Wolframalpha, we have x = $\dfrac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how do I verify that this converges uniformly to $f(x)$? I had to find the fourier series for $f(x)=|x|, -\pi \le x \le \pi$
I got the fourier series as
$$f(x)=\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)$$
but now I have to verify that this converges uniformly to $f(x)$ and I have to evaluate the... | You may find p. 4 here useful.
Anyway,
$$|\frac{\pi}{2}-\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+|\sum_{n=0}^{\infty}\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}|\frac{4}{\pi(2n+1)^2}cos((2n+1)x)|$$
$$\le \frac{\pi}{2}+\sum_{n=0}^{\infty}\frac{4}{\pi(2... | {
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"source": "stackexchange",
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If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$ Problem:
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$
My work:
$a^3b-ab^3=ab(a+b) (a-b)... | You can remove "distinct" to make it slightly stronger (but it's a trivial case). Notice $30=2\cdot 3\cdot 5$ and $2,3,5$ are primes. Your problem is: Among any three integers $a,b,c$, exist at least two $x,y\in\{a,b,c\}$ such that $2,3,5\mid xy(x+y)(x-y)$.
By Pigeonhole principle, either $abc\equiv 0\pmod{5}$ or $a\eq... | {
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"source": "stackexchange",
"question_score": "3",
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The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
| It's also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.
We obtain
\begin{align*}
[x^3](1+x)^3(2+x^2)^{10}&=[x^3]\left(\sum_{k=0}^{10}\binom{10}{k}x^{2k}2^{10-k}\right)(1+x)^3\\
&=[x^3]\left(\binom{10}{0}2^{10}x^0+\binom{10}{1}2^9x^2\right)(1+x)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Solve for $x$, correct to two significant figures, the equation: $4^{x}-2^{x+1}-3=0$ Solve for $x$, correct to two significant figures, the equation:
$$4^{x}-2^{x+1}-3=0$$
My answer: $x\log4=\log3+(x+1)\log2 \Rightarrow 0.602x-0.301x=0.477+0.301 \Rightarrow x = 2.6$ (Conflicting with book answer)
Answer in book: $x=1... | $f(x) = 4^x - 2^{x+1} - 3 = (2^x)^2 - 2^{x+1} - 3 = (2^x)^2 - 2(2^x) - 3$
Let $y = 2^x \Rightarrow f(x) = y^2 - 2y - 3$
$f(x) = 0 \Rightarrow y^2 - 2y - 3 = 0 \Rightarrow (y-3)(y+1) = 0$
The roots for this equation are $y = 3$ or $y = -1$. Since $y = 2^x$, the solutions in terms of $x$ are $2^x = 3$ or $2^x = -1$ Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Sum of an arithmetic progression Hi I cannot for the life of me remember how to use the arithmetic progression formula can someone help?
I just need to find it for this sequence:
$$5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)$$
| We have
\begin{align*}5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)&= \overbrace{5n^2+5n^2+\ldots+5n^2}^{(n+1) \text{ times}}-(1+2+\ldots+n)\\
&= 5n^2(n+1)-(1+2+\ldots+n)\end{align*}
Now, if $S=1+2+\ldots+n$, then
\begin{align*}2S &= (1+2+\ldots+n)+(1+2+\ldots+n)=(n+1)+(n-1+2)+\ldots+\big(n-(n-1)+n\big)\\ &=n(n... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Let z = 1 + i. Find the real and imaginary parts of z^19 Let z = 1 + i. Find the real and imaginary parts of $z^{19}$
| $$\left(1+i\right)^{19}=\left(|1+i|e^{\arg(1+i)i}\right)^{19}=$$
$$\left(\sqrt{2}e^{\frac{\pi i}{4}}\right)^{19}=\sqrt{2^{19}}e^{\frac{19\pi i}{4}}=\sqrt{524288}e^{\frac{19\pi i}{4}}=512\sqrt{2}e^{\frac{19\pi i}{4}}$$
So:
$$\Re\left(512\sqrt{2}e^{\frac{19\pi i}{4}}\right)=512\sqrt{2}\cos\left(\frac{19\pi}{4}\right)=512... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does this determinant calculation work? Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows:
$$
\text{det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_0 \\
-1 & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & ... | If $A=(a_{i,j}) \in M_n(\mathbb{C})$, $$\det(A) = \sum_{k=1}^{n} (-1)^ka_{1,k} \det(\Delta_{1,k}) $$
Where $\Delta_{1,k}$ is A minus the column and the line of $a_{1,k}$
Here the sum only has 2 terms because others are equals to $0$.
