Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integrating the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
Integrate the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
My attempt:
I chose shifted polar coordinates $$\begin{cases}x=2+r\cos\varphi\\ y=... | You can just exploit symmetries without having to do any explicit integral (or have to deal with polar coordinates) to find the answer. Here's an explaination.
Let $D:=\big\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2\le4\big\}$ and $E:=\big\{(x,y) \in \mathbb{R}^2 \mid (x-1)^2+y^2\le1\big\}$. Since $D = \big\{(x,y) \in \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Upper bound of $\sum_{n=1}^N |1-z^n|$ where $|z| \leq 1$ How to derive an upper bound of
$$\sum_{n=1}^N |1-z^n|$$ where $z\in\mathbb{C}$ and $|z| \leq 1$?
A trivial upper bound would be $2N$ since each $|1-z^n| \leq 2$. But I am hoping for tighter bounds. I ran some numerical experiments and believe the bound should be... | Some thoughts:
As already noted, the maximum of
$\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.
Let $z = \mathrm{e}^{\mathrm{i}\theta}$
with $\theta \in [0, 2\pi]$.
We have
$$\sum_{n=1}^N |1 - z^n| = \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|.$$
Conjecture 1: The maximum of $\sum_{n=1}^N 2\left|\sin \frac{n\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did... | $\require{\cancel}$
Your book is incorrect. Since $(a-b)(a+b) = a^2-b^2$ we get for $k>0$
\begin{align}
k - \sqrt{k^2 +1}& = k - \sqrt{k^2 +1} \left(\frac{k + \sqrt{k^2 +1}}{k + \sqrt{k^2 +1}} \right) \\
&= \frac{\cancel{k^2} - (\cancel{k^2} +1)}{k + \sqrt{k^2 +1}} \\
&= - \frac{1}{k + k\sqrt{ 1+\frac{1}{k^2}}} \\
& = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solve the integral $\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$ Question
Solve the integral,$$\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$$
Attempt
I converted the equation in terms of $\sin(x)$ and $\cos(x)$ using the definition of $\tan(x)$ and $\cot(x)$, and then ap... | Clear
$$ \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx=2\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx. $$
Let
$$ I=\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x}dx. \tag1$$
Then
$$ I=\int_{0}^{\frac{\pi}2} \frac{\cos^3x}{\tan^3x+\cot^3x}dx. \tag2$$
Adding (1) to (2) gives
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$ Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$:
$$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$
By some computational arguments, the inequality seems to be true and in particular the difference b... | I agree that the problem has been reduced to proving that for all $m \in \Bbb{Z_{\geq 2}}$,
$$\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m} \geq 1. \tag1 $$
In (1) above, on the LHS, the two rightmost terms may be collapsed into
$$\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{7+m}{m+1} \geq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
How to evaluate the sum of $\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$ I hava an infinite sum
$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$
I factored the denominator
$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$
Then I separated the fraction
$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)... | Hi welcome to MSE:As a hint for telescopic series$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}
\\=\sum_{n=0}^{\infty}\frac{1}{(3n+1)(n+1)}
\\=\sum_{n=0}^{\infty}\frac{3}{(3n+1)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+2)}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+3)-1}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Quadratic With Periodic Coefficents I've come across a problem that results in the equation:
$t^2 -2t\sin t -2\cos t -2 = 0$
I've tried to do this analytically but I can't figure it out. At this point, I just want to know if it's even possible for something like this to be solved exactly. So can an equation $ax^2+bx+c=... | Using graphics or inspection, the zero of function
$$f(t)=t^2 -2t\sin( t) -2\cos (t) -2 $$is just above to $\frac {2\pi}3$
$$f\left(\frac{2 \pi }{3}\right)=-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}=-0.241108\cdots$$
Expanded as series
$$f(t)=\left(-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}\right)+2\pi \left(t-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do you find this limit with a relationship to $e$ using Taylor series? The limit in question is
$$
\lim_{x \to 0}\left(\frac{\sin(x)}{x}\right)^{1/x^2}
$$
When I replace $\sin(x)$ with its Taylor series about $0$ and cancel out the $x$, I get
$$
\lim_{x \to 0}\left(1-\frac{x^2}{6} + \frac{x^4}{5!} \mp \cdots\right)... | To handle the higher power terms, you can use the big O notation.
Note that
$$
\frac{\sin x}{x}=\frac1x\left(x-\frac{x^3}{6}+O(x^5)\right)=1-\frac{x^2}{6}+O(x^4)
$$
and
$$
\ln(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O(y^5)
$$
So
$$
\begin{align}
\left(\frac{\sin x}{x}\right)^{1/x^2}
=&\exp\left(\frac{1}{x^2}\ln(\frac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $ab+1 = r^2$ for $a,b,r \in \mathbb{N},$ how to show that $\gcd(2a(r+b)+1,2b(r+a)+1) = 1?$ Let $a<b$ be positive integers such that $ab+1 = r^2$ for some $r \in \mathbb{N}.$ If $m_1 = 2a(r+b)+1$ and $m_2 = 2b(r+a)+1.$ I want to find the possible values of $\gcd(m_1,m_2).$ I had taken some examples in the range $a,b\... | We are given that $ab+1=r^{2}$.
Note
$$2a(r+b)+1 = 2ar + 2ab+1 = 2ar+2r^2-1$$
$$2b(r+a)+1= 2br+2r^2-1$$
We claim that the gcd of the above two quantities is $1$.
$\textbf{For proof by contradiction}$ suppose that there is some prime $p$ that divides both $2ar+2r^2-1$ and $2br+2r^2-1$. Then note that $ p \not \div 2r, p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ ?
My Attempt : I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le \f... | We shall give a stronger bound. Note that
$$ 2=1+\sum^\infty_{n=1}\frac{1}{n^2+n}<\sum^\infty_{n=0}\frac{1}{n^2+1}<1+\frac{1}{2}+\sum^\infty_{n=2}\frac{1}{n^2}=\frac{1}{2}+\frac{\pi^2}{6}$$
which is a stronger bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've ... | We could also apply the "triple-angle" identities to write $ \ \sin(3 \theta) \ = \ \sin \theta · (3 - 4 \sin^2 \theta) \ = \ 0 \ \ $ and $ \ \cos(3 \theta) \ = \ 4 \cos^3 \theta - 3 \cos \theta \ = \ -1 \ \ . $
From the first of these, we have either $ \ \sin \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ , \ \pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\int \frac{5x^3+2}{\sqrt{x^3+1}}dx$ How do I integrate $\int \frac{5x^3+2}{\sqrt{x^3+1}}dx$ ?
I know that the result is $2x\sqrt{x^3+1}$, but I cannot think of a way to get to it.
| This is a subtle problem with integration by parts.
We begin by noting that integrals with $\sqrt{x^3+1}$ ordinarily require elliptic functions. Our objective is to try to eliminate the integral.
Begin by decomposing the integrand thusly, separating a term that has a higher power ($+1/2$ instead of $-1/2$) of the radic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Given a triangle ABC inscribed in the unit circle , the 3 vertices could be described via 3 complex number, namely, $a$, $b$, and $c$.
Now $AD$ is an altitude, $D$ is the foot of $AD$ on $BC$.
Prove: $$D = \frac{a+b+c}2 - \frac{bc}{2a}$$
So far my progress is --
*
*the circumcentre of triangle $ABC$ is just $O = 0$.
... | line $AH$:
$$\frac{z-a}{b+c} = \overline{(\frac{{z-a}}{b+c})}$$
line $BC$:
$$\frac{z-b}{c-b} = \overline{(\frac{{z-b}}{c-b})}$$
so in the first equation:
$$\bar z = \overline{b+c} \cdot \frac{z-a}{b+c} + \bar a$$
in the second equation:
$$\bar z = \overline{c-b} \cdot \frac{z-b}{c-b} + \bar b$$
now $$a \bar a = b \bar ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the line integral along a parabola
Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$
I need help trying to find a good parameterization for this because what I've done just lands me in a mess.
