Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$ There seems to be a pattern of sorts in the Fibonacci sequence:
The $5$th, $10$th, $15$th & $20$th values are:
$$5, 55, 610, 6765$$
Does this pattern continue ad infinitum? I've tried a few more using Binet's Formula and it seems to hold. So:
Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$?
Bonus Q: Are there any other patterns?
| Let us make a table of the Fibonacci sequence modulo $5$. If we can find two occurrences of the same two terms modulo $5$ with all the $F_{5k}$ (between those two occurence) being $0$ modulo $5$, we can prove this statement.
$$\begin{array}{c|c|c|}
\text{$F_1$ to $F_5$} & \text{1} & \text{1} & \text{2} & \text{3} & \text{0} \\
\text{$F_6$ to $F_{10}$} & \text{3} & \text{3} & \text{1} & \text{4} & \text{0} \\
\text{$F_{11}$ to $F_{15}$} & \text{4} & \text{4} & \text{3} & \text{2} & \text{0} \\
\text{$F_{16}$ to $F_{20}$} & \text{2} & \text{2} & \text{4} & \text{1} & \text{0} \\
\text{$F_{21}$ to $F_{25}$} & \text{1} & \text{1} & \text{2} & \text{3} & \text{0} \\
\end{array}$$
Since $F_1$ and $F_2$ are the same as $F_{21}$ and $F_{22}$, and $F_5$, $F_{10}$, $F_{15}$ and $F_{20}$ are all $0$, then this cycle repeats indefinitely, and hence $F_{5k} \equiv 0 \pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Given $\frac{1}{1^4}+\frac{1}{2^4}...\infty=\frac{\pi^4}{90}$ If $$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}...\infty=\frac{\pi^4}{90}$$
then, $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}...\infty=?$$
Please provide a hint instead of the complete answer. Thanks. [Please edit the tags for me as I have no idea which sector this problem lies in.]
| $$\sum_{n=1}^{\infty}\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}+\sum_{n=1}^{\infty}\frac{1}{(2n)^4}$$
$$\sum_{n=1}^{\infty}(\frac{1}{n^4}-\frac{1}{16n^4})=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}$$
$$(1-\frac{1}{16})\sum_{n=1}^{\infty}\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the limit of the serie $\sum_{k=0}^{\infty}\frac{x^4}{(1+x^2)^k}$ Find the limit of the series :
$$\sum_{k=0}^{\infty}\frac{x^4}{(1+x^2)^k}$$
So, I applied the ratio test and did $\frac{x^4}{(1+x^2)^{n+1}}$*$\frac{(1+x^2)^n}{x^4}$.
Then, I get $1 + \frac{1}{x^2}$. I then multiply $1 + \frac{1}{x^2}$ by $x^4$, to which I get an answer choice of (E) $x^4 + x^2$
I kind of want to know if I'm actually doing the process correctly, and that $x^4 + x^2$ is the limit of the aforementioned series.
These are the following answer choices:
(A) $x^6 + x^4$
(B) $\frac{x^6}{1+x^2}$
(C) $x^6$
(D) $x^4$ + $\frac{x^6}{1+x^2}$
(E) $x^4$ + $x^2$
| hint: You basically have a geometric series with the first term $a_1 = x^4, r = \dfrac{1}{1+x^2}$. The sum is $\dfrac{a_1}{1-r}=...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integration and derivative together I have the function $f(x)=\ln(x+\sqrt{x^2+1})$ and I have already find that $$f'(x)=\frac{1}{\sqrt{x^2+1}}$$ and I want to find the $$\int_0^1{\sqrt{x^2+1}}dx$$ so:
$$\int_0^1{\sqrt{x^2+1}}dx=\int_0^1{\frac{1}{f'(x)}}dx$$
but I am stack and I don't know how to continue. Any ideas?
| Do an integration by parts. Let $I = \int \sqrt{x^2 + 1}\;dx$. Then
\begin{align}
I &= x\sqrt{x^2+1} - \int\frac{x^2}{\sqrt{x^2+1}}\;dx \\
&= x\sqrt{x^2+1} - \int\frac{x^2+1-1}{\sqrt{x^2+1}}\;dx \\
&= x\sqrt{x^2+1} - I + \int\frac{1}{\sqrt{x^2+1}}\;dx
\end{align}
So the integral $I$ is:
$$I = \frac{1}{2}\left(x\sqrt{x^2+1} + \log(x + \sqrt{x^2+1})\right)$$
Now, just plug in your integration limits!