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ ... | Well
$2(x+2)(x-1)^{3}-3(x-1)^{2}(x+2)^{2}=0 \Rightarrow (x-1)^{2}(x+2)[2(x-1)-3(x+2)]=0 \Rightarrow (x-1)^{2}(x+2)(-x-8)=0 $
From here it should be clear why $x=-8$ is also a root.
This is called factoring.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Algebraic way to see why only $n=3$ is a valid coefficient I'm a bit of a sucker for brute force calculations. Say I want to calculate a coefficient with Fourier theory, in my case
\begin{align*}
a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x) dx.
\end{align*}
Clearly, only $a_3$ is nonzero. But if I calculate this integral... | Considering $$a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x)\, dx$$ we can make the calculations faster using $$\sin(a) \cos(b)=\frac 12\big(\sin(a+b)+\sin(a-b)\big)$$ So $$2a_n=\int_0^1 \sin\big((3+n)\pi x\big)\, dx+\int_0^1 \sin\big((3-n)\pi x\big)\, dx$$ which already shows that, for $n=3$, the second integral disappear... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$ I need to find the sum of n terms of the series
$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$
And I've no idea how to move on. It doesn't look like an arith... | It is telescoping. Consider that:
$$ \frac{1}{1\cdot 3} = \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right),\quad \frac{2}{1\cdot 3\cdot 5} = \frac{1}{2}\left(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}\right), $$
$$ \frac{3}{1\cdot 3\cdot 5\cdot 7}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5}-\frac{1}{3\cdot 5\cdot 7}\right),\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Trigonometric equation $\sin x+1=\cos x$ $$\sin x+1=\cos x,\quad x\in[-\pi,\pi]$$
How do you solve by squaring both sides? the solution is $x\in\{-\pi/2,0\}$ so the solutions $\pi$ and $-\pi$ are inadmissible, I do not understand how by subbing $-\pi$ back into both sides of the equations makes them unequal, and the sa... | This is an instance of a “linear equation in sine and cosine”. There are several methods for solving them.
First method.
Write $X=\cos x$, $Y=\sin x$ and consider the system
$$
\begin{cases}
1+Y=X\\
X^2+Y^2=1
\end{cases}
$$
Substitute in the second equation to get
$$
1+2Y+Y^2+Y^2=1
$$
which gives
$$
Y^2+Y=0
$$
so $Y=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
evaluate $\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$
$$\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$$
can I look at $\lim\limits_{x\to \infty} (3^{\frac{1}{3}}x^{\frac{2}{3}}-x+x)$?
| By factoring, the expression becomes $(x^2(3-x))^{1/3}+x=x^{2/3}(3-x)^{1/3}+x$. Now we take a seemingly random detour and multiply the numerator and denominator by $1/x$, giving $$\lim_{x \to \infty}\frac{x^{-1/3}(3-x)^{1/3}+1}{1/x}= \lim_{x \to \infty}\frac{x^{(-1){1/3}}(3-x)^{1/3}+1}{1/x}=\lim_{x \to \infty}\frac{(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding a particular angle from a triangle
The way I have seen a solution is very lengthy to find the value of x. Is there any easy way to get the value of x? I have tried different formula and didn't get any shortcut solution.
|
Place point $E$ on $\overline{AD}$ such that $\angle CED = 60^{\circ}$.