My work so far:
Let $x=t, y=2(t+1)^2$... | Here we apply a substitution to get rid of the square root. Nevertheless the evaluation of the integral is somewhat cumbersome. We have a function
\begin{align*}
&f:\mathbb{R^2}\to\mathbb{R}\\
&f(x,y)=x-y
\end{align*}
and a curve
\begin{align*}
C=\{(x,y):y=2(x+1)^2,-1\leq x\leq 0\}
\end{align*}
where we want to evaluat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Find $\tan\frac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\frac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$. Find $\tan\dfrac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$.
So we have $$\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$$ Then $$\sin^2\alpha+\cos^2\alpha+2\sin\alpha\co... | Hint:
$\sin(x)= \sin(2\frac{x}{2})=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})},$
and $\cos(x)= \cos(2\frac{x}{2})=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}.$
Try to do this in $\sin(\alpha)+\cos(\alpha)=\frac{\sqrt{7}}{2}$,and after some algebraic manipulation you can put $ \tan(\frac{x}{2})=t $ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficult integral from general relativity $\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du$ When solving for the angular deflection of a light ray passing close to a star in general relativity, one needs to compute an elliptic integral. By some clever factoring and application of a relevant approximation, one red... | Hmm. We have
\begin{align*}
-2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{(u-a)(b-u)}} \,du \\
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{-(u^2-u(b+a)+ab)}} \,du \\
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{-\left(\left(u-\frac{b+a}{2}\right)^{\!2}-\frac{(b-a)^2}{4}\right)}} \,du \\
&=2\int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putti... | I calculated it longhand:
$$ \begin{align} x & = 2 + \sqrt 3 \\ x^2 & = 2^2 + 3 + 2 \cdot 2 \sqrt 3 \\ & = 7 + 4 \sqrt 3 \\ x^3 & = 2 \cdot 7 + 4 \cdot 3 + 2 \cdot 4 \sqrt 3 + 7 \sqrt 3 \\ & = 26 + 15 \sqrt 3 \\ x^4 & = 97 + 56 \sqrt 3 \\ x^5 & = 362 + 209 \sqrt 3 \\ x^6 & = 1351 + 780 \sqrt 3 \end{align} $$
We also ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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In how many ways can a number be chosen from 1 to 20 such that it is a multiple of 2 or 3?
In how many ways can a number be chosen from $1$ to $20$ such that it is a multiple of $2$ or $3$?
Numbers that are multiple of $2=\{4,6,8,10,12,14,16,18,20\}$ $=>9$ choices
Numbers that are multiple of $3=\{6,9,12,15,18\}$ $=>... | There are $10$ numbers,$[\frac{20}{2}]=10$, can divisible by $2$
There are $6$ numbers,$[\frac{20}{3}]=6$ can divisible by $3$
There are $3$ numbers,$[\frac{20}{6}]=3$, can divisible by $6=2\times 3$
by inclusion-exclusion princible
$n(A\cup B) = n(A)+n(B)-n(A \cap B)= 10+6-3=13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x... | You have$$\sum_{n=0}^\infty\left(\frac{x^n}{4}+\frac{(-1)^nx^n}{4}+\frac{(-1)^{n+1}(n+1)x^n}2\right)=\sum_{n=0}^\infty\frac{1+(-1)^n+\color{red}2(-1)^{n+1}(n+1)}4x^n$$So, the coefficient of $x^{2020}$ is $-1010$. In other words, $f^{(2020)}(0)=-1010\times2020!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | Why work so hard? From the quadratic formula we have:
$\{ a,b \} = \left\{ \frac{7-\sqrt{41}}{2}, \frac{7+\sqrt{41}}{2} \right\}$ so $a^6 + b^6 = 90585$.
My goodness!!! A downvote on a simple, clear, correct answer.
Well, if the approach demands that the solution be done "by hand"—which was NOT part of the problem s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 2
} |
What purely real analytic techniques are there to evaluate $\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\,\mathrm{d}x$? $\newcommand{\d}{\,\mathrm{d}}$Last night, I evaluated the following integral: $$\begin{align}I:&=\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\d x\\&=\int_{-1}^1\frac{1}{(1+x^4)\sqrt{1-x^2}}\d x\\&=\frac{... | Working the antiderivative
Using your first step
$$\int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du$$
$$\left(u^2+1\right)^2+u^4=2 \left(u^2-u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)
\left(u^2+u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)$$ Partial fraction decomposition
$$ \frac{u^2+1}{\left(u^2+1\right)^2+u^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Given $0\leq a\leq b\leq c$ and $b > 0$, prove that: $2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$ This is an exercise that got me stuck. It's from the book "Functions of Several Real Variables 1.1 - exercise 10" by Martin Moskowitz and Fotios Paliogiannis.
Exercise:
Given $0\leq a\leq b\leq c... | It is easier to prove the two inequalities separately.
*
*$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\quad$ This is just AM-GM $\,\sqrt{uv} \le \frac{u+v}{2}\,$ with $\,u=\frac{b}{c}\,$, $\,v=\frac{a}{b}\,$.
*$\frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}\quad$ This is $\,1 + \frac{a}{c} -\frac{b}{c} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$ I need help in finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$ if it converges.
I can't even prove convergence. I tried comparison test. I tried telescoping or even sandwich the partial sums. But I co... | Without series, we have
$$\sqrt{n^2+1}-\sqrt[3]{n^3+1} = \sqrt{n^2+1}- n + n - \sqrt[3]{n^3+1}$$
which allows us to use those conjugates without appealing to sixth order
$$= \frac{1}{n+\sqrt{n^2+1}} - \frac{1}{n^2+n\sqrt[3]{n^3+1}+(n^3+1)^{\frac{2}{3}}}$$
The divergence comes from the square root term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos t... | Let $u=x^2+1$.
$$I=\frac{1}{2}\int{u^{-\frac{3}{2}}(u-1)^{-\frac{1}{2}}}du=\frac{1}{2}\int{u^{-2}\left(1-\frac{1}{u}\right)^{-\frac{1}{2}}}du$$
Using the general formula $\int{(f(x))^nf'(x)}dx=\dfrac{(f(x))^{n+1}}{n+1}+c$ (which can be verified by differentiating the right side),
$$I=\left(1-\frac{1}{u}\right)^{\frac{1... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Let $n=4\cdot p^t+1.$ Is $r_k=4$ for some $1\le kLet $n=4\cdot p^t+1$, $t\ge1$ and $p\ge5$ prime.
Define sequence $R$ as follows;
$r_0=4,$ $r_{k+1}\equiv r_k^2-2\pmod n$, $0\le r_k<n$ for $k\ge1$.
It appears that whenever $n$ is prime, $r_k=4$ for some positive integer $k<n$. How does one go about proving this res... | I can't seem to find a solution using only elementary number theory. From what I did find, it's obvious that we need
$$r_{k-1}^2-2\equiv 4 \text{ mod } n$$
$$r_{k-1}^2\equiv 6 \text{ mod } n$$
By the Law of Quadratic Reciprocity
$$\left( \frac{2}{n} \right) \equiv (-1)^\frac{n^2-1}{8}\equiv (-1)^{\frac{(4p^k)(4p^k+2)}{... | {
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"url": "https://math.stackexchange.com/questions/4480394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate triple integrals $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$ I am trying to evaluate this integral:
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$$
Honestly, I have no ideas to deal with this.
I hope I can be helped by everyone.