EDIT: If you plug in your limits, you should get:
$$\int_0^1\sqrt{x^2+1}\;dx = \frac{1}{2}\left(\sqrt{2} + \log(1+\sqrt{2})\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Proving that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$ Prove
$$
\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}
$$
My Attempt:
Let $\alpha=\cos^{-1}\frac{4}{5}\implies0<\alpha\leq\pi$ and $\beta=\cos^{-1}\frac{12}{13}\implies0<\beta<\pi$,
thus $0\leq\alpha+\beta\leq2\pi$
$$
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{4}{5}.\frac{12}{13}-\frac{3}{5}.\frac{5}{13}=\frac{33}{65}=\cos\Big[\cos^{-1}\frac{33}{65}\Big]\\
\implies \cos^{-1}\frac{33}{65}=2n\pi\pm(\alpha+\beta)
$$
case 1: If $0\leq\alpha+\beta\leq\pi$,
$$
\cos^{-1}\frac{33}{65}=\alpha+\beta=\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}
$$
case 2: If $\pi<\alpha+\beta\leq 2\pi$,
$$
\cos^{-1}\frac{33}{65}=2\pi-(\alpha+\beta)
$$
Is there any thing wrong with my approach and how do I eliminate case 2 in similar problems ?
| Because $$\cos\left(\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}\right)=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}$$ and $$0^{\circ}<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<90^{\circ}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Roots to an equation, $\frac{3-2x}{x-1}-\sqrt[4]{x^3}$
Find the number of roots to $\frac{3-2x}{x-1}-\sqrt[4]{x^3}=0$.
My method:
$\frac{3-2x}{x-1}-\sqrt[4]{x}$, we multiply $\sqrt[4]{x}$ and $x-1$ to both sides, getting $3\sqrt[4]{x}-2x\sqrt[4]{x}-x(x-1)=0$. We let $\sqrt[4]{x}=k$. Then we have $3k-2xk-x^2+x=0$. Finding the discriminant ($b^2-4ac$), getting $4k^2+8k+1$, evidently greater than $0$, which means there are 2 roots.
Plotting this on Desmos gives 1 root only. How am I wrong?
| write your equation in the form
$$(3-2x)=(x-1)\sqrt[4]{x^3}(x-1)$$ for $x\ne 1$ and raise to the power four:
$$(3-2x)^4=x^3(x-1)^4$$
then you have to solve the equation
$$-{x}^{7}+4\,{x}^{6}-6\,{x}^{5}+20\,{x}^{4}-97\,{x}^{3}+216\,{x}^{2}-
216\,x+81
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \frac{1}{a \sin \theta}$$
However my textbook states that the answer should really be:
$$ -\frac{\sqrt{a^2 - x^2}+x\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})}{x}$$
Where did I mess up?
| To arrive to the textbook formula you need to substitute $v=\dfrac{x}{\sqrt{a^2-x^2}}$ but this is a bit difficult to guess.
In fact proceed to $x=a\tanh(u)$ to exploit $[1-\tanh^2]$.
Then $\displaystyle I=\int \dfrac{\sqrt{a^2-x^2}}{x^2}\mathop{dx}=\int\dfrac {\mathop{du}}{\cosh(u)\sinh(u)^2}$
Now substitute $v=\sinh(u)$
Notice that $v=\sinh(\tanh^{-1}(\frac xa))=\dfrac{\frac xa}{\sqrt{1-(\frac xa)^2}}$ is equal to the change proposed at the start.
$$\begin{align} I &=\int\dfrac{\mathop{dv}}{v^2(1+v^2)}=\int\left(\dfrac 1{v^2}-\dfrac 1{1+v^2}\right)\mathop{dv}=-\dfrac 1v-\arctan(v)+C\\\\&=-\dfrac{\sqrt{a^2-x^2}}x-\arctan\left(\dfrac{x}{\sqrt{a^2-x^2}}\right)+C\end{align}$$
Remark: you can also arrive to the same result with $x=a\sin(u)$ and $v=\tan(u)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Exponentiation of a symmetric $2 \times 2$ matrix Why does the following hold?
$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$
| You can use the split-complex numbers. What you've got on the LHS is the matrix representation of $a+bj, j^2=+1$. So:
$$
\begin{align}
(a+bj)^k &= \sum_{r=0}^k{k \choose r} a^{k-r}b^rj^r\\
&= \sum_{r=0}^k {n \choose r}a^{k-r}b^r[\text{$r$ even}] + j\sum_{r=0}^k {n \choose r}a^{k-r}b^r[\text{$r$ odd}]\\
&= \sum_{r=0}^k {n \choose r}a^{k-r}b^r\frac{1+(-1)^r}{2} + j\sum_{r=0}^k {n \choose r}a^{k-r}b^r\frac{1-(-1)^r}{2}\\
&= \frac{\sum_{r=0}^k {n \choose r}a^{k-r}b^r+\sum_{r=0}^k {n \choose r}a^{k-r}b^r(-1)^r}{2} + j\frac{\sum_{r=0}^k {n \choose r}a^{k-r}b^r-\sum_{r=0}^k {n \choose r}a^{k-r}b^r(-1)^r}{2}\\
&= \frac{(a+b)^k+(a-b)^k}{2} + j\frac{(a+b)^k-(a-b)^k}{2}
\end{align}$$
On the first line we use the binomial theorem.