Now, by the Law of Sines in $\triangle ABD$,
$$\frac{\sin 15^{\circ}}{1}=\frac{\sin 45^{\circ}}{\overline{AD}}$$
$$\therefore \overline{AD} = 1 + \sqrt{3}$$
Since $\triangle CDE$ is equilateral, $\overline{DE} = 2$.
$$\therefore \overline{AE} = \o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached:
| Notice, your mistake $\sqrt{\text{(Hyp)}^2}=\sqrt{16+x^2}\ne 4+x$
you should have $$\tan\theta=\frac{x}{4}\ \ \text{&} \ \ \sec\theta=\frac{\text{Hyp}}{\text{base}}=\frac{\sqrt{16+x^2}}{4}$$
hence substituting the values of $\sec\theta$ & $\tan\theta$, one should get
$$\ln|\sec\theta+\tan\theta|+C=\ln\left|\frac{\sq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility Of $(2^{32} +1)$ Here is my problem:
If $2^{32} +1 $ is completely divisible by a whole number. Which of the following numbers is completely divisible by that number :
(A)($2^{16}+1$)
(B)($2^{16}-1$)
(C)$7*2^{23}$
(D)($2^{96}+1$)
I don't know where to start any help is appreciated. Thank you
| Hint 1: We want to find an option that is divisible by $2^{32} + 1$. Indeed, this is necessary and sufficient: it is necessary since we can choose $2^{32} + 1$ as the factor that divides $2^{32} + 1$ and it is sufficient since if a number is divisible by $r = de$, then it is divisible by $d$ and $e$ as well.
Hint 2: $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Olympiad Trigonometric Inequality
Let $R$ and $r$ be the circumradius and inradius of $\triangle ABC$.
Prove that $$\frac { \cos { A } }{ { \sin }^{ 2 }A } +\frac { \cos { B } }{ { \sin }^{ 2 }B } +\frac { \cos { C } }{ { \sin }^{ 2 }C } \ge \frac { R }{ r }$$
I am not able to get a solution to this inequ... | It seems the following.
Let $a$, $b$, and $c$ be the respective sides of the triangle $\triangle ABC$, $S$ be its area and $p=(a+b+c)/2$ be its semiperimeter. Then $r=S/p$ and $R=abc/4S$. Hence $R/r=abcp/4S^2$. Moreover,
$4S^2=b^2c^2\sin^2 A=c^2a^2\sin^2 B=a^2b^2\sin^2 C$.
Hence the initial inequality is equivalent to... | {
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"url": "https://math.stackexchange.com/questions/1583137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An interesting Sum involving Binomial Coefficients How would you evaluate
$$\sum _{ k=1 }^{ n } k\left( \begin{matrix} 2n \\ n+k \end{matrix} \right) $$
I tried using Vandermonde identity but I can't seem to nail it down.
| The following proof uses complex variable techniques and improves the
elementary one I posted earlier. It serves to demonstrate the method
even though it requires somewhat more of an effort.
Suppose we seek to evaluate
$$\sum_{k=1}^n k {2n\choose n+k}.$$
Introduce
$${2n\choose n+k} =
\frac{1}{2\pi i}
\int_{|z|=\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
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Computing an infinite trigonometric sum $\sum \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$ Let $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$
I'm trying to compute this sum by understanding it as an integral kernel. This question comes from Dym and Mckean Fourier Series and Integrals Ex... | Note:
I am finally correcting my mistake
pointed out by Mark.
For
$G(x,y)
= \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)
$,
since
(here's where my mistake was:
I had cos-cos
instead of the correct
cos+cos)
$\cos(a)\cos(b)
=\frac12(\cos(a-b)+\cos(a+b))
$,
$\begin{array}\\
G(x,y)
&= \frac12\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex number in quadratic equation Find $a,b$ given that a root of $x= 1+2i$ and the equation $ x^2+(a+bi)x+2i-1=0$
I tried finding it by $\Delta$,
which I got $\Delta=a^2+2abi-b^2-8i+4$
I tried substituting the root into the equation but still can't continue.
