Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
| $x^5 \equiv 2 (\mod13)$
$x^{12} \equiv 1(\mod 13)$ (Femat's little theorem.)
$5\cdot5\equiv 1 (\mod 12)\\
(2^5)^5 \equiv 2 (\mod 13)\\
(2^5) \equiv 6 (\mod 13)\\
x\equiv 6 (\mod 13)$
Now use the same logic mod 11
$x^{10} \equiv 1(\mod 11)\\
3\cdot7 \equiv 1(\mod 10)\\
(2^7)^3 \equiv 2(\mod 11)\\
(2^7) \equiv 7(\mod 11)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac... | I think that the reflection formula
$$
\Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin{(\pi z)}} \qquad z \not \in \mathbb{Z},
$$
should work better here, given that you already have computed $\displaystyle{\Gamma\left(k + \frac{1}{2} \right)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Binomial theorem question. Find the value of the constant $k$ $$\left[(k+x)\left(2-\frac{x}{2}\right)\right]^6$$
where the coefficient of $x^{2}$ is $84$.Find the value of the constant $k$.
I tried to expand the equation but got a equation of degree 6 for some reason.
| We have $$\frac{1}{2^6}*(x+k)^6(x-4)^6\\(x+k)^6(x-4)^6=(P(x)+15x^2k^4+6xk^5+k^6)(Q(x)+15x^2*4^4-6x*4^5+4^6)\\15*4^4*k^6-36*4^5*k^5+4^6*15k^4$$ We have isolated the coefficient of $x^2$ so we have $$15*4k^6-36*4^2k^5+15*4^3k^4=84$$ hence
$$5k^6-48k^5+80k^4=7$$
This equation is irreducible and Wolfram gives $$\color{red}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
... | We note that when we are constructing a matrix for a linear map, we need to know the matrix related to which basis. Here, if you consider the vectors
$\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$
as the basis of your vector space, the given matrix by you is not the correct one. Ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it sa... | Replacing $x=m-m^2-2$ in your given polynomial function, we have
$$f(x)=m^8-4 m^7+11 m^6-19 m^5+24 m^4-21 m^3+23 m^2-15 m+28$$ as a result. Now, noticing that $m$ is a cube root of $-1$, this is equal to
$$f(x)=m^2-4 m^1+11 m^0-19 m^2+24 m^1-21 m^0+23 m^2-15 m+28=$$
$$5m^2+5m+18=5(m^2+m+1)+13=\color{blue}{13}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\f... | Your method (what you are trying to do) is correct, but you have a few errors.
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
This is wrong. It should be
$$|x+1|\lt \frac{21}{10}$$
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
This is wrong. It should be
$$|x+3|\lt\frac{41}{10}$$
but we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solutions to the inequality $0>-x^2 +2x+3.$ I am trying to solve an inequality of
$$0>-x^2 +2x+3.$$
I am aware of two different ways of factorizing this.
$$(-x+3)(x+1)\quad\text{ and }\quad(x-3)(-x-1).$$
When I use $(-x+3)(x+1)$, I get the desired solution of $x>3$ and $x<-1$. However when I use $(x-3)(-x-1)$, this gi... | 1) $0 > -x^2 + 2x + 3$ so
$0> (-x+3)(x+1)$
So $(-x + 3)(x+1)$ is negative. So one of the terms is positive and the other is negative.
So either $-x +3 > 0$ AND $x+1 < 0$
OR
$-x + 3 < 0$ AND $x+1 > 0$
If $-x +3 > 0$ AND $x+1 < 0$ then $x < 3$ and $x < -1$. Notice these are redundant statements. If $x < -1$ then OF ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $a, b, c>0$, such that $a+b+c=1$, prove that $\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$ Let $a, b, c>0$, such that $a+b+c=1$, prove that:
$$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$$
| \begin{align}
& A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{{{a}^{2}}}{ab+ac}+\frac{{{b}^{2}}}{bc+ab}+\frac{{{c}^{2}}}{ac+bc} \\
\\
& [(ab+ac)+(bc+ab)+(ac+bc)]A\ge {{(a+b+c)}^{2}} \\
\\
& (2ab+2bc+2ca)A\ge 1 \\
\\
& A\ge \frac{(a+b+c)^2}{(2ab+2bc+2ca)}=\frac{1}{(2ab+2bc+2ca)}\ge \frac{3}{2} \\
\\
& (a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Express the function $ f $ without using absolute value signs $\left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|}$? Good evening to everyone:
This is the equation $$ f(x) = \left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|} $$ What I've tried is: $$ \frac{x-2}{x+3}\ge 0 => x-2 \ge 0 => x \ge 2$$ Then $$ \frac{-x+2}{-x-3} ... | Hint:
1) $$\left|\frac{x-2}{x+3} \right|=\frac{x-2}{x+3} \Leftrightarrow x\in (-\infty;-3) \cup [2;+\infty)$$
2)$$\left|\frac{x-2}{x+3} \right|=-\frac{x-2}{x+3}\Leftrightarrow x\in (-3;2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof of AM-GM inequality for $n=3$: $\frac{a+b+c}{3}\geq\sqrt[3]{abc}$ Sorry for bad formatting, I couldn't mark the 3rd root on the right hand side...
I've figured this out into the point where
(and yeah, the problem is to prove that this applies to all non-negative real numbers)
$(a+b+c)/3\geq (abc)^{1/3}$
$$(a+b+... | The many online proofs for general $n$ and for $n=2$ leave little to do for $n=3$ except maybe to look for an algebraic identity proving the inequality. Let $(a,b,c) = (x^3,y^3,z^3)$, then:
$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx) = (1/2) (x+y+z)(\sum (x-y)^2)$
The conclusion is slightly more prec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What happens when you add $x$ to $\frac{1}{3}x$? I am dealing with an equation that requires me to add $x$ to $\frac{1}{3}x$:
$x + \frac{1}{3}x$ = ??
I know this might be simple to any of you on this site, because you are all asking questions with symbols I have never seen, but this is confusing to me.
I guess one way ... | $$
x + \frac{1}{3} x = 1 \cdot x + \frac{1}{3} x = \left( 1 + \frac{1}{3} \right) x
= \frac{4}{3} x
$$
Your full equation is slighty different, but the same rules (distributivity) apply:
$$
15 = \frac{2}{3} b + b = \left( \frac{2}{3} + 1 \right) b
= \frac{5}{3} b \iff \\
b = \frac{3}{5} \cdot 15 = 9
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
If $x^2+y^2 \equiv 0\pmod{p}$, then $p \equiv 1 \pmod{4}$
Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.
I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure ho... | No. The way the question is worded, $p=2$ also works.
Anyway, $y \neq 0 \pmod p,$ this means $y$ has a multiplicative inverse $\pmod p,$ for no better reason than $\gcd(y,p)=1$ and we have integers with $ys+pt=1.$
$$x^2 + y^2 \equiv 0 \pmod p,$$
$$x^2 \equiv -y^2 \equiv 0 \pmod p,$$
$$ \frac{x^2}{y^2} \equiv -1 \pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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A slightly problematic integral $\int{1/(x^4+1)^{1/4}} \, \mathrm{d}x$ Question. To find the integral of- $$\int {\frac{1}{(x^4+1)^\frac{1}{4}} \, \mathrm{d}x}$$
I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?
| Let $$I = \int\frac{1}{(x^4+1)^{\frac{1}{4}}}dx$$
Put $x^2=\tan \theta,$ Then $2xdx = \sec^2 \theta d\theta$
So $$I = \int\frac{\sec^2 \theta}{\sqrt{\sec \theta}}\cdot \frac{1}{2\sqrt{\tan \theta}}d\theta = \frac{1}{2}\int\frac{1}{\cos \theta \sqrt{\sin \theta}}d\theta = \frac{1}{2}\int\frac{\cos \theta}{(1-\sin^2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
What are the differences between: $\sqrt{(-3)^2}$, $\sqrt{-3^2}$ and $(\sqrt{-3})^2$. First, is $\sqrt{-3}$ is equal to $-3$ or is it imaginary?
