Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Quadrilateral inscribed in semicircle. How to find BC using sin ABD? AB = 3, BD = 5 , tanABD = 0.75
BC is diameter
Question : How to find BC using sin ABD?
I can find sin cos ABD and lenght AD.
I can find BC using Ptolemy's theorem.
| Draw the other diagonal $AC$. Since $BC$ is a diameter of the circumcircle of the quad $BADC$, $$\angle\, BAC = \angle \, BDC = 90^{\circ}$$
Let $P$ be the intersection point of the diagonals $AC$ and $BD$. Then triangle $ABP$ is right-angled, so
$$\frac{AP}{AB} = \frac{AP}{3} = \tan(\angle\, ABP) = \tan(\angle\, ABD)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all points order 3 on an elliptic curve Let $P=(x,y)\in E(\mathbb{F}_p)$ by a Weierstrass equation $y^2=x^3+ax+b$. Show that $3P=\mathcal{O}$ iff $3x^4+6ax^2+12bx-a^2=0$.
I derived that every point in $\{P\in E(\mathbb{F}_p)|3P=\mathcal{O}\}$ is a root of the above equation, then no further clue. I also tried to... | $\textit{Proof:}$
First, note that $3P=\infty$ iff $2P=-P$. Then let $P=\mathbb{F}_p$. If $P$ is the finite point $(x,y)$, then $-P=(x,-y)$; iff $P=\infty$, then $2P=3P=\infty$. If $y=0$, then $2P=\infty\neq-P$. Otherwise, we can calculate $2P=(x', y')$ by
$$\lambda=\frac{3x^2+a}{2y}, x'=\lambda-2x, y'=\lambda(x-x')-y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1... | Hint: The inequality is:
$$\frac{a}2\cdot\frac{b}2\cdot\frac{c}2 \geqslant (1-a)\cdot(1-b)\cdot(1-c)$$
which follows from Karamata's inequality as $\left(1-a, 1-b, 1-c\right) \succ \left(\dfrac{a}2, \dfrac{b}2, \dfrac{c}2\right)$, and $t\mapsto \log t$ is concave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}... | $$ax^2+bx+c=0$$ so $$x_1=\frac{-b + \sqrt{b^2-4ac}}{2a} \; and \; x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$$
From this we have $x_1x_2=\frac{c}{a}$ and $x_1+x_2=-\frac{b}{a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Convergence of the complex power series $\sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n$ I have to study the convergence of
$$S(z)=\displaystyle \sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n,\quad \quad z\in \mathbb{C}$$
I have defined $c_n=\dfrac{1\cdot ... | Using
$$
\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}} = \frac{2}{\pi }\int_0^{\pi /2} {\sin ^{2n} tdt}
$$
and summation together with analytic continuation, we obtain
$$
\left( {\frac{1}{{\sqrt {1 - z} }} -1= } \right)\sum\limits_{n =1}^{N - 1} {\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}}z^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $ without using L'Hopital's rule. Let $ n $ be a positive integer greater than $ 1 $.
Find : $$ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $$
Without L'Hopital's rule or series expansion.
Here is ... | By the binomial formula $$(1+(-1)^kx)^{1/k}=1+\frac{(-1)^k}kx+O(x^2),$$ so that
$$\begin{align}\prod_{k=2}^{2n+1}(1+(-1)^kx)^{1/k}
&=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx+o(x)\right)\\
&=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx\right)+o(x)\\
&=1+x\sum_{k=2}^{2n+1}\frac{(-1)^k}k+o(x)
\end{align}$$
so that the desir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
How to solve the following radical equation?
$$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$
I'm looking for an easy way to solve it.
| If $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
Then:
$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}}}}}=x$
In general:
$x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\ldots}}}}$
Which means that:
$2-(x^2-2)^2=x$
$x^4-4x^2+4+x-2=0$
$x^4-x^3+x^3-x^2-3x^2+3x-2x+2=0$
$(x-1)(x^3+x^2-3x-2)=0$
$(x-1)(x^3+2x^2-x^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Reduction of ${}_{4}F_{3}(\cdots; 1)$ It is suspected that
$$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{8}{\pi} - \frac{4 \, \sqrt{2} \, \Gamma^2\left(\frac{3}{4}\right)}{\Gamma^3\left(\frac{1}{2}\right)} = 1.020959562466698996....$$
The ${}_{4}... | Maple says:
convert(hypergeom([1/4,1/4,3/4,3/4],[1,3/2,3/2],1), `2F1`);
$$ \frac{8}{\pi}+\frac{8}{\pi}{\it EllipticK} \left( i \right) -\frac{8}{\pi}{\it EllipticE}(i)
$$
This could also be written as
$$ \frac{8}{\pi} - \frac{8}{\pi} \int_0^1 \frac{t^2\; dt}{\sqrt{1-t^4}}$$
and Maple says that is indeed
$$ \frac{8}{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What value does the area of the region enclosed by $\sum_{k=1}^m (x^k+y^k)=2$ converge to as $m\rightarrow\infty$? Find the area enclosed by the curve
$$\sum_{k=1}^m (x^k+y^k)=2$$
where $m$ is even and $m \rightarrow\infty$.
That is for a sufficiently large even number $m$, we have:
$$x^1+y^1+x^2+y^2+x^3+y^3+\dots+x^m+... | Doing some obvious stuff.
$\begin{array}\\
2
&=\sum_{k=1}^m (x^k+y^k)\\
&=\sum_{k=1}^m x^k+\sum_{k=1}^m y^k\\
&=\dfrac{x-x^{m+1}}{x-1}+\dfrac{y-y^{m+1}}{y-1}\\
&\to \dfrac{x}{x-1}+\dfrac{y}{y-1}
\qquad\text{as }m \to \infty\\
&= \dfrac{x-1+1}{x-1}+\dfrac{y-1+1}{y-1}\\
&= 1+\dfrac{1}{x-1}+1+\dfrac{1}{y-1}\\
\text{so}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| Let
*
*$f(x) = x^4 + px^3 + qx^2 + rx + 1$.
*$\Lambda = \{ a, b, c, d \}$ be the set of roots of $f(x)$.
*$\Omega = \left\{ \frac{\pm 1 \pm i}{\sqrt{2}} \right\}$ be the set of roots of $x^4 + 1$.
We have
$$\begin{align}\mathcal{P}\stackrel{def}{=}\prod_{\lambda \in \Lambda}(\lambda^4 + 1)
&= \prod_{\lambda \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Stuck trying to sum this series I actually got this question from a book - TMH. Here is the question:
Prove that $\cot^2 \frac{\pi}{2n+1}\cdot \cot^2 \frac{\pi}{2n+1}\cdot \ldots \cdot \cot^2 \frac{\pi}{2n+1}$ are the roots of the equation
$$
x^n - \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} x^{n-1} +\frac{\binom{2n+1}{5}... | You may know the following formula that you will find for example here in paragraph "Sine, cosine, tangent of multiple angles"
$$\displaystyle \tan(n\theta )=\frac {\sum _{k{\text{ odd}=2q+1}}(-1)^{q}{n \choose k}(\tan\theta) ^{2q+1}}{\sum _{k{\text{ even}=2q}}(-1)^{q}{n \choose k}(\tan\theta)^{2q} }\tag{1}$$
Let us ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
| Without loss of generality, we can assume $a\leq b\leq c$. Let's do some remarks. I will suppose $d> 3$, the first cases are easy to check by hands and they will coincides with the solutions in $a=1$.
