Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do I formally show the radius of convergence of the Taylor series of $f(x)=x^6 - x^4 + 2$ at $a=-2$? This is an exercise in Stewart's Calculus (Exercise 19, Section 11.10 Taylor and Maclaurin Series):
Find the Taylor series for $f(x)$ centered at the given value of a. [Assume that f has a power series expansion. Do not show that $R_n(x) \to 0.$ Also find the associated radius of convergence.
Here $f(x)=x^6 - x^4 + 2$ and $a=-2$.
I'm having trouble finding a general formula of this Taylor series and therefore, also having problems finding the radius of convergence since I can't perform the ratio test.
Here is what I know:
\begin{align}
f'(x) = 6x^5 - 4x^3,\quad &f''(x) = 30x^4 - 12x^2,\\
f'''(x) = 120x^3 - 24x,\quad &f^{(4)}(x) = 360x^2 - 24,\\
f^{(5)}(x) = 720x,\quad &f^{(6)}(x) = 720.
\end{align}
And at $a=-2$,
\begin{align}
f(-2) = 50,\quad &f'(-2) = -160,\\
f''(-2) = 432,\quad &f'''(-2) = -912,\\
f^{(4)}(-2) = 1416,\quad &f^{(5)}(-2) = -1440,\\
f^{(6)}(-2) = 720.\quad &
\end{align}
I'm having trouble finding the general formula for each term. Without it, how am I supposed to find the radius of convergence?
Added:
So the general term I have for the n-th derivative of $f$ is:
$$f^{(n)}(x) = \frac{6!x^{6-n}}{(6-n)!}$$
So far the general term I have for the Taylor Series is:
$$\sum_{n=0}^{\infty} \frac{6! 2^{6-n}}{(6-n)!n!}(x+2)^n$$
I can see why the radius of convergence is $\infty$: because for any $x$the series converges.
But how do I show this formally? Can I use the ratio test?
| The Taylor series around $a$ is simply given by $$\begin{align}f(a+x)&=(a+x)^6-(a+x)^4+2\\&=(a^6-a^4+2)+(6a^5-4a^3)x+(15a^4-6a^2)x^2+(20a^3-4a)x^3+(15a^2-1)x^4+6ax^5+x^6\end{align}$$
or
$$f(x)= (a^6-a^4+2)+(6a^5-4a^3)(x-2)+(15a^4-6a^2)(x-a)^2+(20a^3-4a)(x-a)^3+(15a^2-1)(x-a)^4+6a(x-a)^5+(x-a)^6$$
and, as almost all coefficients are $=0$, converges for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find Taylor series of $\sqrt{x}$ centered at $x=4$ and the order 3 Find Taylor series of $\sqrt x$, about $x=4$ and the order 3
I've tried a few timesm but I keep getting a result that does not comply with the answer. Following are the steps I've taken, hopefully I can get a pointer to the factor I'm missing in some of the denominators. I hope all the steps are clear.
Find derivatives of $\sqrt x$
If $f(x) = \sqrt x$, then:
*
*$f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$
*$f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}$
*$f'''(x) = \frac{3}{8}x^{-\frac{5}{2}}$
Write out the Taylor polynomial
$P_3(x) = \sqrt 4 + \frac{1}{2} 4^{-\frac{1}{2}}(x - 4) - \frac{1}{4} 4^{-\frac{3}{2}}(x - 4)^2 + \frac{3}{8} 4^{-\frac{5}{2}}(x - 4)^3$
$P_3(x) = 2 + (\frac{1}{2}) (\frac{1}{\sqrt 4})(x - 4) - (\frac{1}{4}) (\frac{1}{4^{\frac{3}{2}}})(x - 4)^2 + (\frac{3}{8})(\frac{1}{x ^{\frac{5}{2}}})(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{4} \frac{1}{4^{2/2} 4^{1/2}}(x - 4)^2 + \frac{3}{8}\frac{1}{4^{4/2}} \frac{1}{4^{1/2}}(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32}(x - 4)^2 + \frac{3}{256}(x - 4)^3$
But according to the book the series develops like:
$$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2 + \frac{3}{1536}(x - 4)^3$$
So I'm missing some (increasing) factor, I just can't seem to find it. Any hints?
| Where is the factorial ???.....................
$$f(x)=f(x_0)+(x-x_0)f'(x_0)+\frac{(x-x_0)^2}{{\color{Red} 2!}}f''(x_0)+\frac{(x-x_0)^3}{{\color{Red} 3!}}f''(x_0).....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to effectively calculate $\int_0^\pi \sin^3x \sin(nx)dx$, $n\in \mathbb{N}$ How to effectively calculate $$\int_0^\pi \sin^3x \sin(nx)dx$$
I think by the formula: $$\sin 3x=3\sin x -4\sin^3x $$ So we get $$\sin^3x = \frac{3\sin x-\sin 3x}{4}$$
Plug in we obtain $$\frac{1}{4}\int_0^\pi (3\sin x-\sin 3x) \sin(nx)dx$$
Then how do I proceed?
| We proceed exactly as how you started, and then use the product-to-sum formula: $\sin(a) \sin(b) = \frac{1}{2} (\cos(a-b) - \cos(a+b))$. If $n \neq 1, 3$:
$$\begin{aligned}
I_n &= \displaystyle \int_0^{\pi} \sin^3(x) \sin(nx) \; \mathrm{d}x\\
&= \frac{1}{4} \displaystyle \int_0^{\pi} (3 \sin(x) - \sin(3x)) \sin(nx) \; \mathrm{d}x\\
&= \frac{1}{4} \displaystyle \int_0^{\pi} 3 \sin(nx) \sin(x) - \sin(nx) \sin(3x) \; \mathrm{d}x\\
&= \frac{1}{4} \displaystyle \int_0^{\pi} \frac{3}{2} \left ( \cos \left ((n-1) x \right ) - \cos \left ( (n+1) x \right ) \right ) - \frac{1}{2} \left ( \cos \left ( (n-3) x \right ) - \cos \left ( (n+3) x \right ) \right ) \; \mathrm{d}x\\
&= \frac{1}{8} \left [ \frac{3}{n-1} \sin \left ( (n-1) x \right ) - \frac{3}{n+1} \sin \left ( (n+1) x \right ) - \frac{1}{n-3} \sin \left ( (n-3) x \right ) + \frac{1}{n+3} \sin \left ( (n+3) x \right ) \right ]_0^{\pi}\\
&= 0
\end{aligned}$$
Thus, if $n \neq 1, 3$, then the integral evaluates to $0$. If $n=1$, we have:
$$\begin{aligned}
I_1 &= \frac{1}{4} \displaystyle \int_0^{\pi} \frac{3}{2} \left ( \cos \left ((n-1) x \right ) - \cos \left ( (n+1) x \right ) \right ) - \frac{1}{2} \left ( \cos \left ( (n-3) x \right ) - \cos \left ( (n+3) x \right ) \right ) \; \mathrm{d}x\\
&= \frac{1}{8} \displaystyle \int_0^{\pi} 3 - 3\cos(2x)- \cos(-2x) + \cos(4x) \; \mathrm{d}x\\
&= \frac{1}{8} \left [3x - 3 \sin(2x) + \frac{3}{4} \sin(4x) \right ]_0^{\pi}\\
&= \frac{3\pi}{8}
\end{aligned}$$
If $n=3$, we have:
$$\begin{aligned}
I_3 &= \frac{1}{4} \displaystyle \int_0^{\pi} \frac{3}{2} \left ( \cos \left ((n-1) x \right ) - \cos \left ( (n+1) x \right ) \right ) - \frac{1}{2} \left ( \cos \left ( (n-3) x \right ) - \cos \left ( (n+3) x \right ) \right ) \; \mathrm{d}x\\
&= \frac{1}{8} \displaystyle \int_0^{\pi} 3\cos(2x)-3\cos(4x)-1+\cos(6x) \; \mathrm{d}x\\
&= \frac{1}{8} \left [\frac{3}{2} \sin(2x) - \frac{3}{4}\sin(4x) - x + \frac{1}{6} \sin(6x) \right ]_0^{\pi}\\
&= -\frac{\pi}{8}
\end{aligned}$$
Thus, over all cases we have
$$I_n = \begin{cases}
0 \quad &\text{if } n \neq 1, 3\\
\frac{3\pi}{8} \quad &\text{if } n = 1\\
-\frac{\pi}{8} \quad &\text{if } n = 3
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Can't solve system of linear equations (that need simplification first) I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$\frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12}$$
$$\frac{2y+1}{5} + \frac{x-3y}{4} = \frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12\frac{3x+2}{4} - 12\frac{x+2y}{2} = 12\frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20\frac{2y+1}{5} + 20\frac{x-3y}{4} = 20\frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -\frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
| The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Have I found the correct Laurent series expansion? I am supposed to find the Laurent series expansion for
$$f(z) = \frac{z+1}{z^2-4}$$ in the region $1<|z+1|<3$.
My solution:
$$w=z+1 \Leftrightarrow z=w-1 \Rightarrow f(z) = \frac{w-1+1}{(w-1)^2-4} = \frac{w}{(w+1)(w-3)}$$
Partial fractions: $$\frac{w}{(w+1)(w-3)} = \frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3}$$
We have:
\begin{eqnarray}\frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3} &=& \frac{1}{4}\frac{1}{w} \bigg(\frac{1}{1+\frac{1}{w}}\bigg) + \frac{3}{4}\frac{1}{w}\bigg(\frac{1}{1-\frac{3}{w}}\bigg)\\ &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n + \frac{3}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{3}{w}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} + \frac{3}{4} \sum_{n=0}^\infty \frac{3^n}{w^{n+1}}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{w^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z+1)^n} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{(z+1)^{n}}\end{eqnarray}
for the region $1<|z+1|<3$. It feels like there is something wrong here. Any help is greatly appreciated.
| Very nice work! Unfortunately, this is valid on the region $|w|>3,$ rather than the one you want. In order to get $|w|<3,$ we must have $\left|\frac{w}{3}\right|<1,$ so instead, we'll rewrite $$\frac{1}{w-3}=-\frac13\frac1{1-\frac{w}{3}},$$ so that \begin{eqnarray}\frac14\frac1{w+1}+\frac34\frac1{w-3} &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n - \frac{1}{4}\sum_{n=0}^\infty \bigg[\frac{w}{3}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} - \frac{1}{4} \sum_{n=0}^\infty \frac{w^n}{3^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} - \frac{1}{4} \sum_{n=1}^\infty \frac{w^n}{3^n}\\ &=& \sum_{n=1}^\infty \frac{(-1)^{n-1}}{4}w^{-n} - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}\\ &=& \sum_{-\infty}^{n=-1} \frac{(-1)^{-n-1}}{4}w^n - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}\\ &=& \sum_{-\infty}^{n=-1} \frac{(-1)^{n+1}}{4}w^n - \sum_{n=1}^\infty \frac1{3^n\cdot 4}w^{n}.\end{eqnarray}
Thus, we have $$\frac{z+1}{z^2-4}=\sum_{n\in\Bbb Z}a_n(z+1)^n,$$ where $$a_n=\begin{cases}-\cfrac1{3^n\cdot 4} & n\ge0\\-\cfrac{(-1)^n}{4} & n<0.\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$
$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$
$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$
$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$
$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$
$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$
We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x}
\\[5mm] \stackrel{x\ =\ \verts{a}t^{1/4}}{=}\,\,\,&\
{1 \over 4\verts{a}}\int_{0}^{\infty}{t^{\color{red}{7/4} - 1} \over \pars{1 + t}^{2}}\,\dd t
\end{align}
Note that
\begin{align}
{1 \over \pars{1 + t}^{2}} & =
\sum_{k = 0}^{\infty}{-2 \choose k}t^{k} =
\sum_{k = 0}^{\infty}{k + 1 \choose k}
\pars{-1}^{k}\,t^{k}
\\[5mm] & =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{2 + k}}
{\pars{-t}^{k} \over k!}
\end{align}
With the
Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x} =
{1 \over 4\verts{a}}\Gamma\pars{7 \over 4}
\Gamma\pars{2 - {7 \over 4}}
\\[5mm] = &\
{1 \over 4\verts{a}}{3 \over 4}\Gamma\pars{3 \over 4}
\Gamma\pars{1 \over 4} =
{3 \over 16\verts{a}}\,{\pi \over \sin\pars{\pi/4}}
\\[5mm] = &\
\bbx{{3\root{2}\pi \over 16}\,{1 \over \verts{a}}}
\\ &
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Efficiently compute $f'(a)f'(b)f'(c)$ for the roots of a cubic polynomial Let $a,b, c$ be the zeroes of $f(x) = x^3+3x^2-7x+1$. Find $f'(a)f'(b)f'(c)$.
My idea involved substituting in $a,b,c$ into $f'(x)$ then using Vieta's but that would take far too long of a time, especially since I'm training for competitions. What would be a great way to solve this within a time limit?
| According to Vieta's formulas:
$$\begin{align}a+b+c&=-3\\
ab+bc+ca&=-7\\
abc&=-1\\
a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=9+14=23\\
a^2b^2+b^2c^2+c^2a^2&=(ab+bc+ca)^2-2abc(a+b+c)=49-6=43\\
\end{align}$$
Hence:
$$\begin{align}f'(a)f'(b)f'(c)&=(3a^2+6a-7)(3b^2+6b-7)(3c^2+6c-7)=\\
&=27(abc)^2+54abc(ab+bc+ca)-63(a^2b^2+b^2c^2+c^2a^2)+\\
&\ \ \ \ \ \ 108abc(a+b+c)+216abc-252(ab+bc+ca)+294(a+b+c)-343-\\
&\ \ \ \ \ \ 126(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)=\\
&=27+54\cdot 7-63\cdot 43+108\cdot 3+147\cdot 23-216+252\cdot 7-294\cdot 3-343-\\
&\ \ \ \ \ \ 126(ab(-3-c)+ac(-3-b)+bc(-3-a))=\\
&=1724-126(-3\cdot (-7)-3\cdot (-1))=\\
&=1724-126\cdot 24=\\
&=1724-3024=\\
&=-1300.\end{align}$$
WA expansion of $(3a^2+6a-7)(3b^2+6b-7)(3c^2+6c-7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function
Convert trigonometric function into algebraic function.
$$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$
My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma \tag{1}$$
Hence
$$\begin{align}
\cos\theta&=\cot2\gamma \tag{2}\\[4pt]
\cos\theta&=\frac{\cot^2\gamma-1}{2\cot\gamma} \tag{3}\\[4pt]
\cos\theta&=\frac{1-x^2}{2x} \tag{4}
\end{align}$$
The value of $\sin\theta$ is coming in negative.
| \begin{align}
\cot\left(2 \tan^{-1}x\right)
&= \dfrac{1}{\tan\left(2 \tan^{-1}x\right)} \\
&= \dfrac{1-\tan^2(\tan^{-1}x)}{2\tan(\tan^{-1}x)} \\
&= \dfrac{1-x^2}{2x} \\
\hline
\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)
&= 2\sin\left(\dfrac{ 1-x^2}{2x} \right) \cos\left( \dfrac{1-x^2}{2x} \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2)$ for $a, b, c > 0$
For $a,b,c>0$, I have to prove that $$ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2).$$
We have:
$$\begin{align} \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2}
&= \sum_{cyc}^{} \frac {a^4}{a^2+ab+b^2} + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2} \\[4pt]
&\ge \frac{(a^2+b^2+c^2 )^2}{a^2+b^2+c^2+ab+bc+ca} + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2} \\[4pt]
&\ge \frac{1}{2} (a^2+b^2+c^2) + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2}
\end{align}$$
but I can't move on.
| Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$45(u^2-uv+v^2)a^6+18(5u^3-3u^2v+3uv^2+5v^3)a^5+$$
$$+3(23u^4-u^3v-6u^2v^2+59uv^3+23v^4)a^4+$$
$$+3(8u^5+9u^4v-18u^3v^2+34u^2v^3+43uv^4+8v^5)a^3+$$
$$+3(u^6+6u^5v-8u^4v^2+28u^2v^4+12uv^5+v^6)a^2+$$
$$+(3u^5+2u^4v-17u^3v^2+19u^2v^3+20uv^4+3v^5)uva+$$
$$+u^2v^2(u^4-2u^2v-u^2v^2+4uv^3+v^4)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate some induced norms of matrix $ A$
Let
$$A = \begin{pmatrix}
-3 & -4 & -2 \\
5 & 9 & -5 \\
-3 & 8 & -9
\end{pmatrix}$$
Calculate the following norms of matrix $A$.
a. $\|A\|_{1,1}$
b. $\|A\|_{∞,∞}$
c. $\|A\|_{1,∞}$
d. $\|A\|_{2,∞}$
e. $\|A\|_{1,2}$
I know that $||A||_1=21$ and $||A||_\infty=20$, but I'm not sure what to do after that for option a or option b. I haven't done $||A||_2$ yet, but I know how to get that information. I just don't know what do when I have $(1,1)$, but especially letters $c, d, $ and $e$. Any help would be appreciated.
I have the following Lemma:
$$ \textrm{ Lemma 7.22. If } A \in \mathbb{C}^{p \times q} \textrm{ then } $$
$$ \textrm{1.} \| A\|_{1,1} = \max_{1 \leq j \leq q} \Big\{ \sum_{i=1}^{p} |a_{ij}|\Big\} \textrm{ ( maximum column sum ( modulus))}$$
$$ \textrm{2.} \| A\|_{\infty,\infty} = \max_{1 \leq i \leq p} \Big\{ \sum_{j=1}^{q} |a_{ij}|\Big\} \textrm{ ( maximum row sum (modulus) )}$$
$$ \textrm{3.} \| A\|_{2,2} = s_{1} \textrm{ where } s_{1}^{2} \textrm{ is the maximum eigenvalue of the matrix } A^{H}A $$
$$ \textrm{4.} \| A\|_{1,\infty} = \max_{i,j} |a_{ij}| \textrm{ ( maximum (modulus) )} $$
$$ \textrm{5.} \| A\|_{2,\infty} = \max_{1 \leq i \leq p} \Big\{ \bigg( \sum_{j=1}^{q} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of rows)}$$
$$ \textrm{6.} \| A\|_{1,2} = \max_{j} \Big\{ \bigg( \sum_{i=1}^{p} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of columns)}$$
| If you have your matrix $A$ given as the following
$$A = \begin{pmatrix} -3 & -4 & -2 \\ 5 & 9 & -5 \\ -3 & 8 & -9\end{pmatrix} $$
then we have
$$ \textrm{a.} \| A\|_{1,1} = \max_{1 \leq j \leq q} \Big\{ \sum_{i=1}^{p} |a_{ij}|\Big\} \textrm{ ( maximum column sum ( modulus))}$$
This is the maximum sum of the absolute values of the columns
$$ \sum_{i=1}^{p} |a_{i1} | = 3 + 5 + 3 = 11 \\ \sum_{i=1}^{p} |a_{i2}| = 4 + 9 + 8 = 21 \\ \sum_{i=1}^{p} |a_{i3}| = 2 + 5+ 9 = 16 $$
which gives us $\|A\|_{1,1} = 21$
$$ \textrm{b.} \| A\|_{\infty,\infty} = \max_{1 \leq i \leq p} \Big\{ \sum_{j=1}^{q} |a_{ij}|\Big\} \textrm{ ( maximum row sum (modulus) )}$$
which is the maximum row sum
$$ \sum_{j=1}^{q} |a_{1j}| = 3 + 4 + 2 = 9 \\ \sum_{j=1}^{q} |a_{2j}| = 5 + 9 + 5 = 19 \\ \sum_{j=1}^{q} |a_{3j}| = 3+8+9 = 20$$
$$ \|A\|_{\infty, \infty} = 20$$
$$ \textrm{c.} \| A \|_{1,\infty} = \max_{i,j} |a_{ij}| $$
which is the largest absolute value of some entry, $a_{22}$ and $a_{33} $ have absolute value $9$
$$ \| A \|_{1,\infty} = 9 $$
$$ \textrm{d.} \| A\|_{2,\infty} = \max_{1 \leq i \leq p} \Big\{ \bigg( \sum_{j=1}^{q} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of rows)}$$
which is the maximum $2$ norm of the rows
$$ \bigg(\sum_{j=1}^{q} |a_{1j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 3^{2} + 4^{2} + 2^{2} } = \sqrt{17} $$
$$ \bigg(\sum_{j=1}^{q} |a_{2j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 5^{2} + 9^{2} + 5^{2} } = \sqrt{131} $$
$$ \bigg(\sum_{j=1}^{q} |a_{3j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 3^{2} + 8^{2} + 9^{2} } = \sqrt{154} $$
we see that $\|A\|_{2,\infty} = \sqrt{154}$
Finally for the last one
$$ \textrm{e.} \| A\|_{1,2} = \max_{j} \Big\{ \bigg( \sum_{i=1}^{p} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of columns)}$$
is the maximum $2$ norm of the columns
$$ \bigg(\sum_{j=1}^{q} |a_{i1}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{3^{2}+ 5^{2} + 3^{2}} = \sqrt{43} $$
$$ \bigg(\sum_{j=1}^{q} |a_{i2}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{4^{2}+ 9^{2} + 8^{2}} = \sqrt{161} $$
$$ \bigg(\sum_{j=1}^{q} |a_{i3}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{2^{2}+ 5^{2} + 9^{2}} = \sqrt{110} $$
$$ \|A \|_{1,2} = \sqrt{161} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}{(1+\tan x)^2} \,dx$$
| From your last step,
Let, $tanx = t, \ sec^2xdx = dt \ , (1+tan^2x)dx = dt$
$dx = \frac{dt}{1+t^2}$
$I = \int{\frac{2tdt}{(1+t^2)(1+t)^2}}$
Applying partial fractions,
$\frac{2t}{(1+t^2)(1+t)^2} = \frac{1}{1+t^2} - \frac{1}{(t+1)^2}$
$I = \int{\bigg[\frac{1}{1+t^2} - \frac{1}{(t+1)^2}}\bigg]dt = tan^{-1}t + \frac{1}{(t+1)} + c$
$I = x +\frac{1}{tanx+1}+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Converting an equation based on square roots So I was wondering how to convert an equation of the form $\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+...\sqrt{x_n}+k=0$ into a polynomial equation based on each $x_i$.