What is the difference between:
*
*$\sqrt{(-3)^2}$
*$\sqrt{-3^2}$
*$(\sqrt{-3})^2$
Can I write $(\sqrt{-3})^2 = -3$?
And, given the rule that $\sqrt{a^n}$ is equal to ... | $\sqrt{(-3)^2}=\sqrt{9}=3$,
$\sqrt{-3^2}=\sqrt{-9}=\pm 3i$,
$(\sqrt{-3})^2=(\sqrt{-1}\sqrt{3})^2=(\pm i\sqrt{3})^2=-3$,
$\sqrt{a^n}=(\sqrt{a})^n $ iff $ a\geq 0$,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\t... | I think your error comes from the wrong use of the quadratic formula for factoring a second order polynomial. Factoring a polynomial and using the quadratic formula are two different things. Recall that for a quadratic polynomial
$$\color{blue}{ax^2+bx+c = a(x-x_1)(x-x_2)}$$
where you are missing the coefficient $a=3$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$ For $a^2<b$, is there an identity of evaluating the following integral?
$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$
What about:
$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice r... | Let the first integral be $I$ and the second one be $J$, then by putting $x=y-a$ and $y=z\sqrt{b-a^2}$ we have
\begin{align}
I(a,b)&=\int_0^{\infty}\frac{dx}{(x+a)^2+b-a^2}\\[10pt]
&= \int_a^{\infty}\frac{dy}{y^2+b-a^2}\\[10pt]
&=\frac{1}{\sqrt{b-a^2}}\int_{\large\frac{a}{\sqrt{b-a^2}}}^{\infty}\frac{dz}{z^2+1}\\[10pt]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Polynomial identity in positive terms, including AM-GM bound Consider $n$ nonnegative numbers $x_1 \cdots x_n$. An easy consequence of the AM-GM inequality
$$
\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
$$
is a lower bound on a polynomial
$$
(x_1 + x_2 + \cdots + x_n)^n \geq n^n (x_1 x_2 \cdo... | Yes, we can.
We can do it by Bricks Throwing Method.
It's very ugly but it works.
For four variables we need to work with $(a+b+c+d)^4-256abcd$.
$3(a^4+b^4+c^4+d^4)\geq\sum\limits_{cyc}a^3(b+c+d)$ because we throw one brick:
$\sum\limits_{cyc}(3a^4-a^3(b+c+d))=\frac{1}{2}\sum\limits_{sym}(a^4-a^3b-ab^3+b^4)=\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=3:\qu... | You have the right idea. Working mod $7,\,$ by little Fermat $\,2^6\equiv 1\equiv 3^6\pmod 7\,$ so the sequence $\,f_n \equiv 3\cdot 2^n + 2\cdot 3^n \,$ has period $\,6,\,$ i.e $\, f_{n+6}\equiv f_n.\,$ So we need only check the first cycle for $\,n = 0,\ldots, 5.\,$ Using $\,2^3\equiv 1,\ 3^3\equiv -1\,$ we compute ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorization of $x^n + y^n$, what sort of coefficients show up? We know that$$a^2 + b^2 = (a + bi)(a - bi).$$ What are the complete factorizations of $a^3 + b^3$, $a^4 + b^4$, $\ldots$ , $a^k + b^k$, etc.? What sort of coefficients show up?
| The problem simplifies if you divide through by $a$ and lent $\frac{b}{a} = x$. Once you have the factors of $1+x^n$ it is easy to re-introduce $a$ and $b$.
The pattern in these is easy to see. For odd $n$,
$$a^3 + b^3 = (a+b) (a+b\mbox{ cis } (\frac{2\pi}{3})) (a+b\mbox{ cis }(\frac{4\pi}{3})) $$
$$a^5 + b^5 = (a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding an Explicit Formula for $a_n$ from a Recurrence Relation I have the following recurrence relation
$$\beta ^n n(n+2)a_n = \sum_{k=0}^{n-1} \beta^k (\alpha+k+1) a_k , \quad a_0=1, \quad n \ge 1 \tag{1}$$
where $\beta$ and $\alpha$ are positive real numbers.
I know that one can easily find $a_n$ by back substit... | Inspired by the comment of Achille Hui, I noticed that
$$\begin{align}
\beta ^n n(n+2)a_n &= \sum_{k=0}^{n-1} \beta^k (\alpha+k+1) a_k \\
\beta ^{n-1} (n-1)(n+1)a_{n-1} &= \sum_{k=0}^{n-2} \beta^k (\alpha+k+1) a_k
\end{align}$$
and subtracting the above equations will lead to
$$\begin{align}
\beta ^n n(n+2)a_n - \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Proving inequality using Lagrange multipliers I have this question. Prove that for all $ x,y\geq 0 $, $$ \dfrac{x^n+y^n}{2}\geq \bigg(\dfrac{x+y}{2}\bigg)^n $$ using the method of Lagrange Multipliers, via $$ \min \dfrac{x^n+y^n}{2}, \text{where $x+y=s$} $$ for some $s\geq 0$.
This is what I did. I consider $f(x,y)=\df... | Unless you restrict the problem to $n \geq 1$ the statement is false.
Consider $x=1, y=2, n=\frac12$:
$$
\sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ tha... | Hint
The equation you have is that of a circle, $(x-2)^2+y^2= 1$, and you want the maximum of $x^2+y^2=4x-3$. Clearly from the latter we need the extreme possible $x$, which from the circle's equation will be when $x\in \{1,3\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Closed form of function $f(n) = \frac1n \sum\limits _{k=1}^{n-1} f(k)$? Could anyone help me get to the closed form of the function? Here $\mathbb{N}=\{1,2,3,\ldots\}$.
Find all functions $f:\mathbb{N}\to\mathbb{R}$ such that $f(1)=1$ and
$$f(n) = \frac 1 n \sum _{k = 1}^{n-1}f(k)$$
for every integer $n>1$.
So far, ... | This is similar to the first answer, but without rewriting things. Note that
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^n f(x) = \frac{1}{n+1}\sum_{x=1}^{n-1}f(x) + \frac{1}{n+1}f(n).$$
Using the definition for $f(n)$, this becomes
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^{n-1} f(x) + \frac{1}{n(n+1)}\sum_{x=1}^{n-1}f(x) = \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Maximize the product of the partitions of an integer
Let $n>0$ be an integer. Consider all partitions of $n$, i.e. all possible ways of writing $n$ as a finite sum of positive integers,
$$n=n_1+n_2+\cdots+n_k.$$
What partition maximizes the product $n_1n_2\cdots n_k$?
Examples:
*
*$2=2$
*$3=3$
*$4=2+2$
*$5=... | Here is a rough idea. The problem can be divided into two parts. Firstly, given a $n$, for a fixed number of partitions $k$, it is not hard to find that all of the partitions should be equal. However, since we are dealing with integers, we might need to find the nearest integers.
