Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Highschool Exam Question About Cube Factoring Given;
$ a^3 - 3ab^2 = 10 $ and $ b^3 - 3ba^2 = 5$
What is the value of $ a^2 + b^2 $ ?
| \begin{align}
10 &= a^3 - 3ab^2 \\
5 &= b^3 - 3a^2b \\
\hline
100 &= a^6 -6a^4b^2 + 9a^2b^4 \\
25 &= b^6 - 6a^2b^4 + 9a^4b^2 \\
\hline
125 &= a^6 + 3a^4b^2 + 3a^2b^4 + b^6 \\
5^3 &= (a^2 + b^2)^3 \\
5 &= a^2 + b^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using Finite Differences and Integration to prove result If $f(x)$ is a polynomial in $x$ of third degree and:
$$u_{-1}=\int_{-3}^{-1}f(x)dx\ ;\ u_{0}=\int_{-1}^{1}f(x)dx\ ; u_{1}=\int_{1}^{3}f(x)dx$$
then show that
$$f(0) = \frac{1}{2}\Bigg(u_0-\frac{\Delta^2u_{-1}}{12}\Bigg)$$
I attempted this question by assuming th... | This is a straightforward power-rule calculation:
$$
u_{-1} = \int _{-3}^{1} f(x)\,dx = 2 a-4 b+\frac{26 c}{3}-20 d;
$$
$$
u_0 = \int _{-1}^1 f(x)\,dx = 2 a+\frac{2 c}{3};
$$
$$
u_1 = \int _1^3 f(x)\,dx = 2 a+4 b+\frac{26 c}{3}+20 d
$$With your notation,
$$
\Delta^2 u_{-1} = \frac{1}{2}\left(u_{-1}-2u_0+u_1\right) = 8c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $6x^2 −7x−5=0$ I'm trying to factor
$$6x^2 −7x−5=0$$
but I have no clue about how to do it. I would be able to factor this:
$$x^2-14x+40=0$$
$$a+b=-14$$
$$ab=40$$
But $6x^2 −7x−5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?
| Here's an alternative (but you should just learn the quadratic equation)
$6x^2 - 7x - 5= 6(x + a)(x + b)$
So $ab = -5/6$ and $a + b = -7/6$.
$a = \pm 1, \pm 5, \pm 1/2, \pm 5/2, \pm 1/3, \pm 5/3, \pm 1/6, \pm 5/6$
$b = \mp 5/6, \mp 1/6, \mp 5/3, \mp 1/3, \mp 5/2, \mp 1/2, \mp 5, \mp 1$.
So $a + b = \pm 2/3, \pm 29/6, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 9
} |
Evaluate $\int_0^{2\pi}\frac{\sin^2(x)}{a + b\cos(x)}\ dx$ using a suitable contour I need to find a good contour for $\int_0^{2\pi}\frac{\sin^2(x)}{a + b\cos(x)}\ dx$ but I don't know which one to choose. Both a semicircular, and rectangular contour look ugly for this.
I've been looking at a semicircular contour of r... |
Note that the integral diverges for $a\le b$. Therefore, we assume throughout the development that $a>b$.
We can simplify the task by rewriting the integrand as
$$\begin{align}
\frac{\sin^2(x)}{a+b\cos(x)}&=\frac{a}{b^2}-\frac{1}{b}\cos(x)-\left(\frac{a^2-b^2}{b^2}\right)\frac{1}{a+b\cos(x)}
\end{align}$$
Then, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Show that, $(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$ harmonic numbers
$H_n=\sum_{i=1}^{n}\frac{1}{i}$
Euler's constant
$\gamma=\lim_{n\to \infty} [H_n-\ln(n)]$
Factorial
$n!=n(n-1)(n-2)\cdots2\cdot1$; valid for all non-negative integers
Show that,
$... | Here is my two pence worth and is not a complete answer but paves the way I think.
Theorem [Euler 1731]
The limit
$$\gamma = \lim_{n \rightarrow \infty}(H_n-\log n)$$
Is given by the convergent series
$$\gamma = \sum_{n=2}^{\infty}(-1)^{n}\frac{\zeta(n)}{n}$$
We can evaluate this formula using Mercator's expansion
$$\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x... | It is well-known that $\ln (1+z)= z+o(z)$ for small $z$, so plugging in $z=x+x^2+\cdots+x^n$ yields
$$\frac{\ln(1+x+\cdots+x^n)}{x}=\frac{x+\cdots+x^n+o(x+\cdots+x^n)}{x}=\frac{x+o(x)}{x}\to 1$$
as $x\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Complex number and polar coordinates True or false: the polar coordinates of $-1-i$ are $-\sqrt{2}\operatorname{cis}\frac{\pi}{4}$
In my opinion it's true:
$\tan\theta=\frac{-1}{-1}=1\Rightarrow \theta=\frac{5\pi}{4}$, $r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$
Therefore we get: $z=\sqrt{2}\operatorname{cis}\frac{5\pi}{4}=\sqrt... | No, it's false. The $r$ part must be positive and is computed by
$$
r=\sqrt{(-1-i)(-1+i)}=\sqrt{1+1}=\sqrt{2}
$$
This already answers the true/false question. If you want to find the $\theta$ part, you need to find $\theta$ such that
$$
\cos\theta=-\frac{1}{\sqrt{2}},\quad\sin\theta=-\frac{1}{\sqrt{2}}
$$
and this is c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Deadly integral $\int_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$. How to solve this question
$$\int\limits_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$$
. Please help me in solving this short way
my approach is in the answer
Is it correct and can it be solved in a shorter way ?
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{2\cos x}{\cos 2x + 1 }= \sec x$ Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$.
So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$
Where do I go from here?
| It seems to be a widespread practice for students to begin attempted proofs of trigonometric identities by writing things like this:
$$\frac{2\cos x}{\cos 2x + 1 }= \sec x$$
$$\frac{2\cos x}{\cos 2x + 1 }= \frac 1 {\cos x}$$
That's ok in scratchwork, but the finished proof should go like this:
$$
\sec x = \frac 1 {\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Laurent Series of $1/\tan z$ How can we find the Laurent series of the function $$f(z)=\frac{1}{\tan z }$$ around 0.
Thank you very much.
| I read your comment just now (looking for the general form), but I'll leave the answer below: a way to get the series term by term.
Well there' a formula for the coefficients (see wiki), but I suppose you're looking for another way to find the series expansion of $\cot z$.
Dividing the series for $\sin z$ term by term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int{\sqrt{a^2 - x^2}}dx$ I'm trying to solve the following integral, but seems these 2 methods led to different answers. I think one of the methods must be incorrect. But why doesn't one of them work?
Evaluate $\int{\sqrt{a^2 - x^2}}\ dx$
My friend evaluated this way:
First let $x=a\cos{\theta}$, so
$a^2... | $\cos(\pi/2-x)=\sin(x)$ and vice versa. If you haven't seen this before, the geometric explanation is that the sine of one of the acute angles in a right triangle is the cosine of the other.
Therefore $-\cos^{-1}(x)$ and $\sin^{-1}(x)$ are the same up to a constant. Since indefinite integrals are only defined up to a c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove Why $B^2 = A$ exists?
Define
$$A =
\begin{pmatrix}
8 & −4 & 3/2 & 2 & −11/4 & −4 & −4 & 1 \\
2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
−9 & 8 & 1/2 & −4 & 31/4 & 8 & 8 & −2 \\
4 & −6 & 2 & 5 & −7 & −6 & −6 & 0 \\
−2 & 0 & −1 & 0 & 1/2 & 0 & 0 & 0 \\
−1 & 0 & −1/2 & 0 & −3/4 & 3 & 1 & 0 \\
1 & 0 & 1/2 & 0 & 3/4 & −1 & 1... | By putting $A$ into Jordan normal form you have found a non-singular matrix $P$ such that
$$
A = P^{-1}CP
$$
where $C$ is your matrix shown above. It is easy to find a matrix sqare root of $C$, for example, in the upper left start with $\pmatrix{\sqrt{2} &0\\\frac14\sqrt{2}& \sqrt{2}}$
Then $$(P^{-1}DP)^2 = P^{-1}D(P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Computing $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$ how can i compute: $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$?
i tried l'hopital's rule but it's like a loop.
also if it can be done without that rule i'd like to know how.