For example if the equation was $$\sqrt{x_1}+\sqrt{x_2}+k=0$$, then subtracting $\sqrt{x_2}$ from each side and squaring yields:$$x_1+k^2+2k\sqrt{x_1}=x_2.$$
This can then be rearranged to: $$2k\sqrt{x_1}=-x_1-k^2+x_2.$$
Squaring both sides yields: $$4k^2x_1=x_1^2+k^4+x_2^2+2k^2x_1-2x_1x_2-2k^2x_2.$$
Rearranging/simplifying yields:
$$x_1^2+x_2^2+k^4-2k^2x_1-2x_1x_2-2k^2x_2 = 0.$$
How can I find an equation of this form given that $n$ is greater than $4$? I am most interested in when $n = 6$.
| Just for fun, the polynomial drawn from $\sqrt a+\sqrt b+\sqrt c+k=0$, using the method by Somos:
$$a^4+6a^2b^2+6a^2c^2+6a^2k^4-4ab^3+4ab^2k^2+4abc^2-40abck^2+4abk^4-4ac^3+4ac^2k^2+4acb^2+4ack^4-4ak^6+b^4+6b^2c^2+6b^2k^4-4ba^3+4ba^2k^2-4bc^3+4bc^2k^2+4bca^2+4bck^4-4bk^6+c^4+6c^2k^4-4ca^3+4ca^2k^2-4cb^3+4cb^2k^2-4ck^6+k^8-4k^2a^3-4k^2b^3-4k^2c^3=0.$$
Then next challenge is to determine the number of terms as a function of the number of variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from the question, I tried:
Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.
But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
| If $(1+\sqrt{3})(1+\sqrt{7})$ is rational, then
$$\displaystyle \frac{12}{(1+\sqrt{3})(1+\sqrt{7})}=\frac{12(1-\sqrt{3})(1-\sqrt{7})}{(-2)(-6)}=1-\sqrt{3}-\sqrt{7}+\sqrt{21}$$ is also rational.
So, $\displaystyle \frac{1}{2}[(1+\sqrt{3})(1+\sqrt{7})+1-\sqrt{3}-\sqrt{7}+\sqrt{21}]-1=\sqrt{21}$ is rational.
This leads to a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Why is the $2a$ term in the quadratic formula $|2a|$? I am reading through Algebra by Gelfand/Shen. There was a construction of the quadratic formula as follows:
$ax^2 + bx + c = 0$. Dividing by $a$ gives us $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ and we can apply the formula of the equation $x^2 + px + q = 0$ with $p = \frac{b}{a}, q = \frac{c}{a}$, we get:
$x_{1,2} = -\frac{b}{2a} \pm \sqrt{(\frac{b}{2a})^2 - 4\frac{c}{a}} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
My question is if we take $\sqrt{4a^2}$ why do we assume it is always equal to $2a$ instead of $|2a|$? Why does the negative part not matter?
| Because there is a $\pm$ in front of the fraction. Since you are considering either of the signs, both $\frac{\sqrt{b^2-4ac}}{\lvert 2a\rvert}$ and $\frac{\sqrt{b^2-4ac}}{-\lvert 2a\rvert}$ will have to be considered, which is the same as considering both $\frac{\sqrt{b^2-4ac}}{2a}$ and $\frac{\sqrt{b^2-4ac}}{-2a}$. This, as far as real values of $a$ are considered. If $a\in\Bbb C\setminus\Bbb R$, then $\sqrt{a^2}=\lvert a\rvert$ doesn't hold in the first place.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Different ways gives different results - solving $\tan 2a = \sqrt 3 $ Different ways gives different results - solving $\tan 2a = \sqrt 3$
Case 1).
$$ \tan 2a = \sqrt 3 =\tan(\frac{\pi}{3}) $$
$$2a = n\pi + \frac{\pi}{3} $$
$$a = \frac{n\pi}{2} + \frac{\pi}{6},\qquad n \in\mathbb{Z} $$
Case 2)
$$ \dfrac{2\tan a}{1-\tan^2 a} = \sqrt 3$$
Solving above equation gives $ \tan a = \dfrac{1}{\sqrt 3}$ or $-\sqrt 3 $
Thus either
$$ a = n\pi+ \frac{\pi}{6},\qquad n \in \mathbb{Z} $$
Or
$$ a = m\pi - \frac{\pi}{3},\qquad m\in\mathbb{Z} $$
Or either I did something wrong or both these results are same.
If it's the later, I tried a lot to convert one form to other by adding the case 2 results. But that doesn't work.
| If you are in doubt, enumerate the solutions.
Case 1:
$$\cdots-\frac{2\pi}6,\frac\pi6,\frac{4\pi}6,\frac{7\pi}6,\frac{10\pi}6,\frac{13\pi}6,\cdots$$
Case 2:
$$\cdots\frac\pi6,\frac{7\pi}6,\frac{13\pi}6,\cdots$$
and
$$\cdots-\frac{2\pi}6,\frac{4\pi}6,\frac{10\pi}6,\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Polynomial equation for $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}$ I have $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}, k=1,2,3,4$.
I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$
The right answer is $x^4+x^3+x^2+x+1=0$.I tried to replace k with 1,2,3,4 to find the roots and then to use $a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but I didn't get too far.
| Hint:
$$z_k=e^{\frac{2\pi i}{5}}$$
Raising both sides to the 5th power yields $$z_k^5=e^{2\pi i}=1$$ or, $$z_k^5-1=0\tag{1}$$ The polynomial $(1)$ has 5 roots, one of them being $z_k=1$. But $z_k\neq 1$, because $k$ is not a multiple of $5$. Therefore $$\frac{z_k^5-1}{z_k-1}\tag{2}$$ is the polynomial we're after, which has 4 roots. $(2)$ can be simplified, by recognizing it as a geometric series $$\frac{z_k^5-1}{z_k-1}=z_k^4+z_k^3+z_k^2+z_k^1+z_k ^0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Change of variables for double integral Problem: Using the change of variables $$x=\sqrt2u-\sqrt\frac{2}{3}v,y=\sqrt2u+\sqrt\frac{2}{3}v$$
Calculate the double integral $$\iint_Rx^4-2x^3y+3x^2y^2-2xy^3+y^4dA$$ where $R$ is the region bound by $x^2-xy+y^2=2$.
My work so far: The Jacobian is fairly trivial: $\frac{4}{\sqrt3}$
The region $R$ becomes $$(\sqrt2u-\sqrt\frac{2}{3}v)^2-(\sqrt2u-\sqrt\frac{2}{3}v)(\sqrt2u+\sqrt\frac{2}{3}v)+(\sqrt2u+\sqrt\frac{2}{3}v)^2=2$$$$\rightarrow2u^2+2v^2=2$$$$\rightarrow u^2+v^2=1$$
which is the unit disk. And after a lengthy calculation, I believe $$x^4-2x^3y+3x^2y^2-2xy^3+y^4$$$$\rightarrow \frac{8}{3}(-3u^4+6u^2v^2+v^2)$$
So we have
$$\iint_S\frac{8}{3}(-3u^4+6u^2v^2+v^2)\frac{4}{\sqrt3} \ du \ dv$$
where $S$ is the unit disk.
Now what to do? I tried proceeding using polar coordinates, but didn't find it an easy integral to compute. Am I just very bad at polar coordinates, or is there another way? Or have I miscalculated somewhere?
| A comment disputes some of your coefficients, but note that $$\iint_S u^k v^l dudv=\frac{1}{k+l+2}\int_0^{2\pi}\cos^k\theta\sin^l\theta d\theta,$$so you only need to know $$\int_0^{2\pi}\cos^4\theta d\theta=\frac14\int_0^{2\pi}(1+\cos2\theta)^2d\theta=\frac14\int_0^{2\pi}(1+2\cos2\theta+\cos^22\theta)d\theta=\frac{3\pi}{4}$$(and similarly $\int_0^{2\pi}\cos^4\theta d\theta=\frac{3\pi}{4}$) and$$\int_0^{2\pi}\cos^2\theta \sin^2\theta d\theta=\frac14\int_0^{2\pi}\sin^22\theta d\theta=\frac{\pi}{4}.$$For each of these calculations, I use the fact that $\cos^2\phi,\,\sin^2\phi$ both average to $\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution of a quadratic equation with polynomial coefficients Let $a$, $b$ y $c$ be real and positive numbers. If the following quadratic equation in $x$:
\begin{equation*}
(a+b+c)x^2-2(ab+bc+ca)x+ab^2+bc^2+ca^2=0
\end{equation*}
has at least one real solution. Determine the value of $\dfrac{a+5b}{c}$
I tried to apply algebraic identities, but I don't go anywhere.
| The positive numbers $a, b, c$ are very restricted so that the given equation admits a real root.
In effect, since the coefficient of $x ^ 2$ is positive, the minimum of the given quadratic function must be less than or equal to zero. If $f(x)$ is the function we have this minimum at $x=\dfrac{ab+bc+ac}{a+b+c}$ and we must have
$f\left(\dfrac{ab+bc+ac}{a+b+c}\right)\le0$.
Calculation gives the necessary condition $$a^3c+b^3a+c^3b\le abc^2+bca^2+acb^2$$
and it can be shown that for positive numbers this inequality is verified only when $ a = b = c $ (which is left as a secondary problem) whereby the real root is double ($ x = a $).
Thus the required value is equal to $\color{red}6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Taylor series for exp(exp(x)) using just the power series for exp(x) I'm trying to figure out how to calculate the power series for exp(exp(x)) using exp(x) and then to write down the first few terms. I have the answer for the terms but I don't know how they arrived at it
Thanks in advance
For the first few terms we have: exp(exp(x)) = e(1 + x + x^2 + . . .)
| When we put $\cdots$ we'll be ignoring all terms involving $x^n$ for $n \geq 3$. We expand like so:
\begin{align*}
\exp(\exp(x)) &= \exp\left(1 + x + \frac{1}{2}x^2 + O(x^3)\right) \\
&= e \times \exp\left(x + \frac{1}{2} x^2 + O(x^3)\right) \\
&= e\left(1 + \left(x + \frac{1}{2}x^2 + O(x^3)\right) + \frac{1}{2}\left(x + \frac{1}{2}x^2 + O(x^3)\right)^2 + O(x^3) \right) \\
&=e\left(1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^2 + O(x^3) \right) \\
&=e(1 + x + x^2 + O(x^3))
\end{align*}
Normally we'd need to worry about convergence of the series, but in this case the power series for the exponential function has infinite radius of convergence so everything converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Probability of colour with putting back twice. I'm having a bit of trouble with following probability-related question:
A container contains $5$ red and $10$ black balls. Take a ball out of the container at random and note its color. After drawing each ball, the ball gets put back and one extra ball of the same color gets added.
1) Given that the first $n$ drawn balls are all black, calculate the probability (say $a_{n}$) that the $(n+1)$st ball will also be black. What is $\lim_{x\to\infty} a_{n}$ equal to?
2) Given that the second until the $(n+1)$st ball (inclusive) are all black, calculate the probability that (say $b_{n}$) that the first drawn ball was also black. Also calculate $\lim_{n\to\infty} b_{n}$.
I succeeded at answering the first subquestion. As the result I get that $$a_{n} = \dfrac{10+n}{15+n}$$ and thus, $$\lim_{n\to\infty} a_{n} = 1$$
However, I have much more trouble answering the second question, and would like your help.
Thanks!
EDIT:
The end solution of the limit in subquestion (2) is also given, though not the way of getting to that result. The result is: $\lim_{n\to\infty} b_{n} = 1$.
| Let $B_n$ be event "all balls from the second until the $n$-th (inclusive) are black", and $B$ and $R$ be events "first ball was black" and "first ball was red".
Then $b_n = P(B | B_n) = \frac{P(B \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)} = 1 - \frac{P(R \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)}$.
$P(B \cap B_n) = P(B) \cdot P(B_1 | B) \cdot P(B_2 | B, B_1) \cdot \ldots \cdot P(B_n | B, B_{n - 1}) = a_0 \cdot a_1 \ldots \cdot a_n = \prod\limits_{i=0}^n \frac{10 + i}{15 + i} = \frac{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{(n + 11) \cdot (n + 12) \cdot (n + 13) \cdot (n + 14)\cdot(n + 15)}$.
Similarly $P(R \cap B_n) = \frac{5}{15} \cdot \frac{10}{16} \cdot \frac{11}{17} \cdot \ldots \cdot \frac{10 + n - 1}{15 + n} = \frac{5 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{(10 + n) \cdot (11 + n) \cdot (12 + n) \cdot (13 + n) \cdot (14 + n) \cdot (15 + n)}$.
From this we have $\frac{P(R \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)} = \frac{\frac{5}{10 + n}}{1 + \frac{5}{10 + n}} = \frac{5}{15 + n} \to 0$, so $b_n \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for x $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $ $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $
I tried transformation formulas , $ 2\sin a \sin b $ one. I know the value of sin 30° but what about others?
Original problem
In triangle ABC, P is an interior point such that $ \angle PAB = 10°. \angle PBA = 20° PAC = 40° \angle PCA = 30° $ then what kind of triangle it is ? I solved it till I got stuck here.
| Assume AB = 1.
Apply sine law to PAB to get $PB = 2 \sin 10^0$.
Apply sine law to PAC to get $PB = 2 \sin 20^0$ and $PC = 4 \sin 20^0 \sin 40^0$.
Apply cosine law to PBC to get $BC$ in terms of those angles. Note that $\cos 100^0 = - \sin 10^0$.
$BC^2 = 4 \sin^2 10 + 16 \sin^220^0\sin^2 40^0 + 16 \sin 20^0 \sin 40^0 \sin^210^0$
Suggestion:- Convert all angles in $BC^2$to $\sin 10^0$ (or $\sin 20^0$) by compound angle formula. Hope that the result is 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
contour integration $\int_C \frac{\ln^2 z}{\sqrt{z}(1+z^2)}$(1 Short answer) I solved this question and had a try as shown ... i got similar form ... but not getting correct answer.
here is the question:
Question 3.
$\quad$ (a) By considering the integral of $z^{-1/2}\log^2(z)/(1+z^2)$ around a suitably chosen contour in the cut $z$ plane, prove that
$$\int_0^{\infty}\frac{\log^2(t) - \pi^2}{t^{1/2}(1+t^2)}dt = -\frac{\pi^3}{4\sqrt{2}}$$
$\quad$ In your solution, include a diagram that clearly specifies the contour you are using and provide a careful discussion of all required estimates of contributions from contours that you will discard when an appropriate limit is taken.
$\quad$ (b) Using the same integration contour as in part (a) with the integrand $z^{-1/2}\log^2(z)/(1-z)$, find a relation between the integrals
$$\int_0^{\infty}\frac{\log^2(t)}{t^{1/2}(1+t)}dt \quad\text{ and }\quad \int_0^{\infty}\frac{dt}{t^{1/2}(1+t)}.$$
$\quad$Use this relation to deduce, without further contour integration, that
$$\int_0^{1}\frac{\log^2(t)}{t^{1/2}(1+t)}dt = \frac{\pi^3}{2}.$$
Here is what I have so far:
$\quad$ Ans 3. a) $$\int_0^{\infty}\frac{log^2(t) - \pi^2}{\sqrt{t}(1+t^2)}dt$$
$$\begin{multline}
\shoveleft \text{Consider} \quad f(z) = \frac{log^2(z)}{\sqrt{z}(1+z^2)} \\
\shoveleft \text{Consider} \quad \int_Cf(z)dz \qquad \text{along the C as shown below}
\end{multline}$$
$$\begin{multline}
\shoveleft \text{Poles are at } z = 0 \text{ (not lies[sic] in contour)}, \quad z = \pm i \\
\shoveleft \text{Res }f(z=i) \quad \space \space \space = \quad \lim_{z \rightarrow i}\frac{(z-i)log^2(z)}{\sqrt{z}(z-i)(z+i)} \space = e^{i\pi /4}\frac{\pi^2}{8} \\
\shoveleft \text{Res }f(z=-i) \quad = \quad \lim_{z \rightarrow -i}\frac{(z+i)log^2(z)}{\sqrt{z}(z-i)(z+i)} = -e^{3i\pi /4}\frac{\pi^2}{8} \\
\shoveleft \therefore \quad \int_{C}f(z)dz = 2\pi i\left[\frac{\pi^2}{8}(e^{i\pi /4} - e^{3i\pi /4}) \right] = \frac{i\pi^3}{2\sqrt{2}} \\
\shoveleft \\
\shoveleft \text{Thus,} \\
\shoveleft \quad \int_{AB}f(z)dz + \int_{\Gamma}f(z)dz + \int_{FG}f(z)dz + \int_{\gamma}f(z)dz = \frac{i\pi^3}{2\sqrt{2}} \\
\shoveleft \\
\shoveleft \text{On AB: } z = x \quad \text{and on FG: } z = xe^{2\pi i} \\
\shoveleft \therefore \quad \int_{AB}f(z)dz + \int_{FG}f(z)dz = \int_{r}^{R}\frac{log^2x}{\sqrt{x}(1+x^2)}dx + \int_{R}^{r}\frac{\left( log(xe^{2\pi i}) \right)^2e^{2\pi i}}{\sqrt{xe^{2\pi i}}(1+x^2e^{2\pi i})}dx \\
\shoveleft \\
\shoveleft = \int_{r}^{R}\frac{log^2x}{\sqrt{x}(1+x^2)}dx + \int_{R}^{r}\frac{log^2x - 4\pi^2 +4\pi log(x)i}{\sqrt{x}e^{\pi i}(1+x^2)}dx \\
\shoveleft \\
\shoveleft = \int_{r}^{R}\frac{log^2x}{\sqrt{x}(1+x^2)}dx \ + \space \int_{r}^{R}\frac{log^2x - 4\pi^2 + 4\pi log(x)i}{\sqrt{x}(1+x^2)}dx \quad \because \space e^{\pi i} = -1 \\
\shoveleft \\
\shoveleft = \int_{r}^{R}\frac{2log^2x - 4\pi^2 - 4\pi log(x)i}{\sqrt{x}(1+x^2)}dx \\
\end{multline}$$
$$\begin{multline}
\shoveleft \\
\shoveleft \text{For the circle } \gamma : \space z = re^{i \theta} \\
\shoveleft \text{So, } \int_{\gamma}f(z)dz = \int_{2\pi}^{0} \frac{\left( log(re^{i \theta})\right)^2(rie^{i \theta})}{\sqrt{re^{i \theta}}(1 + r^2 e^{2i \theta})}d\theta = \int_{2\pi}^{0}\frac{\sqrt{r}*ie^{i \theta}*log^2(re^{i \theta})}{e^{i\theta /2}(1 + r^2e^{2i\theta})}d\theta \\
\shoveleft \text{Clearly, if } r \rightarrow 0, \space \int_{\gamma}f(z)dz \rightarrow 0 \\
\end{multline}$$
$$\begin{multline}
\shoveleft \\
\shoveleft \text{Now, on the circle } \Gamma : \space z = Re^{i\theta} \\
\shoveleft \text{So, } \int_{\Gamma}f(z)dz = \int_{0}^{2\pi} \frac{\left( log(Re^{i \theta})\right)^2(Rie^{i \theta})}{\sqrt{Re^{i \theta}}(1 + R^2 e^{2i \theta})}d\theta = \int_{0}^{2\pi}\frac{\sqrt{R}*ie^{i \theta /2}*log^2(Re^{i \theta})}{e^{i\theta /2}(1 + R^2e^{2i\theta})}d\theta \\
\shoveleft = \int_{0}^{2\pi}\frac{ie^{i\theta /2}*log^2(Re^{i\theta})}{R^{3/2}\left(\frac{1}{R^2} + e^{2i\theta}\right)}d\theta \\
\shoveleft \text{if } R \rightarrow \infty, \space \int_{\Gamma}f(z)dz \rightarrow 0 \\
\shoveleft \\
\shoveleft \text{So, we have } \\
\shoveleft \quad \int_{0}^{\infty}\frac{2log^2x - 4\pi^2 +4\pi log(x)i}{\sqrt{x}(1+x^2)}dx = \frac{i\pi^3}{2\sqrt{2}} \\
\shoveleft \quad \qquad \implies +\int_{0}^{\infty}\frac{4\pi log(x)}{\sqrt{x}(1+x^2)}dx = \frac{\pi^3}{2\sqrt{2}} \\
\shoveleft \space \space \qquad \quad \qquad \text{or } \int_{0}^{\infty}\frac{log(x)}{\sqrt{x}(1+x^2)}dx = \frac{\pi^2}{2\sqrt{2}}
\end{multline}$$
I don't understand the explanation in the solution, can anyone explain where is my mistake ? How to arrive at the answer?
| You chose the branches of $\sqrt z$ and $\ln z$ corresponding to $\arg z \in [0, 2 \pi)$. Then $\sqrt {-i} = e^{3 \pi i/4}$, $\ln(-i) = 3 \pi i/2$, and, fixing the typo in the $\ln x$ term, you get
$$\int_0^\infty \frac
{2 \ln^2 x + 4 \pi i \ln x - 4 \pi^2}
{\sqrt x \, (1 + x^2)} dx =
2 \pi i \left(
\frac {\ln^2 i} {\sqrt i} \operatorname*{Res}_{x = i} \frac 1 {1 + x^2} +
\frac {\ln^2(-i)} {\sqrt {-i}} \operatorname*{Res}_{x = -i} \frac 1 {1 + x^2}
\right) = \\
-\frac {\left( \frac 5 2 + 2 i \right) \pi^3} {\sqrt 2}.$$
Just take the principal branches and choose a keyhole contour avoiding $(-\infty, 0]$. You'll get
$$\int_{-\infty}^0 \frac {2 \ln^2 |x| - 2 \pi^2} {i \sqrt{|x|} \, (1 + x^2)} dx =
2 \pi i \left( \operatorname*{Res}_{x = -i} + \operatorname*{Res}_{x = i} \right)
\frac {\ln^2 x} {\sqrt x \, (1 + x^2)},$$
which is exactly what is required for a).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Why is this integration method not valid? Let $$I=\int \frac{\sin x}{\cos x + \sin x}\ dx \tag{1}$$
Now let $$u=\frac{\pi}{2} - x \tag{2}$$ so $$I=\int \frac{\sin (\frac{\pi}{2} - u)}{\cos (\frac{\pi}{2} - u)+\sin (\frac{\pi}{2} - u)}\ du \tag{3}$$
$$=\int\frac{-\cos u}{\sin u + \cos u} \ du \tag{4}$$
$$= \int\frac{-\cos x}{\sin x + \cos x} \ dx \tag{5}$$
and hence $$2I=\int\frac{\sin x - \cos x}{\sin x + \cos x} \ dx \tag{6}$$
$$=-\ln\ |\sin x + \cos x| + c \tag{7}$$
$\implies I=-\frac{1}{2}\ln|\sin x + \cos x| + c \tag{8}$
But the actual answer is $$I= \frac{1}{2}x -\frac{1}{2}\ln|\sin x + \cos x| + c \tag{9}$$
according to Wolfram Alpha and supported by a different method.