I just need a hint to process; thank you... | My second solution:
Using the identity ($q > 0, \alpha \in (0, 1)$)
$$q^\alpha = \frac{\alpha}{\Gamma(1 - \alpha)}\int_0^\infty \frac{1 - \mathrm{e}^{-qs}}{s^{1 + \alpha}}\,\mathrm{d} s,$$
we have
$$\sqrt[3]{\ln \frac{1}{xyz}}
= \frac{1}{3\Gamma(2/3)}\int_0^\infty \frac{1 - (xyz)^s}{s^{4/3}}\,\mathrm{d} s.$$
Denote th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Using complex number in integration I know,
$$I_1 = \int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln|x + \sqrt{x^2-1}| + c$$
But if I factor out $i$ from the denominator, I get:
$$I_2 = -i \int \dfrac{dx}{\sqrt{1 - x^2}} = -i \sin^{-1}x + ic$$
Are these 2 expressions equivalent?
| Let $\sin^{-1}x = θ$
$$ \sin θ = x\\
\frac{e^{iθ} - e^{-iθ}}{2i} = x \\ $$
Solving the quadratic,
$$ e^{iθ} = ix \mp \sqrt{1 - x^2} = i(x \mp \sqrt{x^2 - 1})\\ $$
Taking log on both sides and then multiplying the numerator and denominator by the congugate,
$$\begin{align}
iθ &= \ln i + \ln |x \mp \sqrt{x^2 - 1}| \\
i \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}$
If $a+b+c=3$ ,$ a,b,c>0$ and $\lambda \geq 1$,
prove that : $$\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}.$$
my attempt:
using CBC twice and AM_AG inequality
$\fr... | Tangent Line (TL) method:
It suffices to prove that
$$\frac{a}{a^2 + \lambda}
\le \frac{1}{1 + \lambda} + \frac{\lambda - 1}{(1 + \lambda)^2}(a - 1). \tag{1}$$
(The desired result follows by summing cyclically.)
(1) is true since
$$\frac{a}{a^2 + \lambda}
- \frac{1}{1 + \lambda} - \frac{\lambda - 1}{(1 + \lambda)^2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove $\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} dx = \frac{\pi}{2} \sec (\frac{\pi}{2n})$ without complex techniques Prove
$$\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx = \frac{\pi}{2} \sec \left(\frac{\pi}{2n}\right)$$
for all natural numbers $n \ge 2$.
There are several answers (A1 A2) to this integral but they a... | \begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} d x&=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x\\
&= \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{2 n}+\frac{1}{2}\right)-1} x \cos^ {2\left(-\frac{1}{2n}+\frac{1}{2}\right)-1 }x d x\\
&=\frac{1}{2} B\left(\frac{1}{2 n}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The... | The "is $2x - 15$ positive or negative" dichotomy doesn't work in $\Bbb{C}$. What we can say is that, if $-x^2 - 5x - 8 = |2x - 15|$, then $-x^2 - 5x - 8$ must be a non-negative real number, let's call it $k$. Let's start by solving
$$-x^2 - 5x - 8 = k$$
for $x \in \Bbb{C}$. Using the quadratic formula,
$$x = \frac{-5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt$ $$I=\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt=-\frac{4}{3}G $$ G is Catalan's constant.
$$I=\int_0^1 \frac{(1+t^2)^2}{(1+t^2)(1-t^2+t^4)}\ln(t)dt=\int_0^1 \frac{(1+t^2)^2}{(1+t^6)}\ln(t)dt$$
Do series expansion:
$$I=\sum_{n=0}^\infty (-1)^n \int_0^1(1+2t^2+t^4)t^{... | You can arrive to the desired result starting at
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2} \right)$$
Using the Hurwitz zeta function
$$S_a=\sum_{n=0}^\infty \frac{ (-1)^n}{(6n+a)^2}=\frac{1}{144} \left(\zeta \left(2,\frac{a}{12}\right)-\zeta \left(2,\frac{a+6}{12}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Evaluating $\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}$ Does anyone know how to evaluate $\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}$ by hand?
I've been relying on Mathematica for these kinds of integrals, which in this case evaluates to $$\frac{b^2}{2 a^2} e^{\frac{a^2 b^2}{4}... | $$
\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}
$$
is equivalent to
$$
b^3\int_0^\infty d\left(\frac{x}{b}\right) \, \left(\frac{x}{b}\right)^2 e^{-\frac{a^2b^2}{2} \left(\frac{x}{b}\right)^2} \sqrt{\left(\frac{x}{b}\right)^2 + 1} = b^3\int_0^\infty dz \, z^2 e^{-k z^2} \sqrt{z^2 + 1}
$$
where $k = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the $(X - \bar{X})^2$ of first 5 data of dataset given mean and population variance Mean and population variance of the dataset $x_{1},x_{2}..x_{10}$ are $19$ and $49$ respectively. If the value $\sum_{i=6}^{10}x_{i}^{2} = 1900$, what is the value of $\sum_{i=1}^{5}x_{i}^{2} = ?$.
I've solved it as following an... | You don't use the whole definition of the population variance which is
$$s^2 = \frac{\sum\limits_{i=1}^{10} (x_i - \bar{x})^2}{10}=\frac1{10}\cdot \left( \sum\limits_{i=1}^{10} x_i^2 -2 x_i\bar{x} +\overline x^2 \right)$$
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -\sum\limits_{i=1}^{10}2 x_i\bar{x} +\sum\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Region area with double integral I need to find the area $\{(x, y) \in \mathbb{R}^2 : 0 < 2x < y < 3-x^2\}$ as the double integral $\int(\int dx) dy$.
I came up with:
$$\int_0^2\int_0^{\frac{1}{2}y}dxdy + \int_2^3\int_0^\sqrt{3-y}= \int_0^2\frac{1}{2}ydy + \int_2^3\sqrt{3-y}dy = \frac{1}{2}\frac{y^2}{2}\biggr\rvert^2_0... | Your answer is almost correct.
You made a single error when you took the antiderivative of the second integreal.
It should be $- \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $ not $+ \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $
That will give you the correct result of $\frac{5}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of Triangles in this figure How many triangles are in this figure? I suspect there are $45$, but I would like someone to confirm or correct me. I arrived at $45$ after manually counting them. If anyone has a formula by which this number can be obtained, please share.
| I get only $43$ triangles. Here's how I did the counting: consider the collection of sets of points on each segment
$$S=\{ AKDHC, CEB, BFA, ALGJE, CIGMF, BMLK, BGD, BJIH \}$$
(here for simplicity I'm denoting the points of a segment as a concatenated string).
Now make a graph $g$ where vertices are the points $A,\dots,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Two hard integrals: $\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$ and $\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$ I found two integrals that seem hard to evaluate:
$$I_1=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$$
$$I_2=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$$
I am jus... | The routine integrals $\int_0^1 \frac{\ln^2t}{1-t}dt = 2\zeta(3)$,
$\int_0^1 \frac{\ln^2t}{1+t}dt = \frac32\zeta(3)$,
$\int_0^1 \frac{\ln^3t}{1-t}dt = -\frac{\pi^4}{15}$,
$\int_0^1 \frac{\ln^2t}{1+t^2}dt =\frac{\pi^3}{16}$,
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$, and
$\int_0^1 \frac{\ln t}{1+t}dt =-\frac{\pi^2}{12}$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to calculate the determinant of $4 \times 4$ matrix with a variable on the diagonal? I need to calculate the determinant of the following $4 \times 4$ matrix:
\begin{bmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{bmatrix}
I heard there is a way by separating the matrix into blocks, but I couldn't succeed in doing t... | In the special case $$
\left( {\begin{array}{*{20}c}
A & B \\
B & A \\
\end{array}} \right)
$$
where $A,B$ are $2×2$ blocks you can use the formula
$$
\det \left( {\begin{array}{*{20}c}
A & B \\
B & A \\
\end{array}} \right) = \det \left( {A - B} \right)\det \left( {A + B} \right)
$$
In our case you hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| $$ax^2+bx+c=0,~~~a\neq0$$
$$\begin{align}
&\Rightarrow x^2+\frac{b}{a}x+\frac{c}a=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2+\frac{c}{a}-\frac{b^2}{4a^2}=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{4a^2}\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\
\\
&\Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{... | $$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3 \tag{1}$$
In your first line, you miss $x^2$, which is from the chain rule. It should be
$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3x^2$$
Here is a trick, n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$
Show that:
$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$
My attempt:
First , I used Holder's : $$... | Alternative approach using the point of incidence technique.
We will first employ the AM-GM inequality, followed by QM-AM:
\begin{align*}
a+b+c+\frac1a+\frac1b+\frac1c&=a+b+c+\frac13\left(\frac1a+\frac1b+\frac1c\right) + \frac23\left(\frac1a+\frac1b+\frac1c\right)\\
&\geqslant \frac{6}{\sqrt{3}}\sqrt[6]{abc\cdot\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition. For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac... | Your working has no errors as far as I can tell. Sometimes proofs like these have multiple ways to approach it; this is one such instance.