On the second line we decompose into real and imaginary parts using the fact that $j^r=1$ when $r$ is even, and $j^r=j$ when $r$ is odd. We use Iverson Bracket notation.
On the third line we use the fact that $[\text{$r$ even}]=\frac{1+(-1)^r}{2}$ and $[\text{$r$ odd}]=\frac{1-(-1)^r}{2}$.
On the fourth line we distribute the summations.
On the fifth line we use the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find the minimum value of an expression with three variables How can I find the minimum of the following expression:
$(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x})(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy})$?
($x, y, z$ are non-zero real numbers)
The expression can be simplified to $\frac{((xy)^2+(zx)^2+(yz)^2)(x^2+y^2+z^2)}{(xyz)^2}$, but I am not sure that it will help anyhow.
| expanding gives
$$3+\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{x^2}{z^2}+\frac{z^2}{x^2}+\frac{y^2}{z^2}+\frac{z^2}{y^2}$$ and now use that $$A+\frac{1}{A}\geq 2$$ for $$A>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that when $x>0$, $(x+1/2)^3>3x^2+1/8$ without graphing Prove that when $x>0$, $(x+1/2)^3>3x^2+1/8$ without graphing.
The only thing I have tried till now is graphing, from which I proved this, but how would you prove this mathematically?
| \begin{align*}
\left(x+\frac{1}{2}\right)^3-3x^2-\frac{1}{8}&=x^3-\frac{3x^2}{2}+\frac{3x}{4}\\
&=x\left(x-\frac{3}{4}\right)^2+\frac{3x}{16}\\
&>0
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x$ using substitution I know that my result is wrong. What did I do wrong?
Let $u = \sqrt{x-1}$. Then we find first:
$$ u^2 = x - 1 \Rightarrow x = u^2 + 1$$
$$ \Rightarrow \mathrm{d}x = 2u \cdot \mathrm{d}u $$
So:
$$\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x = \int (u^2 + 1) \cdot u \cdot 2u \cdot \mathrm{d}u = 2 \cdot \int (u^4 + u^2) \mathrm{d}u$$
$$ = 2 \cdot (\frac{1}{5} u^5 + \frac{1}{2} u^2) + c = \frac{2}{5} u^5 + u^2 + c$$
$$ = \frac{2}{5} (\sqrt{x-1})^5 + x + c$$
Differentiating this does not give me $ x \cdot \sqrt{x-1}$, so it must be wrong. Which step went wrong and why?
| Because
$$2 \cdot \int (u^4 + u^2) \mathrm{d}u = 2 \cdot \left(\frac{1}{5} u^5 + \color{red}{\frac{1}{3} u^3}\right) + c = \frac{2}{5} u^5 + \color{red}{\frac{2}{3}u^3} + c$$
$$ = \frac{2 (\sqrt{x-1})^5}{5} + \color{red}{\frac{2\sqrt{(x-1)^3}}{3}} + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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power logsine integral $\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ It is not hard to find that $\int_0^\frac{\pi}{2}\ln{(\sin{x})}dx=-\frac{\pi}{2}\ln2$.
https://socratic.org/questions/how-do-you-prove-that-the-integral-of-ln-sin-x-on-the-interval-0-pi-2-is-converg
Moreover wolfram can compute that
$$\int_0^\frac{\pi}{2}\ln^2{(\sin{x})}dx=\frac{\pi^3+3\pi\ln^24}{24}$$
$$\int_0^\frac{\pi}{2}\ln^3{(\sin{x})}dx=-\frac{\pi}{16}(12\zeta(3)+\ln^34+\pi^2\ln4)$$
and $$\int_0^\frac{\pi}{2}\ln^4{(\sin{x})}dx=\pi\zeta(3)\ln{8}+\frac{19\pi^5}{480}+\frac{\pi}{2}\ln^42+\frac{\pi^3}{4}\ln^22$$
How to prove these formulas?
In general, How to evaluate the integral $\displaystyle\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ ?
Thanks in advance.
| $\begin{align}J_n=\int_0^{\frac{\pi}{2}}\ln^n(\sin x)\,dx\end{align}$
Perform the change of variable $y=\sin x$,
$\begin{align}J_n=\int_0^1 \frac{\ln^n(x)}{\sqrt{1-x^2}}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}J_n=\frac{1}{2^{n+1}}\int_0^1 \frac{x^{-\frac{1}{2}}\ln^n(x)}{\sqrt{1-x}}\,dx\end{align}$
Consider,
$\begin{align}B_n(s)&=\frac{1}{2^{n+1}}\int_0^1 x^{-\frac{1}{2}+s}(1-x)^{-\frac{1}{2}}\,dx\\
&=\frac{1}{2^{n+1}}\text{B}\left(\frac{1}{2}+s,\frac{1}{2}\right)\\
&=\frac{1}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1+s)}\\
&=\frac{\sqrt{\pi}}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)}{\Gamma(1+s)}
\end{align}$
$\begin{align}\boxed{J_n=\frac{\partial^n B_n(s)}{\partial s^n}\big|_{s=0}}\end{align}$
NB:
$\text{B}(,)$ denotes the Beta Euler function.