Why does my method not yield the correct result?
| An alternative change:
$$I=\int \frac{\sin \ x}{\cos \ x + \sin \ x}\ dx=\frac1{\sqrt{2}}\cdot \int \frac{\sin \ x}{\frac1{\sqrt{2}}\cos \ x + \frac1{\sqrt{2}}\sin \ x}\ dx=\frac1{\sqrt{2}}\cdot \int \frac{\sin \ x}{\sin \left(x+\frac{\pi}{4}\right)}\ dx.$$
Now let $u=x+\frac{\pi}{4}$, then:
$$I=\frac1{\sqrt{2}}\cdot \int \frac{\sin \left(u-\frac{\pi}{4}\right)}{\sin u}\ du=\frac12\int \frac{\sin u-\cos u}{\sin u}\ du=\frac{x}{2}-\frac12\int \frac{d\sin u}{\sin u}=\\
\frac x2-\frac12\ln |\sin u|+C=\frac x2-\frac12\ln |\sin \left(x+\frac{\pi}{4}\right)|+C=\\
\frac x2-\frac12 \ln |\frac1{\sqrt{2}}(\cos x+\sin x)+C=\frac x2-\frac12\ln |\cos x+\sin x|+\underbrace{\frac14\ln 2+C}_{A}.$$
Now, a simpler example:
$$\int \cos x \ dx \stackrel{x=\pi/2-u}=\int \cos \left(\frac{\pi}{2}-u\right)\cdot (-du)=\\
-\int \sin u \ du\stackrel{\ne -\int \sin x \ dx=\cos x+C}=\cos u+C=\\
\cos \left(\frac{\pi}{2}-x\right)+C=\sin x+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 6,
"answer_id": 5
} |
Prove that $2\tan^{-1}\frac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} $ Prove
$$2\tan^{-1}\dfrac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} \tag{a p u}$$
I find
$$a\sqrt{a^{2}- 2}= \dfrac{\dfrac{\sqrt{a^{2}- 2}}{a}+ \dfrac{\sqrt{a^{2}- 2}}{a}}{1- \dfrac{\sqrt{a^{2}- 2}}{a}\dfrac{\sqrt{a^{2}- 2}}{a}}$$
This may help! How I can use it to solve my problem? Thanks for all the interests!
| First, note that, whenever $\frac{\sqrt{a^2 - 2}}{a}$ is well-defined, we have
$$-1 < \frac{\sqrt{a^2 - 2}}{a} < 1,$$
and since $\tan^{-1}$ is an increasing function,
$$-\frac{\pi}{4} = \tan^{-1}(-1) < \tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) < \tan^{-1}(1) = \frac{\pi}{4}.$$
Therefore,
$$-\frac{\pi}{2} < 2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) < \frac{\pi}{2}.$$
For angles $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we have $\tan^{-1} (\tan \theta) = \theta$. Also note that, for any $\alpha \in \Bbb{R}$, we have $\tan(\tan^{-1}(\alpha)) = \alpha$. Thus, by the double angle formula for $\tan$,
\begin{align*}
2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right) &= \tan^{-1}\left(\tan\left(2\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)\right) \\
&= \tan^{-1}\left(\frac{2\tan \left(\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)}{1 - \tan \left(\tan^{-1}\left(\frac{\sqrt{a^2 - 2}}{a}\right)\right)^2}\right) \\
&= \tan^{-1}\left(\frac{2\frac{\sqrt{a^2 - 2}}{a}}{1 - \left(\frac{\sqrt{a^2 - 2}}{a}\right)^2}\right) \\
&= \tan^{-1}\left(a\sqrt{a^2 - 2}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
| Rearrange:
$$(x^2+6x+7)^2+6(x^2+6x+7)+7=x \iff \\
((x+3)^2-2)^2+6((x+3)^2-2)+10=x+3 \iff \\
(x+3)^4+2(x+3)^2-(x+3)+2=0 \iff \\
(x+3)^4+(x+3-\frac12)^2+(x+3)^2+\frac74=0.$$
The LHS is positive, hence, there is no real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Given $3$ positive reals $a$, $b$ and $c$ such that $a+b+c = 1$, show that $a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \le1$. Good Day! How are you doing?
I was learning about the awesome A.M - G.M. inequality from the Brilliant Wiki. There was a question in the exercises:
Given $3$ positive reals $a$, $b$ and $c$ such that $$a+b+c = 1,$$ show that $$a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \le 1$$.
Unfortunately, I was not able to solve this problem nor do I think that solutions are provided. Also, I was not able to find this question here at MSE.
Here was my thought process:
By the AM-GM inequality, it follows that $a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \ge 3(abc)^\frac {1} {3}$
Also, $1 \ge 3(abc)^\frac {1} {3}$
which yields no help.
I also thought of $3(a^a b^ b c^c)^\frac {1} {3} \le a^a + b^b + c^c$ and similarly to other terms but once again, I got stuck at some messy expressions.
I would really appreciate if you could help me.
Thanks!
| By the weighted AM-GM inequality, $a^ab^bc^c \leq a \times a + b \times b + c \times c$ (by using weights $a$, $b$, $c$). Similar expressions hold for the other terms. So the inequality becomes $a^ab^bc^c + a^bb^cc^a + a^cb^ac^b \leq a^2 + b^2 + c^2 + ab + bc + ca + ac + ba + cb = (a+b+c)^2 = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $5^n + 3^n - 2^{2n+1} > 0$ by induction I am not sure how to deal with the $-2^{2n+1}$ term.
I did the basis proof for n=1
I am stuck at this step:
$$
5^{k+1}+3^{k+1}-2^{2(k+1)+1} = 5\cdot 5^k + 3 \cdot 3^k -2^3 \cdot 2^{2k}
$$
Any advice guys?
| Note that
$$
\begin{align}
5^{k+1}+3^{k+1}-2^{2(k+1)+1}
&=5\cdot5^k+3\cdot3^k-4\cdot2^{2k+1}\\
&=\left(5^k-3^k\right)+4\left(5^k+3^k-2^{2k+1}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
If a rectangle can be split into n and into n+14 squares, then n=18. What is the meaning of this This problem seems to be a geometrical one, but actually it is an algebraic one.
If you split the rectangle into n squares and into n+k squares, then what does that means? What is this supposed to mean?
| The rectangle area must add up exactly to the sums of the areas of the squares. Let $x$ be the side length of each of the $n$ squares and $y$ be the side length of each of the $n+k$ squares. Thus, you get the rectangle area is
$$nx^2 = (n+k)y^2 \tag{1}\label{eq1}$$
This can be adjusted to
$$\frac{n+k}{n} = \frac{x^2}{y^2} \tag{2}\label{eq2}$$
Thus, the left side must be the square of a rational number. In your case, $k = 14$. With $n = 18$, you get that $\frac{n+k}{n} = \frac{32}{18} = \frac{16}{9}$. This gives that $x = 4m$ and $y = 3m$ for some positive integer $m$. From \eqref{eq1}, the area of the rectangle would then be $18 \times (4m)^2 = 288m^2$.
You can do a similar type of check for other values of $k$. To determine a value of $n$ for any given value of $k$, consider $k = 2^a b$ where $a \ge 0$ and $b$ is an odd integer $\gt 1$. Then $n = 2^a\left(\frac{b-1}{2}\right)^2$ will give
\begin{align}
\frac{n+k}{n} & = \frac{2^a\left(\frac{b-1}{2}\right)^2 + 2^ab}{2^a\left(\frac{b-1}{2}\right)^2} \\
& = \frac{b^2 - 2b + 1 + 4b}{4\left(\frac{b-1}{2}\right)^2} \\
& = \frac{b^2 + 2b + 1}{(b-1)^2} \\
& = \frac{(b + 1)^2}{(b-1)^2} \tag{3}\label{eq3}
\end{align}
In the case with $k = 14 = 2 \times 7$, this gives that $b = 7$ and $n = 2 \times \left(\frac{6}{2}\right)^2 = 2 \times 9 = 18$ and \eqref{eq3} gives $\frac{8^2}{6^2} = \frac{4^2}{3^2} = \frac{16}{9}$ as noted above.
For any given $k$ which has a solution, there are at most a finite # of values of $n$ which work, with just the $1$ value of $n = 18$ for $k = 14$. To see this, note that $x$ must be greater than $y$ in \eqref{eq2}, so let $x = y + c$ for some positive integer $c$. Substituting this into \eqref{eq2}, expanding and then simplifying gives
$$\frac{n + k}{n} = \frac{y^2 + 2cy + c^2}{y^2} \\
1 + \frac{k}{n} = 1 + \frac{2cy + c^2}{y^2} \\
\frac{k}{n} = \frac{2cy + c^2}{y^2} \\
ky^2 = 2cny + c^2n \\
ky^2 - 2cny - c^2n = 0 \tag{4}\label{eq4}$$
This is a quadratic equation in $y$, so the quadratic formula gives
\begin{align}
y & = \frac{2cn \pm \sqrt{4c^2n^2 + 4c^2nk}}{2k} \\
& = \frac{cn \pm c\sqrt{n^2 + nk}}{k} \tag{5}\label{eq5}
\end{align}
For $y$ to be an integer requires the part in the square roots in \eqref{eq5} to be a perfect square, i.e.,
$$n^2 + nk = (n + d)^2 \tag{6}\label{eq6}$$
for some positive integer $d$. However, since $n^2 \lt n^2 + 2nk \lt \left(n + \frac{k}{2}\right)^2$, then
$$0 \lt d \lt \frac{k}{2} \tag{7}\label{eq7}$$
Next, expanding the right hand side of \eqref{eq6}, moving the $n$ term over gives
$$n(k - 2d) = d^2 \tag{8}\label{eq8}$$
This requires that $d^2$ be an integral multiple of $k - 2d$. For $k = 14$, using \eqref{eq7}, this requires checking $d$ from $1$ to $6$ for any which also satisfy \eqref{eq8}. In this case, $d = 1$ gives $12n = 1$, $d = 2$ gives $10n = 4$, $d = 3$ gives $8n = 9$, $d = 4$ gives $6n = 16$, $d = 5$ gives $4n = 25$, so only $d = 6$ works to get that $n(2) = 36 \implies n = 18$ is the one and only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Contour integration problem with sin and cos so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$
I thought the sensible thing to do here would be to substitute in $z=e^{ix}$ and take a contour integral around the unit circle, call this path $\ast$ so that my integral becomes $$\frac{1}{2i}Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)$$
Then, letting $f(z)=\frac{1-z^2}{az+bz^2}$, I thought the function had simple poles at $z=0$ with residue $\frac{1}{a}$ and another simple pole at $z=\frac{-a}{b}$ with residue $\frac{a}{b^2}-\frac{1}{a}$ and thus I get the that $$Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)=2i(2\pi i)\frac{a}{b^2}=-4\pi(\frac{a}{b^2})$$ which is not the answer given, that is: $=\frac{2\pi}{b^2}[a-\sqrt{a^2-b^2}]$,but I can't work out why.
Any help appreciated, thank you in advance.
| If you substitute $z = e^{ix}$, then $\sin(x) = (z - \frac{1}{z})/(2i) $ and $\sin^2(x) =(z-\frac{1}{z})^2/(-4) = \frac{(z^2-1)^2}{-4z^2} $
$dz = ie^{ix}dx = izdx$ or $dx = \frac{dz}{zi}$
$a+b\cos x = a + b\frac{z+\frac{1}{z}}{2} = \frac{2az + bz^2 +1}{2z}$
So,
$$I = \int_*\frac{\frac{(z^2-1)^2}{-4z^2}dz}{zi\frac{2az + bz^2 +1}{2z}}$$
$$I = \int_*\frac{i(z^2-1)^2dz}{2z^2(2az + bz^2 +1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove $n^{3/2} \le 2^n/n$ for all $n\geq 8$ via Induction? How to prove $n^{3/2} \le 2^n/n$ for all $n \geq 8$
My attempt:
$\textbf{Base case}$ is $n_0=8$: $\quad \left(8^{3/2}=4\right)\leq \left(32=\frac{2^8}{8}\right) \checkmark$
$\textbf{Induction hypothesis}$: $\exists n\in \mathbb{N}_{\geq 8}:(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1}$
$\textbf{Inductive step}$:
$$
\begin{gather}
(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1} \end{gather} \\
\iff (n+1)^{5/3}\leq 2^{n+1} \\
\iff (n+1)^{5/3}\leq 2^n \cdot 2 \quad |\uparrow ^{3/5} \\
\iff n+1\leq 2^{3n/5}\cdot 2^{3/5}\ $$
This, sadly, doesn't get me anywhere, any ideas?
| Continuing Sonnhard's answer. Hint: $$ (2n^{5/2})^{2} = 4 n^{5} \ge n^{5} + n^{5} + n^{5} + n^{5} \ge n^{5} + 8 n^{4} + 64 n^{3} + 8^{3} n^{2} $$
Or can also
$$n^{5} + n^{5} + n^{5} + n^{5} \ge n^{5} + 16 n^{4} + 64 n^{3} $$
$$ (n+1)^{5} = n^{5} + \binom{5}{1}n^{4} + \binom{5}{2} n^{3} + \binom{5}{4} n^{2} +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$? Could someone explain how to get from: $x-\frac{1}{x}=A$ to $x+\frac{1}{x}=\sqrt{A^2+4}$ ? It is one of the Algebra II tricks.
Thanks.
| Ok, I got it!
$x-\frac{1}{x}=A$
$\Bigl(x-\frac{1}{x}\Bigl)^2=A^2$
$x^2+\frac{1}{x^2}-2=A^2$
$x^2+\frac{1}{x^2}=A^2+2$
$\Bigl(x+\frac{1}{x}\Bigl)^2-2=A^2+2$
$\Bigl(x+\frac{1}{x}\Bigl)^2=A^2+4$
$\sqrt{\Bigl(x+\frac{1}{x}\Bigl)^2}=\sqrt{A^2+4}$
$x+\frac{1}{x}=\sqrt{A^2+4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| For a nice algebraic way to prove it, consider the inequality
$$\sin x\geq \frac{2x}{\pi}, x\in[0,\frac{\pi}{2}]$$
and apply for $A,B,C>\frac{\pi}{4}$:
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq\frac{2}{\pi}(A+B+C-\frac{3\pi}{4})=\frac{1}{2}>0$$
When we assume wlog $A<\frac{\pi}{4}$, we can instead assert that:
$$\sin(A-\frac{\pi}{4})\geq A-\frac{\pi}{4}$$
and then
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 1-\frac{\pi}{4}+(1-\frac{2}{\pi})A>1-\frac{\pi}{4}>0$$
This is not the sharpest bound. Actually one can show that the minimum value this expression is achieved for A=B=C and therefore
$$\require{cancel}\xcancel{\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 3\sin(\frac{\pi}{12})=\frac{3}{2}\sqrt{2-\sqrt{3}}}$$
EDIT: After a while, I finally noticed that, while the solution to the problem satisfactorily addresses the question asked, a mistake above has obscured a clear elementary solution to finding maxima and minima to this function. We want to show that whenever $A,B,C>0$ with $A+B+C=\pi$,
$$\frac{\sqrt{2}}{2}<\Delta=\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin\frac{\pi}{12}$$
In the figure-where the contours of $\Delta(A,B)$ are depicted- one can clearly see that, within the area of interest denoted by a red triangle, the function attains a maximum. The minima are attained on the red triangle itself. As of now I haven't found a satisfactory algebraic/elementary approach to obtain the minima and maxima in the entire triangle interior and boundary (but with calculus it is pretty easy to demonstrate).
To get the maximum in a part of the interior, assume that $A,B,C>\frac{\pi}{4}$. In this case we can apply Jensen's inequality for the concave function $f(x)=\sin(x)~,~ x\in (0,\pi)$ and we obtain
$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin(\frac{A+B+C}{3}-\frac{\pi}{4})=3\sin\frac{\pi}{12}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
| Not an answer, strictly speaking, but an alternate method.
$$J=\int_{-\infty}^{\infty}\frac{x^2 dx}{(1+x^2)^2}=2\int_{0}^{\infty}\frac{x^2 dx}{(1+x^2)^2}$$
We may rewrite this as
$$\begin{align}
J&=2\int_0^\infty \frac{dx}{1+x^2}-2\int_0^\infty \frac{dx}{(1+x^2)^2}\\
&=\pi-2\int_0^\infty \frac{dx}{(1+x^2)^2}\ .
\end{align}$$
From here, we have that if
$$I_n=\int\frac{dx}{(ax^2+b)^n}\qquad a,b>0, n\in\Bbb N$$
then
$$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}\qquad n>1\ ,$$
with the base case
$$I_1=\frac1{\sqrt{ab}}\arctan\sqrt{\frac{a}{b}}x+C\ .$$
Hence we have (with $a=b=n-1=1$)
$$\begin{align}
\int_0^\infty\frac{dx}{(x^2+1)^2}&=\frac{x}{2(x^2+1)}\bigg|_0^\infty+\frac{1}{2}\int_0^\infty\frac{dx}{x^2+1}\\
&=0+\frac12\cdot\frac\pi2\\
&=\frac\pi4
\end{align}$$
So
$$\begin{align}
J&=\pi-2\cdot\frac\pi4\\
&=\frac\pi2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Let $m,n\in \mathbb{Z}$ and $p(x)=x^3+mx+n$ be such that if $107\mid p(x)-p(y)\implies 107\mid x-y$. Prove that $107\mid m$.
Let $m,n\in \mathbb{Z}$ and $p(x)=x^3+mx+n$ be such that for an integers $x,y$ we have: $$107\mid p(x)-p(y)\implies 107\mid x-y$$ Prove that $107\mid m$.
I'm not sure what to do here. I can only deduce that $$107\nmid x^2+xy+y^2+m$$
and since $x\equiv y \pmod {107}$ $$ 107\nmid 3x^2+m$$ Suppose $107\nmid m$ then expressing this with Legendre symbol we have $$\Big({-3m\over 107}\Big)=-1$$
Any sugestion?
| $p(x)-p(y)=x^3-y^3+m(x-y)=(x-y)(x^2 + xy + y^2+m)$
The condition is $$x^2+xy+y^2 +m \not \equiv 0 \pmod{107}, \quad when \quad x\not\equiv y \pmod{107}$$
Setting $y=0$ we get $$ x^2 + m \not \equiv 0 \quad when \quad x\not\equiv 0$$
Since $-1$ is quadratic nonresidue modulo 107 (because $(-1)^{53}\equiv -1$), we have that $$m \text{ is quadratic residue modulo 107}$$
Let $m\equiv a^2 \not \equiv 0 \pmod{107}$
Note Below I will write $\frac{1}{y}$ to mean some integer number such that $y\cdot \frac{1}{y}\equiv 1 \pmod{107}$. Such a number exists iff $y\not\equiv 0$. And $\frac{x}{y}$ means $x\cdot \frac{1}{y}$.
Rewrite the condition in the form
$$ x^2 + xy+ y^2 \not\equiv-a^2 $$
$$\forall b\not\equiv 0 \quad (\frac{x}{y})^2 + \frac{x}{y} +1 \not\equiv -b^2$$
$$\forall b\not\equiv 0 \quad \lambda ^2 + \lambda + 1 \not \equiv -b^2$$
Plugging in $\lambda=2$ yields $7$, which is a quadratic non residue modulo 107.
That means that $2^2 + 2 +1 \equiv -b^2$ for some $b$, multiply that by $\frac{a^2}{b^2}$ to get $\frac{\lambda^2 a^2}{b^2} + \lambda \frac{a^2}{b^2} + \frac{a^2}{b^2} \equiv -a^2$ that is $$ (\frac{\lambda a}{b})^2 + \frac{\lambda a}{b} \frac{a}{b} + (\frac{a}{b})^2 \equiv -a^2$$
since $\lambda = 2 \not \equiv 1$, we have a pair $x=\frac{2 a}{b}$ and $y=\frac{a}{b}$ contradicting the condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Compute without calculator, $ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} - \frac{1}{\cos^{2}(45)} $ Compute without calculator,
$$ \frac{1}{\cos^{2}(10^{\circ})} + \frac{1}{\sin^{2}(20^{\circ})} + \frac{1}{\sin^{2}(40^{\circ})} - \frac{1}{\cos^{2}(45^{\circ})} $$
Attempt:
Let $A = \cos(10) \sin(20) \sin(40)$, then
$$ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} = \frac{\sin^{2}(20) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
notice also
$$2A \sin(10) \sin(40) = \sin^{2}(20) \sin^{2}(40) $$
$$2A \cos(20) \cos(10) = \cos^{2}(10) \sin^{2}(40) $$
so we have
$$\frac{2A \sin(10) \sin(40) + 2A \cos(20) \cos(10)+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[2 \sin(10) \sin(20) \cos(20) + \frac{\sqrt{3}}{2} + \sin(10)\sin(20) \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (1+ \cos(20) + \cos(20)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (3\cos^{2}(10) - \sin^{2}(10)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
How to continue then?
| I believe this is how the problem came into being.
