Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question
| Let it be true for k so for $k^2=\frac{(k)(k+1)(2k+1)}{6}=$..(1). Now we will have to prove for $(k+1)^2$. So we have to prove $(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$. So we start ${(1^2+2^2...k^2+(k+1)^2)}=\frac{(k)(k+1)(2k+1)}{6}.(k^2+2k+1)=\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$. Hope now you know fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Indefinite integral - What am I doing wrong? We have the integral $\displaystyle \int \sqrt{4-x^2} dx$, and I want to evaluate it using integration by parts. So our first step gives us:
$$ \int \sqrt{4-x^2} dx = x \sqrt{4-x^2} + \int \dfrac{x^2}{\sqrt{4-x^2}} dx $$
Now we have to deal with $\displaystyle \int \dfrac{... | Usually for this form of integrand ($\sqrt{a^2-x^2}$) we use a trig sub with $x=a\sin\theta$. In our case $a=2$.
$$x=2\sin\theta, dx=2\cos\theta d\theta$$
Substituting we have
$$\int\sqrt{4-x^2}dx=\int2\cos\theta\sqrt{4-4\sin^2\theta}\,d\theta$$
$$=4\int\cos^2\theta\, d\theta$$
We can rewrite $\cos^2\theta$ as $\cos(2... | {
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"url": "https://math.stackexchange.com/questions/1546594",
"timestamp": "2023-03-29T00:00:00",
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Choosing numbers at random - expected value calculation From set $\{1,2,\ldots,49 \}$ we choose at random 6 numbers without replacing them. Let X denotes quantity of odd numbers chosen. Find $\mathbb{E}X$, how to find that? I have no idea whatsoever.
EDIT:: still looking for the sufficient explanation.
| We consider the $7$ cases and determine how many odd balls are present in each case.
The $7$ cases are:
*
*$6$ even : $\dbinom{24}{6}$
*$5$ even, $1$ odd : $\dbinom{24}{5}\cdot 25$
*$4$ even, $2$ odd : $\dbinom{24}{4}\cdot \dbinom{25}{2}$
*$3$ even, $3$ odd : $\dbinom{24}{3}\cdot \dbinom{25}{3}$
*$2$ even, $4$ o... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How do i solve $\frac{dy}{dx}- \frac {dx}{dy}= \frac {x}{y}-\frac {y}{x}$? I came up with this question in my exam. But i didn't get it right. Can someone show me how to solve this differential equation
$$\frac{dy}{dx}- \frac {dx}{dy}= \frac {x}{y}-\frac {y}{x}$$
| Since $$\frac{dy}{dx}-\frac{dx}{dy}=\frac{dy}{dx}-\frac{1}{\frac{dy}{dx}}=\frac{x}{y}-\frac{y}{x}$$
Let $a=\frac{dy}{dx}$ and $b=\frac{x}{y}$. We are given that $$a-\frac{1}{a}=b-\frac{1}{b}$$
Rearranging gives $$(a-b)(1+\frac{1}{ab})=0$$
Which gives us $$a=b$$ or $$a=-\frac{1}{b}$$
Plugging $a,b$ back, we have $$\frac... | {
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"url": "https://math.stackexchange.com/questions/1548569",
"timestamp": "2023-03-29T00:00:00",
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Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\fr... | Let $G_n=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots $.
Then, we have
$$G_1=1,G_2=2$$
$$G_{n}+G_{n+1}=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots +\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}\cdots $$
$$=\binom{n+1}{0}+\left(\binom n0+\binom{n}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\left(\binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:
Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations
What I've Tried
*
*$ \frac{dx}{dt} = -5\sin(t) $
*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of... | You may like to use a short method to prove this.
Eliminate t to get the Cartesian equation of curve as $f(x,y)=9x^4-225x^2+625y^2$
Now, equate the least degree terms to zero to get tangents at $(0,0)$ which gives $y=3x/5$ and $y=-3x/5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How many 4 digit numbers are divisible by 29 such that their digit sum is also 29? How many $4$ digit numbers are divisible by $29$ such that their digit sum is also $29$?
Well, answer is $5$ but what is the working and how did they get it?
| Given the $4$-digit number $\overline{abcd}$, the sum of digits being divisible by $29$ implies:
$$a+b+c+d=29,$$
because: $29<4\cdot 9<58$.
$\overline{abcd}$ is being divisible by $29$ implies:
$$1000a+100b+10c+d\equiv 14a+13b+10c+d\equiv 0\pmod{29}.$$
Hence:
$$\begin{cases} \ \ \ \ a+\ \ \ \ b+\ \ \ \ c+d=29 \\
14a+13... | {
"language": "en",
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"source": "stackexchange",
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Divisibility of an expression Need some guidance
How to prove that $9\cdot n^9+7\cdot n^7+3\cdot n^3+n$ is divisible by $10$.
I've tried transforming the expression by adding $n^9$ and $-n^9$ in order to make a multiple of 10 but no use. I've even tried math. induction but got stuck, maybe this can not be proven. Are t... | Let $P$ be a polynomial: $P(n)=9n^9+7n^7+3n^3+n$.
$P(n)$ is even for any $n$, because if $n$ is even then you're looking at a sum of $4$ even numbers, if it's odd at a sum of $4$ odd numbers.
Then you only have to prove that $5|P(n)$.
Notice that $n^5\equiv_5 n$ for any $n$. (Proved at the end)
Then $P(n)\equiv_5 -n^9-... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further... | \begin{align}
u & = \arctan \sqrt{\frac x 2} \\[10pt]
du & = \frac{dx/2}{\left(1+ \dfrac x 2\right)2\sqrt{\dfrac x 2}} = \frac{dx}{(2+x)\sqrt{2x}} \\[10pt]
dv & = \frac{dx}{\sqrt{x+2}} \\[10pt]
v & = 2\sqrt{x+2}
\end{align}
\begin{align}
\int u\,dv & = uv - \int v\,du = 2\sqrt{x+2} \arctan \sqrt{\frac x 2} - \int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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$\sum_{n=1} ^\infty \frac{8}{n15^n}$ without calculator I just can't figure out which function to use to get the sum. I tried with ln, but that gives me an alternating series.
| Hint
Consider $$A=\sum_{n=1} ^\infty \frac{8}{n15^n}={8}\sum_{n=1} ^\infty \frac{1}{n15^n}={8}\sum_{n=1} ^\infty \frac{x^n}{n}$$ where $x=\frac 1 {15}$.
You can recongnize that the summation is just Taylor expansion of $-\log(1-x)$ from which the problem becomes simple.
We then have $$A=-8\log(1-\frac 1 {15})=-8\log(\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex equation: $z^8 = (1+z^2)^4$ What's up with this complex equation?
$ z^8 = (1+z^2)^4 $
To start with, there seems to be a problem when we try to apply root of four to both sides of the equation:
$ z^8 = (1+z^2)^4 $
$ z^2 = 1 + z^2 $
which very clearly doesn't have any solutions, but we know there are solutions: ... | $$z^8=(z^2+1)^4\Longleftrightarrow$$
$$z^4=(z^2+1)^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=z^2+1\Longleftrightarrow\space\space\vee\space\space z^2=-1-z^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$0\ne1\space\space\vee\space\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more c... | As a last resort, it amounts to being able to compute
$$I_n=\int\frac{\mathrm d\mkern 1mu x}{(x^2-x+8)^n}$$
This sort of integrals can be calculated recursively. I'll show how to obtain a relation between $I_2$ and $I_3$. $I_1$ is standard, and is an $\arctan$ after a change of variable.