Thanks.
| Hint.
Let $x \to -\dfrac \pi2^-$, then
$$
\begin{align}
\tan x &=\frac{-\cos (x+\frac{\pi}{2})}{\sin (x+\frac{\pi}{2})} =\frac{-1}{x+\frac{\pi }{2}}+O\left(x+\frac{\pi }{2}\right)
\\\\\cos^2 x&=\sin^2 (x+\frac{\pi}{2})\sim\left(x+\frac{\pi}{2}\right)^2
\end{align}
$$ giving, as $x \to -\dfrac \pi2^-$,
$$
\frac{e^{\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
sum of all Distinct solution of the equation $ \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
The sum of all Distinct solution of the equation $\displaystyle \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
Where $x\in (-\pi,\pi)$ and $\displaystyle x\neq 0,\neq \frac{\pi}{2}.$
$\bf{My\; Try::}$ We can write equation a... | Observe that:$$\sqrt{3}\sin x+\cos x = 2\cdot \left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)$$ $$ = 2\cdot \left(\sin\frac{\pi}{3}\sin x+cos\frac{\pi}{3}\cos x\right)$$ $$=2\cos(x\color{red}{-}\frac{\pi}{3})$$
But instead of $2\cos(x\color{red}{-}\frac{\pi}{3})$, you have written $2\cos(x\color{red}{+}\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Check convergence of $\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$ Zoomed version: $$\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$$
So, I've seen similar example at Convergence or divergence of $\sum \frac{3^n + n^2}{2^n + n^3}$
And I liked that answer :
$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\inf... | To begin, notice that for all $n\geq 1$ one has $2^n\geq n^2$
$$\sum\limits_{n=1}^\infty \frac{2^n+n^2}{3^n+n^3}\leq \sum\limits_{n=1}^\infty\frac{2^n+2^n}{3^n+n^3}\leq 2\sum\limits_{n=1}^\infty \frac{2^n}{3^n}= 4$$
Since all terms of the original series are non-negative and the series is bounded above, it must converg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Quaternary numeral system: fractions I have a question related to the expression of a real number in base 4. Consider the table here: it is clear to me how all columns of the table are obtained except the fourth one: how do they get the positional representation in quaternary base?
| The non-repeating ones like $\frac{1}{2},\frac{1}{4}$ are obvious. For the others we repeatedly use $\frac{1}{1-x}=1+x+x^2+x^3+\dots$.
We have $\frac{1}{3}=\frac{1}{4}\frac{1}{1-\frac{1}{4}}=\frac{1}{4}(1+\frac{1}{4}+\left(\frac{1}{4}\right)^2+\dots=\frac{1}{4}+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^2+\dot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 ... | Unit volume assumed:
$$
1 = \frac{4}{3}\pi r^3 = a^3
$$
So we have
$$
r = \sqrt[3]{\frac{3}{4\pi}} \\
\quad a = 1
$$
So the surface ratio is
$$
A_s = 4 \pi r^2
= 4 \pi \left( \frac{3}{4\pi} \right)^{2/3}
= \sqrt[3]{4\pi \cdot 9}
=\sqrt[3]{\pi} \, 6^{2/3} \\
A_c = 6
$$
So we get
$$
A_s : A_c =
\sqrt[3]{\pi} : \sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Check for convergence $$\sum_{n = 2}^\infty (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)$$
I tried to use Maclaurin series, but failed to evaluate little-o.
| Write $\sin x=x-x^3/6+x^3\varepsilon(x)$, where $\varepsilon$ is such that $\lim_{x\to 0}\varepsilon(x)=0$. Consequently,
$$ (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)=(-1)^n\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)-\frac{(-1)^n}6\left(\frac{\sin(3n)}{\sqrt{n}} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
| We only need to prove that for positive $x$ we have $(1+x)^7\gt 7^{7/3}x^4$.
For positive $x$, let
$$f(x)=\frac{(1+x)^7}{x^4}.$$
By using $f'(x)$ we find that $f(x)$ reaches a minimum at $x=4/3$. The minimum value of $f(x)$ is
$$\frac{7^7}{3^34^4,}$$
which is greater than $7^{7/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $636^{369}\equiv x\pmod{126}$
Solve
$$636^{369}\equiv x\pmod{126}$$
My attempt:
$$126=2\times 3^2 \times 7$$
$$\varphi(126)=\varphi(2)\times \varphi(3^2)\times \varphi(7)=36$$
$$\color{gray}{636=6\pmod{126}}$$
$$6^{369}\equiv x \pmod{126}$$
$$2^{369}3^{369}\equiv x \pmod{126}$$
I am stuck here
| $636 \equiv 06 \equiv 6 $ (mod 7)
$\varphi(7)=6$
$369 \equiv 09 \equiv 3$ (mod 6)
So
$636^{369}\equiv 6^3 \equiv 36 \times 6 \equiv 6$ (mod 7)
(Note that $gcd(636,7)=1$)
Also, $636\equiv06 \equiv 6$ (mod 9)
and $6\times6=36\equiv 0$ (mod 9)
So
$636^n\equiv 0$ (mod 9) for $n>1$
Therefore, we have $636^{369}=6+7k\equiv 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x.
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha g... | Another way to look at the problem is to rewrite (assuming $x\neq 0$) $$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)\implies 4 x^2 \left(\frac{\left(n-x^2\right)^2}{4 x^2}-\frac{1}{2}
\left(n-x^2\right)\right)=0$$ Expand and simplify to get $$3 x^4-4n x^2+n^2=0$$ Now, using $y=x^2$, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$ $x,y,z >0$, prove
$$(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$$
The term $\frac{4(x-y)^2}{xy+yz+zx}$ made this inequality tougher. It remains me of this inequality. I... | There is a very nice solution only using SOS, I was found it!
We need to prove:$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) - 9 \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(\frac{x}{y}+\frac{y}{x}-2) + (\frac{y}{z}+\frac{z}{x}-2)+(\frac{z}{x}+\frac{x}{z}-2) \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$\frac{(x-y)^2}{xy}+\frac{(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Problem based on sum of reciprocal of $n^{th}$ roots of unity
Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$
$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$
Now Put $\displaystyle ... | Let $f(x) = \frac{(x^n-1)}{(x-1)} = x^{n-1}+x^{n-2}+...+x+1$
Then $\frac{f'(x)}{f(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2}+ ... + \frac{1}{x-x_{n-1}}$
So what you want is just $\frac{f'(1)}{f(1)}$, which is easy to compute as $\frac{n-1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Evaluating a strange integral... I was given this integral by a friend and have been unable to evaluate it. He said it is easily possible to do it by hand (no calculator or computational aids necessary)
$$\int_{0}^{1} \mathrm{frac}\left(\frac{1}{x}\right)\cdot\mathrm{frac}\left(\frac{1}{1-x}\right)dx $$
frac is a func... | Making the variable change, $u = 1-x,$ we have
$$\int_{1/2}^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = \int_{0}^{1/2} \left\{\frac{1}{1-u} \right\}\left\{\frac{1}{u} \right\} \, du. $$
Hence,
$$\int_0^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = 2\int_0^{1/2} \left\{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a ... | Notice that:
$$\frac1{\sqrt n}\sum_{k=1}^n\frac1{\sqrt k}=\frac1n\sum_{k=1}^n(k/n)^{-1/2}$$
As $n\to\infty$, we get a Riemann sum:
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n(k/n)^{-1/2}=\int_0^1x^{-1/2}~\mathrm dx=2x^{1/2}\bigg|_0^1=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 0
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Limit with root and fraction I have this limit:
$$\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}$$
And I try this
\begin{align}\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}} & = \lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}\frac{\sqrt{1+2x} + e^x}{\sqrt{1+2x} + e^x} \\ & \stackrel{(*)}= \lim_{x\to... | Considering $$A=\frac{\sqrt{1+2x} - e^x}{x\,\tan^{-1}(x)}$$ there are two possibilities.