If $a,b,c\geq 3$ then $3$ divides $a!+b!+c!$, but $3$ does not divide $2^d$. Hence $a=1,2$. (the case $a=0$ is equal t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solve the ODE $2y'' - y' = 2\cos(3t)$ using laplace transform
$2y'' - y' = 2\cos(3t)$
\begin{align*}
2\mathcal{L}[y''] - \mathcal{L}[y'] &= 2\mathcal{L}[\cos3t] \\
2\bigg[-y(0) - sy(0) + s^2Y(s) \bigg] + y(0) - sY(s) &= \frac{2s}{s^2+3^2} \\
\end{align*}
This works out to $$Y(s)(s^2-s) = \frac{2s}{s^2+9} + 3 - 2s$$ a... | You should get:
$$Y(s)(2s^2-s)+y(0)(1-2s))-2y'(0)=\dfrac {2s}{s^2+9}$$
Maybe you are given initial conditions ?
You wrote :
$$Y(s)(\color{red}{2s^2}-s) = \frac{2s}{s^2+9} + 3 - 2s$$
You should keep it this way for simplicity:
$$Y(s) = \frac{2}{(s^2+9)(2s-1)} + \dfrac { 3 - 2s}{s({2s}-1)}$$
It's easier to decompose in s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove thi... | Let $f(n,x)=n-\frac{(n-1)n(n+1)}{6}x^2$ the RHS of the required inequality. Wlog we can ssume $n \ge 3$ as for $n=1,2$ the inequality is trivial ($n=2$ reduces to $2\sin^2(\frac{x}{2}) \le \frac{x^2}{2}$ which follows from $|\sin x| \le |x|$).
Using the well known inequality $|\frac{\sin nx}{\sin x}| \le n$, the requir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluation of $\int^{1}_{0}\frac{x^4}{1+x^8}dx$
How can I Integrate $\displaystyle \int^{1}_{0}\frac{x^4}{1+x^8}dx$?
I have searched in that forum and get the result for $\displaystyle \int^{\infty}_{-\infty}\frac{x^4}{1+x^8}dx$ or for $\displaystyle \int^{\infty}_{0}\frac{x^4}{1+x^8}dx$
Although i know the formula $... | Note
$$\int^{1}_{0}\frac{x^4}{1+x^8}dx
=\frac1{2\sqrt2}\int_0^1 \left(\frac{x^2}{x^4-\sqrt2x^2+1}
- \frac{x^2}{x^4+\sqrt2x^2+1}\right)dx $$
where
$$\begin{align}
I(a)& =\int_0^1 \frac{x^2}{x^4+ax^2+1}dx \\
&=\frac12\int_0^1 \frac{x^2+1}{x^4+ax^2+1}dx -
\frac12\int_0^1 \frac{1-x^2}{x^4+ax^2+1}dx \\
&=\left(\frac1{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem... | Note that we can write
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx&=\frac12\int_{-\infty}^\infty \frac{(x^2+b^2)+(x^2+a^2)}{(x^2+a^2)(x^2+b^2)}\,dx-\frac{1}2\int_{-\infty}^\infty\frac{a^2+b^2}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac12\int_{-\infty}^\infty \frac1{x^2+a^2}\,dx+\frac12\int_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$ Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$
My Attempt :
$$(y-px)^2 (1=p^2)=a^2p^2$$
$$y-px=\frac {ap}{\sqrt {1+p^2}}$$
$$y=px+\frac {ap}{\sqrt {1+p^2}} ... | Both sign variants of the square roots give solutions, you can include this by using also the solutions where $a$ is replaced by $-a$.
In $y=px+f(p)$ the derived equation is $0=y''(x+f'(y'))$ so that any solution will be composed of segments where one of the factors is zero. You already found the lines $y'=p=c$.
lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
y=x^3sinx,what's the 6th derivative of y at x = pi/6? question 1： $y=x^3\sin x$, what's $y^{\left(6\right)}\left(0\right)$ ?
I have solved it use Taylor's Formula in the follow way.
step1:
$y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$
and x_0 = 0 for the questi... | Thanks for @InterstellarProbe's helping me solve this problem.
$y=x^3sinx$
step1:
Let : $x=u+\frac{\pi }{6}$
then
$y=\sum _{n=0}^{\infty }\:\frac{y^{\left(n\right)}\left(0\right)}{n!}u^n$
and
$y=\left(u+\frac{\pi }{6}\right)^3sin\left(u+\frac{\pi }{6}\right)
=\left(u^3+\frac{\pi }{2}u^2+\frac{\pi^2 }{12}u+\frac{\pi \:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $(1+\cos2x)^2+(-2\sin2x)^2=1+2\cos2x+\cos^22x+4\sin^22x$ I'm supposed to find the arc length of this polar curve : $r=1+\cos2x$ where $0\leq x \leq \pi/4$.
I know that I have to use this formula $s=\displaystyle\int_{0}^{\pi/4} \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2}
dx$ My teacher always tells us to simp... | For $r(x) = 1+\cos(2x)$, you have that your expression reads
$$ \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2} = \cos(x) \sqrt{10 - 6 \cos(2 x)} = \cos(x) \sqrt{4 + 12 \sin(x)^2} $$
The integral now can be done by using $y=\sin(x)$ and $dy = \cos(x) dx$. You should obtain
$$
s = \sqrt{5}/2 + \tanh^{-1} \left( \sqrt{3/5} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$ Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$
My Attempt:
The given equation is
$$(D^2+4)y=x\sin^2 x$$
It's auxiliary equation is
$$m^2+4=0$$
$$m^2=-4$$
$$m=\pm 2i$$
$$\textrm {Complementary Function (C.F)}=c_1 \c... | Starting from here:
$$y_p=\frac {x\sin^2 x}{D^2+4}$$
$$2y_p=\frac {x(1-\cos (2 x))}{D^2+4}$$
$$2y_p=\frac {x }{D^2+4}-\frac {x\cos (2 x)}{D^2+4}$$
$$2y_p=\dfrac x 4-(x-\dfrac {2D}{D^2+4})\frac {\cos(2 x)}{D^2+4}$$
More simply:
$$2y_p=\dfrac x 4-x\frac {\cos(2 x)}{D^2+4}$$
$$2y_p=\dfrac x 4-\Re \{x\frac {e^{2 ix}}{D^2+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Computing the index of fixed point for the system of equations $x'=x^2$ and $y'=-y$. Consider the system of equations $$x' = x^2$$ $$y'=-y.$$ Compute the index of fixed point at the origin both visually and computationally, or just write down the integral.
$\textbf{Solution:}$ To find the fixed point, integrate both eq... | To get fixed points, we need $x'=0$ and $y'=0$. So, the fixed points are $x^2 = 0$ and $-y=0$ so $x=0$ and $y=0$. Thus, $(0,0)$ is a fixed point.
Let $f=x^2$ and $g=-y$, then $f_x = 2x, f_y=0, g_x = 0,$ and $g_y=-1.$ Therefore the Jacobian is $$J = \begin{pmatrix} f_x & g_x \\ f_y & g_y \end{pmatrix}$$ $$J_{(0,0)} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +... | Using $\sum $ to denote cyclic sums,
\begin{align}
\sum \sqrt{a^2-ab+b^2} \;+ 9\sqrt[3]{abc} &\leqslant \sum \left(\sqrt{a^2-ab+b^2}+3\sqrt{ab}\right) \tag{*}\\
&\leqslant \sum 2(a+b) \tag{**}\\
&=4\sum a
\end{align}
Notes: * AM-GM inequality ** Jensen's inequality with the concave function $t \mapsto\sqrt{t}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use induction to prove that $3 |(7^n −1)$ for every natural number n. This is what I have so far and now I'm getting lost.
