Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$. Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
So far I have done the following proof, but I am unsure how to complete... | You can factor $(x^2-1)=(x+1)(x-1)$. Since $n$ is prime, $n\mid (x^2-1)$ implies $n\mid x+1$ or $n\mid x-1$. So $x\equiv 1$ mod $n$ or $x\equiv -1$ mod $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of... | If you know $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ then you could simply say $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} $$ $$\lt \frac{\pi^2}{6} - \frac{1}{1^2}- \frac{1}{4^2}- \frac{1}{6^2}- \frac{1}{8^2}- \frac{1}{9^2}- \frac{1}{10^2}- \frac{1}{12^2}- \frac{1}{14^2}- \frac{1}{15^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\sin(4x) = -2\sin(2x)$ solutions in $[0,2\pi)$? My textbook gives the following answer:
$\sin(4x) + 2\sin(2x) = 0$
$2\sin(2x)\cos(2x) + 2\sin(2x) = 0$
$2\sin2x (\cos(2x) + 1) = 0$
$2\sin 2x = 0$
$\sin2x = 0$
$2x = πk$
$x = kπ/2$
In the interval $[0,2π)$ you have the solutions $0,π/2,π$ and $3π/2$. The book then... | Use the following identities
$$\sin(2 x) \equiv 2 \sin(x) \cos(x) $$
$$\cos(2 x) \equiv 2 \cos^2(x)-1 $$
$$\sin(4 x) \equiv \sin(2 x) \cos(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) $$
$$\sin(4 x) +2 \sin(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) + 4 \sin(x) \cos(x) $$
which simplifies to
$$ 8 \sin(x) \cos^3(x) ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following linear congruences For any integers $a, b$, let $N_{a,b}$ denote the number of
positive integers $x<1000$ satisfying $x\equiv a\pmod{27}$ and $x\equiv b\pmod{37}$. Then find $N_{a,b}$.
progress: $x\equiv a\pmod{27}$ and $x\equiv b\pmod{37}$ implies
$x=27m+a$ and $x=37n+b$ for some integers $m,n$.... | The numbers $u$ and $v$ can be found with the extended euclidean algorithm
$37=27\times 1+10$
$27=2\times 10+7$
$10=7\times 1+3$
$7=3\times 2+1$
Giving $1=7-3\times 2=7-(10-7)\times 2=3\times 7-2\times 10=3(27-2\times 10)-2\times 10=3\times 27-8\times 10=3\times 27-8\times (37-27)=11\times 27-8\times 37$
This gives the... | {
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"timestamp": "2023-03-29T00:00:00",
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Comparing $n^n$ and $n!^2$ I tried to prove that if $n>2$ then $(n!)^2>n^n$ but did not managed. That is the trick to compare those as both grows rapidly? Induction seems hard: $((n+1)!)^2=(n+1)^2(n!)^2>(n+1)^2n^n$ but why $(n+1)^2n^n>(n+1)^{n+1}$? I also noted that $n^n=e^{n\ln n}$ and $n!^2=e^{2\ln n!}=e^{2\sum_{i=1}... | Suppose
$(n!)^2 > n^n$.
Then
$((n+1)!)^2
=(n!)^2(n+1)^2
>n^n(n+1)^2
$
so we need
$n^n(n+1)^2
\ge (n+1)^{n+1}
$
which is the same as
$n^n
\ge (n+1)^{n-1}
$
or,
dividing by
$n^{n-1}$,
$n
\ge (1+1/n)^{n-1}
$.
Multiplying by
$1+1/n$,
this becomes
$n+1
\ge (1+1/n)^n
$.
This is true because
$(1+1/n)^n
< e
$,
as has been sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | Try to cancel 0/0 with different numbers each time you will get different solutions for 0/0 . Suppose 1×0/1×0 you get solution 1. Suppose 2×0/0×1 you get solution 2 . Suppose n×0/0×1 you get solution n. ∵0/0 has infinitely many solutions 0/0 is not defined.
| {
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"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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"answer_id": 13
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Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$).
So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{... | Yes.
Another way of doing this would be:
$$\int_{7\pi/6}^{11\pi/6} \int_{0}^{4 + 8\sin \theta} rdrd\theta$$
$$ \ldots = 16\pi - 24\sqrt{3}$$
If you fancy multiple integrals.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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why the limit of this f(x) when x approach infinity is equal to infinity? why the limit of this f(x) when x approach infinity is equal to infinity?:
$$ \lim_{x\to \infty} \frac{x^2 + 4x + 5}{x-1}$$
$$ \lim_{x\to \infty} \frac{1 + \frac{4}{x} + \frac{5}{x^2}}{\frac{1}{x}-\frac{1}{x^2}} = \infty $$
I know that all the... | $$\lim_{x\to\infty}\left(\frac{x^2+4x+5}{x-1}\right)=$$
The leading term in the denominator of $\frac{x^2+4x+5}{x-1}$ is $x$.
So divide the numerator and denominator by this:
$$\lim_{x\to\infty}\left(\frac{x+4+\frac{5}{x}}{1-\frac{1}{x}}\right)=$$
The expressions $\frac{5}{x}$ and $\frac{1}{x}$ both tend to zero as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me evaluate this infinite sum I have the following problem:
For any positive integer n, let $\langle n \rangle$ denote the integer nearest to $\sqrt n$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n \rangle = k$.
(b) Show that $$\sum_{n=1}^\infty{\frac{2^{\langle n \rang... | The idea is to rewrite the sum as a double sum by observing that $$\langle m^2 + k \rangle = m$$ for $k \in \{-m+1, \ldots, m\}$. Therefore, $$\begin{align*} S &= \sum_{n=1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} \\ &= \sum_{m=1}^\infty \sum_{k=-m+1}^{m} \frac{2^m + 2^{-m}}{2^{m^2+k}} \\ &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b... | we have $$\frac{cos\left(\frac{B}{2}\right)}{cos\left(\frac{C}{2}\right)}=\frac{ab+bc}{ac+bc}$$
Applying Componendo and Dividendo we get
$$\frac{cos\left(\frac{B}{2}\right)-cos\left(\frac{C}{2}\right)}{cos\left(\frac{B}{2}\right)+cos\left(\frac{C}{2}\right)}=\frac{a(b-c)}{ab+ac+bc}$$ $\implies$
$$-tan\left(\frac{B+C}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\7... | Good news: the answer is Yes. Moreover, all integer solutions are given by just one formula - the second one. It covers both primitive and non-primitive solutions. That is, for every solution $(x,y,z)$ to $x^2+y^2=z^k$ there exist rational numbers $a,b$ such that $x=a(a^2+b^2)^{(k-1)/2}$, $y=b(a^2+b^2)^{(k-1)/2}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding a convergent majorant series I have a series
$$
\sum_{n=1}^\infty \left( \frac{1}{n^3} \cos(nt) - \frac{1}{(2n+1)^2} \sin(nt) \right)
$$
and I have to find a majorant series to this series.