To get to the line about $\newcommand{\ve}{\varepsilon}
\ve/2$, consider momentarily: we have two terms at this point in the working,
$$\frac 3 2 \cdot \frac{1}{n^2} \text{ and } 2... | {
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"url": "https://math.stackexchange.com/questions/4501404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$
Find (with proof) $\displaystyle\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1}dx$
It's actually not even clear that the integral converges. If there were only sines/cosines in the integral, a standard technique would be to use the trigonometric identity $\sin \th... | $$I=\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx+\int_{\frac{3}2}^{\frac{5}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{5}2}^{\frac{7}2} \frac{\cos(\pi x)}{4x^2-1}dx+\cdots$$
Look at the first term:
$$\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx=\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 2
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Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by... | $$I=\int_0^\pi\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}}x dx=2\sqrt2\int_0^\frac{\pi}{2}\Big(\sqrt{\tan x}-\sqrt{\cot x}\Big)xdx=2\sqrt2(I_1-I_2)$$
Both terms converge, so we can evaluate them separately.
Making the substitution $x=\arctan t$ and $\arctan t=\int_0^1\frac{t}{1+t^2x^2}dx$
$$I_1=\int_0^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
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Eliminate $\alpha$ by combining two integrals I have a function $f(t,x)$:
$$
f(t, x) = | \sin(2 \cdot \pi \cdot (t - x)) - \sin(2 \cdot \pi \cdot t) |
$$
and two values for $x$:
$$
l = \cos(\alpha)\cdot b \\ m = \sin(\alpha)\cdot b
$$
Can I "combine" $f(t, l)$ and $f(t, m)$ via some function $g(u, v)$ such that
$$
r = ... | The first step would be to solve the integral before trying to find such a function $g$.
Since $f(t,x+1) = f(t,x)$, we can assume that $x \in (-\frac{1}{2},\frac{1}{2}]$. Then for $t \in (0, \frac{1}{2}]$ we have $f(t,x) = 0$ if either $ x \in \mathbb{Z}$ or $t = t_* = \frac{1}{4} + \frac{x}{2}$. In particular, you ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, ... | $$I(a)=\int_{0}^{\pi} \log (a\sin (x)+2)\, dx$$
$$I'(a)=\int_{0}^{\pi} \frac{\sin (x)}{a \sin (x)+2}\, dx=\frac \pi a-\frac{2 \pi }{a \sqrt{4-a^2}}+\frac{4 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)}{a \sqrt{4-a^2}}$$
$$\int_0^1 \Bigg[\frac \pi a-\frac{2 \pi }{a \sqrt{4-a^2}}\Bigg]\,da=\pi \log \left(\frac{2+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$
$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$
From some Olympiad problem book.
My attempt:
\begin{align}
P(0, 0): \; & f(0)+f(0)=0, f(0)=0. \\
\ \\
P(x, x): \; & f(-x^2)+f(x^2+x)-x=0. \\
P(x, -x): \; & f(-x^2)+f(x^2-x)+x=0. \\
\therefore \; & f(x^2... | $$\begin{cases}P(x,y):\quad f(-y^2)+f(x^2+y)-y=(x+y)f(x-y)\\
P(-x,y):\quad f(-y^2)+f(x^2+y)-y=(-x+y)f(-x-y)
\end{cases}$$
Then $P(x,y)-P(-x,y)$ is $$(x+y)f(x-y)+(x-y)f(-x-y)=0.$$
Take $(a,b)\in\mathbb{R}^2$, and let $(x,y)$ be the solution of $$\begin{cases}x-y=a\\ x+y=-b \end{cases}$$
so that $P(x,y)-P(-x,y)$ becomes
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.
| Computation using the differentiability of $\sin$ on $\mathbb R$
Let $f(x)=\sin(x^3)$. Then, for every $a\in\mathbb R$ and $x\neq a$, we have
$$
\frac{f(x)-f(a)}{x-a}
= \frac{\sin(x^3)-\sin(a^3)}{x-a}
= \frac{\sin(x^3)-\sin(a^3)}{x^3-a^3} \frac{x^3-a^3}{x-a}
$$
Let $X=x^3$ and $A=a^3$. Then $\frac{\sin(x^3)-\sin(a^3... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove $\int_{0}^{\infty}{e}^{a\cos\left(bx\right)}\sin\left(a\sin\left(bx\right)\right)\frac{\mathrm{d}x}{x}=\frac{\pi}{2}\left({e}^{a}-1\right)$ Show that if $a > 0$ and $b > 0$ that
\begin{align}
\int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} = \frac{\... | Let $bx\to x$ to simplify the integral to\begin{align}
& \int_{0}^{\infty} {e}^{a \cos x}\ \sin (a \sin x)\ \frac{dx}{x} \\
= &\ \Im \int_{0}^{\infty} {e}^{a e^{ix}}\ \frac{dx}{x}
= \Im \int_{0}^{\infty} \sum_{k=0}^\infty\frac{(a e^{ix})^k}{k!}\frac{dx}x\\
= &\sum_{k=1}^\infty\frac{a^k}{k!} \int_{0}^{\infty} \frac{\s... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Exponential equation in $\mathbb{Z}$ Solve the equation:
$$5^{18x}+12=13^{18y+1},$$ where x, y are positive integer.
I notice that $x=y=0$ is the only solution. I tried to reduce modulo primes but I didn't succeed.
| Suppose $x \neq 0$. Modulo $5$ we find that
$$2 \equiv 3^{18y+1} \equiv 3 \cdot 4^y\hspace{10pt} \mbox{ mod } 5$$
since by Euler-Fermat's theorem the exponent can be reduced modulo $\varphi(5)=4$. So looking at the powers of $4$ mod $5$ ($4^2 \equiv 1$) we conclude that $y$ must be odd.
Now, modulo 13 we have
$$5^{18x}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equat... | In general we have
$$\cos x = \sin y \iff y = \frac \pi 2 \pm x + 2k\pi$$
that is in this case
$$\frac \theta 2 = \frac \pi 2 \pm 2\theta + 2k\pi$$
which leads to the general solutions
*
*$\theta_1 = \frac \pi 5+\frac 4 5 k \pi$
*$\theta_2 = -\frac \pi 3+\frac 4 3 k \pi$
and the solution for an acute angle is indee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in
(1)AP
(2) GP
(3) HP
(4) none of these
My approach is as f... | $$b^2-a^2=c^2-b^2=k(\ne0)\text{say}\implies b^2=a^2+k, c^2-a^2=2k$$
$$(c-b)(c+b)=k\iff b+c=\dfrac k{c-b}, \text{similarly } a+b=\dfrac k{b-a}, c-a=\dfrac{2k}{c+a}$$
$$\dfrac1{b+c}+\dfrac1{a+b}=\dfrac{c-b+b-a}k=\dfrac{2k}{k(c+a)}=? $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Advanced Functions, How to simplify $\tan \frac{5\pi}{12}$? I was asked to find the exact value of $\,\tan \left(\frac{5\pi}{12}\right)$, so I did up until this point where I got completely stuck.
I split the ratio into two, so $\;\dfrac{5\pi}{12} = \dfrac{\pi}4 + \dfrac{\pi}6\,.$
using the formula, $\tan(x+y) = \dfrac... | So, you've remembered (or re-derived) that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, and plugging this into your sum-of-angles formula gives:
$$\tan(\frac{5\pi}{12}) = \frac{\tan(\frac{\pi}{4}) + \tan(\frac{\pi}{6})}{1 - \tan(\frac{\pi}{4})\tan(\frac{\pi}{6})} = \frac{1 + \frac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Is there an analytic solution to this set of cubic equations? I have three parameters $(a,b,c)$ that define equations I am looking for an analytical solution.
$$q_1=b/c$$
$$q_2=\frac{(a+b)^3}{3c^2b}$$
$$q_3=\frac{(a+b)^3}{3b^3}$$
I want to solve these equations for $a$, $b$, and $c$ in terms of $q_1$, $q_2$, and $q_3$.... | Slightly rewriting the equations to avoid fractions:
$$q_1 c = b \tag{1}$$
$$3 q_2 bc^2 = (a + b)^3 \tag{2}$$
$$3 q_3 b^3 = (a + b)^3 \tag{3}$$
From (2) and (3), we get $3 q_2 bc^2 = 3 q_3 b^3$, or $q_2 c^2 = q_3 b^2$. But from (1), $b = q_1 c$, so plugging that in gives $q_2c^2 = q_3 q_1^2 c^2$, or $q_2 = q_1^2 q_3$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate the integral of $\frac{1}{x^2 +x + \sqrt x}$ How to correctly calculate the integral:
$$\int_0^\infty \frac{1}{x^2 +x + \sqrt x}dx$$
Edit: I tried to figure out if the limit exists:
Step 1: break the integral to two parts: from 0 to 1, from 1 to infinity.