Integrate $I_2$ by parts, sett... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Finding the (complex) solutions $z$ to $z^4 = w$
What are possible solutions to $$z^4 = w$$ where $z \in \mathbb{C} $ and $w \in \mathbb{R}$?
Here is my attempt:
Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.
Working with the conventions of $r \geq 0$ and $0 \leq \th... | If $w < 0$, then the equation $r = w^{1 / 4}$ is not even sensible, as any fourth root of $w$ is nonreal (and anyway in this case there is not a preferred choice among these) but $r$ is a real variable. The fix is to write any $w < 0$ in polar form as $$w = |w|e^{\pi i} = |w|(\cos \pi + i \sin \pi).$$ Then, regardless ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| First of all: you have to be carefull with your notation because you mean:
$$\int\left(\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\right)\space\text{d}x$$
And Partial fractions is the most easy way:
$$\int\frac{x-1}{(x+3)(x^2+1)}\space\text{d}x=$$
$$\int\left(\frac{2x-1}{5(x^2+1)}-\frac{2}{5(x+3)}\right)\space\text{d}x=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find Least Squares Regression Line I have a problem where I need to find the least squares regression line. I have found $\beta_0$ and $\beta_1$ in the following equation
$$y = \beta_0 + \beta_1 \cdot x + \epsilon$$
So I have both the vectors $y$ and $x$.
I know that $\hat{y}$ the vector predictor of $y$ is $x \cdot \b... | Sequence of $m$ measurements:
$$\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$$
Model:
$$
y(x) = \beta_{0} + \beta_{1} x
$$
Linear system:
$$
\begin{align}
%
\mathbf{A} \, \beta &= y \\
% A
\left[ \begin{array}{cc}
1 & x_{1} \\
1 & x_{2} \\
\vdots & \vdots \\
1 & x_{m}
\end{array} \right]
% beta
\left[ \begin{array}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of trignometric identity Could you help me prove this? I've gotten stuck, need some help..
$$\sin^2\Theta + \tan^2\Theta = \sec^2\Theta - \cos^2\Theta$$
Here's what I've done so far:
Left Side:
$$\sin^2\Theta + \frac{\sin^2\Theta}{\cos^2\Theta}=\frac{\sin^2\Theta\cos^2\Theta+\sin^2\Theta}{\cos^2\Theta}$$
Right ... | Am I the only one here who was required to work on only one side when doing trig proofs?
The strategy for all such proofs is to work on both sides until you can get them to be equal, then retrace your steps and put them all on a single side.
To wit,
$$ \sin^2x + \tan^2x = \sec^2x - \cos^2x$$
Add $\cos^2 x$ to both si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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An exercise concerning complex numbers Assume that $| z + 1 | > 2$. Show that $|z^3 + 1| > 1$.
My try was:
$$|z^3 + 1| = |z + 1| |z^2 - z + 1| > 2 |z^2 - z + 1| $$
but I'm stuck proving that $|z^2 - z + 1| > \frac 1 2$
| Write $z = -1+ w$, so $|z+1|>2$ means $|w| > 2$. Then
$z^3 + 1 = w^3 - 3 w^2 + 3 w = w (w^2 - 3 w + 3)$. The claim is that
if $|w| > 2$, $|w^2 - 3 w + 3| > 1/2$. If $w = r \exp(i\theta)$,
$$|w^2 - 3 w + 3|^2 = r^4 - 6 r^3 \cos(\theta) + (3 + 12 \cos(\theta)^2) r^2 - 18 r \cos(\theta)+ 9$$
Call that $F(r,\theta)$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of a Series With Denominators of the form $(2^i) (3^j)(5^k)$ Can anyone solve this?
Find the sum of the series $1 + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \cdots,$ where the denominators are of the form $(2^i) (3^j)(5^k)$?
The test came with the next answer ... | We have an Euler product:
$$ \prod_{p\leq 5}\left(1-\frac{1}{p}\right)^{-1} = \prod_{p\leq 5}\left(1+\frac{1}{p}+\frac{1}{p^2}+\ldots \right) = \sum_{n\in A}\frac{1}{n} $$
where $A$ is the set of positive integers whose prime divisors are $\leq 5$.
It follows that our series equals:
$$ 2\cdot \frac{3}{2}\cdot\frac{5}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559946",
"timestamp": "2023-03-29T00:00:00",
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Linear Algebra - seemingly incorrect result when looking for a basis
For the following matrix $$ A = \begin{pmatrix}
3 & 1 & 0 & 0 \\
-2 & 0 & 0 & 0 \\
-2 & -2 & 1 & 0 \\
-9 & -9 & 0 & -3
\end{pmatrix} $$
Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value.
First step is to find the ... | If you calculate $\det(A-\lambda I)$ by the first row (and all the minors too) you get
$$
\det(A-\lambda I)=(3-\lambda)\lambda(1-\lambda)(-3-\lambda)+2(1-\lambda)(-3-\lambda)=(3+\lambda)(1-\lambda)(\lambda^2-3\lambda+2)
=(3+\lambda)(1-\lambda)(\lambda-1)(\lambda-2).
$$
So the eigenvalues are $1$ (with multiplicity $2$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\g... | All of them follow from Holder's inequality: for all $a_{ij}>0$:
$$\prod_{i=1}^k\sum_{j=1}^m a_{ij}^k\ge \left(\sum_{j=1}^m\prod_{i=1}^k a_{ij}\right)^k$$
$k=2$ gives Cauchy-Schwarz inequality. In this case, let $k=8, m=2, a_{ij}=1$ for all $i\in\{1,2,\ldots,7\}, j\in\{1,2\}$ and $a_{81}=a, a_{82}=b$:
$$(1^8+1^8)^7(a^8... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \inf... | For $x=-1/3$, we can separate the even $n$ and odd $n$ terms and obtaining:
$$A=\sum_{n = 1}^{\infty} \left((-1)^n + n \left(-\frac 1 3\right)^n\right)
=\sum_{k = 1}^{\infty} \left((-1)^{2k} + (2k) \left(-\frac 1 3\right)^{2k}\right)
+\sum_{k = 1}^{\infty} \left((-1)^{2k+1} + (2k+1) \left(-\frac 1 3\right)^{2k+1}\right... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral of $\sqrt{R^2-x^2}$ Is there any way to compute the integral
\begin{equation}
\int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R
\end{equation}
without using trig substitution or integration by parts?
I'm thinking to relate this to area of circle, but I couldn't find the relationship.
Thank you.
| Without trigonometric substitution, integration by parts, or appeal to the geometric interpretation of the integral, we need to devise some quite fortuitous manipulations. We proceed, therefore, and write
$$\begin{align}
\sqrt{R^2-x^2}&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)\\\\
&=\frac12 \left(\sqrt{R^2-... | {
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"timestamp": "2023-03-29T00:00:00",
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Fun Q6: Side length of the pentagon in a five sided star? Consider a regular pentagon of side length $a$. If you form a 5-sided star using the vertices of the pentagon, then you'll get a pentagon inside that star. What is the side length of that pentagon?