The convergent majorant series I was supposed to find is
$$
\sum_{n=1}^\infty \left( \frac{1}{n^3} + \frac{1}{(2n+1)^2} \r... | Hint. You may use
$$
|a+b|\leq |a|+|b|,\qquad a,b \in \mathbb{R},
$$ giving, for $n=1,2,3 \cdots,$
$$
\left| \frac{1}{n^3} \cos(nt) - \frac{1}{(2n+1)^2} \sin(nt) \right|\leq \left| \frac{1}{n^3} \cos(nt)\right|+\left| - \frac{1}{(2n+1)^2} \sin(nt)\right|\leq \frac{1}{n^3} + \frac{1}{(2n+1)^2}
$$ since $|\cos (nt)|\leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hidden Markov Models and Viterbi Algorithm: Fair and Biased Die So following is the problem that I am trying to solve using Viterbi algorithm and HMM:
Before attempting to write a program, I want to do this problem by hand for the first 3 observations($651$). Based on the question, I understand that:
$P(i | Fair) = \f... | Your figures look like you're doing it right. I get the following, skipping the first fictitious state, which is probably not necessary.
Iteration $1$:
\begin{align}
\nu_1(F) &= P(X_1\mid F)P(F) = P(6\mid F)P(F) = \dfrac{1}{6}\dfrac{1}{2} = \dfrac{1}{12} \\
& \\
\nu_1(B) &= P(X_1\mid B)P(B) = P(6\mid B)P(B) = \dfrac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If x is real , the numbers $ 5^{1+x} + 5^{1-x} $ , a/2 , $ 25^x + 25^{-x} $ form an A.P. then to which interval must ' a ' belong to? A) [ 1 , 5 ]
B) [ 2 , 5 ]
C) [ 5 , 12 ]
D) [ 12 , $ \infty $ ]
My progress so far :
Let $ 5^{x} = y $
Since they are in AP so ,
=> $ 5y $ + $ 5\over y $ + $ y^2 $ + $ 1\over (y^2)... | As they are in AP, we have
$$
a = 5^{1+x} + 5^{1-x} + 25^{x} + 25^{-x}.
$$
So say $5^x = t$, then,
$$
a = 5t + \frac{5}{t} + t^2 + \frac{1}{t^2}.
$$
Now
$$
5t+\frac{5}{t} \geq 10
\qquad\text{and}\qquad
t^2+\frac{1}{t^2} \geq 2
$$
by the AM-GM inequality.
So $a \ge 12$, are the values that $a$ can take.
Hope that helps... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd ... | Assume that $a_n$ can be expressed as $a_{n+1}=Sa_n+Ta_{n-1}$.
Then $a_n$ is expressed as
$$a_n=A\alpha^n+B\beta^n$$
Then
$$a_{n+1}a_{n-1}=(A\alpha^{n+1}+B\beta^{n+1})(A\alpha^{n-1}+B\beta^{n-1})$$
$$=A^2\alpha^{2n}+AB\alpha^{n-1}\beta^{n+1}+AB\alpha^{n+1}\beta^{n-1}+B^2\beta^{2n}$$
And
$$a_n^2+5=A^2\alpha^{2n}+2AB\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the equation of tangent line to the given curve
Find the equation of tangent line to $(x^2+y^2)^{3/2} = 2xy$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$
I just can't figure out how to take the derivative of this function!
| You can use the implicit function theorem, which states that (in particular)
$$F(x,y)=0, \quad \text{and} \quad F_y(x,y) \neq 0 \quad \Rightarrow \quad \frac{dy}{dx}(x,y)=-\frac{F_x(x,y)}{F_y(x,y)}.$$
First define
$$ F(x,y) \equiv (x^2+y^2)^{3/2}-2xy $$
then calculate
\begin{align}
F_x(x,y) &= \frac{3}{2}(x^2+y^2)^{1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to take an integral using half angle trigonometric substitution. So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$
and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this probl... | This substitution is used for integrals involving only trigonometric expressions. This method is very useful as it transforms the trigonometric integral into just rational integral. You should know how to write $\sin x, \cos x, \tan x$ in terms of $\tan \frac{x}{2}$
(try proving)
$\sin x=\dfrac{2\tan \frac{x}{2}}{1+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $abc$.
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.
In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?
| Hint:
Let $[xyz]$ denote a thre-digit number. Then you want $a,b,c$ such that $[abc]\cdot[cba]=396396$, with $a>c$.
Write $[abc]=100\cdot a+100\cdot b+1\cdot c$ and $[cba]=100\cdot c+10\cdot b+1\cdot a$, then
$$[abc]\cdot[cba]=(100\cdot a+100\cdot b+1\cdot c)\cdot(100\cdot c+10\cdot b+1\cdot a)$$
$$=10\ 000\cdot ac+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
| Note that:
$sin(3\theta)=3sin(\theta)-4sin^{3}(\theta)$
Let $\theta =\frac{\pi}{3}$ and let $sin(\frac{\pi}{3})=x$
Then:
$0=x(3-4x^{2})$.
Hence either $x=0$ or $x^{2}=\frac{3}{4}$
But since $sin(0)=0$ and the $sine$ function has a period of $2\pi$ then we must conclude that $sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 6
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Help me with the result of this determinant.. $$
D =
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
2 & 1 & 1 & \dots & 1 & 0 \\
3 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n-1 & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=n*1*(-1)^\frac{n(n-1)}{2}
$$
I do... | To compute
$$\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}$$
You can proceed like that: If we swap the $k$-th row with the $n-k$-th row ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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(Perhaps) An Easy Combinatorics Problem: (Perhaps not so easy...)
Still have some difficulty with problems like this.
Suppose I have Box with various buttons inside. For the sake of example, let the box contain: $$[ 4 \textrm{ Red}, 5 \textrm{ Blue}, 7 \textrm{ Green} ]$$ How many distinct tuples of, say, 5 buttons ca... | The generating function approach is to write:
$$(1+x+x^2+x^3+x^4)(1+x+\cdots + x^5)(1+x+\cdots + x^7) = \frac{(1-x^5)(1-x^6)(1-x^8)}{(1-x)^3}$$
The coefficient of $x^n$ is the number of ways of taking $n$ buttons out.
Now, $$\frac{1}{(1-x)^3} = \sum_{k=0}^\infty \binom{k+2}{2}x^k$$
And: $$(1-x^5)(1-x^6)(1-x^8)=1-x^5-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What are the valors of $4x_1-x_1^2+x_3^2$? $x_1$ and $x_2$ $x_3$ real numbers such that
$x_1<x_2<x_3$ are solutions of the equation :
$x^3-3x^2+(a+2)x-a=0$ where a is real
What are the valors of $4x_1-x_1^2+x_3^2$
After factorizing i get :
$(x-1)(x(x-2)+a)=0$
The problem is that i don't know relation of solutions l... | From what you get $(x−1)(x(x−2)+a)=0$, we can find the other two roots, which satisfy $$x^2-2x+a=(x-1)^2-(1-a)=0$$ Since all roots are real, hence we have $a\leq 1$ ($a\neq1$ since the roots are distinct) then $x_1=1-\sqrt{1-a},x_3=1+\sqrt{1+a}$.
Hence, $$4x_1−x^2_1+x^2_3=4(1-\sqrt{1-a})-(1-\sqrt{1-a})^2+(1+\sqrt{1-a})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570951",
"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
| Let $a \le b \le c$.
If $m \ge 2$ m! is even so if $a = 1$ (odd), $b = 1$ and $c! = 2^n - 2 = 2(2^{n-1} - 1)$ so $c = 2; n=1$ or $c = 3; n = 2$. Two answers so far. (1,1,2,1)(1,1,3,2)
If $a > 1$ then $a! + b! + c! = a!(1 + b!/a! + c!/a!) = 2^n$ If $a \ge 3$ $3|a!$ and so $3|2^n$ which is impossible so $a = 2$.