The first one is to use L'Hôpital's rule just as Workaholic commented, setting $$u=\sqrt{1+2x} - e^x\qquad , \qquad v=x\,\tan^{-1}(x)$$ and you will probaly need to apply it more than once.
The second one would involve Taylor expa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing $\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2$ Showing
$$I=\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2\tag1$$
$$I=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)+\int_{0}^{\infty}{1-x^2\over x^2}\ln(1+x^2)dx\tag2$$
Let $$J=\int_{0}^{\in... | The Taylor series approach is quite straightforward, too. We have:
$$ 2\log(1+x^2)-\log(1-x^2)=\sum_{n\geq 1}\frac{1-2(-1)^n}{n}x^{2n} \tag{1}$$
so by multiplying the RHS by $\frac{1}{x^2}-1$ we get:
$$ \frac{1-x^2}{x^2}\left(2\log(1+x^2)-\log(1-x^2)\right)=3+\sum_{n\geq 1}\frac{-1+(2+4n)(-1)^n}{n(n+1)}x^{2n} \tag{2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Equation involving Wilson's theorem Find all primes $p$ such that
$$(p-1)!=p^k-1.$$
Where $k$ is a natural number.
| this is a pretty known one:
for $p=2$, we have $2^k-1=1$ which gives $k=1$.
for $p=3$, we have $3^k-1=2$ which also gives $k=1$.
for $p=5$, we have $5^k-1=4!=24$ which gives $k=2$.
For $p>5$:
note that $p-1|(p-2)!$ for every prime $p>5$. Why?
Because $p-1=2\frac{p-1}{2}$ and given that $2$, and $\frac{p-1}{2}$ are smal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort ... | Take $x=4\sec\left(u\right)
$. We get $$I=\int\frac{x^{2}}{\sqrt{x^{2}-16}}dx=16\int\sec^{3}\left(u\right)du
$$ and now for the reduction formula for powers of $\sec\left(u\right)
$ we have $$I=8\tan\left(u\right)\sec\left(u\right)+8\int\sec\left(u\right)du
$$ $$=8\tan\left(u\right)\sec\left(u\right)+8\log\left(\ta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions to a quadratic diophantine equation $x^2 + xy + y^2 = 3r^2$. Let $k,i,r \in\Bbb Z$, $r$ constant. How to compute the number of solutions to $3(k^2+ki+i^2)=r^2$, perhaps by generating all of them?
| There are infinitely many solutions to this Diophantine equation.
I change your variables. We have the Diophantine equation$$3(a^2+ab+b^2)=c^2$$Let's assume $c$ is not constant and find all possible solutions! $c=0$ gives $a=b=0$. W.L.O.G. suppose $c>0$. It's easy to see that $c=3d$ for some positive integer $d$. Equat... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding solution of the irrational equation Given equation $\sqrt{x + 3 - 2\sqrt{x + 2}} + \sqrt{x + 27 - 10\sqrt{x + 2}} = 4$, find its solution(s).
At first, finding the domain of the function. Noting that $\sqrt{x + 2} \geq 0 \implies x \geq -2$. Then, solving inequalities for $x + 3 - 2\sqrt{x + 2} \geq 0$ and $x +... | Put $t=\sqrt{x+2}$, so we require $t\ge0$. Then $\sqrt{x+3-2t}=\sqrt{t^2+1-2t}=\sqrt{(t-1)^2}=|t-1|$. Similarly, $\sqrt{x+27-10t}=\sqrt{(t-5)^2}=|t-5|$. So we have $|t-1|+|t-5|=4$. That holds for $t\in[1,5]$.
So we must have $1\le\sqrt{x+2}\le5$ and hence $1\le x+2\le25$, so $-1\le x\le 23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of infinite series with Irrational terms
The sum of the series $\displaystyle 1+\frac{\sqrt{2}-1}{2\sqrt{2}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{2}}+\frac{17-12\sqrt{2}}{80}+.....\infty$
$\bf{My\; Try::}$Here $3-2\sqrt{2} = (\sqrt{2}-1)^2$ and $5\sqrt{2}-7 = (\sqrt{2}-1)^3$
similaryly $17-12\sqrt{2... | From the suggestion by @Winther, express the series as
$$S = 1+\sum_{n=1}^{\infty} a_n x^n $$
where $a_n = \frac1{n (n+1)}$ and $x=1-\frac1{\sqrt{2}} $. Then
$$\begin{align}S &= 1+\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1}\\ &= 1-\log{(1-x)} - \frac1x\left (-\log{(1-x)}-x \right )\\ &= 2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start.
Thanks in advance.
| I always start by writing a couple of powers, in this case modulo $50$:
$$\begin{align}3^1\equiv 3&\mod 50\\
3^2\equiv 9&\mod 50\\
3^3\equiv 27&\mod 50\\
3^4=81\equiv 31&\mod 50\\
3^5\equiv 93\equiv 43&\mod 50\\
3^6\equiv 129\equiv 29&\mod 50\\
3^7\equiv 87\equiv 37&\mod 50\\
3^8\equiv 111\equiv 11&\mod 50\\
3^9\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Cauchy sequence $x_n=\sqrt{a+x_{n-1}}$ I have to show that this sequence
$$
x_n=\sqrt{a+x_{n-1}} \hbox{ with } x_1=\sqrt{a}
$$
is a Cauchy sequence for every $a>0$. I have done the following calculations:
$$
\left| x_{n+2}-x_{n+1} \right|=\left| \sqrt{a+x_{n+1}}-x_{n+1}\right|=\left|\frac{a+x_{n+1}-x_{n+1}^2}{\sqrt{a+x... | Claim. The given sequence is bounded from above by $$ \frac{1 + \sqrt{4a + 1}}{2} = \alpha $$
Proof. We proceed by induction. For $ k = 1 $ we have that
$$ \sqrt{a} = \frac{2\sqrt{a}}{2} = \frac{\sqrt{4a}}{2} < \frac{1 + \sqrt{4a + 1}}{2} $$
Now, assuming that the statement is true for $ k = n $, we have
$$ x_{n+1} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the order of the pole How would you go about finding the order of the pole at $z=0$ of the following function?
$$f(z)=\frac{1}{(2\cos(z)-2+z^2)^2}$$
I feel like you might need to rewrite $\cos(z)$ as a Maclaurin series but I'm not entirely sure what you'd do next.