Proof - We prove by induction. Let $P(n)$ be the statement "$3|(7^n - 1)$". Since $3|6$, we see that $P(1)$ holds. Suppose that is true for $n = k$. We must show the result is true for $n = k + 1$... | Your proof is valid if we assume that we know that $a^{m+1} - b^{m+1}=(a^m + ba^{m-1} + ... + b^{m-1}a + b^m)$.
But if we do know that then your proof is one line: $7^n - 1=(7-1)(7^{m-1}+...+1) = 6*(7^{m-1}+...+1)$ and no need for induction at all.
So I'm assuming you aren't supposed to know that result. But could yo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to
Let an angle between the vectors $a$ and $b$ be $\frac{2\pi}{3}$. If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to $\frac23/\frac25/\frac13/\frac15$?
My attempt:... | Remember, $a$ and $b$ are vectors, not scalars, so $a^2 = 4b^2$ doesn't make a lot of sense. Much of your working suffers from the same issue, where it is unclear what is a vector, what is a scalar, and hence what can even be added together.
Our information is:
*
*The angle between the two vectors is $\frac{2\pi}{3}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal. Let $z=a+b\omega + c\omega^2$
$$z=a+b\omega -c (1+\omega)$$
$$z=a-c+\omega (b-c)$$
Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$
How should I proceed?
| Note that:
$$|a+b\omega+c{\omega}^2|^2=(a+b\omega+c{\omega}^2)(a+b{\omega}^2+c\omega)$$
This is because $\bar{\omega}={\omega}^2$, and $\bar{{\omega}^2}=\omega$. If you multiply out the terms, you'll get $a^2+b^2+c^2-ab-bc-ca$, which is equal to $\frac {(a-b)^2+(b-c)^2+(c-a)^2}{2}$. Since $a,b,c$ are not all equal, two... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and ... | Clearly, $x=0$ is not a root. WLOG, assume that $|b| \ge |a|$ (otherwise, letting $x = \frac{1}{y}$, we have
$y^4 + by^3 + 2y^2 + ay + 1 = 0$; then swap $a$ and $b$).
Let $x_1 \le x_2 \le x_3 \le x_4$ be the four real roots. We split into two cases.
1) If $x_1x_4 > 0$, then either $x_i>0, \forall i$ or $x_i < 0, \for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{n=1}^{\infty}((n+\frac{1}{2})\ln(1+\frac{1}{n})-1)=1-\ln(\sqrt{2\pi})$ I am looking for a derivation of the following sum:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$
My current derivation(s) uses the zeta function at negative integers (an... | One has
\begin{align*}
& \sum\limits_{n = 1}^\infty {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - t}}{{n + t}}dt} } = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3... | ==== old answer ===
Assume the answer is $M$.
Then there are $k$ numbers less than or equal to $M$ that are not divisible by $n$.
So $k = f(M)$ where $f(M)$ is the function that tells you how many number less than or equal to $M$ are not divisible by $n$.
So if we can "undo" $f(M)$ to get $f^{-1}(k) = M$ we can solve f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate... | Let $n = 15$ and $P(x) = \frac{x^n-1}{x-1} = \sum\limits_{k=0}^{n-1} x^k$. Last part of your question suggest you already know:
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = - \frac12 \left.\frac{d^2}{dx^2} \frac{P'(x)}{P(x)}\right|_{x=1} = -\frac12\left.\frac{d^3}{dx^3}\log P(x)\right|_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$ Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$
I have proven the first half of the inequality, which is t... | Because $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2019}{2020}=\sqrt{\frac{1\cdot3}{2^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{5\cdot7}{6^2}\cdot...\cdot\frac{2017\cdot2019}{2018^2}\cdot\frac{2019}{2020^2}}<$$
$$<\sqrt{\frac{2019}{2020^2}}<\frac{1}{\sqrt{2020}}<\frac{1}{44}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integrate $\int \frac {dx}{x\sqrt{x^2-49}}\,$ I'm trying to integrate the following:
$$\int \frac {dx}{x\sqrt{x^2-49}}\,$$
using the substitution $x=7\cosh(t)$
This is as far as I've gotten:
$\int \frac {dx}{x\sqrt{x^2-49}}\,$ = $\int \frac {7\sinh(t)dt}{7\cosh(t)7\sinh(t)}\,$ = $\int \frac {dt}{7\cosh(t)}\,$ = $\int \... | Note
$$\frac{1}{7} \arctan(\sinh t)
=\frac{1}{7} \arctan\sqrt{\cosh^2t -1}\\
=\frac{1}{7} \arctan\sqrt{\frac{x^2}{49}-1}
=\frac{1}{7} \arctan\frac{\sqrt{x^2-49}}7\\
= \frac{1}{7} \text{arccot } \frac7{\sqrt{x^2-49}}
= \frac{1}{7}(\frac\pi2- \text{arctan} \frac7{\sqrt{x^2-49}})\\
=- \frac{1}{7}\text{arctan} \frac7{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$... | Hint. That's equivalent to extremising $$\sqrt{6-5x}+\sqrt {1+5x}$$ on the interval $[0,1].$ Can you now continue?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theor... | Here's a more routine answer. First, to find critical points in the interior, compute the gradient:
$$
\nabla f = (4x^3-4yz,4y^3-4xz,4z^3-4xy)
$$Certainly the origin is a critical point, and if any of the variables equal zero, the other two do as well. Otherwise, we have
$$
\begin{cases}
x^3-yz &=0\\
y^3-xz &=0\\
z^3-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I ... | There is also a straight-forward solution with Lagrange multipliers. It is more convenient to examine the (continuous) function
$$
f(x, y, z) = (2-x)(2-y)(2-z)
$$
on the (compact) set $K = \{ (x, y, z) \in \Bbb R^3 \mid x^2+y^2+z^2 = 3 \}$ first, without the restriction to positive arguments.
$f$ attains both minimum ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Computing $\int_Q \frac{xy}{x^2+y^2}dxdy$ I'm asked to compute the following integral:
$$\int_Q \frac{xy}{x^2+y^2}dxdy \qquad Q=[0,1]^2$$
Solution:
First I'm going to study if the integral is convergent. To do this, we notice that
$$\int_Q \frac{xy}{x^2+y^2}dxdy < \infty \iff \int_S \frac{xy}{x^2+y^2}dxdy<\infty$$
wher... | As you wrote it, it looks like you claim
$$ I = \frac{1}{2} \int_0^1 y \ln (y^2) dy = 0.$$
But this is not the case:
\begin{align*}
\frac{1}{2} \int_0^1 y \ln(y^2) dy &= \frac 1 4 \int_{0}^1 \ln(y^2)2y dy \\
&= \frac 1 4 \int_0^1 \ln(u) du.
\end{align*}
Using the fact the $u \ln u - u $ is a primitive of $\ln u$ you ge... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Boundaries of a triple integral between $x^2+y^2=2z$ and $x+y=z$ I am calculating the integral $\iiint_V z\,dV$ between $x^2+y^2=2z$ and $x+y=z$, but I am not sure how to find boundaries for this case. I tried with cylindrical coordinates, but I was unable to solve the problem.
I would also like to know if there is any... | Here I present a way to solve the integral. Maybe there's an easier way to integrate, because now you get some troubling terms in the last integral.
Since we have a function of $z$, we would like to integrate over $z$ first. The boundaries of $z$ evaluates to
$$\frac{x^2 + y^2}2 \le z \le x + y$$
since that's the only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The area of a triangle determined by two diagonals at a vertex of a regular heptagon
In a circle of diameter 7, a regular heptagon is drawn inside of it. Then, we shade a triangular region as shown:
What’s the exact value of the shaded region, without using trigonometric constants?