So $1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find all pairs $(x,y)$ of integers such that $y^2 = x(x+1)(x+2)$? Here $y^2$ is divisible by $12.$ And satisfying all those conditions I think $y=0$ is the only solution. But I can't show it mathematically.
| Assume $y \ne 0$.
If $m | y^2$ $m>2$ then $m$ divides exactly 1 of $x, x + 1, x + 2$ so $m^2$ divides exactly 1 of $x, x+1, x +2$ so each $x, x+1, x + 2 = 2^nm^2$ for some (maybe 0) power of two and some $m$ odd.
Suppose $m, n$ odd and $n^2 = m^2 + 2$.
Then $(n + m)(n - m) =2$. But this has no integer solution.
So $x$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$ Take the following integral over the specified region:
${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$
(i'm answering my own question because I was writing it and then found out my e... | This is the intersection betweet the paraboloid $z = \frac{x²+y²}{3}$ and the sphere $x²+y²+z²=4$. Their intersection forms a circle with radius $\sqrt{3}$. Therefore we just need to find the height of the region by integrating with respect to $z$, from the paraboloid to the sphere:
$$\int_{\frac{x²+y²}{3}}^{\sqrt{4-x²... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can you simplify this term? $$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$
| $$\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}=
\frac{r^2}{r^2}\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}=
\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}=\\
\frac{(1+r)^{T+1}}{(1+r)^{T+1}}\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of $\lim_{x\to 0} (\frac{1+\tan x}{1+\sin x})^{\csc^3x}$ I failed to find the limit of:lim(x->0) $(\frac{1+tan(x)}{1+sinx})^{\frac{1}{sin^3(x)}}$?
as X approches 0
How do I find the answer for this?
Thanks in advance. the answer supposed to be sqr(e). but my answer was 1.
Can anyone please help me find... | $$\\ \lim _{ x\rightarrow 0 }{ { \left( \frac { 1+tan(x) }{ 1+sin\left( x \right) } \right) }^{ \frac { 1 }{ sin^{ 3 }(x) } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \right) }^{ \frac { 1 }{ sin^{ 3 }(x) } } } =\lim _{ x\rightarrow 0 }{ { \le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Symmetry Group $S_3$ in a function I was considering functions $f: \Bbb{C} \rightarrow \Bbb{C}$ and I defined the following instrument (I call it the Symmetry Group of a function)
$$ \text{Sym}(f) = \left< m(x)|f(m(x))=f(x) \right> $$
An intuitive example is to consider $\text{Sym}(e^x)$ and observe that
$$m(x)... | There's a little ambiguity as to the type of functions you are considering. For example, you seem ok with allowing the function to have some singularities at zero as your $z+1/z$ example indicates. The group $S_3$ acts naturally on $\mathbf{C} \cup \{\infty\}$ via the following rational functions:
$$\Sigma = \left\{z, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$ Let $x,y \in \mathbb{R} $ such that;
If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1 \leq y^2 +1 -1 < 2\d... | Usually, before trying to attack this kind of problems with the definition of uniform continuity, it is much easier and much faster to try to prove that the functions involved are Lipschitz-continuous.
In your case,
$$\left| \sqrt{x^2 + 1} - \sqrt {y^2 + 1} \right| = \left| \frac {x^2 - y^2} {\sqrt{x^2 + 1} + \sqrt {y^... | {
"language": "en",
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why $ 1 - \cos^2x = \sin^2x $? I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working:
\begin{array}\\
\lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1... | It's a Pythagorean identity and comes from $$\sin^2 x + \cos ^2 x = 1$$
Just subtract $\cos ^2 x$ from both sides and you have your answer. Now, as to where $\sin^2 x + \cos ^2 x = 1$ comes from:
Let's say you have a right triangle with legs $a$ and $b$. By the Pythagorean theorem, the hypotenuse is $$\sqrt {a^2 + b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\... | Since it is of the form $\frac{\infty}{\infty}$ L'Hopital's rule gives \begin{align}\lim_{x\to \infty}\frac{(\ln{(1+4^x)})'}{(\ln{(1+3^x)})'}&=\frac{\ln 4}{\ln 3}\cdot\lim_{x\to \infty}\frac{4^x(1+3^x)}{3^x(1+4^x)}=\\&=\frac{\ln4}{\ln3}\cdot\lim_{x\to \infty}\frac{12^x\left(\frac1{3^x}+1\right)}{12^x\left(\frac{1}{4^x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1... | Hint:
Write
$$x = \sin X \qquad y = \sin Y \qquad z = \sin Z$$
Where $X + Y + Z = \pi$ (meaning that those are the angles of a triangle).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the volume between two paraboloids Find the volume of the solid enclosed by the paraboloids $z = 1-x^2-y^2$ and $z = -1 + (x-1)^2 + y^2$.
Using triple integrals, it is known that $V = \iiint_R \mathrm dx\,\mathrm dy\,\mathrm dz$, and I will have to change variables. But I can't just say that $r^2 = x^2 + y^2$, bec... | Since $z=1-x^2-y^2$ is a downward paraboloids and $z=-1+(x-1)^2+y^2$ is upward , the limits for $z$ are:
$$
-1+(x-1)^2+y^2 \le z \le 1-x^2-y^2
$$
The projection on the $xy$ plane of the curve of intersection between the two paraboloids is the circumference:
$$
x^2+y^2-x-\frac{1}{2}=0
$$
so the limits for $y$ are:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute $(235432_7 \cdot 2551_7) \pmod{311_7} = N_7 = ?$ This is for my assembly language class. I am finding different answers. My answer was $15_7$. But a friend got 2824. Can someone please explain the correct way to do it if $15_7$ is wrong?
$$(235432_7 \cdot 2551_7) \pmod{311_7} = N_7 = ?$$
| $235432$ in base 7 is $2+3\cdot 7 + 4\cdot 7^2+5\cdot 7^3+3\cdot 7^4+2\cdot 7^5 = 42751$ in base 10
$2551$ in base 7 is $1+5\cdot 7+5\cdot 7^2+2\cdot 7^3 = 967$ in base 10
$311$ in base 7 is $1+1\cdot 7 + 3\cdot 7^2 = 155$ in base 10
$42751\cdot 967\pmod{155}\equiv 12$
$12$ in decimal is $5+1\cdot 7$ is $15$ in base 7
... | {
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"url": "https://math.stackexchange.com/questions/1585848",
"timestamp": "2023-03-29T00:00:00",
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How is this limit being solved? I can't grasp it I am going over limits for my finals as I notice this example in my schoolbook discribing limits of the undefined form $0\over0$ in the shape of an irrational fraction.
$$\lim\limits_{x \to 1} {\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = {0\over0} = \lim\limits_{x \to 1} {(\sqrt... | $${\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = \frac{\sqrt[3]{x}-1}{\sqrt{2x-1}-1}\cdot \frac{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$=\frac{\left((\sqrt[3]{x}-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)\right)(\sqrt{2x-1}+1)}{\left((\sqrt{2x-1})^2-1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586052",
"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\frac{1}{\sin^{2k}(x)}+\frac{1}{\cos^{2k}(x)} = 8$
Solve the equation $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = 8$ where $k$ is an integer and $x$ is a real number.