Also, once you think you've worked out the order o... | Consider the expansion of $f(z)=\cos(z)$ at $z=0$ namely
$$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}+\mathcal{O}(z^6)$$
We can put this into the expression,
\begin{align*}
\frac{1}{(2\cos(z)-2+z^2)^2}&=\frac{1}{\left(2\left(1-\frac{z^2}{2}+\frac{z^4}{4!}+\mathcal{O}(z^6)\right)-2+z^2\right)^2}\\
&=\frac{1}{\left(2\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this homogeneous recurrence relation of 2nd order? I have this homogeneous recurrence relation: $x_n = 3x_{n-1} + 2x_{n-2}$ for $n \geq 2$ and $x_0 = 0$, $x_1 = 1$. I form the characteristic polynomial: $r^2 - 3r -2 = 0$ which gives the roots $r = \frac{3}{2} - \frac{\sqrt{17}}{2}$, $\frac{3}{2} + \frac{\s... | Name the roots $r$ and $s$ and solve for the initial conditions
$$ar^0+bs^0=0=a+b,\\ar^1+bs^1=1=ar+bs.$$
Then
$$a=-b=\frac1{r-s}.$$
The difference between the roots being $\sqrt{17}$, the "enormous fraction" is
$$x_n=\frac{\left(\frac{3+\sqrt{17}}2\right)^n-\left(\frac{3-\sqrt{17}}2\right)^n}{\sqrt{17}}=\frac{(3+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to apply Fubini's theorem? I was asked to show the equality of these integrals
$$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dydx
=\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dxdy\tag{1}$$
Which can be answered by using Fubini's theorem but in order to use Fubini here is what I ... | Hint: change to polar coordinates.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Pell equation in ${\mathbb Q}(x)$ Is it known whether the equation $A^2-(x^2+3)B^2=1$ has a solution $A,B\in{\mathbb Q}(x)$ with $B\neq 0$ ?
My thoughts : I think that there is no solution, as the fundamental solution of $A^2-(x^2+3)B^2=1$ for $x\in {\mathbb Z}$ seems to vary uncontrollably.
| $$ \left( 2x^2 + 3 \right)^2 - \left( x^2 + 3 \right) \left( 2x \right)^2 = 9 $$
$$ \left( \frac{2}{3}x^2 + 1 \right)^2 - \left( x^2 + 3 \right) \left( \frac{2}{3}x \right)^2 = 1 $$
From $ \color{red}{\mbox{Jyrki's}} $ comment, material I had not realized applied here: we get a matrix that takes a solution ( as a col... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there an elegant way to evaluate $ I={ \int \sqrt[8]{\frac{x+1}{x}} \ \mathrm{d}x}$? Is there an elegant way to evaluate the following integral?
$$ I={ \int \sqrt[8]{\dfrac{x+1}{x}} \ \mathrm{d}x}$$
This seems to me a very lengthy question, yet it was given in my weekly worksheet, so there must be an elegant ... | Just to rework @H. R.'s solution by hand. First off setting
$$t=\frac{x+1}x\ge0$$
was definitely a good step. Solving for $x$,
$$x=\frac1{t-1}$$
Note that this implies that either $0\le t<1$ or $t>1$, a property which we will use later to simplify the result. Then
$$\int\sqrt[8]{\frac{x+1}x}dx=-\int\frac{t^{1/8}}{(t-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x... | Ok, so, I had some mistakes in my first approach to the problem, specially with signs, but I already proved it, so, here it is.
Hypothesis :
$$F(x) = 1+2q+3q^2+...+xq^{x+1}=\frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}$$
Proof:
$P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}+(x+1)q^x = \frac{1-(x+2)q^{x+1}+(x+1)q^{x+2}}{(1-q)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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Prove that the square of an integer $a$ is of the form $a^2=3k$, or $a^2=3k+1$, where $k\in \mathbb{Z} $ Here's my attempt to prove this. I'm not sure about it, but i hope i'm correct.
Let $a=λ, λ\in \mathbb{Z}$. Then $a^2=λ^2=3\frac{λ^2}{3}$.
When $λ^2$ is divided by 3 there are three possible remainders $0,1,2$. So
... | By the division algorithm
\begin{equation}
x=3q+r, \quad r \in \{0, 1, 2\}
\end{equation}
writing
\begin{align}
x^{2} &= 9q^{2}+r^{2} +6qr \\
&= 3(3q^{2}+2qr)+r^{2}
\end{align}
The result follows if for a given $x$, $r=0, 1$. If $r=2$ then clearly $r^{2}=4=1+3$ and thus
\begin{align}
x^{2} &=3 \lambda + 1 + 3 \\
&= 3(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding an invertible matrix.
I want to find an invertible matrix $P$ where $P^tAP$ is a diagonal matrix.
$$A=\begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 1 \end{pmatrix} $$
I have calculated the eigenvalues of $A$: $0,-2,4$ so the diagonal of $A$ should be $$D=\begin{pmatrix} -2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & ... | The characteristic polynomial of $A$ is $q(x)=x^3-2x^2-8x$, hence the spectrum is $\{-2,0,4\}$.
The nullspace of $A+2I$ is generated by $(1,-2,1)$, the nullspace of $A$ is generated by $(1,0,-1)$ and the nullspace of $A-4I$ is generated by $(1,1,1)$, hence with your $D$
$$ P = \begin{pmatrix}\frac{1}{\sqrt{6}} & \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826025",
"timestamp": "2023-03-29T00:00:00",
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Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x... | Hint the series converges only if $|r|<1$ so your second proof is wrong as $|r|=2$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$ Evaluate $$\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}.$$
The first thing i noticed was that $$\cos \frac{\pi}{7}=\frac{\zeta_{14}+\zeta_{14}^{-1}}{2},$$ where $\zeta_{14}=e^{2\pi i/14}$ is the 14-th root of unity.
Substituting this into... | Multiply by $8\sin\tfrac{\pi}{7}$ and simplify using $2\sin x \cos x = \sin 2x$ three times:
$$8\sin\tfrac{\pi}{7} \cos \tfrac{\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7}
= 4 \sin\tfrac{2\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7}
= 2 \sin\tfrac{4\pi}{7} \cos \tfrac{4\pi}{7}
= \sin\tfrac{8\pi}{7}$$
But you ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of solutions in non-negative integers question (Stars and bars) Q How many solutions are there in non-negative integers $a, b
, c, d$ to the equation:
$$ a + b + c + d = 79 $$
with the restrictions that $a \geq 10$, $b \leq 40$ and $20 \leq c \leq 30?$
If anyone could point me to any notes regarding the topic th... | Using generating functions, we want to find the coefficient of $x^{79}$ in the product
$\displaystyle(x^{10}+x^{11}+x^{12}+\cdots)(1+x+\cdots+x^{40})(x^{20}+x^{21}+\cdots+x^{30})(1+x+x^2+\cdots)$
$\displaystyle=\left(\frac{x^{10}}{1-x}\right)\left(\frac{1-x^{41}}{1-x}\right)\left( x^{20}\cdot\frac{1-x^{11}}{1-x}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficulty in finding appropriate $\delta$ I'm trying to prove that for every $\epsilon > 0$ there exists $\delta = $___
s.t for every $0 < \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$ :
$\lvert x^2y\rvert+\lvert y \rvert + 1 < \epsilon$
Can you help me find such $\delta$?
This is what I got so far:
$\lvert x^2y\r... | I'm going to assume this is related to this question Is $f$ continuous at $(0, 1)$ , $(1,1)$?.
The similarities are too great.
So that question was to show
$f(x,y) =
\left\{
\begin{array}{ll}
x^2y & \mbox{if } x \in \mathbb{Q} \\
y & \mbox{if } x \notin \mathbb{Q}
\end{array}
\right.$
show that $f$ is continuou... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Bounds for $\sum_{n\geq 2}\frac{n+2}{(n^2-1)(n+3)}$ Use the integral test to prove this inequaliy
I calculated the integral $\int_{2}^{\infty}\frac{2+x}{(x^2-1)(x+3)}dx$
How can I use the integral test to show that $0.45 < \sum_{n=2}^{\infty}\frac{2+n}{(n^2-1)(n+3)} < 0.75$ ?