Having the first part, we would like to... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove: $\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$ Show that if $m$ and $n$ are integers with $0\leq m<n$ then $$\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$$
Attempts:
*
*$(-1)^{k}\binom{n}{k}$ is the coefficient of $x^{k}$ in the expansion of $(1-x)^{n}$
*And $\b... | Following the hint of Arthur, note that $$\frac{1}{m!}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{k-1}=\frac{\left(1-x\right)^{n}-1}{xm!}
$$ and if we differentiate the LHS $m$ times, we get $$\frac{1}{m!}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(k-1\right)\left(k-2\right)\cdots\left(k-m\right)x^{k-m-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4... | You're almost there! Note that $$\frac{x}{x^2 + 4x + 8} =\frac{1}{2}\left( \frac{2x + 4 - 4}{x^2 + 4x + 8}\right) = \frac{1}{2}\left(\frac{2x + 4}{x^2 + 4x + 8}\right) - \frac{2}{x^2 + 4x + 8}$$
So that we have $$10\int\frac{x}{(x^2+4x+8)}dx +40\int \frac{1}{x^2+4x+8}dx = 5 \int \frac{2x + 4}{x^2 + 4x + 8} \, \mathrm{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Need help with the proof of the following theorem of the equation $x^4+y^4=z^4$ Theorem 7-2. The equation $x^4+y^4=z^4$ is not solvable in nonzero integers.
Proof: It suffices to show that there is no primitive solution of the equation $$x^4+y^4=z^2$$ (why?)
Suppose that $x,y,z$ constitute such solution; with no loss... | If $(x,y,z)$ is a solution to $x_1^4+x_2^4=x_3^4$, then $(x,y,z^2)$ is a solution to $x_1^4+x_2^4=x_3^2$. This is the same thing as saying if $x_1^4+x_2^4=x_3^2$ has no solution, then neither does $x_1^4+x_2^4=x_3^4$.
$1\equiv x^2 \equiv a^2-b^2 \equiv 0-1 \equiv -1 \mod 4$, because $a$ is even and $b$ is odd, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
| \begin{align*}
\sqrt{\cos x} & > \cos \sin x \\
\cos x & > \cos^2 \sin x \\
1 - 2 \sin^2 \frac x2 & > 1 - \sin^2 \sin x \\
\sin^2 \sin x & > 2 \sin^2 \frac x2 \\
\sin \sin x & > \sqrt 2 \sin \frac x2 \\
\end{align*}
Note that for $0<x<\frac \pi 4$ we have
$$\frac \pi 2 > \frac 1{\sqrt 2} = \sin \frac \pi 4 > \sin x = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Solve for $x,y,z$ from the linear equations. The main question is :
$$\begin{align}
(b+c)(y+z)-ax &= b-c \tag{1} \\
(c+a)(z+x)-by &= c-a \tag{2} \\
(a+b)(x+y)-cz &= a-b \tag{3}\\
\end{align}$$
Solve for $x,y,z$ if $a+b+c\ne0$
My method :
Opening all brackets, we get,
$$by+bz+cy+cz-ax=b-c$$
$$cz+cx+az+ax-by=c-a$$
$$a... | You're almost there; nice work!
Plug $x+y+z = 0$, in the form $(y+z) = -x$, into your first equation to get $(-x)(a+b+c) = b-c,$ so $$x = \frac{c-b}{a+b+c}.$$
Do similar things to the other two equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
| Note that your polynomial is a geometric series with first term $-x$, common ratio $-x^2$ so that it can be written as $$\frac{x(x^{16} - 1)}{x^2 + 1} = 7$$
So $$x^{16} = \frac{7(x^2 + 1)}{x} + 1$$
And now, all you need to do is show that $\frac{x^2 + 1}{x} = x + \frac{1}{x} > 2$, which is not hard.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equa... | I will only look for solutions on $[0, 2\pi)$
We need to make $2$ observations:
$1)$ The maximum value of $\sin \theta$ is $1$. Therefore, if any of $\sin x$, $\sin y$ is negative, $\sin x + \sin y < 1$. It follows that both $x$ and $y$ must be $\in [0, \pi]$.
$2)$ $\cos x= -\cos y \iff \cos x = \cos (\pi - y)$ . (You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
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Why does $\sum_{n=1}^{\infty} \frac{2^n+1}{5^n+1}$ converge?
Why does $\sum_{n=1}^{\infty} \frac{2^n+1}{5^n+1}$ converge?
I've tried by using the ratio test but I don't get so far, I'm a little lost with it. Any help will be really aprecciated.
| We have $a_n=\frac{2^n+1}{5^n+1}$.
\begin{align*}
\frac{a_{n+1}}{a_{n}}=&\frac{2^{n+1}+1}{2^{n}+1}\cdot\frac{5^{n}+1}{5^{n+1}+1}\\
&=\frac{2+\frac{1}{2^n}}{1+\frac{1}{2^n}}\cdot\frac{1+\frac{1}{5^n}}{5+\frac{1}{5^n}}
\end{align*}
When $n$ tends to $+\infty$, $|\frac{a_{n+1}}{a_{n}}|$ approaches $2/5<1$. Hence the serie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
I do not know how to approach this problem and would appreciate advice how to proceed.
| In $ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$, put $x=0, y=1$ to get $c+1 = 0$ and hence $c=-1$. Put $y=0, x=-1$ to get $a-2=0$ and hence $a=2$. Putting $x=1, y = 2$, we get $a+2b+4c+5-4+3 = 0$ and hence $b=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem ... | Rewrite your equation as
$$
\frac{1}{\cos x}+\frac{\sin x}{\cos x}=4
$$
that becomes
$$
\frac{1+\sin x}{\cos x}=4
$$
This reminds the formula for the tangent of the half angle
$$
\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}
$$
but sine and cosine are mixed up. Not a problem: set $x=\pi/2-t$ and take the recipro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 12,
"answer_id": 7
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Number theory problem, fractions and gcd The problem says:
If $a$ and $b$ are positive integers such that $\dfrac{a+1}{b}+\dfrac{b+1}{a}$ is an integer, then show that $\sqrt{a+b}\ge \gcd(a,b)$.
Adding $\frac{2ab}{ab}$ to $\frac{a+1}{b}+\frac{b+1}{a}$ yields that ab divides (a+b+1)(a+b), but I haven´t been able to co... | If $\gcd(a,b)=1$, then clearly $\sqrt{a+b} \geq \gcd(a,b)$. Now assume $\gcd(a,b)=d>1$. Then if $\frac{a+1}{b}+\frac{b+1}{a}=n$, then by your simplification, we have $$(a+b+1)(a+b)=(n+2)ab$$Note that $d^2$ divides the righthand side, so it must also divide the lefthand side. Since $d$ divides $a$ and $d$ divides $b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to solve this using power series method $\left(x^2+2\right)y''\:+\:xy'\:-\:y=0$ $\left(x^2+2\right)y''\:+\:xy'\:-\:y=0$
What's next after this?
$\sum _{n=2}^{\infty }\:n\left(n-1\right)a_nx^n+2\:\sum _{n=0}^{\infty \:}\left(n+2\right)\left(n-1\right)a_{n+2}x^n+\sum _{n=2}^{\infty \:\:}na_n^{\:}x^n\:-\:\sum _{n=0}^{... | Restarting from scratch, we assume that the DE $$ (x^2+2)y''+xy'-y=0 \tag{1}$$
has an analytic solution in a neighbourhood of zero, given by
$$ y(x) = \sum_{n\geq 0} a_n x^n \tag{2} $$
In terms of the coefficients, $(1)$ translates into:
$$ (x^2+2)\sum_{n\geq 2} n(n-1)a_n x^{n-2} + x\sum_{n\geq 1}na_n x^{n-1} - \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac... | Your teacher has left out some justification, which is probably why you are confused.