Step 2: use limit comparison test for both of the inte... | We have
\begin{align*}
I &= \int_0^\infty \frac{1}{x^2 + x + \sqrt{x}} \ \mathrm{d}x\\
&= \int_0^\infty \frac{2}{y^3 + y + 1} \ \mathrm{d}y && \text{using $y = \sqrt{x}$}\\
\end{align*}
Clearly, this only has one real root, denoted $r$, where $r < 0$. Since $r^3 + r + 1 = 0$, we have $r^{-1} = -r^2 - 1$, so $y^3 + y + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Cartesian equation for a transcendental / trigonometric curve
Hello! Please see figure above. I am searching for the cartesian equation for the curve in green, similar to how the equation for a semicircle is $f(x) = √(1 - x^2)$. I'm not sure if this is even possible, so any feedback is appreciated. My several attempts... | Relabelling your figure a bit, I'm taking center $O$, and points $A$ and $B$ on the circle with $M$ the foot of the perpendicular from $B$ to $\overline{OA}$. If the circle meets the $x$-axis at $R$, then define $\alpha :=\angle ROA$ and $\beta :=\angle AOB$.
If the radius of the circle is $r$, then we can write the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the... | Using the triangle inequality like this:
$$|f(x,y)|\le \frac{|2x^2y^4|+|-3xy^5|+|x^6|}{(x^2+y^2)^2}$$
Notice that
$$\frac{|x^6|}{(x^2+y^2)^2}\le \frac{|x|^6}{x^4}=\frac{x^6}{x^4}=x^2$$
$$\frac{|2x^2y^4|}{(x^2+y^2)^2} \le \frac{2x^2y^4}{y^4}=2x^2$$
$$\frac{|-3xy^5|}{(x^2+y^2)^2} \le \frac{3|x||y|y^4}{y^4}=3|x||y|$$
It i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Trouble with the Binomial sum I guess the following sum is equal to $1$ since you can compute the baby cases for $t=1,2$ straightforward to get the desired result. So, in general, we should have the following sum to be $1$, I think. But, I am stuck in the last equality. Any comments or advice would be appreciated.
\beg... | $$\frac tn\sum_{i=0}^{n-t}\frac{\binom{n-t}i}{\binom{n-1}i}=\frac1{\binom nt}\sum_{i=0}^{n-t}\binom{n-1-i}{t-1}=\frac1{\binom nt}\sum_{j=0}^{n-t}\binom{t-1+j}{t-1}=1.$$
The first equality is obtain by using factorials, the second one by letting $j=n-t-i,$ and the last one is the hockey-stick identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532046",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A precalculus solution for $x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$
Using algebra (precalculus) and suggest the solution method for the polynomial $$x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$$
I'm solving problems on polynomials. I'm stuck here.
My attempts.
First, I tried the Rational root theorem, then I failed.
... | We can rewrite the given polynomial as
$$p(x) = x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 = \\
(x^6 + 2x^3 + 1) - 3x (x^3 - x + 1) = \\
(x^3 + 1)^2 - 2x (x^3 + 1) + 2x^2 - x (x^3 - x + 1) = \\
(x^3 - x + 1)^2 - x (x^3 - x + 1) + x^2.$$
Therefore, if $x$ is a root of $p$, then either
$$x^3 - x + 1 = -\omega x$$
or
$$x^3 - x... | {
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"url": "https://math.stackexchange.com/questions/4532181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\... | We have the bound
$$\sum_{n=2}^{2015}\frac{1}{n^2}>\int_2^{2016}\frac{1}{x^2}dx=\frac{1}{2}-\frac{1}{2016}>\frac{1}{2}-\frac{1}{806}=\frac{402}{806}=\frac{201}{403}$$
by thinking about the graph of the function $y=1/x^2$ and noticing that if you draw rectangles of base length $1$ with four vertices $(n,0),(n+1,0),(n,1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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An alternating sum I ran into an alternating sum in my research and would like to know if the following identity is true:
$$
\sum_{i = 0}^{\left\lfloor \left(n + 1\right)/2\right\rfloor} \frac{\left(n + 1 - 2i\right)^{n + 1}}{2^{n}\left(n + 1\right)!}\binom{n + 1}{i}\left(-1\right)^{i} = 1\quad
\forall\ \mbox{positive ... | We seek to show that
$$\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
= 2^{n-1} n!.$$
Observe that
$$(n-2q)^n {n\choose q} (-1)^q
= \frac{1}{2}
\left[ (n-2q)^n {n\choose q} (-1)^q
+ (2q-n)^n {n\choose n-q} (-1)^{n-q} \right].$$
Hence we get for our sum
$$\frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-... | We have $$(\sqrt{1}-\sqrt{2})^k = \sqrt{a_k^2} - \sqrt{2b_k^2}$$
where $a_k,b_k$ obey the following recurrence relationships:
$$a_{k+1}=a_{k}+2b_{k}\\
b_{k+1}=a_{k}+b_{k}$$
with $a_{1}=1$ and $b_1 = 1$. This can be proved inductively. To understand the recurrence relationship we can recast it as a matrix equation:
$$... | {
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"url": "https://math.stackexchange.com/questions/4540918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How to find n for $n = 2^{\frac {n^2}{16}}$ I have this equation $n = 2^{\frac{n^2}{16}}$
I need to find n.
I tried using the Lambert W function, but I don't know how to get one side into the form $W(xe^x)=x$ without having an $n$ on the other side.
I have done:
$n = e^{\frac{n^2ln2}{16}}$
$1 = \frac{1}{n}e^{\frac{n^2l... | $n = 2^{(\frac {n}{4})^2}$
We need to get the right hand side into the form ${f(n)}e^{f(n)} = c$
Then $f(n) = W(c)$
$1 = n^{-1}e^{(\frac {n}{4})^2\ln 2}$
$1^{-1} = ne^{-(\frac {n}{4})^2\ln 2}$
$\frac 14 = \frac {n}{4}e^{-(\frac {n}{4})^2\ln 2}$
$\frac 1{16} = (\frac {n}{4})^2e^{-2(\frac {n}{4})^2\ln 2}$
$\frac {\ln 2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Finding all $x \in \mathbb{R}$ with $|x+1| + |x-1| < 4$ We want to find all $x \in \mathbb{R}$ with
$$|x+1| + |x-1| < 4$$
I tried the following, but I'm uncertain if this is correct:
Critical points: $x= -3$ and $x = 1$.
Case $1$:
$$x \in (-\infty, -3) \\
|x+1| = -(x+1) = -x-1 \\
|x-1| = -(x-1) = -x+1 \\
|x+1| + |x-1|... | Sol 1)
This image shows the graph of $y=|x+1|+|x-1|$ and $y=4$.
From this graph, we can easily get the answer: $x \in (-2, 2).$
Sol 2)
Case 1. $x < -1.$
$|x-1|+|x+1|=1-x-x-1=-2x < 4.$
$\therefore \;-2 < x < -1.$
Case 2. $-1 \leq x \leq 1.$
$|x-1|+|x+1|=1-x+x+1<4.$
$\therefore \; -1 \leq x \leq 1. $
Case 3. $1 <x. $
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sum of Fibonacci Number and Product of Fibonacci Number are equal? Let $a,b,c,d$ be nonnegative integers. Consider the equation
$$F_a+F_b=F_cF_d$$
WLOG, I can assume that $a \geq b$ and $c \geq d$. I try to solve for all subscripts $a,b,c,d$ in the above equation.