My attempt
I tried to solve it wit... | It boils down to
$$I= \sin\frac\pi7 \sin\frac{2\pi}7 \sin\frac{3\pi}7= \frac{\sqrt7}8
$$
Let $a =\frac\pi7$ and evaluate
\begin{align}
I^2 & = (1-\cos^2 a ) (1-\cos^2 2a ) (1-\cos^2 3a )\\
& =\frac18 (1-\cos 2a ) (1-\cos 4a ) (1-\cos 8a )\\
& =\frac18 (1-\cos 2a \cos 4a \cos 8a )\tag1\\
\end{align}
where the fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}... | To continue from
$\tag 1 \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) = \dfrac{(ad+bc)^2}{abcd}$
set
$\quad u = ad$
and
$\quad v = bc$
Then substituting in the rhs of $\text{(1)}$, we have
$\quad \dfrac{(u+v)^2}{uv} \ge 4 \text{ iff } (u-v)^2 \ge 0$
Note that if we let $a,b,c,d \in... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$ For $x,y,z\geqq 0$ and $x+y+z=1.$ Prove that$:$
$$2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \leqq \left(x^2+y^2+z^2+3xyz\right)^2.$$
Let $y=\hbox{mid} \{x,y,z\}$ and
$\text{P}= \left[\left(x^2+y^2+z^2\right)(x+y+z)+3xyz\r... | Also, BW helps.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that:
$$\left(\sum_{cyc}(x^3+x^2y+x^2z+xyz)\right)^2\geq4\sum_{cyc}(x^2+xy)\sum_{cyc}(x^3y+x^2yz)$$ or
$$16(u^2-uv+v^2)x^4+8(3u^3-5u^2v+4uv^2+3v^3)x^3+$$
$$+(17u^4-34u^3v+3u^2v^2+38uv^3+17v^4)x^2+$$
$$+2(3u^5-6u^4v-3u^3v^2+5u^2v^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any... | Once we have proved that all of $x,y,z$ are, if the gcd is 1, cubes themselves, we arrive at, using
$$ \delta = \frac{-1 + i \sqrt 3}{2} $$
so that $\delta^2 + \delta + 1 = 0$ and $\delta^3=1,$
$$ ( u^3 + v^3 + w^3)^3 - 27 u^3 v^3 w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + ... | {
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"url": "https://math.stackexchange.com/questions/3697234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that for an integer $x \ge 7$, it follows that $x\# > x^2+x$ Is the following argument sufficient to show that for an integer $x \ge 7, x\# > x^2 + x$.
Please let me know if I made a mistake or if there is a more straight forward way to make the same argument.
Let:
*
*$p_n$ be the $n$th prime
*$p\#$ be the pr... | What you've done looks correct. However, one small thing to note in your step $(1)$ base case, you only need to show $10^2 + 10 = 110 \lt 210$ since $11$ is prime and, as such, will be handled later by induction as $p_5$.
Also, I believe a somewhat simpler way to proceed after your step $(5)$ is to then show, for all $... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solutions to $\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2$. $$\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2.$$
How can such an equation be solved? I don't know where to start?
My attempt: $$(3 - \frac{\sin^2 3 y}{\cos^2 3 y}) \cdot 2 \cos y \sin y = (3 - 2 \sin^2 y)^2$$, where $y = \frac{x}{2}... | By transferring $\cos x$ to the RHS and squaring, we have
$$(3 - \tan^2\tfrac32x)\sin ^2x=(2+\cos x)^2.$$
Using the fact that $\cos^2x+\sin^2x=1$, some rearrangement gives
$$(\tan\tfrac32x\sin x)^2+(1+2\cos x)^2=0.$$
A real solution to this equation would simultaneously require, modulo $2\pi$, that $x=0$ and $x=\pm\fra... | {
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"url": "https://math.stackexchange.com/questions/3699346",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that for integer n greater than 2 and is coprime to 10, the decimal expansion of $\frac{1}{n}$ repeats with period of ${e_n(10)}$ Prove that for integer n greater than 2 and is coprime to 10, the decimal expansion of $\frac{1}{n}$ repeats with period of ${e_n(10)}$ (the order of 10 mod n). I tried computing $\fra... | Note: $\frac 1{a-1} = \frac a{a(a-1)} = \frac {a-1}{a(a-1)} + \frac 1{a(a-1)} = \frac 1a + \frac1{a(a-1)}=\frac 1a + \frac 1a(\frac 1a + \frac 1{a(a-1)})=\frac 1a + \frac 1{a^2} + \frac 1{a^2(a-1)}=....$.
And by induction $\frac 1{a-1} = \frac 1a + \frac 1{a^2} + \frac 1{a^3} + ....$
So $\frac 1{10^m -1} = \frac 1{10^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ I am trying to understand each step in order to get from $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the... | If $x^2 - 4$ can be factored then there exist $a, b \in \Bbb R$ such that
$\quad x^2 - 4 = (x-a) (x-b)$
Now multiplying out the rhs and collecting terms,
$\quad (x-a) (x-b) = x^2 - (a+b)x + ab$
So we need to solve the system of equations
$\tag 1 a + b = 0$
$\tag 2 ab = -4$
From equation $\text{(1)}$ we have
$\quad a = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then ... | Solution of AOPS member (don't remember name)
Setting
$$x=\sqrt{(a+b)(a+c)}-a, \quad y=\sqrt{(b+c)(b+a)}-b, \quad z=\sqrt{(c+a)(c+b)}-c,$$
we have
$$ab+bc+ca=xy+yz+zx,$$
Inequality become
$$x+y+z \geqslant \sqrt{3(xy+yz+zx)}.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Continuity of multivariable function $\frac{xy}{\sqrt{16-x^2-y^2}}$ How can I define where is the following function continuous? The function being: $$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$$
Is it enough with getting the domain by doing $16-x^2-y^2>0$. How does it change if instead of $xy$ I have $\frac{y}{\sqrt{16-x^2... | For $\sqrt{x}$ to be defined, we must have $x\geq 0$. For $\frac{1}{\sqrt{x}}$ to be defined, we must have $\sqrt{x}\neq 0$, i.e. $x\neq 0$. Hence, for the function
$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$ to be defined, we must have $16-x^2-y^2>0$, just as you said.