In general, we have the following identity for all real $x$:
$$\sqrt{x^2}=|x|.\tag{1}$$ Also, for nonnegative real $a$ and positive real $b,$ we have $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac ab}.\tag{2}$$ Putting $(1)$ and $(2)$ togeth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to ... | Using Minkowski Inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$$
and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$
So $$\sqrt{x^2+(1-x)^2}+\sqrt{(1-x)^2+(1+x)^2}\geq \sqrt{[x+(1-x)]^2+[(1-x)+(1+x)]^2}=\sqrt{5}$$
and Equality hold when $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is it possible to $\int \sqrt{\cot x}$ by hand $$\int \sqrt{\cot x}{dx}$$
$$\int \sqrt{\frac{\cos x}{\sin x}}{dx} $$
Using half angle formula
$$\int \sqrt{\frac{1-\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}}}{dx}$$
But I am not getting any lead from here .I think it is not possible to integrate $\sqrt{\cot x}$ by hand .
I c... | Using Ted Shifrin Hint
$$I = \int \sqrt{\cot x}dx$$ Put $\cot x= t^2\;,$ Then $$\csc^2 xdx = -2tdt\Rightarrow dx = -\frac{2t}{1+t^4}dt$$
So $$I = -\int\frac{2t^2}{1+t^4}dt = -\int \frac{\left[(t^2+1)+(t^2-1)\right]}{t^4+1}dt$$
Now Let $$J = \int \frac{t^2+1}{t^4+1}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also ... | Let $$I = \int\frac{1}{\sin^4 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx$$
$$I =\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx= \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+\left(\sqrt{2}\right)^2}dx$$
Now Put $\tan x-\cot x = t\;,$ Then $(\sec^2 x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational... | The rational root theorem tells you that any rational root $\frac{p}{q}$ will have $p\mid 1$ and $q \mid 1$ so $\frac{p}{q} = \pm 1$, you can then verify that only $x=1$ is a root and be done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:... | When you cancel out the terms from LHS and RHS, you drop the solutions when these terms are 0 or do not exist (because the denominator is 0). A trivial example would the $\theta=0$, which certainly is a solution, but you did not find it because of the canceled terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know tha... | @Michael Freimann
What follows is not an answer. It is just an aside comment, that might interest some readers.
The kind of matrices considered in this question can be described by the following formula for their coefficients
$$A_{ij}=a_{min(i,j)}$$ for a given sequence $a_1,a_2,...a_n$, this sequence being here {a,b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Answer mismatch in definite integral question:$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$
I faced a problem while evaluating the following integral.
$$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$
MY ATTEMPT:
Let $$I=\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$
By King's Rule $$I=-\int_{0}^{2\pi}... | Symmetry is a bit more evident if we set $x=z+\pi$. Since $\sin(z+\pi)=-\sin(z)$,
$$ I = -\int_{-\pi}^{\pi}\frac{(z^2+2\pi z+\pi^2)\sin(z)}{8+\sin(z)^2}\,dz=-2\pi\int_{-\pi}^{\pi}\frac{z\sin(z)}{8+\sin(z)^2}=-4\pi\int_{0}^{\pi}\frac{z\sin z}{8+\sin(z)^2}\,dz $$
and since $\sin(\pi-z)=\sin(z)$,
$$ I = -4\pi^2\int_{0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 ... | So $$x = 81 \cdot a + 64 \cdot b$$ is what you need; where $81a \equiv 59\pmod{64}$ which you can do by finding $k$ such that $81k \equiv 1 \pmod{64}$ and then multiplying both sides by $59$.
Then you also need $64b \equiv 32\pmod{81}$ which you can do by finding $m$ such that $64m \equiv 1 \pmod{81}$ and multiplying ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove the Inequality proves true, three variables Prove that for all positive real numbers $a,b,c$ we have:
$$\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\geq\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$
So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that t... | Also we can use $uvw$ here.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $\frac{3v^2}{4w^3}+\frac{\sum\limits_{cyc}(3u+a)(3u+b)}{\prod\limits_{cyc}(3u+a)}\geq\frac{\sum\limits_{cyc}(a+b)(a+c)}{9uv^2-w^3}$ or
$\frac{v^2}{4w^3}+\frac{15u^2+v^2}{54u^3+9uv^2+w^3}\geq\frac{3u^2-v^2}{9uv^2-w^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About $(x^2+y^2)dy-y^2dx=0$ differential equation Let us show $(x^2+y^2)dy-y^2dx=0$ differantial equation is homogeneous then solve the equation.
Firstly I did show we can also write $\frac{y^2}{x^2+y^2}=\frac{dy}{dx}$ equation depending on the first equation. Let $v=y/x$. Then we get $\frac{v^2}{v^2+1}=\frac{dy}{dx}x+... | First of all you have $y=vx$ and $\frac{dy}{dx}=x\frac{dv}{dx}+v$, so we need to solve
$$\frac{v^2}{1+v^2}=x\frac{dv}{dx}+v$$
$$\frac{v^2-v(1+v^2)}{1+v^2}=x\frac{dv}{dx}$$
or
$$\frac{1+v^2}{-v^3+v^2-v}dv=\frac{1}{x}dx.$$
The right is easy to integrate; the left you should be able to tackle with partial fraction:
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why do we find vertical asymptote of functions (in rational fractions) by turning the denominator into $0$? Like for $$f(x) = \frac{1}{x-2}$$ we can find its vertical asymptote by finding what $x$ can be in order to turn the denominator into zero.
e.g. $x=2$
But why is that? Why does the vertical asymptote correspond ... | A function $f$ has a vertical asymptote at $x = a$ if $|f(x)|$ increases without bound as $x$ approaches $a$, that is, $f$ has a vertical asymptote if one or more of the following conditions holds:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty\\
\lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$. Question:
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$.
My attempt:-
$$\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx$$
$$\frac{dy}{1+y}=-\frac{cos(x)}{2+sin(x)}dx$$
Integrating both s... | You made a mistake by antiloging. The line $ 1+y=\frac{1}{2+\sin(x)}+C $ is wrong.
It would be easier if you would write $\log C$ instead $C$ in the line before. So in the next step you would get:
$$1+y=\frac{1}{2+\sin(x)} \cdot C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ wh... | $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac} \ge \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}=\frac{(a+b+c+d)^2}{(a+c)(b+d)+2ac+2bd}$$
From Cauchy.
$$(a+b+c+d)^2=(a+c)^2+(b+d)^2+2(a+c)(b+d) \ge 4ac +4bd+2(a+c)(b+d)$$
Since $(x+y)^2 \ge 4xy$.
So $$\sum_{cyc}\frac{a}{b+c} \ge 2 >\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
| The center:
find the minimum of $(5x^2 + 6xy + 2y^2 + 7x + 2y)$
$\frac {\partial}{\partial x}(5x^2 + 6xy + 2y^2 + 7x + 2y) = 10x + 6y + 7= 0\\
\frac {\partial}{\partial y}(5x^2 + 6y + 2y^2 + 7x + 2y) = 6x + 4y + 6= 0$
Solve this system of equations
$x = 2, y = -4.5$
$2(x-2)^2 + 6(x-2)(y+4.5)+5(y+4.5)^2 = 0.5$
$4(x-2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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How to prove a fomula is real or not? The formula is:
$$f(u):=\sqrt{1+i}\arctan\left(\dfrac{\sqrt{2}}{\sqrt{(-1+i)u}}\right)+\sqrt{1-i}\arctan\left(\frac{(-1)^{3/4}\sqrt{(-1-i)u}}{u}\right)\tag{1}$$
I obtained it from the following definite integral:
$$f(u):=\int_u^{\infty } \frac{\sqrt{x}}{x^2-2 x+2} \, dx\quad {\text... | Inspired by How to simplify $\Re\left[\sqrt2 \tan^{-1} {x\over \sqrt i}\right]$?
