Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Rationalize $\left(\sqrt{3x+5}-\sqrt{5x+11} -\sqrt{x+9}\right)^{-1}$ I was trying to find if there a method similar to multiplying and dividing by the conjugate $$\frac{1}{\sqrt{3x+5}-\sqrt{5x+11} - \sqrt{x+9}},$$ but that doesn't seem to work here. Also, is there a method of multiplying by a conjugate for roots other than the square root? Such as $(ax+b)^\frac1n ± (cx+d)^\frac1m$
| You can apply it twice:
$$\frac1{\sqrt a-\sqrt b-\sqrt c}=\frac{(\sqrt a-\sqrt b)+\sqrt c}{(\sqrt a-\sqrt b)^2-c}=\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}},$$
then
$$\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}}=\frac{(\sqrt a-\sqrt b+\sqrt c)(a+b-c+2\sqrt{ab})}{(a+b-c)^2-4ab}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.
| Another one just for fun ;-)
Set $x=\frac{1}{\sqrt 2}\sinh(u)$ then,
$$\int^t\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt 2 t)}\frac{\cosh(x)\sinh(u)}{\sqrt{1+\sinh^2(u)}}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\frac{\cosh(u)\sinh(u)}{\cosh u}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\sinh(u)du=\frac{1}{2}\cosh\left(\text{argsinh}(\sqrt 2 t)\right)+C=\frac{1}{2}\sqrt{2t^2+1}+C.$$
But it's not the shortest way to calculate it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^{2}+16\, \left( -1 \right) ^{2\,{\it n}}
\left( {2}^{2\,{\it n}} \right) ^{2} \right) }{{2}^{2\,{\it n}}
\left( 1+i\sqrt {3} \right) ^{2\,{\it n}+2}}},
$$
Сalculation for small $n$ are as follows $I_1=-1, I_2=0,I_3=1, I_4=-1, \ldots$
and leads to a hypothese:
$
I_n= -1 \text{ for } n=3k+1, =0, \text{ for } n=3k+2,=1, \text{ for } n=3k.
$
But how to prove it?
| Generating Functions
Let's compute the generating function of $\displaystyle\sum_{k=0}^n(-1)^k\binom{n-k}{k}$:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty(-1)^kx^k\sum_{n=k}^\infty\binom{n-k}{k}x^{n-k}\\
&=\sum_{k=0}^\infty(-1)^kx^k\frac{x^k}{(1-x)^{k+1}}\\
&=\frac1{1-x}\frac1{1+\frac{x^2}{1-x}}\\
&=\frac1{1-x+x^2}\tag{1}
\end{align}
$$
What the question asks is for the odd terms, which we get by taking the odd part of $(1)$:
$$
\frac12\left(\frac1{1-x+x^2}-\frac1{1+x+x^2}\right)=\frac{x}{1+x^2+x^4}\tag{2}
$$
which implies that $\displaystyle\sum_{k=0}^n(-1)^k\binom{2n+1-k}{k}$ satisfies the recurrence
$$
a_n=-a_{n-1}-a_{n-2}\tag{3}
$$
and starts out: $(1,-1,0,1,-1,\dots)$ and $(3)$ ensures that it will follow this pattern. That is,
$$
\boxed{\displaystyle\bbox[5px]{
\sum_{k=0}^n(-1)^k\binom{2n+1-k}{k}=\left\{\begin{array}{rl}
1&\text{if }n\equiv0\pmod{3}\\
-1&\text{if }n\equiv1\pmod{3}\\
0&\text{if }n\equiv2\pmod{3}\\
\end{array}\right.}}\tag{4}
$$
Solving the Recurrence $\boldsymbol{(3)}$
We can also get a solution by solving the recurrence $(3)$.
Since the roots of $x^2+x+1$ are $\frac{-1\pm i\sqrt3}{2}=e^{\pm i2\pi/3}$ and $a_0=1$ and $a_1=-1$, we get the general solution to be
$$
\begin{align}
a_n
&=\frac{\left(\frac{-1+i\sqrt3}2\right)^{n+1}-\left(\frac{-1-i\sqrt3}2\right)^{n+1}}{i\sqrt3}\\[6pt]
&=\frac{e^{i2\pi(n+1)/3}-e^{-i2\pi(n+1)/3}}{i\sqrt3}\\[4pt]
&=\frac2{\sqrt3}\sin\left(\frac{2\pi(n+1)}{3}\right)\tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
commutative matrix multiplication of nxn matrices? If there are two matrices A and B that are both nxn matrices, will AB = BA always?
Is there a way to have those two matrices so that AB = 0 but BA ≠ 0?
| No. I can give a simple counterexample:
$\begin{pmatrix}1 & 2 \\ 3 &4\end{pmatrix}$ * $\begin{pmatrix}4 & 3\\2 & 1\end{pmatrix}$ = $\begin{pmatrix}8 & 5\\20 & 13\end{pmatrix}$
while
$\begin{pmatrix}4 & 3\\2 & 1\end{pmatrix}$ * $\begin{pmatrix}1 & 2\\3&4\end{pmatrix}$ = $\begin{pmatrix}16 & 20\\8 & 8\end{pmatrix}$.
Two matrices are rarely commutative. Even square ones.
Check this question here for more on commutativity with matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/938020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sums of binomial coefficients Does anyone know something about the following sums?
$$
S_m(n)=\sum\limits_{k=o}^n(-1)^k{mn\choose mk}
$$
Notice that $S_m(n)=0$ for odd $n$, so we only consider $S_m(2n)$. It holds that $S_0(2n)=1$, $S_1(2n)=0$, $S_2(2n)=(-4)^n$, $S_3(2n)=\frac{2}{3}(-27)^n$, but $S_4(2n)$ is no longer a geometric progression.
| Start by restating the problem: we seek to evaluate
$$S_m(n) = \sum_{k=0}^n (-1)^k {nm\choose km}
= \sum_{k=0}^n (-1)^k {nm\choose nm-km}.$$
Introduce the integral representation
$${nm\choose nm-km}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm-km+1}} \; dz.$$
This gives the following for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k=0}^n \frac{(-1)^k}{z^{-km}} \; dz.$$
We may extend the sum to infinity because the integral correctly
represents the binomial coefficient being zero for $k>n$ to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k\ge 0} \frac{(-1)^k}{z^{-km}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\sum_{k\ge 0} (-1)^k z^{km} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}}
\frac{1}{1+z^m}\; dz.$$
Extracting coefficients we obtain
$$[z^{nm}] (1+z)^{nm} \frac{1}{1+z^m}
= \sum_{q=0}^{nm} {nm\choose nm-q} [z^q] \frac{1}{1+z^m}
= \sum_{q=0}^{nm} {nm\choose q} [z^q] \frac{1}{1+z^m}.$$
Introduce $\rho_k = e^{i\pi/m + 2\pi ik/m}$ so that using partial
fractions by residues on simple poles we have
$$\frac{1}{1+z^m}
= \sum_{k=0}^{m-1}
\frac{\mathrm{Res}\left(\frac{1}{1+z^m}; z=\rho_k\right)}{z-\rho_k}.$$
This is
$$\sum_{k=0}^{m-1}
\frac{\frac{1}{m} \rho_k^{-(m-1)}}{z-\rho_k}
= \sum_{k=0}^{m-1}
\frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1}.$$
Substituting this into the sum formula yields
$$\sum_{q=0}^{nm} {nm\choose q} [z^q]
\sum_{k=0}^{m-1}
\frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1}
= - \sum_{q=0}^{nm} {nm\choose q}
\sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-m} \rho_k^{-q}$$
Now $\rho_k^{-m} = -1$ so this yields
$$\sum_{q=0}^{nm} {nm\choose q}
\sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-q}
= \frac{1}{m} \sum_{k=0}^{m-1}
\sum_{q=0}^{nm} {nm\choose q} \rho_k^{-q}
\\ = \frac{1}{m} \sum_{k=0}^{m-1}
\left(1 + \frac{1}{\rho_k}\right)^{nm}
= \frac{1}{m} \sum_{k=0}^{m-1}
\left(1 + \rho_k\right)^{nm}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$ \begin{align}
& \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \\
& \text{ordering} \\
& a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)\cdot 1}{2\cdot 4\cdot 6...(2n-2)(2n)(2n+2)}= \\
& \underbrace{\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{4} \right)...\left( 1-\frac{1}{2n} \right)}_{n\text{ times}}\frac{1}{(2n+2)} \\
& \text{clearly} \\
& \left( 1-\frac{1}{2} \right)^{n}\frac{1}{(2n+2)}\le a_{n}\le \left( 1-\frac{1}{2n} \right)^{n}\frac{1}{(2n+2)} \\
& \text{Root test} \\
& \sqrt[n]{\left( 1-\frac{1}{2} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}}\le \sqrt[n]{a_{n}}\le \sqrt[n]{\left( 1-\frac{1}{2n} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}} \\
& n\to \infty \\
& \frac{1}{2}\le \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{a_{n}}\le 1 \\
& \text{nothing :(} \\
& \text{Ratio test} \\
& \frac{a_{n}}{a_{n-1}}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)}{2\cdot 4\cdot 6...(2n)(2n+2)}\centerdot \frac{2\cdot 4\cdot 6...(2(n-1)+2)}{1\cdot 3\cdot 5...(2(n-1)-1)}=\frac{2n-1}{2n+2} \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=1 \\
& \text{nothing again} \\
\end{align}
Any suggestions?
| An explicit computation is even better. We have:
$$\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}$$
hence your series is telescopic and we have:
$$\sum_{n=1}^{N}\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=\frac{1}{2}+O\left(\frac{1}{\sqrt{N}}\right).$$
In order to prove the last asymptotics, notice that:
$$\frac{1}{4^n}\binom{2N}{N}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)^{2N}dx\leq\frac{2}{\pi}\int_{0}^{\pi/2}e^{-Nx^2}\,dx\leq\frac{1}{\sqrt{\pi N}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Which one is true? (CSIR) Let $a,b,c$ be a positive real number such that $b^2+c^2<a<1$. Let
$A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$. Then
(1) all eigen values of $A$ are positive
(2) all eigenvalues of $A$ are negative
(3) all eigenvalues of $A$ are either positive or negative
(4) all eigenvalues of $A$ are nonreal complex number
Since $A$ is symmetric, all eigen values are real. Hence option (4) is not true.
| Using submatrices:
$$\Delta_1=\det\{[1]\}=1>0$$
$$\Delta_2=\det\left\{\left[\begin{array}{cc}1&b\\b&a\end{array}\right]\right\}=a-b^2>a-(b^2+c^2)>0$$
$$\begin{array}{rcl}
\Delta_3&=&\det\left\{\left[\begin{array}{ccc}1&b&c\\b&a&0\\c&0&1\end{array}\right]\right\}\\
&=&(a+0+0)-(ac^2+b^2+0)\\
&=&a-ac^2-b^2>a-c^2-b^2>0
\end{array}$$
Then, all subdeterminants are positive, so $A$ is a positive definite matrix and all its eigenvalues are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
finding a function from given function here is a function for:
$f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$
what is the $f(x)$?
I calculate $f(x)$ as follows:
$$\begin{align}
x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\
3f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=(1/3)\sin\frac{5\pi}{6}
\end{align}$$
$f(x)=(1/3)\sin\frac{5x}{2}$
| Shift the argument by $\pi/2$:
$$f(x)=\sin(x+\frac\pi2)-2f(\frac{\pi}3).$$
Then plug $\pi/3$:
$$f(\frac{\pi}3)=\sin(\frac{5\pi}6)-2f(\frac{\pi}3).$$
Solve for $f(\pi/3)$:
$$f(x)=\sin(x+\frac\pi2)-\frac23\sin(\frac{5\pi}6).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$.
For variance of $Y$, I need to find $E(Y^2)$ first.
This is what I have with U-substitution: $E(Y^2) = \frac{n\theta_{2} - n \theta_{1} + 2n \theta_{1} \theta_{2} - n\theta_{1}^{2} + \theta_{1}^{2}}{(n+1)}$.
But with this the $Var(Y)= E(Y^2) - \left [E(Y) \right ]^{2} \neq n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$
My mistake is definitely the $E(Y^2)$ part, so I would appreciate it if someone here can go over it with me.
Edit:
$$\begin{eqnarray}
\mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$ from below and also my own work.
But now how do I simplify $$Var(Y) = \mathbb{E}\left(Y^2\right) - \left [ \mathbb{E}\left(Y\right) \right ]^{2}$$ to $$Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}?$$
| Instead of $\mathbb{E}(Y^2)$ consider
$$
c_2 =\mathbb{E}\left(\left(Y-\theta_1\right)^2\right) = \int_{\theta_1}^{\theta_2} \frac{n}{\left(\theta_2-\theta_1\right)^n} \left(y-\theta_1\right)^{n+1} \mathrm{d} y = \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2
$$
Similarly
$$
c_1 = \mathbb{E}\left(\left(Y-\theta_1\right)\right) = \frac{n}{n+1} \left(\theta_2-\theta_1\right)
$$
Now clearly
$$\begin{eqnarray}
\mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trig limit without L'Hospital Rule: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$ I'm really getting stuck on this and would appreciate some help:
$$
\lim_{x\ \to\ 0}\left[\,\tan\left(\,x\,\right) - \sin\left(\,x\,\right) \over x^{3}\,\right]
$$
I know I need to change $\tan\left(\,x\,\right)$ into $\sin\left(\,x\,\right)/\cos\left(\,x\,\right)$ and turn $\sin\left(\,x\,\right)$ into $1 - \cos^{2}\left(\,x\,\right)$.
But then I get stuck.
| $$\begin{align}\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}&=\lim_{x\to 0}\frac{\sin x-\sin x\cos x}{x^3\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{\sin^2x}{x^2(1+\cos x)}\cdot\frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x}\cdot\frac{1}{\cos x}\\&=1\cdot 1^2\cdot \frac 12\cdot \frac{1}{1}\\&=\frac 12.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $a_{n+1}=a_n+\frac1{a_n}$, then $a_n/n$ converges to $0$ Let $a_{n+1}=a_n+\dfrac1{a_n}$, with $a_n=1$.
Prove $\lim \limits_{n\to \infty }\left(\dfrac{a_n}{n}\right)=0$.
Now I already know that it is monotonically increasing and that $a_n\to \infty$ as $n\to \infty$.
I thought of using Cauchy here, but I don't know how exactly.
NOTE: Stolz–Cesàro theorem is forbidden in this question.
Could anyone help me with this, please?
| If you know that $a_n \to \infty,$ you can apply Stolz theorem to obtain
$$\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{(n+1)-n} = \lim_{n \to \infty} \frac{1}{a_n} = 0.$$
Edit: I'm adding solution which doesn't use Stolz theorem.
Let $\varepsilon > 0$. Since $a_n \to \infty$, there is such $N$ that $a_n \geqslant \frac{2}{\varepsilon}$ for $n \geqslant N$. Thus
$$a_{N+k+1} = a_{N+1} + \frac{1}{a_{N+1}} + \ldots + \frac{1}{a_{N+k}} \leqslant a_{N+1} + k \cdot \frac{1}{a_{N+1}} \leqslant a_{N+1} + k \cdot \frac{\varepsilon}{2}$$
so
$$\frac{a_{N+k+1}}{N+k+1} \leqslant \frac{a_{N+1}}{N+k+1} + \frac{k}{N+k+1} \cdot \frac{\varepsilon}{2} \leqslant \frac{a_{N+1}}{N+k+1} + \frac{\varepsilon}{2}$$
For large $k$ we have $\frac{a_{N+1}}{N+k+1} < \frac{\varepsilon}{2}$ and therefore $a_{N+k+1} < \varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.
How can I proceed?
| $$y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C$$
$$\implies\cos^2C-\cos(A-B)\cos C+y=0$$ which is Quadratic Equation in $\cos C$
As $C$ is real $\implies\cos C$ is real,
the discriminant $\cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Simplest way to integrate $\int \frac{1}{1+\tan x}dx,$ $$\int \frac{1}{1+\tan x}dx,$$
A substitution like $t = \tan x, \;dt = (1+t^2)dx$ etc. immediately comes to mind, but I find this method a bit lengthy with the partial fractions. Is there a more concise solution to this?
| Generalization:
For $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx,$ where at least one of $a,b$ is non-zero
write $a\cos x+b\sin x=A(c\sin x+d\cos x)+B\dfrac{d(c\sin x+d\cos x)}{dx}\ \ \ \ (1)$
So, $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx=A\int\ dx+B\dfrac{d(c\sin x+d\cos x)}{(c\sin x+d\cos x)dx}dx$
$=Ax+\int\dfrac{d(c\sin x+d\cos x)}{(c\sin x+d\cos x)}=Ax+B\ln|c\sin x+d\cos x|+K$
Now from $(1)$,
$a\cos x+b\sin x=A(c\sin x+d\cos x)+B(c\cos x-d\sin x)=(Ac-Bd)\sin x+(Ad+Bc)\cos x$
and equating the coefficients of $\cos x,\sin x;$
$Ac-Bd=b,Ad+Bc=a$
This simultaneous equation can be easily solved for $A,B$ in terms of $a,b,c,d$(given)
Can you recognize $a,b,c,d$ here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? I only got that $3^n+5^n=8k$ when $n$ is odd. How to solve this one?
| Let $A_n=5^n+3^n,$
If $d$ divides $A_m,A_{m-1}$
$d$ will divide $A_m-3A_{m-1}=2\cdot5^{m-1}$
$d$ will divide $5A_{m-1}-A_m=2\cdot3^{m-1}$
So, $d$ will divide $(2\cdot3^{m-1},2\cdot5^{m-1})=2(3,5)^{m-1}=2$
So, the necessary condition for $A_{m-1}\mid A_m$ is $: A_{m-1}$ must divide $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$ when $\sin 2x = \cos x$ where $ x$ is in the domain $ [0, 2\pi]$ Quick question on trig (which I haven't dealt with in a long time):
since $\sin 2x = 2\sin x\cos x $
$2\sin x\cos x = \cos x$
$2\sin x\cos x/\cos x = 1$
$\sin x = 1/2$
since $\sin x = 1/2$ in quadrants $1$ and $2$, $x = \pi/6$ and $x = 5\pi/6$
Is this correct? If not, hint please.
| Generalized Solution :
$$\sin ax=\cos bx$$ for example, $\sin\dfrac{3x}2=\cos5x$ can also be addressed this way
Method $\#1:$
$$\cos bx=\sin\left(\frac\pi2-bx\right)\implies\sin ax=\sin\left(\frac\pi2-bx\right)$$
$$\implies ax=n\pi+(-1)^n\left(\frac\pi2-bx\right)$$ where $n$ is any integer
If $n$ is odd, $=2m+1$(say), $ax=(2m+1)\pi-\left(\dfrac\pi2-bx\right)\implies x=\dfrac{(4m+1)\pi}{2(a-b)}$
If $n$ is even, $=2m$(say), $ x=\dfrac{(4m+1)\pi}{2(a+b)}$
Now we need to find the values of $m$ such that $0\le x\le2\pi$
Method $\#2:$
$$\sin ax=\cos\left(\frac\pi2-ax\right) \implies\cos bx=\cos\left(\frac\pi2-ax\right)$$
$$\implies bx=2r\pi\pm\left(\frac\pi2-ax\right)$$ where $r$ is any integer
Check for '+', '-' sign one by one
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Intuition for Euler's Partition Theorem Euler's Partition Theorem states the following:
Every number has as many integer partitions into odd parts as into distinct parts.