Observe that
$\dfrac1{\cos^210^\circ}=1+\tan^210^\circ$
$\dfrac1{\sin^220^\circ}=\dfrac1{\cos^270^\circ}=1+\tan^270^\circ$
$\dfrac1{\sin^240^\circ}=\dfrac1{\cos^250^\circ}=1+\tan^2(-50^\circ)$
Observe that the angles differ by $60^\circ$
Now if $\tan3x=\tan3A,3x=180^\circ n+3A$ where $n$ is any integer
$x=60^\circ n+A$ where $n=-1,0,1$ or more generally $n\equiv-1,0,1\pmod3$
Again, $\tan3A=\tan3x=\dfrac{3t-t^3}{1-3t^2}$
$\implies t^3-(3\tan3A)t^2-3t+\tan3A=0$ where $t=\tan x$
whose roots are $t_r=\tan(60^\circ r+A)$ where $r=-1,0,1$
$\displaystyle\implies S(A)$
$\displaystyle=\sum_{r=-1}^1t_r^2 =\left(\sum_{r=-1}^1t_r\right)^2-2(t_{-1}t_0+t_0t_1+t_1t_{-1}) =\left(\dfrac{3\tan3A}1\right)^2-2\left(\dfrac{-3}1\right)=6+9\tan^23A$
Here $3A=30^\circ\equiv-150^\circ\equiv210^\circ\pmod{180^\circ}$
So, $\tan3A=\tan30^\circ=\dfrac1{\sqrt3}$
$\implies S(10^\circ)=6+9\left(\dfrac1{\sqrt3}\right)^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$
Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$
I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain
$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$
I am stuck here. Any hints on solving the R.H.S will be appreciated.
|
$\sin 75^\text o= \dfrac{\sqrt{3}+1}{2\sqrt2}$
Can you find now ?
PROOF :
$\sin (x+y) = \sin x\cos y +\cos x \sin y $
$\sin(75^\text o) = \sin(30^\text o+45^\text o)\\\qquad\quad= \sin 30^\text o\cos 45^\text o +\cos 30^\text o \sin 45^\text o\\\qquad\quad = \dfrac{1}{2\sqrt2}+\dfrac{\sqrt3}{2\sqrt2} = \dfrac{\sqrt3+1}{2\sqrt2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous method of only using a characterization of $e$ with inequalities since we thought it was impractical to do it. Let's return instead to analysis.
*
*$f(1) = 2 \ln(\frac 5 3) > 1$
*$f'(x) = \ln(1+ \frac 2 {2x+1}) - \frac {4(x+1)}{(4x^2+8x+3)} < 0$ on $[1,+\infty[$
*$\lim f(x) = 1$
However how do I show formally 2.?
| $\quad$The question currently asks for a formal demonstration that
$$\ln\left(1+ \dfrac 2 {2x+1}\right) < \dfrac {4(x+1)}{(4x^2+8x+3)} \text{ for } x\in[1,+\infty).$$
$\quad$To show it's true for $x=1$, we must show $\ln\left(1+\dfrac23\right)<\dfrac{4(2)}{4+8+3}=\dfrac8{15}.$
By exponentiating both sides, this is equivalent to $\dfrac53<\exp\left(\dfrac8{15}\right).$
Well, $\dfrac53=\dfrac{375}{225}<\dfrac{377}{225}=1+\dfrac8{15}+\dfrac12\left(\dfrac{8}{15}\right)^2<\exp\left(\dfrac8{15}\right).$
$\quad$Furthermore, it can be shown that
$$\lim_{x\to\infty}\ln\left(1+ \dfrac 2 {2x+1}\right) = \lim_{x\to\infty}\dfrac {4(x+1)}{(4x^2+8x+3)} =0.$$
$\quad$Finally, it can be shown that $\ln\left(1+ \dfrac 2 {2x+1}\right)$ and $\dfrac {4(x+1)}{(4x^2+8x+3)}$ are monotonically decreasing in the relevant interval. If $0\le x _0<x_1,$ then $2x_0+1<2x_1+1,$ so $\dfrac1{2x_1+1}<\dfrac1{2x_0+1},$
so $\ln\left(1+ \dfrac 2 {2x_1+1}\right) < \ln\left(1+ \dfrac 2 {2x_0+1}\right),$
and $4x_1x_0^2+4x_0^2+5x_0<4x_0x_1^2+4x_1^2+5x_1; $
adding $8x_0x_1+3x_1+3x_0+3$ to both sides,
$4x_1x_0^2+8x_0x_1+3x_1+4x_0^2+8x_0+3<4x_0x_1^2+8x_0x_1+3x_0+4x_1^2+8x_1+3$
i.e., $4(x_1+1)(4x_0^2+8x_0+3)<4(x_0+1)(4x_1^2+8x_1+3),$ so $$\dfrac{4(x_1+1)}{(4x_1^2+8x_1+3)}<\dfrac{4(x_0+1)}{(4x_0^2+8x_0+3)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential equation :
$$ (x^2-y^2-y)dx - (x^2-y^2-x)dy=0$$
My attempt:
The coefficients of $dx$ and $dy$ are not homogenous functions. Further,
$
\begin{align}
P(x,y) &= x^2 - y^2 - y \\
\frac{\partial P(x,y)}{\partial y}&=-2y-1 \\
Q(x,y) &= -(x^2-y^2-x)\\
\frac{\partial Q(x,y)}{\partial x}&=-(2x-1) \\
\therefore \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
\end{align}
$
The given differential equation is not exact. We have :
$\begin{align}
& \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \\
=& -2y-1+(2x-1)\\
=&(2x-2y-2)
\end{align}$.
Moreover,
$\begin{align}
& yQ-xP\\
= & -y(x^2-y^2-x)-x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy - x^3 + xy^2 + xy\\
= & y^3 - x^3 + xy - xy(x - y - 1)
\end{align}$
It doesn't look like $(\partial P / \partial y - \partial Q / \partial x)/(yQ-xP)$ will be a function of $u=xy$ alone.
Also,
$\begin{align}
& yQ+xP\\
= & -y(x^2-y^2-x)+x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy + x^3 - xy^2 - xy\\
= & y^3 + x^3 - x^2 y - x y^2 \\
= & (y + x)(y^2 + x^2 - xy) - xy(y + x) \\
= & (y + x)(y^2 + x^2 - 2xy)\\
= & (y + x)(y - x)^2
\end{align}$
It doesn't look like $y^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=x/y$ or $x^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=y/x$ alone.
| $(x^2-y^2-y)~dx-(x^2-y^2-x)~dy=0$
$(-x^2+y^2+x)~dy=(-x^2+y^2+y)~dx$
$(y^2-x^2+x)\dfrac{dy}{dx}=y^2-x^2+y$
You can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=169
The general solution can express as $\begin{cases}x=C|t|e^{4t}-t\\y=C|t|e^{4t}+t\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must be solved without using the complex numbers. With complex numbers i found that the result is $1$.
Is there any way to solve this in $\Bbb R$ without entering $\Bbb C$?
| A bit late answer but I think worth mentioning it.
Here a direct calculation which uses only once the fact
*
*$\frac{2}{x}=2-x \stackrel{x\neq 0}{\Leftrightarrow} \boxed{x^2 = 2x-2} \quad (\star)$
\begin{eqnarray*} \left[x^9-(x^4+x^2+1)(x^6+x^3+1)\right]^3
& = & \left[x^9-\frac{x^6-1}{x^2-1}\frac{x^9-1}{x^3-1}\right]^3\\
& = & \left[x^9-\frac{x^3+1}{x^2-1}(x^9-1)\right]^3\\
& = & \left[x^9-\frac{x^2-x+1}{x-1}(x^9-1)\right]^3 \\
& \stackrel{(\star)}{=} & \left[x^9-\frac{2x-2-x+1}{x-1}(x^9-1)\right]^3 \\
& = & \left[x^9-(x^9-1)\right]^3 \\
& = & 1 \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\int_3^4 \sqrt {x^2-3x+2} \, dx$ using Euler's substitution
Calculate $\int_3^4 \sqrt {x^2-3x+2}\, dx$ using Euler's substitution
My try:
$$\sqrt {x^2-3x+2}=x+t$$
$$x=\frac{2-t^2}{2t+3}$$
$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$
$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$
$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$ However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.Can you help me?P.S. I must use Euler's substitution because that's the command.
| Hint: Make another substitution $z=2t+3$. However ugly it turns out to be, you only need to calculate integrals of the form $x^{\alpha}$ for $\alpha$ integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\frac{1}{\sqrt{2-x^2}}$ I know that the integral of $\frac{1}{\sqrt{1-x^2}} = arcsin(x)$, but what is the the integral of $\frac{1}{\sqrt{2-x^2}}$? Is this how you do it?
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2}(1-\frac{x^2}{\sqrt{2}})} = \frac{1}{\sqrt{2}} \arcsin\left(\frac x{\sqrt{2}}\right)$$
| $$\int \frac{1}{\sqrt{2-x^2}}dx=\int\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}}dx=\int\frac{\frac{1}{\sqrt{2}}}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}}dx=\sin^{-1}\frac{x}{\sqrt{2}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Comparison Test for Series I am trying to prove that the series below converges by the comparison test.
\begin{align}
\sum_{n=1}^\infty\frac{2n^3+5^n+5\log(n)}{13-n+8^n}
\end{align}
How do I show that:
\begin{align}
\frac{2n^3+5^n+5\log(n)}{13-n+8^n} \le 16\left(\frac{5}{8}\right)^n
\end{align}
| Simple with asymptotic equivalence of functions:
*
*$2n^3+5^n+5\log n\sim_\infty 5^n$,
*$13-n+8^n\sim_\infty 8^n$,
so that
$$\frac{2n^3+5^n+5\log n}{13-n+8^n}\sim_\infty\frac{5^n}{8^n}=\Bigl(\frac 58\Bigr)^{\!n},\quad\text{a convergent geometric series.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2$
Prove $$\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2$$
Out of boredom, I decided to play with some integrals and Inverse Symbolic Calculator and accidentally found this to my surprise
$$\int_0^\infty\Big(\arctan \frac1x\Big)^2 \mathrm d x = \pi\ln 2 \quad (\text{conjectural}) \,\,\, {\tag{1}} $$
Here is Wolfram Alpha computation which shows (1) to be true to 50 digits. Is (1) true and how to prove it?
I can calculate
$$\int_0^\infty\arctan \frac{1}{x^2}\mathrm d x = \frac{\pi}{\sqrt2}$$
easily by expanding $\arctan$ into Maclaurin series. But how to proceed with $\arctan^2$?
| Let
$$ I(a,b)=\int_0^\infty\left(\arctan \frac ax\right)\left(\arctan \frac bx\right) \mathrm d x. $$
Then
\begin{eqnarray}
\frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_0^\infty\frac{x^2}{(x^2+a^2)(x^2+b^2)}\mathrm d x\\
&=&\frac{1}{a^2-b^2}\int_0^\infty \bigg(\frac{a^2}{x^2+a^2}-\frac{b^2}{x^2+b^2}\bigg)\mathrm d x\\
&=&\frac{1}{a^2-b^2}\frac\pi2(a-b)\\
&=&\frac{\pi}{2}\frac{1}{a+b}
\end{eqnarray}
and hence
$$ I(1,1)=\frac{\pi}{2}\int_0^1\int_0^1\frac{1}{a+b}\mathrm d a\mathrm d b=\frac\pi2\int_0^1(\ln(b+1)-\ln b)\mathrm d b=\pi\ln2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 7,
"answer_id": 5
} |
How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
| $\frac{1}{(x+1)(x+2)^2(x+3)^3} = \frac {A}{x+1} + \frac {B}{x+2} + \frac {C}{(x+2)^2} + \frac {D}{x+3} + \frac {E}{(x+3)^2} + \frac F{(x+3)^3} $
Here is a little trick.
Multiply trough by $(x+1)$
$\frac{(x+1)}{(x+1)(x+2)^2(x+3)^3} = A + \frac {B}{x+2}(x+1) + \frac {C}{(x+2)^2}(x+1) + \frac {D}{x+3}(x+1) + \frac {E}{(x+3)^2}(x+1) + \frac F{(x+3)^3}(x+1) $
And take the limit as x approaches $-1$
$\lim_\limits{x\to -1} \frac{(x+1)}{(x+1)(x+2)^2(x+3)^3} = \frac {1}{(1^2)(2^3)} = \frac 18 = A$
We can do something similar to quickly find $C, F$
$\lim_\limits{x\to -2} \frac{(x+2)^2}{(x+1)(x+2)^2(x+3)^3} = C$
$\lim_\limits{x\to -3} \frac{(x+3)^3}{(x+1)(x+2)^2(x+3)^3} = F$
That leaves B, D, E
Multiplying through by $(x+2)^2$ and simplifying the LHS
$\frac{1}{(x+1)(x+3)^3} = \frac {A}{x+1}(x+2)^2 + B(x+2) + C + \frac {D}{x+3}(x+2)^2 + \frac {E}{(x+3)^2}(x+2)^2 + \frac F{(x+3)^3}(x+2)^2$
If we take the derivative of both sides and take the limits of as x approaches $- 2$
$\lim_\limits{x\to -2}\frac {d}{dx}\frac{1}{(x+1)(x+3)^3} = B\\
\lim_\limits{x\to -2}\frac{- 3(x+1) - (x+3)^3}{(x+1)^2(x+3)^4} = B\\
2 = B$
$\lim_\limits{x\to -3}\frac {d}{dx}\frac{1}{(x+1)(x+2)^2} = E$
And we take a second derivative to find $D$
$\lim_\limits{x\to -3}\frac {d^2}{dx^2}\frac{1}{(x+1)(x+2)^2} = 2D$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this limit doesn't exists (for example we can take $x,y = {1\over n},{1\over n}$ and $x,y = {2\over n},{1 \over n}$
| You may use AM-GM (inequality between arithmetic and geometric mean) and the mean value theorem to get reasonable bounds:
*
*AM-GM: $a+b \geq 2\sqrt{ab}$ with $a =2x^2, b= y^2$ and equality iff $a=b \Leftrightarrow 2x^2 = y^2$
*MVT: For $y \neq 0, y>-1$ you have $\frac{\ln (1+y)}{y} = \frac{1}{1+\eta}$ with $\eta$ between $0$ and $y$
Consider $(x,y)$ with $xy\neq 0$ (otherwise the expression is equal to $0$ anyways). For convenience assume further $|x|,|y| < 1$, since we want to study the behaviour of the expression around $(0,0)$:
\begin{eqnarray*}\left| \frac{x \ln(1+y)}{2x^2+y^2}\right|
& \stackrel{AM-GM}{\leq} & \frac{|x||\ln(1+y)|}{2\sqrt{2x^2y^2}} \\
& = & \frac{1}{2\sqrt{2}}\cdot \left| \frac{\ln(1+y)}{y}\right| \\
& \stackrel{MVT}{=} & \frac{1}{2\sqrt{2}}\cdot\frac{1}{1+\eta} \\
& \leq & \begin{cases}\frac{1}{2\sqrt{2}}\cdot\frac{1}{1+0} & y>\eta> 0 \\ \frac{1}{2\sqrt{2}}\cdot\frac{1}{1-|y|} & -1 < y < 0 \end{cases}\\
& \stackrel{e.g. \color{blue}{|y|<\frac{1}{2}}}{\leq} & \begin{cases}\frac{1}{2\sqrt{2}} & y> 0 \\ \frac{1}{\sqrt{2}} & \color{blue}{-\frac{1}{2}<y<0} \end{cases}\\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
sum of series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$
Sum of $n$ terms of the series
$$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots \cdots$$
Plan
$$\frac{x}{1-x^2}=\frac{1}{2}\frac{2x}{1-x^2}=\frac{1}{2}\bigg[\frac{1}{1-x}-\frac{1}{1+x}\bigg]$$
$$\frac{x^2}{1-x^4}=\frac{1}{2}\frac{2x^2}{1-x^4}=\frac{1}{2}\bigg[\frac{1}{1-x^2}-\frac{1}{1+x^2}\bigg]$$
Did not get any pattern to convert into Telescopic sum
How do i solve it Help me please
| Statement:
$$S_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots+\frac{x^{2^{n-1}}}{1-x^{2^{n}}}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}\tag{1}$$
How did I invent formula (1)? By adding a few items by hand and looking into the pattern that started to emerge. Let us prove it by induction:
For $n=1$:
$$S_1=\frac{x-x^2}{(1-x)(1-x^2)}=\frac{x}{1-x^2}$$
So the statement is true for $n=1$.
Now the induction step:
$$S_{n+1}=S_n+\frac{x^{2^{n}}}{1-x^{2^{n+1}}}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}+\frac{x^{2^{n}}}{1-x^{2^{n+1}}}$$
$$S_{n+1}=\frac{x-x^{2^n}}{(1-x)(1-x^{2^n})}+\frac{x^{2^{n}}}{(1-x^{2^{n}})(1+x^{2^{n}})}$$
$$S_{n+1}=\frac{(x-x^{2^n})(1+x^{2^n})+x^{2^{n}}(1-x)}{(1-x)(1-x^{2^n})(1+x^{2^n})}$$
$$S_{n+1}=\frac{x+x^{2^n+1}-x^{2^n}-x^{2^{n+1}}+x^{2^n}-x^{2^n+1}}{(1-x)(1-x^{2^n})(1+x^{2^n})}$$
$$S_{n+1}=\frac{x-x^{2^{n+1}}}{(1-x)(1-x^{2^{n+1}})}$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum\limits_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} = 1$. Prove
$$\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} = 1.$$
I couldn't find this problem solved online and I haven't reviewed series in a long time. I thought maybe squeeze theorem could help? A related question asks to prove
$$ \sum_{x=0}^\infty \frac{x}{(x+ 1)(x+2)} = +\infty.$$
| $$S_\infty =\sum_{x=0}^\infty \frac{1}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{(x+2)-(x+1)}{(x+ 1)(x+2)} $$
$$=\sum_{x=0}^\infty \frac{1}{(x+1)}- \frac{1}{(x+2)} $$
Which if you will expand and cancel
$$ S_\infty=1- \frac{1}{2} +\frac{1}{2}-\frac{1}{3}+\frac{1}{3}.... \infty$$
$$=1$$
a few terms , you will see that except 1 all get cancelled and you are left with 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Finding $\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\cot^{-1}\bigg(\frac{n^2(r^2+r+1)+2n+1}{n^2+n}\bigg)$
Finding $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\cot^{-1}\bigg(\frac{n^2(r^2+r+1)+2n+1}{n^2+n}\bigg)$$
Plan
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\tan^{-1}\bigg(\frac{n^2+n}{1+n^2r^2+nr+n^2+2n}\bigg)$$
I am trying to convert it into $$\tan^{-1}\bigg(\frac{x-y}{1+xy}\bigg)=\tan^{-1}(x)-\tan^{-1}(y)$$
But not find any way
How do i solve it Help me please
| Doing some basic manipulation we get the following terms $cot^{-1}(\frac{n(r^2+r)}{n+1}+\frac{(n+1)}{n})$ . Now I let $\frac{n}{n+1}=a$ so we have $cot^{-1}(a(r^2+r)+\frac{1}{a})=\arctan(\frac{a}{a^2(r^2+r)+1})$ we have $y=ar+a,x=ar$ thus it becomes $\arctan(ar+a)-\arctan(ar)$ thus sum is $\sum_{r=1}^{\infty}\arctan(\frac{n}{n+1}(r+1))-\arctan(\frac{nr}{n+1})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\frac {1+\frac {2^2}{2!} +\frac {2^4}{3!}+\frac {2^6}{4!} +\dots}{1+\frac {1}{2!}+\frac {2}{3!}+\frac {2^2}{4!}+\dots}$ Evaluate the given series
$$\dfrac {1+\dfrac {2^2}{2!} +\dfrac {2^4}{3!}+\dfrac {2^6}{4!} +....}{1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....}$$
If we factor out $\dfrac {1}{2^2}$ from the numerator we are left with
$$\dfrac {2^2}{1!}+\dfrac {(2^2)^{2}}{2!} + \dfrac {(2^2)^{3}}{3!}+.....$$ which is equal to $(e^2)^{2} -1$. But I couldn't manipulate the denominator.
| $${1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....} = \frac{1}{2^2}\bigg\{4+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\}$$
$$=\frac{1}{2^2}\bigg\{1+1+2+\dfrac {2^2}{2!}+\dfrac {2^3}{3!}+\dfrac {2^4}{4!}+....\bigg\} = \frac{1}{4}\bigg(1+e^2\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Systems of equations involving linear and quadratic terms Can we solve for $y$ in this system using algebra?
$$\left\{
\begin{aligned}
x^2 - yz &= 3 \\
y^2 - xz &= 4 \\
z^2 - xy &= 5
\end{aligned}
\right.$$
I’ve tried to evaluate it using elimination and it just gives another equation with unknowns.
First I've tried to multiply the first equation by $y$, second by $z$ and third by $x$. I get $x^2 - y^2z = 3y, y^2z - xz^2 = 4z,$ and $z^2x - x^2y=5x$. Simplifying I get $5x + 4z + 3y = 0$. I've tried it again by multiplying the 1st and 3rd equation by $z, x$ and $y$ respectively. I get $5y + 4x + 3z = 0$. I don't know where to get my third equation.
| Multiply each equation by $2$ and add all of them to get
$$(x-y)^2+(y-z)^2+(z-x)^2=24.$$
Let $u=x-y, v=y-z, w=z-x$. Then you have
\begin{align*}
u+v+w&=0\\
u^2+v^2+w^2&=24
\end{align*}
Now in your original system of equations, subtract equ 2 from equ 1, equ 3 from equ 2 and equ 1 from equ 3 to get
\begin{align*}
(x-y)(x+y+z) &=-1\\
(y-z)(x+y+z) &=-1\\
(z-x)(x+y+z) &=2
\end{align*}
Clearly $x+y+z \neq 0$, so from here we can conclude that
\begin{align*}
x-y&=y-z \implies u=v\\
z-x&=-2(y-z) \implies w=-2v\\
\end{align*}
So now plug this in the $u,v,w$ equation s above to get
$$(v)^2+(v)^2+(-2v)^2=24 \implies v =\pm 2.$$
Hopefully now you can solve the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\lim\limits_{n\rightarrow \infty} \frac{1}{n^2}\sum\limits_{k=1}^{n}(n \bmod k)=1-\frac{\pi^2}{12}$? I learnt that
$$\lim_{n\rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^{n}(n \bmod k)=1-\frac{\pi^2}{12}$$
where $ (n{\bmod {k}})$ is the remainder upon division of $n$ by $k$.
However, I am not sure how to arrive at this equation.
I understand that
$$\infty \bmod 1 = 0$$
But I don't see how infinity can mod anything else. For example,
$$\infty \bmod 2 = ~?$$
$$\infty \bmod 3 = ~?$$
I am thinking about either converting the limit into an integral, but I haven't figured out how. Any hints on how to evaluate the limit?