I manage to write a primitive of $\frac{\sqrt{x}}{x^2-2 x+2}$ using real functions. Let
$$F(x):=\frac{1}{\sqrt{2\sqrt{2}-2}}\cdot
\arctan\left(\frac{\sqrt{2\sqrt{2}-2}\cdot \sqrt{x}}{x-\sqrt{2}}\right)$$
$$G(x):=\frac{1}{\sqrt{2\sqrt{2}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to determine the matrix in the following case Say we have a vector $\textbf{b}$ and $\textbf u$ such that:
$$A \mathbf b= \mathbf u$$
Where $A$ is a square matrix.
If $\mathbf b$ and $\mathbf u$ are known and $A$ is the unknown, How to get the matrix $A$ (perhaps it is not unique but how can one proceed to get it... | $$\begin{pmatrix}a_{11} &a_{12} \\a_{21} &a_{22}\end{pmatrix}\begin{pmatrix}b_1\\ b_2\end{pmatrix}=\begin{pmatrix}u_1\\ u_2\end{pmatrix}$$
Is equivalent to
$$\begin{pmatrix}a_{11}b_1+a_{12}b_2 \\a_{21}b_1+a_{22}b_2 \end{pmatrix}=\begin{pmatrix}u_1\\ u_2\end{pmatrix}$$
or
$$\begin{pmatrix}b_1 &b_2 &0 &0\\0 &0 &b_1 &b_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proof verification: $a_1=2$, $a_{n+1}=3+\frac{a_n}{2}$ is increasing and bounded For base case, $a_2=3+\frac{a_1}{2}=4>a_1$
For $n=k$, let $a_{k+1}>a_k$
Adding 3 on both sides after dividing by 2,
$$3+\frac{a_{k+1}}{2}>3+\frac{a_k}{2}$$
$$a_{k+2}>a_{k+1}$$
Hence the sequence is increasing.
How can I show that the seque... | Let $a_n=b_n+6$. Then
$$ b_{n+1}+6=3+\frac{b_n}{2}+3 $$
is equivalent to $b_{n+1}=\frac{b_n}{2}$ or to $b_n=\frac{b_1}{2^{n-1}}$. Since $a_1=2$ we have $b_1=-4$, hence
$ b_n = -\frac{8}{2^n} $ and
$$\boxed{ a_n = \color{red}{6-\frac{8}{2^n}} }$$
Now the claim is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35?
Also, are there any standard m... | \begin{align}
n \equiv 5 &\pmod{6} \\
n \equiv 2 &\pmod{11} \\
n \equiv 31 &\pmod{35}
\end{align}
The idea is to find three integers, $A, B,$ and $C$ such that
\begin{array}{|l|l|l|}
A \equiv 1 \pmod{6} &
B \equiv 0 \pmod{6} &
C \equiv 0 \pmod{6}
\\
A \equiv 0 \pmod{11} &
B \equiv 1 \pmod{11}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
| Start by squaring both sides of $\frac{x}{y}+\frac{y}{x}=3$. What happens to the cross terms?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving a system of linear equations with undefined number of equations I need to solve this system of linear equations:
$k_0=1+k_1$
$k_1=1+\frac{1}{3} k_1 +\frac{2}{3} k_2$
$k_2=1+\frac{1}{3} k_1 +\frac{2}{3} k_3$
$k_3=1+\frac{1}{3} k_1 +\frac{2}{3} k_4$
$...$
$k_{N-1}=1+\frac{1}{3} k_1 +\frac{2}{3} k_N$
$k_N=0$
I nee... | You have done all the work already. Now simply use $[3]$ for $i=2$ together with the second equation in your question to - after reducing a bit - obtain
\begin{align}
k_1&=\frac{9}{2}\left(\frac{3}{2} \right)^{N-2}-3 \\
k_0&=1+k_1=\frac{9}{2}\left(\frac{3}{2} \right)^{N-2}-2
\end{align}
EDIT: Or use Johannes Kloos muc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
Aft... | Since $\sum\limits_\text{cyc}\,\frac{a}{b+c}=1$, we have $$a+b+c=(a+b+c)\,\left(\sum_\text{cyc}\,\frac{a}{b+c}\right)=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\right)\,.$$
That is,
$$a+b+c=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)+(a+b+c)\,.$$
Hence,
$$\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)=0\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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integral $\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$ Evaluate , if possible in a closed form, the integral:
$$\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$$
Basically, I have not done that much. I broke the integral
\begin{align*}
\int_{0}^{\pi} \left ( \f... | Too long for a comment. We have the following representations:
\begin{align*}
\int_{0}^{\pi} \left(\frac{\pi}{2} - x\right) \frac{\tan x}{x} \, dx
&= \frac{\pi^2}{4} + \frac{\pi}{2}\int_{0}^{1} \psi\left(\frac{1+x}{2}\right) \sin(2\pi x) \, dx \\
&= \frac{\pi^2}{4} - \pi \int_{0}^{\infty} \frac{dt}{(1+t^2)(e^{2\pi t} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Decomposing $\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)}$ I got stuck at this expression and couldn't decompose it further. The book I am referring to has the answer $$\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)} = \frac5{4(1-z)} + \frac1{2(1-z)^2} - \frac3{1-2z} - \frac2{(1-2z)^2} + \f... | Was able to decompose it using following method
$let \frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2(1−3z)}=\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}+\frac{E}{(1-3z)}
$
Multiply both sides by $(1−z)^2(1−2z)^2(1−3z)$
$1−8z+27z^2−35z^3+14z^4 = A(1-z)(1-2z)^2(1-3z)+
\\B(1-2z)^2(1-3z)+
\\C(1-z)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Value of $4\sin^2(\alpha)$ If the point on $y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$ ,$\alpha>0$ where the tangent is parallel to $y=x$ has an ordinate $\frac{u^2}{4a}$ then value of $4\sin^2(\alpha)$ is?. If we substitute $y$ in the given equation we get a quadratic in $x$ whose roots are $-\tan(\alpha)\pm\sqr... | Let $P \equiv \left(x,\frac{u^2}{4a}\right) $
Then by $$y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
we get $$\frac{u^2}{4a}=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
$$\frac{u^2}{4}=ax\tan(\alpha)-\frac{(ax)^2}{2u^2\cos^2(\alpha)}\rightarrow (1)$$
And $$\frac{dy}{dx}=\tan(\alpha)-\frac{ax}{u^2\cos^2(\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Help with this integral substitution This came up in my physics exercise:
$$\int_{\theta=0}^{\pi}\frac{\cos\left ( \theta \right )\sin\left ( \theta \right )}{\sqrt{R^{2}+r^{2}-2Rr\cos\left ( \theta \right )}}d\theta$$
I've tried the substitution $u=$ everything under the square root but it didn't worked.
I don't like ... | Let $u=R^2+r^2-2Rr\cos\theta,\;$ so $du=2Rr\sin\theta d\theta$ and $\cos\theta=\frac{R^2+r^2-u}{2Rr}$.