I played around with small examples (I wrote out the partitions into odd/distinct parts for integers 1 through 10), but can't develop an intuition for why this is true. Any help?
| You can set up a one-to-one correspondence (aka a bijection) between the partitions of $N$ into odd parts and the partitions of $N$ into distinct parts. I will just illustrate with an example; I leave it to you to formalize it and verify that it works.
Here is one way to partition the number $N=863$ into odd parts: partition it into $22$ ones, $49$ threes, $48$ fives, $25$ sevens, and $31$ nines. From this we derive a partition of $863$ into distinct parts, as follows:
$863$
$=22\cdot1+49\cdot3+48\cdot5+25\cdot7+31\cdot9$
$=(16+4+2)\cdot1+(32+16+1)\cdot3+(32+16)\cdot5+(16+8+1)\cdot7+(16+8+4+2+1)\cdot9$
$=16\cdot1+4\cdot1+2\cdot1+32\cdot3+16\cdot3+1\cdot3+32\cdot5+16\cdot5+16\cdot7+8\cdot7+1\cdot7+16\cdot9+8\cdot9+4\cdot9+2\cdot9+1\cdot9$
$=16+4+2+96+48+3+160+80+112+56+7+144+72+36+18+9$
$=160+144+112+96+80+72+56+48+36+18+16+9+7+4+3+2$.
If we apply this procedure to the partitions of $7$ we get the following correspondence:
$7\ \to\ 7$
$5+1+1\ \to\ 5+2$
$3+3+1\ \to\ 6+1$
$3+1+1+1+1\ \to\ 4+3$
$1+1+1+1+1+1+1\ \to\ 4+2+1$
The relevant arithmetical facts here are unique factorization (every number factors uniquely into a power of $2$ and an odd number) and base $2$ notation (every number is uniquely expressible as a sum of distinct powers of $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(ac+bd)^2 + (ad-bc)^2 \geq 144$ , if $a+b=4, c+d=6$. I got that $(ac)^{2}+(bd)^{2}+(ad)^{2}+(bc)^{2} \geq 144$. How is this one solved?
| $$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$$
Method $\#1:$
Assuming $a,b$ are real,
$(a-b)^2\ge0\iff a^2+b^2\ge2ab\iff2(a^2+b^2)\ge(a+b)^2=4^2$
Similarly, for $c^2+d^2$
Method $\#2:$
Assuming $a,c$ are real,
$a^2+b^2=a^2+(4-a)^2=2(a^2-4a+8)=2[(a-2)^2+4]\ge2\cdot4$
Similarly, for $c^2+d^2$
See also: Brahmagupta-Fibonacci Identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt$
Now Using Partial fraction,
$\displaystyle \frac{1}{(1+t)\cdot (1+t^2)} = \frac{A}{1+t}+\frac{Bt+C}{1+t^2}\Rightarrow 1=A(1+t^2)+(Bt+C)(1+t)$
Now put $(1+t)=0\Rightarrow t=-1\;,$ We get $\displaystyle 1=2A\Rightarrow A = \frac{1}{2}.$
Now Put $(1+t^2)=0\Rightarrow t^2=-1\;,$ We Get $1=Bt^2+(B+C)t+C$
So $\displaystyle 1=\left(-B+C\right)+(B+C)$. So Solving equation...$B+C=0$ and $-B+C=1$
So $\displaystyle B=-\frac{1}{2}$ and $\displaystyle C=\frac{1}{2}$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt = \int\frac{1}{1+t}dt+\int\frac{-t+1}{1+t^2}dt$
So $\displaystyle I = \frac{1}{1+t}dt-\frac{1}{2}\int\frac{2t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$
So $\displaystyle I = \ln \left|1+t\right|-\frac{1}{2}\ln \left|1+t^2\right|+\tan^{-1}(t)+\mathcal{C}$
So $\displaystyle I = \ln \left|1+\tan \frac{x}{2}\right|-\frac{1}{2}\ln \left|1+\tan^2 \frac{x}{2}\right|+\frac{x}{2}+\mathcal{C}$
Can we solve it without using partial fraction? If yes then please explain to me.
Thanks
| Set $t=\dfrac{x}{2}$, then multiply the integrand by $\cos t$, we get
\begin{equation}
\frac{2\cos t}{\cos t+\sin t}
\end{equation}
then let
\begin{equation}
I=\int\frac{2\cos t}{\cos t+\sin t}dt
\end{equation}
and
\begin{equation}
J=\int\frac{2\sin t}{\cos t +\sin t}dt
\end{equation}
Find $I+J$ and $I-J$, where $I-J$ can be found by using
\begin{equation}
\int\frac{f'(x)}{f(x)}dx=\ln\left|f(x)\right|+C
\end{equation}
We have
\begin{equation}
I+J = 2t+C
\end{equation}
and
\begin{equation}
I-J = 2\ln|\cos t +\sin t|+C
\end{equation}
So
\begin{equation}
I =\int\frac{1}{\tan \frac{x}{2}+1}dx = \frac{x}{2}+\ln \left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|+C
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
The convergence of an infinite seqeunce Suppose that
$$
a_n = \prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right),
$$
for $n \geq 2$. How can we show that
$$
\lim_{n \to \infty} a_n = \lim_{n \to \infty}\prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right) = 1?
$$
Thanks very much.
| Consider the following.
Method 1
The product is
\begin{align}
\prod_{k=n}^{\infty} \left(1 - \frac{1}{k^2}\right) = \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) }
\end{align}
for which
\begin{align}
\lim_{n \rightarrow \infty} \, \prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right) &= \lim_{n \rightarrow \infty} \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) } \\
&= \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) } = 1
\end{align}
Method 2
It can be determined that
\begin{align}
\prod_{k=2}^{m} \left(1 - \frac{1}{k}\right) &= \frac{1}{m} \\
\prod_{k=2}^{m}\left(1 + \frac{1}{k}\right) &= \frac{m+1}{2}
\end{align}
such that
\begin{align}
\prod_{k=2}^{m} \left( 1 - \frac{1}{k^{2}} \right) = \frac{m+1}{2 m}.
\end{align}
Now,
\begin{align}
\lim_{n \rightarrow \infty} \prod_{k=n}^{\infty} \left(1 - \frac{1}{k^2}\right) &= \lim_{m,n \rightarrow \infty} \prod_{k=n}^{2m} \left(1 - \frac{1}{k^2}\right) \\
&= \lim_{m,n \rightarrow \infty} \frac{ \prod_{k=2}^{2m} \left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) } \\
&= \lim_{m,n \rightarrow \infty} \, \frac{ \frac{2m+1}{4m} }{ \frac{n}{2(n-1)} } \\
&= \lim_{m,n \rightarrow \infty} \left( 1 + \frac{1}{2m} \right) \left( 1 - \frac{1}{n} \right) = 1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove using induction principles $$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$
For any fixed value of $a > 1$.
Induction step:
$$\sum_{k=1}^{2^{n+1} - 1} \frac{1}{k^a} = (\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a} \leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a}$$
I need help from Induction step and on. So if someone would help me, that would be greatly appreciated! People on this website keep putting this problem on hold even though I have clarified it as much as I can.
I need to prove P(n+1) is true:
$$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^{n+1}-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{(n+1)(1-a)}}{1-2^{1-a}}\right)$$
| Ellaborating on the hint:
Notice that $$\frac{1 - 2^{n(1-a)}}{1 - 2^{1-a}} = 1 + 2^{1-a} + \ldots + 2^{(n-1)(1-a)}$$
and that
$$\frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \ldots + \frac{1}{(2^{n+1} - 1)^a} \leq 2^n \cdot \frac{1}{(2^n)^a}.$$
Now
$$\sum_{k=1}^{2^{n+1}-1}\frac{1}{k^a} = \sum_{k=1}^{2^{n}-1}\frac{1}{k^a} + \sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k^a}$$
By the induction hypothesis,
\begin{align*}
\sum_{k=1}^{2^{n+1}-1}\frac{1}{k^a} &\leq \frac{1 - 2^{n(1-a)}}{1 - 2^{1-a}} + \sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k^a} \\
&\leq 1 + 2^{1-a} + \ldots + 2^{(n-1)(1-a)} + 2^{n(1-a)}.
\end{align*}
Now what does this final expression evaluate to? It is the sum of a certain geometric progression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 + \log_6 256\\
x + \log_6 4 &= x\log_64 + \log_6 256\\
x-x\log_6 4 &= \log_6 256 - \log_6 4\\
x(1-\log_6 4) &= \log_6 256 - \log_6 4\\
x &= \dfrac{\log_6 256 - \log_6 4}{1-\log_6 4}\\
x &= 10.257}$$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..
| I think the problem with your solution might be here:
$2^2x * 3^x => 2^2 * 2*x * 3^x$
$4^{3x+1} => 4^3 * 4^x * 4$
since $a^{bc} = (a^b)^c$ and not $a^b * a^c$. So for instance:
$4^{3x+1} => 4^x * 4^x * 4^x * 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$ What is the inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$?
Using the residues, I can calculate the residues at $s_n=2n\pi i$, but I have problem in calculating residue at $s=0$.
The final answer should be:
$f(t)=\frac{1}{6}x(x^2-1)+xt+\frac{2}{\pi ^3}\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}e^{-n^2\pi ^2 t}\sin{n\pi x}$
which I could get the last term from the residues at $s_n=2n\pi i$, but I have problem for the first two terms.
| First of all, the poles you are thinking of are at $s=-n^2 \pi^2$. ($s=0$ is a removable pole of the sinh terms.)
Also note that, even though there are square roots in the sinh terms, we need not worry about taking branches because any odd phase behavior will cancel in the fraction. Therefore, what we have is a straightforward Bromwich contour encircling the poles.
The residue from the pole at $s_n=-n^2 \pi^2$ is simply
$$\frac1{s_n^2} \frac{\sinh{x \sqrt{s_n}}}{\cosh{\sqrt{s_n}}} 2 \sqrt{s_n} e^{s_n t} = \frac{2 (-1)^{n+1}}{n^3 \pi^3} e^{-n^2 \pi^2 t} \sin{n \pi x}$$
The residue at $s=0$ is a little trickier but still straightforward; it is equal to
$$\begin{align}\left [\frac{d}{ds} \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} e^{s t} \right ]_{s=0} &= \left [\frac{d}{ds} \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} \right ]_{s=0} + t \left [ \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} \right ]_{s=0}\\ &= \lim_{s\to0} \frac{x \sinh{\sqrt{s}} \cosh{x\sqrt{s}}-\sinh{x \sqrt{s}} \cosh{\sqrt{s}}}{2 \sqrt{s} \sinh^2{\sqrt{s}}} + t x \\ &= \lim_{s\to0} \frac{\frac{x}{2} \left (1+\frac{s}{6} \right ) \left (1+\frac{x^2 s}{2} \right )-\frac{x}{2} \left (1+\frac{x^2 s}{6} \right ) \left (1+\frac{s}{2} \right )}{s}+ t x \\ &= \frac16 (x^3-x) + x t \end{align}$$
The sought-after result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/965257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$?
$\sqrt{n^2-4} \in \mathbb{Q}
\\
\sqrt{n^2-4} = \frac{p}{q}
\\
(\sqrt{n^2-4})^2 = \left(\frac{p}{q}\right)^2
\\
q^2\left( n^2-4\right)=p^2
\\
\text{p is divisible by} \left (n^2-4 \right)
\Rightarrow p=k\left (n^2-4 \right)
\\
q^2 \left (n^2-4 \right)=k^2 \left(n^2-4 \right)^2
\\
q^2=k^2 \left (n^2-4 \right )\Rightarrow
\text{it follows that q is also divisible by} \left (n^2-4 \right)
\\
\text{Therefore p and q are not co-prime therefore} \Rightarrow \sqrt{n^2-4} \notin \mathbb{Q} \
\square
$
| The fault is when you deduce that $p$ is divisible by $n^2-4$ from the fact that
$$
q^2(n^2-4)=p^2
$$
which only implies that $p^2$ is divisible by $n^2-4$.
This would be true if $n^2-4$ is prime, but it isn't in general. For instance, if $n=4$ and $p=6$, $p^2=36$ is divisible by $n^2-4=12$, but $6$ is not divisible by $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/965390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \frac{x^2 - x}{x^2 +x + 1}dx$ I know the answer is $x - \ln|x^2 + x + 1|$ but I don't understand how to get it. Its in the partial fraction decomposition section of homework. The way the homework worked it is like this...
$$
\int \frac{2x+1}{x^2 + x + 1} \, dx
$$
I see they split it up and got the derivative of $x^2 + x + 1$ in the numerator, but I don't understand how they did that or where the other interal dx came from.
| Suppose you have
$$
\frac{\text{some polynomial}}{x^2 + x + 1}.
$$
Do long division and get
$$
\text{some polynomial} + \frac{\text{some first-degree polynomial}}{x^2+x+1}.
$$
In this case you get
$$
1 + \frac{-2x-1}{x^2+x+1}.
$$
First degree is the derivative of second degree, so $\dfrac{d}{dx}(x^2+x+1) = 2x+1$. So
\begin{align}
u & = x^2 + x + 1, \\[6pt]
du & = (2x+1)\,dx.
\end{align}
Then we have
$$
\int \left(1 - \frac 1 u \right) du.
$$
Somewhat more typically, one might have something like
$$
\frac{3x+5}{x^2+x+1}.
$$
Then one would write
$$
3x+5 = \frac 3 2 \left(2x + \frac{10}3\right)= \frac 3 2 \left((2x+1) + \frac 7 3\right)
$$
so we get
$$
\int\frac{3x+5}{x^2+x+1} \,dx=\frac 3 2 \int \frac{du}{u} + \frac 7 2 \int \frac{dx}{x^2+x+1}.
$$
That last integral would then need to get done by a different method, whose first step is completing the square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the oblique asymptote of: Given $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}$$ how would you find the oblique asymptote of that?
| When $x$ is large, $x^2+1 \approx x^2$ and $\sqrt{x+1} \approx \sqrt{x}$. So, for large values of $x$, $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}\approx x^{3/2}$$ I suppose that you consider "oblique" in a very extended way.
More precise answers could be obtained writing $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}=\frac{x^2+1}{\sqrt{x} \sqrt{1+\frac1x}}$$ and considering the Taylor expansion $$\frac{1}{\sqrt{1+y}}=1-\frac{y}{2}+\frac{3 y^2}{8}+O\left(y^3\right)$$ Replace $y$ by $\frac{1}{x}$ and obtain, for large values of $x$ $$f(x)=\frac{x^2+1}{(x+1)^{\frac{1}{2}}}=x^{3/2}-\frac{\sqrt{x}}{2}+\frac{11 \sqrt{\frac{1}{x}}}{8}-\frac{13}{16}
\left(\frac{1}{x}\right)^{3/2}+O\left(\left(\frac{1}{x}\right)^{5/2}\right)$$ So the asymptotic curve is $$g(x)=x^{3/2}-\frac{\sqrt{x}}{2}$$ and it is approached from above when $x$ tends to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry equation with arctan Solve the following equation: $\arctan x + \arctan (x^2-1) = \frac{3\pi}{4}$.
What I did
Let $\arctan x = \alpha, \arctan(x^2-1) = \beta$, $\qquad\alpha+\beta = \frac{3\pi}{4}$
$\tan(\alpha+\beta) = \tan(\frac{3\pi}{4}) = -1$
$$\frac{\tan\alpha + tan\beta}{1-\tan\alpha\tan\beta} = \frac{x+x^2-1}{1-x(x^2-1)} = -1$$
$\begin{align}
x^2+x-1 &= -(1-(x^3-x)) = -1+x^3-x \\
\iff x^2 + x &= x^3-x \\
\iff x(x+1) &= x(x^2-1) \qquad\implies \boxed{x_1 = 0}\\
\implies x+1 &= x^2-1 \\
\iff x^2-x-2 &= 0 \\
\end{align}$
$\therefore x_1 = 0,\quad x_2 = 2,\quad x_3 = -1$
However, the equation only works for $x=2$.
I wonder
Did I do this in an efficient manner? Is there any easy way to find $x$ where there's no fake solutions?
| Liek Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$,
$$\arctan x+\arctan(x^2-1)=\begin{cases} \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) &\mbox{if } x(x^2-1)<1 \\\pi+ \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) & \mbox{if } x(x^2-1)> 1. \end{cases} $$
Now, $-\dfrac\pi2\le\arctan(z)\le\dfrac\pi2$
$\implies-\dfrac\pi2\le\arctan x+\arctan(x^2-1)\le\dfrac\pi2$ if $x(x^2-1)<1$ which is true if $x=0,-1$
Then $\arctan x+\arctan(x^2-1)\ne\dfrac{3\pi}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx $ I wonder if this integral $$\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx \quad n=1,2,3,...$$ admits a general formula for integers $n$. I've found $$\int_{0}^{1}\frac{\ln{(1+x)}}{1+x} dx = \left[ \frac{1}{2}\left(\ln(1+x)\right)^2 \right]_0^1=\frac{1}{2}\left(\ln2\right)^2. $$ Using integration by parts twice,
$$\begin{align}
\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^2 dx &= \left[ \frac{-1}{(1+x)}\left(\ln(1+x)\right)^2 \right]_0^1-\int_{0}^{1}\frac{\ln{(1+x)}}{\left(1+x\right)^2} dx \\
&=-\frac{1}{2}\left(\ln2\right)^2+ \left[ \frac{1}{(1+x)}\ln(1+x) \right]_0^1-\int_{0}^{1}\frac{1}{\left(1+x\right)^2} dx\\
&=-\frac{1}{2}\left(\ln2\right)^2+\frac{1}{2}\ln2-\frac{1}{2}.
\end{align}$$
Could you find a general formula? Any help is welcome! Thank you.
| Result:
$$\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x=\frac{n!}{(n-1)^{n+1}}-\frac{n!}{2^{n-1}}\sum^n_{j=0}\frac{\ln^{j}{2}}{j!(n-1)^{n-j+1}}$$
$\text{for $n\in \mathbb{N}$, $n\geq2$}.$
Derivation:
\begin{align}
\small{\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x}
&\small{=\int^2_1\frac{\ln^n{x}}{x^n}{\rm d}x\tag1}\\
&\small{=(-1)^n\int^1_{\frac{1}{2}}x^{n-2}\ln^n{x}\ {\rm d}x\tag2}\\
&\small{=(-1)^n\frac{\partial^n}{\partial n^n}\int^1_{\frac{1}{2}}x^{n-2}{\rm d}x}\\
&\small{=(-1)^n\frac{\partial^n}{\partial n^n}\left[\frac{1}{n-1}-\frac{1}{2^{n-1}(n-1)}\right]}\\
&\small{=(-1)^n\left[\frac{(-1)^nn!}{(n-1)^{n+1}}-\sum^n_{j=0}\binom{n}{j}\frac{(-1)^{n-j}(n-j)!}{(n-1)^{n-j+1}}\frac{(-1)^{j}\ln^{j}(2)}{2^{n-1}}\right]\tag3}\\
&\small{=\frac{n!}{(n-1)^{n+1}}-\frac{n!}{2^{n-1}}\sum^n_{j=0}\frac{\ln^{j}{2}}{j!(n-1)^{n-j+1}}}
\end{align}
Explanation:
$(1):\text{Let $x\mapsto x-1$}$
$(2):\text{Let $x\mapsto x^{-1}$}$
$(3):\text{Apply Leibniz Generalised Product Rule}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to get tangent of inverse of curve?? Ok so my question is. Let $ f(x)=(1/7)x^3+21x-1.$ and let y=g(x) be the inverse function of f. Determine all points on the graph of the inverse function g so that the tangent line is perpendicular to the straight line $ y=-42x+4. $There are two point $P(x1,y1)$ and $Q(x2,y2)$ where y1
So i put the equation into wolfram and got a very long inverse.