It is definitely solvable. For example, we have $F_d=1$... | All solutions to $F_a+F_b=F_cF_d$ with $a \geq b$, $c \geq d$ are given by ($n \geq 1$):
\begin{align}
(a,b,c,d) \in \{(n+5,n+1,n+3,4),(n+3,n,n+2,3),(n+1,n,n+2,2),(n+1,n,n+2,1),(n+2,n+2,n+2,3),(1,1,3,1),(1,1,3,2),(2,2,3,1),(2,2,3,2),(3,1,4,1),(3,1,4,2),(4,2,3,3),(5,1,4,3),(6,1,4,4)\}.
\end{align}
To prove this we use Z... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$ Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$
My attempt:
If the minimal polynomial is $(t-1)^2(t+1)(t-2)$ then we only have the following invaria... | It will be simpler to consider $3 \times 3$ block matrices.
In the spirit of the answer of Masacroso here, here is "prototypical case" helping (so I think) to understand what is going on is as follows:
$$M=\begin{bmatrix}
B & 0 & 0\\
0 & B & 0 \\
0 & 0 & D
\end{bmatrix} \ \text{where} \ B:=\begin{bmatrix}
\color{red}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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find the smallest positive integer $n$ so that there exist $n\times n$ matrices $A,B$ with $(AB)^k = 0$ and $(BA)^k\neq 0$
For any positive integer k, find the smallest positive integer $n$ so that there exist $n\times n$ matrices $A,B$ with $(AB)^k = 0$ and $(BA)^k\neq 0$.
I know that by the rank nullity theorem, fo... | If $(AB)^k = 0$ then $B(AB)^k A = (BA)^{k+1} = 0$. So $BA$ is nilpotent with nilpotence degree exactly $k+1$. An $n \times n$ matrix is nilpotent iff its characteristic polynomial is $t^n$, meaning its nilpotence degree is at most $n$; this gives that $n \ge k+1$.
Here is a construction showing that we can take $n = k+... | {
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"url": "https://math.stackexchange.com/questions/4548480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$
Evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$ where $\ln$ denotes the natural logarithm.
Let $A_N = \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot... | Assume that $N > 4$. Let
$$A_N = \frac{\ln^2 N}{N}\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}.$$
Let $f(x) := \frac{1}{\ln x \ln (N - x)}$. Note that $f(x)$ is strictly decreasing on $[2, N/2]$,
and strictly increasing on $[N/2, N-2]$. We have
\begin{align*}
\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}
&\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $(x+1)^4=2(1+x^4)$ Solve the equation $$(x+1)^4=2(1+x^4)$$ The most intuitive approach for me was to use the formula $$(a+1)^4=a^4+4a^3+6a^2+4a+1,$$ so our equation is $$x^4+4x^3+6x^2+4x+1-2-2x^4=0\\-x^4+4x^3+6x^2+4x-1=0$$ $\pm1$ aren't solutions, so this equation does not have whole roots.
Another t... | Based on the coefficients and the polynomial $ x^4 - 4 \, x^3 - 6 \, x^2 - 4 \, x + 1 = 0$ try a product of the type
$$( x^2 - 2 \, a \, x + 1)(x^2 - 2 \, b \, x + 1) = 0$$
which gives
$$ a + b = 2 \hspace{5mm} \text{and} \hspace{5mm} a \, b = -2.$$
The roots of the original polynomial take the form
$$ x \in \left\{ a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.
Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\... |
$u^3+3u=108+76\sqrt{2}$
A few posts here assumed $u = a + b\sqrt{2}$, then search for rational $(a, b)$
Let "slope", $\displaystyle \;m_1 = b/a$
$\displaystyle u^2 = (a^2+2b^2) \;+\; (2ab)\,\sqrt{2}\qquad\; →m_2 = \frac{2\,m_1}{1+2\,m_1^2}$
Note that $m_2$ has same sign as $\displaystyle m_1 \quad →m_∞ = \frac{sgn(m1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
so, integrate $\int \sin^3 x \cos^2 x dx = \int \sin x (1- \cos^2 x) \cos^2 (x) dx $
let $u = \cos x$
$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $
Therefore $ \frac... | First of all, Both answers $I=-\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+C= \frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C$ are correct, which can be confirmed by the following proof:
Let $z=\cos x+i\sin x$, then $\sin x= \frac{1}{2i}(z-\frac{1}{z})$ and $\cos x= \frac{1}{2}(z+\frac{1}{z})$.
Expressi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can we select $6$ cards from a pack of $52$ cards such that we get all $4$ suits? Number of ways to select $1$ card from each suit $= 13^4=28561$
Total cards remaining $=52-4=48$
Cards remaining in each suit $=13-1=12$
Now I have to choose $2$ more cards and this can happen in two ways. First, either b... | JMoravitz has provided you with two nice solutions to the problem. To understand where you made your mistakes, we will consider how your method compares with solving by cases.
Solution:
There are two cases:
*
*three cards are drawn from one suit and one card apiece is drawn from each of the other three suits
*two c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Simplifying $\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)}$ using Lagrange’s polynomial I have a question of symplifying this expression
$$
A=\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \tag{1}
$$
where $a$, $b$ and $c$ are distinct nonzero real numbers. There is ... | You want to write down a quadratic polynomial $p$ for which $p(a)=a^4$, $p(b)=b^4$, and $p(c)=c^4$.
We see that $p(x)=x^4$ whenever $x\in\{a,b,c\}$. This means that
$$(x-a)(x-b)(x-c)\mid (x^4-p(x)).$$
In particular, we can reduce $x^4$ modulo $(x-a)(x-b)(x-c)$, and we'll have the polynomial we need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$, $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\f... | $$\lim_{h\to 0}\frac{4\sqrt{4+h}-8}{h}+\lim_{h\to 0}\frac{2e^{\tfrac{3h}{6+h}}-2}{h}$$$$=4\lim_{h\to 0}\frac{\sqrt{4+h}-\sqrt4}{h}+2\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{h}$$Rationalising the first limit and adjusting the denominator of the second limit, $$ =4\lim_{h\to 0}\frac{h}{h\cdot(\sqrt{4+h}+\sqrt4)}+2\times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
My Attempt:
$$\cos\left(\frac\pi4(n^2+2n)\right)=\cos\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies\frac\pi4... | It follows from $2n^2+3n=8p+1$ that $8p=2n(n+1)+(n-1)$. Since $2\mid n(n+1)$, we have $4\mid 2n(n+1)$, hence $4\mid(n-1)=8p-2n(n+1)$. Let's assume that $n=4k+1$ for some integer $k$, then we have
$$8p=4(2k+1)(4k+1)+4k\\\implies 2p=(2k+1)(4k+1)+k.$$
Since $2p$ is even and $(2k+1)(4k+1)$ is odd, we know that $k=2p-(2k+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Use the substitution $u=\frac{1}{\sqrt{3}}\tan x$ to find $\int\frac{dx}{3-2\sin^2x}$ Question:
Use the substitution $u=\frac{1}{\sqrt3}\tan x$ to find $$\int\frac{dx}{3-2\sin^2x}$$
My Working:
Let $u=\frac{1}{\sqrt3}\tan x$, then
\begin{align}
\frac{du}{dx}&=\frac{1}{\sqrt3}\cdot\sec^2x+0=\frac{\sec^2x}{\sqrt3}\\
du&=... | Hint: $\sec^{2}x=1+\tan^{2}x$ and
$$3-2 \sin^{2}x=3-2 (1-\cos^{2}x)=3-2 \left(1-\frac{1}{\sec^{2}x}\right)=3-2 \left(1-\frac{1}{1+\tan^{2}x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\th... | Rather than have to guess to use a special trick like multiplying by $\frac{i}{i}$, just multiply top and bottom by the conjugate of the denominator. This can be done in trig form, or exponential form, or just like this:
$$\frac{1+z}{1-z}\times\frac{1-\bar{z}}{1-\bar{z}}$$
$$=\frac{1+z-\bar{z}-z\bar{z}}{1-z-\bar{z}+z\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Proof that $\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer. Let $k_1 < k_2 < ... < k_n$ will be integers. Prove that the quotient
$\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer.
where $V_n(x_1,...,x_n)$ is vandermonde determinant
My idea: Show that the numerator is divisible by the denominator
Can anyone giv... | Without loss of generality, we can assume that $k_1\geq n-1$ since $V_n(k_1,\ldots,k_n) = V_n(k_1+p,\ldots,k_n+p)$ for any $p\in\mathbb{Z}$.