Can you help me?
| If you specifically want to do this via the quadratic method (which is not the easiest method as you could use the sum and product of roots to get two expressions that can be solved simultaneously).
But continuing with your method anyway:
Using the quadratic formula for a general quadratic $ax^2+bx+c$ which is
$$x=\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Nice way to solve $\int\int \frac{1}{1-(xy)^2} dydx$? This is something I've been thinking about lately;
$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triang... | You can also do this, and this one does not involve any fancy tricks:
Consider the double integral: $$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(1+x)(x+y^2)}dydx.$$
Integrate with respect to $y$ first and recognize this is :
$$I=\int_{0}^{\infty} \frac{1}{1+x}\lim_{y \rightarrow \infty} \frac{\arctan{\frac{y}{\sqr... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to prove $2\sqrt{2+\sqrt{3}}=\sqrt{2}+\sqrt{6}$? My calculator and I were arguing one day about the cosine of some number.
The calculator said "$\cos(\frac x2)=\sqrt{2}+\sqrt{6}$".
I said "That's absurd because $\cos(\frac x2)=\sqrt{\frac{1+\cos(x)}2}$, which evaluates to $2\sqrt{2+\sqrt{3}}$ for this particular $x... | $$\color{gray}{2+\sqrt 3=\frac 1 2+\sqrt 3+\frac 3 2=\frac{1+2\sqrt3 +3}{2}=\frac{1+2\sqrt 3+(\sqrt 3)^2}{2}=\frac{(1+\sqrt3)^2}{2}}\\2\sqrt{\frac{(1+\sqrt 3)^2}{2}}\\ \color{gray}{\sqrt{\frac 1 2(1+\sqrt 3)^2}=\frac{\sqrt{(1+\sqrt 3)}}{\sqrt 2}}\\ 2\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}\\ \color{gray}{\sqrt{(1+\sqrt 3)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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minimum $x^2 + y^2$ on $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1 $ ellipse Given $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
T... | HINT: use the Lagrange Multiplier Method
$$F(x,y,\lambda)=x^2+y^2+\lambda((x-12)^2/16+(y+5)^2/25-1)$$
solve the system
$$2\,x+\lambda\, \left( x/8-3/2 \right) =0$$
$$2\,y+\lambda\, \left( {\frac {2}{25}}\,y+2/5 \right)=0 $$
$$1/16\, \left( x-12 \right) ^{2}+1/25\, \left( y+5 \right) ^{2}-1=0$$
| {
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"url": "https://math.stackexchange.com/questions/1587380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\frac{AC \times BC}{AD \times BD}$
$AC$ is $2004$. $CD$ bisects angle $C$. If the perimeter of $ABC$ is $6012$, find $\dfrac{AC \times BC}{AD \times BD}$.
Attempt
Let $c = AD+BD$.
We have that $\dfrac{AC}{AD} = \dfrac{BC}{BD}$. Thus, $$\dfrac{BD}{AD}+1 = \dfrac{BC}{AC}+1 \implies \dfrac{BD+AD}{AD} = \dfrac{AC+... | If instead of numerical value of $AC$, numerical value of $AB$ was given then the required expression would have a unique answer.
Let $a, b, c$ represent $BC, AC, AB$ respectively.
By, ange bisector theorem we have, $\frac{AD}{BD}=\frac{b}{a}$.
So, $AD=\frac{bc}{a+b}$ and $\frac{ca}{a+b}$.
$\frac{AC \times BC}{AD \tim... | {
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Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.