I also found a more general form of the limit (not sure if this will help with the original question):
For $a>0,~b\geq1$,
$$\lim_{n\rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^{an}(an \bmod bk)=a^2\left(1-\frac{\pi^2}{12b}\right)$$
| Using division with remainder, if $n=q_kk+r_k$ s.t. $0\leq r_k < k$, then $r_k=n \pmod{k}$. Also $q_k=\left \lfloor \frac{n}{k} \right \rfloor$ and
$$r_k=n-k\left \lfloor \frac{n}{k} \right \rfloor$$
Then
$$\frac{1}{n^2}\sum_{k=1}^{n}(n \bmod k)=
\frac{1}{n^2}\sum_{k=1}^{n}r_k=
\frac{1}{n^2}\sum_{k=1}^{n}\left(n-k\left \lfloor \frac{n}{k} \right \rfloor\right)=\\
1-\frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\left \lfloor \frac{n}{k} \right \rfloor \to
\color{red}{1-\int\limits_{0}^{1}x\left \lfloor \frac{1}{x} \right \rfloor dx}, n\to\infty \tag{1}$$
Now
$$\int\limits_{0}^{1}x\left \lfloor \frac{1}{x} \right \rfloor dx=
\int\limits_{0}^{\frac{1}{2}}x\left \lfloor \frac{1}{x} \right \rfloor dx + \int\limits_{\frac{1}{2}}^{1} xdx=
\int\limits_{0}^{\frac{1}{3}}x\left \lfloor \frac{1}{x} \right \rfloor dx +
\int\limits_{\frac{1}{3}}^{\frac{1}{2}} 2xdx +
\int\limits_{\frac{1}{2}}^{1} xdx=\\
\int\limits_{0}^{\frac{1}{4}}x\left \lfloor \frac{1}{x} \right \rfloor dx +
\int\limits_{\frac{1}{4}}^{\frac{1}{3}} 3xdx +
\int\limits_{\frac{1}{3}}^{\frac{1}{2}} 2xdx +
\int\limits_{\frac{1}{2}}^{1} xdx=\\
\sum_{n=1}n \int\limits_{\frac{1}{n+1}}^{\frac{1}{n}} xdx=
\sum_{n=1}\frac{n}{2} \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=
\frac{1}{2}\sum_{n=1}\frac{2n+1}{n(n+1)^2}=\\
\frac{1}{2}\sum_{n=1}\frac{n+n+1}{n(n+1)^2}=
\frac{1}{2}\sum_{n=1}\left(\frac{1}{(n+1)^2}+\frac{1}{n(n+1)}\right)=\\
\frac{1}{2}\sum_{n=1}\left(\frac{1}{(n+1)^2}+\frac{1}{n}-\frac{1}{n+1}\right)=\\
\color{red}{\frac{1}{2}\sum_{n=1}\frac{1}{(n+1)^2}}+\color{blue}{\frac{1}{2}\sum_{n=1}\left(\frac{1}{n}-\frac{1}{n+1}\right)}=\\
\color{red}{\frac{1}{2}\sum_{n=1}\frac{1}{n^2}-\frac{1}{2}}+\color{blue}{\frac{1}{2}}=\frac{1}{2} \cdot \frac{\pi^2}{6}$$
which indeed is $\frac{\pi^2}{12}$, because of this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The sum of an infinite series containing a finite series in each denominator
Evaluate
$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\cdots+\frac{1}{\left[\frac{k(k+1)}{2}\right]^2}+\cdots$$
to $\infty$, where $k$ is the $k$th term of the series.
Using Microsoft Excel, I found that the sum of the first $100$ terms $=1.15947\dots$.
Does the convergence value have an exact form? If yes, what is it?
| Using fractional decomposition we get:
$$
\frac 1{\left(\frac{k(k+1)}{2}\right)^2} = \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k
$$
So, your sum simplifies to one telescoping sum, and the sum
$$\sum_{k=1}^\infty \frac 1 {k^2} =\frac{\pi^2} 6$$
So, all together we get:
$$
\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\dots+\frac{1}{[\frac{k(k+1)}{2}]^2}+\dots\\ =
\sum_{k=1}^\infty \frac 1{\left(\frac{k(k+1)}{2}\right)^2}
\\=
\sum_{k=1}^\infty
\frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k
\\ =
4\left( \sum_{k=1}^\infty \frac 1{(k + 1)^2}
\right) +
4\left( \sum_{k=1}^\infty \frac 1 {k^2}
\right)+
8\left( \sum_{k=1}^\infty \frac 1{(k + 1)}- \frac 1 k
\right)
$$
In the first sum, you do an index shift to the left and add the first summand (as a zero).
$$
=4\left(-1+ \sum_{k=1}^\infty \frac 1{k^2}
\right) +
4\left( \sum_{k=1}^\infty \frac 1 {k^2}
\right)+
8\left( \sum_{k=1}^\infty \frac 1{(k + 1)}- \frac 1 k
\right)
\\=
-4+4\cdot \frac{\pi^2} 6 + 4\cdot \frac{\pi^2} 6 -8
=
4·(\pi^2 - 9)/3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do we solve $(w^2-w^{-2})(2w^2+2w^{-2} -1)-\sqrt{3}i(w^2+w^{-2})=0$? How do we solve the following equation?
$$(w^2-w^{-2})(2w^2+2w^{-2} -1)-\sqrt{3}i(w^2+w^{-2})=0$$
where $w = e^{ix}$.
UPDATE:
I converted the trigonometric equation $2\sin(4x) -\sin(2x) - \sqrt{3}\cos(2x) = 0$ to euler.
| We have
$$
\sin(4x)=\sin\left(2x+\frac{\pi}{3}\right)
$$
So
$$
4x = n\pi + (-1)^n \left(2x+\frac{\pi}{3}\right)
$$
and hence
$$
x = \frac{n\pi + (-1)^n \frac{\pi}{3}}{2(2-(-1)^n)}
$$
If you don't want to use Euler, note that the given equation is
$$
2 w^8 - (1+i \sqrt{3}) w^6 - (i \sqrt{3}-1) w^2 - 2 = 0
$$
The LHS has an obvious factor $2w^2-(1+i\sqrt{3})$, and so
$$
[2w^2-(1+i\sqrt{3})] \left( w^6 - \frac{1+i \sqrt{3}}2\right)=0
$$
so you know the eight roots of $w$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the solutions of the next congruence using Chinese Remainder Theorem Find the solutions of the congruence using Chinese Remainder Theorem:
$2x^2 - 3x -2 \equiv 0\mod21$
By now I've done this:
$$
\left\{
\begin{array}{c}
2x^2-3x-2\equiv 0\mod 7\\2x^2-3x-2\equiv 0\mod3
\end{array}
\right.
$$
$$
2x^2-3x-2\equiv 0\mod 7 \\ \Delta = 25 = 5^2 \\x_1,x_2 = (2a)^{-1}*(-b \pm \sqrt\Delta) \\ x_1=4^{-1}*2 = 2*2 = 4 \mod 7 \\ x_2 = 4^{-1}*(-8)= 6*2 = 5 \mod 7
$$
Then I made $\Delta$ for the (mod 3) equation and got $x_3=1 \mod 3, x_4=2 \mod 3$
So I got:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 4 \mod 7 \\ x \equiv 5 \mod 7 \\
$$
And I don't know what to do further. I should apply Chinese Remainder Theorem on it or it would be wrong?
Edit: Thanks for the help! The correct form was:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7 \\ x \equiv 3 \mod 7
$$
And I understand how to solve it.
| Solving the quadratic equation mod. $3$ and mod. $7$, you obtain:
*
*mod. $3$: the equation i s equivalent to $2(x^2-1)=0$, and since $2$ is a unit mod. $3$, this means $x\equiv\pm 1\bmod 3$.
*mod $7$: the discriminant is $\Delta=9+16=25\equiv 4\bmod 7$, so the roots are
$$x\equiv(3\pm 2)\cdot 4^{-1}\equiv (3\pm 2)\cdot 2\equiv 2,3\mod 7.$$
There remains to combine these solutions in 4 pairs $(\alpha,\beta)$. Modulo $21$, you find the corresponding values via the inverse isomorphism of the Chinese remainder theorem. Explicitly, start from a Bézout's relation between $7$ and $3$, say $\;7-2\cdot 3=1$. Then
$$\begin{cases}
x\equiv \alpha\mod 3,\\
x\equiv \beta\mod 7
\end{cases}\iff x\equiv\alpha \cdot 7-2\beta \cdot3\mod 21.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$
$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is
$$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$
My Try
I tried this using the conventional method,$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$
But then I got long answers for coefficients $A$, $B$ & $C$.
Is there a better way to solve this? Can anyone give me a hint to work this?
| Call the purported remainder $\,r(x).$ $\,f(x)-r(x)$ has roots $\,x = a,b,c\,$ so by the Factor Theorem $\,f(x)-r(x) = (x\!-\!a)(x\!-\!b)(x\!-\!c) q(x),\,$ so $\deg r < 3\,$ $\Rightarrow\ r(x) =$ remainder by its uniqueness.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximize $(x-1)2y$ subject to $x^3+y^2=3$ I have to do with lagrange multiplier
f = $(x-1)2y + \lambda (x^3+y^2-3) = 0 $
I get
$f_x= y + \lambda x = 0 $
$f_y = y\lambda + x - 1 = 0 $
$x^3+y^2=3$
and
how to solve this
| Correcting your error in computing $f_x$ the problem amounts to solution of the equation system:
$$
\begin{array}{rcl}
x^3+y^2&=3\\
3 \lambda x^2+2 y&=0 \\
2x+2 \lambda y&=2 \\
\end{array}
$$
Multiplying the second equation by $\lambda$ and subtracting the third one we obtain:
$$
\lambda^2=\frac{2(x-1)}{3x^2}.
$$
Substituting the obtained value into the second equation one obtains:
$$4y^2=9\lambda^2x^4\implies y^2=\frac32x^2(x-1),$$
which after substitution into the first equation gives rise to:
$$
5x^3-3x^2-6=0.
$$
The equation has one real
$$x_r=\frac{1+\omega+\omega^{-1}}5, \text{ with }
\omega=\left(76+5\sqrt{231}\right)^{1/3}
$$
and two complex roots. From the context of your question the complex roots can be disregarded. Since the real root is larger than $1$ the pair
$$(x,y)=\left(x_r,x_r\sqrt{\frac{3(x_r-1)}2}\right)\approx(1.304821,0.882306)$$
is the only candidate for the solution (you should however check if the stationary point is indeed the global maximum).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| $\overbrace{ x^4\!-\!x^3\!+\!2x^2\!-\!x\!+\!2}^{\large \bbox[5px,border:1px solid #0a0]{\!\!(\color{#c00}{x^2})^2-x(\color{#c00}{x^2})+2\color{#c00}{x^2}\!}\color{#0a0}{-x+2}\ \ }\!\!\! = a(x^2\!+\!2)^2\! + (bx\!+\!c)(x^2\!+\!2)(x\!-\!1) + (dx\!+\!e)(x\!-\!1)\,$ via clear denoms
$x=1 \,\Rightarrow\, 3=9a\,\Rightarrow\,\bbox[5px,border:1px solid #c00]{a=1/3}\ \ $ Compare lead coef's $\,\Rightarrow\, 1 = a\!+\!b=1/3+b\iff \bbox[5px,border:1px solid #c00]{b= 2/3}$
$\!\!\!\!\left.\begin{align}\bmod\ x^2\!+2\ \\ {\rm so}\,\ \color{#c00}{x^2\equiv -2}\ &\end{align}\!\!\right\}\! $
$\!\begin{align} \bbox[5px,border:1px solid #0a0]{4\!+\!2x\!-\!4\!}\!\color{#0a0}{-\!x\!+\!2}&\equiv (d\color{#c00}x+e)(\color{#c00}x-1)\\ \iff\ x\!+\!2 &\equiv (e\!-\!d)x\!-\!e\!\color{#c00}{-\!\!2}d\end{align} $
$\!\!\!\iff\!\!\!\!\! \begin{align} e-\,d &=1\\ -e\!-\!2d& =2\end{align}$
$\!\!\!\iff\!\!\!\!\!\begin{align}-3d&=3\\ 3e&=0\end{align}$
$\!\!\iff\!\!\bbox[5px,border:1px solid #c00]{\!\!\!\begin{align}&d=-1\\ &e\ =\ 0\end{align}\!\!}$
$x=0 \,\Rightarrow\, 2 = 4a\!-\!2c\!-\!e = 4/3-2c\iff 2c=-2/3\iff \bbox[5px,border:1px solid #c00]{c=-1/3}$
Remark $ $ The modular calculation is the higher degree Heaviside cover-up method described here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Prove that : $\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}$ Prove that :
$$\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}.$$
I know how to use the definition of the Beta and Gamma functions so my problem here is when I divide the integral I find:
$$\int_0^{1}\frac{1}{x^2}dx=\infty!!$$
(divergent)
I tried using $y=\frac{1}{x}$ but didn't get the answer.
I also tried $y=x^{4}$ and got the same problem (a divergent integral).
| Substitute $u = x^4$ to get
$$ \int_{0}^{1} \frac{1-\sqrt{1-x^4}}{x^2\sqrt{1-x^4}} \, \mathrm{d}x
= \frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u. $$
Now we regularize the integral by replacing $u^{5/4}$ by $u^s$ for $s < 1$. Then
\begin{align*}
\frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{s}(1-u)^{1/2}} \, \mathrm{d}u
&= \frac{1}{4}\left(B\left(1-s,\tfrac{1}{2}\right)-\frac{1}{1-s}\right) \\
&= \frac{1}{4}\left(\frac{\Gamma(1-s)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}-s\right)}-\frac{1}{1-s}\right).
\end{align*}
Note that the left-hand side defines an analytic function for $\operatorname{Re}(s) < 2$, while the right-hand side is a meromorphic function on all of $\mathbb{C}$ with the removable singularity at $s = 1$. So both sides give rise to analytic functions on $\operatorname{Re}(s) < 2$, and so, they are equal on this region by the principle of analytic continuation. Therefore we can simply plug $s = 5/4$ to get
\begin{align*}
\frac{1}{4} \int_{0}^{1} \frac{1-(1-u)^{1/2}}{u^{5/4}(1-u)^{1/2}} \, \mathrm{d}u
&= 1 + \frac{\Gamma\left(-\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{4\Gamma\left(\frac{1}{4}\right)} \\
&= 1 - \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{4}\right)},
\end{align*}
which simplifies to the desired expression by Euler's reflection formula $\Gamma(z)\Gamma(1-z) = \pi\csc(\pi z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| Hint:
It's $\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{y^9-8}=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y^3-2)(y^6+2y^3+4)}$
$$=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y-\sqrt[3]2)(y^2+\sqrt[3]2y+\sqrt[3]2^{2})(y^6+2y^3+4)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Integral $\int_0^{1}\frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx$
Find: $$I=\int_0^{1}\frac{(1+x^2)\ln(1+x^4)}{1+x^4}\,dx$$
I need to use the definition of do Trigamma function but I don't see how.
My try was as following :
Let $y=x^4$ then $dy=4x^{3}\,dx$ so $\;dx=\frac{dy}{4y^{\frac{3}{4}}}$.
$$I=\int_0^{1}\frac{(1+x^{\frac{3}{4}})\ln(1+x)}{x^{\frac{3}{4}}+x^{\frac{7}{4}}}dx$$
Use : $\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n}$
We obtain
$$J=\int_0^{1}\frac{x^n+x^{(3+4n)/4}}{x^{3/4}+x^{7/4}}dx$$
Then I use definition of digamma function
$$\int_0^{1}\frac{x^{a-1}\ln x}{1+x}\,dx=\beta'(a)$$
Where: $$2\beta(a)=\psi\left(\frac{1+a}{2}\right)-\psi\left(\frac{a}{2}\right)$$
But I find sum related with digamma function.
So I don't have other ideas to approach it!
| $$\small \boxed{\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt 2}\ln 2-\frac1{32}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)}$$
To see that write the integral as:
$$I=\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx\overset{x\to \frac{1}{x}}=\int_1^\infty \frac{(1+x^2) (\ln(1+x^4)-\ln(x^4))}{1+x^4}dx$$
$$\Rightarrow 2I=\int_0^\infty \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx-4\int_1^\infty \frac{(1+x^2)\ln x}{1+x^4}dx$$
For the first integral consider:
$$I(a)=\int_0^\infty \frac{(1+x^2)\ln(a+x^4)}{1+x^4}dx\Rightarrow I'(a)=\int_0^\infty \frac{1+x^2}{(a+x^4)(1+x^4)}dx$$
$$=\frac{1}{a-1}\int_0^\infty \frac{1+x^2}{1+x^4}dx-\frac{1}{a-1}\int_0^\infty \frac{1+x^2}{a+x^4}dx$$
$$\int_0^\infty \frac{1+x^2}{1+x^4}dx=\int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}=\frac{1}{\sqrt 2}\arctan\left(\frac{x^2-1}{\sqrt 2 x}\right)\bigg|_0^\infty =\frac{\pi}{\sqrt 2}$$
$$\int_0^\infty \frac{1+x^2}{a+x^4}dx\overset{x=\sqrt[4]at}=\sqrt[4]a\int_0^\infty \frac{1+\sqrt at^2}{a(1+t^4)}dt=a^{-3/4}\int_0^\infty \frac{dt}{1+t^4}+a^{-1/4}\int_0^\infty \frac{t^2dt}{1+t^4}$$
$$\int_0^\infty \frac{1}{1+t^4}dt\overset{t\to \frac{1}{t}}=\int_0^\infty \frac{t^2}{1+t^4}dt=\frac12 \int_0^\infty \frac{1+t^2}{1+t^4}dt=\frac{\pi}{2\sqrt 2}$$
$$\Rightarrow I'(a)= \frac{\pi}{2\sqrt{2}}\left(\frac{2-a^{-1/4}-a^{-3/4}}{a-1}\right) $$
We are looking here for $I(1)$, but $I(0)=0$, (put $x=\frac{1}{x}$ to see that), so:
$$\Rightarrow I(1)= \frac{\pi}{2\sqrt{2}}\int_0^1 \left(\frac{1-a^{-1/4}+1-a^{-3/4}}{a-1}\right)da=\frac{3\pi}{\sqrt 2}\ln 2 $$
Above follows by splitting into two parts and using this.
For the second integral we will use the trigamma function as you mentioned in your question.
$$\int_1^\infty \frac{(1+x^2)\ln x}{1+x^4}dx\overset{x\to \frac{1}{x}}=-\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx$$
$$=\frac1{64} \left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)$$
One can use trigamma's reflection formula in order to simplify two more terms, but I think it looks nice this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$. Let $x$ be an odd natural number. Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$.
For the first part I would assume
$$x \equiv 1 \mod 2$$
$$x^2 \equiv 1^2 \mod 2^2$$
$$x^2 +2 \equiv 1^2+2 \mod 2^2$$
$$x^2 +2 \equiv 3 \mod 4$$
Can someone briefly tell me if this is correct and if it is a rule that squaring the number on the left means you square both numbers on the right?
Also I don't know how to start with the second part.
Note: This question is to do with rings in general.
| Assuming $x$ is odd, $x\equiv1\pmod2$ so $2|x-1$ so $2|(x-1)+2=x+1,$
so $4|(x-1)(x+1)=x^2-1=x^2+2-3, $ so $ x^2+2\equiv3\pmod4$.
Let $p$ be a factor of $x^2+2$. $p$ must be odd because $x$ and therefore $x^2+2$ is.
If all such factors were $\equiv1\pmod4$ then their product would be $\equiv1\pmod4$, a contradiction.
So $x^2+2$ has a prime factor $\equiv3\pmod4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find every $a,b \in \mathbb{C}$ such that dividing $P=3X^{120}+\sqrt{2}X^{19}+aX^{11}+\sqrt{2}X^9+3X^2+b$ by $Q=X^2+1$ has remainder $R=5X+2$ We know that, for some $H\in\mathbb{C}[x]$,
$$P = QH+R$$
Hence,
$$P-R=QH$$
Given that $Q$'s roots are $\{i,-i\}$, the Remainder Theorem for polynomial rings guarantees that evaluating $(P-R)(\pm i)=0$. So we have the following system of equations
$$\begin{cases} 3i^{120}+\sqrt{2}i^{19}+ai^{11}+\sqrt{2}i^9+3i^2+5i+b+2 =0 \\ 3(-i)^{120}+\sqrt{2}(-i)^{19}+a(-i)^{11}+\sqrt{2}(-i)^9+3(-i)^2+5(-i)+b+2 =0 \end{cases}$$
which entails that
$$a=-5-\sqrt{2},b=2$$
Is this more or less correct and properly justified?
| Right method, but there are sign errors:
$$\begin{cases} 3i^{120}+\sqrt{2}i^{19}+ai^{11}+\sqrt{2}i^9+3i^2\color{red}-5i+b\color{red}-2 =0 \\ 3(-i)^{120}+\sqrt{2}(-i)^{19}+a(-i)^{11}+\sqrt{2}(-i)^9+3(-i)^2\color{red}-5(-i)+b\color{red}-2 =0 \end{cases} \Rightarrow \\
\begin{cases} \require{cancel}\cancel{3}-\cancel{\sqrt{2}i}-ai+\cancel{\sqrt{2}i}-\cancel{3}\color{red}-5i+b\color{red}-2 =0 \\ \cancel3+\cancel{\sqrt{2}i}+ai-\cancel{\sqrt{2}i}-\cancel{3}+5i+b\color{red}-2 =0 \end{cases} \Rightarrow a=-5,b=2.$$
Alternatively, note:
$$3x^{120}+3x^2\equiv 0\pmod{x^2+1}\\
\sqrt{2}x^{19}+\sqrt{2}x^9 \equiv 0\pmod{x^2+1}\\
ax^{11}+b\equiv 5x+2\pmod{x^2+1} \Rightarrow \\
ax^{11}+b=g(x)(x^2+1)+5x+2 \Rightarrow \\
ai^{11}+b=g(i)\cdot 0+5i+2\Rightarrow \\
-ai+b=5i+2 \Rightarrow \\
a=-5,b=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is taking square root of both parts of equation in this way is an equivalent transformation of the equation? Solve equation
$$(2x+7)^2=(2x-1)^2$$
$t=2x-1 $, so equation becomes $$ (t+8)^2 = t^2 $$
Now let's make a "prohibited" - take a square root from both parts (minding that $\sqrt{x^2} = \lvert x\rvert$), so we get a union of solutions of two equations
$$
t+8 = t \\
t+8 = - t
$$
$-(t+8) = t, -(t+8) = -t $ are same as the 2 equations above, so no need to solve them separately.