Since
$$V_n(1,\ldots,n) = 1!2!\cdots (n-1)!.$$
We have that
$$\frac{V_n(k_1,\ldots,k_n)}{V_n(1,\ldots,n)}
= \det\begin{pmatrix}
1 & \cdots & 1 \\
\frac{k_1}{1!} & \cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$? How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$?
What I have done so far:
$\lim\limits_{x\to\inf... | First note that $$\frac{x^3+x}{2+4x^3}-\frac{1}{4} = \frac{4(x^3+x)-(2+4x^3)}{4(2+4x^3)} = \frac{4x-2}{16x^3+8} = \frac{\frac{1}{4x^2}-\frac{1}{8x^3}}{x^3+\frac{1}{2}}$$
and so $$\left|\frac{x^3+x}{2+4x^3}-\frac{1}{4}\right| \leq \frac{\left|\frac{1}{4x^2}\right|+\left|\frac{1}{8x^3}\right|}{\left|x^3+\frac{1}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C... | To add some more insight, there is indeed a geometric intepretation of this inequality.
As shown in the picture, $AI$ is an angle bisector. $$\sin{A\over 2}={BH\over AB}\leq {BI\over AB} = {CI\over AC} = {BC\over AB+AC}$$
Therefore, with similar argument for $B$ and $C$ angles,
$$\sin{A\over 2}\sin{B\over 2}\sin{C\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculating $\sum_{k = 1}^{\infty} k^{-8}$ using Parseval´s identity I calculated the Fourier series of $x^2$ getting
$$x^2=\frac{\pi^2}{3}+4\sum_{k=1}^\infty \frac{(-1)^k}{k^2} \cos(kx)$$
Then, integrating this equation two times i got
$$\frac{x^4-2\pi^2x^2}{12}=4\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^4} \cos(kx)$$
Usi... | Starting with
$$ \frac{\theta^2}{4} - \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \, \cos(n \theta) $$
then, by integration, the series
$$ \frac{\theta^3}{12} - \frac{\zeta(2) \, \theta}{2} + c_{1} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \, \sin(n \theta) $$
is obtained. The constant of integration, $c_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1... | By using $$ \lim_{n \to \infty} \, \left(1 + \frac{x}{n}\right)^n = e^{x}, $$ $n^2 + 5 n + 3 = (n+a_{1})(n+a_{2}),$ and $n^2 + n + 2 = (n+b_{1})(n+b_{2})$ then
\begin{align}
L &= \lim_{n \to \infty} \, \left( \frac{n^2 + 5 n + 3}{n^2 + n + 2} \right)^n \\
&= \lim_{n \to \infty} \, \left( \frac{(n+a_{1})(n+a_{2})}{(n+b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0 ,y=mx$ Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0, y=mx$
The limits of integration which I set went as follows:
$${\int \int}_{D} \int_... | From $ \dfrac{x^2}{a^2} + \dfrac{z^2}{c^2} = 1 $, take
$x = a u \cos t$
$z = c u \sin t$
$y = y$
where $ 0 \le t \le 2 \pi $ and $ 0 \le u \le 1 $
The Jacobian of the above transformation is
$ J = \dfrac{\partial(x, y, z) }{\partial(u, t, y)} = \big| \begin{vmatrix} a \cos t && - a u \sin t && 0 \\ 0 && 0 && 1 \\ c \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is... | If you only have one throw left, then the expected amount you will have after the throw, if you do use it, is $\ \frac{7}{2}\ $. Therefore, if your previous throw was $3$ or less, you should use your last throw, but if it was $4$ or more you should not use your last throw, but accept the value of the previous one.
Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Will the perpendicular bisector between the line connecting two cubic roots of the same arc never intersect its turning point? The quadratic graph: $$ f(x) = (x+2)(x+1)$$
would have a midpoint between its roots at $x = -1.5$. This line would intersect its turning point.
However the cubic graph: $$ f(x) = (x+1)(x-2)(x... | We can take as a requirement that the perpendicular bisector of two adjacent zeroes of a polynomial pass through the "turning point" [relative extremum] between them and see what polynomials may satisfy it. For a quadratic polynomial $ \ x^2 + a_1 x + a_0 \ $ (it suffices to use a monic polynomial, since a "leading c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\... | Starting with the integral
$$ \int_{0}^{1} (1-x)^m \, x^{-t} \, dx = B(m+1, 1-t), $$
a case of the Beta function, then
\begin{align}
\partial_{m} \int_{0}^{1} (1-x)^m \, x^{-t} \, dx &= \partial_{m} \, B(m+1, 1-t) \\
\int_{0}^{1} \ln(1-x) \, (1-x)^m \, x^{-t} \, dx &= B(m+1, 1-t) \, \left(\psi(m+1) - \psi(m-t+2) \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $(x^2+xy+y^2)$ divide $(x+y)^n-x^n-y^n$ if and only if n has no prime factors less than 5 I noticed that for $n$ with prime factors greater than 5, that $xy(x+y)(x^2+xy+y^2)$ always seems to divide $(x+y)^n-x^n-y^n$. The $xy(x+y)$ factors seem fairly obvious, but I can't figure out where the $x^2+xy+y^2$ term come... | By homogeneity, it’s enough to show that $x^2+x+1$ divides $(x+1)^n-x^n-1$. Because these polynomials are rational and $x^2+x+1$ is irreducible over $\mathbb{Q}$, it’s enough to show that $(j+1)^n-j^n-1=0$, where $j=e^{2i\pi/3}$ is a primitive third root of unity.
Now, $(j+1)^n-j^n-1=(-j^2)^n-j^n-1=-(1+j^n+j^{2n})$. No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sove the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$
Consider the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$. Find a closed form expression for $a_n$.
I have computed the OGF $A(z)$ of $(a_n)_n$ as
$$A(z) = \frac{z^2}{1-5z+... | Characteristic equation is $r^3-5r^2+8r-4=(r-1)(r-2)^2$
So $a_n=a\,1^n+(bn+c)\,2^n$
Applying initial conditions give $\begin{cases}a_0=a+c=0\\a_1=a+2(b+c)=0\\a_2=a+4(2b+c)=1\end{cases}\iff a=1,\ c=-1,\ b=\frac 12$
$$a_n=2^n(\tfrac n2-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving a system of equations with logarithms in exponents I am interested in what approach could be applied to solve such a system:
$$ 2^{x-\log_2{y}} - 8^{x+\log_2{y}} = 5y - 2^{x-\log_2{y}} $$
$$ \frac{2^{x-\log_2{y}}}{8^{x+\log_2{y}}} = \frac{5y}{2^{x-\log_2{y}}} $$
| Notice that the first equation is
$$\frac{2^x}y-8^x\cdot y^3=5\cdot y-\frac{2^x}y$$
Multiply this equation for $y$ and get
$$8^x\cdot y^4+5\cdot y^2-2^{x+1}=0$$
so you have a way to express $y$ purely in function of $x$. Now, the second equation is
$$\frac{2^x}{8^x\cdot y^4}=\frac{5\cdot y^2}{2^x}$$
so you can rewrite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to prove identity $\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$? Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$
How can one prove this?