Then $\displaystyle\int_{0}^{\pi}\frac{\cos\left ( \theta \right )\sin\left ( \theta \right )}{\sqrt{R^{2}+r^{2}-2Rr\cos\left ( \theta \right )}}d\theta=\frac{1}{2Rr}\int_{(R-r)^2}^{(R+r)^2}\frac{\frac{R^2+r^2-u}{2Rr}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Probability of a 7 card sequence (same suit) in a 32 card game Deck of four suits of 8 cards each (32 cards total). Two Players with 12 cards each.
The probability of 8 sequential cards is supposably:
$$\binom{4}{1} \times \frac{\binom{8}{8}\binom{24}{4}}{\binom{32}{12}} $$
What is the probability of getting 7 sequent... | There are four suits, with 8 cards each.
If one player has 12 cards out of those 32, then the probability is indeed
$\frac{\color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} }{\color{blue}{\binom{32}{12}}}$, where $\color{red}{4\cdot 1}$ is the number of ways to choose a sequence of 8 colors, $\color{green}{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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An inequality concerning three positive numbers If $a, b, c$ belong to positive real numbers then prove that
$$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$
I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.
| Applying the inequality between the Arithmetic Mean and the Harmonic Mean we get:$$\frac{2}{(a+b)}\le\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\tag{1}$$
$$\frac{2}{(b+c)}\le\frac{\frac{1}{b}+\frac{1}{c}}{2}=\frac{b+c}{2bc}=\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using trig identity to solve a cubic equation I need to use the identity $\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) $ to solve the cubic equation $t^3+pt+q=0$ when $27q^2+4p^3<0$.
I believe I need to let $t=Acos(\phi)$ when $A=2\sqrt{\frac{-p}{3}}$, so that my equation looks more like the identity and I have $A^3\co... | For a general cubic,
$$ax^{3}+bx^{2}+cx+d=0$$
use what is known as the Tschirnhaus-Vieta approach: let $x=t+B$, then the cubic becomes
$$t^3+\left(3B+\frac{b}{a}\right)t^2+\left(3B^2+\frac{c+2bB}{a}\right)t+\left(B^3+\frac{bB^2+cB+d}{a}\right)=0$$
By setting $B=-\frac{b}{3a}$, the coefficient of $t^2$ becomes zero a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$ for $n\in \mathbb N^+$
$$\frac{1}{2n}\le \frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+n}}{n}\le\frac{1}{n}$$
I tried math induction and I tried take integral but I want to solve this with most elementary methods please give me hint or just show that. Th... | It holds when $k=1$
$$\frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \ge \overset { n }{ \overbrace { \frac { 1 }{ 2n } +\frac { 1 }{ 2n } +...\frac { 1 }{ 2n } } } =n\frac { 1 }{ 2n } =\frac { 1 }{ 2 } \\ \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \le \overset { n }{ \overbrace {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Show that $\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$ If $\sin\alpha+\sin\beta+\sin\gamma=\cos\alpha+\cos\beta+\cos\gamma=0$, then show that $$\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\ga... | Let $z_1 = \cos\alpha + i\sin\alpha, z_2 = \cos \beta + i\sin \beta, z_3 = \cos\gamma + i\sin\gamma$. Given that $z_1+z_2+z_3 = 0$. Since the orthocenter of the triangle with vertices $z_1, z_2, z_3$ is $z_1+z_2+z_3$ it follows that for this triangle, orthocenter and circum center coincide and hence the triangle is equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\fra... | Note $\tan^2(x)$ is convex, so by Jensen's inequality
$$\sum_{cyc} \tan^2 \frac{A}2 \geqslant 3\tan^2 \frac{A+B+C}{2\cdot3} = 1$$
Further as $\tan(x)$ is unbounded, clearly we can have the sum unbounded by allowing one of the angles to approach $\pi$.
| {
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"url": "https://math.stackexchange.com/questions/1915376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
proof using the mathematical induction Prove $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer for every positive integer k.
In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While so... | We have, for instance, $(k+1)^7/7 = (\sum_{i=0}^7 \binom{7}{i}k^i)/7$. Note that the only times the coefficient of $k^i$ in the numerator is not divisible by $7$ is when $i=0,7$. Therefore, $(k+1)^7/7 = k^7/7 + 1/7 + \text{integer}$. Applying the same logic to the other terms, we find
$$\begin{align}\frac{(k+1)^7}{7}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Easy way to compute $\int_{-\pi}^\pi \cos^4(x)dx$. Is there an easy way to compute $$\int_{-\pi}^\pi \cos^4(x)dx\ \ ?$$
What I did is using $$\cos^4(x)+1-1=1+(\cos^2(x)-1)(\cos^2(x)+1)=-\sin^2(x)(1+\cos^2(x))$$
and then using formula $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
and $$\sin^2(x)=\frac{1-\cos(2x)}{2},$$
but at the... | The integral has four-fold symmetry, so it is equivalent to:
$$\huge 4\int_0^\frac{\pi}{2} \cos^{4}(x) \, \text{d}x = 2 \,\text{B}\left(\frac{5}{2},\frac{1}{2}\right)$$
$$\huge = 2 \, \frac{\Gamma{(\frac{5}{2})}\Gamma{(\frac{1}{2})}}{\Gamma{(3)}} = 2 \, \frac{ \frac{3}{2} \times \frac{1}{2} \times \pi}{2} = \frac{3\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roo... | We must have $f(2) = \lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})$ for $f(x)$ to be continuous at $x=2$. Since plugging in $2$ gives us $\frac{0}{0}$, we can use L'hopital's Rule and differentiate the top and bottom:
$$\lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})=\lim_{x \to 2}(\frac{2\frac{1}{2}(2x+5)^{-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$
I tried AM-GM, which onl... | It must be
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}\geq\frac{9+3\sqrt3}{2\sqrt{a+b+c}}$$
Let $a+b+c=3$.
Hence,
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}-\frac{9+3\sqrt3}{2\sqrt3}=\sum\limits_{cyc}\frac{\sqrt3+\sqrt{a}}{3-a}-\frac{3\sqrt3+3}{2}=$$
$$=\sum\limits_{cyc}\left(\frac{1}{\sqrt3-\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
| Using what you already did ( the discriminant is no-negative and thus $\;a^2-4b\ge8\;$) then
$$a^2+b^2\ge8+4b+b^2=(b+2)^2+4$$
So the above rightmost expression is at least...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is there a less trickier way to solve this differential equation? Solve the following differential equation:
$$ y^3 dy+(x + y^2)dx = 0$$
Solution:
The following solution uses the substitution $y^2=tx$.
$$y^3\frac{dy}{dx}+x+y^2=0\tag1$$
Differentiating $y^2 = tx$,
$$ 2y\frac {dy}{dx}= t + x\frac{dt}{dx}$$
Now substituti... | The substitution presented does seem somewhat unmotivated.
Given
\begin{equation}
(x+y^2)dx+y^3dy=0
\end{equation}
one would be inclined to first try the substitution $x=uy^2$ so that the $y^2$ might factor out of the first term.
This gives $dx=y^2du+2uy\,dy$ yielding after a bit of algebra the separable equation
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | This is straightforward when you have another identity in your pocket, namely
$$\sec\theta \cdot \csc\theta = \tan\theta + \cot\theta \tag{$\star$}$$
This implies
$$\sec^2\theta \csc^2\theta = ( \tan\theta + \cot\theta )^2 = \tan^2\theta + 2 \tan\theta\cot\theta + \cot^2\theta = \tan^2\theta + 2 + \cot^2\theta$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
General term of recurrence relation Find the general term of the following recurrence relation:
$$a_{1} = 2$$
$$a_{n+1} = \frac{2a_{n} - 1}{3}$$
I've tried to find the first few terms:
$$a_{1} = 2$$
$$a_{2} = \frac{2 \cdot 2 - 1}{3} = 1$$
$$a_{3} = \frac{2 \cdot 1 - 1}{3} = \frac{1}{3}$$
$$a_{4} = -\frac{1}{9}$$
$$a_{... | Following your (@Jack D'Aurizio) suggestion, here is what I understood:
The given recurrence relation: $3a_{n} = 2a_{n-1} - 1$, with initial conditions: $a_{1} = 2$ is a $1^{st}$ degree linear non-homogeneous recurrence relation.