$-(7 2^(1/3) 7^(2/3))/(1+x+sqrt(9605+2 x+x^2))^(1/3)+(7/2)^(1/3) (1+x+sqrt(9605+2 x+x^2))^(1/3) $
now i took the derivative of this
$(14 2^(1/3) 7^(2/3) + 2^(2/3) 7^(1/3) (1 + x + sqrt[9605 + 2 x + x^2])^(2/3))/(6 Sqrt[9605 + 2 x + x^2] (1 + x + Sqrt[9605 + 2 x + x^2])^(1/3))$
i equated the derivative to $1/42$
and solved for x and got $x=195$ and $x=-197$ is this correct so far? and how do i solve for the rest?
| The curve $y=g(x)$ has equation that can be written in implicit form as
$$\frac{y^3}{7}+21y-1=x.$$
Differentiate. We get
$$\frac{dy}{dx}\left(\frac{3y^2}{7}+21\right)=1.\tag{1}$$
We want the tangent line to $y=g(x)$ to be perpendicular to a line with slope $-42$. So we want the tangent line to $y=g(x)$ to have slope $\frac{1}{42}$.
Set $\frac{dy}{dx}=\frac{1}{42}$ in (1) and solve for $y$. We get after a while $y=\pm 7$. Now we can find the appropriate values of $x$.
Remark: The curve $y=g(x)$ is obtained by reflecting the curve $y=f(x)$ in the line $y=x$. So if a certain tangent line to $y=g(x)$ has slope $\frac{1}{42}$, then after reflection the correspnding tangent line to $y=f(x)$ has slope $42$, the reciprocal. It is easy to find the points $(a,b)$ on $y=f(x)$ where the tangent line has slope $42$. Now reflect back, and we obtain the appropriate points for $y=g(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$, prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$ Given $\sqrt{1+x} < 1 +0.5x$ for all $x>0$,
prove that $\sqrt{1+x} > 1 + 0.5x - 0.125x^2$
Im thinking along the lines of binominal expansion as $\sqrt{1+x} = 1 + 0.5x - 0.125x^2 + ...$
But how im not sure how to continue
| You do not have a calculus tag, so the first solution below may be inappropriate.
First Solution: Let
$$f(x)=\sqrt{1+x}-(1+0.5x-0.125x^2).$$
We want to show that $f(x)\gt 0$ if $x\gt 0$. Note that $f(0)=0$. We will show that $f(x)$ is increasing.
We have
$$f'(x)=\frac{1}{2\sqrt{1+x}}-(0.5-0.25 x).$$
We will show that $f'(x)\gt 0$ if $x\gt 0$.
Since $\sqrt{1+x}\lt 1+0.5x$, we have
$\frac{1}{2\sqrt{1+x}}\gt \frac{1}{2+x}$. We will show that
$$\frac{1}{2+x}-(0.5-0.25x)\gt 0.$$
Bring to common denominator. We need to show that
$$1-(2+x)(0.5-0.25x)\gt 0.$$
This is clear, for $1-(2+x)(0.5-0.25x) =0.25x^2$.
Second Solution: If $1+0.5x-0.125x^2$ is negative the result is obvious.
Note that if $x\ge 8$, then $1+0.5x-0.125x^2$ is negative. For then $1+0.5x-0.125x^2\le 1+0.5x-x\le -3$.
So we may assume that $1+0.5x-0.125x^2$ is positive, and also that $x\lt 8$. It is then enough to show that $(1+0.5x-0.125x^2)^2\lt 1+x$. Do the squaring. We get
$$(1+x/2-x^2/8)^2=1+x+x^2/4-2(x^2/8)-2(x^3/16)+x^4/64.$$
We want to show this is $\lt 1+x$, so we need to show that
$$x^4/64-x^3/8\lt 0.$$
This is clearly true if $x\lt 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Property of greatest integer function I came across the following mathematical statement in a proof. Can somebody tell me which property of greatest integer function makes it possible?
$x + y - \lfloor x + y \rfloor + z - \lfloor x + y - \lfloor x + y \rfloor + z \rfloor = x + y - \lfloor x + y \rfloor + z - \lfloor x + y + z \rfloor + \lfloor x + y \rfloor$
How do we get $ \lfloor x + y - \lfloor x + y \rfloor + z \rfloor
= \lfloor x + y + z \rfloor - \lfloor x + y \rfloor$?
Thanks in advance...
| In general, if $N$ is an integer, then
$$\lfloor N+\alpha\rfloor=N+\lfloor\alpha\rfloor.$$
Proof : Let $N+\alpha=M+\beta$ where $M$ is an integer and $0\le \beta\lt 1$. Then, since we have
$$\alpha=M-N+\beta\Rightarrow \lfloor\alpha\rfloor=M-N,$$
we have
$$\begin{align}\lfloor N+\alpha\rfloor&=\lfloor M+\beta\rfloor\\&=M\\&=N+(M-N)\\&=N+\lfloor\alpha\rfloor.\end{align}$$
Here, setting $N=-\lfloor x+y\rfloor,\alpha=x+y+z$ gives us
$$\lfloor -\lfloor x+y\rfloor+(x+y+z)\rfloor=-\lfloor x+y\rfloor+\lfloor x+y+z\rfloor.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that a random variable has a Poisson distribution? If $X$ is a random variable whose PMF, $p_X(x)$, is positive on and only on non-negative integers. How can we show that $X$ has a Poisson distribution if
$$\begin{align*}
p_X(x) = \frac{3}{x} p_x(x-1) && x = 1, 2, 3, \ldots
\end{align*}$$
| $$
p_X(4) = \frac 3 4 p_X(3) = \frac 3 4 \cdot \frac 3 3 p_X(2) = \frac 3 4 \cdot \frac 3 3\cdot \frac 3 2p_X(1) = \frac 3 4 \cdot \frac 3 3 \cdot \frac 3 2 \cdot\frac 3 1p_X(0) = \frac{3^4 p_X(0)}{4!}.
$$
As with $4$, so with other positive integers (or a bit less informally: prove by induction that $p_X(x)= \dfrac{3^x p_X(0)}{x!}$).
To show that this is a Poisson distribution with expected value $3$, it remains only to show that $p_X(0)=e^{-3}$. So
$$
1=\sum_{x=0}^\infty \frac{3^x p_X(0)}{x!} = p_X(0)\sum_{x=0}^\infty \frac{3^x}{x!} = p_X(0) e^3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ Show $\displaystyle\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ for $|z|<1$.
This is an additional problem for my complex analysis class and I've attempted it for a few hours but ended up taking wrong routes. All of my attempts I haven't used complex analysis at all and I don't see how I could here.
edit: this is meant for a BEGINNER complex analysis course so please try keep the solutions to that (if you could)
Any help would be great
| $$\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots$$
Add $\displaystyle \frac{-z}{1-z}$ to both sides. It's Telescoping series:
$$\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\\ =\frac{-z-z^2}{1-z^2}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^2}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^4-z^2}{1-z^4}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\frac{-z^4}{1-z^4}+\frac{z^4}{1-z^8}+ \cdots$$
$\textbf{Edit:}$ If you want to have finite sum rather than infinity. You can show that (like above - by cancelling terms) :
$$\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z^{2^{m}}}{1-z^{2^{m}}}$$
Next calculate limit for $m \to \infty$ both sides.
$$\lim_{m \to \infty}\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}$$
Finally (because $|z|<1$):
$$\lim_{m \to \infty}\frac{-z^{2^{m}}}{1-z^{2^{m}}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
| $$a^3+3a^2(ib)-3ab^2+(ib)^3=8$$
Equating the real & the imaginary parts,
$$a^3-3ab^2=8;3a^2b-b^3=0\iff b(3a^2-b^2)=0$$
Case $\#1:b=0\implies a^3=8\implies a=2$
Case $\#2:3a^2-b^2=0\iff b^2=3a^2$
and we have $a(a^2-3b^2)=8\implies a[a^2-3(3a^2)]=8\iff a^3=-1\implies a=-1$ as $a$ is real
$\implies b^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 3
} |
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
| I'm sorry but AM-GM Inequality is all that I know,
So, By using AM-GM Inequality for $ (a+b) , (b+c) , (c+a)$ and
$\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$
We get
$$\frac{(a+b)+(b+c)+(c+a)}{3}
\ge\sqrt[3]{(a+b)(b+c)(c+a)}\tag{1}$$
$$\frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3}
\ge\sqrt[3]{\frac{1}{(b+c)(c+a)(a+b)}}\tag{2}$$
Multiplying $(1)$ and $(2)$ We get.
$$\frac{1}{9}\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)}}$$
$$\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9$$
After Expanding
$$3+\frac{a+b}{b+c}+\frac{c+a}{b+c}+
\frac{a+b}{c+a}+\frac{b+c}{c+a}+
\frac{b+c}{a+b}+\frac{c+a}{a+b}
\ge\ 9$$
By rearrangement
$$\left(\frac{a+b}{b+c}+\frac{c+a}{b+c}\right)+
\left(\frac{a+b}{c+a}+\frac{b+c}{c+a}\right)+
\left(\frac{b+c}{a+b}+\frac{c+a}{a+b}\right)
\ge\ 6$$
$$\frac{2a+b+c}{b+c}+
\frac{a+2b+c}{c+a}+
\frac{b+2c+a}{a+b}
\ge\ 6$$
$$\frac{2a}{b+c}+\frac{b+c}{b+c}+
\frac{2b}{c+a}+\frac{a+c}{c+a}+
\frac{2c}{a+b}+\frac{b+a}{a+b}
\ge\ 6$$
$$3+\frac{2a}{b+c}+
\frac{2b}{c+a}+
\frac{2c}{a+b}
\ge\ 6$$
$$\frac{2a}{b+c}+
\frac{2b}{c+a}+
\frac{2c}{a+b}
\ge\ 3$$
$$\frac{a}{b+c}+
\frac{b}{c+a}+
\frac{c}{a+b}
\ge\ \frac{3}{2}$$
As Simple as that
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Hermite Polynomials and Brownian motion I am asked to prove the following :
Let $B_t$ be a standard brownian motion.
The $n$th Hermite polynomial is $\displaystyle H_n(t,x)=\frac{(−t)^n}{n!} e^{x^2/(2t)} \frac{d^n}{dx^n}e^{-x^2/(2t)}$.
Show that the $H_n$ play the role that the monomials $\dfrac{x_n}{n!}$ play in ordinary calculus.
In other words show : $dH_{n+1}(t, B_t) = H_n(t, B_t)$
Now my guess is that we can take the integral of both sides and use ito integration ? But I am not sure how to proceed from there.
| First of all, the statement
$$dH_{n+1}(t,B_t) = H_n(t,B_t)$$
doesn't make sense. Or can you explain what you mean by it? I guess, it should read
$$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t;$$
at least that's what I prove in this answer.
Lemma 1 $$H_{n+1}(t,x) = \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1} H_{n-1}(t,x). \tag{0}$$
Proof: Denote by $$G_n(y) := (-1)^n \exp \left( \frac{y^2}{2} \right) \frac{d^n}{dy^n} \exp \left(- \frac{y^2}{2} \right)$$ the "standard" Hermite polynomial. Then it follows from the chain rule that $$H_n(t,x) = t^{n/2} \frac{1}{n!} G_n \left( \frac{x}{\sqrt{t}} \right). \tag{1}$$ Using the recursion relation $$G_{n+1}(y) = y G_n(y)- n G_{n-1}(y) \tag{2}$$ we find $$\begin{align*} H_{n+1}(t,x) &\stackrel{(1)}{=} t^{(n+1)/2} \frac{1}{(n+1)!} G_{n+1} \left( \frac{x}{\sqrt{t}} \right) \\ &\stackrel{(2)}{=} t^{(n+1)/2} \frac{1}{(n+1)!} \left[ \frac{x}{\sqrt{t}} G_n \left( \frac{x}{\sqrt{t}} \right) - n G_{n-1} \left( \frac{x}{\sqrt{t}} \right) \right] \\ &\stackrel{(1)}{=} \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1}H_{n-1}(t,x). \end{align*}$$
Lemma 2 $$\frac{\partial}{\partial t} H_n(t,x) = - \frac{1}{2} H_{n-2}(t,x).$$
Proof: It follows straight from the product rule that
$$\begin{align*}
\frac{\partial}{\partial t} H_n(t,x) &= - \frac{(-t)^{n-1}}{(n-1)!} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \exp \left(-\frac{x^2}{2t} \right)\\
&\quad - \frac{(-t)^n}{n!} \frac{x^2}{2t^2} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \exp \left(-\frac{x^2}{2t} \right) \\
&\quad + \frac{(-t)^n}{n!} \frac{1}{2t^2} \exp \left(\frac{x^2}{2t}\right) \frac{d^n}{dx^n} \bigg( x^2 \exp \left(-\frac{x^2}{2t} \right)\bigg). \tag{3}
\end{align*}$$
Applying Leibniz' product rule
$$\frac{d^n}{dx^n} (f \cdot g) = \sum_{k=0}^n {n \choose k} f^{(k)}(x) g^{(n-k)}(x)$$
we get
$$\begin{align*}
\frac{d^n}{dx^n} \bigg( x^2 \exp \left(-\frac{x^2}{2t} \right)\bigg) &= x^2 \frac{d^n}{dx^n} \exp \left( - \frac{x^2}{2t} \right) + 2nx \frac{d^{n-1}}{dx^{n-1}} \exp \left( - \frac{x^2}{2t} \right) \\ &\quad + (n-1) n \frac{d^{n-2}}{dx^{n-2}} \exp \left( - \frac{x^2}{2t} \right) \tag{4}
\end{align*}$$
Plugging $(4)$ into $(3)$ yields$$\begin{align*} \frac{\partial}{\partial t} H_n(t,x) &= - \frac{(-t)^{n-1}}{(n-1)!} \exp \left( \frac{x^2}{2t} \right) \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right) \\ &\quad + \frac{1}{2} \frac{(-t)^{n-2}}{(n-2)!}\exp \left( \frac{x^2}{2t} \right) \frac{d^{n-2}}{dx^{n-2}} \exp \left(- \frac{x^2}{2t} \right) \\ &= \frac{n}{t} H_n(t,x) + \frac{1}{2} H_{n-2}(t,x) \\ &\stackrel{(0)}{=} \frac{n}{t} \left( \frac{x}{n} H_{n-1}(t,x) - \frac{t}{n} H_{n-2}(t,x) \right)+ \frac{1}{2} H_{n-2}(t,x) = - \frac{1}{2} H_{n-2}(t,x). \end{align*}$$
Lemma 3 $$\frac{\partial}{\partial x} H_{n}(t,x) = H_{n-1}(t,x).$$
Proof: $$\begin{align*} \frac{\partial}{\partial x} H_n(t,x) &= \frac{(-t)^n}{n!} \exp \left( \frac{x^2}{2t} \right) \bigg[ \frac{x}{t} \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right) + \frac{d^{n+1}}{dx^{n+1}} \exp \left(- \frac{x^2}{2t} \right) \bigg] \\ &= \frac{x}{t} H_n(t,x) + \frac{n+1}{-t} H_{n+1}(t,x) \\ &\stackrel{(0)}{=} H_{n-1}(t,x). \end{align*}$$
Theorem $$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t$$
Proof: The claim follows directly from Itô's formula and Lemma 2+3. Note that Lemma 3 implies $$\frac{\partial^2}{\partial^2 x} H_{n+1}(t,x) = \frac{\partial}{\partial x} H_n(t,x) = H_{n-1}(t,x). $$
Remark Note that this shows in particular that $(H_n(t,B_t))_{t \geq 0}$ is a martingale (see also this question).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
$$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$
But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$
Where did I make a error in my calculation?
Thanks!
| You failed in $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{18}{3}$, actually $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{2}{3}\cdot3 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$.
Also write the identity used
| use the sum of cubes of continuous integers: $\dfrac{n^2 (n+1)^2}{ 4}$ , this is the same as $1^3 + 2^3 +\cdots+n^3$. You can use this. But you need to have coefficients of 1 in front of your cubes, so split the $(2)4^3=4^3+4^3$. Leave the negatives at the end and do the same thing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Understanding the Cholesky decomposition I'm attempting to understand the Cholesky decomposition via the following site: http://en.wikipedia.org/wiki/Cholesky_decomposition
If I have a matrix, say
$$A = \begin{bmatrix}
2 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 2\end{bmatrix},$$ then I'd like to use the Cholesky Algorithm to find a matrix $D$ such that $A = LDL^{T}$.
I left blank the parts of this example that I do not understand how to find.
The Cholesky Algorithm for this example follows:
Skipping a few steps since we don't need to derive how they get $L$, we know that $L = L_{1}L_{2}L_{3}$ since $n=3$.
For $i=1$,
$$L_1 = \begin{bmatrix}
I_{1-1} & 0 & 0\\
0 & \sqrt{a_{1,1}} & 0\\
0 & b_1/\sqrt{a_{1,1}} & I_{3-1}\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & \sqrt{2} & 0 & 0\\
0 & b_1/\sqrt{2} & 1 & 0\\
0 & 0 & 0 & 1\end{bmatrix}$$
For $i=2$,
$$L_ = \begin{bmatrix}
I_{2-1} & 0 & 0\\
0 & \sqrt{a_{2,2}} & 0\\
0 & b_2/\sqrt{a_{2,2}} & I_{3-2}\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 \\
0 & \sqrt{2} & 0\\
0 & b_2/\sqrt{2} & 1\end{bmatrix}$$
For $i=n = 3$,
$$L_3 = \begin{bmatrix}
I_{3-1} & 0 & 0\\
0 & \sqrt{a_{3,3}} & 0\\
0 & b_3/\sqrt{a_{3,3}} & I_{3-3}\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & \sqrt{2} & 0\\
0 & 0 & b_3/\sqrt{2} & 0\end{bmatrix}$$.
Thus,
\begin{align}
L &:= L_1\cdot L_2\cdot L_3\\\notag
&= \begin{bmatrix}
0 & 0 & 0 & 0\\
0 & \sqrt{2} & 0 & 0\\
0 & b_1/\sqrt{2} & 1 & 0\\
0 & 0 & 0 & 1\end{bmatrix} \cdot
\begin{bmatrix}
1 & 0 & 0 \\
0 & \sqrt{2} & 0\\
0 & b_2/\sqrt{2} & 1\end{bmatrix} \cdot
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & \sqrt{2} & 0\\
0 & 0 & b_3/\sqrt{2} & 0\end{bmatrix}\\\notag
&= ?
\end{align}
I guess $b_i$ and $b_i^*$ come from $B$. But how do I find the non-Hermitian matrix $B$ to finish this process?
| Wikipedia’s probably not the ideal place to learn about the Cholesky decomposition or of the various algorithms that can be used to generate the decomposition. I’d recommend reading a textbook on numerical linear algebra or general book on numerical mathematics. That being said, I’ve gone ahead and fixed up your steps in the outer-product Cholesky algorithm example that you’ve chosen to follow.
For the record, I’m using the notation from the following snapshot of the Wikipedia article on the Cholesky decomposition:
https://en.wikipedia.org/w/index.php?title=Cholesky_decomposition&oldid=711015429
Before I run through the algorithm, however, note that $I_i$ denotes the $i \times i$ identity matrix. $I_0$, in particular, refers to a $0 \times 0$ matrix, so when it occurs in an algorithm step it gets omitted from the matrix block partition. Also, all matrices $A^{(i)}$ and $L_i$ should have the same size as the original matrix $A$. Last, all elements $a_{ii}$, $b_i$, and $B$ are pulled directly from matrix $A^{(i)}$ using the partitioning described at the Wikipedia article.