Attempt
We have that $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = \dfrac{\sin^{2k}(x)+\cos^{2k}(x)}{\cos^{2k}(x)\sin^{... | We have
$$8=\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)} \geq \dfrac{2}{\left\vert\sin^k(x)\cos^k(x)\right\vert} = \dfrac{2^{k+1}}{\left\vert\sin^k(2x)\right\vert} \geq 2^{k+1}$$
Hence, we have $k+1 \leq 3 \implies k \leq 2$.
If $k=0$, we have $\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)}=2$.
If $k < 0$, we have $$\sin^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586233",
"timestamp": "2023-03-29T00:00:00",
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How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$? How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( shou... |
In order to calculate the surface of the cone $\mathcal{C}$
\begin{align*}
\mathcal{C}:x^2+y^2=z^2
\end{align*}
between the planes
\begin{align*}
z=0\qquad\text{ and }\qquad
z=\frac{3-x}{2}\tag{1}
\end{align*}
we consider a parameter representation $\Phi(t,\varphi)$ of $\mathcal{C}$
\begin{align*}
\Phi(t,\va... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove this inequality $\sum \cos{A}\ge\frac{1}{4}(3+\sum\cos{(A-B)})$ Prove that in any triangle $ABC$ the following inequality holds
$$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
And I have gotten
$$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
$$2(\cos{(A... | use
$$\sum\cos{A}=\dfrac{R+r}{R},\sum\cos{A}\cos{B}=\dfrac{s^2+r^2-4R^2}{4R^2},\sum\sin{A}\sin{B}=\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow \dfrac{4R+4r}{R}\ge 3+\dfrac{s^2+r^2-4R^2}{4R^2}+\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow 4R^2+6Rr\ge s^2+r^2$$
use Gerrentsen inequality
$$s^2\le 4R^2+4Rr+3r^2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589245",
"timestamp": "2023-03-29T00:00:00",
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$2^m+3^n$ is a square Determine all pairs ($m,n$) of positive integers such that $2^m+3^n$ is a square.
One such pair is ($4,2$) and I think that there are no other solutions. Am I correct? If I am can somebody please give me a hint to prove it?
| If $m=1$ , then we have $2+3^n$, which cannot be a square because the residue modulo $3$ is $2$.
If $m>1$, then $n$ cannot be odd because the residue of $2^m+3^n$ modulo $4$ would be $3$. So, $2^m$ and $3^n$ must both be squares. So, we need a pythagorean triple $(u,v,w)$ , where $u$ is a power of $3$ and $v$ is a powe... | {
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Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$ for any $p,q$ with $p+q = 1$.
Should I prove this us... | $\Rightarrow$ We will consider triangle $OAB$ wher $A,\ B$ are
vectors on $\mathbb{R}^2$ Define $$a=|A|,\ b=|B|,\ c=|A-B|$$ Then
$$ |A|^2+|B|^2-2A\cdot B=|A-B|^2 $$
Then $pA+qB,\ p+q=1$ is in a line passing through $A,\ B$. Then
\begin{align} |pA+qB|^2&=p^2a^2+q^2b^2+2pqA\cdot B \\&=
p^2a^2+q^2b^2+pq (a^2+b^2-c^2)
\\& ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I se... | First of all note that the RHS of your original equation can be written as $4(x-2)+3(x+1)$. Now transfer the terms on either side of the equation obtaining $$2(x-2)\big[\sqrt {5-x^2}-2\big] = (x+1)\big[3-\sqrt{5+x^2}\big].$$ The LHS vanishes for $\pm 1$ and $2$. The RHS vanishes for $-1$ and $\pm2$. Two of the root ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590132",
"timestamp": "2023-03-29T00:00:00",
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Range of function $ f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ Where $x>0$
Find the range of the function $\displaystyle f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ where $x>0$
$\bf{My\; Try::}$ Let $\sqrt{x}=t\;,$ Then $\displaystyle f(t) = t^3+\frac{1}{t^3}-4\left(t^2... | We have
$$\color{blue}{x\sqrt{x} + \dfrac1{x\sqrt{x}}-4\left(x+\dfrac1x\right) = \underbrace{\dfrac{\left(1+\sqrt{x}\right)^2\left(x-3\sqrt{x}+1\right)^2}{x^{3/2}}}_{\text{Is non-negative}}-10}$$
Hence, the minimum is $-10$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right... | The denominator 'factorises', giving us this:
$$f(x) = {x(x^2-1) \over (x^2-1)^2 + x^2}
$$
We can establish this is less than $1/2$ fairly easily:
$${uv \over {u^2+v^2}} \le {1 \over 2}
\iff u^2 + v^2 - 2uv \ge 0
\iff (u-v)^2 \ge 0
$$
($u,v$ not both 0, but this holds for us.)
And equality is obtained when $u=v$, i.e.... | {
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"question_score": "3",
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Inequality, Cauchy Schwarz and Schur For $a,b, c>0$, prove that $$\frac{a^3}{a^3+b^3+abc}+\frac{b^3}{b^3+c^3+abc}+\frac{c^3}{c^3+a^3+abc}\geq 1$$
I tried the following $$\sum_{cyc}\frac{a^3}{a^3+b^3+abc}\cdot \sum_{cyc}a^3+b^3+abc\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
where cyc stands for cyclic sum of $a,b,c$, then we cou... | Let $x=\dfrac{b}{a},y=\dfrac{c}{b},z=\dfrac{a}{c},\Longrightarrow xyz=1$
then
$$\dfrac{a^3}{a^3+b^3+abc}=\dfrac{1}{1+x^3+\frac{x}{z}}=\dfrac{1}{xyz+x^3+x^2y}=\dfrac{xyz}{xyz+x^3+x^2y}=\dfrac{yz}{yz+x^2+xy}$$
and use Cauchy-Schwarz inequality we have
$$\left(\sum_{cyc}\dfrac{yz}{yz+x^2+xy}\right)\sum_{cyc}yz(yz+x^2+xy)\... | {
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"url": "https://math.stackexchange.com/questions/1591020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k ... | Rewrite your first step as
$$4(k^2 + k + 1) \cdot 4(k^2 + k + 2) = 16 (k^2 + k + 1)(k^2 + k + 2)$$
Now notice that $k^2 + k + 1$ and $k^2 + k + 2$ have opposite parities....
| {
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Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
|
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}dx$$
Hints:
$$(1)\int\bigg({1-\frac{6x+6}{x^2 + 6x +10}}\bigg)dx$$
$(2)$rewrite the integrand
${\frac{x+1}{x^2 + 6x +10}}$ as $\frac{2x+6}{2(x^2+6x+10)}-\frac{2}{x^2 + 6x +10}$
$(3)$ for inegrand $\frac{2x+6}{2(x^2+6x+10)}$ substitute $u=x^2+6x+10$ and $du=(2x+6)dx$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \frac{x^2}{x^2 -6x + 10}\,dx$ Evaluate $$\int \frac{x^2}{x^2 -6x + 10} \, dx$$
I'd love to get a hint how to get rid of that nominator, or make it somehow simpler.