In general, for a n-sided star, what is the side length of the ... | Let $x$ be the length of small n-sided regular polygon in the star & $a$ be the distance between two adjacent vertices, then the angle of spike of regular star polygon is given as
$$\alpha=\frac{\pi}{\text{number of vertices (points) in the star}}=\frac{\pi}{n}$$
Now, draw a perpendicular from one vertex of star to t... | {
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find the matrix representation of $T$ relative to $B,B'$ where $T(p(x)) = (x+1)p'(x^2)$ Hi I really need help with this linear algebra question.
Let $T:P_2\to P_3$ be defined by $T(p(x)) = (x+1)p'(x^2)$ and let $B = (1,x+1,x^2+x)$ and $B' = (x^3,x^3+x,x^2+x,x+1)$ be ordered basis for $P_2$ and $P_3$ respectively.
a.) f... | So, first of all, the expression $T(p(1))$ doesn't make sense. The symbol $p(x)$ is supposed to represent a polynomial. So your computation would look like this:
If $p(x) = 1$, then $p'(x) = 0$, and so $p'(x^2) = 0$. So $T(1) = (x+1)(0) = 0$.
If $p(x) = x + 1$, then $p'(x) = 1$, and so $p'(x^2) = 1$. So $T(x + 1) = (x ... | {
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"answer_id": 0
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How to take an integral using half angle trigonometric substitution. So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$
and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this probl... | By letting $u=\tan\frac{x}2$, we get
$$\int\frac{\sin x}{(6\cos x-2)(3-2\sin x)}d x=\int\frac{u d u}{(1-2u^2)(3u^2-4u+3)}\\
=\frac{8}{49}\int\frac{(9u+4)du}{1-2u^2}+\frac{12}{49}\int\frac{(9u-8)du}{3u^2-4u+3},$$
the later two are both integration of rational functions.
$$\int\frac{(9u+4)du}{1-2u^2}=9\int\frac{udu}{1-2u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Triangle of numbers sum What is the total of all these numbers?
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
. . . . . .
1 2 ........N
The answer should be a single binomial coefficient. It looks pretty easy, but I cannot com... | The total of the $1$'s is $N$. The total of the $2$'s is $2(N-1)$, the toal of the $3$'s is $3(N-2)$,..., the total of the $N$'s is $N(N-(N-1))=N$.
Throwing this into a sum we find:
$$\sum_{i=1}^N i \cdot (N-(i-1)) = \sum_{i=1}^N i \cdot N - \sum_{i=1}^N i^2 + \sum_{i=1}^N i$$
$$= N \frac{N(N+1)}{2} - \frac{N(N+1)(2N+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help on solving the equation $\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$ Could you give me some help on finding the roots (if any) of the following equation:
$$
\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}
$$
I tried to apply some classic approaches, ... | $$\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$$
$$\frac{\sqrt{1+\frac xa}}{\sqrt{1}+\sqrt{1+\frac xa}}=\frac{\sqrt{1-\frac xa}}{\sqrt{1}-\sqrt{1-\frac xa}}$$
$$\frac xa \mapsto y$$
$$\frac{\sqrt{1+y}}{1+\sqrt{1+y}}=\frac{\sqrt{1-y}}{1-\sqrt{1-y}}$$
$$\sqrt{1+y}-\sqrt{1-y^2}=\sqrt{1-y}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The probability that each child gets at least one chocolate is?
5 different chocolates are to be distributed among 4 children.The
probability that each child gets at least one chocolate is ?
Total number of ways is $4^5$.Got that.After that what to do?
| We count the number of "favourables." The numbers involved are very small, so we use a counting procedure that would be inefficient for larger numbers.
The lucky child can be chosen in $\binom{4}{1}$ ways. For each such way, the chocolates she gets can be chosen in $\binom{5}{2}$ ways. And for every way of doing this, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit exists theoritically,but graphically there is a vertical asymptote there.Why is it so. In finding the limit of $\lim_{x\to 0}\frac{2^x-1-x}{x^2}$
I used the substitution $x=2t$
$\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\lim_{t\to 0}\frac{2^{2t}-1-2t}{4t^2}=\frac{1}{4}\lim_{t\to 0}\frac{2^{2t}-2\times2^t+2\times2^t+1-2-2t... | $\lim_{x\to 0}\frac{2^x-1-x}{x^2}
$
For small $x$,
$2^x
=e^{x \ln 2}
\approx 1+x \ln 2+O(x^2)
$
so
$\frac{2^x-1-x}{x^2}
=\frac{1+x \ln 2 +O(x^2)-1-x}{x^2}
=\frac{x (\ln 2-1) +O(x^2)}{x^2}
=\frac{(\ln 2-1) +O(x)}{x}
\to \infty
$
as
$x \to 0$.
In general,
$\lim_{x \to 0}\frac{a^x-1-x\ln a}{x^2}
$
exists
and is
$\begin{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Her... | The numbers are small, and one could find the answers without the full CRT machinery. But we will go ahead and use a "general" procedure.
The idea is that to solve $x\equiv a\pmod{7}$, $x\equiv b\pmod{13}$, you use
$$x\equiv (C)(13)(a)+(D)(7)(b)\mod{91},$$
where $C$ is the inverse of $13$ modulo $7$ and $D$ is the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Green's Theorem on Line Integral I am asked to find the line integral for the following field:
$$F = (e^{y^2}-2y)i + (2xye^{y^2}+\sin(y^2))j$$
On the line segment with points $(0,0),(1,2)$ and $(3,0)$. I have to do it with Greens theorem. This is the setup I have so far.
$$\int_C F \cdot dr = \iint_D \frac{\partial Q}{... | The problem lies in two places: the first one in $$\frac{\partial Q}{\partial x} = 2ye^{y^2},$$ but this is only algebraic, even though it made your life much harder.
The second one is much more sensible. Look at Green's Theorem
$$\oint_C (P dx + Q dy) = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if ... | You are forgetting that there is a relation between $x$ and $y$. Your answer and the official "correct" answer are the same. One way to see this: make the substitution $x^2=\frac{x+y}{x-y}$ into your answer and:
$$\begin{align}
\frac{2x(x-y)-1+x^2}{1+x^2}&=\frac{2x(x-y)-1+\frac{x+y}{x-y}}{1+\frac{x+y}{x-y}}\\
&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The value of double integral $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$? Given double integral is :
$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$$
My attempt :
We can't solve since variable $x$ can't remove by limits, but if we change order of integration, then
$$\int _0^1\int _0^{\frac{1}{x}... | Right way is:
$$\int\limits_0^1x\:dx\int\limits_0^\dfrac1x \dfrac{dy}{1+y^2} = \int\limits_0^1\arctan y\:\Biggl.\Biggr|_0^{\dfrac1x} x\:dx =\int\limits_0^1x\arctan \dfrac1x\:dx = \int\limits_0^1\left(\dfrac{\pi}2-\arctan x\right)x\:dx =$$$$ \left.\dfrac{\pi}2\dfrac{x^2}2\right|_0^1 -\int\limits_0^1\arctan x\: d\dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
why $ 1 - \cos^2x = \sin^2x $? I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working:
\begin{array}\\
\lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1... | Let $F(x)=\sin ^2 x + \cos ^ 2 x$.
$$F'(x)=2 \sin x\cos x- 2 \cos x\sin x=0$$
Since $F(0)=1$ and $F$ is constant, we get
$$\sin ^2 x + \cos ^ 2 x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
For give permutation $\sigma\in S_{13}$ solve equation $x^3=\sigma$ We have this permutation
$$\sigma =\left({\begin{array}{*{20}c}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\ 3 & 2 & 1 & 11 & 5 & 8 & 13 & 10 & 9 & 12 & 4 & 6 & 7\end{array}}\right)\ $$
Find $x^3=\sigma\in S_{13}$
The cycles are (1 3)(2)(4 1... | The permutation with the cycles $(1\ 3)(2)(4\ 11)(5)(6\ 12\ 10\ 8)(9)(13\ 7)$ does the job.