$A = A^{\left( 1 \right)} = \left( {\begin{array}{*{20}c}
2 & { - 1} & 0 \\
{ - 1} & 2 & { - 1} \\
0 & { - 1} & 2 \end{array}} \right)$, so $a_{11} = 2$, $b_1^T = \left( {\begin{array}{*{20}c} { - 1} & 0\end{array}} \right)$, and $B^{(1)} = \left( {\begin{array}{*{20}c}
2 & { - 1} \\
{ - 1} & 2 \end{array}} \right)$.
Therefore, $L_1 = \left( {\begin{array}{*{20}c}
{\sqrt 2 } & 0 & 0 \\
{\frac{{ - 1}}{{\sqrt 2 }}} & 1 & 0 \\
0 & 0 & 1\end{array}} \right)$ and $A^{\left( 2 \right)} = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & {\frac{3}{2}} & { - 1} \\
0 & { - 1} & 2\end{array}} \right)$.
Continuing on, we see that $a_{22} = \frac{3}{2}$, $b_2 = -1$, and $B^{(2)} = 2$, so
$L_2 = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & {\sqrt {\frac{3}{2}} } & 0 \\
0 & { - \sqrt {\frac{2}{3}} } & 1\end{array}} \right)$ and $A^{\left( 3 \right)} = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & {\frac{4}{3}}\end{array}} \right)$.
From $A^{\left( 3 \right)}$ we immediately write down $L_3 = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & {\sqrt {\frac{4}{3}} }\end{array}} \right)$.
Therefore, $L = L_1 L_2 L_3 = \left( {\begin{array}{*{20}c}
{\sqrt 2 } & 0 & 0 \\
{\frac{{ - 1}}{{\sqrt 2 }}} & {\sqrt {\frac{3}{2}} } & 0 \\
0 & { - \sqrt {\frac{2}{3}} } & {\sqrt {\frac{4}{3}} }\end{array}} \right)$ which you can confirm satisfies $LL^T = A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/986642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the roots of $(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$
Write down, in any form, all the roots of the equation $z^5 − 1 = 0$
Hence find all the roots of the equation
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
and deduce that none of them is real
My Try:
I know how to do the first part:
$$z^5=1=cos 2\pi k + i sin 2\pi k$$
$$z= cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
where $k=0,1,2,3,4$
Please help me to do the second part. Thanks.
Attempt:
$$z^5-1=(z-1)(z^4+z^3+z^2+1)=0$$
Have to find:
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
Let $w=z+1$
$$z^4+z^3+z^2+z+1=0$$
Multiply by $(z-1)$ each side:
$$(z-1)(z^4+z^3+z^2+z+1)=0 (z-1)=0$$
$$z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0$$
$$z^5-1=0$$ (back to Inital result)
Then
$$z=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
Since $w=z+1$ So $z=w-1$
$$w-1=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
$$w=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}+1$$
This seems weird, is this correct?
| Your approach looks good: $z^5-1=0$ has five solutions that are evenly distributed around the unit circle, as defined by your trigonometric solution. These are the black points in the following figure:
As you say, the left hand side factors into
$$z^5-1 = (z-1)(z^4+z^3+z^2+z+1).$$
If we divide off the first order term and substitute $z=w-1$ into the remaining we get
$$z^4+z^3+z^2+z+1 = (w-1)^4 + (w-1)^3 + (w-1)^2 + (w-1) + 1.$$
Thus, the roots of the polynomial in $w$ are exactly the (non-real) roots of the polynomial shifted to the right, as shown in red. Hence, they cannot be real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^{24}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^8$ I'm not sure how to go about doing this. Do I find the ways to add up to 24 using the exponents with repetition? Is the multinomial theorem useful here? I also have a feeling that generating functions might be useful here, but I can't see how. Any help would be appreciated.
| $\bf{My\; Solution::}$ Let $S = 1+x+x^2+x^3+........+x^5......(1)$
Multiply both side by $x\;,$ We get
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;xS = x+x^2+x^3+..............x^6........(2)$
Now Subtract $(1)$ and $(2)\;,$ we get
$\Rightarrow \displaystyle S(1-x) = 1-x^6\Rightarrow S = \frac{(1-x^6)}{(1-x)}$
So we have to find Coeff. of $x^{24}$ in $\displaystyle (1-x^6)^8\cdot (1-x)^8$
Now Using Binomial Theorem, For $(+ve)$ Integral Index
$\displaystyle (1-y)^n = \binom{n}{0}-\binom{n}{1}y+\binom{n}{2}y^2+..............+(-1)^n\binom{n}{n}y^n$
and Using Binomial Theorem, For $(-ve)$ Integral Index
$\displaystyle (1-y)^{-n} = 1+ny+\frac{n(n+1)}{2}y^2+\frac{n(n+1)(n+2)}{3}y^3+.........$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
| Here I have taken the compact answer by /u/Quanto (to whom all credit is due) and simply filled in some intermediate steps. His trick was to (twice) differentiate inside the integral ( a special case of the Leibniz integral rule ).
Using the known definite integral:- $\int_0^{2\pi}\frac{1}{E+F\cos x}\mathrm{d}x= \frac{2\pi}{\sqrt{E^2-F^2}}$ with $E=1$, $F=-a$ and $x=\theta$, we can write:-
$$
\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}
$$
We can multiply both sides by $a$ (so long as $a$ is not a function of $\theta$) to give:-
$$
\int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2a\pi}{\sqrt{1-a^2}}
$$
now differentiate both sides (within the integral on the LHS) with respect to $a$:-
$$\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta
=\frac{2\pi}{(1-a^2)^{3/2}}
$$
again, differentiate both sides (within the integral on the LHS) with respect to $a$:-
$$\int_{0}^{2\pi} \frac{2\cos \theta}{(1-a \cos \theta)^3} d\theta
=\frac{6a\pi}{(1-a^2)^{5/2}}
$$
Hence:-
$$\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta
=\frac{3a\pi}{(1-a^2)^{5/2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Beautiful triangle problem Circle, inscribed in $ABC$, touches $BC, CA, AB$ in points $A', B', C'$. $AA' BB', CC'$ intersect at $G$. Circumcircle of $GA'B'$ crosses the second time lines $AC$ and $BC$ at $C_A$ and $C_B$. Points $ A_B, A_C, B_C,B_A, C_A, C_B$ are concyclic.
Looks straightforward, but I'm struggling to get something out of the given. I tried proving that two points and one line and two points and some other are concyclic and then combining three similar resul together, but not much yet.
| This is a coordinate-based approach, making heavy use of tools from projective geometry.
Without loss of generality, you can choose the coordinate system in such a way that the inscribed circle is the unit circle. On that you can use a rational parametrization, i.e. choose $a,b,c\in\mathbb R$ such that $A'=(1-a^2,2a)/(1+a^2)$ and likewise for $B'$ and $C'$. Or, even better, use homogeneous coordinates for these:
$$
A'=\begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix}\qquad
B'=\begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix}\qquad
C'=\begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix}
$$
From that, everything else follows. The tangent at a point on the circle is simply its polar line, which you get by multiplication with the matrix of the unit circle, namely the matrix
$$U=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$
You can join points using lines, and intersect lines to obtain points, simply by computing the cross product.
$$A=(U\cdot B')\times(U\cdot C')$$
and likewise for $B$ and $C$, the other two corners of the triangle. Then you get
$$G=(A\times A')\times(B\times B')$$
A circle through three points can be constructed as a conic through these points and the special points $I=(1,i,0)$ and $J=(1,-i,0)$ which have complex coordinates and lie on the line at infinity. Since they also lie on every circle, they are often called the ideal circle points. So we construct the matrix of the circle $\bigcirc GA'B'$ in several steps:
\begin{align*}
M_1 &= (A'\times I)\cdot(B'\times J)^T &
M_2 &= (A'\times J)\cdot(B'\times I)^T \\
M_3 &= M_1 + M_1^T & M_4 &= M_2 + M_2^T \\
M_5 &= (G\cdot M_3\cdot G)M_4 - (G\cdot M_4\cdot G)M_3 &
M_C &= iM_5
\end{align*}
$M_1$ and $M_2$ describe degenerate conics through $A',B',I,J$. $M_3$ and $M_4$ are the same except using symmetric matrices. $M_5$ is a linear combination which also passes through $G$, so that's the circle. $M_C$ is a real matrix describing the same circle, avoiding all the purely imaginary entries of $M_5$.
To intersect that circle with $AC$ (which is the same line as $B'C$) you compute
$$C_A=(C^T\cdot M_C\cdot C)B'-2(B'^T\cdot M_C\cdot C)C$$
This is still a homogeneous coordinate vector of a point, e.g. some $(x,y,z)$. Dehomogenize that to $(x/z, y/z)$ then take the norm of that:
\begin{align*}
\lVert C_A\rVert =& \frac{\sqrt s}{t} \\
s =& \phantom+ (a^4b^4 + a^4c^4 + b^4c^4) \\&
- 2\,abc\,(a^3b^2 + a^2b^3 + a^3c^2 + b^3c^2 + a^2c^3 + b^2c^3) \\&
+ 3\,a^2b^2c^2\,(a^2 + b^2 + c^2) \\&
+ 11\,(a^4b^2 + a^2b^4 + a^4c^2 + b^4c^2 + a^2c^4 + b^2c^4) \\&
+ 16\,abc\,(a^2b + ab^2 + a^2c + b^2c + ac^2 + bc^2) \\&
- 20\,(a^3b^3 + a^3c^3 + b^3c^3) \\&
- 20\,abc\,(a^3 + b^3 + c^3) \\&
- 42\,a^2b^2c^2 \\&
- 2\,(a^3b + ab^3 + a^3c + b^3c + ac^3 + bc^3) \\&
+ 3\,(a^2b^2 + a^2c^2 + b^2c^2) \\&
+ (a^4 + b^4 + c^4) \\
t =&\phantom+ (a^2 + b^2 + c^2) \\&
- (ab + ac + bc) \\&
+ (a^2b^2 + a^2c^2 + b^2c^2) \\&
- abc\,(a + b + c)
\end{align*}
This is the radius for one of the six points of your claimed circle. The other five can be obtained using the same computation, starting from a permutation of the three initial points. So the result will be the same except for a permutation of the parameters $a,b,c$. But the formula stated above is invariant under such a permutation, therefore all six points lie on a circle as claimed. Its center is the center of the coordinate system, i.e. the incenter of the triangle. Its radius will be the fraction described by the lengthy expressions stated above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Double Integration with change of variables I am having trouble with the following double integral:
$$\iint\limits_D(x^2+y^2) \;dA$$
where $D$ is given by the region enclosed by the curves
*
*$xy=1$
*$xy=2$
*$x^2-y^2 =1$
*$x^2-y^2 =2$
I have tried changing variables to $u=x^2+y^2$, $v=x^2-y^2$ (and other changes of variables) but the Jacobian looks grim.
| Consider the variables
\begin{align*}
u = xy, && v = x^2-y^2.
\end{align*}
We obtain that $x=\frac{u}{y}$. Therefore:
$$y^2=x^2-v,$$
$$y^2 -\frac{u^2}{y^2}=-v,$$
$$\frac{y^4-u^2}{y^2} =-v,$$
$$y^4+vy^2-u^2=0,$$
$$y^2 = \frac{-v\pm \sqrt{v^2-4(-u^2)}}{2} = \frac{-v \pm \sqrt{4u^2+v^2}}{2}.$$
From which we obtain:
$$x^2=\frac{-v \pm \sqrt{4u^2+v^2}}{2}+v.$$
Thinking now on the integrand:
$$x^2+y^2 = \frac{-v\pm \sqrt{4u^2+v^2}}{2}+\frac{-v\pm \sqrt{4u^2+v^2}}{2}+v = \pm\sqrt{4u^2+v^2}.$$
Last equation gives us 2 options for the integrand
$i)$$\displaystyle \frac{v+\sqrt{4u^2+v^2}}{2}+\frac{v+\sqrt{4u^2+v^2}}{2}-v=\sqrt{4u^2+v^2}.$
$ii)$$\displaystyle \frac{v-\sqrt{4u^2+v^2}}{2}+\frac{v-\sqrt{4u^2+v^2}}{2}-v= -\sqrt{4u^2+v^2}.$
Of course, the volume of the solid must be positive, hence we take the first case. Then we have that:
$$\iint\limits_{D}(x^2+y^2)\;dA=\iint\limits_{S}\sqrt{4u^2+v^2}\;|\textbf{J}(T(u,v))|d\hat{A},$$
where $T(u,v)$ is the transform that takes us from $S$ to $D$. It is clear that $S$ is the rectangle:
\begin{align*}
S:\begin{cases}
u=1\\
u=2\\
v=1\\
v=2
\end{cases}.
\end{align*}
Now let's calculate the Jacobian:
$$\textbf{J}(T)= \left| {\begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{array}}\right|. $$
Para calcular $\displaystyle \frac{\partial x}{\partial u}$ considere las variables $u=xy$ y $v=x^2-y^2$, de donde se obtiene que:
\begin{align*}
0=\frac{\partial x}{\partial u}y+ \frac{\partial y}{\partial u}x, && 1 = 2x\frac{\partial x}{\partial u}-2y\frac{\partial y}{\partial u}.\\
\end{align*}
Multiplying the first equation times $2y$ and the second one times $x$ and adding the equations, we obtain:
\begin{align*}
2y=2y^2\frac{\partial x}{\partial u} +2xy \frac{\partial y}{\partial u}, && 0=2x^2\frac{\partial x}{\partial u}-2xy\frac{\partial x}{\partial u},
\end{align*}
$$2y = \frac{\partial x}{\partial u}(2y^2+2x^2),$$
$$\frac{\partial x}{\partial u} = \frac{2y}{2y^2+2x^2}.$$
In a similar fashion we obtain the rest of the partial derivatives:
$$\textbf{J}(T)= \left| {\begin{array}{cc}
\frac{y}{y^2+x^2} & \frac{x}{2x^2+2y^2}\\
\frac{x}{x^2+y^2} & \frac{-y}{2x^2+2y^2}
\end{array}}\right| = -\frac{y^2}{2(y^2+x^2)^2}-\frac{x^2}{2(y^2+x^2)^2} = -\frac{y^2+x^2}{2(y^2+x^2)^2} = -\frac{1}{2\sqrt{4u^2+v^2}}. $$
Therefore, the original integral is now:
$$\iint\limits_{S}\sqrt{4u^2+v^2}\left(\frac{1}{2\sqrt{4u^2+v^2}}\right)\;d\hat{A} = \iint\limits_{S}\frac{1}{2}\;d\hat{A}.$$
Because of Fubini's Theorem we can write:
$$\int_1^2 \left(\int_1^2 \frac{1}{2}\;du\right)dv=\int_1^2 \left[\frac{1}{2}u\right]^2_1dv =\int_1^2 \frac{1}{2}\;dv =\boxed{\frac{1}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$
I think that the sum should be $\frac{1}{2}$ since the limit of :
$$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$
But that wrong. Any ideas?
| $$-\frac{1}{2}[(\frac{1}{3} - 1) + (\frac{1}{4} - \frac{1}{2}) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{6} - \frac{1}{4})\ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2} + (\frac{1}{3} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{4}) + (\frac{1}{5} - \frac{1}{5}) \ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2}] = \frac{3}{4} $$
Notice that I am just rearranging to cancel all the positive terms as it is a telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergent series and comparison test Show that the series converges, but not absolutely: $\displaystyle \sum_{n=1}^\infty\Bigg(\exp\Bigg(\frac{(-1)^n}{n}\Bigg)-1\Bigg)$.
This is what I did so far:
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\frac{1}{n})-1>0$ when $n$ is even
$\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1=\exp(\frac{-1}{n})-1<0$ when $n$ is odd
So, $\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\exp(\frac{1}{n})$. I was going to use comparison test but now stuck. Any hints?
| Use the fact that $\exp(x) = 1 + x + x^2/2! + x^3/3!+ \cdots$ for any $x$, so
$$
\exp(\frac{(-1)^n}{n}) = 1 + \frac{(-1)^n}{n} + \frac{1}{2!}\left(\frac{(-1)^n}{n}\right)^2 + \dots.
$$
Then
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\frac{2}{n} \to 0.
$$
Indeed
$$
\Bigg|\exp\Bigg(\dfrac{(-1)^n}{n}\Bigg)-1\Bigg|\leq\frac{1}{n} + \Bigg|\frac{1}{2!}\left(\frac{(-1)^n}{n}\right)^2 + \frac{1}{3!}\left(\frac{(-1)^n}{n}\right)^3 + \dots\Bigg|
\leq \frac{1}{n} + \Bigg|\left(\frac{1}{n}\right)^2 + \left(\frac{1}{n}\right)^3 + \dots\Bigg| \leq \frac{1}{n} + \frac{1}{n^2}\frac{n}{n-1} \leq \frac{2}{n}.
$$
Let's prove monotony. I'm going to show that $|a_{2n+1}|<|a_{2n}|<|a_{2n-1}|$.
For $2n$ we have
$$
|a_{2n}| = \frac{1}{2n} + \frac{1}{2!}\left(\frac{1}{2n}\right)^2 + \frac{1}{3!}\left(\frac{1}{2n}\right)^3 + \dots
$$
Then
$$
|a_{2n+1}| \leq\frac{1}{2n+1} + \frac{1}{2!}\left(\frac{1}{2n+1}\right)^2 +\dots < \frac{1}{2n} + \frac{1}{2!}\left(\frac{1}{2n}\right)^2 +\dots < \frac{1}{2n-1}+\frac{1}{2}\frac{1}{(2n-1)^2} \leq |a_{2n-1}|.
$$
Now by Leibniz's test the series converges.
As mvggz noted the series doesn't converge absolutely because $|a_n|\sim \frac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $ab-h^2 =0 $ represent a parabola,
$ab-h^2>1$ represent an ellipse
$ab-h^2 <0$ represent a hyperbola
Here $ab-h^2 <0$ therefore equation (1) represents hyperbola and equation (2) is parabola. ( where a,b, h are coefficients of $x^2,y^2,xy$ respectively.
Now how to get the minimum distance between the two curves please suggest .. thanks.
| When the shortest distance is achieved, the normal to the two curves coincide.
We are lucky that the first curve degenerates in $x+y=0$ and $2x-y=0$, giving two normal directions $(1,1)$ and $(2,-1)$.
Then, taking the gradient of the second expression, we express parallelism:
$$(-1,2y)\times(1,1)=-1-2y=0,$$
$$y=-\frac12,x=\frac94,$$
and compute the distance to the line
$$d=\frac{|\frac94-\frac12|}{\sqrt2}=\frac7{4\sqrt2}.$$ Similarly,
$$(-1,2y)\times(2,-1)=1-4y=0,$$
$$y=\frac14,x=\frac{33}{16},$$
$$d=\frac{|2\frac{33}{16}-\frac14|}{\sqrt5}=\frac{31}{8\sqrt5}.$$
Finally,
$$\sqrt\frac{49}{32}<\sqrt{\frac{961}{320}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Evaluate the sum. Evalute the following sum: $ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$.
I tried doing it but I keep getting the wrong answer. I've used known sums to solve it.
| Assuming $n$ is even
$$S=1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$S=(1+2+3+\cdots+(n-1)+n)+(2+4+6+\cdots+n)$$
$$S=\sum_{k=1}^{n} k +\sum_{k=1}^{n/2} 2k$$
$$S=\frac{n(n+1)}{2}+\frac{n}{2}\left(\frac{n}{2}+1\right)=
\frac{2n(n+1)}{4}+\frac{n(n+2)}{4}
=\frac{3n^2+4n}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.