Before posting this, I've looked into: Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$
And I've not understood how they simplied the n... | Hint : $$\int\frac{x^2}{x^2-6x+10}dx=\int1+\frac{6x-10}{x^2-6x+10}dx$$
$\dfrac{d}{dx} \ln{(x^2-6x+10)}=\frac{2x-6}{x^2-6x+10}$ , therefore make
$$\frac{6x-10}{x^2-6x+10}=\frac{3(2x-6)}{x^2-6x+10}+\frac{8}{x^2-6x+10}=3\frac{2x-6}{x^2-6x+10}+8\frac{1}{(x-3)^2+1}$$
$$I=x+3\ln(x^2-6x+10)+8\arctan{(x-3)}+constant$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If a and b are non-negative real numbers then demonstrate inequality If $a$ and $b$ are non-negative real numbers then show that
$$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$
Inequality is less common. To demonstrate the usual instruments (means inequality, .... | Well... Notice that by the AM-GM inequality:
$$ 3a+\frac{4}{a+1}=\left(a+1\right)+\frac{2\left(a^2+1\right)}{a+1}+1 \ge 2\sqrt{2\left(1+a^2\right)}+1 $$
Then using the Cauchy–Bunyakovsky–Schwarz inequality, we get
$$ LHS \ge \left(2\sqrt{2\left(1+a^2\right)}+1+\frac{8}{\sqrt{2\left(1+b^2\right)}}\right)\left(\frac{8}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit of the sequence $T_1=0,T_2=1, T_n=\frac{T_{n-1}+T_{n-2}}{2}$ Given that $T_1=0$, $T_2=1$ and $T_n=\frac{T_{n-1}+T_{n-2}}{2}$, show that the sequence converges to $\frac{2}{3}$.
| I am quite a fan of the following solution: We fist show that the sequence converges for arbitrary two starting values and then derive the limit indirectly from its dependence on the starting values.
Let $a,b \in \mathbb{R}$ and define the sequence $(c_n)_{n \in \mathbb{N}}$ by $c_0 = a$, $c_1 = b$ and $c_{n+2} = (c_n ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A Locus in the complex plane I am having trouble understanding exactly what the following set represents:
The set of all complex numbers $z\neq 5$ such that $\frac{\lvert z -3\rvert }{\lvert z-5\rvert}=\frac{\sqrt2}{2}$
I have a feeling that it represents a circle of some sort, but I do not know how to prove it. I have... | We can rewrite this as
$$|z-3|=\frac{\sqrt 2}2|z-5|$$
which can be interpreted geometrically as the locus of all points that has a distance from the point $3+0i$ that is $\frac{\sqrt 2}2$ times the distance from the point $5+0i$. From geometry we know that is a circle.
If you can solve this geometrically, you can find ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A trigonometric identities with the ratio of four terms like $1+(\frac{\tan x}{\sin y})^2$ Prove:
$$\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}$$
I started by opening the brackets and squaring... | $$\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\dfrac{\sin x}{\tan y}\right)^2}{1+\left(\dfrac{\sin x}{\tan z}\right)^2}$$
$$\iff\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\sin x}{\tan y}\right)^2}=\frac{1+\left(\dfrac{\tan x}{\sin z}\right)^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate the square root of the ratio of two quadratic polynomials $$\int \sqrt{\frac{x^2+x-1}{x^2-1}} dx$$
I have been trying to find this integral for a while and I just can't. Does it even have a closed form?
| Mathematica gives:
$\frac{\sqrt{\frac{x^2+x-1}{x^2-1}} \left((x+1)
\left(x^2+x-1\right)+\frac{\sqrt{\frac{x+1}{1-x}} \sqrt{\frac{2 x+\sqrt{5}+1}{1-x}}
(x-1)^2 \left(\sqrt{\frac{1+\sqrt{5}}{x-1}+2 \left(2+\sqrt{5}\right)} \left(\left(14-6
\sqrt{5}\right) F\left(\sin ^{-1}\left(\sqrt{-2+\sqrt{5}}
\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Is $|9x-1|^3$ differentiable at $1/9$? I think it is, but I'm stuck at showing it. I know that it is equal to:
$$|9x - 1| \cdot |9x - 1|^2.$$
$|9x - 1|$ is not differentiable at $1/9$, and I am pretty sure that $|9x - 1| ^2$ is differentiable there.
| We can look separately at the limits of $\frac{f(x)-f(1/9)}{x-1/9}$ (for $x\neq 1/9$) on both sides of $1/9$: the goal is to see if they exist, and (if so) if they are the same. If this is the case, then $f$ is differentiable at $1/9$, and $f^\prime(1/9)$ equals this common limit.
See below for more details. The hidden... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$
For positive real numbers $a,b,c$ with $abc = 8$ prove that
$$
\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}
+ \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}
+ \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}}
\geq \frac{4}{3}.... | By AM-GM $\sqrt{a^3+1}\leq\frac{a+1+a^2-a+1}{2}=\frac{a^2+2}{2}$.
Hence, it remains to prove that $\sum\limits_{cyc}\frac{a^2}{(a^2+2)(b^2+2)}\geq\frac{1}{3}$, which is
$\sum\limits_{cyc}(a^2b^2+2a^2)\geq72$, which is AM-GM again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| Since $x^2 + y^2 = 1$ draws a circle on the Euclidean plane, we have $(x, y) = (\cos \theta , \sin \theta)$ for some $\theta$. Now let $\tan (\theta/2) = t$, then $\tan \theta = \dfrac{2t}{1-t^2}$, $\cos\theta = \dfrac{1-t^2}{1+t^2}$, $\sin\theta = \dfrac{2t}{1+t^2}$ by trigonometry.
Another way to see this :
Let $(0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometric progression in an inequality Problem: Show that if $a>0$ and $n>3$ is an integer then $$\frac{1+a+a^2 \cdots +a^n}{a^2+a^3+ \cdots a^{n-2}} \geq \frac{n+1}{n-3}$$
I am unable to prove the above the inequality.
I used the geometric progression summation formula to reduce it to proving $\frac{a^{n+1}-1}{a^2(... | In fact, that identity is useful. In details, by AM-GM inequality, $$ \left(n-3\right)\left(1+a^n\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ and $$ \left(n-3\right)\left(a+a^{n-1}\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ Thus, $$ LHS = \frac{1+a+a^{n-1}+a^{n}}{a^2+\cdots+a^{n-2}}+1 \ge \frac{4}{n-3}+1 =RHS $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the no. of non-negative integral solutions to $x+y+2z=33$.
Finding the no. of non negative integral solutions to $x+y+2z=33$.
I tried many pure combinatorial approaches (I don't like making individual cases, too long). But they went in vain. I was just pondering, when randomly, a thought came to my mind.
$$x... | Clearly the number of non-negative integral solutions to the equations $x + y + 2z = 33$ is same as the number of similar solutions to $i + j + 2k = 33$. This is obviously equal to the coefficient of $x^{33}$ in the expansion $$(1 + x + x^{2} + \cdots + x^{i} + \cdots)(1 + x + x^{2} + \cdots + x^{j} + \cdots)(1 + x^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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inequality with three variables and condition If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?