Solution:
We are searching for solution1 of the form: $a_{n} = b_{n} + h_{n}$ (1), where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
connection between two distinct prime numbers $p,q$ such that $p^n=q^2+q+1$ Let $p,q$ be two distinct prime number such that $p^n=q^2+q+1$, where $n \in \Bbb{N}$. I want to proof that, this forces $n=1$.
| I borrowed two books on contest problems; this one is An Introduction to Diophantine Equations, by Titu Andreescu, Dorin Andrica, and Ion Cucurezeanu. Most of what is needed for your problem, probably all, is in section 4.2, The Ring of Integers of $\mathbb Q[\sqrt d].$ This is mentioned above by Jack, suggesting worki... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
show that $ab+bc+ca|(b^2-ac)^2$ Let $a,b,c$ be postive integers,and such $(a,b)=(b,c)=(c,a)=1$,if
$$ab+bc+ca|(a^2-bc)^2$$
show that
$$ab+bc+ca|(b^2-ac)^2$$
Own idea,maybe we look for some identities?
| $$0\equiv (a^2-bc)^2 \equiv (a^2-(-ab-ac))^2 = a^2(a+b+c)^2 \bmod (ab+bc+ca).$$
However, $\gcd(a,bc)=1 \implies \gcd(a,ab+bc+ca)=1,$ so
$$ab+bc+ca | (a+b+c)^2.$$
$$ab+bc+ca | (b^2+ab+bc)^2.$$
$$ab+bc+ca | (b^2-ca)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| $$ x^2 - 2 x + 3 = ( x + 1) ( x -3) $$
$ x= -1, x = 3 $ should satisfy the polynomial. So,
$$ -a + 8 -b + 6 =0 ;\, a \,3^3 + 8 \,3^2 - 3 \,3 + 6 =0 ;$$
Siimplify and solve for $a,b. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 4
} |
Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
| Lengthy but not impossible. We have for sure
$$\begin{eqnarray*} f(x)&=&\frac{1}{(x+3)(x+4)^2(x+5)^3}\\&=&\frac{A}{x+3}+\frac{B}{x+4}+\frac{C}{x+5}+\frac{D}{(x+4)^2}+\frac{E}{(x+5)^2}+\frac{F}{(x+5)^3}\end{eqnarray*}\tag{1}$$
where
$$ F=\lim_{x\to -5}(x+5)^3 f(x) = \lim_{x\to -5}\frac{1}{(x+3)(x+4)^2} = -\frac{1}{2},$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Non-trivial solution to a homogeneous system of linear equations. I have equations:
\begin{cases}
2x + y - z = 0\\
x - 2y - 3z =0\\
-3x - y + 2z =0
\end{cases}
After I put this in matrix row-reduced echelon form I get solutions of $x=0, y=0, z=0$. But my book says it has a non-trivial solution. Could some... | We can write the LHS of your system as
$$
\left[
\begin{array}{rrr}
2 & 1 & - 1 \\
1 & -2 & - 3 \\
-3 & -1 &2
\end{array}
\right]
$$
and apply Gauß elimination:
$$
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
2 & 1 & - 1 \\
-3 & -1 &2
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
0 & 5 & 5 \\
0 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2... | Using the following trigonometric identitiy:
$$(1+\cot x)^2 =\frac{ 2 \tan\left(\frac{\pi}{4}+x\right)}{\tan(2 x) \, \tan(x)},$$
We have
\begin{align}2 \int_0^{\pi/8} \ln(1+ \cot x) \, dx=
& \ln2 \int_0^{\pi/8} dx-\int_0^{\pi/8} \ln \tan 2 x \, dx
\\ -
& \int_0^{\pi/8} \ln \tan x \, dx+\int_0^{\pi/8} \ln \tan\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatoric... | You are just showing that the Pythagorean theorem is a consequence of some property of the (complex) exponential function, there is no circularity in such argument. For instance, we may define, for any $z\in\mathbb{C}$,
$$ f(z)=e^{z}=\sum_{n\geq 0}\frac{z^n}{n!} \tag{1}$$
and prove through a combinatorial argument that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 2
} |
$X\sim\mathcal N(0,1);\ Y = \sqrt{|X|}$; find $f_Y(y)$
Let $X\sim\mathcal N(0,1)$ be a normal distribution and $Y = \sqrt{|X|}$. Find $f_Y(y)$, the probability density function of $Y$.
I figure trying to find the expected value of $Y$ and the variance is a good start, but I have no idea how to integrate $\frac{2}{\sq... | \begin{align}
f_Y(y) & = \frac d {dy} \Pr(Y\le y) = \frac d{dy} \Pr( -y^2 \le X \le y^2) \\[10pt]
& = \frac d {dy}\int_{-y^2}^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx = 2 \frac d {dy} \int_0^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx \\[10pt]
& = 2 \frac d {dy} \int_0^u \cdots = 2 \frac{du}{dy} \cdot \frac d {du} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
what is remainder when in the following scenario? What is the remainder when $x^{2016} - 1$ is divided by
$ x^{5}+x^{4}+x^{3}+x^{2}+x+1 ?$
how to solve this kind of problem.
| $$x^{2016}-1=(x-1)(x^{2015}+x^{2014}+\cdots+x+1)$$
When you divide $x^{2015}+x^{2014}+\cdots+x+1$ by $x^5+x^4+x^3+x^2+x+1$ by long division, $6$ terms cancel out from the dividend at each step. For example, in the first step, the quotient term is $x^{2010}$, which on multiplication with $x^5$ cancels out $x^{2015}$, $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding a closed form for a recurrence relation $a_n=3a_{n-1}+4a_{n-2}$ Consider the sequence defined by
$$
\begin{cases}
a_0=1\\
a_1=2\\
a_n=3a_{n-1}+4a_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $a_n$.
I tried listing out examples, but I don't see any common pattern between them. All solutions ... | This recurrence relation can be described with the following matrix:
$\operatorname A = \left[\begin{array}{cc}
3 & 4 \\
1 & 0 \\
\end{array} \right]$
This means that the equation, $$ \operatorname A\left[\begin{array}{cc}
a_{n-1} \\
a_{n-2} \\
\end{array} \right] = \left[\begin{array}{cc}
3a_{n-1} + 4 a_{n-2} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right ... | Another way to solve this problem is to define $a_n = (3 + \sqrt{5})^n + (3-\sqrt{5})^n$ and note that the characteristic equation of this sequence is $x^2 - 6x + 4 = 0$. Therefore $a_{n+2} = 6a_{n+1} - 4a_n$
Obviously we have that $2 \mid a_1$ and $4 \mid a_2$. Now assume that $2^n \mid a_n$ for all $n \le k$ for some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Squeeze theorem inequality: $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$ show that for $n>1$, $$\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$$
by using $\frac{1}{n} < \ln n $
and $\ln(1+n) < n$ for $n>1$
I am unable to prove $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n $
| $$\frac{1}{\frac{n+1}{n}}=\frac{n}{n+1}<\ln(\frac{n+1}{n})=\ln(1+\frac{1}{n})<\frac{1}n$$
For $n>1$, we have $0<\ln(1+\frac{1}{n})<1, $ so
$$\ln(1+\frac{1}{n})^n<1$$
Also, $$1>\ln(1+\frac{1}{n})^n=n*\ln(1+\frac{1}{n})>n(\frac{n}{n+1})>\frac{n}{n+1}\text{ , for }n>1$$
finishes the proof
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Equation with double absolute value I have problem with solving the following equation.
$$2x - |5-|x-2|| = 1$$
How to handle absolute value in absolute value?