I know I can use the integral test but I do... | Why do you need the integral test? The series can be computed through partial fraction decomposition:
$$\frac{n+2}{(n-1)(n+1)(n+3)}=\frac{3}{8}\cdot\frac{1}{n-1}-\frac{1}{4}\cdot\frac{1}{n+1}-\frac{1}{8}\cdot\frac{1}{n+3}\tag{1}$$
together with:
$$ \sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) = \frac{3}{2},$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find range of $\frac{\sqrt{1+2x^2}}{1+x^2}$? How to find range of $$\frac{\sqrt{1+2x^2}}{1+x^2}$$ ?
I tried put it equal to $y$ and squaring but I'm getting $4$th degree equation.
| You indeed get a fourth degree equation, but $x$ only appears with even exponent:
$$
y=\frac{\sqrt{1+2x^2}}{1+x^2}
$$
means that $y>0$ and that
$$
(1+x^2)^2y^2=1+2x^2
$$
Expanding and reordering gives
$$
y^2x^4+2(y^2-1)x^2+y^2-1=0
$$
and the usual quadratic formula provides the value for $x^2$; it's common to advise se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
... | Denote
$$A=\begin{pmatrix} a & b \\ c & d\end{pmatrix} $$
and solve the system of $4$ equations
$$A\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $$
$$A\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$$
for the unknowns $a,b,c$ and $d$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the action of T on a general polynomial given a basis I am given a question as follows:
Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is
$$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \... | First determine the matrix representation relative to the basis, $T(1), T(x),T(x^2)$.$$T(x^2)=\frac{T(4x^2+2x+2)-T(2x^2+2x+2)}{2}\\=(\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix}-\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix})/2 =\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}$$
$$T(x)=\frac{T(2x^2+2x+2)-2T(x^2)-2T(1)}{2}=(\begin{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way... | We can try to simplify the fraction using trigonometric formulae :
$$\cos (x) + 5\cos (3x) + \cos (5x)=\cos(x) (2 \cos(2 x)-1) (2 \cos(2 x)+5)$$
and $$\cos (6x) + 6\cos(4x) + 15\cos(2x) + 10=32 \cos^6(x)$$
So $$f(x)=\frac{(2 \cos(2 x)-1) (2 \cos(2 x)+5)}{32\cos^5(x)}=\frac{4\cos^2(2x)+8\cos(2x)-5}{32\cos^5(x)}$$
$$f'(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum value of $4a+b$ Let $ax^2+bx+8=0$ be an equation which has no distinct real roots then what is the least value of $4a+b$ where $a,b\in \Bbb R$.
My Try:
I differentiated the given function to get $f'(x)=2ax+b$ now if we put $x=2$ in this we get the required value . Now it's given that equation has no distinct r... | This is the solution of DeepSea, just a bit more detailed. Since the constraint given in the question has to do with the roots of the polynomial, try solving it by completing the square (assuming $a > 0$)
$$
ax^2 + bx + 8 = 0 \iff x^2 + \frac ba x + \frac 8a = 0 \iff \left(x + \frac{b}{2a}\right)^2 + \frac 8a - \frac{b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.
Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 ... | In polar coordinates the expression is
$$r\frac{\cos^2t\sin^3t}{\cos^4 t + \sin^4 t}.$$
The denominator in this fraction has a positive minimum; thus the fraction is a bounded function of $t,$ and the $r$ in front guarantees a limit of $0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent i... | You're almost there! We want to find the equation of line $OP$. We know its slope, and (since it passes through the origin) we know its $y$-intercept, so its equation is:
$$
y = \tfrac{-1}{3}x
$$
We can now find the intersection point of the radius and tangent by solving the system of equations. Equating, we obtain:
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that...
Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that
$M(g,B) = \begin{pmatrix} 2&0&3&0\\ 1&2&0&3\\0&0&2&0\\ 0&0&1&2 \end{pmatrix}$, where $B=(\begin{pmatrix} 0&1\\ 1&1 \end{pmatrix},\begin{pmatrix} 1&0... | Let $\alpha=\{E_1,E_2,E_3,E_4\}$ where
\begin{align*}
E_1 &=
\left[\begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}\right]
&
E_2 &=
\left[\begin{array}{rr}
1 & 0 \\
1 & 1
\end{array}\right]
&
E_3 &=
\left[\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}\right]
&
E_4 &=
\left[\begin{array}{rr}
1 & 1 \\
1 & 0
\end{array}\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding arc length of the curve $6xy=x^4+3$ from $x=1$ to $x=2$ Looking at this as a graph of a function of $y$ is more convenient
$$
y=\frac{x^4+3}{6x}\Rightarrow \frac{dy}{dx}=\frac{x^3-1}{2x^2}\Rightarrow \left( \frac{dy}{dx} \right)^2=\frac{x^6-2x^3+1}{4x^4}
$$
And the integral for arc length:
$$
\int_1^2 \sqrt{1... | HINT:
$$1+y'^2(x)=\frac14\left(x^2+\frac{1}{x^2}\right)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differential equation $\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$ I am not sure which type of differential equation this falls into:
$$\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$$ any hints? P.S. I first tried reornazing it so I have $y'$ alone, and hoping that I would get a homogeneous equation, but no such luck.... | Divide $x^2$ from both sides and then let $z=\dfrac{y}x$.
Note that $y'=z+xz'$.
The new equation becomes $(1+z)y'=\sqrt{1-z^2}+z+z^2$.
$y'=\sqrt{\dfrac{1-z}{1+z}}+z$
$x\dfrac{\mathrm dz}{\mathrm dx}=\sqrt{\dfrac{1-z}{1+z}}$
$\dfrac{\mathrm dx}{x}=\sqrt{\dfrac{1+z}{1-z}}\ \mathrm dz$
$\ln(x)=\sqrt{\dfrac1{1-z}}\left((z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove inequality $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$
Let $a,b,c,d$ positive real numbers, such that $$\frac1a+\frac1b+\frac1c+\frac1d=4.$$
Prove inequality
$$\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\... | $\sum\limits_{cyc}\left(2a-\sqrt[3]{\frac{a^3+b^3}{2}}\right)\geq\sum\limits_{cyc}\left(a+b-\frac{a^2+b^2}{a+b}\right)=2\sum\limits_{cyc}\frac{ab}{a+b}=2\sum\limits_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\geq\frac{32}{\sum\limits_{cyc}\left(\frac{1}{a}+\frac{1}{b}\right)}=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression
$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$
My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\... | hint: Alternatively, using Lagrange Multiplier we have:
$x = \cos \theta, y = \sin \theta\implies f(x,y) = x^2-6xy+3y^2+2, x^2+y^2 = 1\implies f(x,y) = (x^2+y^2)+2y^2+2 - 6xy = 3-6xy+2y^2 \implies f_x = -6y= 2\lambda x, f_y = 4y-6x= 2\lambda y\implies4\lambda y - 6\lambda x = 2\lambda^2y\implies 4\lambda y+18y=2\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2 ... | Wolfram answer:
$y = 2c + 1$
$x = c + 3d + 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can someone suggest a way to simplify $x_1(y_1 - x^Ty) + x_1(w_1 - x^Tw)^2 - x_1^2(w_1 - x^Tw)^2 + x_1x_2(w_1 - x^Tw)^2$ Let $x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$, $y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$, $w =\begin{bmatrix} w_1 \\ w_2 \end{bmatrix}$
I have the following vector:
$V = \begin{bmatrix} x_... | $$B \begin{array}[t]{l} = \begin{bmatrix} x_1(w_1 - \mathbf x^T\mathbf w)^2 - x_1^2(w_1 -
\mathbf x^T\mathbf w)^2 + x_1x_2(w_1 - \mathbf x^T\mathbf w)^2 \\ x_2(w_2 - \mathbf x^T\mathbf w)^2
- x_2^2(w_2 - \mathbf x^T\mathbf w)^2 + x_2x_1(w_2 - \mathbf x^T\mathbf w)^2 \end{bmatrix}
\\[3ex]
=\begin{bmatrix} (w_1 - \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Smallest number whose $\sin(x)$ in radian and degrees is equal Question:
What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?