| By C-S and Vasc we obtain $\sum\limits_{cyc}\frac{a^2}{a^2+c}=\sum\limits_{cyc}\frac{a^4}{a^4+a^2c(a+b+c)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3c+a^2b^2+a^2bc)}\geq$
$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^2b^2+a^2bc)+\frac{(a^2+b^2+c^2)^2}{3}}=\frac{3(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^4+5a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$
For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$
Attempt
Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \g... | Very similar to Nesbitt's inequality. If we set $A=b+2c,B=c+2a,C=a+2b$, we have $ 4A+B-2C = 9c $ and so on, and the original inequality can be written as:
$$ \frac{4B+C-2A}{9A}+\frac{4C+A-2B}{9B}+\frac{4A+B-2C}{9C} \geq 1 $$
or:
$$ \frac{4B+C}{A}+\frac{4C+A}{B}+\frac{4A+B}{C} \geq 15 $$
that follows from combining $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of ... | See my previous answer here: How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$
For an elegant solution, apply the so-called Purkiss Principle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I pref... | Let's use the elementary techniques to solve the simple limit
\begin{align}
L &= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - (1 + \cos x)^{2}}\notag\\
&= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - 4\cos^{4}(x/2)}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometry Identity (Proof): $ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $
Question: Prove that $$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
What I have attempted (Usually I start of with the complex side)
So starting with the LHS
$$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \... | Alternatively, you may just square the identity
$$
\cos^2\theta +\sin^2 \theta=1
$$
giving
$$
\cos^4\theta+2\cos^2\theta\sin^2\theta +\sin^4 \theta=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Modular arithmetic problem (mod $22$) $$\large29^{2013^{2014}} - 3^{2013^{2014}}\pmod{22}$$
I am practicing for my exam and I can solve almost all problem, but this type of problem is very hard to me. In this case, I have to compute this by modulo $22$.
| This number is obviously even so let's look at it modulo $11$ :
$$29^{2013^{2014}}-3^{2013^{2014}} \equiv 7^{2013^{2014}}-3^{2013^{2014}} \pmod{11}$$
Now look at the powers $7^x$ modulo $11$ and notice that : $$7^{10} \equiv 1 \pmod{11}$$ (this follows also from Fermat's little theorem )
So we need to look at $2013^{20... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove if $2|x^{2} - 1$ then $8|x^{2} - 1$ I have seen this question posted before but my question is in the way I proved it. My books tells us to recall we have proven
if $2|x^{2} - 1$ then $4|x^{2} - 1$
Using this and the fact $x^{2} - 1 = (x+1)(x-1)$ and a previous proof in which we have shown if$a|b$ and $c|d$ then... | If $2\vert x^2-1$, then $x$ must be odd, so write $x=2y+1, y\in\mathbb{Z}$.
Then $x^2-1=4y^2+4y=8\frac{y(y+1)}{2}$.
Note that at least one of $y,y+1$ is even, so the fraction on the right is always an integer.
Thus we have that $8\vert x^2-1$.
| {
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In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$.Find $\frac{a^2+b^2+c^2}{S}$ In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$
Find the value of $\frac{a^2+b^2+c^2}{S}$,where $S$ is area of the triangle $ABC$and find the value of $1+\frac{R}{r}$ wh... | HINT: use that $$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$ holds
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea h... | Start by trying to simplify $\sqrt{6-4\sqrt{2}}$. Let's assume there is some number $p+q\sqrt{2}$ for which
$$\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$$
Squaring both sides gives
$$6-4\sqrt{2} = (p + q\sqrt{2})^2 = p^2+2q^2 + 2pq\sqrt{2}$$
Comparing coefficients gives $6=p^2+2q^2$ and $-4=2pq$, i.e. $-2=pq$.
We need to solv... | {
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"url": "https://math.stackexchange.com/questions/1613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\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}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfra... | Assume $$\dfrac{a}{b}=x,\dfrac{b}{c}=y,\dfrac{c}{a}=z$$
So for instance
$$\dfrac{a+c}{b+c}=\dfrac{1+xy}{1+x}=x+\dfrac{1-x}{1+y}$$
And the problem would be transformed to:
$$\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1}\ge0$$
$\equiv(x^2-1)(z+1)+(y^2-1)(x+1)+(z^2-1)(y+1)\ge0$
$\equiv \sum{x^2z}+\sum{x^2}\ge\sum{x}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prime factorization of Gaussian integers
I want to find $a, b\in\mathbb{Z}[i]$ such that $a(2+3i)+b(5+5i)=1$.
I don't know how to do this, but my first thought was to do something with the norm or otherwise factoring $5+5i=(2+i)(2-i)(1+i)$, but I don't see how to use this further.
The second question is for which $x\... | Using the Extended Euclidean Algorithm as implemented in this answer, modified for Gaussian integers, we get
$$
\begin{array}{r}
&&2&-1+2i&1-i\\\hline
1&0&1&1-2i&2+3i\\
0&1&-2&-1+4i&-5-5i\\
5+5i&2+3i&1-i&1&0
\end{array}\tag{1}
$$
This says that
$$
(1-2i+(2+3i)k)\color{#C00000}{(5+5i)}+(-1+4i-(5+5i)k)\color{#C00000}{(2+... | {
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"url": "https://math.stackexchange.com/questions/1615801",
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If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfi... | Two circles have at least one common point only if the distance between their centers is equal to or smaller than the sum of their radii.
All you need to do is rewrite equations as circles, extract the center, the radius of each circle and check.
$$(x-2)^2+(y+1)^2=4, a_{0}=2, b_{0}=-1, r_{0}=2$$
$$(x-5)^2+(y-3)^2=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$
We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me
| $\bf{I\; have \;Solved\; Like \;This \; Way}$
$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0\Rightarrow 512x^9+(9\cdot 64)x^6+(27\cdot 8)x^3-512x+219=0$$
So $$\underbrace{(8x^3)^3+3(8x^3)^2\cdot 3+3(3^2)\cdot 8x^3+3^3}-512x+219-3^3=0$$
So $$(8x^3+3)^3=512x-219\Rightarrow (8x^3+3)^3=512\left(x-\frac{192}{51... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Substitution in a geometric series: $\sum_{n=0}^{N} x^n = \frac{x^{N+1} -1}{x-1}$ To find a formula for the sums of square of first n natural no, the following method is applied in generating functionology, we know
$$\sum_{n=0}^{N} x^n = \frac{x^{N+1} -1}{x-1}$$
So we apply the $\left\lbrace xD \right\rbrace^2$ opera... | Hint. I would set $$\epsilon:=x-1$$ then, as $\epsilon \to 0$, use the binomial theorem to get
$$
x^N=(1+\epsilon)^N=1+N\epsilon+\frac{1}{2} N(N-1)\:\epsilon^2+\frac{1}{6} N(N-1)(N-2)\:\epsilon^3+o(\epsilon^3). \tag1
$$ Inserting $(1)$ in your identity gives the sought result.
In fact, by applying $\left\lbrace (1+\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Not getting the same solution when using the rule sin(x)\x=1 on a limit There is a rule in limits that when $x$ approaches zero:
$$\frac{\sin\left(x\right)}{x}=1$$
So I used this rule on the following exercise:
Evaluate
$$
\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}
$$
I... | I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\lim_{x \to 0}\frac{\sin x}{x} = 1 \tag{1}$$ you don't mean $(\sin x)/x = 1$ when $x \to 0$. Hence it is just not possible to replace $\sin x$ by $x$. However whenev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | A nice way to simplify this problem, using nothing more than elementary trigonometry and algebra, is this:
As noted in another answer, the given expression simplifies to:
$$2\sec^2\theta +2\csc^2\theta - 1$$
Finding $\theta$ to minimize that expression is the same as finding $\theta$ to minimize
$$\begin{align}
\sec^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left ... | It is a well-known formula that
$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
So, in your case,
\begin{align*}
\sum_{i=1}^n (i-1)^2 &= 0^2 + 1^2 + \cdots + (n-1)^2 \\
&= 1^2 + \cdots + (n-1)^2 \\ \\
&= \frac{(n-1)n(2n-1)}{6}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) \ge 10 + 8\cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$
For the positive real numbers $a, b, c$ prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$$
I did the following:
$$\begin{split}\dfrac{a^2+b^2+c^2}{... | Note that we must prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\bigg(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\bigg) \\ $$
Adding $6$ gives us
$$\dfrac{3(a+b)(b+c)(c+a)}{abc} \ge \dfrac{8(a+b+c)^2}{ab+bc+ca}$$
However, since $(a+b)(b+c)(c+a) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca)$, it is enough to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Volume of Cone By Integration I am trying to find the volume of a cone by integrating it in spherical coordinates, but elementary geometry suggests that my approach is incorrect.