Is the answer
$$
\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?
$$
I understand the concept of Matrix Transformation, I don't think I'm doing it right.
| The columns of the matrix tell you where it sends the standard basis vectors. For instance if I am interested in the third column then I need to determine what the action of our linear operator is on the column vector ,
$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right].$$
This vector can be written as a linear combination of the vectors used to define the linear operator,
$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =
\left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right]
-\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]
-\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$
Multiplying both sides by our linear operator $M$ we get,
$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =
M\left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right]
-M\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]
-M\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$
Note that we know what $M$ does to the vectors on the right so we can just substitute those values in and add,
$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =
\left[ \begin{array}{c}1 \\ 0 \\ 1 \end{array}\right]
-\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]
-\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right]
=\left[ \begin{array}{c}0 \\ -2 \\ 0 \end{array}\right].$$
The resulting vector is the third column of our matrix,
$$ M = \left[\begin{array}{ccc}
\ & \ & 0 \\
\ & \ & -2 \\
\ & \ & 0
\end{array}\right].$$
A similar process will yield the other collumns.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
constructive counting math problem about checkers on a checkerboard In how many ways can we place anywhere from 0 to 9 indistinguishable checkers on a 3x3 checkerboard (no more than one checker per square), such that no row or column contains exactly 1 checker?
I tried reducing the checkerboard to a 2x2 checkerboard and solving it that way, but it didn't work out and now I'm not sure what to do/how to re-start.
| We can also do this using inclusion-exclusion. There are $\binom3j\binom3k$ ways to choose $j$ particular rows and $k$ particular columns for which the restriction is violated. Due to the symmetry with respect to $j$ and $k$, we have $4(4+1)/2=10$ cases to consider:
$j=0$, $k=0$, $2^9=512$ arrangements
$j=0$, $k=1$: $3\cdot2^6=192$ arrangements
$j=0$, $k=2$: $3^2\cdot2^3=72$ arrangements
$j=0$, $k=3$: $3^3=27$ arrangements
$j=1$, $k=1$: $(1+2\cdot2)\cdot2^4=80$ arrangements
$j=1$, $k=2$: $(2\cdot2+2\cdot2)\cdot2^2=32$ arrangements
$j=1$, $k=3$: $3\cdot2\cdot2=12$ arrangements
$j=2$, $k=2$: $(1+2+4)\cdot2^1=14$ arrangements
$j=2$; $k=3$: $3!=6$ arrangements
$j=3$, $k=3$: $3!=6$ arrangements
Thus by inclusion-exclusion there are
$$
\binom30\binom30\cdot512-2\cdot\binom30\binom31\cdot192+2\cdot\binom30\binom32\cdot72-2\cdot\binom30\binom33\cdot27+\binom31\binom31\cdot80-2\cdot\binom31\binom32\cdot32+2\cdot\binom31\binom33\cdot12+\binom32\binom32\cdot14-2\cdot\binom32\binom33\cdot6+\binom33\binom33\cdot6\\=512-1152+432-54+720-576+72+126-36+6\\=50$$
admissible arrangements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
| $$\sum_{k=1}^n\frac{1}{k(k+1)} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=\frac{n}{n+1}$$
So
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k(k+1)}= \lim_{n\to \infty} \frac{n}{n+1}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplify $f(x) = \arctan(2x) + \arctan(3x)$
Simplify $f(x) = \arctan(2x) + \arctan(3x)$
I had a go at it and this is what I got to :
We have: $-\pi<\arctan(2x)+\arctan(3x)<\pi$
Let $a=\arctan(2x)$ and $b=\arctan(3x)$
Then I cut it into different intervals :
$-\pi<a+b<-\pi/2$
$-\pi/2<a+b<\pi/2$
$\pi/2<a+b<\pi$
In the Interval $(-\pi/2, \pi/2)$ :
I got : $$\tan(a+b) = \frac{a+b}{1-\tan(a)\tan(b)} = \frac{5x}{1-6x^2}$$
So: $$\arctan(\tan(a+b)) = \arctan\left(\frac{5x}{1-6x^2}\right)$$
$$\implies a+b = \arctan\left(\frac{5x}{1-6x^2}\right)$$
So that means that : $$\arctan(2x)+\arctan(3x) = \arctan\left(\frac{5x}{1-6x^2}\right) \in \ (-\pi/2, \pi/2)$$
And I get stuck here not knowing what to do. Can I please get some help on how I can simplify this better and/or the correct way?
| Drawing 2 right triangles with common side of length 1 and having adjacent sides of lengths 2x and 3x, respectively, the Law of Cosines gives
$25x^2=(1+4x^2)+(1+9x^2)-2\sqrt{1+4x^2}\sqrt{1+9x^2}\cos\theta\;\;\;$ where $\theta=\arctan 2x+\arctan 3x$,
so $\displaystyle\cos\theta=\frac{1-6x^2}{\sqrt{(1+4x^2)(1+9x^2)}}$. $\hspace{.2 in}$Since $\arctan(-x)=-\arctan x$,
$\arctan 2x+\arctan 3x=\displaystyle\cos^{-1}\left(\frac{1-6x^2}{\sqrt{(1+4x^2)(1+9x^2)}}\right)\cdot\text{sgn}(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
continuous functional calculus; spectrum of an self adjoint element in a c*algebra Let A be a C$^*$-Algebra, $a\in A$ selfadjoint and $\|a^2-a\|<\frac{1}{4}$. The claim is: $\sigma(a)\subseteq (-\frac{1}{2},\frac{1}{2}) \cup (\frac{1}{2},\frac{3}{2})$ and there is a projection $p\in A$ such that $\|a-p\|<\frac{1}{2}$.
I have found out: $a^2-a$ is selfadjoint too, $\sigma(a^2-a)\subseteq (-\frac{1}{4},\frac{1}{4})$, and i think i have to use the functional calculus to find out $\sigma(a)\subseteq (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2},\frac{3}{2})$, but i dont know how. If $\Phi:C_0(\sigma(a))\to A$ is the continuous functional calculus, $\Phi(z\mapsto z^2-z,\;\sigma(a)\to \mathbb{C})=a^2-a=f(a)=\sigma(f(a))$. Can you help me what to do next? Regards If $z^2-z=-1/4\ $ I get $z=1/2$ and if $z^2-z=1/4$, I get $z=1/2+\sqrt{2}$ and $z=1/2-\sqrt{2}$..
| What you did is correct, $||a^2 - a || < \frac{1}{4}$ means $\sigma(a^2 - a) \subset (-1,1)$, that is the function $t^2 - t$ takes $\sigma(a)$ to the subset $(-1,1)$, and so, $\sigma(a) \subset (\frac{1-\sqrt{2}}{2},\frac{1}{2}) \cup (\frac{1}{2} , \frac{1+\sqrt{2}}{2}) \subset (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2})$.
Define $f\colon (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2}) \to \mathbb{R}$, $f(x) = x$ and $g\colon (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2}, \frac{3}{2}) \to \{0,1\}$, $g(x) = $ closest integer to $x$ ( no ties on this set). We have $|f(x) - g(x)| < 1$ for all $x$. Hence on the compact set $\sigma(a)$ we have $\sup |f-g| < 1$. Define $p = g(a)$ ( functional calculus). We have $p^2 = p$ and $|a-p| < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?
| Let's prove this first: x>0, : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $
This is true iif : $ (2*\sqrt{x} +\frac{1}{\sqrt{x}})^2 > (2*\sqrt{x+1})^2 $
You get : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $ <=> $ 4x + \frac{1}{x} +4 > 4*(x+1) $ <=> $ \frac{1}{x} > 0 $ which is true, so the inequality is true as well :) .
Now for the induction step :
$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} $
=> $2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$
But we have from the first part of the answer : $2*\sqrt{n+1} +\frac{1}{\sqrt{n+1}} -2 > 2*\sqrt{n+1+1} -2 = 2*\sqrt{n+2} -2$
Hence you get: $ 2*\sqrt{n+2} -2 < 2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$ Induction holds :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold...
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos x} = 0$$
$$\frac{2 \sin x \cos^2x - 3 \sin x \cos 2x}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 (\cos^2 x - \sin^2 x))}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3\cos^2 x + \sin^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{(\sin^2 x - \cos^2 x) \cos x} = 0$$
$$\frac{\sin x}{\cos x} = 0$$
Looks much simpler. Now solving for x, since $\frac{\sin x}{\cos x} = 0 $ when $\sin x = 0$ and $\sin x = 0$ for every half rotation, the answer must be $k\pi$.
Alas, according to my answer sheet, I'm missing two values: $\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$. But since $\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\sqrt3}{3}$, I'm not sure where these answers come from.
Furthermore, this is the kind of mistake I'm making all over these exercises, I'd like to avoid that, but how can I be sure I have ALL the answers needed?
| By your method, $$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$
Either $\sin x=0\implies x=n\pi$ where $n$ is any integer
else $2 \cos^2x - 3 \cos 2x=0\iff 1+\cos2x-3\cos2x=0\iff\cos2x=\dfrac12=\cos\dfrac\pi3$
The rest is like my other answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Getting $(1-3x^6 + 3x^{12} - x^{18}) \sum_{i=0}^{\infty} \binom{i+2}{2} x^{i}$ from $(\frac{1-x^{6}}{1-x})^3$ using generating functions I'm not sure how to get $(1-3x^6 + 3x^{12} - x^{18}) \sum_{i=0}^{\infty} \binom{i+2}{2} x^{i}$ from $(\frac{1-x^{6}}{1-x})^3$.
I know the following series.
$$\frac{1}{1-x}=(1+x+x^2 + x^3 + x^4 + ...)$$
$$\frac{1}{1-x^6}=(1+x^6+x^{12} + x^{18} + x^{24} + ...)$$
| Hint. From
$$
\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i,\quad |x|<1,
$$
by differentiating, you get
$$
\frac{1}{(1-x)^2}=\sum_{i=1}^{\infty}ix^{i-1},\quad |x|<1,
$$
or, changing $i-1$ to $i$,
$$
\frac{1}{(1-x)^2}=\sum_{i=0}^{\infty}(i+1)x^{i},\quad |x|<1,
$$
differentiating again and changing $i-1$ to $i$ again :
$$
\frac{2}{(1-x)^3}=\sum_{i=0}^{\infty}(i+1)(i+2)x^i,\quad |x|<1,
$$ then dividing by $2$ observing that
$$
\binom{i+2}{2} =\frac{(i+1)(i+2)}{2}
$$ leads to the desired result, taking into account that $$ \left(1-x^6\right)^3=1-3 x^6+3 x^{12}-x^{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to integrate the dilogarithms? $\def\Li{\operatorname{Li}}$
How can you integrate $\Li_2$? I tried from $0 \to 1$
$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$
$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$
$$(An + B)(n+1) + D(n^2) = 1$$
Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$
$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$
$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$
The $1/n$ is the problem, it is the harmonic series, which diverges.
| $$
\begin{align}
\int_0^1\mathrm{Li}_2(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k^2}\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac1{(k+1)k^2}\\
&=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1k+\frac1{k+1}\right)\\
&=\zeta(2)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\
&=\frac{\pi^2}{6}-1
\end{align}
$$
where
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\
&=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\[12pt]
&=1
\end{align}
$$
is a telescoping series.
Note that we only break $\left(\dfrac1k-\dfrac1{k+1}\right)$ into $\dfrac1k$ and $\dfrac1{k+1}$ when we are looking at finite sums. When considering infinite sums, it is kept together, and is equal to $\dfrac1{k(k+1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Volume of rotated region (integration) Let $T$ be a right-angled triangular region with vertices $(0,−b)$,$(1,0)$ and $(0,a)$ where $a$ and $b$ are positive numbers. When $T$ is rotated about the line $x=2$, it generates a solid with volume $V=\dfrac{410\pi}{27}$
Find $a$ and $b$.
Really having trouble with this one. Is $r=(2-x)$ and $h=(a+b)(1-x)$, and how do I integrate that?
Then, how do I display the result as a result of '$a$' and '$b$'?
| Using Pappus's Theorem, $V=A(2\pi\rho)=\frac{1}{2}(a+b)(2\pi(2-\frac{1}{3}))=\pi(a+b)\frac{5}{3}=\frac{410}{27}\pi$,
$\hspace{.3 in}$ so $a+b=\frac{82}{9}$.
By the Pythagorean Theorem,
$a^2+1+b^2+1=(a+b)^2,\;\;\;$ so $2=2ab\implies ab=1\implies b=\frac{1}{a}$.
Then $a+\frac{1}{a}=\frac{82}{9}\implies \frac{a^2+1}{a}=\frac{82}{9}\implies9a^2-82a+9=0\implies(9a-1)(a-9)=0$,
so $a=\frac{1}{9}$ and $b=9,\;\;\;$ or $a=9$ and $b=\frac{1}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show for $\alpha\in (0,1)$, any $f\in C^\alpha([0,1]/{\sim})$ has a Fourier series $S_nf$ uniformly converging to $f$ Technically homework(a midterm) but its over and I'm itching to know the solution. I know how to show it for $\alpha>1/2$ (the Fourier series will converge absolutely), but apparently its true for any $\alpha$; the question guided me as follows:
*
*Show that if a equicontinuous sequence of functions ($f_n$) converges pointwise to $f$, then $f_n$ converges uniformly to $f$.
*Show for $f∈ C^\alpha([0,1]/{\sim})$ that $S_nf → f$ pointwise.
*Show that the sequence $(S_nf)$ is equicontinuous and conclude.
1 and 2 posed no problems to me but I could not do 3. Any help? In addition, I would not mind other ways to prove the result.
| Suppose that $|f(x)|\le C$ and $|f(x)-f(y)|\le C|x-y|^\alpha$.
Express the Difference Using the Dirichlet Kernel
Using the Dirichlet Kernel, we get
$$
\begin{align}
|S_nf(x)-f(x)|
&=\left|\,\int_{-1/2}^{1/2}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|\\
&=\left|\,\sum_{k=-n}^n\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|\tag{1}
\end{align}
$$
Estimate each Integral Using the smoothness of $\boldsymbol{f}$
Since $\left|\,\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}\,\right|\le\frac{2n+1}{\big|2|k|-1\big|}$ and each interval is $\frac1{2n+1}$ wide, we can bound
$$
\begin{align}
\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\frac{\sin((2n+1)\pi y)}{\sin(\pi y)}[f(x-y)-f(x)]\,\mathrm{d}y\,\right|
&\le\frac{C}{\big|2|k|-1\big|}\left(\frac{2|k|+1}{4n+2}\right)^\alpha\tag{2}
\end{align}
$$
Estimate each Integral Using Cancellation from $\boldsymbol{\sin((2n+1)\pi x)}$
For $|y|\le\frac12$, we have $|2y|\le|\sin(\pi y)|\le|\pi y|$, and because
$$
\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\,\mathrm{d}y=0\tag{3}
$$
and
$$
\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}|\sin((2n+1)\pi y)|\,\mathrm{d}y=\frac2{(2n+1)\pi}\tag{4}
$$
if we let $m_k$ be the middle of the range of $\frac{f(x-y)-f(x)}{\sin(\pi y)}$ on $\left[\frac{2k-1}{4n+2},\frac{2k+1}{4n+2}\right]$, for $k\ne0$, we can bound
$$
\begin{align}
&\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\frac{f(x-y)-f(x)}{\sin(\pi y)}\,\mathrm{d}y\,\right|\\
&=\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\left[\frac{f(x-y)-f(x)}{\sin(\pi y)}-m_k\right]\,\mathrm{d}y\,\right|\\
&\le\frac1{(2n+1)\pi}\frac{\overbrace{\pi\frac{2|k|+1}{4n+2}}^{\sin(\pi y)}\overbrace{C(2n+1)^{-\alpha}\vphantom{\frac{|}2}}^{\Delta (f(x-y)-f(x))}+\overbrace{2C\vphantom{()^1}}^{f(x-y)-f(x)}\overbrace{\pi(2n+1)^{-1}}^{\Delta\sin(\pi y)}}{\underbrace{\frac{4k^2-1}{(2n+1)^2}}_{\sin^2(\pi y)}}\\
&=\frac{C(2n+1)^{-\alpha}}{4|k|-2}+\frac{2C}{4k^2-1}\tag{5}
\end{align}
$$
Use each Estimate in its Proper Place
If we use estimate $(2)$ for $k\le m=n^{\frac{\alpha}{\alpha+1}}$ and estimate $(5)$ for $k\gt m$, then we get
$$
\begin{align}
\sum_{|k|\le m}\frac{C}{\big|2|k|-1\big|}\left(\frac{2|k|+1}{4n+2}\right)^\alpha
&\le\frac{C}{(4n+2)^\alpha}\left[1+6\sum_{k=1}^m(2k+1)^{\alpha-1}\right]\\
&\le\frac{C}{(4n+2)^\alpha}\frac3\alpha(2m+1)^\alpha\\
&\sim\frac{3C}{\alpha2^\alpha}n^{-\frac\alpha{\alpha+1}}\tag{6}
\end{align}
$$
and
$$
\begin{align}
\sum_{m\lt|k|\le n}\frac{C(2n+1)^{-\alpha}}{4|k|-2}
&\le\frac{C}{2^{\alpha+1}}\frac{H_n}{n^\alpha}\\
&\sim\frac{C}{2^{\alpha+1}}\frac{\log(n)}{n^\alpha}\\
&=o\left(n^{-\frac{\alpha}{\alpha+1}}\right)\tag{7}
\end{align}
$$
and
$$
\begin{align}
\sum_{m\lt|k|\le n}\frac{2C}{4k^2-1}
&\le C\sum_{k=m}^\infty\frac1{k^2-1}\\
&=\frac{C}{2}\sum_{k=m}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)\\
&=\frac{C}{2}\left(\frac1{m-1}+\frac1m\right)\\
&\sim Cn^{-\frac{\alpha}{\alpha+1}}\tag{8}
\end{align}
$$
Put Everything Together
Therefore, we have uniform convergence:
$$
|S_nf(x)-f(x)|\le\left(1+\frac3{\alpha2^\alpha}\right)Cn^{-\frac{\alpha}{\alpha+1}}\tag{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Product series general formula? For a series like :
$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3$$
There is a general formula : $(n(n+1)/2)^2$
My question: Is there any general formula possible for following series :
$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdots n^n$$
| let $$s_n=\prod_{r=1}^nr^r$$
Write$$s_n=1^1 \times2^2 \times 3^3 \times 4^4 \times\cdots \times n^n$$
Take $\log$ on both sides
$$\ln (s_n)=\ln(1^1 \times2^2 \times 3^3 \times 4^4 \times\cdots \times n^n)$$
$$\ln (s_n)=\ln(1^1)+\ln(2^2)+\ln(3^3)+\ln(4^4)+\cdots +\ln(n^n)$$
$$\ln (s_n)=1\ln(1)+2\ln(2)+3\ln(3)+4\ln(4)+\cdots +n\ln(n)$$
And then See This Answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
| For the equation:
$$4c^2b^2+4d^2a^2-c^2a^2-4d^2b^2=3t^2$$
Can specify any number : $d,c$ . Then decisions will be.
$$a=(d^2-c^2)p^2+3s^2$$
$$b=(d^2-c^2)p^2-3cps-3s^2$$
$$t=c(d^2-c^2)p^2+4(d^2-c^2)ps-3cs^2$$
$p,s$ - any integer asked us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate this improper integral? I got stuck when evaluating these two improper integrals:$$
\int_a^b\frac{dx}{\sqrt{(b-x)(x-a)}}
$$
and$$
\int_0^1\frac{dx}{\sqrt{x-x^3}}
$$
How to evaluate them? Thank you!
| \begin{align*}
\int_a^b{\frac{\mathrm dx}{\sqrt{(b-x)(x-a)}}}&=\int_a^b\frac{\mathrm d x}{\sqrt{-ab+(a+b)x-x^2}}\\
&=\int_a^b\frac{\mathrm dx}{\sqrt{-ab+\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}\\
&=\int_a^b\frac{\mathrm dx }{\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}
\end{align*}
Let's put $$\frac{b-a}{2}\sin t=x-\frac{a+b}{2},\quad\mathrm dx=\frac{b-a}{2}\cos t\mathrm dt$$
where $$x=a\implies \frac{b-a}{2}\sin t_1=-\frac{b-a}{2}\implies t_1=-\frac{\pi}{2},$$
$$x=b\implies \frac{b-a}{2}\sin t_2=\frac{b-a}{2}\implies t_2=\frac{\pi}{2}$$
It follows
\begin{align*}
\int_a^b\frac{\mathrm d x}{\sqrt{(b-x)(x-a)}}&=\int_{-\pi/2}^{\pi/2}\frac{\frac{b-a}{2}\cos t \,\mathrm dt}{\frac{b-a}{2}\cos t}\\
&=\pi
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Positive integers $x$,$y$,$z$ If $x$ , $y$ and $z$ are positive integers and $3x = 4y = 7z$, then calculate the smallest possible value for $x+y+z$.