You only have to invert the $4$-cycle. The other cycles remain. You can also take the cycle $(2\ 5\ 9)$ to get another solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find gcd polynomials? How to find gcd of polynomials $gcd(x^3+x^2-x-1,3x^2+2x-1)$ ??
I divide of polynomials. It worked like this $\frac 13 x - \frac19$,$ R\left( x\right) =-\frac89x -\frac89$
| You want the gcd of $f(x) = x^3+x^2-x-1$ and its derivative. Now
$$f(x) = x^2 (x+1) - (x+1) = (x^2-1)(x+1) = (x-1)(x+1)^2$$ so
$$\gcd(f(x), f'(x)) = x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | No one's pointing out my favorite result that $n^2$ is the sum of the first odd numbers?
$(n + 1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1$.
and the result follows...
Case 1: $k=2m$ is even.
$1 - 2^2 + 3^2 - 4^2 + .... + (k-1)^2 - (k)^2 = -(2^2 -1) - (4^2 - 3^2) - ...- (k^2 - (k-1)^2)=$
$-(3) - (-7)-....-(2k + 1) = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Prove that $\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca} \leq 1.$ The following problem was on the IMO 1996 shortlist :
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that
$$\dfrac{ab}{a^5+b^5+ab}+\dfrac{bc}{b^5+c^5+bc}+\dfrac{ca}{c^5+a^5+ca} \leq 1.$$
I tried factoring out thing... | since
$$a^5+b^5\ge a^2b^3+a^3b^2$$
so
$$\sum\dfrac{ab}{a^5+b^5+ab}\le\sum\dfrac{1}{ab^2+a^2b+1}=\sum\dfrac{abc}{ab^2+a^2b+abc}=\sum\dfrac{c}{a+b+c}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
find the maximum possible area of $\triangle{ABC}$ Let $ABC$ be of triangle with $\angle BAC = 60^\circ$
. Let $P$ be a point in its interior so that $PA=1, PB=2$ and
$PC=3$. Find the maximum area of triangle $ABC$.
I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle... | Let $\mathcal{A}$ be the area of $\triangle ABC$.
Let $\theta$ and $\phi$ be the angles $\angle PAC$ and $\angle BAP$ respectively.
We have $\theta + \phi = \angle BAC = \frac{\pi}{3}$.
As functions of $\theta$ and $\phi$, the side lengths $b$, $c$ and area $\mathcal{A}$ are:
$$
\begin{cases}
c(\theta) &= \cos\theta + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:
Let $x$, $y$, $z$ be real
numbers. Prove that
$$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$
The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square ... | For all the equations of symmetrical type, the extreme value, if exists, is achieved when variables are equal.
It is not difficult to prove it. If you have $f(x,y,z)$ symmetrical over $x,y,z$ then all their derivatives are the same, so whatever condition is needed for $x$ is the same for $y$ and the same for $z$. This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proving a trigonometric inequality I've been having difficulty with the following,
Prove that,
$[\sin^{n+1}(x)]^{2}+[\cos^{n+1}(x)]^{2} \geq (\frac12)^{n}$ where $x$ is real and $n$ is a non-negative integer.
I've tried an inductive approach but have struggled with the inductive step.
| use Holder inequality
$$[(\cos^2{x})^{n+1}+(\sin^2{x})^{n+1}][1+1]^{n} \geq (\cos^2{x}+\sin^2{x})^{n+1}=1$$
or
Use AM-GM inequality we have
$$(\sin^2{x})^{n+1}+\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}\ge (n+1)\cdot\sqrt[n+1]{\dfrac{1}{2^{n(n+1)}}(\sin^2{x})^{n+1}}=\dfrac{n+1}{2^{n}}\sin^2{x}$$
so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k ... | insert $a=2b+1$ and you get
$((2b+1)^2 + 3)((2b+1)^2 + 7) /32= 1/2 (b^2+b+1) (b^2+b+2)$
either $(b^2+b+1)$ or $(b^2+b+2)$ will be even so the RHS is an integer for all b.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | For positive $x$, $y$, $z$, $a$, $b$, and $c$, note that that
\begin{align*}
(x+y+z)^2 &= \left(\frac{x}{\sqrt{a}}\sqrt{a} + \frac{y}{\sqrt{b}}\sqrt{b} +\frac{z}{\sqrt{c}}\sqrt{c}\right)^2\\
&\le \left(\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c}\right)(a+b+c).
\end{align*}
That is,
\begin{align*}
\frac{x^2}{a} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| \begin{align}
\int \frac{x^2+4}{x^2+6x+10}dx&=\int \frac{x^2+6x+10-(6x+18)+12}{x^2+6x+10}dx\\
&=\int1dx-3\int \frac{2x+6}{x^2+6x+10}dx+12\int \frac{1}{x^2+6x+10}dx\\
&=x-3\ln(x^2+6x+10)+12\int \frac{1}{(x+3)^2+1}dx\\
&=x-3\ln(x^2+6x+10)+12\int \frac{\sec^2u}{\tan^2u+1}du\\
&=x-3\ln(x^2+6x+10)+12\int 1du\\
&=x-3\ln(x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem :
Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$
My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8... | Using formulas $\sin2a = 2\sin a\cos a,\quad \cos a \cos b = \frac12(\cos(a+b)-\cos(a-b))$, have:
$$\dfrac{\sin 8x}{\sin x} = 8\cos 4x\cos 2x\cos x = 4\cos4x(\cos3x+\cos x) = 2(\cos7x+\cos 5x+\cos 3x +\cos x),$$so
\begin{align}
\dfrac {105}{19}\int_0^\limits\dfrac\pi2\dfrac{\sin 8x}{\sin x} &= \dfrac {210}{19}\int_0^\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
| For $b \in \{0,\ldots,7\}$, consider the function $G(b) := \sum_{n \geq 1} \frac{e^{-2\pi i nb/8}}{n^3}$. Observe that
\begin{equation*}
G(b) = \sum_{0 \leq a \leq 7} e^{-2\pi i ab/8} \sum_{n \geq 1} \frac{1}{(8n-a)^3}.