It looks like AM-GM should be used here but the square roots make it difficult. So maybe Cauchy-Schwarz works?
| This is simply 2 applications of $ab\leq\frac{1}{2}(a^2+b^2)$:
$$
x\sqrt{y^2+z^2}\leq\frac{1}{2}(x^2+y^2+z^2),\quad y\sqrt{x^2+z^2}\leq\frac{1}{2}(y^2+x^2+z^2)
$$
Now, sum the above 2 inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$... | Hints: For the first inequality, note that $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ because this is equivalent to $(a+b)^2\geq 4ab$. For the second, apply AM-GM to
$$
(a+b) + (b+c) + (c+a)\quad\text{and}\quad\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}
$$
and then multiply the resulting two inequalities.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The cubic equation $x^3 - 4 x^2 + x + 1 =0$
*
*The cubic polynomial
$P(x) = x^{3} - 4x^{2} + x + 1$ has discriminant $\Delta = 169 = 13^{2}$ which tells us that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is any root of the equation $P(x) = 0$.
*Therefore, given one root $a$, one can find the other... | Since $P(x) = x^3 - 4 x^2 + x + 1$ is normal, $P(x)$ splits in $\mathbb Q(\alpha)$.
This means that the polynomial will factor into linear factors over $\mathbb Q(\alpha)$. One of the factors will be $(x - \alpha)$. The others will be $(x - (\alpha^2 - 4 \alpha + 2))$ and $(x - (-\alpha^2 + 3 \alpha + 2))$
You can calc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solving equation involving complex numbers
(a) Find real numbers $a$ and $b$ such that $(a+bi)^2 = -3-4i.$
(b) Hence solve the equation: $z^2+i\sqrt{3}z+i = 0$.
Original Image
In the above question, I have solved part (a), with $a=\pm1$ and with $b=\mp2$, but I am not sure how to use this information to solve (b). I ... | Notice, $$(a+bi)^2=-3-4i$$
$$(a^2-b^2)+2iab=-3-4i$$
comparing real & imaginary parts, one should get $$a^2-b^2=-3\tag 1$$ $$2ab=-4\tag 2$$
solve (1) & (2) for the values of $a$ & $b$.
b) $$z^2+i\sqrt 3z+i=0$$
$$\left(z^2+2i\frac{\sqrt3}{2}z+\left(\frac{i\sqrt{3}}{2}\right)^2\right)-\left(\frac{i\sqrt{3}}{2}\right)^2+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | If you want to use induction:
$5^{n+1} + 2\cdot3^n +1 = 5^n\cdot5 + 2\cdot3^{n-1}\cdot3 + 1 = 5^n + 3^{n-1} + 1 + 4\cdot5^n + 4\cdot3^{n-1} = 8K + (4\cdot5^n + 4\cdot3^{n-1})$.
Suffices to show $4\cdot5^n + 4\cdot3^{n-1}$ is divisible by 8. It's clearly divisible by 4. So it suffices to show $5^n + 3^{n-1}$ is even. ... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\... | Hint: use $f(x)=\ln x$ in $[a,a+1]$.
| {
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Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?
| As an expansion of what I said in the comments:
When we multiply the equation $\frac{2}{3} > -4y - \frac{25}{3}$ by $3$ we get the following equation
$\frac{2\times3}{3} > (-4y)\times3 - \frac{25\times3}{3}$. Simplifying the following we get
$\frac{6}{3} > -12y - \frac{75}{3}$.