The specifications of the cone are $0\le R \le 5$, $0 \le \theta \le \frac{\pi}{3}$, and $0 \le \phi \le 2\pi$. My reasoning was as follows:... | Your problem is simple. The cone you described is an ice cream cone ie a cone with the a spherical cap since $R $ goes from 0 to 5, and it's volume cannot be given by $V=\frac{1}{3}\pi r^2 h $ (which is the volume of a cone without a cap), but by the integration you did.
Draw it and it will become more obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding Eigenvectors of a 3x3 Matrix (7.12-15) Please check my work in finding an eigenbasis (eigenvectors) for the following problem. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons).
For reference the following id... | Eigenvectors are never unique. In particular, for the eigenvalue $2$ you can take, for example, $x_2=-1$ which gives you the answer in the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equi... | WLOG, $C = \tan^{-1}(3) \approx 72^{\circ}, A = \tan^{-1}(1) = 45^{\circ}, B = \tan^{-1}(2) \approx 63^{\circ}$. This shows that $\triangle ABC$ is scalene !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Series expansion: $1/(1-x)^n$ What is the expansion for $(1-x)^{-n}$?
Could find only the expansion upto the power of $-3$. Is there some general formula?
| I realize this is an old thread, but I wanted to expand on the above answers on how to derive the formula for anyone else that might come along. Starting with the geometric series and taking successive derivatives:
$$
\begin{align}
\dfrac{1}{(1-x)} &= 1+x+x^2+x^3+x^4+x^5\dots+x^m+\dotsm\\
\dfrac{1}{(1-x)^2} &= 1+2x+3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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Integration by parts - hint I'm stuck on a passage on my textbook:
$$ \int \frac{1}{(1+t^2)^3} dt = \frac{t}{4(t^2+1)^2}+\frac{3}{4} \int \frac{1}{(t^2+1)^2} dt$$
I know that it should be easy but I just can't figure out what is the product of functions considered in this integration by parts..
can you help me?
thanks ... | $$\int \dfrac {1}{(t^2+1)^3}\,dt = \int \dfrac {(1+t^2) - t^2}{(t^2+1)^3}\, dt = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\int t\, d(t^2+1)^{-2}$$$$ = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} - \dfrac{1}{4}\int \dfrac{1}{(t^2+1)^2}\,dt $$$$ = \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} + \dfrac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| We have
$$
\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1} \tag1
$$
Multiplying $(1)$ by $s$ and making $s \to \infty$ gives
$$
1=A+B
$$ from which $\color{red}{B=0}$.
Making $s=0$ in $(1)$ gives
$$
0=A+C
$$ from which $\color{red}{C=-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:
*
*Equality is attained at $x=0$ and near $x=0.8$.
*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how t... | For our interval, the inequality is equivalent to $1-x\ge e^{-2x}$. (We squared and flipped.)
This inequality can be proved using differential calculus. Let $f(x)=1-x-e^{-2x}$. Then $f'(x)=2e^{-2x}-1$. So $f(x)$ is increasing until $x=\frac{\ln 2}{2}\approx 0.34$ and then decreasing. Thus all we need to do is check it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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The order of pole of $\frac{1}{(2\cos z -2 +z^2)^2} $at $z=0$ What is the order of the pole of :
$$\frac{1}{(2\cos z -2 +z^2)^2}$$
at $z=0$
This is what I did :
$$\cos z = \frac{e^{iz}-e^{-iz}}{2} $$
then :
$$ \implies \frac{1}{(e^{iz}-e^{-iz}+z^2-2)^2} $$
How do I continue?
| Since
$$
\cos z = 1 - \frac12 z^2 + \frac1{4!}z^4 + \cdots
$$
we have
$$
(2\cos z - 2 + z^2)^2 = (\frac1{12}z^4 + \cdots)^2 = \frac{1}{144}z^8 + \cdots
$$
where $\cdots$ denote higher order terms. This shows that your function has an $8$-fold zero in the denominator, so the order of the pole is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do I solve $\int4\cos^2(x) dx$? I have the basic idea of how to work out the integral of a trig function, but am having trouble in applying the concept. Would really appreciate it if someone could help me. Thanks!
| $\int 4 \cos^2(x) dx$
$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$
$=4\int\cos^2(x) dx$
$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$
$=4\int \frac{1+\cos \left(2x\right)}{2}dx$
$\mathrm{Take\:the\:cons... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\lef... | Hint:
\begin{align}
\lim_{x\to\infty}\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)&=\left(\lim_{x\to\infty}\frac{(2x^3+x)(x+1)}{(x^2+1)(4x^2+3)}\right)\lim_{x\to\infty}\frac{\sin\left(\frac{x+1}{4x^2+3}\right)}{\frac{x+1}{4x^2+3}}\\
\end{align}
Now, by making $\color{blue}{t=\frac{x+1}{4x^2+3}}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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... | Jonas produced the solution first, but I wanted to write up a solution for my own edification.
For $x>1$ we can write the following inequalities,
$$x < x+a\sqrt{x} < (1+a)x\qquad (a>0),$$
This allows us to come up with an upper and lower bound for the numerator,
$$\sqrt{x+\sqrt{x+\sqrt{x}}} < \sqrt{x+\sqrt{2x}}< \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to evaluate $\lim\limits _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{... | $$\lim_{x\to\infty}\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}} =\lim_{x\to\infty}\dfrac{\dfrac1{x^{4/3}}+\left(1+\dfrac1{x^4}\right)^{1/3}}{\dfrac1{x^{4/3}}+\left(9+\dfrac3{x^8}\right)^{1/6}}=\dfrac1{9^{1/6}}$$
which is $<1$
Hence the given limit should converge to $0$
Keep in mind $$\lim_{n\to\infty}\left(1+\dfrac1n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) +... | Without induction you can do it easily taking three common so $3(1+3+..(2n-1))$ now we know that sum of odd terms is a perfect square so we can write as $3n^2$ or sum of AP=$\frac{n}{2}(2+(n-1)2)=n^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
This sigma to binom? Can you please show me how to get from the left side to the right side?
$$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k} = \binom{100}{20}$$
| $$(x+1)^{50}(1+x)^{50}=(1+x)^{100}$$
$$(\sum_{k=0}^{50}\binom{50}k x^k)\cdot(\sum_{k=0}^{50}\binom{50}kx^{50-k})=\sum_{k=0}^{50}\binom{100}r x^r$$
Consider the coefficients of $x^{20}$
$$\binom{100}{20}$$
$$=\binom{50}0\cdot\binom{50}{20}+\binom{50}1\cdot\binom{50}{19}+\cdots+
\binom{50}{19}\cdot\binom{50}1+\binom{50}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the sum $\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$ Find the sum $$\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$$
When I asked my teacher how can I solve this question he responded it is very hard, you can't solve it. I hope you can help me in solv... | \begin{align}
\sum\limits_{k=1}^{100} \frac {k\cdot k!}{100^k} \frac{100!}{k!(100-k)!} &= \frac{100!}{100^{100}} \sum\limits_{k=1}^{100} \frac{k\cdot100^{100-k}}{(100-k)!}\\
&= \frac{100!}{100^{100}} \sum\limits_{k=0}^{99}\frac{(100-k)\cdot 100^k}{k!}\\ &=\frac{99!}{100^{99}} \sum\limits_{k=0}^{99}
\left( \frac {100^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ Let $\alpha =\left(\frac{1+\sqrt{5}}{2}\right)$ and $\beta = \left(\frac{1-\sqrt{5}}{2}\right)$. Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ where $L_n$ denotes the Lucas numbers.