I have tried multiple times to solve it but I get no solution.
| No "easy way out". Solve the first absolute value $$|x-2|=\begin{cases}-x+2, & x<2\\\phantom{-}x-2, & x\ge 2 \end{cases}$$ to take cases
*
*Case 1: $x<2$. Then $$2x-|5-|x-2||=2x-|5-(-x+2)|=2x-|x+3|$$ And now take subcases (but do not forget that you are in $x<2$) depending on the second absolute value
$$|x+3|=\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Schrodinger equation to find general wave function I am trying to answer this question:
I have tried solving the Schrodinger equation using separation of variables.
However in the later time wave function i can not seem to get the exp(-3i...)
My current workings are:
| Expanding
\begin{align*}
\Psi(x,0) &=
\frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+
\frac{2}{\sqrt{a}} \sin \frac{\pi x}{a} \cos \frac{\pi x}{a} \\
&=
\frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+
\frac{1}{\sqrt{a}} \sin \frac{2\pi x}{a} \\
\omega_{n} &= \frac{n^2 \pi^2 \hbar}{2ma^2} \\
\psi_{n} (x,t) &= \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational. Using the fact that $\sqrt{2}$ is irrational, prove the following:
There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational.
My attempt:
Let $x= \sqrt{2}$ and $y= \... | Your argumentation is fine.
However, one has to be careful that the statement "So, $x+y$ is irrational if $x$ and $y$ are irrational." does not hold for all $x,y \in \mathbb{R}$ (but, of course, for your choice).
An easier choice should be $x = y = \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$
Let $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$. Find $f'(x)$.
Is that integral undefined/nonexistant, or just impossible to integrate?
In this case does $f'(x)$ exist and can be solved normally?
Is it correct that $f'(x)=\frac{\sin(x^2)}{x^2}2x$?
| The answer provided by @Clarinetist is the most efficient way to solve this problem. Here is another way.
Using the series definition of the sine function we have
\begin{equation}
\sin(z) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}
\end{equation}
and
\begin{align}
f(x) &= \int\limits_{0}^{x^{2}} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \bino... | Generating Functions
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}x^n\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^k\binom{n}{k}x^{n+k}\\
&=\sum_{n=0}^\infty x^n(1-x)^n\\
&=\frac1{1-x(1-x)}\\
&=\frac1{1-x+x^2}\\
&=\frac{1+x}{1+x^3}\\[6pt]
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
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How to integrate this function over this surface? I'm solving a physics problem, and at some point I need to solve this integral
$$\iint\limits_S \frac{1}{x+2a} dxdy$$
where
$$S=\{(x,y,z):x^2+y^2\leq a^2, \ x\geq0, \ z=0\}$$
So $S$ is the right semicircle corresponding to the circle centered at the origin with radius ... | Integrating $y$ first and using the trig substitution $x = a\sin\theta$ leads to
$$
\int_{0}^{a} \frac{2\sqrt{a^{2} - x^{2}}\, dx}{2a + x}
= \int_{0}^{\pi/2} \frac{2a\cos^{2}\theta\, d\theta}{2 + \sin\theta}.
\tag{1}
$$
The tangent half-angle substitution $t = \tan\frac{\theta}{2}$ gives
$$
\cos\theta = \frac{1 - t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find $a,b,c$ where $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ Find integers $a,b,c$ such that $1<a<b<c$ and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ .I tried it solve using elementary number theory but I can't proceed . Somebody help me.
| Not completely different from Oleg517's. write $x=a-1$, $y=b-1$, $z=c-1$, we have $1\le x < y <z$ and $$\frac{abc-1}{(a-1)(b-1)(z-1)} = 1 + \frac{x+y+z + xy + yz + zx}{xyz}$$ is an integer.
On the other hand, $$0< x+y+z + xy + yz + zx \le xyz + (xy+yz+zx) = xyz + x(y+z) + yz < xyz + xyz + yz < 3xyz,$$ so $x+y+z + xy + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find Min of function $x+\frac{2}{x}$ for x>0 with following solution? I want to find Minimum of function $x+\frac{2}{x}$ for $x>0$ with following solution. My problem is how can we say we found the answer with following solution.
Solution:
Let Minimum be something like c. We have:
$x+\frac{2}{x} \ge c \to \frac{x^2+2}{... | Let's look at the original inequality $$x + \frac{2}{x} \ge c$$ after you have concluded that $$-2\sqrt{2} \le c \le 2\sqrt{2}.$$ If we suppose $c = 0$, then is the original inequality true? Yes: $x + 2/x \ge 0$ is a true statement for $x > 0$. Indeed, any value of $c$ in the interval you obtained, when substituted... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$
After substitution $a=\frac{y}{x}$... I tried C-S, but without success.
| I proved this inequality!!!
Let $a=\frac{x}{z}$, $b=\frac{y}{x}$ and $c=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}\geq\frac{3}{\sqrt2}$.
Now by AM-GM $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}=\sum\limits_{cyc}\frac{2\sqrt2x^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to show $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$ I am new to the subject of rings of formal power series. The problem is to show that $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$, where $\mathbb{Z_6}[[x]]$ is the ring of formal power series with coefficients in $\mathbb{Z}_6$, also known as $ \mathbb{Z }/6\mathbb{Z}$.
M... | First at all $1+3x$ is an element of $\mathbb{Z_6}[[x]]$in which all the coefficients except $a_0=1$ and $a_1=3$ are nuls. Your expand of $\frac{1}{1+3x}$ modulo $6$ is correct. It follows, changing $3$ by its equivalent modulo $6$, which is $-3$.
$$(1+3x)(1-3x+3x^2-3x^3+....\pm3x^n\mp3x^{n+1}\pm.......)$$ This multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determina... | $${\begin{vmatrix}1&a & b &c\\1 &b &c &d \\1 &c &d &a\\1&d &a &b &\end{vmatrix}}=-{\begin{vmatrix}1&a & b &d\\1 &b &c &a \\1 &c &d &b\\1&d &a &c &\end{vmatrix}}$$
because the second determinant is the same as the first by cyclically permuting the columns $2,3$ and $4$, an cyclically permuting the four rows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
A circle inside an ellipse Consider an ellipse with semi-axes $a$ and $b$, taller than it is wide with a small circle of radius $r$ inside. Assume the circle falls to the lowest point possible while staying inside the ellipse.
If $2r\le a-c$ then the circle and ellipse will meet at a single point at the bottom. If $2r... | I got $$w =b - \sqrt{\frac{(b^2-a^2)(a^2-r^2)}{a^2} }-r $$
Start with polar coordinates for the ellipse $$ \begin{pmatrix}x \\ y \end{pmatrix} = \tfrac{a b}{\sqrt{b^2-(b^2-a^2) \cos^2 \varphi}} \begin{pmatrix} \sin \varphi \\ -\cos\varphi \end{pmatrix}$$
where $(x,y)$ are the contact point coordinates relative to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Fractions in Questions and Answers
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