\end{equation*}
Now, we know that if $a,a' \in \{0,\ldots,7\}$ then
\begin{equation*}
\frac{1}{8}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series represen... | Starting with $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, the sum simplifies to
\begin{align*}
z=\frac{1}{2}\sum_{i=1}^\infty \frac{i(i+1)}{i!}&=\frac{1}{2}\sum_{i=1}^\infty \frac{i+1}{(i-1)!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\frac{i+2}{i!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\left(\frac{i}{i!}+\frac{2}{i!}\right)\\
&=\frac{1}{2}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ Wanting to calculate the integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ it will certainly already known to many of you that an interesting way to attack it is to refer to the method of integration and differentiation with respec... | I thought it might be instructive to present two approaches that begin with the Feyman "trick" for differentiating under the integral. We write
$$I(a)=\int_0^1\frac{\arctan (ax)}{x\sqrt{1-x^2}}\,dx$$
Then, we differentiate with $I(a)$ to find
$$I'(a)=\int_0^1\frac{1}{(1+x^2a^2)\sqrt{1-x^2}}\,dx$$
can be evaluated by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 0
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For all $n >0$, $ \left(1+\frac{1}{n} \right)^n = 1+ \sum_{k=1}^n\bigl[ \frac{1}{k!} \prod_{r=0}^{k-1}(1-\frac{r}{n}) \bigr]$ I am working through some problems on induction and I have been stuck on this one for a while. If anyone has any hints. I can show it is true for the $n=1$ and $n=2$ case but I am having difficu... | While the following development is not an induction proof, I thought it would be instructive to present a direct proof of the equality of interest. To that end, we have from the binomial theorem
$$\begin{align}
\left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\\\\
&=\sum_{k=0}^n\frac{n!}{k!(n-k)!n^k}\\\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Modular arithmatic $x^a \mod N= a\cdot b$ $x^3 \equiv 2 \pmod{15}$
How to solve it mod $3$ and mod $5$?
And how does the Chinese remainder theorem help?
I want a general method to follow in case the modulus is composite number.
Thanks in advance!
| $x^3\equiv 2\pmod{15}\Rightarrow \begin{cases}x^3\equiv 2\pmod{3}\\x^3\equiv 2\pmod{5}\end{cases}$
It is clear that if $x\equiv 0$ or $x\equiv 1\pmod{3}$ that $x^3\not\equiv 2\pmod{3}$ so we know that $x\equiv 2\pmod{3}$. Checking confirms this since $2^3=8\equiv 2\pmod{3}$
For the second implication, it should be cle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$
Find $\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$
I tried $\displaystyle \lim_{x \to \infty} \frac{10+3^x}{20-3^x} =\lim_{x \to \infty} \frac{\frac{d}{dx}10+3^x}{\frac{d}{dx}20-3^x}= -\lim_{x \to \infty}\frac{3^x \log{3}}{3^x \log{3}} = -1.$
However, this i... | Notice, $$\lim_{x\to \infty}\frac{10+3^x}{20+3^x}$$
$$=\lim_{x\to \infty}\frac{3^x\left(\frac{10}{3^x}+1\right)}{3^x\left(\frac{20}{3^x}-1\right)}$$
$$=\lim_{x\to \infty}\frac{\frac{10}{3^x}+1}{\frac{20}{3^x}-1}$$
$$=\frac{0+1}{0-1}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
\ge 1$.
My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of ... | It is not true that $a^2\geq a$ for all $a\geq0$ (e.g. for $a=\frac12$), nor that $3b^2\geq b$ or $5c^2\geq c$.
Proof
Because $c\geq a$, it suffices to prove $3a^2+3b^2+3c^2=1$. This follows from AM-QM: $\frac{a^2+b^2+c^2}3\geq\left(\frac{a+b+c}3\right)^2$.
Or by using $3x^2\geq2x-\frac13$ (indeed, this is equivalen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$
Do you know if there are other integer solutions to
$$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$
besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?
... | I found that, through solving the equations, we will have following (if a>b):
$$a=\frac{cd+\sqrt{c^2d^2-4(c+d)}}{2}$$
and
$$b=\frac{cd-\sqrt{c^2d^2-4(c+d)}}{2}$$
I think it will be useful for you to find all of the solutions.
Only you should find the solutions for:
$$c^2d^2-4(c+d)=m^2$$
where $m$ is any positive integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
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If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| We show that except for the case $a=-1$, $b=0$, there always is a $t$ satisfying the conditions of the OP. The calculation is geometric. A similar calculation is of importance at the beginning of the theory of elliptic curves.
Let $P=(a,b)\ne(-1,0)$ be on the unit circle $x^2+y^2=1$. Suppose that the line joining $(-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0}\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$
I tried L'hopital, but the denominator gets more and more complicated.
How does one calculate this limit?
| Recall that $\cos(2t) = 1-2\sin^2(t)$. Hence, we have
$$\cos(2x^2)-1 = -2\sin^2(x^2)$$
Hence, we have
$$\lim_{x \to 0} \dfrac{\cos(2x^2)-1}{x^2\sin(x^2)} = \lim_{x \to 0} \dfrac{-2\sin^2(x^2)}{x^2\sin(x^2)} = -2 \lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
limit as n goes to infinity of $\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$
How do you go about solving $$\lim_{n\to\infty}\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$$
I know that I have to fix the top so that it is not $(\infty - \infty$),
but if I multiple it by
$$\frac{\sqrt{n^3-3}+\sqrt{n^3+2n^2+... | This is a straightforward calculation: Expanding the fraction by $\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}$ gives
\begin{align*}
&\, \frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}} \\
=&\, \frac{(\sqrt{n^3-3}-\sqrt{n^3+2n^2+3)}(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})}{\sqrt{n+2} (\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})} \\
=&\, \frac{(n^3-3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\... | $$y-4=z(x-7)\tag{1}$$
is an equation of all the lines that pass $(7,4)$ with slope $z$. When $z$ is at its minimum or maximum then the line touches the unit circle. Equation of tangent lines with slope $z$ to a unit circle is
$$y=zx\pm\sqrt{z^2+1}\tag{2}$$
Comparing $(1)$ and $(2)$,
$$-7z+4=\pm\sqrt{z^2+1}$$
$$(7z-4)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to properly set up partial fractions for repeated denominator factors I was just trying to solve a problem that had the following item which I needed to split into separate generating functions:
$$\frac{x}{(1-2x)^2(1-5x)}$$
I had assumed I needed to split it into:
$$\frac{A}{1-2x} + \frac{B}{1-2x} + \frac{C}{1-5x}$... | Consider the simplest of cases. $$\frac{\xi}{x^3}$$ Assuming $\xi$ is some polynomial, one could carry on long division to determine the quotient and the remainder. Thus one could get $$\frac{\xi}{x^3}=q(x) + \frac{r(x)}{x^3}$$ Now what can we say for certain about $r(x)$ ? We can say that it is the remainder thus of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
integrate $\int \cos^{4}x\sin^{4}xdx$
$$\int \cos^4x\sin^4xdx$$
How should I approach this?
I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$
| we now that
$$ \sin 2x=2\cos x\sin x$$
$$\cos x\sin x=\frac{\sin 2x}{2}$$
so...
$$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$
$$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it ... | Rewrite, $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$
as $n\left(\left(1+\frac{1}{n^{\frac{5}{2}}} \right)^{\frac{1}{3}} -\left(1-\frac{1}{n^3} \right)^{\frac{1}{3}} \right)\times \sqrt{3}n^{\frac{3}{2}}\left(1+\frac{1}{3n^3}\right)^{\frac{1}{2}}$
Now use the Binomial theorem for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Prove the inequality $\tan{\frac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\frac{\pi\cos{x}}{4\cos{\alpha}}} > 1$
Prove the inequality $$\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} > 1$$
for any $x, \alpha$ with $0 \leq x \leq \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$... | Since $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$, we have $\sin \alpha > \dfrac{1}{2}$ and $\cos \alpha > \dfrac{1}{2}$.