How do you do this? Can someone please give me a hint?
| $k=3x=4y=7z\implies x=\dfrac{k}{3},y=\dfrac{k}{4},z=\dfrac{k}{7}\implies x+y+z=k\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{7}\right)=k\left(\dfrac{4\cdot7+7\cdot3+3\cdot4}{84}\right)\geq \dfrac{61}{84}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Need explain how to find sum of series Can someone explain me how to find sum of next series:
$\sum_{n=1}^\infty n^4 \tan^{n-1}(x)$
Thanks for answers in advance.
| We can write $n^4$ as a sum of combinatorial polynomials of degree $4$ or less:
$$
n^4=24\binom{n}{4}+36\binom{n}{3}+14\binom{n}{2}+\binom{n}{1}\tag{1}
$$
Next, for $|x|\lt1$, we can use the Generalized Binomial Theorem and negative binomial coefficients to show
$$
\begin{align}
\sum_{n=k}^\infty\binom{n}{k}x^{n-1}
&=\sum_{n=k}^\infty\binom{n}{n-k}x^{n-1}\\
&=\sum_{n=k}^\infty\binom{-k-1}{n-k}(-1)^{n-k}x^{n-1}\\
&=\sum_{n=0}^\infty\binom{-k-1}{n}(-1)^{n}x^{n+k-1}\\
&=\frac{x^{k-1}}{(1-x)^{k+1}}\tag{2}
\end{align}
$$
Therefore, using $(1)$ and $(2)$, we get
$$
\begin{align}
\sum_{n=1}^\infty n^4x^{n-1}
&=\frac{24x^3}{(1-x)^5}+\frac{36x^2}{(1-x)^4}+\frac{14x}{(1-x)^3}+\frac1{(1-x)^2}\\
&=\frac{1+11x+11x^2+x^3}{(1-x)^5}\tag{3}
\end{align}
$$
Now, just substitute $x\mapsto\tan(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof By Induction Fibonacci Numbers How do I prove that
$$ f_{ 2n+1 } = 3f_{ 2n } + 1 - f_{ 2n-3 } $$
I'm not sure how to prove it using the defining recurrence of Fibonacci numbers.
| Let's do this the most basic way:
$f_{2n+1} = 3f_{2n} + 1 - f_{2n-3}$
$f_{2n} + f_{2n-1} = 3f_{2n} + 1 - f_{2n-3}$
$f_{2n-1} = 2f_{2n} + 1 - f_{2n-3}$
$f_{2n-1} + f_{2n-3} = 2f_{2n} + 1$
$f_{2n-1} + f_{2n-3} = f_{2n} + f_{2n-1} + f_{2n-2} + 1$
$f_{2n-3} = f_{2n} + f_{2n-2} + 1$
$f_{2n-3} = f_{2n} + f_{2n-3} + f_{2n-4} + 1$
$0 = f_{2n} + f_{2n-4} + 1$
Right side is strictly positive, which shows that the first equation didn't hold at first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding roots in marginally stable system modeled by complex number A system can be modeled by $(z + 3)(z + 2)(z + 1) + C = 0$, where $C > 0$, and $z = x + iy$. When it is marginally stable $Re(z) = 0$.
What are the values of the roots in marginally stable condition? At what values of $C$ do they occur?
Please give me help, as I don't know how to go about this question.
Thanks in advance,
Adithya
| If $(z + 3)(z + 2)(z + 1) + C = 0$ is the characteristic equation for a transfer function (usually denoted in powers of $s$), we can apply the Routh-Hurwitz criterion. Unless there is more to the question, I will assume this is the case. We have that
$$
(z + 3)(z + 2)(z + 1) + C = z^3 + 6z^2 + 11z + 6 + C
$$
Then
\begin{array}{ccc}
z^3 & 1 & 11\\
z^2 & 6 & 6 + C\\
z^1 & \frac{6 + C - 66}{6} & 0\\
z^0 & 6 + C
\end{array}
For the system to be stable, all the coefficients in the second column must have the same sign. Therefore, $\frac{6 + C - 66}{6} > 0$ and $6 + C > 0$.
$$
C > -6
$$
and
$$
C > 60
$$
Thus, $C > 60$. Now, if $C = 60$, $(z + 3)(z + 2)(z + 1) + C = z^3 + 6z^2 + 11z + 6 + C$, will have two roots on the imaginary axis. Using the second row of the Routh-Hurwitz table, we see that $z = \pm i\sqrt{11}$ are the two roots on the imaginary axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
I already know that:
\begin{equation}
dom f = \{ f(x,y) ∈ \mathbb{R}^2\backslash(x^2y^2-4xy^2+3y^2=0) : \sqrt{4x-x^2-y^2} \geq 0 \}
\end{equation}
But of course this needs to be written in a more simpler form.
During class we solve simpler functions like without fractions and roots, so I don't have anything that can help me.
After this I also need to find:
*
*b) zeros of the function
*c) Calculating Algebraically, the range of the function:
\begin{equation}
T(x,y) = \sqrt{4x-x^2-y^2}
\end{equation}
*d) Extreme values of the function T
I'm of course not expecting the complete solution but something like a kick start.
| we have for the domian two cases
$4x-x^2-y^2\geq 0$ and $y^2(x^2-4x+3)>0$
or
$4x-x^2-y^2\le 0$ and $y^2(x^2-4x+3)<0$
note that $x^2-4x+3=0$ gives $x_1=1$ or $x_2=3$ and $4x-x^2-y^2=4-(y^2+(x-2)^2)$ thus we get for the first case
$4>y^2+(x-2)^2$ and $x>3$ and $y\neq 0$
or
$4>y^2+(x-2)^2$ and $x<1$ and $y \neq 0$
and for the second case
$(x-2)^2+y^2\geq 4$ and $y\neq 0$ and $1<x<3$
sorry i thougt the problem was $\sqrt{\frac{4x-x^2-y^2}{x^2y^2-4xy^2+3y^2}}$
b) the zeros of our functions are $(x-2)^2+y^2=4$ and $y^2\neq 0$ and $x\neq 3$ and $x\neq 1$
c) we have to solve
$T_x=\frac{2-x}{\sqrt{4x-x^2-y^2}}=0$
and
$T_y=\frac{y}{\sqrt{4x-x^2-y^2}}=0$
a) for the domain we have $(x-2)^2+y^2\le 4$ and $y\neq 0$ and $x\neq 3$ and $x\neq 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Showing that $a_n$ is a contracting sequence I have a sequence defined by $a_1=\frac{1}{2}$ and $a_{n+1}=\frac{1}{2}(2+a_n-a_n^2)=\frac{1}{2}(\frac{9}{4}-(a_n-\frac{1}{2})^2)$
I have shown that $\frac{1}{2}\le a_n\le\frac{9}{8}$ for all $n$ and now need to show that $a_n$ is a contracting sequence. I know the definition of a contracting sequence to be there exists $0\le\lambda<1$ such that $|a_{n+2}-a_{n+1}|\le\lambda|a_{n+1}-a_n|$, $\forall n$ in $N$.
Then I considered $a_{n+2}-a_{n+1}$ and eventually rearranged to $\frac{1}{2}((a_n-1/2)^2-(a_{n+1}-1/2)^2)$ and then also $\frac{1}{2}(a_n^2-a_n-a_{n+1}^2+a_{n+1})$.
I am unsure what to do from here, I know I somehow need to compare a form of this expression to $a_{n+1}-a_n$ but how? Am I on the right lines?
| So simplify further and factor the difference:
$a_{n+2}-a_{n+1}= \frac{1}{2}(a_n^2-a_n-a_{n+1}^2+a_{n+1}) =
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}^2-a_n^2)) =
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}-a_n)(a_{n+1}+a_n)) =
\frac{1}{2}(a_{n+1}-a_n)(1-(a_{n+1}+a_n))$.
Note $a_{n+1}-1=\frac{a_n(1-a_n)}2$. Also, if $a>1$ then $a-a^2<0$ so $\frac12(2+a-a^2)<\frac12\cdot2=1$, and similarly if $0<a<1$ then $\frac12(2+a-a^2)>1$. Since $a_1=\frac12<1$ it follows that $a_2=\frac98>1$ so $a_3<1$.
I will use this to prove by induction that $|a_{n+1}-1|\le\frac12$, i.e. $\frac12\le a_n\le\frac32$, and that $a_n>1$ if $n$ is even, and $a_n<1$ if $n$ is odd. True for $n=1$. If $\frac12\le a_n\le1$ (when $n$ is odd) then $0\le1-a_n\le\frac12$ so $|a_{n+1}-1|=\frac{a_n(1-a_n)}2\le\frac14$.
If $1\le a_n\le\frac32$ then $|1-a_n|\le\frac12$ so $|a_{n+1}-1|=\frac{a_n|1-a_n|}2\le\frac38$.
The above proves $\frac12\le a_n\le\frac32$ and that the $a_n$ alternate on different sides of $1$. It follows that
$|a_{n+1}-a_n|\le1$ and $\frac32\le a_{n+1}+a_n\le\frac52$, so $|1-(a_{n+1}+a_n)|\le\frac32$. Note $a_2-a_1=\frac58$. Prove by induction that both $|a_{n+1}-a_n|\le\frac58$ and $|a_{n+2}-a_{n+1}|\le\frac34\cdot|a_{n+1}-a_n|$ (so $\lambda=\frac34$ works).
We have
$|a_{n+2}-a_{n+1}|=\frac{1}{2}|(a_{n+1}-a_n)(1-(a_{n+1}+a_n))|\le$
$\le\frac12|(a_{n+1}-a_n)|\frac32 = \frac34|(a_{n+1}-a_n)|\le\frac34\cdot\frac58
\le \frac58$.
This could all be written in a better way and perhaps simplified, but I think it is right.
The following older answer applies to the original statement,
which was that $a_1=1$ (which was later changed to $a_1=\frac12$).
I get that $a_2=1$, so $a_n=1$, does that help? $|a_{n+1}-a_n|=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int_0^{\pi/4}\sqrt{1+\left( \tan x\right)^2}dx$ I would like to understand all the steps to find out this integral $$ \int_0^{\pi/4} \sqrt{1+\left( \tan x\right)^2} dx$$ Wolfram Alpha returns: $$ \frac12 \log(3+2 \sqrt2) = 0.881373587019543...$$ Why does it give this value? Thank you.
| Here are different steps.
$$ \int_0^{\pi/4} \sqrt{1+\left( \tan x\right)^2} \:dx = \ln\left(1+\sqrt{2}\right).$$
Proof.
For $0<x<\pi/4$, set $\displaystyle t:=\tan\left(\frac{x}{2}\right)$ giving $$\cos x = \frac{1-t^2}{1+t^2},\quad x=2 \arctan t, \quad dx=\frac{2}{1+t^2}dt.$$
Then
$$
\begin{align}
\int_0^{\pi/4} \sqrt{1+\left( \tan x\right)^2} \:dx & = \int_0^{\pi/4} \frac{\sqrt{\cos^2 x+\sin^2 x}}{\cos x} \:dx \\\\
& = \int_0^{\pi/4} \frac{1}{\cos x} \:dx \\\\
& = \int_0^{\tan(\pi/8)} \frac{1+t^2}{1-t^2}\frac{2 \:dt}{1+t^2} \\\\
& = \int_0^{\tan(\pi/8)} \left( \frac{1}{1-t}+\frac{1}{1+t}\right)dt \\\\
& = \ln \left( \frac{1+\tan(\pi/8)}{1-\tan(\pi/8)}\right)\\\\
&= \ln\left(1+\sqrt{2}\right)\\\\
& \approx \color{blue}{0.88137358701954 \dots}\\
\end{align}
$$ where we have used $\displaystyle \tan(\pi/8)=\sqrt{2}-1$ obtained from $\displaystyle \tan(2x)=\frac{2\tan x}{1-\tan^2 x}$ with $x=\pi/8$ knowing that $\displaystyle \tan(\pi/4)=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Derivation of formula involving Gamma function? I'm trying to prove that:
$$\prod_{n=1}^{\infty}\frac{n(n+a+b)}{(n+a)(n+b)} = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}$$
whenever $a$ and $b$ are positive.
I know that
$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} = \frac{\int_0^{\infty}e^{-s}s^ads \int_0^{\infty}e^{-t}t^adt}{\int_0^{\infty}e^{-n}n^{a+b}dn}$$
but am confused as to where to proceed from here... Should I use the product formula for $1/\Gamma$ instead? Any direction would be appreciated. Thanks.
| Another hint (bit more rigorous perhaps): The Beta Function can be written as
\begin{equation}
B(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1},
\end{equation}
the right hand side of the above formula can be expanded
\begin{eqnarray}
B(x,y) &=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1} \\
&=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)+xy}{n(x+y+n)} \right)^{-1} \\
&=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{xy+n(x+y)+n^{2}}\right) \\
&=& \frac{x+y}{x y} \prod_{n=1}^\infty \left( \frac{n(x+y+n)}{(n+x)(n+y)}\right)
\end{eqnarray}
Now; The Beta function has many other forms such as
\begin{equation}
B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}
\end{equation}
From which a relationship between your left hand side and your right hand side could become readily found once you use properties of the gamma fuction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$ Let $a,b,c,d,e$ be integers such that $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$. Prove that $a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$.
I'm reminded of the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. But for $5$th degree, how can I find a factorization for $a^5+b^5+c^5+d^5+e^5-5abcde$?
| Hint: firts note that; if $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$, then $$(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2$$
Now take $P(x)=x^5+kx^4+rx^3+sx^2+tx+u$, with roots $a,b,c,d,e$ then from Viète’s
Relations;
$\boxed{ u=-abcde}$,
$\boxed{k=-(a+b+c+d+e)}$,
$\boxed{r=ae+be+ce+de+ab+ac+ ad+bc+bd+cd=0}$
Take $m=a^4+b^4+c^4+d^4+e^4$ and How $P(a)+P(b)+P(c)+P(d)+P(e)=0$, then:
$${a^5+b^5+c^5+d^5+k(m)+s(a^2+b^2+c^2+d^2+e^2)+t(a+b+c+d+e)+5u=0}$$
then
$${a^5+b^5+c^5+d^5-5u=-k(m)-s(a^2+b^2+c^2+d^2+e^2)-t(a+b+c+d+e)}$$
$$\implies{a^5+b^5+c^5+d^5-5abcde=(a+b+c+d+e)\cdot M}$$
$$\implies(a+b+c+d+e)|(a^5+b^5+c^5+d^5+e^5-5abcde)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do after that...
| Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+abc\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$
$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Series convergence and limit I am trying to solve an exercise that asks me this:
Prove using the $ε–N$ method that the sequence
$a(n) = \frac{n^2 + n - 1}{n^2 + n}$
converges and state the limit.
My attempt is the following:
First, I guess that the $\lim \limits_{n\to\infty} a(n) = 1$.
Then we require
\begin{equation}\begin{split}
|a(n)-1| < ε &\iff \left| \frac{n^2 + n - 1}{n^2 + n} - 1 \right| < ε \\
&\iff \frac{1}{n^2 + n} < ε \\
&\iff \frac{1 }{ ε} < n^2 + n \\
&\iff n^2 + n > \frac{1}{ε}
\end{split}\end{equation}
Now, we take another number $2n^2$ which is greater or equal than $n^2 + n$ and say:
$2n^2 \geq n^2 + n > \frac{1}{\epsilon}$
$\implies 2n^2 > \frac{1}{\epsilon}$
$\implies n > \frac{1}{\sqrt{2ε}}$
Therefore $N(ε) = \left\lceil\frac{1}{\sqrt{2\epsilon}}\right\rceil$
Is this correct? Can we just take another number greater than or equal to $n^2 + n$ and just continue to find the solution?
Any help is welcomed. Thanks! :)
| $2n^2 > 1/\epsilon$ will not guarantee $n^2+n > 1/\epsilon$.
You need to find a nice function of $n$, that lies between $n^2+n$ and $1/\epsilon$, which will allow you to solve for $n$ in terms of $\epsilon$. We can in fact choose $n>1/\epsilon$, since this will ensure that $n^2+n > 1/\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Two roots of the polynomial $x^4+x^3-7x^2-x+6$ are $2$ and $-3$. Find two other roots. I have divided this polynomial first with $(x-2)$ and then divided with $(x+3)$ the quotient. The other quotient I have set equal to $0$ and have found the other two roots. Can you explain to me if these actions are correct and why?
| $(x-2)(x+3)=x^2+x-6$
$x^4+x^3-7x^2-x+6\\=(x^4+x^3-6x^2)-(x^2+x-6)\\= x^2(x^2+x-6)-1(x^2+x-6)\\=(x^2-1)(x^2+x-6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| Hint: Let $f(x) = \cot^{-1} x$ then $f'(x) = -\frac{1}{1+x^2}$ and if $g(x) = \frac{1-x}{1+x}$ then $g'(x) = -\frac{2}{(x+1)^2}$
Now use the chain rule $$[f(g( x ))]' = f'(g(x))\ g'(x) $$
Notice that $$-\frac{1}{1+\Big(\frac{1-x}{1+x}\Big)^2} = -\frac{(1+x)^2}{(1+x)^2+(1-x)^2} = -\frac{(1+x)^2}{1+2x + x^2 +1-2x+ x^2} = -\frac{1}{2}\frac{(1+x)^2}{(1 + x^2)} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$ (3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n] $ Proof that for all $n\in \mathbb{N}$ :
$$
(3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n]
$$
| Another view on this:
$\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$
Is the trace of the $n$-th power of
$\left(\begin{array}{cc} 2 & 1 \\ 1 & 1\end{array}\right)$
And another:
$\dfrac{3+\sqrt 5}{2}$ and $\dfrac{3-\sqrt 5}{2}$ are algebraic integers (they're roots of $x^2 - 3x + 1$). Then since $\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$ is rational, it's an integer.
All of these methods are closely related.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute$\int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$. Compute: $I=\displaystyle \int\limits_{0}^{2} \sqrt{x^2-2x+2}\ln(2+x)dx$.