Now $\frac{6}{3} =2$ and $\frac{75}{3} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | $$\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx= \frac{2}{\sqrt{23}}\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
| See you need $6.5$ in reduced form . So we can just manipulate an get solution . So we need a cubic term to get real number so $x\sqrt{35-x^3}=6$ and $x=3$ gives you $6$ also $x+\sqrt{35-x^3}=5$ where $3$ is the solution thus we have one real answer which is 3 other way is solving by Vieta method.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \... | Note that
$$
A=PDP^{-1}
$$
where
\begin{align*}
P &=
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}
&
D &=
\begin{bmatrix}
\cos\theta+i\sin\theta & 0 \\
0 & \cos\theta-i\sin\theta
\end{bmatrix}
\end{align*}
De Moivre's formula then implies
\begin{align*}
A^{2010}
&= PD^{2010} P^{-1} \\
&= P
\begin{bmatrix}
\cos(2010\,\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
| Let
$$S_n = \ln\prod\limits_{k=1}^n\dfrac{k+a+1}{k+a} - \ln\prod\limits_{k=1}^n\dfrac{k+b+1}{k+b}= \ln\dfrac{k+a+1}{a+1} - \ln\dfrac{k+b+1}{b+1}$$$$ = \ln\dfrac{b+1}{a+1} + \ln\dfrac{k+b+1}{k+a+1}.$$
$$\sum_{k=1}^{\infty} \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right) = \lim\limits_{n\to \infty} S_n =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to evaluate the following integral, $\int\frac{x \, dx}{x^2+2x+17}$? I am new to integration. This function is kinda tricky for me : $$\int\frac{x \, dx}{x^2+2x+17}$$
I came up with following three approaches:
*
*Partial fraction decomposition, but I can't factor the denominator into different parts.
*Substitu... | Variations on this one come up here from time to time.
There is a standard technique in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term.
$$
u = x^2+2x+17,\qquad du = (2x+2)\,dx, \qquad \frac{du} 2 = (x+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show that $(\frac{S_1}{S_n+1},\frac{S_2}{S_n+1},...\frac{S_n}{S_n+1})=_d (U_{(1)},U_{(2)},...,U_{(n)})$. Let $(X_1, X_2,...,X_n) \in \mathbb R^n$ have density function $p(x)$.
(1) Find the density of $(U_{(1)},U_{(2)},...,U_{(n)})$, the order statistics from a sample of iid $\mathbb U[0,1]$ (uniform distributions) vari... | If we let
$$ \begin{align*} Y_k & = \frac {S_k} {S_{n+1}}, k = 1, 2, \ldots, n \\
Y_{n+1} & = S_{n+1}
\end{align*}$$
Then it is not hard to find out the inverse transform:
$$ \begin{align*} E_1 & = Y_1Y_{n+1} \\
E_k &= (Y_k - Y_{k-1})Y_{n+1}, k = 2, 3, \ldots, n \\
E_{n+1} & = Y_{n+1} - Y_nY_{n+1} \end{align*}$$
And ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Testing the convergence of cube root of some function of n I have to test the convergence of the following series:-
$$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$
My approach is as follows :-
$$n^3+1>1=\sqrt[3]{n^3+1}>1=\sqrt[3]{n^3+1}-n>1-n$$
Now since$\sum 1-n$ diverges, the series under consideration diverges.
Is this right... | Hint
Let $a=\sqrt[3]{n^3+1}$ and $b=n$. Then,
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{1}{a^2+ab+b^2}.$$
Observe that $a \geq b$, therefore
$$\sqrt[3]{n^3+1}-n=a-b \leq \frac{1}{3b^2}=\frac{1}{3n^2}.$$
Now use comparison to claim convergence.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficult Integral I have a problem with solving this integral:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$
I tried to use substitution but I got stuck. Can anyone help me?
| You can also use one of the Euler's substitutions:
$$\begin{align}
\sqrt{x^2+1} &= t-x \\
x &= \frac{1}{2}\left(t - \frac{1}{t}\right) \\
dx &= \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt \\
\sqrt{x^2+1} &= \frac{1}{2}\left(t + \frac{1}{t}\right) \\
\frac{dx}{\sqrt{x^2+1}}&=\frac{dt}{t}\\
2x^2+3x+1&= ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.
Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you ... | HINT: the left-hand side minus the right-hand side is equal to
$${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a
bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2}
\geq 0$$
can you proceed? (sum of squares!)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2... | I have given a detailed proof using an easier method. Please click on this link to see the proof
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$
For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$
What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$
Then it suffices to prove that $$(abc... | Among numbers $a-1, b-1, c-1$ there are two having the same sign, say $a-1, b-1$. In other words, $(a-1)(b-1)\ge 0.$ Multiplying both sides by a positive number $(a+1)(b+1)$ we get
\begin{align*}
\left(a^2-1\right)\left(b^2-1\right)&\ge 0 \\
a^2b^2-a^2-b^2+1&\ge 0 \\
a^2b^2+2a^2+2b^2+4&\ge 3\left(a^2+b^2+1\right) \\
\l... | {
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"url": "https://math.stackexchange.com/questions/1602810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\d... | In the spirit as Ian Miller's answer, the first equation leads to $$\cos(x)=\frac{2 b^2+b-4}{2 \sqrt{4-b^2}}$$ In the second equation, replace $$\cos{(x+\frac{\pi}{18})}=\cos \left(\frac{\pi }{18}\right) \cos (x)-\sin \left(\frac{\pi }{18}\right) \sin (x)$$ which allows to extract $\sin(x)$ given by $$-\frac{\csc \left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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What is the correct way to approach $\int\frac{\cos x}{6+2\sin x-\cos^2x}dx$? I have $$\int\frac{\cos x}{6+2\sin x-\cos^2x}dx .$$
Online solvers have problems with it so what I need are some general guidelines on how to proceed with such as to whether do I need to apply some goniometric identity, or do I need to split ... | $$\int\frac{\cos(x)}{6+2\sin(x)-\cos^2(x)}\space\text{d}x=$$
Use $\cos^2(x)=1-\sin^2(x)$:
$$\int\frac{\cos(x)}{5+2\sin(x)+\sin^2(x)}\space\text{d}x=$$
Substitute $u=\sin(x)$ and $\text{d}u=\cos(x)\space\text{d}x$:
$$\int\frac{1}{u^2+2u+5}\space\text{d}u=$$
$$\int\frac{1}{(u+1)^2+4}\space\text{d}u=$$
Substitute $s=... | {
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"url": "https://math.stackexchange.com/questions/1603342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqr... | Hint: Implicitly differentiate each, then set the derivatives equal to one another.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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General technique for finding minimal polynomial? I always have a lot of trouble with these problems, "find the minimal polynomial of {number} over {field}"
What are the general procedures for solving problems of this format?
Thank you for your help
| I'll assume you're working over $\mathbb Q$.
As an example, take $\alpha = 1+\sqrt{2}$. Notice that $\alpha^2 = 3+2\sqrt{2}$, and so we can write
$$\left[\begin{array}{c} \alpha \\ \alpha^2\end{array}\right] =
\left[\begin{array}{cc} 1 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2}\end{array}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Evaluate $ \lim_{x \rightarrow - \infty} \frac{\sqrt{9x^6-x}}{x^3+1} $ I have started learning limits in calculus and I came across this question:
Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ .
I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\df... | Notice, one can easily change limit as $x\to +\infty$ as follows $$\lim_{x\to -\infty}\frac{\sqrt{9x^6-x}}{x^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9(-x)^6-(-x)}}{(-x)^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9x^6+x}}{1-x^3}$$
$$=\lim_{x\to +\infty}\frac{|3x^3|\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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What is the inverse of $f(x) = \dfrac{x}{x^2 - 1}$ I need to find a continuous inverse of $$f(x) = \dfrac{x}{x^2 - 1}$$
Let $y = f(x) = \dfrac{x}{x^2 - 1} \Rightarrow y(x^2-1) = x \Rightarrow yx^2-x = y$
How should I proceed from here?
| Let $f(x) = y = \frac{x}{x^{2} - 1}.$ To find the inverse of $f(x),$ we switch $x$ and $y$ in the equation and solve for $y.$ This yields $x = \frac{y}{y^{2} - 1},$ which we can rearrange to get $xy^{2} - y - x = 0.$ This is a quadratic with respect to $y.$ Using the quadratic equation, we get that $y = \boxed{f(x) = \... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$
Let $ \triangle ABC$ be an acute-angled triangle. Prove that
$ \sum\limits_\text{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2 $
Attempt
Since $\triangle{ABC}$ is acute, we may say that $A,B,C \in (0, \frac{\pi}{2})$. Now, we have that by a result for ... | Let $\cot\alpha=a$, $\cot\beta=b$ and $\cot\gamma=c$.