Now, $xy$ is a continuous function, and $\sqrt{16-x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$ I want to calculate the following improper integral:
$$\int_A \frac{z}{y^2}\\
A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$$
First I noticed that the conditions imply $x^2<2x\rightarrow x<2;1<x^2<y^2\rightarrow y>1$, and thus $... | The conditions $z\ge 0$, $x\ge 1$, $y\ge 0$, $x^2+z^2 \lt 2x$, and $x^2+z^2 \lt y^2$ define the intersection of two solids. Solid one is the quarter of the cylinder $(x-1)^2 + z^2 = 1$ to the right of the plane $x=1$ and above the plane $z=0$. Solid two is the portion of the cone $x^2 + z^2 = y^2$ to the right of the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of "equations cannot have an integral solutions" using modular arithmetic. We a got an equation $$15x^2 -7y^2 =9$$ and we have to prove that this equation cannot have an integral solution. So, this what we will do:
$7y^2 = 15x^2 -9$, and since $3| 15x^2 -9 \implies 3|7y^2$. As 3 cannot divide 7 therefore, it must... | Given
$$15x^2 -7y^2 =9\quad\implies\quad x = ± \frac{\sqrt{7 y^2 + 9}}{\sqrt{15}}\quad\land\quad y = \sqrt{\frac{3}{7}} \sqrt{5 x^2 - 3}$$
we can see that there are no rational solutions, let alone integers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivation of natural log inequality While looking at a proof of this inequality
$(1+\frac{1}{x})^x < e < (1+\frac{1}{x})^{(x+1)}$
The authors take the natural log of on both sides and get
$xln(1+\frac{1}{x}) < 1 < (x+1)ln(1+\frac{1}{x})$
But then they write this last inequality as follows (which I don't understand)
$\... | Let's start from the known equation
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
This is, in fact, two different results:
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 \qquad \text{and} \qquad 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
When you divide both sides of the first equa... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r... | The conversion to polar coordinates has not failed. Note what happens when $r=\tan^2(\theta)\sec(\theta)$, $|\theta| > \pi/2$. Then,
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=\frac{\sin^4(\theta)}{2\sin^4(\theta)}=\frac12$$
And of course, if $\theta =0$ or $\theta =\pi/2$, then
$$\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I ... | For the first inequality, using the substitution of $ a = \frac{y}{x}, b = \frac{z}{y},c = \frac{x}{z}$, (which is a good approach to nomarlize these inequalities), we WTS (see Michaels' solution)
$$ \sum y \times \frac{1}{ (x+y+z) - y } \geq \sum x \times \frac{ 1}{(x+y+z) - y }.$$
This is just rearrangement inequa... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Given $S =\{ x-y \mid (x,y) \in \Bbb R , x^2+y^2=1\}$, find $\max S$ My solution:
$x^2+y^2=1$ is the equation of a circle.
Let $x-y=k$. That becomes the equation of a line. Since $(x-y) \in S$,
the point $(x,y)$ satisfies both the circle and the line equation. So we know that the graph of the circle and the line have a... | Well,... let $y = \pm \sqrt {1-x^2}$ and let $d_1(x) = x-y= x-\sqrt{1-x^2}$ when $y\ge 0$ and $d_2(x) =x-y= x+\sqrt{1-x^2}$ when $y < 0$.
Then $d_1'(x) = 1 -\frac 12\frac 1{\sqrt{1-x^2}}*(-2x)= 1+\frac x{\sqrt{1-x^2}}$
And $d_2'(x) = 1-\frac {x{\sqrt{1-x^2}}$.
Solving for $d_1'(x) = 0$ and $d_2'(x)=0$ we get:
$d_1'(x)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$... | In other words you want to show that $$x^4-x^3+x^2-x+1>0$$ identically.
This is clearly true if $x=0.$
So for $x\ne 0$ factor out $x^2>0$ to get $$x^2\left(x^2-x+1-1/x+1/x^2\right),$$ and then we only need consider the expression in parentheses. This may be written as $$x^2+1/x^2-(x+1/x)+1,$$ and now we set $u=x+1/x,$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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What went wrong in the evaluation of $\int \frac{1}{3-2\sin(x)}dx$? I tried to evaluate the following integral:
$$\int \frac{1}{3-2\sin(x)}dx\,\, $$ with universal substitution, using the fact that
$t(x):=\tan\left(\frac{x}{2}\right)$
$\sin(x)=\frac{2t(x)}{1+t(x)^2}$ and $t'(x)=\frac{1+t(x)^2}{2}$
$\displaystyle\int \... | Use the fact that, if $\color{blue}{\int f(x)dx = F(x)}$, then $\color{blue}{\int f(ax+b)dx = \frac{1}{a} F(x)}$
So, you'd have
$$\frac{1}{\sqrt3}\cdot\left[\frac{2\sqrt3}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)\right] =\frac{2}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Does the following integral converge? $ \int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx $ Does the following integral converge?
$$
\int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx
$$
I haven't solved such problems for a while. So, I would really appreciate it if someone gave me a hint.
Or maybe my solution is correct?
$$
\sin... | Just for the fun of it !
The problem of the convergence being solved, there are analytical solution for this kind of integrals (and antiderivatives; have a look here.
Since @Von Neumann wrote an answer where complex numbers do appear, I wondered what would give the $1,400$ years old approximation
$$\sin(x) \simeq \frac... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Solving $\sin^2x + 3\sin x\cos x + 2\cos^2x=0$, for $0\leq x\leq 2\pi$
Solve
$$\sin^2x + 3\sin x\cos x + 2\cos^2x=0$$
for $0\leq x\leq 2\pi$.
My answers are
$$x=2.03, 5.18 \qquad\text{or}\qquad
x=\frac{3\pi}{4},\frac{7\pi}{4} \qquad\text{or}\qquad
x=\frac{\pi}{2}, \frac{3\pi}{2},$$
but the answer states $x=2.03, 5.18... | Use that $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}
$$ and
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove: $\int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = \frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$ I'm trying to prove
$$I := \int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = \frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$$
which is around $2.5063328837$.
Using the $u$-substitution $u = \sqrt{-1 + \sqrt{... | The explanation for simplicity of the result follows the same reasoning as in my answer here.
You obtained
$$\frac{I}{6} = \int_{0}^{\infty}{\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \,du}$$
The denominator factors over $\mathbb{Q}(i\sqrt[4]{3})$ as $$u^6 (u^2 + 2)^3 + 1 = (u^2+1)^2 h(x) \overline{h}(x)$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Solving $\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$
I need some tips with solving this algebraic equation
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$
I've tried subtracting: $$\frac{8x}{x+2}$$
and also setting $$\sqrt x = t$$
Where "t" is a substitution to make things simpler.
This is how it ends up:
$$x + 2 = 6\sqrt x - \fr... | Middle term splitting will always remain my favorite method for factorising polynomials!
\begin{align*}
&\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x\\
\iff &\dfrac{(x+2)^2+8x}{x+2}=6\sqrt x\\
\iff &(x+2)^2-6\sqrt x(x+2)+8x=0\\
\iff &(x+2)^2-4\sqrt x(x+2)-2\sqrt x(x+2)+8x=0\\
\iff &(x+2-4\sqrt x)(x+2-2\sqrt x)=0
\end{align*}
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Suppose that $x$ and $y$ are real numbers. Prove that if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. Not a duplicate of
Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$
Prove that for any real numbers $x$ and $y$ if $x \neq 0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.
This is e... | You could have said this:
if $y=\dfrac{3x^2+2y}{x^2+2}$, then $y(x^2+2)=3x^2+2y$, so $(y-3)x^2=0$, so $x=0$ or $y=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\f... | Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
As $\dfrac{12}5\cdot\dfrac{3}4>1,$
$$\tan^{-1}\dfrac{12}5+\tan^{-1}\dfrac{3}4=\pi+\tan^{-1}\left(\dfrac{\dfrac{12}5+\dfrac{3}4}{1-\dfrac{12}5\cdot\dfrac{3}4}\right)=?$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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What maths rule allows this expression with powers to be rewritten as below? I have been reading through a programming book and the author asked to calculate the following expression in the program.
$$4*5^3+6*5^2+7*5+8$$
I approached this by expanding the expression like so:
$$(4*5*5*5)+(6*5*5)+(7*5)+8$$
$$500+150+35+8... | The main rule used is the distributivity rule which states that for real $a,b,c$, $$a(b+c)=ab + ac.$$
This rule can easily be used on more than $2$ summands, so say you have a real number $a$ and $n$ real numbers $b_1,\dots, b_n$.