Since $0 \le x \le \dfrac{\pi}{2}$, we have $0 \le \sin x \le 1$ and $0 \le \cos x \le 1$.
Hence, $0 \le \dfrac{\pi\sin x}{4\sin \alpha} < \dfrac{\pi}{2}$ and $0 \le \dfrac{\pi\cos x}{4\cos \alpha} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why is $\sqrt {12} = 2 \sqrt 3$? Why $\sqrt {12} = 2 \sqrt 3$? It is obvious? If we considered the function $f(s) = s^2 $ it is injective on positive numbers so we obtain the conclusion. But in the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots.
| Let's assume (unless that's the part you want proven) that there is only 1 positive number $x$ with the property $x^2=12$. We'll denote that number by $\sqrt{12}$. Likewise let $\sqrt{3}$ be the positive $y$ such that $y^2=3$.
We know that $\Bbb R$ is a field and thus multiplication is commutative. We also know that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Is there an easy way to compute the determinant of matrix with 1's on diagonal and a's on anti-diagonal? \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Thanks
| We have the matrix
$$A= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$
Elininating the $a$s below the diagonal by adding multiples of the first, second and third line of $A$, we obtain the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Probability that $2^a+3^b+5^c$ is divisible by 4
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.
For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=... | Let's consider the remainders when dividing by $4$. If $a=1$ then $2^a=2$ has the remainder $2$. Otherwise, $2^a$ has the remainder zero. If $b$ is odd then $3^b$ has remainder $3$, but if $b$ is even $3^b$ has remainder $1$. Whatever $c$ is, $5^c$ will have remainder $1$. (All of these statements are provable by induc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Find prob. that only select red balls from $n$ (red+blue) balls There are 4 blue balls and 6 red balls(total 10 balls). $X$ is a random variable of the number of selected balls(without replacement), in which
$$P(X=1)=0.1$$
$$P(X=2)=0.5$$
$$P(X=3)=0.2$$
$$P(X=4)=0.1$$
$$P(X=10)=0.1$$
Then, what is probability of only se... | Let $A = \{\text{Choose only red}\}.$ Let $X$ be the number of balls drawn, and let $R$ be the number of red balls drawn. Then
\begin{align*}
P(A) &=\sum_{k = 0}^{6} P(R = k|X = k)P(X = k)\tag 1\\
&=\sum_{k = 0}^{6} \frac{\binom{6}{k}}{\binom{10}{k}}p_k\\
&=\frac{\binom{6}{1}}{\binom{10}{1}}(.1)+\frac{\binom{6}{2}}{\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Compute this integral (Is there a trick hidden to make it eassier?) I need some tips to compute this integral:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx $$
What I did was express the denominator in the following form:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx = \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{8x... | $$I=\int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx$$
Let $u=\displaystyle\frac{\sqrt{x^2-1}}{\sqrt{9x^2-1}}$, which means that $\displaystyle\frac{du}{dx}=\frac{8x}{(x^2-1)^\frac{1}{2}(9x^2-1)^\frac{3}{2}}$
$$\Rightarrow I=\int\,\dfrac{u}{x^5}\,\frac{dx}{du}du$$
$$=\frac{1}{8}\int\,\dfrac{1}{x^6}\frac{\sqrt{x^2-1}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea h... | $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{4^2}\sqrt{2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{16}\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{16\cdot2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{6^2}-\sqrt{32}}+\sqrt{2}=$$
$$\sqrt{\sqrt{36}-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{4\cdot9}-\sqrt{4\cdot8}}+\sqrt{2}=$$
$$\sqrt{2\sqrt{9}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfra... | Note that, $0=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}=\ln\dfrac{a+b}{a+c}+\ln\dfrac{b+c}{b+a}+\ln\dfrac{c+a}{c+b}$ and taking $\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$ WLOG gives us $ac\ge b^2$ ,$\;$ $a^2\ge bc$ $\;$ and $\;$ $ab\ge c^2$ also because $(a+c)^2=a^2+c^2+ac+ac\ge \dfrac{(a+c)^2}{2}+b^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Solution to $p^3-p+1=a^2$ What are the solutions to $p^3-p+1=a^2$ where $p$ is prime and $a$ is natural?
I found the solutions:
$p=3$ and $a=5$
$p=5$ and $a=11$
and one solution when $p$ is not a prime:
$p=56$ and $a=419$, i think
I don't know what to do else. I tried to find limits for $a$ or limits for $p$, I tried t... | If the equation is true, than $a^2 \equiv 1 \pmod p$. This implies that $a \equiv 1$ or $-1 \pmod p$, since $p$ is prime.
If $a \equiv 1 \pmod p$, then $a=pk+1$. This implies that $p^2-pk^2-(2k+1)=0$. This implies that $2k+1$ is divisible by $p$. $k=\frac{pb-1}{2}$ for some natural number $b$. Putting this into our e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 ... | Hint: a common series that is used for computing log of any real number is
$$
\log\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\dots\right)
$$
$u=\frac{1+x}{1-x}\iff x=\frac{u-1}{u+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| Using geometric series
Consider $S_k = 1 + 2\sum_{i=0}^{k-1}4^i$. It follows that $S_k$ is in $\mathbb{N}^{+}$. However,
$$S_k = 1 + 2\cdot\frac{4^k - 1}{4 - 1} = \frac{2^{2k+1}+1}{3}$$
QED
Using recurrence
For $k\in\mathbb{N}$, consider the recurrence $J_{k+1} = 4J_{k} - 1$, with $J_0 = 1$.
Since $J_0 = 1$, it follows... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 3
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To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin... | Your given matrix is : $\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
Determinant of the given matrix is :
$\implies b^2c^2[ca^2 + abc - abc + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$
$\implies b^2c^2[ca^2 + a^2b] - bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left ... | $$\sum_{i=1}^n (i-1)^2$$
$$=\sum_{i=1}^n (i^2-2i+1)$$
$$=\sum_{i=1}^n i^2-2\sum_{i=1}^ni+\sum_{i=1}^n1$$
$$=\frac{n(n+1)(2n+1)}{6}-2\cdot \frac{n(n+1)}{2}+n$$
$$=\frac{(n^2+n)(2n+1)}{6}-n(n+1)+n$$
$$=\frac{2n^3+3n^2+n}{6}-n^2$$
$$=\frac{2n^3-3n^2+n}{6}$$
$$=\frac{n(n-1)(2n-1)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Over an integral arising from Kepler's problem [also: generally useful integral, NOT DUPLICATE!] This post might appear as a duplicate of the following:
Over an integral arising from Kepler's problem [also: generally useful integral]
So recalling quickly:
$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\the... | There is an elementary path:
$$I = \frac{1}{2\pi}\int_0^{2\pi}\frac{\sin^2\theta}{(1+\varepsilon\cos\theta)^2}\,d\theta =\frac{1}{2\pi\varepsilon}\int_0^{2\pi}\sin\theta\,d\left(\dfrac{1}{1+\varepsilon\cos\theta}\right)$$
By parts:
$$ I = \frac{1}{2\pi\varepsilon}\dfrac{\sin\theta}{1+\varepsilon\cos\theta}\,\biggr|_0^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.
| Hint:
$4x^4+25$ factors as the product of two irreducible quadratic polynomials:
$$4x^4+25=(2x^2+5)^2-20x^2=(2x^2-2\sqrt 5x+5)(2x^2+2\sqrt 5x+5,$$
whence the partial fractions decomposition:
$$\frac{x^2}{4x^4+25}=\frac{Ax+B}{2x^2-2\sqrt 5x+5}+\frac{Cx+D}{2x^2+2\sqrt 5x+5}.$$
As $\dfrac{x^2}{4x^4+25}$ is an even functio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I evaluate this series? How do I evaluate this series:
\begin{equation}
\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots
\end{equation}
I wanted to use the Comparison test to show convergence, but I didn't know what to compare it to s... | Here are hints to show that the series converges. (Actually computing the value will take more work.)