My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides... | One degree is $\frac{\pi}{180}$ radians, so what we want here is
$$\sin(x)=\sin(\frac{\pi x}{180})$$
And so
$$x = \frac{\pi x}{180}+2\pi k, \qquad \text{or} \qquad x = \frac{-\pi x}{180}+\pi (2k+1)$$
Where $k$ is any integer (since $\sin(a) = \sin(b)$ iff $a-b$ is an integer multiple of $2\pi$, or if $a+b$ is an odd mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
| You can apply $u=x^2$ :
$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx=\frac{1}{2}\int \frac{x^2-1}{x^4\sqrt{2x^4-2x^2+1}} 2xdx=\frac{1}{2}\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du$$
Now note that :
$$\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du=\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du$$
we can substitute $v=\frac{1}{u}$ :
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How do I evaluate this integral using cauchy's residue theorem. $$\int_0^{2\pi} \dfrac{\cos 2 \theta}{1+\sin^2 \theta}d\theta$$
$$=\dfrac{-2}{i}\oint_{|z|=1} \dfrac{z^4+1}{z(z^4-4z^2-2z+1)}dz $$
I am stuck on how to use Cauchy's residue theorem since the bottom does not factor nicely. I know $z=-1$, $z=0.3111$ and $z=0... | Using trig. identities and eulers formula and long division:
$$
\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{1+\sin^2 \theta} d \theta=\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{1+\frac{1}{2}(1-\cos 2\theta)}$$
$$=2\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{3-\cos 2\theta}$$
$$= 2 \int_0^{2 \pi} - d\theta + 6\int_0^{2\pi}\dfrac{d \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find limit $\lim \limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $ I need to find this limit $\lim\limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $.
Give a hint please.Thanks
| Hints:
$$\frac{\pi x}{2x+1} = \frac{\pi}{2}-\frac{\pi}{4x+2}\tag{1}$$
$$\tan\left(\frac{\pi x}{2x+1}\right) = \cot\left(\frac{\pi}{4x+2}\right)\tag{2}$$
$$\frac{1}{x}\,\log\cot\left(\frac{\pi}{4x+2}\right)=\frac{\log\frac{4}{\pi}+\log x}{x}+O\left(\frac{1}{x^2}\right)\text{ as }x\to+\infty.\tag{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find whether the sum of a geometric series is prime or divisible by 3, 13 or 125? Let $A = 1+5+5^2+\dots+5^{99}$, then $A$ is:
*
*A prime number
*not divisible by 3
*divisible by 13
*divisible by 125
I know this is a sum of a Geometric Progression, so $ A = (5^{100}-1)/4$ but I cannot find $5^{100}$ So I... | I will assume that you mean $1+5+5^2+\cdots +5^{99}$. Let's deal with the choices one at a time.
Of course 4.) is false, all the terms except one are multiples of $5$. So the sum cannot be a multiple of $5$.
1.) is clearly false, we have an even number of odd numbers. The sum is therefore even, and thus not prime.
2.)... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $f'(c)$ using the derivative definition for $f(x)=\frac{1}{x^2}$ So $f:\mathbb R \backslash\{0\}\rightarrow\mathbb R$ by $f(x)=\frac{1}{x^2}$
I'm need to use the definition of the derivative to find $f'(c)$ for $\frac{1}{x^2}$. If I use standard differentiation techniques I get:
$-\frac{2}{c^3}$
However, using th... | set $x-c=h$ then $$f^{ \prime }\left( c \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( c+h \right) -f\left( c \right) }{ h } } =\lim _{ h\rightarrow 0 }{ \frac { \frac { 1 }{ { \left( c+h \right) }^{ 2 } } -\frac { 1 }{ { c }^{ 2 } } }{ h } } =$$ $$ \\ \\ =\lim _{ h\rightarrow 0 }{ \frac { { c }^{ 2 }-{ \left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix equation $A^2+A=I$ when $\det(A) = 1$ I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.
| Consider the Jordan Canonical Form for $A$; that is, $A = PJP^{-1}$ for some invertible $P$ and block diagonal matrix $J$ whose blocks are either diagonal or Jordan (same entry on the diagonal, have $1$s on the the diagonal above the main diagonal, and $0$s elsewhere).
Then, the equation reduces to $J^2 + J = I$. Looki... | {
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"timestamp": "2023-03-29T00:00:00",
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solve $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ Help with this excercise.. :)
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
the book says it is an exact differential equation, but how?
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)^{-1/2}dx+y(x^2+2y^2)dy=0$$
$M=x(x^2+y^2)^{-1/2}$
$N=y(x^2+2y^2)$
$$\frac{\partial ... | Obviously $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ isn't an exact differential equation.
So, there is a typo. Moreover the equation is certainly simple. The simplest guess is :
$$x(x^2+y^2)+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)dx+y(x^2+2y^2)dy=0$$
If it is the exact differential of $F(x,y)$ then :
$\begin{cases}
\frac{\p... | {
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"url": "https://math.stackexchange.com/questions/1852278",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4... | If $d$ is the common divisor of $4n-5$ and $60-12n$, then it should also divide $3(4n-5)+(60-12n) =45$. Thus when numerator and denominator are not divisible by 3 and 5, the fraction is irreducible. Hence $n \not \equiv 2 \pmod 3$ and $n \not \equiv 0 \pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The shortest distance from surface to a point Many have asked the question about finding the shortest distance from a point to a plane. I have checked those questions and answers and haven't found what I am looking for. Might still have missed some question though.
I have a problem where I am to calculate the shortest... | Set
$$f(x,y,z)=(x-5)^2+y^2+(z-1)^2$$
$$g(x,y,z)=z-x^2-3y^2=0$$
By application of Lagrange method, we have
$$\nabla f=\lambda \nabla g$$
$$(2x-10\,,\,2y\,,\,2z-2)=\lambda(-2x\,,\,-6y\,,\,1)$$
therefore
$$\left\{ \begin{align}
& y=0\,\,\,\,\quad\Rightarrow \,\,\,z={{x}^{2}} \\
& \lambda =-\frac{1}{3}\,\,\,\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off an... | Your proposition is not true.
try $n = 2$
$1\cdot 2 + 2\cdot3 \ne \frac 13 3\cdot4$
Which is why you are having a tough time proving this by induction.
Induction is usefull to prove things, but it doesn't always have a whole lot of insight why things are the way they are.
This is probably what you should have:
$\sum_\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get a whole number from $y = \frac{1}{x + 2}$ How to come up with a whole number for y.
I keep coming up with fractions from $y = \frac{1}{x +2}$
I've tried numerous numbers, as in, $ 1, 2, 3 ,4 , 5, -1, -2, -3, -4, -5$.
For example, $y = \frac{1}{1 + 2} = \frac{1}{3}$.
It's suppose to make two separate identic... | Because the numerator is $1$ in this case, using whole numbers for $x$ the only way that the fraction $\frac{1}{x+2}$ can be a whole number is when the denominator is either $1$ or $-1$. Knowing this, the only solutions are when $x$ is either $-1$ or $-3$. This gives us $y$ values of $1$ and $-1$ respectively.