Then,
$$a(b_1+\cdots + b_n) = ab_1 + ab_2 + \cdots + ab_n$$
This means that
$$\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to show $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0$ have five different roots? I have to prove this fact when determining the elementary divisors of a matrix. The numerical solution confirms that this equation indeed has five different roots (one real root, two pairs of conjugate complex roots), but how to show it theo... | To tell if $f$ has any multiple zeros, take the GCD of $f$ and $f'$. (Any multiple zero of $f$ is also a zero of $f'$.)
In this case
$$
\gcd(f,f') = 1 ,
$$
so there are no multiple zeros, so there are $5$ simple zeros.
Euclidean algorithm computation.
$f(x) = x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$,
$f'(x) = 5x^4+20x^3+12x... | {
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"url": "https://math.stackexchange.com/questions/3732742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{... | Think geometrically: The equation $3x+2y+z=7$ describes a plane in $\mathbb{R}^3$ with normal vector proportional to $(3,2,1)$. The closest point to the origin is therefore of the form $(a,b,c)=(3r,2r,r)$. From $3a+2b+c=7$ we have $9r+4r+r=7$, so $r=1/2$, and thus $a^2+b^2+c^2=(9+4+1)/4=7/2$.
(Rohan Nuckchady posted th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+... | The solution is due to @Calvin Lin.
Problem: Find the best constant $C$
such that $x^2+y^2 - 1 \ge C(1-x)(1-y)$ holds for all
$x, y \ge 0$ with $x^3+y^3 = 1$.
Solution: The best constant is $C = 2^{4/3} + 2^{2/3} + 2$. Let us prove it.
Let $w = x + y$. Since $x^3+y^3 \le (x+y)^3 \le 4(x^3 + y^3)$, we have $1\le w \le \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Taylor series for 1/(1-x). I am trying to write a Taylor series for $$ f(x)=\frac{1}{1-x}, \ x<1 \ .$$
In most sources, it is said, that this function can be written as a Taylor series, if $$ \left| x \right|<1. $$
However, I don't get the same condition for x.
Because $$ f^{\left(n\right)}\left(x\right)=\frac{n!}{\le... | It is a well-known and indisputable (even elementary) fact that for all $x\ne1$ and all natural $n$
$$\sum_{k=0}^n x^k=\frac1{1-x}-\frac{x{^{n+1}}}{1-x},$$
implying that $\forall|x|<1$,
$$\lim_{n\to\infty}\sum_{k=0}^n x^k=\frac1{1-x}.$$
This falsifies your claim.
Needless to say,
$$\left.\left(\frac1{1-x}\right)^{(k)}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find maximum point of $f(x,y,z) = 8x^2 +4yz -16z +600$ with one restriction I need to find the critical points of $$f(x,y,z) = 8x^2 +4yz -16z +600$$ restricted by $4x^2+y^2+4z^2=16$.
I constructed the lagrangian function $$L(x, y, z, \lambda ) = 8x^2 +4yz -16z +600 - \lambda (4x^2+y^2+4z^2-16) $$
but I'm very confused ... | Using Lagrange Multipliers, we wish to find the points $(x, y, z)$ such that $\nabla f(x, y, z) = \lambda \nabla g(x, y, z),$ where $g(x, y, z) = 4x^2 + y^2 + 4z^2 = 16$ and $\lambda$ is some constant. Observe that the gradients are given by $\nabla f = \langle f_x, f_y, f_z \rangle = \langle 16x, 4z, 4y - 16 \rangle$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integral $\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx$ For $a>0$ and $D>0$, I want to evaluate the integral
$$\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx.$$
Here $i$ is the imaginary unit. This integral arises when evaluating a Feynman diagram in physics.
| It does not seem to be the most pleasant integral. Focusing on the antoderivative, let first $x=a y$ and $k=\frac{2D}a$to make
$$I=\int \frac{1}{x-ia} \log \left(\frac{(x-2D)^2+a^2}{x^2+a^2}\right)\,dx=\int\frac{1}{y-i}\log \left(\frac{\left(y-k\right)^2+1}{y^2+1}\right)\,dy$$ Deloping the logarithm of the ratio, you t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and ... | Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ be points on the $xy$-plane. Then the points that divide the line segment $\overline{AB}$ in the ratio $m:n$ are
$$ P = \left( \frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \right), \quad P = \left( \frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n} \right). $$
The point $P$ is the internal d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}... | Hint: Write the function of interest as
$$
(1+\sin x)^\frac{1}{x}\,\frac{e^{\frac{1}{x}\log(1+\frac{\sin^2 x}{1+\sin x})}-1}{x}
$$
and remember that
$$
\lim_{x\to0}\frac{\sin x}{x}=1\,,\qquad \lim_{x\to0}\frac{\log(1+x)}{x}=1\,,\qquad \lim_{x\to0}\frac{e^{x}-1}{x}=1\,.
$$
You should find that the limit is $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integer solutions of $2a+2b-ab\gt 0$ Let $a\in\mathbb{N}_{\ge 3}$ and $b\in\mathbb{N}_{\ge 3}$. What are the solutions of the Diophantine inequality
$$2a+2b-ab\gt 0?$$
By guessing, I found 5 solutions:
$$\text{1)}\, a=3,\, b=3$$
$$\text{2)}\, a=3,\, b=4$$
$$\text{3)}\, a=4,\, b=3$$
$$\text{4)}\, a=5,\, b=3$$
$$\text{5)... | $2a+2b-ab>0$ is equivalent to $(a-2)(b-2)<4$, or to $(a-2)(b-2) \le 3$. If $a \ge 3$ and $b \ge 3$, then $(a-2)(b-2) \ge 1$. Thus, $(a-2)(b-2) \in \{1,2,3\}$.
$\bullet$ If $(a-2)(b-2)=1$, then $a-2=b-2=1$, so that $(a,b)=(3,3)$.
$\bullet$ If $(a-2)(b-2)=2$, then $\{a-2,b-2\}=\{1,2\}$, so that $\{a,b\}=\{3,4\}$.
$\bulle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
System of quadratic homogeneous Diophantine equations Is there a method for determining if a system of quadratic diophantine equations has any solutions?
My specific question is:
\begin{align*}
a^2+b^2&=c^2\\[4pt]
2a^2+b^2&=d^2\\
\end{align*}
I want to know if there are any positive integers $a, b, c, d$ which satisfy... | Suppose $(a,b,c,d)$ is a solution to the system
$$
\left\lbrace
\begin{align*}
&a^2+b^2=c^2\\[4pt]
&2a^2+b^2=d^2\\[4pt]
\end{align*}
\right.
$$
where $a,b,c,d$ are positive integers, and where $d$ is as small as possible.
If $a,b$ have a common factor $g > 1$, then $c$ and $d$ must also be divisible by $g$, hence
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Differentiate $\left(x^6-2x^2\right) \ln\left(x\right) \sin\left(x\right)$ Differentiate
$$\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)$$
with respect to $x$
My work so far
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)\... | Your method is correct. If you have an derivative of a function that is the product of three other functions, it looks like this:
$$h(x)=f(x)\cdot g(x)\cdot y(x)$$
$$h'(x)=f'(x)\cdot g(x)\cdot y(x)+f(x)\cdot g'(x)\cdot y(x)+f(x)\cdot g(x)\cdot y'(x)$$
This derivative is just particularly algebraically intense, and at t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the closed form of $\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:\mathrm{d}x$ I solved a similar case which is also a very well known integral
$$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x=\frac{\pi }{4}\ln \left(2\right)+G$$
My teacher gave me a hint which was splitting the integr... | You can evaluate this integral with Feynman's trick,
$$I\left(a\right)=\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:dx$$
$$I'\left(a\right)=\int _0^{\infty }\frac{x}{\left(1+x^2\right)\left(1+ax\right)}\:dx=\frac{1}{1+a^2}\int _0^{\infty }\left(\frac{x+a}{1+x^2}-\frac{a}{1+ax}\right)\:dx$$
$$=\frac{1}{1+a^2}\:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{... | You got that we need to prove that:
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2.$$
Now, by C-S $$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}=\sqrt{(2x^2-2xz+2z^2)(2y^2-2yz+2z^2)}=$$
$$=\sqrt{((x-z)^2+x^2+z^2)(z^2+y^2+(y-z)^2)}\geq$$
$$\geq(x-z)z+xy+z(y-z)=xy+xz+yz-2z^2$$ and we are done!