The first equation is a contradiction ( 0 = 8) so it has no solutions.
The second equation give $t = -4 \Rightarrow 2x-1 = -4 \Rightarrow x = -3/2$ and this is the only solution of the original quadratic equation, and it need not to be checked (substituted into original equation) because all transformations all along the way were equivalent.
May I go ahead and take square roots in that manner at my exams, is it OK?
| Yes, when the square root is taken from both sides, the equation results in two equivalent equations.
In fact, that is how the quadratic formula is obtained:
$$ax^2+bx+c=0 \iff \\
x^2+\frac bax+\frac ca=0\iff \\
\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac ca=0 \iff \\
\left(x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2} \iff \\
x+\frac b{2a}= \frac{\pm\sqrt{b^2-4ac}}{2a} \iff \\
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
| We had one root $x=-\frac{1}{\sqrt{2}}$ as already said in answers and comments. After long division remains $$2 x^4+4 \sqrt{2} x^3+16 x^2+12 \sqrt{2} x+8=0$$
Make $x=y-\frac{1}{\sqrt{2}}$ to get
$$2 y^4+10 y^2+\frac{5}{2}=0$$ which is a quadratic in $y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the area between functions $x^3=2y^2;x=0;y=-2$
Take the element of area parallel to the y axis:
$$x^3=2y^2;x=0;y=-2$$
First, I isolated in terms of $y$,
$$y= \pm \sqrt{x^3\over2}
\\ = \pm{\sqrt2\over 2}x^{3 \over 2}$$
Since bounded by $y=-2$, consider the negative portion:
$$= -{\sqrt2\over 2}x^{3 \over 2}$$
and let
$$f(x)= -{\sqrt2\over 2}x^{3 \over 2}$$
$$g(x) = -2$$
since $f(x)$ and $g(x)$ intersect at $x=2$ and bound by $x=0$
So,
$$ \int_0^2 [-g(x) -f(x)]dx
\\ = \int_0^2 (2 + {\sqrt2\over 2}x^{3 \over 2} ) dx
\\ = 2x + {\sqrt2\over 5}x^{5 \over 2}\Bigg]^2_0
\\ = 4 + {\sqrt2\over 5}2^{5 \over 2}$$
However the answer i supposed to be $12\over 5$
And I do not understand what the question means by "The element of area parallel to y axis"
| The area should be
$$\int^2_0[f(x)-g(x)]dx$$ as $f(x)\ge g(x)$ in the given interval.
So,$$\sqrt2\cdot2^{5/2} = 2^{1/2}2^{5/2} = 2^{3} = 8 \implies 4- \frac{8}{5} = \frac{12}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)dx$
Calculate$$\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)\mathrm dx$$
My try:
$$\sin(x+\pi/3)=\sin x\cos(\pi/3)+\sin(x+\pi/3)\cos x=\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x$$
$$\int_{0}^{\pi/3}\ln^4\left(\frac{2\sin x}{\sin x+\sqrt{3}\cos x}\right)\mathrm dx$$
$$\int_{0}^{\pi/3}\ln^4\left(\frac{2}{1+\sqrt{3}\cot x}\right)\mathrm dx$$
$u=\frac{2}{1+\sqrt{3}\cot x}$
$u^{'}=\frac{2\sqrt{3}}{\sin^2 x(1+\sqrt{3}\cot x)^2}$
$$\frac{1}{2\sqrt{3}}\int_{0}^{1}\sin^2 x(1+\sqrt{3}\cot x)^2\ln^4(u)\mathrm du$$
$$\frac{2}{\sqrt{3}}\int_{0}^{1}\sin^2 x u^{-2}\ln^4(u)\mathrm du$$
$\cot^2 x=\frac{(2-u)^2}{3u^2}$
using $1+\cot^2 x=\frac{1}{\sin^2 x}$
$$\frac{2}{\sqrt{3}}\int_{0}^{1}\frac{3u^{2}}{3u^2+(2-u)^2} u^{-2}\ln^4(u)\mathrm du$$
$$\frac{6}{\sqrt{3}}\int_{0}^{1}\frac{1}{3u^2+(2-u)^2}\ln^4(u)\mathrm du$$
$$\frac{3}{\sqrt{3}}\int_{0}^{1}\frac{1}{u^2-u+1}\ln^4(u)\mathrm du$$
I am not sure what to do next...
| $$I=\int_{0}^{\pi/3}\ln^4\left(\frac{\sin(x+\pi/3)}{\sin x}\right)dx=\int_0^\frac{\pi}{3}\ln^4 \left(\frac12 +\frac{\sqrt 3}{2}\cot x\right)dx=\frac{\sqrt 3}{2}\int_0^1 \frac{\ln^4 x}{x^2-x+1}dx$$
Above follows by the substitution $\frac12+\frac{\sqrt{3}}{2}\cot x\to x$. We also have for $t\in(0,\pi), x\in (-1,1)$:
$$\frac{\sin t}{x^2-2x\cos t+1}=\sum_{n=1}^\infty x^{n-1}\sin(nt)$$
$$\Rightarrow I=\sum_{n=1}^\infty \sin\left(\frac{n\pi}{3} \right)\int_0^1 x^{n-1} \ln^4 x dx=24\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3} \right)}{n^5}$$
Similarly to here, we have for $x\in(0,2\pi)$:
$$\frac{\pi-x}{2}=\sum_{n=1}^\infty\frac{\sin(nx)}{n}$$
Integrating the above with respect to $x$ gives:
$$\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}=\frac{(\pi-x)^2}{4}+C_1$$
Setting $x=\pi$ gives $C_1=-\frac{\pi^2}{12}$ and
integrating again produces:
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(\pi-x)^3}{12}-\frac{\pi^2}{12}x+C_2$$
Putting $x=\pi $ yields $C_2= \frac{\pi^3}{12}$. One more time:
$$\sum_{n=1}^\infty \frac{\cos(nx)}{n^4}=-\frac{(\pi-x)^4}{48}+\frac{\pi^2}{24}x^2-\frac{\pi^3}{12}x+C_3$$
For $x=\pi \Rightarrow C_3=\frac{23\pi^4}{720}$. And finally, one more similar step gives for $x \in(0,2\pi)$:
$$S(x)=\sum_{n=1}^\infty \frac{\sin(nx)}{n^5}=\frac{(\pi-x)^5}{240}+\frac{\pi^2}{72}x^3-\frac{\pi^3}{24}x^2+\frac{23\pi^4}{720}x-\frac{\pi^5}{240}$$
And well, the value of the integral is just $24S\left(\frac{\pi}{3}\right)$. Doing the algebra yields:
$$\boxed{\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)dx=\frac{17\pi^5}{243}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Polynomial expansion (plus/minus trick in statistics) Suppose the random variables $X_1, ..., X_n$ are independent and identically distributed. Let $\mu_x$ denote the expected value of $X$, and let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$. I often see the following plus/minus trick used in statistics, i.e.
\begin{align*} \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 &= \frac{1}{n}\sum_{i=1}^n(X_i - \mu_X + \mu_X - \bar{X})^2 \\
&= \frac{1}{n}\sum_{i=1}^n(X_i - \mu_x)^2 - (\bar{X} - \mu_x)^2\end{align*}
Can someone formulate this into an identity for me? That is, something like
$(a - b)^2 = (a - c + c - b)^2 = (a - c)^2 + (c - b)^2 + \text{some cross term}?$
What exactly is that cross term?
On a similar token, I've also seen
\begin{align*} \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})(Y_i - \bar{Y}) &= \frac{1}{n}\sum_{i=1}^n(X_i - \mu_X )(Y_i - \mu_Y) - (\bar{X} - \mu_x)(\bar{Y} - \mu_Y) \end{align*}
What polynomial expansion identity is at play here?
| The cross term is as you would expect from $(x+y)^2=x^2+y^2+2xy$:
\begin{align}
(a-b)^2 &= ((a-c)+(c-b))^2 \\
&=(a-c)^2 + (c-b)^2 + 2(a-c)(c-b)
\end{align}
So for any constant $\alpha \in \mathbb{R}$ you get
\begin{align}
\left(X_i-\overline{X}\right)^2 &= \left(X_i-\alpha + \alpha - \overline{X}\right)^2 \\
&= (X_i-\alpha)^2 + \left(\alpha - \overline{X}\right)^2 + 2\left(X_i-\alpha\right)\left(\alpha-\overline{X}\right)
\end{align}
But since $(\alpha - \overline{X})$ does not depend on $i$, when you sum the last term over $i \in \{1, ..., n\}$ and then divide by $n$, you get
\begin{align}
\frac{1}{n}\sum_{i=1}^n 2(X_i-\alpha)(\alpha-\overline{X})&= 2(\alpha-\overline{X})\frac{1}{n}\sum_{i=1}^n(X_i-\alpha) \\
&= 2\left(\alpha-\overline{X}\right)\left(\overline{X}-\alpha\right)
\end{align}
Thus, indeed we get:
$$ \boxed{\frac{1}{n}\sum_{i=1}^n \left(X_i-\overline{X}\right)^2 = \frac{1}{n}\sum_{i=1}^n(X_i-\alpha)^2 - \left(\overline{X}-\alpha\right)^2 \quad \forall \alpha \in \mathbb{R}}$$
Similarly it can be shown
\begin{align}
&\frac{1}{n}\sum_{i=1}^n(X_i-\overline{X})(Y_i-\overline{Y}) \\
&= \frac{1}{n}\sum_{i=1}^n(X_i-\alpha)(Y_i-\beta) - (\overline{X}-\alpha)(\overline{Y}-\beta) \quad \forall \alpha, \beta \in \mathbb{R}
\end{align}
This is similar to the following identities:
\begin{align}
Var(X) &= Var(X-\alpha) \quad \forall \alpha \in \mathbb{R}\\
Cov(X,Y) &= Cov(X-\alpha, Y-\beta) \quad \forall \alpha, \beta \in \mathbb{R}
\end{align}
where we recall
\begin{align}
Var(X) &= E[(X-E[X])^2] = E[X^2] - E[X]^2\\
Cov(X,Y)&= E[(X-E[X])(Y-E[Y])] = E[XY]-E[X]E[Y]
\end{align}
For example, for all $\alpha \in \mathbb{R}$ we get
\begin{align}
E[(X-E[X])^2] &= Var(X) \\
&= Var(X-\alpha)\\
&= E[(X-\alpha)^2] - (E[X]-\alpha)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$? I encountered the following hard problem in a math olympiad book:
Evaluate $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}.$$
I tried to evaluate $\sum^{k}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$ where $k=1$ to $10$ and got $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{5}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{10}{5}$ respectively. How can I prove that the pattern continues?
| Hint:
$$n(n+1)(n+2)(n+3)$$
$$=n(n+3)\cdot (n+1)(n+2)$$
$$=(n^2+3n)(n^2+3n+2)$$
$$=(n^2+3n-1)(n^2+3n+1)$$
$$=n^4+6n^3+9n^2-1$$
Let $n^4+6n^3+9n^2-1=f(n+1)-f(n)$ where $f(m)=a_0+a_1m+a_2m^2+\cdots$
Clearly, $a_r=0\forall r\ge6$
$n^4+6n^3+9n^2-1$
$=a_1+a_2(2m+1)+a_3(3m^2+3m+1)+a_4(4m^3+6m^2+4m+1)+a_5(5m^4+10m^3+10m^2+5m+1)$
Compare the coefficients of $n^4,n^3,n^2,n,n^0$
$5a_5=1$
$4a_4+10a_5=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$
Find
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$$
for $x≥0$.
At first, I use a partial fraction
$$S=\displaystyle\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2(x+1)(n+1)}-\frac{(-1)^{n}}{2(x+1)(n+2x+3)}\right)
=\frac{1}{2(x+1)}(I-J),$$
where
$$I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}
=\ln 2$$
and
$$J=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+3+2x}$$
I think to use:
$$\ln (1+y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}y^{n+1}}{n+1}$$
Is my work correct? How to complete this work ?
| $$S(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2x+3)}=\sum_{n=0}^{\infty}
\frac{(-1)^n}{n+1} \int _{0}^{1} t^{n+2x+2} dt=\int_{0}^{1} t^{2x+1} \ln(1+t) dt ~~~(1).$$ $$\Rightarrow S(x)=\left .\ln(1+t) \frac{t^{2x+2}}{2x+2}\right |_{0}^{1}-\frac{1}{2x+2}\int_{0}^{1} \frac{t^{2x+2}dt}{1+t}=\frac{\ln 2}{2x+2} - \frac{\beta(2x+3)}
{2x+2}~~~~(2).$$
Here we have used 3.222 of Garadshteyn and Ryzhik where $$\int_{0}^{1}\frac{x^{{\mu}-1}}{1+x} dx= \beta(\mu).$$
Next $$\beta(x)=\frac{1}{2}[\psi(\frac{x+1}{2})-\psi(\frac{x}{2})]~~~~(3)$$
and $\psi(x)$ is Digamma function definded as $\psi(x)=\frac{\Gamma'(x)}
{\Gamma(x)}= -C -\sum \left( \frac{1}{x+k}-\frac{1}{k+1} \right),$ $C$ is Eular constant equa to =$-\psi(1)=0.5772....$
Using (3), we find that $$ S(0)=0.25, S(1)=0.1458..=7/48, S(2)=0.1027=37/360, S(3)=0.7931=533/6720.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Finite Riemann-like sum and integration For all $x<0$, $y>0$, let's define the following function on $[-1,1]$ : $$ f_{x,y}(t) = \frac{-2xy}{((y-x) - t*(x+y))^2} $$
The denominator is never zero, so it is well defined. It is specially designed so that (you may verify if you want) : $$ \int_{t=-1}^1 f_{x,y}(t) dt = 1$$
My question is : Does there exist a finite set of abscissae $\{t_0,\dots,t_n\}\in[-1,1]^n$ and positive weights $\{w_0,\dots,w_n\} \in \mathbb{R}^n$ (independent of $x$ and $y$) such that for all values $x<0$ and $y>0$ we have : $$ 1 = \int_{t=-1}^1 f_{x,y}(t) dt = \sum_{i=1}^n w_i f_{x,y}(t_i)$$
Any tip, partial of full solution is welcome.
| WLOG, we assume $x+y\ne 0$.
\begin{align}
\int_{-1}^1\frac{-2xy}{((y-x)-t(x+y))^2}\,dt
&=
\frac{-2xy}{(x+y)^2}\int_{-1}^1\frac{dt}{(t+\frac{x-y}{x+y})^2}
\\ &=
\frac{-2xy}{(x+y)^2}\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
\end{align}
Put $x_i=-1+\frac{x-y}{x+y}+\frac{2i}{n}$ and $\xi_i=\sqrt{x_{i=1}x_i}$ . Note that
\begin{align}
\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
&=
\lim_{n\to \infty} \sum_{i=1}^n \frac{x_i-x_{i-1}}{x_{i-1}x_i}
\\ &=
\lim_{n\to \infty} \sum_{i=1}^n \left( \frac{1}{x_{i-1}}-\frac{1}{x_{i}} \right)
\\ &=
\lim_{n\to \infty} \left( \frac{1}{x_{0}}-\frac{1}{x_{n}} \right)
\\ &=
\frac{1}{-1+\frac{x-y}{x+y}}-\frac{1}{1+\frac{x-y}{x+y}}
\end{align}
Thus
\begin{align}
\int_{-1}^1\frac{-2xy}{((y-x)-t(x+y))^2}\,dt
&=
\frac{-2xy}{(x+y)^2}\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
\\ &=
\frac{-2xy}{(x+y)^2}\frac{2}{(\frac{x-y}{x+y})^2-1}
\\ &=
1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum of the function $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$ Find the minimum of the function $$y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$$
By computer I found $\min_y=3;$ then I will prove $y\ge 3$. After squaring we got $$(x+2)(178-37x)\ge 0\quad \forall -2\le x\le 5.$$
My idea is not beautiful so I need some other solution. Thanks!
| Let
\begin{align*}
f_1(x)&=\sqrt{-x^2+4x+21}\\
f_2(x)&=\sqrt{-x^2+3x+10}.
\end{align*}
An examination of the domains of each of these functions separately shows that
\begin{align*}
\mathcal{D}(f_1)&=[-3,7]\\
\mathcal{D}(f_2)&=[-2,5].
\end{align*}
If we define $f(x)=f_1(x)+f_2(x),$ then the above forces the domain $\mathcal{D}(f)=[-2,5].$ Therefore, we must evaluate the function $f$ at $-2$ and $5$. However, there may be critical points inside the domain. Differentiating $f$ yields
$$f'(x)=\frac{2-x}{\sqrt{-x^2+4x+21}}+\frac{3/2-x}{\sqrt{-x^2+3x+10}}. $$
Setting this equal to zero yields the following:
\begin{align*}
(2-x)\sqrt{-x^2+3x+10}&=(x-3/2)\sqrt{-x^2+4x+21}\\
(2-x)^2(-x^2+3x+10)&=(x-3/2)^2(-x^2+4x+21)\\
&\vdots\\
51x^2-104x+29&=0\\
x&=\frac{29}{17},\;\frac{1}{3},
\end{align*}
both of which are in $\mathcal{D}(f).$ However, plugging $1/3$ into $f'(x)$ shows that we picked up a spurious root by squaring. Evaluating, then, we have
\begin{align*}
f(-2)&=3 \\
f\left(\frac{29}{17}\right)&=6\sqrt{2}\approx 8.49 \\
f(5)&=4.
\end{align*}
This proves that the minimum occurs at $x=-2,$ and has a value of $y=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\lim_{x\to 0+}\left[\frac{1}{x^{3/2}}-\frac{1}{x^{1/2}\sin(x)}\right]=0$? This exercise $$\lim_{x\to 0+}\left[\frac{1}{x^\frac{3}{2}}-\frac{1}{x^\frac{1}{2}\sin(x)}\right]$$ was in my calculus III test at college, after trying hard to solve it I was not really able to do so, I know intuitively that this limit equals 0 but did not find an appropriate way to prove it, as the limit gives an indeterminate form $\infty - \infty$. I worked out the fractions and got $$\lim_{x\to 0+}\frac{x^\frac{1}{2}\sin(x) - x^\frac{3}{2} }{x^2\sin(x)}$$ which gives me an indeterminate form $\frac{0}{0}$, then using L'hopitals rule, the limit starts getting uglier as more indeterminate forms keep showing up. My college test is over, but, still I want to find out how to solve this limit, any help would be highly appreciated.
| I wouldn't rationalise the denominator. Instead, stick with
$$\frac{\sin x - x}{x^{\frac{3}{2}}\sin x}.$$
Applying L'Hopital's rule,
\begin{align*}
\frac{\cos x - 1}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x} &= \frac{(\cos x - 1)(\cos x + 1)}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\
&= \frac{-\sin^2 x}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\
&= \frac{-x^{\frac{1}{2}}\frac{\sin^2 x}{x^2}}{\left(\frac{3}{2}\frac{\sin x}{x} + \cos x\right)(\cos x + 1)} \\
&\to \frac{-0 \cdot 1}{\left(\frac{3}{2} \cdot 1 + 1\right)(1 + 1)} = 0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a matrix associated with a given domain and target space of a transformation For the following matrix $A$, first find a nonzero vector $X$ in the nullspace of $A$.Then compute the matrix of $T_A$using the ordered basis $\mathcal{B}=\{[1, 0]^t,X\}$ for the domain and the ordered basis $\bar{\mathcal{B}}=\{A_1, [0, 1, 0]^t, [0, 0, 1]^t\}$ for the target space where $A_1$ is the first column of $A$.
$$A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix}$$
I was able to find the vector $X$ as $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$, and thus obtained that the ordered basis $\mathcal{B}$ is given by $$\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \end{bmatrix} \right\}$$
and $$\bar{\mathcal{B}} = \left\{ \begin{bmatrix} 1 \\ 2\\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$$
but am unable to continue. I believe that the transformation $T_A: \mathbb{R}^2 \rightarrow \mathbb{R}^3$, but I don't know how to find the matrix that corresponds to this.
| To find the matrix corresponding to linear transformation A from X to Y, in a given basis for X and a given basis for Y, apply A to each basis vector for X then write the result as a linear combination of the basis vectors for Y. The coefficients form a column in the matrix.
Here the basis vectors for X are $u= \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $v= \begin{bmatrix}1 \\ -2\end{bmatrix}$. $Au= \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$ and $Av= \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}-3 \\ -6 \\ -9 \end{bmatrix}$.
Write those in terms of the basis vectors for Y, $p= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$, $q= \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$, and $\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$.
So we want to find A, B, and C so that $A\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ B\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}+ C\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}A \\ 2A+ B \\ 3A+ C \end{bmatrix}= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$ which is easily solved: A= 1, B= C= 0.
And we want to find A, B, and C so that $A\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ B\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}+ C\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}A \\ 2A+ B \\ 3A+ C \end{bmatrix}= \begin{bmatrix}-3 \\ -6 \\ -9 \end{bmatrix}$. That is almost as easy to solve: A= -3 so 2A+ B= -6+ B= -6 so B= 0 and 3A+ C= -9+ C= -9. C= 0. A= -3 and B= C= 0.
Using those as columns the matrix we want is $\begin{bmatrix}1 & -3 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^2 \equiv 4\pmod 5 \\ 3^3\equiv 7\pmod 5 \\ 3^4\equiv 1\pmod 5 \\ 3^5\equiv 3\pmod 5$
Also,
$8\equiv 3\pmod 5 \\ 8^2 \equiv 4\pmod 5 \\ 8^3\equiv 7\pmod 5 \\ 8^4\equiv 1\pmod 5 \\ 8^5\equiv 3\pmod 5$
Adding the congruence up (since the same cycle repeats after the 4th power) none of them are divisible by 5 or equal to 0.
But I need a proof by Induction.