If you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove/disprove : $\left(1+\frac{(-1)^{n}}{\sqrt{n}} \right)^{-1}=\left(1-\frac{(-1)^{n}}{\sqrt{n}}+o\left(\frac{1}{n}\right) \right)$ Why we have that
$$\begin{align*}
\frac{(-1)^n}{\sqrt{n}}\left(1+\frac{(-1)^n}{\sqrt{n}} \right)^{-1} & =\frac{(-1)^n}{\sqrt{n}}\left(1-\frac{(-1)^n}{\sqrt{n}}+O\left(\frac{1}n\right) \... | By Taylor expansion,
\begin{align*}
\frac{1}{1+\frac{(-1)^n}{\sqrt{n}}} &= 1+ \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + o\left(\frac{1}{n}\right)\\
&=1+ \frac{(-1)^n}{\sqrt{n}} + O\left(\frac{1}{n}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Showing that $2^6$ divides $3^{2264}-3^{104}$
Show that $3^{2264}-3^{104}$ is divisible by $2^6$.
My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$.
I have no idea how to show divisibility by $2^6$.
| A variant with some details:
By *Euler's theorem, $\;3^{\varphi(2^6)}\equiv 3^{32}\equiv=1\mod 2^6$. So
$$3^{2264}-3{104}\equiv3^{2264\bmod\varphi(2^6)}-3^{104\bmod\varphi(2^6)}\equiv 3^{24}-3^{8}\equiv3^8(3^{16}-1)\mod 2^6,$$
so it is enough to prove $3$ has order $16$ mod $2^6=64$.
Let's use the fast exponentiation ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible that $a^2+b^2+c^2 = d^2+e^2+f^2$?
Let $a,b,c$ be nonnegative integers such that $a \leq b \leq c, 2b \neq a+c$ and $\frac{a+b+c}{3}$ is an integer. Is it possible to find three nonnegative integers $d,e,$ and $f$ such that $d \leq e \leq f, f \neq c,$ and such that $a^2+b^2+c^2 = d^2+e^2+f^2$?
I have t... | Short answer (explained in the comments above):
$$ (4x)^2+(4x^2-1)^2 = (4x^2+1)^2 $$
can be "interlaced" with
$$ (4x^2+1)^2 + (8x^4+4x^2)^2 = (8x^4+4x^2+1)^2 $$
to get:
$$ (4x)^2 + (4x^2-1)^2 + (8x^4+4x^2)^2 = 0^2 + 0^2 + (8x^4+4x^2+1)^2 $$
or:
$$ (12x)^2 + (12x^2-3)^2 + (24x^4+12x^2)^2 = 0^2 + 0^2 + (24x^4+12x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$
*
*Let $k$ be a positive integer.
*Define $u_0 = 0\,,\ u_1 = 1\ $ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\
n \geq 2$.
*Show that for each integer $n$, the number
$u_{1}^{3} + u_{2}^{3} + \cdots + u_{n}^{3}\ $ is a multiple of
$\ ... | fiddling with small $k.$ Take $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$
CONCLUSION: for $k \geq 2,$
$$ \color{blue}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},} $$
while $\color{blue}{x \equiv 0,1 \pmod {k+1}}.$
When $k=2,$ $$ y = x^2. $$ This comes up pretty often, the sum of the consecutive cubes (starting with $1$) is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
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Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?
| It's a well-known fact that $$\log(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3+\underset{x\to 0}{o}(x^3)$$
Hence we get $$\dfrac 1x\log(1+x)=1+u(x)$$ where $$u(x)=-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)$$
Notice that $$\lim_{x\to 0} u(x)=0$$
We can write $$(1+x)^{1/x}=ee^{u(x)}$$
But we know that $$e^u=1+u+\dfrac{u^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\lim\limits_{x\to\infty} f(x)$ exists if $f'(x) = \frac{1}{x^2+f(x)^2}$ and $f(1)=1$
Let $f(x)$ be a real differentiable function defined for $x\geq 1$ such that $f(1)=1$ and $f'(x)=\dfrac{1}{x^2+f(x)^2}.$ Show that $$\lim_{x\to \infty}f(x)$$ exists and is less than $1+\frac{\pi}{4}$
I have no idea how to ... | Notice that $f'(x)$ is always positive, so $f$ is increasing. Hence for $x > 1$ we have $f(x) > 1$. Consequently, for $x > 1$, we have $f'(x) < \frac{1}{x^2 + 1}$.
We also have that $f'(x)$ is continuous for $x > 1$ since it's the quotient of continuous functions and the denominator is not zero. (The numerator is the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any... | The general solution of the above Diophantine Equation is: $x = m^2 - n^2, y = 2mn, z = m^2+n^2$. Its better if you start at this point. Then your analysis yields the followings: $m^2 = \pm 1 \pmod 5, n^2 = \pm 1 \pmod 5$. Since $5 \nmid xy\implies 5 \nmid x, 5 \nmid y\implies m^2 = 1\pmod 5, n^2 = -1\pmod 5$ or $m^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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nature of the series $\sum (-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}$ I would like to study the nature of the following serie:
$$\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $$
we can use simply this question :
Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \rig... | You made a mistake in
\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern... | Summation by parts gives:
$$ \sum_{n=1}^{N}\frac{n}{2^n} = N\left(1-\frac{1}{2^N}\right)-\sum_{n=1}^{N-1}\left(1-\frac{1}{2^k}\right)=1-\frac{N}{2^N}+\left(1-\frac{1}{2^{N-1}}\right)=\color{red}{2-\frac{N+2}{2^N}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem ... | $$\tan x +\sec x =\frac{1+ \sin x}{ \cos x}$$
$$=\frac{(\cos\frac{x}{2}+ \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}$$
$$=\tan \left(\frac{\pi}{4}+\frac{x}{2} \right)$$
$$\frac{\pi}{4}+\frac{x}{2} =n\pi +tan^{-1}4$$
$$ x =2 n\pi +2 tan^{-1}4 -\frac{\pi}{2}$$
For solution to be in $[0,2\pi]$
$$ n =0 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone explain this part of the definition of a linear combination of column vectors to me? This is what I need explained: "The system Ax=b is consistent iff b can be expressed as a linear combination, where the coefficients of the linear combination are a solution of the system."
I thought a linear combination of... | $$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}3\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&3\\2&4\end{pmatrix}}_A\underbrace{\begin{pmatrix}x\\y\end{pmatrix}}_{\bf x}$$
If we change this slightly to $$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}2\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&2\\2&4\end{pma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inner Product Examples, what is the points? Example:
For $ -\pi<x<\pi$,
$$x =-2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$$
and
$$x^3 =-2 \sum_{n=1}^{\infty} \left( \frac{\pi^2}{n}-\frac{6}{n^3} \right)(-1)^n \sin(nx)$$
by using inner products of these two functions, the value of
$$\sum_{n=1}^{\infty} \frac{1}... | HINT:
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4} $$
FURTHER EXPLANATION:
You have two functions $f(x)=x$ and $g(x)=x^3$. The coefficients of the expansion of $f$ in the basis $\sin(nx)$ are of the form $$a_n=-2\times \frac{(-1)^n}{n}$$
and the coefficients for $g(x)$ are: $$b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871562",
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"source": "stackexchange",
"question_score": "2",
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Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need th... | $(1−x^2)\frac{dy}{dx}−xy−1=0$
differentiate
$-2x \frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2}−y - x\frac{dy}{dx}=0\\
(1+x^2)\frac{d^2y}{dx^2} -3x \frac{dy}{dx} - y=0$
This covers the base case $n=0$
Suppose,
$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$
(this is the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also ... | $$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\sin^4 x + \cos ^4 x =\frac{1}{4}\left[(1- \cos2x)^2 +(1+ \cos2x)^2 \right]
=\frac{1}{2}((1+ \cos^2 2x)$$
$$\implies 2 \int \frac{dx}{1+ \cos^2 2x}$$
$$= 2 \int \frac{\sec^2 2x}{\sec^2 2x+ 1}{dx}$$
$$t= \tan2x$$
$$\implies 2 \int \frac{dt}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
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Evaluation of $\int^{\pi/2}_{0} \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$ Evaluate the given integral:
$$\int^{\pi/2}_0 \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$$
I multiplied and divided by $\sec(x)+\tan(x)$ to get denominator as $1$ but In calculation of integral, $x$ is creating problem. Is there any way to eliminate $x$ her... | Hint: Try the sub $x \mapsto \frac{\pi}{2} - x$ to get $$\int_0^{\pi/2} \frac{(\pi/2 - x) \cot x}{\csc x + \cot x} \, \mathrm{d}x = \int_0^{\pi/2} \frac{\pi/2 - x}{1 + \sec x} \, \mathrm{d}x$$
Then the rest is do-able using $$\frac{1}{\sec x + 1} = \frac{\cos x}{\cos x + 1} = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874176",
"timestamp": "2023-03-29T00:00:00",
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How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?