I tried to : $I=\displaystyle \int \limits_{-1}^{1}\sqrt{t^2+1}\ln(3+t)dt$
set $t=\tan u\Rightarrow dt=(1+\tan^2u)du$
and $I=\displaystyle \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfrac{(\ln(4+\tan^2u))du}{\cos^3u}$
| Hint:
*
*${x+2=e^z \implies dx=e^zdz}$
$\displaystyle \int \sqrt{x^2-2x+2}\ln(2+x)dx\\=\displaystyle \int ze^z\sqrt{e^{2z}-6e^z+10}\ dz\\=z\displaystyle \int e^z\sqrt{e^{2z}-6e^z+10}\ dz-\displaystyle \int \left(\displaystyle\int e^z\sqrt{e^{2z}-6e^z+10}\ dz\right)dz$
$\\$
*$e^z=y \implies e^zdz = dy$
$\therefore\displaystyle\int e^z\sqrt{e^{2z}-6e^z+10}\ dz=\displaystyle\int \sqrt{y^2-6y^2+10}\ dy=\displaystyle\int \sqrt{(y-3)^2+1}\ dy$
$\\$
*$y-3=t \implies dy=dt$
$\therefore \displaystyle\int \sqrt{(y-3)^2+1}\ dy=\displaystyle\int \sqrt{t^2+1}\ dt=\dfrac{t\sqrt{t^2+1}}{2}+\dfrac{1}{2}\operatorname{arsinh}t$
$\\$
*$\displaystyle\int\operatorname{arsinh}t\ dt=t \operatorname{arsinh}t -{\sqrt{t^2+1}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$? I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?
I ask this, because I'm following this answer, and I get lost at how they arrive at $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$.
Thanks.
EDIT: To further clarify what I'm not understanding, please read below:
I don't understand how the author goes from the line:
$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$
to the line:
$\frac{\sqrt{k^2 + k} + 1}{\sqrt{k + 1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$
I would really appreciate it if someone could lead me through to that line and explain exactly what is being done on each line.
Thanks again.
| It cannot hold obviously. When $k=\color{red}1$, $\sqrt {k} + \frac{1}{\sqrt{k+1}}=\color{red}{1+\frac{\sqrt2}{2}}$, however, $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}=\color{red}{\sqrt2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$ To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 - 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ is irreducible, then $N(a) = p^f$, where $a$ lies over the prime $p$ and $f$ is the inertia degree of $K$ over $p$. Since the congruences $x^2 \equiv -2 \pmod 3$ and $x^2 \equiv -2 \pmod{11}$ are easily seen to be solvable (since $x^2 \equiv m \pmod p$ is solvable if and only if $m^{\frac{p-1}{2}} \equiv 1 \pmod p$), the primes $3$ and $11$ should split in $\mathbb{Z}[\sqrt{-2}]$. It was pretty easy to figure out that $$3 = (1 + \sqrt{-2})(1 - \sqrt{-2})$$ and $$11 = (3 + \sqrt{-2})(3 - \sqrt{-2})$$ and $17$ also splits as $(3 + 2 \sqrt{-2})(3 - 2 \sqrt{-2})$. So none of the primes $3, 11,$ or $17$ have inertia. Let $\sigma: K \rightarrow K$ be the unique nonidentity automorphism which is determined by $\sigma(\sqrt{-2}) = - \sqrt{-2}$.
If $\alpha_1 = 1 + \sqrt{-2}, \alpha_2 = 3 + \sqrt{-2}, \alpha_3 = 3 + 2\sqrt{-2}$, what I'm pretty sure should happen is that $55 - 88 \sqrt{-2}$ should be equal to some unit in $\mathbb{Z}[\sqrt{-2}]$, times either $\alpha_1^2$ or $\sigma \alpha_1^2$, times either $\alpha_2^2$ or $\sigma \alpha_2^2$, times either $\alpha_3$ or $\sigma \alpha_3$.
But how do I figure out which combination is right?
| I think that only after doing several problems like this one do certain "obvious" basic facts really begin to feel truly obvious and basic.
The first such fact, which Zoe already mentioned, is that if for $a + b \sqrt d$ with $a$ and $b \in \mathbb Z$ we have $\gcd(a, b) > 1$, then $a + b \sqrt d$ is divisible by some purely real rational integer. In this case that would be 11. Then, like Zoe already said, $$\frac{55 - 88 \sqrt{-2}}{11} = 5 - 8 \sqrt{-2}.$$
The second "obvious" fact is that since we're working in an imaginary quadratic ring, smaller $a$ and $b$ guarantee a smaller norm. So $N(5 - 8 \sqrt{-2}) = 153$, which is more manageable than $N(55 - 88 \sqrt{-2}) = 18513$.
The two important "obvious" facts specific to $\mathbb Z[\sqrt{-2}]$ are first that there are only two units, 1 and $-1$, and that purely real positive integers congruent to 1 or $3 \pmod 8$ are composite in $\mathbb Z[\sqrt{-2}]$ even if prime in $\mathbb Z$.
So, as you already figured out, 3, 11, 17 all split. Obviously, going by the norms, $$(1 - \sqrt{-2})^2 (1 + \sqrt{-2})^2 (3 - 2 \sqrt{-2})(3 + 2 \sqrt{-2}) = 153.$$ This means six prime factors (with some repetition).
The problem now is: which of those factors do you delete to get $5 - 8 \sqrt{-2}$? Dividing, we get $$\frac{153}{5 - 8 \sqrt{-2}} = 5 + 8 \sqrt{-2}.$$ Oops, sorry, that was no help.
One of the "obvious" general facts is telling us that the combination we're looking for can't have conjugates. So we must remove either $(1 - \sqrt{-2})^2$ or $(1 + \sqrt{-2})^2$, and either $3 - 2 \sqrt{-2}$ or $3 + 2 \sqrt{-2}$.
Instead of complicating your life with Greek letters, since we're dealing with numbers of small norm, why not just try multiplying a few combinations and see what you get?
$$(1 - \sqrt{-2})^2 (3 - 2 \sqrt{-2}) = -11 - 4 \sqrt{-2}$$
Wrong quadrant of the complex plane.
$$(1 - \sqrt{-2})^2 (3 + 2 \sqrt{-2}) = 5 - 8 \sqrt{-2}$$
Ding, ding, ding!
Of course it could very well have happened that I tried all the wrong combinations before hitting on the right one. Then I would have looked even less clever. But looking clever has never been my primary concern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What am I doing wrong? (Trigonometric Identity) $$\frac { \cos\theta }{ 1-\sin\theta } =\frac { \sin\theta -\csc\theta }{ \cos\theta -\cot\theta } $$
Steps I took:
$$\frac { \sin\theta -\frac { 1 }{ \sin\theta } }{ \cos\theta -\frac { \cos\theta }{ \sin\theta } } $$
$$\frac { \frac { \sin^{ 2 }\theta -1 }{ \sin\theta } }{ \frac { \sin\theta \cos\theta -\cos\theta }{\sin\theta } } $$
$$\frac { \sin^{ 2 }\theta -1 }{ \sin\theta } \cdot \frac { \sin\theta }{ \sin\theta \cos\theta -\cos\theta } $$
$$\frac { (\sin^{ 2 }\theta -1) }{ (\cos\theta )(\sin\theta -1) } $$
Now where do I go from here? As far as I can tell, $\sin^{ 2 }\theta -1=-\cos^2\theta $
so this makes no sense to me.
| $\dfrac{\sin^2\theta-1}{\cos \theta(\sin\theta-1)}=\dfrac{-\cos^2\theta}{\cos \theta(\sin\theta-1)}=\dfrac{-\cos\theta}{\sin\theta-1}=\dfrac{\cos\theta}{1-\sin\theta}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How prove this inequality $\prod_{1\le i
let $z_{1},z_{2},z_{3},z_{4},z_{5}$ are complex numbers,and such
$$|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2+|z_{5}|^2=5$$
$$A=\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix}$$
show that
$$\det{(AA^{*})}\le 5^5$$
my idea:
we must prove
$$|z_{1}-z_{2}|^2|z_{1}-z_{3}|^2|z_{1}-z_{4}|^2|z_{1}-z_{5}|^2|z_{2}-z_{3}|^2|z_{2}-z_{4}|^2|z_{2}-z_{5}|^2|z_{3}-z_{4}|^2|z_{3}-z_{5}|^2|z_{4}-z_{5}|^2\le 5^5$$
then let $$LHS=f(z_{1},z_{2},z_{3},z_{4},z_{5})$$
maybe can use this indentity
$$\sum_{i=1}^{n}|z_{i}-z|^2=n\sum_{i=1}^{n}|z_{i}|^2-|\sum_{i=1}^{n}z_{i}|^2$$
where
$$z=\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}$$
But following I can't
| Not the exact solution, but maybe a way to go: From the AM-GM inequality we have that
$$
\prod_{i<j}|z_i-z_j|^2\leq\frac{1}{5^5}\left(\sum_{i<j}|z_i-z_j|^2\right)^5.
$$
By the parallelogram identity we have $|z-w|^2=2(|z|^2+|w|^2)-|z+w|^2\leq2|z|^2+2|w|^2$ for each $z,w\in\mathbb C$. Thus,
$$
\prod_{i<j}|z_i-z_j|^2\leq\frac{2^5}{5^5}\left(\sum_{i<j}(|z_i|^2+|z_j|^2)\right)^5=\frac{2^5\cdot 4^5}{5^5}\left(\sum_{i}|z_i|^2\right)^5=8^5.
$$
Maybe one can use this to refine the argument a little...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Riemann sum with circles! My friend and I were talking about Riemann sum and friend suggested that all rectangles would be replaced by circles this is the formula I came up with:
$$
\lim _{n\ \to\ \infty}\sum_{i\ =\ 1}^{n}\pi\,{\rm f}^{2}\left(\, i\,\right)
$$
Is this formula right and if so does this formula work for all Riemann integrable functions ?. I can visualize it failing for a straight line. What is the error bound for this formula in terms of the number of circles ?. I am very confident that it is related to the maximum curvature somehow. I should note here that circles may not stack on each other. This would mean that for a straight line it would probably fail.
| I've just written a paper on this subject for class. My answer is a bit different from yours, because I am allowing circles to stack on each other.
Rectangular Riemann sums take the form $$\sum_{i=0}^n f(a+i\Delta x)\Delta x $$ when given a function $f(x)$, divided into $n$ partitions, bounded between $ a \text{ and } b. \Delta x = \frac{b-a}{n} $. Multiplying the function value and the width of the partition gives us the area of the rectangle that is being created.
For circles, we can set the diameter of the circle equal to the partition width, and then stack the circles up to the height of the function value. We can then get the area of the stack by finding the number of circles which will fit under the curve and then multiplying by the area of a partition-diameter circle.
$$ A_{stack} = \frac{f(a+n_i\Delta x)}{\Delta x} \times \frac{\pi \Delta x^2}{4} $$
Substituting $ \Delta x = \frac{b-a}{n} $ back in, and summing up to the number of partitions, we eventually come to this simplification:
$$ A_{total} = \sum_{i=0}^n \frac{f(a+i(\frac{b-a}{n}))(b-a)\pi}{4n} $$
Any area generated by this sum is, interestingly, 78% of the value given by a rectangular sum. This is because a circle has 78% of the area of a square with the same radius (a line from the middle of the rectangle to a side, which is perpendicular to that side). This calculation is with givens $ f(x)=x^2+1, [a,b] = [0,1], n = 10 $.
$$ A_{circle} = \displaystyle \sum_{i=0}^{10} \frac{f(\frac{i}{10})\pi}{40} = \frac{\pi}{40} + \frac{1.01\pi}{40} + \dots + \frac{2\pi}{40} = \frac{297}{800}\pi \approx 1.166 $$
$$ A_{rectangle} = \displaystyle \sum_{i=0}^{10} \frac{f(\frac{i}{10})}{10} = \frac{1}{10} + \frac{1.01}{10} + \dots + \frac{2}{10} = 1.485 $$
$$ \frac{1.166}{1.485} \approx 0.7851 $$
When you take the sum to infinity, it converges at a value that is 78% of the real value.
$$ A_{total} = \displaystyle \sum_{i=0}^n \frac{f\left(\frac{i}{n}\right)\pi}{4n} = \frac{(\left(\frac{0}{n}\right)^2+1)\pi}{4n} + \dots + \frac{(\left(\frac{n}{n}\right)^2+1)\pi}{4n} $$
$$ = \frac{\pi}{4n}\left[\left(\left(\frac{0}{n}\right)^2 + 1 \right) + \left(\left(\frac{1}{n}\right)^2 + 1 \right) + \dots + \left(\left(\frac{n}{n}\right)^2 + 1 \right)\right] $$
$$ = \frac{\pi}{4n}\left[\left(\left(\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)^2 + \dots + \left(\frac{n}{n}\right)^2\right) + n \right] $$
$$ = \frac{\pi}{4n}\left[\left(\frac{0}{n}\right)^2 + \left(\frac{1}{n}\right)^2 + \dots + \left(\frac{n}{n}\right)^2 \right] + \frac{\pi}{4} $$
$$ = \frac{\pi}{4n^3}\left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{\pi}{4} $$
$$ = \frac{8\pi n^2 + 3\pi n + \pi}{24n^2} = \frac{8\pi}{24} + \frac{3\pi}{24n} + \frac{\pi}{24n^2} $$
$$ = \displaystyle\lim_{n \to \infty}\left(\frac{8\pi}{24} + \frac{3\pi}{24n} + \frac{\pi}{24n^2}\right) = \frac{1}{3}\pi \approx 1.0472 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
| Sum of the first and last terms = $1 + (3n-2) = 3n-1$
Sum of 2nd and (n-1)th terms = $4 + (3n-5) = 3n-1$
Sum of 3rd and (n-2)th terms = $7 + (3n-8) = 3n-1$
$...$
Sum of (n-1)th and 2nd terms = $(3n-5) + 4 = 3n-1$
Sum of n-th (last) and 1st terms = $(3n-2) + 1 = 3n-1$
Add both sides up .
$(1+4+...+(3n-2)) + (1+4+...+(3n-2)) = n(3n-1)$
which means:
$2(1+4+...+(3n-2)) = n(3n-1)$
or
$1+4+...+(3n-2) = n(3n-1)/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 5
} |
Inequality problem involving Cauchy Schwarz inequality If $a+b+c=3$, prove that $\sum \frac{a^2}{b^2-2b+3} \geq 3/2$. How to prove it using the Cauchy-Schwarz inequality? Denote the expression with $P$. What I got was $P$ is greater than or equal to $\frac{9}{\sum a^2-2a-2b-2c+9}$. Is it correct if I found the minimum value of the denominator as $6$ and plugged it in and got the desired result. I mean I got $6$ when equality holds, so I will automatically get the result, wouldn't I?
| By C-S $\sum\limits_{cyc}\frac{a^2}{b^2-2b+3}=\sum\limits_{cyc}\frac{a^4}{a^2b^2-2a^2b+3a^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)}$.
Hence, it remains to prove that $2(a^2+b^2+c^2)^2\geq3\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)$, which after homogenization gives
$\sum\limits_{cyc}(a^4+a^2b^2-2a^3c)\geq0$, which is $\sum\limits_{cyc}b^2(a-b)^2\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find orthogonal matrices Let $A=\begin{bmatrix} 1 & -1/2&-1/2 \\ -1/2 & 1& -1/2\\ -1/2&-1/2 &1 \end{bmatrix}$. Is it possible to find explicitly orthogonal matrices $P, Q$ such that $2A=PJP^t+QJQ^t$? Here $J$ is a matrix of all entries one.
| Note that $A$ has eigenvalues $0$, $3/2$, $3/2$, with null space spanned by
$[1,1,1]^t$. On the other hand, $J$ has eigenvalues $0,0,3$ with $[1,1,1]^t$ its eigenvector for $3$. Consider an orthonormal basis
$$u = \dfrac{1}{\sqrt{3}}\pmatrix{1\cr 1\cr 1\cr}, \ v = \dfrac{1}{\sqrt{2}} \pmatrix{1\cr -1\cr 0\cr},\ w = \dfrac{1}{\sqrt{6}} \pmatrix{1\cr 1\cr -2\cr}$$
We take $P$ so that $P v = u$, $P u = v$ and $P w = w$, and
$Q$ so that $Qu = w$, $Qv = v$, $Qw = u$.
$$ P = \dfrac{1}{6} \pmatrix{1 + 2\sqrt{6} & 1 & -2 + \sqrt{6} \cr
1 & 1 - 2 \sqrt{6} & -2 - \sqrt{6}\cr
-2 + \sqrt{6} & -2 - \sqrt{6} & 4\cr},
\ Q = \dfrac{1}{18} \pmatrix{2 \sqrt{18} + 9 & 2 \sqrt{18} - 9 & -\sqrt{18}\cr
2 \sqrt{18} - 9 & 2 \sqrt{18} + 9 & -\sqrt{18}\cr
-\sqrt{18} & -\sqrt{18} & -4 \sqrt{18}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
| Using the polynomial division algorithm, rewrite the base as
$$\frac {x^2 + 2}{x^2 - 4} = 1 + \frac 6{x^2 - 4} = 1 + \frac 1 {\frac {x^2 - 4}{6}}$$
We will make use of the standard limit:
$$\lim_{x \to +\infty} \left(1 + \frac 1x \right)^x = e$$
So:
$$\begin{align}
\lim_{x \to +\infty} \left(\frac {x^2 + 2}{x^2 - 4}\right)^{9x^2} &= \lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6} \cdot \frac 6 {x^2 - 4}}\right]^{9x^2}=\\
&=\lim_{x \to +\infty} \left[\left(1 + \frac 1 {\frac {x^2 - 4}{6}}\right)^{\Large\frac {x^2 - 4}{6}}\right]^{\Large\frac {6\cdot9x^2}{x^2 - 4}}=\\
&=e^{54}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| here is how i explain this: look at the numbers $9, 99, 999, 9999, \cdots$ in base ten. they are $9 = 10 -1, 99 = (10-1)*11 = 10^2 - 1, 999 = (10-1)*111 = 10^3 - 1, 9999 = (10-1)*1111 = 10^4 - 1$ and the left hand side has the factor $9 = (10-1)$. now you can rewrite the string of equations in the form $$(10 -1) = 1(10 -1),\\ (10^2 - 1) = (10 + 1)(10-1),\\ (10^3 - 1) = (10^2 + 10 + 1)(10-1),\\ (10^4 - 1) = (10^3 + 10^2 + 10 + 1)(10-1), \cdots.$$
now to get your identity think of the polynomials as numbers expressed in base $x,$ that is replace $10$ in the above equations by $x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 7
} |
Find the minimum possible value of $x(1-z)+y(1-x)+z(1-y)$ It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$
Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$
Using the AM-GM inequality concepts, I can write that the value is minimum when
$$x(1-z)=y(1-x)=z(1-y)$$
What else can I conclude from the given information?
| Making the substitutions $\displaystyle \frac{1-x}{x} = a, \frac{1-y}{y} = b$ and $\displaystyle \frac{1-z}{z} = c$,
Then, $abc = 1$ and
$\begin{align}\displaystyle \sum\limits_{cyc} x(1-z) &= \sum\limits_{cyc} \frac{c}{(1+c)(1+a)} \\&= \frac{\sum\limits_{cyc} c(1+b)}{(1+a)(1+b)(1+c)} \\ &= \frac{\sum\limits_{cyc} c(1+b)}{1+\sum\limits_{cyc} a + \sum_{cyc} ab + abc} = \frac{\sum\limits_{cyc} a + \sum_{cyc} ab}{2+\sum\limits_{cyc} a + \sum_{cyc} ab}\end{align}$
Assume the minimum value of $\sum\limits_{cyc} x(1-z)$ is $m$ (a minumum exists since $\sum\limits_{cyc} x(1-z) > 0$).