Any help will be appreciated.
| Alternatively,
$$3^n + 8^n \equiv 3^n + 3^n \equiv 2*3^n \not\equiv 0 \bmod 5$$
because $2*3^n$ does not contain a factor $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
The closed-form solution of the family $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)}$? (The results below extend this post.) Given the Clausen function $\operatorname{Cl}_n\left(z\right)$. And,
$$\begin{aligned}
\operatorname{Cl}_2\left(\frac\pi2\right) &= \text{Catalan's constant}\\
\operatorname{Cl}_2\left(\frac\pi3\right) &= \text{Gieseking's constant}\\
\operatorname{Cl}_2\left(\frac\pi4\right) &= \text{unnamed}\\
\operatorname{Cl}_2\left(\frac\pi6\right) &= \tfrac23\,\operatorname{Cl}_2\left(\frac\pi2\right)+\tfrac14\,\operatorname{Cl}_2\left(\frac\pi3\right)
\end{aligned}$$
Then we have the closed-forms,
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)} &=& \frac{11}{8} \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)} &=& \frac{5}{3} \zeta(3) -\frac{2}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(4n+m)} &=& \frac{67}{32} \zeta(3) -\frac{1}{2}\pi\, \operatorname{Cl}_2\left(\frac\pi2\right) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(6n+m)} &=& \frac{73}{24} \zeta(3) -\frac{8}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(8n+m)} &=& \frac{515}{128} \zeta(3) -\frac{3}{8}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{red}4}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(12n+m)} &=& \frac{577}{96} \zeta(3) -\frac{7}{6}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\frac{19}{18}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\end{eqnarray*}
where for $p=12$ we could have used $\operatorname{Cl}_2\left(\frac\pi2\right)$ and $\operatorname{Cl}_2\left(\frac\pi6\right)$. As the OP from the other post points out, note that,
$$I(p)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)} =\int_0^1 \frac{\ln(1-z) \ln(1-z^p)}{z} dz$$
Q: The results above suggest a family. Can we find the closed-form of the integral $I(p)$ for $p=5$ and others?
$\color{red}{\text{Update July 24}}$: Thanks to Zacky's answer which provided the clue that more than one Clausen function with argument $\frac{m\,\pi}p$ may be needed, after some tinkering, I managed to find a closed-form for $I(p)$, namely,
$$I(p)= \frac{p^3+3}{2p^2}\zeta(3)-\frac{\pi}p\sum_{k=1}^{\lfloor(p-1)/2\rfloor}(p-2k)\operatorname{Cl}_2\left(\frac{2k\pi}p\right)$$
with floor function $\lfloor x\rfloor$. I found this using odd $p$, but it seems to work for even $p$ as well. However, a rigorous proof is needed to show it holds true for all $p$.
| We may apply a discrete Fourier transform to the following generating function
$$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$
since
$$ I(p) = \sum_{n\geq 1}\frac{H_{p n}}{pn^2}. $$
The only term leading to a non-elementary contribution is the sum of $\operatorname{Li}_3(1-x)$ over the $p$-th roots of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Number of nine digits numbers whose sum of the digits is even I am reading Mathematical Circle. Problem $48$ in chapter two says that
How many nine-digit numbers have an even sum of their digits?
I am trying in this way, that we can divide the problem in four cases.
*
*$1$ even digit and $8$ odd digits
*$3$ even digits and $6$ odd digits
*$5$ even digits and $4$ odd digits
*$7$ even digits and $2$ odd digits
For the first case we get $4 \cdot 5^8 +5\cdot 5^7 \cdot 5$ number of solution. Because if the even digit is placed in first place (left to right) then we get $4\cdot 5^8$ ways to write the number and if an odd digit is placed in first place then we get $5\cdot 5^7 \cdot 5$ ways to write the number. Similarly for the second case we get $4\cdot 5^8+5^9$ , for the third case we get $4\cdot 5^8+5^9$ and for the fourth case we get $4\cdot 5^8+5^9$ ways to write the number. So total number is $ 4 \cdot (4\cdot 5^8+5^9)$.
The answer is different. So Where I have made a mistake?
Thanks.
| There are $900000000$ nine-digits numbers and exactly half of them have an even sum of digits (because every number can be paired with another of the opposite parity by changing the last digit).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?
| First you must solve
$x=y-\frac{1}{y}$
$y^2-xy-1=0$
So
$y=\frac{x+\sqrt{x^2+4}}{2}$
Then
$h(x)=h(y-\frac{1}{y})=y^2-\frac{1}{y^2}=(y-\frac{1}{y})(y+\frac{1}{y})=$
$=(\frac{x+\sqrt{x^2+4}}{2}-\frac{2}{x+\sqrt{x^2+4}})(\frac{x+\sqrt{x^2+4}}{2}+\frac{2}{x+\sqrt{x^2+4}})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?
| Here is the evaluation based on familiar trigonometric identities. Let $\theta = \pi/7$ and express $H$ in terms of cosine functions
$$H=\frac{1}{\cos2\theta} + \frac{1}{\cos4\theta} + \frac{1}{\cos6\theta}$$
or, in the form of common denominator,
$$H =\frac{\cos2\theta + \cos4\theta + \cos6\theta}{\cos2\theta \cos4\theta \cos6\theta}=\frac ND \tag{1}$$
where we used $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ and the relationships $\cos 4\theta = \cos 10\theta$ and $\cos 6\theta = \cos 8\theta$. To compute the denominator, apply $\sin 2x = 2\sin x \cos x$ to the denominator three times,
$$ D = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{\sin 16\theta}{8\sin 2\theta} = \frac{1}{8} \tag{2} $$
To compute the numerator, write it as
$$ N =\frac{1}{\sin 2\theta} \left({\sin 2\theta\cos2\theta + \sin 2\theta\cos4\theta + \sin 2\theta\cos6\theta} \right)$$
and apply $\sin(x+y)+\sin(x-y)=2\sin x\cos y$ and $\sin 6\theta = -\sin 8\theta$ to get
$$ N = \frac{\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)}{2\sin 2\theta}=-\frac12\tag 3$$
Finally, plugging (2) and (3) into (1), we arrive at
$$H=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Prove that if $x$ is odd, then $x^2$ is odd
Prove that if $x$ is odd, then $x^2$ is odd
Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\frac{x}{2} = x(a+0.5) = xa + \frac{x}{2}$$
$xa \in \mathbb Z$ and $\frac{x}{2} \notin \mathbb Z$, hence $xa + \frac{x}{2}$ is not a integer. And since $xa + \frac{x}{2} = \frac{x^2}{2}$, it follows that $x^2$ is not divisible by two, and thus $x^2$ is odd.
Is it correct?
| Your proof is correct.
You could also prove by contrapositive:
if $x^2$ is even, then $2|x^2=x\cdot x,$ so by Euclid's lemma $2|x,$ so $x$ is even.
If you don't like that, you could say $x$ is odd, and the product of two odd numbers is odd,
so $x^2=x\cdot x$ is odd.
If you don't like that, you could argue as follows:
if $x$ is odd, then $x-1=2k$ and $x+1=2k+2$ with $k \in\Bbb Z$,
so $x^2=(x-1)(x+1)+1=(2k)(2k+2)+1=2(k(2k+2))+1$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Where is the mistake in the equality of roots? I have this statement:
If $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}}, g(x) =
> \frac{\sqrt[6]{8x^2}}{2}$, with $x < 0$ is $f(x) = g(x)$ ?
My attempt was:
$(1)$ $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$
$(2)$ $g(x) = \frac{\sqrt[6]{8x^2}}{2}$ using the property of $\sqrt[n]{k^m} = k^{\frac{m}{n}}$ and since $x^2 \geq 0$, thus:
$(3)$ $g(x) = \frac{\sqrt[6]{2^3}\sqrt[6]{x^2}}{2} = \frac{2^{\frac{3}{6}}\cdot (x^2)^{\frac{1}{6}}}{2} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$, that is equal to $f(x)$ from the step $(1)$
But according to the guide, this is false. So, what is wrong?
| If $x \geq 0$, then
$$ \frac{\sqrt[6]{8x^{2}}}{2} = \frac{\sqrt[6]{2^{3}x^{2}}}{2} = \frac{2^{1/2}x^{1/3}}{2} = \frac{\sqrt[3]x}{\sqrt{2}}. $$
If $x < 0$,
$$ \frac{\sqrt[6]{8x^{2}}}{2} = \frac{\sqrt[6]{2^{3}x^{2}}}{2} = \frac{2^{1/2}|{x}\mid^{1/3}}{2} = \frac{\sqrt[3]|x|}{\sqrt{2}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction $2^n \gt \frac{n(n-1)(n-2)}{6}$ I am looking at the problem $2^n \gt \frac{n(n-1)(n-2)}{6}$
skipping the base step this post to save time:
$2^{n+1} \gt \frac{n(n+1)(n-1)}{6}$
No matter how I rearrange the right hand side it does nothing. Moving terms back and forth has done nothing for me as well. The only insight I have is that for $P(n)$ the numerator represents terms of a factorial. Since it stops at $n-2$. $3!$ comes to mind but I'm pretty sure that has nothing to do with the equation. I feel like there is an obvious equivalency for the right side of $P(n)$. A google search on the expression yielded nothing. I prefer to avoid using binomial theorems.
On the second part of the question I am asked to either prove or find a counter example to the following problem:
if $a \lt 0 \lt b$ then $(a+b)^n \leq b^n \space \space \forall n \in \mathbb{N}$
Well if we let $n=1$ we get:
$a+b \leq b$
$a \leq 0$
which is a contradiction to our hypothesis. Also how can I prove statements like this without using binomial theorem and just induction?
| $\frac {n(n+1)(n-1)}6= [\frac {n(n-1)(n-2)}6]\cdot \frac {n+1}{n-2}$ if $n > 2$
And $[\frac {n(n-1)(n-2)}6]\cdot \frac {n+1}{n-2}< 2^n\frac {n+1}{n-2}$
All you have to do is show $\frac{n+1}{n-2} \le 2$ for $n> 2$ which is true if $n\ge 5$.
Base cases: If $n =1$ or $2$ we have $\frac {n(n-1)(n-2)}6 = 0 < 2^{n}$.
If $n = 3,4$ we have $\frac {n(n-1)(n-2)}6 = 1,4 < 2^3, 2^4$.
.....
Or you could note $\frac {n(n-1)(n-2)}6 < \frac {n^3}{6}$ and for $n \le 4$ $\frac {n^3}6 \le \frac {2^6}6 = \frac {2^5}3 < 2^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to solve $\int\frac{\cos(2x)}{\cos x-\sin x}dx$? $$\int\frac{\cos(2x)}{\cos x-\sin x}dx$$
$\cos(2x) = \cos^2(x) - \sin^2(x)$ thus the integral becomes:
$$\int\frac{\cos^2(x)}{\cos x-\sin x} -\int\frac{\sin^2(x)}{\cos x-\sin x} $$
I am not sure what to do next, I'd appreciate any kind of help.
| Note that $\cos(2x) = \cos^2(x) - \sin^2(x) = (\cos(x)+\sin(x))(\cos(x)-\sin(x))$. Thus, we have
$$\begin{aligned}
\displaystyle \int \dfrac{\cos(2x)}{\cos(x) - \sin(x)} \; \mathrm{d}x &= \displaystyle \int \dfrac{(\cos(x)+\sin(x))(\cos(x)-\sin(x))}{\cos(x) - \sin(x)} \; \mathrm{d}x\\
&= \displaystyle \int \cos(x)+\sin(x) \; \mathrm{d}x\\
&= \sin(x) - \cos(x) + C
\end{aligned}$$
Therefore, the answer is $\sin(x) - \cos(x) + C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| Use that $$\sin(2x)=2\sin(x)\cos(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Getting bounds from Riemann sums $y=\sum_{i=0}^9\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}$ and $x=\sum_{i=1}^{10}\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}$. Prove that
*
*$x<\frac{\pi}6<y$
*$\frac{x+y}2<\frac{\pi}6$
My try: I have shown that $x,y$ are respectively lower and upper Riemann sum of the function $$I=\sqrt 3 \int_0^1\frac{dx}{3+x^2}=\frac{\pi}6.$$
So we are done with part a). How to deal with part b). Here $$x+y=\frac34\frac{1}{10\sqrt 3}+2\sum_{i=1}^{9}\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}+\frac{1}{10\sqrt 3}$$
| Write $f(x) = \dfrac{1}{3+x^2}.$ Then part (b) actually follows from the fact
$$0 < x < 1 \implies f''(x) = \frac{6(x^2-1)}{(3+x^2)^3} < 0$$
Graphically, this is because $\frac{x+y}{2}$ can be reinterpreted as the "trapezoid rule" for the integral $\int_0^1 f(x)\,dx.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why can't $\int x(x-3)^8\ dx$ be integrated by parts? I've been given the task of integrating the following equation $$\int x(x-3)^8dx$$ but they've all been saying that it must be done via substitution and is impossible to do view integration by parts. I'm wondering why that is.
Video of solution by substition:
https://youtu.be/CXgsorgesS0?t=109
Attempt to solve it by parts:
$$\int x(x-3)^8dx$$
$$ v=x,u'= (x-3)^8 $$
$$ u'=1, u = 1/9(x-3)^9$$
$$\int x(x-3)^8dx = (1/9)x(x-3)^9 -\int 1/9(x-3)^9$$
$$=(1/9)x(x-3)^9 -(1/90)(x-3)^{10}+C$$
| Yes, it can be done by parts, treating $(x-3)^8$ as the expression to integrate and $x$ as the expression to differentiate:
$$\int x(x-3)^8\,dx=x\cdot\frac19(x-3)^9-\int\frac19(x-3)^9\,dx=\frac19x(x-3)^9-\frac1{90}(x-3)^{10}+K=\frac1{30}(3x+1)(x-3)^9+K$$
To get to the book answer:
$$\frac1{30}(3x+1)(x-3)^9+K=\frac{3x-9+10}{30}(x-3)^9+K=\left(\frac{x-3}{10}+\frac13\right)(x-3)^9+K=\frac{(x-3)^{10}}{10}+\frac{(x-3)^9}3+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx$ I am trying to solve by a different approach the fourth sum from here, namely:
$$S= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(4n+m)} =\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx= \frac{67}{32} \zeta(3) -\frac{\pi}{2}G$$
One way to solve it is similarly to my answer from there:
$$S=\int_0^1 \frac{\ln(1-x)\ln(1-x^2)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx$$
From here we know that:
$$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$
Thus by setting $m=0,n,p,q=1$ in the first integral we get that:
$$S=\frac{11}{8}\zeta(3)+\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx=\frac{11}{8}\zeta(3)+I$$
$$I=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 x^{2n-1} \ln(1-x)dx=\frac12\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{n^2}=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$
And the result for $S$ follows. The last sum appears to be known, see $(659)$ from here, or alternatively since $I=2\Re\left( S(i)\right)$ just use the following identity:
$$S(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$
However I am trying to find a different method since the result is quite nice and I believe there's a nicer way to solve the integral without using such sums.
Thus I would appreciate to get some help with the following problem:
Prove without using Euler's sum or polylogs that
$$\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$
I also tried to consider the following integral:
$$J=\int_0^1 \frac{\ln(1+x)\ln(1+x^2)}{x}dx$$
$$\Rightarrow I+J=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{x}dx\overset{x^2=t}=\frac12 \int_0^1\frac{\ln(1-t)\ln(1+t)}{t}dt=-\frac{5}{16}\zeta(3)$$
So now I am after the following integral:
$$I-J=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln(1+x^2)}{x}dx=\frac74 \zeta(3)-\pi G$$
| First let's calculate the integral $\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy$:
$$\int_0^1\frac{x\text{Li}_2(x)}{1+x^2}dx=\int_0^1\frac{\text{Li}_2(x)}{x}dx-\int_0^1\frac{\text{Li}_2(x)}{x(1+x^2)}dx$$
$$=\zeta(3)-\int_0^1\frac{1}{1+x^2}\left(\int_0^1\frac{-\ln(y)}{1-xy}dy\right)dx$$
$$=\zeta(3)-\int_0^1\ln(y)\left(\int_0^1\frac{-dx}{(1+x^2)(1-yx)}\right)dy$$
$$=\zeta(3)-\int_0^1\ln(y)\left(\frac{y\ln(1-y)}{1+y^2}-\frac{\ln(2)y}{2(1+y^2)}-\frac{\pi}{4(1+y^2)}\right)dy$$
$$=\zeta(3)-\int_0^1\frac{y\ln(y)\ln(1-y)}{1+y^2}dy-\frac1{16}\ln(2)\zeta(2)-\frac{\pi}{4}G$$
$$\small{\overset{IBP}{=}\zeta(3)-\frac12\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy+\frac12\int_0^1\frac{\ln(1-y)\ln(1+y^2)}{y}dy-\frac1{16}\ln(2)\zeta(2)-\frac{\pi}{4}G. \quad(*)}$$
On the other hand, by integration by parts we have
$$\int_0^1\frac{x\text{Li}_2(x)}{1+x^2}dx=\frac12\ln(2)\zeta(2)+\frac12\int_0^1\frac{\ln(1-x)\ln(1+x^2)}{x}dx.\quad (**)$$
Combining $(*)$ and $(**)$ gives
$$\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy=2\zeta(3)-\frac98\ln(2)\zeta(2)-\frac{\pi}{2}G.$$
This integral can be converted to $\sum_{n=1}^\infty (-1)^n \frac{H_{2n}^{(2)}}{n}$ by expanding $\ln(1+x^2)$ in series, which can be related to $\sum_{n=1}^\infty (-1)^n \frac{H_{2n}}{n^2}$ through the Cauchy product of $-\ln(1-x)\text{Li}_2(x)$, which is related to your integral as you showed in your question body. this way, we avoided using the generating function of $\sum_{n=1}^\infty\frac{H_n}{n^2}x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Factoring $x^4 + 5x^3 + 4x^2 + 2x - 3$
So I have to factor this polynomial
$$x^4 + 5x^3 + 4x^2 + 2x - 3$$
I got $(x^2 + 2x -3)(x^2 + 3x + 1)$
but when I multiplied it, I got a different equation:
$$x^4 + 5x^3 + 4x^2 - 7x - 3$$
I don’t really understand how to find the factors. I only found numbers that could add up to 5 and multiply to -3 but they’re wrong. What did I do wrong and what can I do to fix this?
In the format of:
($x^2$+ __ x +__ ) ($x^2$ + __ x + __ )
or
$$(x^2 + ax + b) (x^2 + cx + d)$$
Also, how does comparing coefficients help factor this polynomial?
| As said in comments, there must be a typo in the question since it does not factorize.
If you write
$$(x^2 + ax + b) (x^2 + cx + d)=x^4 + 5x^3 + 4x^2 + 2x - 3$$ expanding and grouping terms, we end with
$$(b d+3)+x (a d+b c-2)+x^2 (a c+b+d-4)+x^3 (a+c-5)=0$$ and each coefficient must be zero.
So, trying, we have $d=-\frac 3b$, $c=5-a$ but the remaining equations are nightmares since we get
$$a=\frac{b (5 b-2)}{b^2+3}$$ and what is left to solve for $b$ is
$$-27-36 b-39 b^2+47 b^3+13 b^4-4 b^5+b^6=0$$ which does not show any obvious roots (in fact, there are two real non-rational solutions).
If you had $b$, then all the other are available.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Establishing identity: $\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ I need some help in trying to prove the following identity $$\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$$
I'm not sure where to start really. I thought it was going to be as simple as just adding the two terms together but nothing comes of that....
| You may know already that $\sum_{k=1}^\infty\frac{1}{k^2} = \frac{\pi^2}{6}$ and $\sum_{k=1}^\infty\frac{1}{(2k + 1)^2} = \frac{\pi^2}{8}$.
So
$$\frac{1}{4}\frac{\pi^2}{6} + \frac{\pi^2}{8} = \frac{\pi^2}{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$f(f(x)) = 1 + x^2$, then what is f(1)? I get $f(f(a)) = a^2 + 1 = f(f(-a))$, and so $f(a)^2 + 1 = f(a^2 + 1) = f(-a)^2 + 1$, so $f(a) = f(-a)$ or $f(a) = -f(-a)$, but then I donot know what to do next. Thanks for any help.
| A piece of information is still required
$f(f(x)) = 1+x^2$
Let's say $y = f(x)$ is the function, then $y = g(x)$ is the inverse function
$f(x) = g(1+x^2)$, $f(1) = g(2)$
$f(f(x)) = 1+x^2 = f(f(-x))$
$f(f(x)) = f(f(-x))$
Therefore $f(x) = f(-x)$
$f(x)$ is a even function
$f(f(x)) = 1+x^2$, say $x → g(x)$
$f(x) = 1+g(x)^2$, $f(1) = 1+g(1)^2$
$1+g(x)^2 = g(1+x^2)$
Say $x → f(x)$
$1+x^2 = g(1+f(x)^2)$, $f(1+x^2) = 1+f(x)^2$
$f(x) = $
$1+f(x)^2 = f(1+x^2)$
I can't find $f(1)$ unless I knew $f(n)$
$f(1+x^2) = 1+f(x)^2$, Remember $f(x) = 1+g(x)^2$
$f(1+x^2) = 1+(g(x)^2+1)^2$, $f(1+x^2) = g(x)^4+2*g(x)^2+2$
Recall $f(x) = g(1+x^2)$, say $1+x^2→x$
$g(x) = f(\sqrt(x-1))$
$f(1+x^2) = f(\sqrt(x-1))^4+2*f(\sqrt(x-1))^2+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Characterize all real-valued $2\times 2$ matrices with eigenvalues $\pm c$, for $c > 0$.
Characterize all real-valued $2\times 2$ matrices that have as eigenvalues $\lambda_1 = c$ and $\lambda_2 = −c$, for $c > 0$. Use your result to generate a matrix that has its eigenvalues $-1$ and $1$ and does not contain any zero elements.
Where do I even start with this? I know how to compute eigenvalues/vectors and everything, but am I finding the matrix that these eigenvalues came from like matrix $A$ from $(A-\lambda I)x=0$? Or am I finding $\lambda_i$?
| I think the problem is asking us to find a general expression for such matrices in terms of parameters yet to be identified; finding these parameters is part of the challenge.