Hence, $ab+ac+bc=1$ and we need to prove that $\sum\limits_{cyc}\sqrt{a+b}\geq2\sqrt2$ or
$a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}\geq4$, which is true because by C-S
$a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}=a+b+c+\frac{1}{2}\sum\limits_{cyc}\sqrt{(1+3)(a^2+1)}\geq$
$\geq a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Integers represented by $x^2 + 3 y^2$ vs. integers represented by $x^2 + x y + y^2$. How does one show that the quadratic forms $x^2 + 3 y^2$ and $x^2 + x y + y^2$ represent the same set of integers?
I think it relates to a classical result of Euler about primes of form $6k+1$. In fact a positive integer $n$ is of t... | Others have already answered well; the principal form $x^2 + xy + k y^2$ always represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For $k=1,$ the two sets agree. Same for $k = -1,$ so $x^2 + xy - y^2$ represents the same numbers as $x^2 - 5 y^2.$
It is always the case that $x^2 + xy + 2k y^2$ alway... | {
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"source": "stackexchange",
"question_score": "7",
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Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequalit... | Why is $a^2+b^2+c^2\geq 3$?
I think you need to apply somewhere that you are dealing with a triangle.
Maybe something like this?
If you sum up the following inequalities,
$$ab+bc=b(a+c)> b^2$$
$$ab+ac=a(b+c)> a^2$$
$$ac+bc=c(a+b)> c^2$$
you obtain $a^2+b^2+c^2\leq 6$ which by your argument gives you
$(a+b+c)^2<12.$... | {
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"source": "stackexchange",
"question_score": "9",
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If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even. We say an integer $p>1$ is prime when its only positive divisors are $1$ and $p$. Let $a$ and $b$ be natural number not both $1$. Prove that if $a^4+4^b$ is prime, then $a$ is odd and $b$ is even.
I'm also given a hint:
Consider the expression $(x^2-2xy+2y^2)(x^... | It's clear that $a$ must be odd, as otherwise the number would be divisible by $2$.
Suppose $b$ is odd. Then we can write our number as $a^4 + 4\cdot 4^{2n} = a^4 + 4 \cdot 2^{4n}$ for some $n$. You were given the key factorization,
$$ (a^4 + 4 b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2).$$
here, that means
$$ a^4 + 4 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Dealing with a difficult sum of binomial coefficients, $\sum_{l=0}^{n}\binom{n}{l}^{2}\sum_{j=0}^{2l-n}\binom{l}{j} $
I am interested in finding an upper bound for the sum
$$F(n)= \sum_{l=0}^{n}\binom{n}{l}^{2}\;\sum_{j=0}^{2l-n}\binom{l}{j}$$
Ideally it should be possible to evaluate it exactly using some combin... | By way of an intermittent progress report I submit several integral
representations of the sum
$$\sum_{k=0}^n {n\choose k}^2 \times
\sum_{j=0}^{2k-n} {k\choose j}.$$
They were obtained using the Egorychev method.
The reader is cordially invited to comment, verify and prove these.
Saddle point asymptotics could b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 1
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double integral getting different results I am trying to calculate the double integral $$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx$$ If you plug this into wolfram, you get $-\frac{1}{2}$ and if you plug it into symbolab you get $\frac{1}{2}$ I will show you my steps, I just want to make sure I got th... | I may be wrong but I think the correct result is $0$.
Where I think you got it wrong is in the second equality, where you exchange limit and the integral for one term.
Here I sketched some more detailed computations:
\begin{align}
\lim_{b \to 0}
\int_b^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx
{}={} &
\lim_{b \to 0}
... | {
"language": "en",
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"source": "stackexchange",
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