I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\le... | A proof by induction is possible, if a bit messy. For $n\in\Bbb N$ let $$s_n=\sum_{k=0}^n\binom{n+k}k\frac1{2^k}\;.$$ Clearly $s_0=1=2^0$. Suppose that $s_n=2^n$ for some $n\in\Bbb N$. Then
$$\begin{align*}
s_{n+1}&=\sum_{k=0}^{n+1}\binom{n+1+k}k\frac1{2^k}\\\\
&=\binom{2n+2}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\left(\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 15,
"answer_id": 9
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To find area enclosed within curve $x^4+y^4=x^2+y^2$ Find area enclosed by the curve
$x^4+y^4=x^2+y^2$
My attempt:
How to describe $y$ as function of $x$.
Only after that we can integrate.
Or can this curve be described parametrically.
| $$y^4-y^2+x^4-x^2=0$$
$$r^2(\cos^4(\theta)+\sin^4(\theta))=1$$
$$r^2=\frac 1 {\cos^4(\theta)+\sin^4(\theta)}$$
$$A/4=\frac 12\int_0^{\pi/2}\frac 1 {\cos^4(\theta)+\sin^4(\theta)}d\theta$$
$$A/4=\frac 12\int_0^{\pi/2} \frac 1{\cos^4 \theta}\frac{1}{1+\tan^4\theta}d\theta $$
Let $s=\tan(\theta)$ then $$A/4=\frac 12\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \su... | If $n>1$ we have $n+n^2+\dots + n^n=\frac{n^{n+1}-1}{n-1}-1\leq\frac{n^{n+1}}{n-1}\leq 2n^{n}$
So $\frac{n+n^2+\dots + n^n}{n^{n+2}}\leq n^{-2}$
Therefore $\sum_{n=2}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq\sum_{n=2}^\infty n^{-2}<\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878090",
"timestamp": "2023-03-29T00:00:00",
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Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes... | I'll give you a derivation which I don't think is all that known. You just need to know four things:
(1) A version of the umbral Taylor series
Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How many solutions are there to $x_1 + x_2 + ... + x_5 = 21$? How many solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$
where $x_1, i = 1,2,3,4,5$ is a nonnegative integer such that $0 ≤ x_1 ≤ 3, 1 ≤ x_2 < 4$, and $x_3 ≥ 15$?
I have correctly completed the previous parts of the question but am... | An approach using generating functions:
In order to count up the number of solutions that sum to $n$, we can look at the coefficient of $x^n$ in the generating function $f(x)$:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{3}x^n\right)\left(\sum_{n=1}^{3}x^n\right)\left(\sum_{n=15}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Integral $\int_0^1 \frac{\ln (2-x)}{2-x^2} \, dx$ Here is an integral that I am trying to solve for quite some time. Find a closed form for the integral:
$$\mathcal{J}=\int_0^1 \frac{\log (2-x)}{2-x^2} \, {\rm d}x$$
Here is what I've done.
\begin{align*}
\int_{0}^{1}\frac{\log (2-x)}{2-x^2} \, {\rm d}x &= \int_{0}^{1}... |
Here's a way to calculate the integral that circumvents the use of polylogarithms entirely.
Let $I$ denote the value of the integral,
$$I:=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x.$$
Using a clever choice of substitution, we find
$$\begin{align}
I
&=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Can this systems of equations be solved? There is a system of equations as follow:
$$\left[\begin{matrix}x\\ y\\ \end{matrix} \right] = \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]\left[\begin{matrix}\frac{1}{x}\\ \frac{1}{y}\\ \end{matrix} \right]$$
Given $ \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right... | This is basically asking for $\frac{a}{x}+\frac{b}{y}=x$ and $\frac{b}{x}+\frac{c}{y}=y$. From the first equation, $y=\frac{xb}{x^2-a}$. You can substitute this into the second equation to get: $$\frac{b}{x}+\frac{c(x^2-a)}{bx}=\frac{xb}{x^2-a}$$
This should be translatable into a polynomial. However, the smallest poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Decide if the series converges and prove it using comparison test: $\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
Decide if the series converges and prove it using comparison test:
$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
$$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4... | IMHO this could be written more precisely. For example the second inequality does not look right. I would suggest the following.
$$\frac{3k^2+k+1}{k^4+k^3+4}=\frac{k^2}{k^4} \cdot \frac{3+1/k +1/k^2}{1+1/k} \frac{1}{k^2} \cdot \frac{3 + 1+ 1}{1+1}. $$
Now one can conclude that each of the entries of the sum is smaller ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\lef... | Your summation should be
$$15\times \sum \limits^{\infty }_{k=2}\left( \frac{1}{5} \right) ^{k}=15\times \frac{\left( \frac{1}{5} \right) }{1-\frac{1}{5} } ^{2}=15\times \frac{1}{20} =\frac{3}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\int_0^{2\pi} \frac{dt}{3+\sin(t)}$? Got stuck with the integral.
I rewrote it as $$I=\int_0^{2\pi} \frac{2idt}{6i+e^{it}-e^{-it}},$$
then I took $z:=e^{it}\implies dz=ie^{it}dt\implies dt = -\frac{idz}{z}$, so I'd get:
$$\int_\gamma \frac{2dz}{6i+z-\frac{1}{z}}=\int_\gamma \frac{2zdz}{6iz+z^2-1}.$$
It ca... | A real-analytic way is to break the integration range into sub-intervals with length $\frac{\pi}{2}$ to get
$$ I = \int_{0}^{\pi/2}\left(\frac{1}{3+\sin z}+\frac{1}{3+\cos z}+\frac{1}{3-\sin z}+\frac{1}{3-\cos(z)}\right)\,dz \tag{1}$$
from which:
$$ I = \int_{0}^{\pi/2}\frac{12}{9-\cos^2 z}\,dz =\int_{0}^{+\infty}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Convex Quadrilaterals Let $n>4$.
In how many ways can we choose $4$ vertices of a convex $n$-gon so as to form a convex quadrilateral, such that at least $2$ sides of the quadrilateral are sides of the $n$-gon?
Explain your answer, which should be expressed in terms of $n$.
| We need to count the number of ways to choose 4 vertices so that at least two pairs of vertices are adjacent.
There are $\dbinom{n-3}{4}-\dbinom{n-5}{2}$ ways to choose them so that none are adjacent,
and there are $n\dbinom{n-5}{2}$ ways to choose them so that exactly two are adjacent;
so this gives a total of $\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the expected number of red apples left when all the green apples are picked? We have 4 green apples and 60 red apples. Each time we pick one out without replacement. Then what is the expected number of red apples left when all 4 green apples are picked?
| Suppose we have $G$ green apples and $R$ red apples and ask about the
number of red apples left when all green apples have been picked.
The probability of the last green apple being picked as apple number
$q$ is
$${R+G\choose R}^{-1} {q-1\choose G-1}.$$
We now verify that this is indeed a proability. We use the Ego... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
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.