Now we see that the starting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | Here is another easier way to integrate as follows
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac14\frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac14\int \frac{x}{\sqrt{4x^2+9}}dx-\frac14\int \frac{9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac1{32}\int \frac{d(4x^2+9)}{\sqrt{4x^2+9}}-\frac9{32}\int \frac{d(4x^2+9)}{(4x^2+9)^{3/2}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Solve Complex Equation $z^3 = 4\bar{z}$ I'm trying to solve for all z values where $z^3 = 4\bar{z}$.
I tried using $z^3 = |z|(\cos(3\theta)+i\sin(3\theta)$ and that $|z| = \sqrt{x^2+y^2}$ so:
$$z^3 = \sqrt{x^2+y^2}(\cos(3\theta)+\sin(3\theta))$$ and $$4\bar z = 4x-4iy = 4r\cos(\theta)-i4r\sin(\theta)$$
but I have no id... | If $z^3=4\overline z$, then $z^4=4z\overline z=4|z|^2$. So, $|z|^4=|z^4|=4|z|^2$, and therefore $z=0$ or $|z|=2$. So, unless $z=0$, $z$ can be written as $2(\cos\theta+i\sin\theta)$, in which case$$z^3=4\overline z\iff8\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)=8(\cos\theta-i\sin\theta).$$Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is c... | You can also note the the singularity at $x=0$ is removable and that $f$ is in fact at least 4 times differentiable. Taylor's expansion gives you the answer...
$$
f(x)=f(0)+f'(0)x + \frac 12 f''(0) x^2 + O(x^3)
$$
where
$$
f(0)=\lim_{x\to 0}f(0) = 1, \quad f'(0) = \lim_{x\to 0}f'(x)=0, \quad f''(0)=\lim_{x\to 0}f''(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the... | $$\int \frac{1}{\sqrt{2x}-\sqrt{x+4}}\ dx=\int \frac{(\sqrt{2x}+\sqrt{x+4})}{(\sqrt{2x}-\sqrt{x+4})(\sqrt{2x}+\sqrt{x+4})}\ dx=$$
$$=\int \frac{\sqrt{2x}+\sqrt{x+4}}{x-4}\ dx$$
$$=\int \frac{\sqrt{2x}\ dx}{x-4} + \int \frac{\sqrt{x+4}\ dx}{x-4}$$
$$=\int \frac{\sqrt{2}\ xd(\sqrt{x})}{x-4} + \int \frac{2(x+4)d(\sqrt{x+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$. Let $a,b$ be coprime integers. I am trying to find all integers $x,y$ such that:
$$x^3+5y^3=(a^3+5b^3)^3$$
What I have tried:
$$5y^3=(a^3+5b^3-x)[(a^3+5b^3)^2+(a^2+5b^3)x+x^2 ]$$
There are 2 Cases depending which factor of $5y^3$ is divisible by $5$:
Case 1: $5$... | As noted in the other answer, this is a question about rational points on elliptic curves:
Claim: There are no solutions to $x^{3}+y^{3} = 5$ with $x,y \in \mathbb{Q}$.
This follows from the following Proposition:
Proposition: Set $R := \mathbb{Z}[t]/(t^{2}+t+1)$. Let $A,B,C$ be elements of $R_{\mathbb{Q}}$ such tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Is it possible to justify these approximations about prime numbers? A recently closed question asked for a possible closed form of the infinite summation
$$f(a)=\sum _{i=1}^{\infty } a^{-p_i}$$ for which I already proposed a first simple but totally empirical approximation.
Since we quickly face very small numbers, I t... | These estimates are correct within a reasonable degree of accuracy. Below is the explanation for $f(a)$; the case for $h(a)$ can be dealt similarly. We have
$$
f(a) = \frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{a^5} + O\bigg(\frac{1}{a^7}\bigg)
$$
whereas
$$
\frac{2a^2}{(a-1)(2a^3 + 2a - 1)} = \frac{1}{a^2} + \frac{1}{a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
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What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i d... | Okay, so you have shown the expression to be equal to $20\cdot 2^n\cdot 25^n+18\cdot3^n\cdot4^n=20\cdot 50^n+18\cdot12^n$ and as @mwt as shown in comments , write $20=19+1$ and $18=19-1$ to get the expression equal to $19(50^n+12^n)+50^n-12^n$. Now we know that $a^n-b^n$ is divisible by $a-b$ for any natural number $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Contradiction on an Inequality The problem on which I have a problem is this-
Let a, b, c be non-negative real numbers. Prove that $$ \sum_{cyc} {a^2-bc \over 2a^2+b^2+c^2} \ge 0 $$
While solving and after some resolution, we get $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le 3 $$
And by C-S, we have,
$$ \sum_{cyc} {... | Your first step was not so strong, which gave a wrong inequality.
SOS helps here:
We need to prove that $$\sum_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{2a^2+b^2+c^2}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a+c}{2a^2+b^2+c^2}-\frac{b+c}{2b^2+a^2+c^2}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(c^2-(a+b)c+a^2-ab+b^2)(2c^2+a^2+b^2)\geq0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding $\frac{\cot\gamma}{\cot \alpha+\cot\beta}$, given $a^2+b^2=2019c^2$ This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:
In a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, it is known that $$a^2+b^2=2019c^2$$ Find $$\frac{\cot\gamma}{\cot\alpha+\cot\beta} $... | Here is the complete solution from AoPS.
Continuing from Aqua's answer, we know that $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Furthermore, as you have found, $\cos C = \frac{1009c^2}{ab}$. Substituting these values in gives:
$$\frac{\cot\gamma}{\cot \alpha+\cot\beta} = \frac{\cos\gamma \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Combination to find integers satisfying a condition
Let $n$ and $k$ be positive integers such that $n\ge\frac{k(k+1)}{2}$.
The number of solutions $(x_1,x_2,\dots,x_{k})$, with $x_1\ge1$, $x_2\ge2$,..., $x_{k}\ge k$ for all integers satisfying $x_1+x_2+\dots+x_{k}=n$ is?
I substituted the last equation in the first i... | No, the sequence $x_1,x_2,\dots,x_{k}$ is not necessarily increasing, so your conclusion is not correct.
Let $t_k=x_k-k\geq 0$, then we have to count the number of non-negative integer solutions of
$$t_1+t_2+\dots+t_k=n-\frac{k(k+1)}{2}$$
which is, by Stars-and-Bars, given by
$$\binom{n-\frac{k(k+1)}{2}+k-1}{k-1}=\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)... | Substitute $x = \sin\theta$ and let's restrict our attention to $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ since that covers the whole range of $\sin\theta$:
$$\sin^{-1}\left(\frac{\sin\theta+\cos\theta}{\sqrt{2}}\right) = \frac{\pi}{4}+\sin^{-1}(\sin\theta)$$
$$\implies \sin^{-1}\left(\sin\left(\theta+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}... | $$\frac{2^n\cdot n^5}{n!} = \frac{6^n}{n!} \cdot \left(\frac{2}{3}\right)^n<\frac{6^n}{n!} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | Consider
$P(x)= ax^9+bx^8+1$ and $\alpha$ one of the numbers$\frac{1 \pm \sqrt 5}{2}.$
We have equalities:
$\alpha^2$ = $\alpha$ + 1, $\alpha^4$ = 3$\alpha$ + 2,
$\alpha^8$ = 21$\alpha$ + 13, $\alpha^9$ = 34$\alpha$ + 21.