Example:
Let $n=1$.
$\begin{align}J_1=\frac{\sqrt{\pi}}{4}\frac{\Gamma^\prime\left(\frac{1}{2}\right)\Gamma(1)-\Gamma\left(\frac{1}{2}\right)\Gamma^\prime(1)}{\Gamma^2(1)}\end{align}$
But,
$\begin{align}\Gamma^\prime\left(\frac{1}{2}\right)=-(\gamma+2\ln 2)\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(1\right)=1\end{align}$
$\begin{align}\Gamma^\prime\left(1\right)=-\gamma\end{align}$
Therefore,
$\begin{align}J_1&=\frac{\sqrt{\pi}}{4}\times \frac{-(\gamma+2\ln 2)\sqrt{\pi}+\gamma\sqrt{\pi}}{1}\\
&=\boxed{-\frac{1}{2}\pi\ln 2}
\end{align}$
Addendum:
To compute $J_n$ one needs to be able to evaluate $\Gamma^{(k)}(1),\Gamma^{(k)}\left(\frac{1}{2}\right)$ for $0\leq k\leq n$.
Fact 1):
$\begin{align} \Gamma^{(m+1)}(z)=\sum_{k=0}^{m} \binom{m}{k}\Gamma^{(k)}(z)\psi^{(m-k)}(z)\end{align}$
(following from the definition of Digamma function and the use of Leibniz-Newton formula)
Fact 2):
For $0<|z-1|<1$,
$\begin{align}\psi(z)=-\gamma+\sum_{k=1}^{\infty}(-1)^{k+1}\zeta(k+1)(z-1)^k\end{align}$
Therefore,
For $k\geq 1$,
$\displaystyle \psi^{(k)}(1)=(-1)^{k+1}k!\zeta(k+1)$
Fact 3):
For $0<\Re(z)<1$,
$\displaystyle \psi(z)+\psi(1-z)=\pi\cot(\pi x)$
(allowing to compute $\displaystyle \psi^{(k)}\left(\frac{1}{2}\right)$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
\begin{align}
\pars{{p \over q} = z \implies
\left\{\begin{array}{lcl}
\ds{p + q \over q} & \ds{=} & \ds{z + 1}
\\[1mm]
\mbox{and}&&
\\[1mm]
\ds{p - q \over q} & \ds{=} & \ds{z - 1}
\end{array}\right.}
\implies \bbx{{p \over q} = z \implies {p + q \over p - q} =
{z + 1 \over z - 1}}\label{1}\tag{1}
\end{align}
Then,
\begin{align}
&\arctan\pars{\root{1 + x^{2}} - \root{1 - x^{2}} \over
\root{1 + x^{2}} + \root{1 - x^{2}}} = \alpha
\implies
{\root{1 + x^{2}} - \root{1 - x^{2}} \over
\root{1 + x^{2}} + \root{1 - x^{2}}} = \tan\pars{\alpha}
\end{align}
With the identity \eqref{1}:
\begin{align}
&{2\root{1 + x^{2}} \over
-2\root{1 - x^{2}}} = {\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1}
\implies
{1 + x^{2} \over 1 - x^{2}} = \bracks{\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1}^{2}
\\[5mm] \stackrel{\mrm{see}\ \eqref{1}}{\implies}\,\,\ &
{2 \over 2x^{2}} = {\braces{\bracks{\tan\pars{\alpha} + 1}/
\bracks{\tan\pars{\alpha} - 1}}^{\,2} + 1 \over
\braces{\bracks{\tan\pars{\alpha} + 1}/
\bracks{\tan\pars{\alpha} - 1}}^{\,2} - 1} =
{2\tan^{2}\pars{\alpha} + 2 \over 4\tan\pars{\alpha}}
\\[5mm] \implies &\
x^{2} =
{2\tan\pars{\alpha} \over \tan^{2}\pars{\alpha} + 1} =
{2\tan\pars{\alpha} \over\sec^{2}\pars{\alpha}} =
2\sin\pars{\alpha}\cos\pars{\alpha} \implies \bbx{x^{2} = \sin\pars{2\alpha}}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2644104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Set of values of $a$ satisfying an inverse trigonometric equation.
Find the set of values of a so that the equation $(\arcsin x)^3+ (\arccos x)^3= a\pi^3$ has a solution.