Hint1: The series is $\frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots$.
Hint2: One attempt is to cancel the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving trig identity using De Moivre's Theorem
Question: Prove $$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x) $$ by using De'Moivres Theorem
So far (learning complex numbers at the moment) that De Moivre's theorem states that
if $z$ $=$ $r\text{cis}(\theta)$ then $z^n = r^n\text{cis}(n\theta)$
so with this question I ... | The solution can be completed in this manner:
We know, by De-Moivre's Theorem,
$$(\cos x + i \sin x)^3=\cos 3x + i \sin 3x$$
Therefore, we can write
$$\cos^3 x + 3i\cos^2x\sin x + 3i^2\cos x\sin^2 x + i^3 \sin^3 x=\cos 3x + i \sin 3x$$
or, $$\cos^3 x + 3i\cos^2x\sin x - 3\cos x\sin^2 x - i \sin^3 x=\cos 3x + i \sin 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving the equation $\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$ for $x$ I have the following equation:
$$\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$$
I know that $x=2$, but I don't know steps to ... | Take lcm & simplify as follows $$\frac{x+\sqrt 3}{\sqrt x+\sqrt{x+\sqrt 3}}+\frac{x-\sqrt 3}{\sqrt x-\sqrt{x-\sqrt 3}}=\sqrt x$$
$$\frac{(x+\sqrt 3)(\sqrt x-\sqrt{x-\sqrt 3})+(x-\sqrt 3)(\sqrt x+\sqrt{x+\sqrt 3})}{(\sqrt x+\sqrt{x+\sqrt 3})(\sqrt x-\sqrt{x-\sqrt 3})}=\sqrt x$$
$$2x\sqrt x-x\sqrt{x-\sqrt 3}-\sqrt 3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit $\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$ I have to calculate the following limit:
$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$
I believe the limit equals $1$, and I think I can prove it with... | This is the final answer I got, thanks to all the help:
For every $n>0$,$\frac{n}{\sqrt{n^2+n}}\le \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right) \le \frac{1}{\sqrt{n^2+n}}$
Using the squeeze theorem, we calculate the middle expression's limit.
$\lim_\limits{n \to \infty} \frac{n}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac... | Hint: You need to use the Taylor series expansion for the square root as well as the cosine. Expanding $\sqrt{1-u}$ around $u=0$ we have that $$\sqrt{1-u}=1-\frac{u}{2}-\frac{u^2}{8}+O(u^3)$$ and so as $x\rightarrow \infty$ $$x^2\sqrt{1-\frac{1}{x}}=x^2-\frac{x}{2}-\frac{1}{8}+O\left(\frac{1}{x}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\s... | $\dfrac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = \sqrt{\dfrac{x+\sqrt{x+\sqrt{x}}}{x^2}}\\
= \sqrt{\dfrac{1}{x} + \dfrac{\sqrt{x+\sqrt{x}}}{x^2}}
= \sqrt{\dfrac{1}{x} + \sqrt{\dfrac{1}{x^3}+\sqrt{\dfrac{1}{x^7}}}}$
then the limit when $x \rightarrow \infty$ is clearly $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^... | See one root is complex so other os obviously its conjugate as $b^2-4ac<0$ so the equations have both roots in common and as $a,b,c$ have no common factor thus $min(a+b+c)=1+3+5=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Working out expression values What is the value of this expression?
$$\frac{1}{\dfrac{1}{\frac{1}{2} + \frac{1}{3}} + \dfrac{1}{\frac{1}{4} + \frac{1}{5}}}$$
I thought I'd start by working out 1/2 + 1/3,which is 5/6, and then working out 1/4 + 1/5, which is 9/20. Then I added them together, which is 1 17/60. But how d... | As you've noticed, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ and $\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$. Therefore:
$$\begin{align}
\cfrac{1}{
\cfrac{1}{ \cfrac{1}{2}+\cfrac{1}{3} } + \cfrac{1}{ \cfrac{1}{4}+\cfrac{1}{5} }
}
& =\cfrac{1}{ \cfrac{1}{ \cfrac{5}{6} }+\cfrac{1}{ \cfrac{9}{20} } }
\\[2ex] & =\cfrac{1}{\cfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation. I have a question.
Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$
I don't know how to proceed. Any help would be appreciated.
| Just to be different.
If $\alpha$ is a solution of $ax^2 + bx + c = 0$ Then
$a\alpha^2 + b\alpha + c = 0$
So
$\quad \alpha^2 = -\dfrac{b\alpha + c}{a}$
$\quad \alpha^3 = -\dfrac{b\alpha^2 + c\alpha}{a}
= -\dfrac{b\left( -\dfrac{b\alpha + c}{a} \right)+ c\alpha}{a}
= -\dfrac{ -b^2\alpha -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Can I use GAP to show block structures in a multiplication group clearer $\ $? The usual output of GAP for the multiplication table of the group $S3$ is
$$\pmatrix{1&2&3&4&5&6\\2&1&4&3&6&5\\3&6&5&2&1&4\\4&5&6&1&2&3\\5&4&1&6&3&2\\6&3&2&5&4&1}$$
The table
$$\pmatrix{1&2&3&4&5&6\\2&3&1&6&4&5\\3&1&2&5&6&4\\4&5&6&1&2&3\\5&6... | Let N be a normal subgroup of G. Then
MultiplicationTable(Flat(List(RightCosets(G, N),i->List(i))));
produces a table with block structure.
For example:
gap> G:=QuaternionGroup(8);
<pc group of size 8 with 3 generators>
gap> N:=NormalSubgroups(G)[5]; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can you find the maximum or minimum of an equation without calculus? Without using calculus is it possible to find provably and exactly the maximum value
or the minimum value of a quadratic equation
$$ y:=ax^2+bx+c $$
(and also without completing the square)?
I'd love to know the answer.
| This is almost the same as completing the square but .. for giggles.
If $a = 0$ we know $y = xb + c$ will get "extreme" and "extreme" positive and negative values of $x$ so no max or minimum is possible.
So if $ax^2 + bx + c = a(x^2 + x b/a)+c := a(x^2 + b'x) + c$ So finding the max/min is simply a matter of finding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find a thousand natural numbers such that their sum equals their product The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :
I worked on this question for lesser cases :
\begin{align}
&2 \times 2 = 2 + 2\\
&2 \times 3 \times 1 = 2 + 3 + 1\\
&3 \times 3 \time... | Do we have an alternative to fill up with ones?