Then, $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge m\left(2+\sum\limits_{cyc}a + \sum\limits_{cyc} ab\right)$
i.e. $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge \frac{2m}{1-m}$.
But, we know $\displaystyle \sum\limits_{cyc} a \ge 3\sqrt[3]{abc} = 3$ and $\displaystyle \sum\limits_{cyc} ab \ge 3\sqrt[3]{a^2b^2c^2} = 3$ (from Am-Gm inequality and $abc = 1$)
Thus, $\displaystyle \frac{2m}{1-m} = 6 \implies m = \frac{3}{4}$.
2nd approach:
$\displaystyle xyz = (1-x)(1-y)(1-z) \implies \sum\limits_{cyc} x(1-z) = \sum\limits_{cyc} x - \sum\limits_{cyc} xy = 1-2xyz$
and, $\displaystyle xyz = \frac{1}{\prod\limits_{cyc} \left(1+\frac{1-x}{x}\right)} \le \frac{1}{\prod\limits_{cyc} 2\sqrt{\frac{1-x}{x}}} = \frac{1}{8}$ (by Am-Gm inequality)
Thus, $\displaystyle \sum\limits_{cyc} x(1-z) \ge \frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$ Prove that the difference between the product of the first 2000 even numbers and the first $2000$ odd numbers is a multiple of $2001$. Please show the method.
I have started with the following process:
$$(2\cdot 4 \cdot 6 \cdots 4000)-(1\cdot 3 \cdot 5 \cdots 3999)$$
How we can proceed it to find that it is a multiple of $2001$?
| We have $$2001=3\times 23 \times 29.$$
Now note that $$\begin{align}2\cdot 4 \cdot 6 \cdot ... \cdot 4000&=(2\times 1)\cdot (2 \times 2) \cdot (2 \times 3) \cdot ... \cdot (2\times 2000)\\
&=2^{2000}\times 1\cdot2\cdot3\cdot... \cdot23 \cdot ... \cdot 29 \cdot....\cdot2000
\end{align}$$
Therefore $2\cdot 4 \cdot 6 \cdot ... \cdot 4000$ is a multiple of $2001$. Now since the other number is the multiplication of odd numbers including $3$, $23$, and $29$, it is also a multiple of $2001$. Therefore the difference is also a multiple of $2001$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ $f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.
$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$
If $f'(x) = \displaystyle\frac{1}{1-x}$, then $f(x) = \displaystyle\int\frac{1}{1-x}dx$
But $\displaystyle\int \frac{1}{1-x}dx = -\ln(1-x) + C$, and so $f(x) = -\ln(1-x) + C$
Do we need a definite integral here so there will be no $C$ and will agree with how $f(x)$ was originally defined? If so what would the definite integral be? Or maybe there is no definite integral at all?
| There are various things going on here. One is the issue of the $+C.$ What you wrote might better be put
If $f'(x) = \displaystyle\frac{1}{1-x}$, then by definition of the indefinite integral $\displaystyle\int\frac{1}{1-x}dx=f(x)+C$
Also , by definition of definite integrals
If $f'(x) = \displaystyle\frac{1}{1-x}$, on $[a,b]$ then $ \displaystyle\int_a^b\frac{1}{1-t}dt=f(b)-f(a)$
Note that an arbitrarily chosen constant of integration $C$ will cancel out.
I changed the variable of integration so that I could say that for $|x| \lt 1,$ $$\int_0^x\frac1{1-t}dt=-\ln(1-t)\mid_0^x=\left(-\ln(1-x)\right)-(-\ln(1-0))=\ln\left(\frac1{1-x}\right)-0$$
There is also the matter of term by term integration and differentiation of power series. Notice that this is one of the interesting cases where
$f(x) = = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$ and also $x=-1$ but not $x=1.$
$f'(x) = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$ but not for $x= -1$ or $x=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
| First, show that this is true for $n=2$:
*
*$\frac{2^{3\cdot2-1}+5\cdot3^1}{11}=7\in\mathbb{N}$
Second, assume that this is true for $n$:
*
*$\frac{2^{3n-1}+5\cdot3^n}{11}=k\in\mathbb{N}$
Third, prove that this is true for $n+2$:
*
*$\frac{2^{3(n+2)-1}+5\cdot3^{n+2}}{11}=\frac{2^{3n+5}+5\cdot3^{n+2}}{11}$
*$\frac{2^{3n+5}+5\cdot3^{n+2}}{11}=\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}$
*$\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}=\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$
*$\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$
*$\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}$
*$\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}=\frac{55\cdot2^{3n-1}+9\cdot11k}{11}$ assumption used here
*$\frac{55\cdot2^{3n-1}+9\cdot11k}{11}=\frac{11(5\cdot2^{3n-1}+9k)}{11}$
*$\frac{11(5\cdot2^{3n-1}+9k)}{11}=5\cdot2^{3n-1}+9k\in\mathbb{N}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 3
} |
ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
then total no. of Subest of $X = 2^5 = 32$
(Which also contain $\phi$)
$\bullet\; $ If $A = \{1\}\;,$ Then $B=\left (\{2\}\;,\{3\}\;,\{4\}\;,\{5\}\;,\{2,3\}\;,\{2,4\}\;,\{2,5\}\;,\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{2,3,4\}\;,\{2,3,5\}\;,\{3,4,5\}\;,\{2,4,5\}\;,\{2,3,4,5\}\right )$
Similarly for $A=\{2\}\;,A=\{3\}\;,A=\{4\}$ and $A=\{5\}\;,$ we get $15$ ordered pair for each single
element ed set $A$
$\bullet\; $ If $A = \{1,2\}\;,$ Then $B=\left(\{3\}\;,\{4\}\;,\{5\},\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{3,4,5\}\right)$
Similarly for $A=\{1,3\}\;,A=\{1,4\}\;,A=\{1,5\}\;,A=\{2,3\}\;,A=\{2,4\}\;,A=\{2,5\}\;,A=\{3,4\}\;,A=\{3,5\}\;,A=\{4,5\}$ we get $7$ ordered pair for each double elemented set $A$
$\bullet\; $ If $A=\{1,2,3\}\;,$ Then $B=\left(\{4\}\;,\{5\}\;,\{4,5\}\right)$
So for $10$ ordered pair of $A\;,$ we get $3$ ordered pair of $B$
$\bullet \;$ If $A = \{1,2,3,4\}\;,$ Then $B = \{5\}$
So for $5$ ordered pair of $A\;,$ we get $1$ ordered pair of $B$
So Total ordered pair of $\left(A,B\right)$ is $ = \left(5\times 15\right)+\left(10 \times 7\right)+\left(10 \times 3\right)+\left(5 \times 1\right) = 75+70+30+5 = 180$
If Question is How can we solve using Combination (selection) way.
plz explain me, Thanks
|
Hi. You might want to look about Stirling numbers of the second kind. You want the number of ways you can take a surjective function from $[5]=\{1,2,3,4,5\}$ to $[2]=\{1,2\}$ plus the number of surjective functions from $[5]=\{1,2,3,4,5\}$ to $[3]=\{1,2,3\}$.(in $3$ to represent elements that are not taken) In your case $2!*S_{5,2}+3!*S_{5,3}=180$.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does Wolfram Alpha come up with the substitution $x = 2\sin u$? Integration/Analysis I have to integrate
$$
\int_0^2 \sqrt{4-x^2} \, dx
$$
I looked at the Wolfram Alpha step by step solution, and first thing it does is it substitutes
$x = 2\sin(u)\text{ and } \,dx = 2\cos(u)\,du$
How does it know to substitute $2\sin(u)$ for $x$?
| Theóphile's comment is a good answer. I will give another.
The $\sqrt{4-x^2}$ and indeed anything of the form $\sqrt{a^2-b^2}$ --- think $\sqrt{x^2-b^2}$ or $\sqrt{a^2-x^2}$ --- or $x^2+b^2$ can be considered --- via Pythagoras --- as sides of a right-angled triangle.
In this example we have $\sqrt{2^2-x^2}=b\Rightarrow b^2+x^2=2^2$ so that we have a right-angled-triangle with sides $x$, $b=\sqrt{2^2-x^2}$ and hypotenuse $2$.
Draw this triangle. Now choose an angle $\theta$ in the triangle so that you can write
$$\sec/\tan/\sin\theta=\frac{x}{2}.$$
In this case we have $\sin \theta=\frac{x}{2}$.
Now, using the triangle you sketched,
$$\cos\theta=\frac{\sqrt{4-x^2}}{2}\Rightarrow \sqrt{4-x^2}=2\cos\theta.$$
We can also get a handle on $dx$ no problem.
When we eventually antidifferentiate, our answer will be in terms of $\theta$. We can find any of the trig ratios in terms of $x$ using our triangle and $\theta$... well $\theta$ is than angle whose sine is $\frac{x}{2}$ so
$$\theta=\sin^{-1}\left(\frac{x}{2}\right).$$
... not that it arises in this question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Consider the linear transformation $T: \mathbb{R}^{2\times 2} \rightarrow \mathbb{R}^{2\times 2}$ such that $T(A)=A-A^T$
*
*Find a basis for $\ker T$.
*Find a basis for $\operatorname{im} T$.
*Prove that $T$ is diagonalizable. Find an eigenbasis $\mathcal{B}$ for $T$ and find the $\mathcal{B}$-matrix $[T]_\mathcal{B}$.
My attempt:
For 1:
If $A = \begin{pmatrix} a&b\\c&d \end{pmatrix}$, then $T(A) = \begin{pmatrix} a&b\\c&d \end{pmatrix}-\begin{pmatrix} a&c\\b&d \end{pmatrix} = \begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}$. Setting this equal to $0$, we have that $b = c$. Since $\mathbb{R}^{2\times 2}$ is isomorphic to $\mathbb{R}^4$, we can write $\begin{pmatrix} a\\b\\b\\d \end{pmatrix}$ as the kernel, and find a basis: $\begin{pmatrix} a\\b\\b\\d \end{pmatrix} = a\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}+b\begin{pmatrix} 0\\1\\1\\0 \end{pmatrix}+c\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix}$. So a basis for $\ker T = \{ \begin{pmatrix} 1&0\\0&0 \end{pmatrix}, \begin{pmatrix} 0&1\\1&0 \end{pmatrix}, \begin{pmatrix} 0&0\\0&1 \end{pmatrix} \}$.
For 2:
We found that the $\operatorname{im} A$ has matrices of the form $\begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}$, so that the diagonals are negatives of each other. A basis for $\operatorname{im} A$ would thus be $\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}$.
For 3:
Now I don't know how we can prove that $T$ is diagonalizable. I know we have to find an eigenbasis for it to be diagonalizable, but I don't know how to find the eigenvalues.
| You already know three eigenvectors. Since your space is four-dimensional, you just need to find a fourth. Here's a hint: if $Tv = \lambda v$ and $\lambda \neq 0$, this means that $v$ must be in $T$'s....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help with inverse matrix problem? (Specific problem in description) \begin{equation}
\text{If}
\begin{vmatrix}A\end{vmatrix}
\text{=}\frac{1}{24}
\text{, solve }
\begin{vmatrix}
\begin{pmatrix}\frac{1}{3}A\end{pmatrix}^{-1} - 120 \text{ }A^*
\end{vmatrix}
\end{equation}
My attempt so far comes from the definition of an inverted matrix, namely
\begin{equation}
A^{-1} = \frac{A^{*}}{\begin{vmatrix}A\end{vmatrix}}
\end{equation}
So substituting gets
\begin{vmatrix}
\begin{pmatrix}\frac{1}{3}\frac{A^{*}}{\begin{vmatrix}A\end{vmatrix}}\end{pmatrix}^{-1} - 120 \text{ }\frac{A^{-1}}{\begin{vmatrix}A\end{vmatrix}}
\end{vmatrix}
Simplifying
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}A^{*}
\end{pmatrix}^{-1} - 5 \text{ }A^{-1}
\end{vmatrix}
Substituting using the inverse matrix definition again
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}A^{-1}|A|\end{pmatrix}^{-1} - 5A^{-1}
\end{vmatrix}
Using
\begin{equation}
(AB)^{-1} = B^{-1}\cdot A^{-1}
\end{equation}
And I get this?
\begin{vmatrix}
\begin{pmatrix}\frac{1}{72}|A|\end{pmatrix}^{-1}
\begin{pmatrix}A^{-1}\end{pmatrix}^{-1} - 5A^{-1}
\end{vmatrix}
I think I went about this the wrong way completely, but I'm not getting any new ideas for now - woud appreciate any tips or hints regarding this. Thanks.
| Use the $A^{-1}$ only in second part or first part of the problem not both
\begin{equation}
A^{*} = \frac{1}{24}{A^{-1}}
\end{equation}
sub this where $A^*$ is in second part and don't change $A^{-1}$ until fully simplified
$=|\frac13 A^{-1} - 120 (\frac{1}{24} A^{-1})|$
$=|\frac13 A^{-1} - 5 A^{-1})|$
$=(\frac{-14}{3})^n |A^{-1}|$
$=(\frac{-14}{3})^n (24)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about converting a polar equation to a rectangular equation $$\sec\theta =2$$
So I went through all the steps and got:
$$\cos\theta =\frac { 1 }{ 2 } $$
$$\sin\theta =\pm \sqrt { 1-\frac { 1 }{ 4 } } $$
$$\sin\theta =\pm \frac { \sqrt { 3 } }{ 2 } $$
$$y=\pm {\sqrt { 3 } }$$
Now why is it that the correct answer has an $x$ in it and why can't I give the answer $x^2+y^2=4$?
| From $\sec\theta=2$, draw a triangle in either the first or fourth quadrant with angle $\theta$, adjacent side $1$, opposite side $\sqrt 3$, hypotenuse $2$. Then
\begin{align}
\sec\theta=2\implies \tan\theta&=\pm\sqrt{3}\\
{r\sin\theta\over r\cos\theta}&=\pm \sqrt{3}\\
{y\over x}&=\pm \sqrt{3}\\
y&=\pm \sqrt{3}\,x.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
using power expansion to find limit I am preparing for my final exam, and stuck on this question.
Using power series expansion, evaluate $$\lim_{x\to 0} \frac{x\cos(x)
-\sin(x)}{x^2-x\ln(1+x)}$$
I have no idea how to proceed. Any help would be highly appreciated!
| We have: $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} +\cdots $,
$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} +\cdots $,
$\log(1+x) = x -\dfrac{x^2}{2} +\dfrac{x^3}{3} + \cdots $
Thus:
\begin{align}
x\cos x - \sin x
&= \left(x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!}+\cdots \right) - \left(x -
\dfrac{x^3}{3!} + \dfrac{x^5}{5!} + \cdots\right) \\
&= x^3\left(-\dfrac{1}{3} + \dfrac{x^2}{30}+ O(x^4)\right) \\
&= x^3\left(-\dfrac{1}{3} + O(x^2)\right)\tag{1}
\end{align}
\begin{align}
x^2 - x\log (1+x)
&= x^2 - x\left(x - \dfrac{x^2}{2} +\dfrac{x^3}{3}+\cdots \right) \\
&= \dfrac{x^3}{2} - \dfrac{x^4}{3} + \dfrac{x^5}{4} + \cdots \\
&= x^3\left(\dfrac{1}{2} - \dfrac{x}{3}+\dfrac{x^2}{4}\right) \\
&= x^3\left(\dfrac{1}{2} + O(x)\right)\tag{2}.
\end{align}
$(1), (2) \Rightarrow \displaystyle \lim_{x\to 0} \dfrac{x\cos x - \sin x}{x^2-x\log(1+x)} = \dfrac{-\dfrac{1}{3}}{\dfrac{1}{2}} = -\dfrac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.
| \begin{align}
\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}&=\int_0^\infty\frac{1}{x}\int_a^b \frac{\mathrm d}{\mathrm dy}\ln \left(x^2+2kx\cos y+k^2\right) \;\mathrm dy\,\mathrm dx\\[10pt]
&=-\int_0^\infty\frac{1}{x}\int_a^b \frac{2kx\sin y}{x^2+2kx\cos y+k^2} \;\mathrm dy\,\mathrm dx\\[10pt]
&=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{x^2+2kx\cos y+k^2} \;\mathrm dy\\[10pt]
&=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2-k^2\cos^2y} \;\mathrm dy\\[10pt]
&=2k\int_b^a\sin y\underbrace{\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2\sin^2y}}_{\large\color{blue}{(k\sin y)\, u\,=\,x+k\cos y}} \;\mathrm dy\\[10pt]
&=2\int_b^a\int_{\cot y}^\infty \frac{\mathrm du}{u^2+1} \;\mathrm dy\\[10pt]
&=2\int_b^a\left(\frac{\pi}{2}-\arctan\left(\cot y\right)\right)\;\mathrm dy\\[10pt]
&=2\int_b^ay\;\mathrm dy\\[10pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large a^2-b^2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 2
} |
What is the required radius of the smaller circles around a larger circle so they touch? I am trying to determine how to calculate the required radius of the smaller circles so they touch each other around the larger circle. (red box)
I would like to be able to adjust the number of smaller circles and the radius of the larger circle.
As an example:
$$\begin{align}
R&=1.5\\
n&=9\\
r&=\,?
\end{align}$$
| Suppose that the center $c_k,k=1,\ldots,n$ of the small circles are placed equidistantly on the bigger circle.
Then we have $c_k=\left(R\sin(2\pi\frac{k}{n}),R\cos(2\pi\frac{k}{n})\right), k = 1,\ldots,n$. So for $k=1,\ldots,n$ we must have $r=\frac{\|c_{k}-c_{k+1}\|_2}{2}$, since two circles with same radius are tangent if their radius is the half of the distance between their centers. In particular $$r=\frac{\|c_{n-1}-c_{n}\|_2}{2}=\frac{1}{2}\sqrt{\big(R\sin(2\pi)-R\sin\big(2\pi\frac{n-1}{n}\big)\big)^2+\big(R\cos(2\pi)-R\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ =\frac{R}{2}\sqrt{\sin\big(2\pi\frac{n-1}{n}\big)^2+\big(1-\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ = \frac{R}{2}\sqrt{2-2\cos\big(2\pi\frac{n-1}{n}\big)} =\frac{R}{2}\sqrt{2\Big(1-\cos\big(2\pi\frac{n-1}{n}\big)\Big)}\\
=\frac{R}{2}\sqrt{4\sin\big(2\pi\frac{n-1}{n}\big)^2}=R\sin\big(\pi\frac{n-1}{n}\big)=R\sin\big(\frac{\pi}{n}\big).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$
$\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers.
Any help is appreciated. Thanks in advance.
| We have $16\triangle=(b+c+a)(b+c-a)$
$\iff16r\cdot s=2s\cdot2(s-a)$ where $r$ is the in-radius & $2s=a+b+c$
Using this, $\tan\dfrac A2=\dfrac r{s-a}=\dfrac14$
Finally use $\tan2x$ formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam
By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:
$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx-\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$
$$\Longleftrightarrow\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$$
Now I'm stuck. WolframAlpha says the indefinite integral of $\dfrac{\sin x\cos x}{\sin^4 x+\cos^4x}$ evaluates nicely to $-\frac12\arctan(\cos(2x))$.
I already factored $\sin^4 x+\cos^4 x$ into $1-\left(\frac{\sin(2x)}{\sqrt{2}}\right)^2$, but I don't know how to continue.. I suggest a substitution $u=\frac{\sin(2x)}{\sqrt{2}}$?
Could someone provide me a hint, or maybe an easier method I can refer to in the future?
| $1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.