Cons... | If $\frac{2n\theta}{\pi}$ is an odd integer, one $\tan^2$ value in the sum is undefined (or "infinity"), so assume that's not the case.
$$ z = \tan \left(\theta + \frac{k\pi}{n}\right); 0 \leq k \leq n-1 $$
satisfy $\arg(z+i) = \frac{\pi}{2} + \theta + \frac{k\pi}{n}$. Defining $\alpha = \theta + \frac{\pi}{2}$ just fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of Sequence For the second time I've had trouble calculating this limit
$$\lim\limits_{n\rightarrow\infty}\frac{(n＋2)(n＋5)\cdot\ldots\cdot(n+3r-1)\cdot\ldots\cdot(4n－1)}{(n＋1)(n＋4)\cdot\ldots\cdot(n+3r-2)\cdot\ldots\cdot(4n－2)}$$
This limit can be calculated by limitation method but it is not a smart method.
When... | Using
$$ \Gamma(x) \approx \sqrt{2 \pi} \, x^{x - 1/2} \, e^{-x} \, \left(1 + \frac{1}{12 \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) $$
then it can be shown that
$$ \frac{\Gamma\left(\frac{n+1}{3}\right)}{\Gamma\left(\frac{n+2}{3}\right)} \approx \left(\frac{n}{3}\right)^{-1/3}. $$
Now,
\begin{align}
L &= \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:
In any triangle $\triangle ABC$, sho... | Here is a rather simple geometrical approach that does not use the hint
Let the incenter of the triangle be $I$. Draw a perpenduclar line from $I$ to $BC$ and call the foot of the perpendicular $D$. Now, $IB$ is the angle bisector of $\angle CBA$ and $\triangle DBI$ is a right-angled triangle, thus,
$$ \cot\frac{B}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof of a tighter inequality than Cauchy-Schwarz inequality A few days ago, I found this post which discuss about some tighter versions of Cauchy-Schwarz (or some might prefer the name AM-GM) inequality. User @Michael Rozenberg proposed a very interesting inequality at the comment section
Let $a$, $b$ and $c$ be three... | *
*Let $c=\min\{a,b,c\}$,$a=c+u$,$b=c+v$ and $u^2+v^2=2tuv$.
Thus, $$(a^4+b^4+c^4)(3(a^2+b^2+c^2)-(a+b+c)^2)-25\prod_{cyc}(a-b)^2=$$
$$=((c+u)^4+(c+v)^4+c^4)\sum_{cyc}(a-b)^2-25\prod_{cyc}(a-b)^2\geq$$
$$=(u^4+v^4)(u^2+v^2+(u-v)^2)-25(u-v)^2u^2v^2=$$
$$=u^3v^3((4t^2-2)(4t-2)-25(2t-2))=u^3v^3(16t^3-8t^2-58t+54)=$$
$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? We are given
$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$
We can write
$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$
It can ... | To check where it went wrong , let us use Series Expansion :
$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $
$ L = \lim _{x \to 0} \frac { x ( 1-x^2/2!+x^4/4!-x^6/6! \cdots ) - ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } {x^2 ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } $
$ L = \lim _{x \to 0} \frac { ( x-x^3/2!+x^5/4!-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find the area of Quadrilateral $ABCD$. A puzzle for 10th graders. As the title suggests, the problem in this post was meant to be a puzzle for 10th graders, so claims the person who posted this on a language exchange platform: The problem is as follows:
Given a Quadrilateral $ABCD$ with internal point $P$, where $AP=1... | I'm pretty bad at geometry so I just spammed a bunch of algebra.
First, note that the area is clearly $3k^2.$ Then, draw a line through $P$ perpendicular to $AD$ and $BC,$ which intersects them at $X$ and $Y,$ respectively. Let $PX=n,$ so $PY=2k-n.$ Similarly, let $AX=BY=m$ so $CY=2k-m.$ Then, note that $m^2+n^2=1.$ Al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proofs about a recursive sequence. Let $x_1=a>2$, and $$x_{n+1}=\frac{x_n^2}{2(x_{n}-1)}, n=1,2,...$$
Prove
(i) If $a\leq 3$, we have $x_n\leq 2+\frac1{2^{n-1}},(n=1,2...)$
(ii) If $a> 3$, then when $n\geq\frac{lg(\frac a3)}{lg \frac 43 }$, we have $x_{n+1}<3$.
My attempts:
I can use induction to show that $x_n>2$, and... | The claimed inequality $$x_n \le 2 + \frac{1}{2^{n-1}}$$ suggests defining an auxiliary sequence $$y_n = x_n - 2,$$ hence the original recursion may be written
$$y_{n+1} = x_{n+1} - 2 = \frac{(x_n-2 + 2)^2}{2(x_n - 2 + 1)} - 2 = \frac{(y_n + 2)^2 - 4(y_n + 1)}{2(y_n + 1)} = \frac{y_n^2}{2(y_n + 1)}. \tag{1}$$ In turn,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find Laurent series of $\frac{x^2}{(1+x^2)^2}$ I would like to find the Laurent series of the function
$$
f(x)=\frac{x^2}{(1+x^2)^2}.
$$
I already know that there are two poles of order $2$, namely $x=\pm i$. And I also know that, by partial fraction decomposition,
$$
\frac{x^2}{(1+x^2)^2}=\frac{1}{x^2+1}-\frac{1}{(1+x... | Here are two variations to calculate the residue of $f$ at $x=i$. We already know the pole of $f$ at $x=i$ is of order $2$. A common way is to use the
Limit formula for higher order poles:
We obtain
\begin{align*}
\color{blue}{\mathrm{res}_{x=i}f(x)}&=\mathrm{res}_{x=i}\frac{x^2}{\left(1+x^2\right)}\\
&=\lim_{x\to i}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\cot(\beta)$, knowing $\sin(\alpha+\beta)=4/5$ and $\sin(\alpha-\beta)=3/5$ Let's assume that for $0<\beta<\alpha<\frac{\pi}{2}$, $\sin(\alpha+\beta) = \frac{4}{5}$, and $\sin(\alpha-\beta) = \frac{3}{5}$. Then, how could we find $\cot(\beta)$?
$$\sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin(\alpha)\cos(\beta)... | $\displaystyle sin( a+b) =4/5\ \ \ sin( a-b) =3/5.$
$\displaystyle So\ cos( a+b) \ =\ 3/5$ but that means $\displaystyle sin\left(\frac{\pi }{2} -( a+b)\right) =sin( a-b) =3/5$
So $ $$\displaystyle \frac{\pi }{2} =a+b+a-b$
Or $\displaystyle a\ =\ \frac{\pi }{4}$
So $cot(a) = 1 $
And since you have already figured out $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$
Find the value of the expression $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$$
where $\alpha$ is a positive number.
It is my first time seeing a question like this. Normally, we just put $n=\inf... | Actually the person who posted the problem has also given its solution, that I would like to share.
The limit equals $1$ when $0<\alpha<1$, $\ln2$ when $\alpha=1$ and $0$ when $\alpha>1$
$$\frac{1}{n+n^{\alpha}}<\frac{1}{n+1^{\alpha}}+\frac{1}{n+2^{\alpha}}+\cdots+\frac{1}{n+n^{\alpha}}<\frac{n}{n+1}$$ and hence when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(\sum x^2)^3\ge9\sum x^4yz$
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.
The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to g... | Hint :
Due to homogeneization we have :
$$\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)$$
We need to show for $x\ge 1$ and WLOG $c\geq 1$
$$f(x)=\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)\geq 0$$
The function is convex on $(0,\infty)$ because we have :
$$f''(x)=6(c^{6}+6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to show $k \cdot a + c \cdot b \ge \frac{1}{2} \cdot (k+c) \cdot (a+b)$? Assume k, c, a, b $\gt$ 0 and that $k \ge c$ and $a \ge b$. Prove $k \cdot a$ + $c \cdot b$ $\ge$ $\frac{1}{2}\cdot (k+c) \cdot (a+b)$. This is what I have so far:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\... | $$
\begin{aligned}
2(k a+c b)-(k+c)(a+b)
= & 2 k a+2 c b-k a-c a-k b-b c \\
= & -k a+c b-c a-k b \\
= & (k-c) a+(c-k) b \\
= & (k-c)(a-b)\\ \geqslant & 0 \\
\end{aligned}
$$
Therefore $$k a + cb \ge \frac{1}{2} (k+c)(a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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.