Attempt:
Let $y = \arcsin x$ (just for convenience to avoid typing/ writing $\arcsin$ over and over again )
Note that $y + \arccos x = \pi /2$
We have
$(y+\arccos x)(y^2 + (\frac\pi 2-y)^2 - y (\frac \pi2 - y)= a\pi^3$
$\implies (\sqrt 3y- \dfrac{\sqrt3}{4}\pi)^2+\dfrac{\pi^2}{16}= 2a\pi^2$
$-\frac \pi 2\le y \le \frac\pi 2$
$\implies -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4\\ \implies 0\le 2a\pi^2- \dfrac{\pi^2}{16}\le \dfrac{3}{16}\pi^2 \implies a\in [\dfrac{1}{32}, \dfrac{1}{16}]$
I get the right lower limit of $a$ but the upper limit given in the answer is $7/8$.
I'll be grateful if someone could point out my error and if that has been done, provide alternative, easier ways to solve this problem.
| The question is equivalent to finding the range of the function $$x\mapsto \frac{(\arcsin x)^3+ (\arccos x)^3}{\pi^3}$$
defined on $[-1,1]$.
You can do that by studying the variations of the function via its derivative, and the data of extremal values in $x=-1$ and $x=1$.
Edit
The problem is here:
$ -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4$
from here, the largest value for the square will come from squaring the LHS, not the RHS.
| {
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How to show $\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}$ I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help?
(I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},z_0=e^\frac{\pi i}{3} \right )$
Thanks!
| Here is an approach that makes use of Euler's Beta function. Enforcing a substitution of $x \mapsto \sqrt[3]{x}$ one has
\begin{align*}
\int_0^\infty \frac{dx}{x^3 + 1} &= \frac{1}{3} \int_0^\infty \frac{x^{-2/3}}{x + 1} \, dx\\
&= \frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3} - 1}}{(x + 1)^{\frac{1}{3} + \frac{2}{3}}}\\
&= \frac{1}{3} \text{B} \left (\frac{1}{3}, \frac{2}{3} \right ) \tag1\\
&= \frac{1}{3} \frac{\Gamma \left (\frac{1}{3} \right ) \Gamma \left (\frac{2}{3} \right )}{\Gamma (1)} \tag2\\
&= \frac{1}{3} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\
&= \frac{1}{3} \cdot \frac{\pi}{\sin (\pi/3)} \tag3\\
&= \frac{1}{3} \cdot \frac{2\pi}{\sqrt{3}}\\
&= \frac{2\pi}{3 \sqrt{3}}.
\end{align*}
Explanation
(1) Since $\displaystyle{\text{B}(x,y) = \int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Since $\text{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) From Euler's reflection formula.
| {
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Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
| Keep going! $$\begin{align}\frac{1}{x + \sqrt{x^2 - 1}}\left(1 + \frac{2x}{2\sqrt{x^2 - 1}}\right)&=\frac{1}{x + \sqrt{x^2 - 1}}\left(\frac{x+\sqrt{x^2-1}}{\sqrt{x^2 - 1}}\right)\\&=\frac1{\sqrt{x^2-1}}\end{align}$$
| {
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Prove that $\int{\frac{x^2}{(x^2 + a^2)^n}}dx = \int{\frac{1}{(x^2 + a^2)^n}}dx - a^2\int{\frac{1}{(x^2 + a^2)^{n+1}}}dx$ I am trying to derive the following reduction formula for integrating $\int{\frac{1}{(x^2 + a^2)^n}}dx, n,a>0$:
$$I_n = \frac{1}{2a^2(n-1)}\left(\frac{x}{(x^2 + a^2)^{n-1}} + (2n-3)I_{n-1} \right)$$
So far I got $$\int{\frac{1}{(x^2 + a^2)^n}}dx = \frac{x}{(x^2 + a^2)^n} + 2n\int{\frac{x^2}{(x^2 + a^2)^{n+1}}}dx$$
and
$$I_n = \frac{x}{(x^2 + a^2)^n} + 2nI_n - 2na^2I_{n+1} \Rightarrow I_n = \frac{1}{2a^2(n-1)}\left(\frac{x}{(x^2 + a^2)^{n-1}} + (2n-3)I_{n-1} \right)$$
but I'm struggling with the step in between: $$\int{\frac{x^2}{(x^2 + a^2)^n}}dx = I_n - a^2I_{n+1}$$
| $$\frac{1}{(x^2+a^2)^{n}}-\frac{a^2}{(x^2+a^2)^{n+1}}=$$
$$\frac{x^2+a^2}{(x^2+a^2)^{n+1}}-\frac{a^2}{(x^2+a^2)^{n+1}}=$$
$$\frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}}=$$
$$\frac{x^2}{(x^2+a^2)^{n+1}}\neq\frac{x^2}{(x^2+a^2)^{n}}$$
| {
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Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$
| Convince yourself that
$$\frac1{n^2+n}+\frac2{n^2+n}+\cdots+\frac n{n^2+n}\le \frac1{n^2+1}+\frac2{n^2+2}+\cdots+\frac n{n^2+n} \le$$
$$\le \frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2},$$
and remember that $1+2+\cdots+n=\tfrac {n(n+1)}2$. Can you apply the Sandwich rule now?