We write the matrix
$A \in M_{2 \times 2}(\Bbb R) \tag 1$
in terms of its entries
$A = \begin{bmatrix} a_1 & b_1 \\ b_2 & a_2 \end{bmatrix}, \tag 2$
and recall the eigenvalues satisfy the characteristic polynomial
$\chi_A(x) = \det(A - xI) = \det \left ( \begin{bmatrix} a_1 - x & b_1 \\ b_2 & a_2 - x \end{bmatrix} \right ) = (a_1 - x)(a_2 - x) - b_1b_2$
$= x^2 - (a_1 + a_2)x + (a_1a_2 - b_1b_2) = x^2 - \text{Tr}(A)x + \det A; \tag 3$
now if the eigenvalues of $A$ are
$\pm c, \; c > 0, \tag 4$
then
$\chi_A(x) = (x - c)(x + c) = x^2 - c^2; \tag 5$
comparing (3) and (5) we find that
$\text{Tr}(A) = c + (-c) = 0, \tag 6$
whilst
$\det A = -c^2; \tag 7$
it follows then that
$a_1 + a_2 = \text{Tr}(A) = 0, \tag 8$
i.e., we may write
$a_1 = a = -a_2 \tag 9$
for some
$a \in \Bbb R, \tag{10}$
and also
$a_1a_2 - b_1b_2 = \det A = -c^2, \tag{11}$
which in the light of (9) yields
$-a^2 - b_1b_2 = -c^2, \tag{12}$
or
$b_1b_2 = c^2 - a^2. \tag{13}$
Based upon this equation, we may now derive the specific forms $A$ may take. The simplest case is
$b_1 = 0 = b_2, \tag{14}$
whence via (12)
$a^2 = c^2 \Longrightarrow a = \pm c, \tag{15}$
and
$A = \begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}; \tag{16}$
if
$b_1 \ne 0, \tag{17}$
$b_2 = \dfrac{c^2 -a^2}{b_1}, \tag{18}$
so that
$A = \begin{bmatrix} a & b_1 \\ \dfrac{c^2 -a^2}{b_1} & -a \end{bmatrix}; \tag{19}$
likewise, when
$b_2 \ne 0, \tag{20}$
the corresponding results are had:
$b_1 = \dfrac{c^2 -a^2}{b_2}, \tag{21}$
$A = \begin{bmatrix} a &\dfrac{c^2 -a^2}{b_2} \\ b_2 & -a \end{bmatrix}; \tag{22}$
we have shown that the forms (16), (19), and (22) are necessary if the eigenvalues of $A$ are $\pm c$; they are also sufficient; this is self-evident in the case (16); in the case (19), we see that the characteristic polynomial is
$\chi_A(x) = \det \left ( \begin{bmatrix} a - x & b_1 \\ \dfrac{c^2 -a^2}{b_1} & -a -
x \end{bmatrix} \right )$
$= -(a - x)(a + x) - (c^2 - a^2) = x^2 - c^2, \tag{23}$
the zeroes of which are $\pm c$; a similar calculation applies to (22).
As a final observation, the diagonal form (16) is a one-parameter family of matrices depending solely on $a$, whereas (19), (22) are two-parameter families hinging on $a$ and $b_1$ or $b_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| Know that
$$\ln{\left(1+x\right)}=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$$
Then,
$\quad\left(1+3x\right)\ln{\left(1+x\right)}\\=\ln{\left(1+x\right)}+3x\ln{\left(1+x\right)}\\=\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\right)+3x\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\right)\\=x+\dfrac{5}{2}x^2-\dfrac{7}{6}x^3+\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Order of factors in partial decomposition Is there a protocol for deciding which denominator fraction goes under A and which goes under B during partial decomposition?
Doing this question: integral $(5x-5)/(3x^2-8x-3)$ I factored the denominator and got $(3x+1)(x-3)$, so I moved on with $A/(3x+1) + B/(x-3)$.
This came out as $A=2, B=1$, so my final answer answer was $2\ln|x-3| + 1/3\ln|3x+1|$.
However the solution in the textbook had it reversed, with $A=1, B=2$, and the final answer was therefore $\ln|x-3| + 2/3 \ln|3x+1|$.
Are the two answers the same? And thus just need some natural log algebra to reflect this? Or is there, as asked, a definite way of saying which factor goes under A and which B?
| As JG123's comment stated, you just flipped the $A$ and $B$ in your calculations. Note you have
$$\begin{equation}\begin{aligned}
\frac{5x-5}{3x^2 -8x -3} & = \frac{A}{3x+1} + \frac{B}{x-3} \\
\frac{5x-5}{3x^2 -8x -3} & = \frac{A(x-3) + B(3x + 1)}{(3x + 1)(x-3)} \\
5x-5 & = A(x-3) + B(3x + 1)
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
This leads to
$$5x = (A + 3B)x \implies 5 = A + 3B \tag{2}\label{eq2}$$
$$-5 = -3A + B \tag{3}\label{eq3}$$
Multiplying \eqref{eq2} by $3$ and adding \eqref{eq3} gives
$$15 - 5 = 9B + B \implies B = 1 \tag{4}\label{eq4}$$
Using this in \eqref{eq3} gives $A = 2$.
Thus, using this in the integration, the final answer will then match what the book gave of $\ln|x-3| + 2/3 \ln|3x+1|$.
As for which partial fraction to use for $A$ and which for $B$, this is arbitrary as both $A$ and $B$ are dummy variables. The book just seems to have used them in the opposite order to what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you
| It is:
$$\frac{x^{2015}-x^{2014}}{(x-1)^3}=\frac{(x-1+1)^{2015}-(x-1+1)^{2014}}{(x-1)^3}=\\
=\small{\frac{\left[S(x)+{2015\choose 2}(x-1)^2+{2015\choose 1}(x-1)+1\right]-\left[T(x)-{2014\choose 2}(x-1)^2-{2014\choose 1}(x-1)-1\right]}{(x-1)^3}}=\\
=\frac{S(x)-T(x)+2014(x-1)^2+(x-1)}{(x-1)^3}=Q(x)+\frac{(x-1)(2014x-2013)}{(x-1)^3}.$$
Note: It is more expanded and direct version of lab bhattacharjee's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Trigonometric inequality $\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}$ for all $x>1/2$ In Hobson's book "A treatise on plane trigonometry", the following exercise is to be found
If $x>1/2$ then
$$\frac{1}{1+x+x^2}<\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}.$$
The first inequality is easy as it follows directly from $\alpha<\tan(\alpha)$ and the fact that $x$ is positive.
For the second one can produce the even tighter bound (on the given interval, checked only graphically)
$$\tan(\alpha)<\frac{\alpha}{1-\frac{\alpha^2}{2}}$$
but from here I cannot get to the exact given inequality.
I am aware of the fact that $\tan(\alpha)\sim \alpha$ for small $\alpha$, but I cannot manage to find the exact bound given in the exercise.
Any hints are welcome.
| Taking arctangent of both sides (and using its monotonicity in a given range):
$$f_1(x)=\frac{1}{1+x^2} \\ f_2(x)=\arctan \frac{1}{1-x+x^2}$$
$$f_2(x)= \int_0^1 \frac{1-x+x^2}{(1-x+x^2)^2+t^2} dt$$
We need to prove that:
$$\int_0^1 \frac{(1-x+x^2)(1+x^2)}{(1-x+x^2)^2+t^2} dt >^? 1$$
$$\int_0^1 \frac{1-x+2x^2-x^3+x^4}{(1-2x+3x^2-2x^3+x^4)+t^2} dt >^? 1$$
$$1+\int_0^1 \frac{x(1-x+x^2)-t^2}{(1-x+x^2)^2+t^2} dt >^? 1$$
Now we need to prove that:
$$f_3(x)=\int_0^1 \frac{x(1-x+x^2)-t^2}{(1-x+x^2)^2+t^2} dt >^? 0, \quad x> \frac12$$
Since $1-x+x^2 \geq 1$ for $x \geq 1$, the inequality follows trivially in this case, so we only need to consider the range:
$$\frac12 <x<1$$
$$f_3'(x)=\int_0^1 \frac{(1-x^2)(1-x+x^2)^2-(1-4 x+3 x^2-4 x^3) t^2}{((1-x+x^2)^2+t^2)^2} dt$$
We have:
$$(1-x^2)(1-x+x^2)^2-(1-4 x+3 x^2-4 x^3)=x (2-x) (1+x^2)^2 >0, \quad 0<x<2$$
So $f_3(x)$ increases monotonely in the range of interest, which means it's enough to prove that:
$$f_3 \left( \frac{1}{2} \right) >^? 0$$
$$f_3 \left( \frac{1}{2} \right)=\int_0^1 \frac{\frac{3}{8}-t^2}{\frac{9}{16}+t^2} dt=\frac{3}{4} \int_0^{4/3} \frac{2/3-u^2}{1+u^2} du= \\ = \frac{5}{4} \arctan \frac{4}{3}-1 >0$$
This finishes the proof.
Note:
$$\arctan \frac{4}{3}= 2 \arctan \frac{3}{4} \left(\sqrt{1+\frac{16}{9}}-1 \right)=2 \arctan \frac{1}{2}$$
$$f_3 \left( \frac{1}{2} \right)=\frac{5}{2} \arctan \frac{1}{2}-1>\frac{5}{2} \left(\frac12-\frac{1}{24} \right)-1=\frac{7}{48}$$
Thus we don't need any numerical checks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Write the quartic equation $4x^4+12x^3-35x^2-300x+625$ as the product of two quadratic expressions. Write the quartic expression
$4x^4+12x^3-35x^2-300x+625$
as the product of two quadratic expressions with real coefficients.
I would like to know the way of solving it other than solving with the simultaneous equation.
| By the Ruffini's theorem, I have that $x=\frac{5}{2}$ is a rational zero of the polynomial. So, dividing by $x-\frac{5}{2}$, I have: $P(x)=(x-\frac{5}{2})(4x^3+22x^2+20x-250)=(x-\frac{5}{2})$. Using another time Ruffini's theorem, I have $\frac{5}{2}$ a zero, so $P(x)$ becomes: $$P(x)=2(x-\frac{5}{2})^2(x^2+8x+25)$$ $x^2+8x+25$ is irriducibile in $R$, so the factorization stop.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Summation of fractional parts of $x/n$ Let us consider the sums $$\displaystyle T_1(x)=\sum_{n\geq a\sqrt{x}}^{b \sqrt{x}} \left\{ \frac{x}{n} \right\} \\T_2(x)=\sum_{n\geq a\sqrt{x}}^{b \sqrt{x}} \left\{ \frac{x}{2n} \right\}
$$ where $x$ is a positive integer, $0\leq a<b\leq 1$, $n$ runs over all integers in the interval $[a \sqrt{x}, b\sqrt{x}]$, and $\{ \}$ indicates the fractional part. Because of the equidistribution of such fractional parts, the plots of $T_1(x)$ and $T_2(x)$ versus $x$ show oscillations around the line $\frac{b-a}{2} \sqrt{x}$.
By experimental calculations, I noted that the average value of the difference between $T_1(x)$ (or $T_2(x)$) and $\frac{b-a}{2} \, \sqrt{x}$, calculated over all positive integers $x \leq N$, is $O(1)$. In particular, we have
$$\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{x=1}^{N} \left( T_1(x)- \frac{b-a}{2} \sqrt{x} \right)= K_1(a,b)\\
\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{x=1}^{N} \left( T_2(x)- \frac{b-a}{2} \sqrt{x} \right)= K_2(a,b)$$
where $K_1(a,b)$ and $K_2(a,b)$ are constant terms that depend on $a$ and $b$, indicating that the distributions of such differences are biased.
The problem can be simplified by considering the case where $b=1$ (once solved this case, the general case with $b<1$ can be solved by difference). In this case, the biases are determined only by $a$. For example, analyzing the distributions of differences over all integers $N$ up to $2500$, setting $a=0.5$ we have $K_1(0.5,1)\approx -0.370...$ and $K_2(0.5,1)\approx -0.169...$, with rather slow convergence rates. I wonder how these terms are generated and whether they could be exactly calculated. Again based on experimental results, it seems that $K_1$ and $K_2$ have a logarithmic relation with $a$, with values near to $\frac{1}{2} \log(a)$ and $\frac{1}{4} \log(a)$, respectively.
This question is somewhat linked to this other one, for which a nice answer was previously given.
EDIT (after Mathworker21's answer): A rough numerical computation until $N=10000$ for the case $a=0.5$, $b=1$ seems to confirm the above estimates obtained with $N$ up to $2500$. Here is the plot of the differences $T_1(x)- \frac{b-a}{2} \sqrt{x}$ versus $x$, followed by that of the average value of these differences calculated over the first $N$ integers. As show in this second plot, $K_1(0.5,1)$ seems to converge to $\approx -0.37$. The dotted black line in the first graph is the best fitting line, whose intercept is compatible with such value. Based on a visual assessment, large departures from this value (as suggested by the provided answer) for higher $N$ seem unlikely:
Similar considerations can be made for $K_2(0.5,1)$, which seems to converge to $\approx -0.17$:
| Sigh, I hate fourier analysis. It turns people (especially me) into mindless, computing zombies.
The way to obtain $K_1(a,b)$ is literally just to interchange sums. I don't know why I didn't see this days ago. I don't have a fully rigorous proof, but I think it's just as rigorous as the fourier argument given in the other answer. This answer shows that $K_k(a,b) = \frac{-1}{2k}\log(\frac{b}{a})-\frac{k}{24}(b^2-a^2)$.
The intuitive reason the $\log$ appears is that $\frac{1}{2}$ is not the average value of $\{\frac{x}{m}\}$ for fixed $m$ as $x$ ranges. Instead, the average is $\frac{1}{m}\left[0+\frac{1}{m}+\dots+\frac{m-1}{m}\right] = \frac{m-1}{2m} = \frac{1}{2}-\frac{1}{2m}$. This "$\frac{-1}{2m}$" is the reason for the log. The reason for the $-\frac{k}{24}(b^2-a^2)$ is a bit harder to directly understand.
We assume $\frac{1}{a^2},\frac{1}{b^2}$ are integers that have the same residue modulo $k$. The result for any $a,b$ such that $\frac{1}{a^2}$ and $\frac{1}{b^2}$ are integers follows from an easy extension (to $k > 2$) of the reasoning given in the other answer.
$$\frac{1}{N}\sum_{x=1}^N \sum_{m=a\sqrt{x}}^{b\sqrt{x}} \left(\{\frac{x}{km}\}-\frac{1}{2}\right) = \frac{1}{N}\sum_{m=1}^{a\sqrt{N}}\sum_{x=m^2/b^2}^{m^2/a^2}\left(\{\frac{x}{km}\}-\frac{1}{2}\right)+\frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}}\sum_{x=m^2/b^2}^N \left(\{\frac{x}{km}\}-\frac{1}{2}\right)$$ Since we repeatedly cycle through the residues mod $km$, the first term is $$\frac{1}{N}\sum_{m=1}^{a\sqrt{N}} \frac{-1}{2}\left(\frac{\frac{m^2}{a^2}-\frac{m^2}{b^2}}{km}+O(1)\right) = \frac{-1}{4k}\left(1-\frac{a^2}{b^2}\right)+O\left(\frac{1}{\sqrt{N}}\right)$$ For each $m$, writing $N = q_m km+r_m$ for $0 \le r_m \le km-1$, the second term is $$\frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}}\sum_{x=m^2/b^2}^{q_m km} \left(\{\frac{x}{km}\}-\frac{1}{2}\right) + \frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}} \sum_{x=q_m km+1}^{q_m km+r_m} \left(\{\frac{x}{km}\}-\frac{1}{2}\right)$$ $$ = \frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}} \frac{-1}{2}\left(q_m-\frac{m}{b^2k}+O(1)\right) + \frac{1}{N}\sum_{a=\sqrt{N}}^{b\sqrt{N}} \left[\left(\frac{1}{km}-\frac{1}{2}\right)+\dots+\left(\frac{r_m}{km}-\frac{1}{2}\right)\right]$$ $$ = \frac{-1}{2k}\log(b/a)+\frac{1}{4k}(1-\frac{a^2}{b^2})+\frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}} \left(\frac{r_m(r_m+1)}{2km}-\frac{r_m}{2}\right)$$ The sum can be written as $$\frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}} \frac{mk}{2}(\frac{r_m}{km})^2-\frac{mk}{2}(\frac{r_m}{mk}),$$ and since $\frac{r_m}{mk} = \{\frac{N}{mk}\}$ should be equidistributed, and since $\int_0^1 x^2-x dx = \frac{-1}{6}$, we heuristically get $$\frac{1}{N}\sum_{m=a\sqrt{N}}^{b\sqrt{N}} \frac{mk}{2}(-\frac{1}{6}) = \frac{-k}{24}(b^2-a^2).$$ Putting everything together gives the desired $$\lim_{N \to \infty} \frac{1}{N}\sum_{x=1}^N \sum_{m=a\sqrt{x}}^{b\sqrt{x}} \left(\{\frac{x}{km}\}-\frac{1}{2}\right) = \frac{-1}{2k}\log\left(\frac{b}{a}\right)-\frac{k}{24}\left(b^2-a^2\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
Limit development for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. I'm trying to solve this problem: Find a limit development of order
7 for the function $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. Where
we use the next definition:
Definition. Let $f\colon I\to\mathbb{R}$ be a function and $x_{0}\in I$. We
say that $f$ has a limit development of order $n$ in $x_{0}$ provided
that there exist $a_{0},a_{1},\ldots,a_{n}\in\mathbb{R}$ such that
for $x\in I$
$$
f(x)=a_{0}+a_{1}(x-x_{0})+\ldots+a_{n}(x-x_{0})^{n}+o((x-x_{0})^{n}) \text{ (small $o$)}.
$$
Using Taylor series for $\arctan x$ at 0, we have that $\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots$,
and therefore
\begin{align*}
\arctan(x^{2}) & =x^{2}-\frac{(x^{2})^{3}}{3}+\frac{(x^{2})^{5}}{5}-\frac{(x^{2})^{7}}{7}+\ldots\\
& =x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots
\end{align*}
So, replacing for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$ I have
that:
\begin{align*}
f(x) & =\frac{x}{x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots}-\frac{1}{x}\\
& =\frac{x}{x\left(x-\frac{x^{5}}{3}+\frac{x^{9}}{5}-\frac{x^{13}}{7}+\ldots\right)}-\frac{1}{x}\\
& =\frac{1}{x}\cdot\left(1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots\right)^{-1}-\frac{1}{x}
\end{align*}
But I couldn't obtain the form that is necessary for the limit development
because I have that part with the inverse of $1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots$.
Could you help me or give me some suggestion?
Thanks.
| We may start with few terms of the Maclaurin series of $\frac{\tan x}{x}$ (it is not that painful to compute a few derivatives of the tangent function at the origin)
$$ \frac{\tan x}{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O(x^6)$$
then replace $x$ with $\arctan(z)=z-\frac{z^3}{3}+\frac{z^5}{5}+o(z^6)$, giving
$$ \frac{z}{\arctan z} = 1+\frac{z^2}{3}-\frac{4z^4}{45}+O(z^6)$$
(this is related to Gregory coefficients, by the way) and, by setting $z=t^2$,
$$ \frac{t^2}{\arctan(t^2)} = 1+\frac{t^4}{3}-\frac{4t^8}{45}+O(t^{12})$$
$$ \frac{t}{\arctan(t^2)} = \frac{1}{t}+\frac{t^3}{3}-\frac{4t^7}{45}+O(t^{11})$$
then finally
$$ \frac{x}{\arctan(x^2)}-\frac{1}{x} = \frac{x^3}{3}-\frac{4x^7}{45}+O(x^{11}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Chapter 2 Section 17 Problem 13 - Mary L Boas.
Find real $x$ and $y$ for which $|z+3| = 1 - iz$, where $z=x+iy$.
My attempt:
$\require{cancel}\begin{align} |z+3|\;&=\;1-iz \\ \sqrt{(x+3)^2+y^2}\;&=\;1-iz \\ (x+3)^2+y^2\;&=\;(1-iz)^2 \\ x^2 +6x+9+y^2 &= 1-2iz-z^2 \\ x^2 +6x+9+y^2 &= 1-2i(x+iy)-(x+iy)^2 \\ x^2 +6x+9+y^2 &= 1-2ix+2y-x^2-2xiy+y^2 \\ x^2 +6x+9+\cancel{y^2} &= 1-2ix+2y-x^2-2xiy+\cancel{y^2} \\ (x^2 +6x+9) &= (1+2y-x^2)+i(-2x-2xy)
\end{align}$
Is this the correct way to keep going further or initially errors were made?
| $$iz=1-|z+3|$$ which is real, so $z$ must be purely imaginary
Set $z=iy$ where $y$ is real
$$1+y=|3+iy|\ge0\implies y\ge-1$$
$$(1+y)^2=3^2+y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the following differential equation $(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$ I'd love your help with solving this following differential equation: $$(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$$
I am trying to use the method of finding integrating factor, that is $\frac{1}{Mx+Ny}$. But it is not coming.
Any suggestions?
| The integrating factor $\mu = xy$ turns the ODE into the exact one
$$(xy^4 - 2y^2x^3)\mathrm d x + (2x^2 y^3 - x^4y) \mathrm d y = 0.$$
You can derive this the following way: If the expression
$$\frac{h_x - g_y}{xg - yh} $$
is a function of $xy$ alone (say $f(xy)$), where $g(x, y) = y^3 - 2yx^2$ and $h(x,y) = 2xy^2 - x^3$, then $\mu = \mathrm e^{\int f(u) \mathrm du}$ with $u = xy$. Here,
$$\frac{h_x - g_y}{xg - yh} = \frac{2y^2 - 3x^2 - 3y^2 + 2x^2}{xy^3-2yx^3 -2xy^3+x^3y} = - \frac{x^2 + y^2}{ -(x^3y + xy^3)} = \frac{x^2 + y^2}{ xy(x^2 + y^2)} = \frac{1}{xy}, $$
so when $\frac{1}{xy} = \frac{1}{u}$, one has the integrating factor
$$\mu = \mathrm e^{\int \frac{\mathrm d u}{u} } = \mathrm{e}^{\ln(u)} = xy.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Algebraic manipulation solving for $x+y$ There are real solutions $x,y$ to
$x^3+2016x+1=0$
$y^3+2016x-1=0$.
Find $x+y$.
My thinking: adding them we get
$x^3+y^3+2016(x+y)=0$
$(x+y)(x^2-xy+y^2)+2016(x+y)=0$
$(x+y)(x^2-xy+y^2+2016)=0$
Then, either
$x+y=0$
or,
$x^2-xy+y^2+2016=0$.
I’m not sure how to proceed with the second equation and the first seems a little odd.
| Note that
$$x^2-xy+y^2+2016= \left( x-\frac{1}{2}y \right)^2 +\frac{3}{4}y^2 + 2016>0$$
So, the second equation
$$x^2-xy+y^2+2016=0$$
has no real solutions. Thus, $x+y=0$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.