(The coefficients to the right of the equations are terms of the Fibonacci sequence)
From $P(\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)... | Guide:
Case $1$: If $z=1$. Check what happens here.
Case $2$: If $z \ne 1$, then $x+y \ne 0$,
$$x^2-xy+y^2=1+z=1+(1-(x+y))$$
$$x^2+y^2-xy+x+y = 2$$
$$(x^2-xy+x)+(y^2+y)=2$$
$$(x^2-x(y-1))+(y^2+y)=2$$
$$\left(x - \frac{y-1}2\right)^2-\left(\frac{y-1}2 \right)^2 + (y^2+y)=2$$
$$(2x-y+1)^2 -(y^2-2y+1) + 4y^2+4y=8$$
$$(2x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Taylor/Maclaurin series of $\arctan(e^x - 1)$ Right, so this keeps bugging me, and I'm probably stuck in some tunnel trying the same thing over and over again.
Give the Maclaurin series of the function $\arctan(e^x - 1)$. up to terms of degree three. Since I try to be as lazy as possible and differentiating this thing ... | The correct expansion should be $e^x-1=\sum_{i=1}^\infty \frac{x^i}{i!}$.
Hence substituting it, we get
\begin{align}
\arctan (e^x - 1) &\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{(x+\frac{x^2}{2}+\frac{x^3}{6})^3}{3} \\
&\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{x^3}{3}\\
&= x+\frac{x^2}2 - \frac{x^3}6
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | Another way.
We need to prove that $$\prod_{cyc}(3+3a^2)\geq8\cdot27$$ or
$$\prod_{cyc}(ab+ac+bc+3a^2)\geq8(ab+ac+bc)^3$$ or
$$\sum_{sym}(3a^4b^2+a^3b^3+3a^4bc-6a^3b^2c-a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0\succ(3,2,1),$ $(4,1,1)\succ(3,2,1)$ and $(3,3,0)\succ(2,2,2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\... | No need a sledgehammer to crack a nut.
\begin{align}
J&=\int_0^{\sqrt{2}-1} \frac{\text{arctanh } x}{x}\,dx\\&=-\frac{1}{2}\int_0^{\sqrt{2}-1}\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}\,dx\\
&\overset{y=\frac{1-x}{1+x}}=-\int_{\sqrt{2}-1}^1 \frac{\ln y}{1-y^2}dy\\
&=\int_0^{\sqrt{2}-1} \frac{\ln y}{1-y^2}dy-\int_0^1 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\... | $$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!} $$
Proof:
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-1x} = 1 +\sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1 = \sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1 = \sum_{n=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
convergence of $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1}\left\{\frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ (new methods) The series $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ is said to converge when $0<s<1,$ which seem... | In $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\} \right]$, thinking of the subtraction may result in $\frac{1}{n^\sigma}$ where $\sigma\geq1>s$, I add up the fractions and then get
$$
\frac{u_{n+1}}{u_n}=\frac{1-s-(\frac{n+1}{n+2})^s(n+2)+n+1}{1-s-(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The integral of $ \int_{0}^{1} 2^{x^2 +x} dx$ So, What I tried was,
$$ I(b) = \int_{0}^{1} 2^{x^2 + x +b} dx$$
And hence,
$$ I'(b) = \ln(2) I$$
Hence,
$$ I = C_o 2^{b}$$
or,
$$ C2^{b} = \int_{0}^{1} 2^{x^2 +x + b} dx$$
Now I'm trying to find an easy 'b' to evaluate the right side integral at, so as to figure out my c... | If you do not want to use special functions, build a series expansion around $x=\frac 12$ to get for the integrand
$$2^{3/4} \left(1+2 L \left(x-\frac{1}{2}\right)+L (2 L+1) \left(x-\frac{1}{2}\right)^2+\frac{2}{3}
L^2 (2 L+3) \left(x-\frac{1}{2}\right)^3+\frac{1}{6} L^2 (4 L^2+12L+3)
\left(x-\frac{1}{2}\right)^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} ... | Near $0$, you have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x} = \frac{1 + \frac{x^2}{2} + o(x^3)}{x + \frac{x^3}{6} + o(x^4)} - \frac{1}{x} = \frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\left( 1 + \frac{x^2}{6} + o(x^3) \right)^{-1}-1\right]$$ $$ =\frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Finding $ \mathop {\lim }\limits_{n \to \infty } \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$ I tried to find the limit $$\mathop {\lim }\limits_{n \to \infty } \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$$, but I couldn't find a conclusive argument.
By assuming that the sequence $\displaystyle... | Substitute $x=t\ln(n)$, then the integral becomes
$$\lim_{n\to\infty}\int_0^{\ln(n)}\ln\left(1+e^{-x}\right)\,dx=\int_0^{\infty}\ln\left(1+e^{-x}\right)\,dx.$$
Now
$$\int_0^\infty \ln(1+e^{-x})dx=\int_0^\infty\left(e^{-x}-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}-\frac{e^{-4x}}{4}+\cdots\right)dx $$
$$=\int_0^\infty e^{-x}dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given $U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$, find $\frac{100U_{10}-1}{U_8}$
If
$$U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$$
Find $\frac{100U_{10}-1}{U_8}$
Answer: $90$
My Attempt:
I tried applying Integration By Parts, and when that failed, I tried the substitution $x\rightarrow \frac{\pi}{2} -x$ , only to establish a... | Note
$$U_{n-2} -U_n =\int_0^\frac{\pi}{2} x\sin^{n-2}\cos^2x dx
\overset{IBP}=\frac1{n-1} \int_0^\frac{\pi}{2} x\cos xd(\sin^{n-1}x)\\
= \frac1{n-1}\left( U_n -\frac1n \int_0^\frac{\pi}{2} d(\sin^{n}x)\right)= \frac1{n-1}\left(U_n - \frac1n\right)
$$
Thus, $U_n = \frac{n-1}nU_{n-2} + \frac1{n^2}$ and
$$\frac{100U_{10}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Correct result for this integral $\int \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}}\, dx$ Wolfram|Alpha and its CAS, Wolfram Mathematica are, as far as I know, the only website and software that give the correct solution to this integral,
$$ f(x) = \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 ... | These are the steps to get to the given result:
$$f(x)=\dfrac1{\sqrt x}\sqrt{2+\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}
=\dfrac1{\sqrt x}\,\dfrac{\sqrt{2-\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}}$$
$$=\dfrac1{\sqrt x}\,\dfrac{\sqrt{2-2\cos(5\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
$2$ fair coins, $1$ biased coin $-$ probability of getting a $3^{\text{rd}}$ head after $2$ consecutive heads? I solved this problem but my solution is apparently incorrect.
Let $A$ and $B$ denote the event that we chose the fair and biased (both sides are heads) coin, respectively. $P(A) = \frac{2}{3}$, $P(B) = \frac... | You need to be precise describing the experiment. It looks like you are saying that (1) a single coin is chosen at random from three coins and that coin is flipped three times, and (2) two of the three coins are fair, and the third always comes up heads. And you want to know, given that the first two flips were heads... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.