| {
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"url": "https://math.stackexchange.com/questions/2648984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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MIT Integration Bee $21$ Question:
$$\int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx$$
How to solve these integrals. I have no idea.
| Use the substitution $x\mapsto 1/x$:
\begin{align*}
I &= \int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx \\
&= \int_0^\infty {{1}\over{(1+x^{-2015})({1+x^{-2}}})} \frac{1}{x^2}dx\\
&= \int_0^\infty {{x^{2015}}\over{(1+x^{2015})({1+x^{2}}})} dx
\end{align*}
Then,
\begin{align*}
2I &= \int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx+\int_0^\infty {{x^{2015}}\over{(1+x^{2015})({1+x^2}})}dx\\
&=\int_0^\infty {{1+x^{2015}}\over{(1+x^{2015})({1+x^2}})}dx\\
&=\int_0^\infty \frac{1}{1+x^2}dx = \frac{\pi}{2}
\end{align*}
So $I=\pi/4$.
| {
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Prove $15\mid 4^{2n}-1$ without induction
Prove without induction that $\forall n \in\mathbb Z$, $15\mid4^{2n}-1$.
$4^{2n} = (4^2)^n = 16^n$.
If $n=1,$ then $(4^2)^n-1=15$, and for $n=2$, it is $255$, which is divisible by $15$.
Using the congruence arithmetic, any $n$ in $4^{2n}-1$ can be expressed as a product of prime factors. Also, there is only one even prime, and any odd prime will be having for any positive integer $k$, an addition of $1+ k.$(even terms).
Say, for $n=5$, with $5=2^2+1$.
And it already proved that any even power of $n$ leads $4^{2n}-1$ to be a multiple of $15.$
edit Based on the selected answer, have observations, that are possibly implied in that answer, but not explicitly stated.
$4^{2n} -1 = (2^n -1)(2^n+1)(2^{2n}+1) = (2^n -1)(2^n+1)(4^n+1)$
Need consider both even and odd cases separately, for values w.r.t. two modulo $3,5$.
(A) $n$ is even :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv 1 \pmod 3 \implies (2^n-1) \equiv 0 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 2 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv 0 \pmod 3)(\equiv 2 \pmod 3)(\equiv 1 \pmod 3)$
As the first term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: Take $n = 2n', 2^2 \equiv -1 \pmod 5\implies$$ 4^{n'} \equiv 1 \pmod 5 $ $\implies (2^n-1) \equiv 0 \pmod 5$
(b) $ 2^n+1$ :: Take $n = 2{n'}, 4^{n'} +1 \equiv 2 \pmod 5$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n \equiv 1 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(\equiv 0 \pmod 5)(\equiv 2 \pmod 5)(\equiv 1 \pmod 5)$
As the first term of the product is a factor of $5$, so divisible by $5$.
(B) $n$ is odd :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv -1 \pmod 3 \implies (2^n-1) \equiv -2 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 0 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv -2 \pmod 3)(\equiv 0 \pmod 3)(\equiv 1 \pmod 3)$
As middle term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(b) $ 2^n+1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n +1\equiv 0 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(....)(...)(\equiv 0 \pmod 5)$
As the last term of the product is a factor of $5$, so divisible by $5$.
Don't know how to handle the last case's , sub cases (a), (b) for checking w.r.t. modulo $5$.
| $$4^{2n}-1=16^n-1=(16-1)(16^{n-1}+16^{n-2}+\dots+1)$$
| {
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Prove the inequality $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$
Let $a,b,c$ are nonnegative real numbers such that $a^2+b^2+c^2=1$.
Prove the inequality
$$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$$
I tried the method of Lagrange multipliers and Jensen's inequality but I have not been proved this inequality
| Substitute $a^2, b^2, c^2$ by $x, y, z$. Then we will try to bound $\sqrt{x}/(1+x)$ by a linear form of $x$.
Consider a line tangent to the expression when the equality occurs i.e. when $x=1/3$. Evaluating $f’(1/3)=3\sqrt{3}/16$, we find that such line is $y=\frac{3\sqrt{3}(x+1)}{16}$. By simple algebraic manipulation,
$$\frac{\sqrt{x}}{1+x}\leq\frac{3\sqrt{3}(x+1)}{16}$$
is equivalent to $(3x-1)^2(3x^2+14x+27)\geq0$, which is obviously true. The desired result follows immediately.
| {
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Factoring High School Level Olympiad Problem Factor $x^2 - 3xy + 2y^2 + x -8y - 6$
Attempt at a solution:
I have factored these and don't know how to continue...