I managed to solve the base case:
$$L_2 = \left(\frac{1+... | The crucial fact is that $\alpha$ and $\beta$ are the roots of $x^2=x+1$ and so $\alpha+\beta=1$. This makes induction work easily:
$L_0 = 2 = 1+1 = \alpha^0+\beta^0$
$L_1 = 1 = \alpha+\beta = \alpha^1+\beta^1$
$L_{n+2}=L_{n+1}+L_n
= \alpha^{n+1}+\beta^{n+1}+\alpha^n+\beta^n
= \alpha^{n+1}+\alpha^n+\beta^{n+1}+\beta^n
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding char polynomial in $Z_3$ $ K=Z_3 $
$ A \in K_{(4 \times 4)} $
$$A=
\begin{bmatrix}
a & -1 & -2 & -2 \\
0 & a-1 & -2 & 0 \\
-2 & 0 & a & 0 \\
-2 & -1 & 0 & a-2 \\
\end{bmatrix}
$$
I need to find $\det(A)$.
I came up with $\det(A)= (a)(a-1)(a)(a-2) - (a-2)(-2)(a-1)(-2)$
If I rewrite this I get:
$a^4 - 2a^3... | Keeping in mind the patterns $\left|\begin{array}{cccc}+&-&+&-\\-&+&-&+\\+&-&+&-\\-&+&-&+\end{array}\right|$ and $\left|\begin{array}{cccc}+&-&+\\-&+&-\\+&-&+\end{array}\right|$, expanding along the last column we get:
$$-(-2)\left|\begin{array}{cccc}0&a-1&-2\\-2&0&a\\-2&-1&0\end{array}\right|
+(a-2)\left|\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Decompose $\frac{x^4 + 5}{x^5 + 6x^3}$ (partial fraction decomposition)
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.
$$\frac{x^4 + 5}{x^5 + 6x^3}$$
So I factored the denominator to be $x^3(x^2+6)$ and here is the answer i got:
$\f... | The correct way is
$$\frac{x^4 + 5}{x^3(x^2 + 6)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+6}$$
and
$$\frac{5}{(x^2 − 16)^2}=\frac{A}{x+4}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-4}+\frac{D}{\left(x-4\right)^2}$$
In general, if the denominator has degree $n$, then the numerator is given a degree $(n-1)$.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we h... | $b=0$ is a solution for any $a$. We can then divide it out, looking for non-zero solutions. If we just feed it to the quadratic equation, we get $$a=\frac {3b-1 \pm \sqrt{(1-3b)^2-8(b+3)b^2}}{2(b+3)}$$ You must have $b \lt 0$ to have the discriminant positive. A quick search finds $(-1,-1), (9,-6), (8,-10), (9, -21)... | {
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"answer_id": 2
} |
If $P(x) = (x^4+x^3-3x^2+4x-4)\cdot q(x) + (2x^3-5x^2+7x-3)$ find $P(2)$
If for the polynomial $P(x)$ is true that $$P(x) = (x^4+x^3-3x^2+4x-4)\cdot q(x) + (2x^3-5x^2+7x-3)$$ find $P(2)$
I assumed that the polynomial $(x^4+x^3-3x^2+4x-4)$ has $(x-2)$ as one of its factors, but it turns out that this is not true.
By t... | I give you what I think is the only chance that your problem has a meaning.
There are many possibilities for a such polynomial: you can choose for $g(x)$ any arbitrary polynomial and the resultant one $$P(x) = (x^4+x^3-3x^2+4x-4)\cdot g(x) + (2x^3-5x^2+7x-3)$$ satisfy in fact the property of your statement.
In partic... | {
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"answer_count": 1,
"answer_id": 0
} |
Prove that the sequence of combinations contains an odd number of odd numbers
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+... | $$
\overbrace{\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}} + \overbrace{\binom n {\frac{n+1} 2} + \cdots + \binom n {n-1} + \binom n n} = 2^n.
$$
The two sums under the $\overbrace{\text{overbraces}}$ are equal; hence
$$
\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}... | {
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2})... | The quadratic formula only gives the roots of the equation, which are unaltered by the constant factor.
Let $\ P(x)=3x^{2}-x-2 \ \text{ and } \ Q(x) = x^2 -\frac{x}{3}-\frac{2}{3}$
The roots of both the polynomials are the same, as
$$\begin{align}3x^{2}-x-2&=0 \\ \implies x^2 -\frac{x}{3}-\frac{2}{3}&=0\end{align}$$
... | {
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"answer_count": 5,
"answer_id": 2
} |
How to solve $\int \frac{1}{1-y^2}$ with respect to $y$? I was solving an A Level paper when I came across this question. I tried substitution, but I'm not getting the answer with that. Would appreciate it if someone would help me.
| Step 1: Factorize the denominator
To compute the integral, you start of by using the third binomial theorem
$$\int \frac{1}{1-y^2}\ \mathrm{d}y = \int \frac{1}{1-y}\frac{1}{1+y}\ \mathrm{d}y$$
Step 2: Partial fractions
Use a technique called partial fractions to write the integrant as $\frac{A}{1-y}+\frac{B}{1+y}$. $A... | {
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Integration of $\frac{dx}{(6x-4x^2)^{1/2}}$ and completing square. Integrate $$\frac{dx}{\sqrt{6x-4x^2}}$$
While completing the square of $6x-4x^2$, I want to know where did $9/16$ come from in the following after taking $4$ out as the common factor $$-4\bigg(x^2 -\frac{3x}{2} +\frac{9}{16}\bigg) +\frac{9}{4}$$. I want... | $$-4x^2+6x = -4(x^2-\frac{3}{2}x)$$
$$ = -4\bigg(\big(x-\frac{3}{4}\big)^2-\frac{9}{16}\bigg) $$
$$ = -4\big(x-\frac{3}{4}\big)^2+\frac{9}{4}$$
Do you see it now?
| {
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How to simplify this equation to solve for m? It has been way too many years since high school. How can I simplify this equation to solve for m:
$\frac{x}{c+pm}=m$
I got to
$x = cm + pm^2$
and I don't know how to get any further. I wish this stuff stayed in my head over the last 20 years.
| Your equation
$$
x = c\, m + p\, m^2 \quad (1)
$$
is a quadratic equation of the form
$$
y = a \, x^2 + b \, x + c
$$
(where $x$ is the unknown, in your case $m$).
If $p \ne 0$, then we can divide both sides of equation $(1)$ by $p$ to get
$$
m^2 + \frac{c}{p} m = \frac{x}{p}
$$
we now try to align the left hand side f... | {
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"source": "stackexchange",
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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.