Let $N=1000$, then
\begin{align}
\sum_{i=1}^N n_i &= \prod_{i=1}^N n_i \iff \\
N n_a &= n_g^N
\end{align}
where $n_a$ is the arithmetic mean and $n_g$ the geometric mean of the numbers $n_i$.
Those numbers on the left and right hand side of the equation
drift apart very ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 2
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Find the value of $x$ which is correct I have one exercise which is $$(x+2013)(x+2014)(x+2015)(x+2016)+1=0$$
I tag $A=x+2013$ or other for many ways but still can not find the first $x$ value. please help.
| $a(a+1)(a+2)(a+3)+1=0$ thus $(a^2+3a)(a^2+3a+2)+1=0$ thus let $a^2+3a=x$ thus $b^2+2b+1=0 so b=-1,-1$ thus $a^2+3a+1=0$ thus $a=\frac{-3\pm(\sqrt{5})}{2}$ and accordingly $x$ will be approximated
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637361",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Image of a family of circles under $w = 1/z$ Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.
However, I am having difficu... | Hint: This transformation can be seen as successive transformations $$Z = \frac{z}{|z|^2} , \,\,\,w = \overline {Z}$$
since $z \cdot \overline{z} = |z|^2$.
Then $z = \frac{1}{w} = \frac{\overline w}{|w|^2}$ and $$x = \frac{u}{u^2 + v^2} \,\,\,\text{and} \,\,\,\,y = \frac{-v}{u^2+v^2}$$
where $w = u + iv$.
Edit: Now s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$ Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?
| Note: As pointed out by Hans, this answer uses the fact that $\zeta(2) = \dfrac{\pi ^2}{6}$ and therefore isn't what the OP asked for.
$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3}...$$
$$\int\frac{\ln(1+x)}{x}dx = x - \frac{x^2}{4} + \frac{x^3}{9}...+C$$
$$\int\limits^{0}_{1}\frac{\ln(1+x)}{x}dx = -1 + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$27$ balls into $3$ cells Spreading $15$ white balls and $12$ black balls into $3$ cells, each of which can contain any number of balls.
$(A.)$ Find the probability that in each cell there will be exactly $5$ white balls.
$(B.)$ Find the probability that in each cell there will be exactly $5$ white balls and exactly $4... | $(A)$ You distributed all of the balls in the denominator but only distributed the white balls in the numerator. $|A|$ should be
$$
\binom{15}{5}\binom{10}{5}\binom{5}{5}3^{12}
$$
$(B)$ There is a similar error.
$(C)$ If you take away $3$ white balls, $12$ remain not $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$? I have tried possibly all approaches.
I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ a... | $$\tan20^\circ-\tan60^\circ=-\dfrac{\sin(60-20)^\circ}{\cos20^\circ\cdot\cos60^\circ}=-\dfrac{2\sin40^\circ}{\cos20^\circ}$$
$$\tan40^\circ+\tan80^\circ=\dfrac{\sin(40+80)^\circ}{\cos40^\circ\cos80^\circ}$$
Adding $(1),(2)$
$$\dfrac{\sin120^\circ}{\cos40^\circ\cos80^\circ}-\dfrac{2\sin40^\circ}{\cos20^\circ} =\dfrac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + ... | Base case:
For $n=2$,
$$LHS=(1-\frac{1}{4})=\frac{3}{4}$$
$$RHS=\frac{(2+1)}{2\cdot 2}=\frac{3}{4}$$
Hence for $n=2$, the equality holds.
Induction hypothesis:
Let it hold for some $n=k$.
Then,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right) = \frac{k + 1}{2k}$$
For $n=k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$ - show that $a_n$ is convergent sequence Problem:
Show that $a_n$ is convergent sequence and find a limit of $a_n$.
$$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$$
I tried to look at this as normal limit problem so I wrote this:
$$\lim_{n \to... | You can invert the function $y = {x \over \sqrt{x^2 + 1}}$ as follows.
$$y^2 = {x^2 \over x^2 + 1} = 1 - {1 \over x^2 + 1}$$
$$1 - y^2 = {1 \over x^2 + 1}$$
$${1 \over 1 - y^2} = x^2 + 1$$
$${1 \over 1 - y^2} - 1 = x^2$$
So we have
$$x^2 = {y^2 \over 1 - y^2}$$
Seeing that $x$ and $y$ must have the same sign, we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$ I am trying to re-learn some basic math and I realize I have forgotten most of it.
Evaluate $$\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$$
Call the terms $S_n$ and the total sum $S$.
$$S_n < \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1} = S < \infty... | Notice, use partial fractions as follows $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$$$=\sum_{n=1}^{\infty}\frac{n}{(n^2-n+1)(n^2+n+1)}$$
$$=\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$
$$=\frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Solving Trigonometric Equation.
Solve for $\theta$ $[0°<\theta<180°]$
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
My solution is here:
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
After using the transformation formula, I got
$$\sin3\theta\cos\theta=\cos3\theta\cos\theta.$$
I could not procee... | \begin{array}{rcl}
\sin 2\theta+\sin 4\theta &=& \cos \theta+\cos 3\theta \\
2\sin 3\theta \cos \theta &=& 2\cos 2\theta \cos \theta \\
\cos \theta \, (\sin 3\theta-\cos 2\theta) &=& 0 \\
\cos \theta \, (3\sin \theta-4\sin^{3} \theta+2\sin^{2} \theta-1)
&=& 0 \\
\cos \theta \, (1-\sin \theta)(4\sin^{2} \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from... | Separate the integral into two integrals:
$\displaystyle\int\frac{u}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} + \displaystyle\int\frac{\frac{3}{2}}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}}$
Use another $u$-substitution on the first integral and then use trig substitution on the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve without involving hyperbolic function. How to solve this integral without involving hyperbolic functions?
$$\int \frac{1}{4-5\sin^2 x}dx$$
The answer is $\frac{1}{4}(\ln (\sin x+2 \cos x)-\ln(2\cos x-\sin x))+c$
| \begin{align}I&= \int \frac{\sec^2 x}{4\sec^2 x -5\tan^2 x}\, \mathrm dx \\&= \int\frac{\mathrm du}{4 - u^2} \,\,\,\;\;\; [\textrm{substituting} \;\tan x = u\; \textrm{and using the differential}\; \mathrm du = \sec^2 x\,\mathrm dx \,.]\\& = \frac{1}{4} \ln \left|\frac{2+ u}{2-u}\right| +\rm C\,\,\,\;\;\; \left[\textrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac... | Given $a+b+c=3$ and $a^2+b^2+c^2 =5$ and $a^3+b^3+c^3=7$
Using $$ab+bc+ca = \frac{1}{2}\left[(a+b+c)^2-(a^2+b^2+c^2)\right] = 2$$
and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]=9$$
So $$7-3abc=9\Rightarrow abc=-\frac{2}{3}$$
Now Let $(t-a)\;,(t-b)\;,(t-c)$ be the root of cubic equation in terms of $t\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to g... | Let $f(x)=x^5-x^3+x-2$.
*
*First step. Differenciate the polynomial to obtain:
$$f'(x)=5x^4-3x^2+1$$
which has no zeros. Hence, $f$ has exactly one root.
*Second step. We have that $f(1)=-3<0$ and $f(2)=24>0$. Then the root lies in the interval $(1,2)$.
*Third step. Let $a$ be the root of $f$. We have
$$a^6=a\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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.