$x^2-3xy +2y^2 = (x - y) (x-2y)$
$x^2 + x -6 = (x + 3) (x - 2)$
$2y^2 - 8y + 6 = 2 (y - 3)(y - 1)$
| Look for in form:
$$x^2-3xy+2y^2+x-8y-6=(x+Ay+B)(x+Cy+D)
$$
Plug $y=0$:
$$x^2+x-6=(x+B)(x+D) \Rightarrow B=3; D=-2.$$
Plug $x=0$:
$$2y^2-8y-6=(Ay+3)(Cy-2) \Rightarrow \begin{cases} AC=2 \\ -2A+3C=-8 \end{cases}$$
Can you finish?
Appendix: Note that the found parameters will not be suitable. So this method may not always work.
| {
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Solve a system of equation
Solve this system of equations.
$$\begin{cases}
x^{2} + y^{2} = 4 \\
z^{2} + t^{2} = 9 \\
xt + yz = 6
\end{cases}$$
My attempts:
*
*$(x^{2}+y^{2})(z^{2} + t^{2})=36 \xrightarrow{\text{by applying Lagrange formula}} (xt + zy)^{2} + (zx -ty)^{2}=36 \xrightarrow{xt + yz = 6} 36 + (zx -ty)^{2} =36\Rightarrow (zx -ty)^{2} = 0 \Rightarrow zx = ty \Rightarrow z = \frac{ty}{x}$
*$xt + yz = 6 \xrightarrow{z = \frac{ty}{x}} xt + \frac{ty^{2}}{x} = 6 \Rightarrow x^{2}t + y^{2}t = 6x \Rightarrow t(x^{2} + y^{2})=6x \xrightarrow{x^{2} + y^{2} = 4} t=\frac{3}{2}x$
*$z= \frac{ty}{x} \xrightarrow{t=\frac{3}{2}x} z = \frac{3}{2}y$
Can we go further and determine the exact value of variables?
| $$\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2=1$$
$$\left(\frac{t}{3}\right)^2+\left(\frac{z}{3}\right)^2=1$$
Let $x_1 = \frac{x}2, y_1=\frac{y}2, t_1=\frac{t}{3}, z_1 = \frac{z}3$.
Then $P=(x_1, y_1)$ and $Q=(t_1, z_1)$ are points on the unit circle.
Also, from the third constraint, we know that $\cos \angle POQ= 1 $
Hence $\frac{x}2=\frac{t}{3}$ and $\frac{y}{2}=\frac{z}{3}$ and they need to satisfy $\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2=1$. The answer is not unique.
We can charactherize it by a parameter $\theta$ where $\cos(\theta)=\frac{x}2=\frac{t}{3}$ and $\sin(\theta)=\frac{y}{2}=\frac{z}{3}$
| {
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Prove the inequality $\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge\left(1+\frac{1}{p(1+p)}\right)^k$
Let $a_1, a_2,...,a_k$ are any positive real numbers. Prove the inequality
$$\left(1+\frac{1}{a_1(1+a_1)}\right)\left(1+\frac{1}{a_2(1+a_2)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge$$
$$\ge\left(1+\frac{1}{p(1+p)}\right)^k$$
where $\sqrt[k]{a_1a_2...a_k}=p$.
My work so far:
I used Jensen's inequality.
Let $f(x)=\ln\left(1+\frac{1}{x(1+x)}\right)$. If $x\ge0$ then $f''(x)\ge0$. Then
$$\ln\left(1+\frac{1}{a_1(1+a_1)}\right)+...+\ln\left(1+\frac{1}{a_k(1+a_k)}\right)\ge k\ln\left(1+\frac{1}{q(1+q)}\right)$$
where $q=\frac{a_1+a_2+...a_k}{k}$.
Then $$\ln\left(\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\right)\ge \ln\left(1+\frac{1}{q(1+q)}\right)^k$$
$$\left(1+\frac{1}{a_1(1+a_1)}\right)...\left(1+\frac{1}{a_k(1+a_k)}\right)\ge \left(1+\frac{1}{q(1+q)}\right)^k$$
But $$\left(1+\frac{1}{q(1+q)}\right)\not \ge\left(1+\frac{1}{p(1+p)}\right)$$
I can not finish the proof of this inequality
Addition 1:
Addition 2:
| Hint: Set $b_i = \ln(a_i)$ and apply Jensen's inequality to
$$\ln\left(1+\frac{1}{e^x(1+e^x)}\right).$$
| {
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Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$ I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as
$$p(x,y)=(y+a)(y^2+by+c) = y^3 + (a+b)y^2 + (ab+c)y + ac$$
for some $a,b,c \in \mathbb{Z}[x]$.
Comparing coefficients we have
\begin{align}
&a+b=x \\
&ab+c = x^2+x \\
&ac = 4x^3+2x = 2x(2x^2+1)
\end{align}
From the last equation we have $a = 2x$ and $c=2x^2+1$ or alternatively $a = 2x^2+1$ and $c=2x$.
If $a=2x$, $c=2x^2+1$ then from the first equation, $2x+b=x$ so $b=-x$. From the second equation, $-2x^2+x-1=x^2+x$, but this doesn't hold for all $x$...
If $a=2x^2+1$, $c=2x$ then from the first equation, $2x^2+1+b=x$ so $b=-2x^2+x-1$. From the second equation, $(2x^2+1)(-2x^2+x-1) = x^2+x$, but again the l.h.s. and r.h.s. don't match since without even computing we see the l.h.s. is degree $4$.
Is there a better way to prove this?
| Rewrite:
$$p(x,y)= y^3+xy^2+x(x+1)y+2x(2x^2+1), $$
and as $p(x,y)\in \mathbf Z[x][y]$, we can use the generalisation of Eisenstein's criterion directly:
$(x)$ is a prime ideal in the U.F.D. $\mathbf Z[x]$, and
*
*$x$ divides divides all coefficients, but the leading coefficient;
*$x^2$ does not divide the constant term.
Hence $p(x,y)$ is irreducible in $\mathbf Z[x][y]\simeq\mathbf Z[x,y]$.
| {
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Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ in $x\in(0,2\pi)$
solution i try $1+\sin 2x =\sin x+\sin^2 x$
$5+4\sin 2x=4\sin^2 x+4\sin x+1$
$5+\sin 2x=(2\sin x+1)^2$ How i solve it after that point
| Use the tan-half-angle substitution ($t \rightarrow \tan\left( \frac{x}{2} \right)$) to turn the equation into a polynomial.
$$ \begin{cases}
\sin(x) = \frac{2 t}{1+t^2} \\
\cos(x) = \frac{1-t^2}{1+t^2} \\
x =2 \tan^{-1}(t)
\end{cases} $$
Since $\sin(2 x) = 2 \sin(x) \cos(x)$ the expression is
$$ 1 + 2 \frac{2 t}{1+t^2} \frac{1-t^2}{1+t^2} = \frac{2 t}{1+t^2} + \left( \frac{2 t}{1+t^2} \right)^2 $$
or
$$ \left. \frac{2 t (3 t^2+2 t-1)}{(1+t^2)^2} = 1 \right\} (1+t^2)^2 -2 t ( 3 t^2+2 t-1) =0$$
This polynomial is of 4th order with two real solutions
$$ \begin{cases} t = 0.647989325215799 & \rightarrow x = 2.82499129600076
\\
t = 6.26423704911644 & \rightarrow x = 1.14992088410486
\end{cases} $$
| {
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"answer_id": 3
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Number of roots of $z^4+z^3+4z^2+2z+3=0$ in each quadrants This question has been asked but I am stuck with my method.
I have shown that the roots can not be on the real and imaginary axis. Since the coefficients are real and thus the roots must be in conjugate pairs. So If there is no root in first quadrant (or right half plane), then we have 2 roots in second quadrant and third quadrant respectively. (this is the answer to this question)
So I am going to show that there is no root in right half plane. Consider the integral along the semi-circle in right half plane with radius $R$ $$\int_{C}\dfrac{f'}{f}dz=\int_{-Ri}^{Ri}+\int_{-\pi/2}^{\pi/2}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz$$
For the second part, $$\int_{-\pi/2}^{\pi/2}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz=i\int_{-\pi/2}^{\pi/2}\dfrac{4z^4+3z^3+8z^2+2z}{z^4+z^3+4z^2+2z+3}d\theta=4\pi i$$
as $R\rightarrow \infty$.
For the first part,
\begin{align*}
&\int_{-Ri}^{Ri}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz\\
&=\int^R_{-R}\dfrac{3y^2-2+i(4y^3-8y)}{y^4-4y^2+3-i(y^3+2y)}idy\\
\\
&=\int^R_{-R}\dfrac{(3y^2-3)(y^4-4y^2+3)-(4y^3-8y)(y^3-2y)+i[(4y^3-8y)(y^4-4y^2+3)+(y^3-2y)(3y^2-2)]}{(y^4-4y^2+3)^2+(y^3+2y)^2}idy\\
&=\int^R_{-R} i(even \,\,part)-(odd \,\,part) dy\\
&=\int^R_{-R} i(even \,\,part) \,\,dy
\end{align*}
It is expected that $\int^R_{-R} i(even \,\,part) \,\,dy=-4\pi i$, then we are done, but how?
Also is there any method using Rouche theorem?
| The coefficients are positive, so the polynomial has no positive real roots.
We check that $f$ has no purely imaginary roots. Let $z=yi$, then $$f=y^4-4y^2+3+i(2y-y^3).$$ If $f=0$, we have $y=0$ or $\pm\sqrt{2}$ from the imaginary part, but $f\neq 0$ by substituting these values of $y$ into the real part of $f$.
We check that there is no negative real roots. On the real axis, we have $$f'=4z^3+3z^2+8z+2 \text{ and } f''=12z^2+6z+8>0.$$ From these, we see that $f$ is convex and have local minima. Let $x_1$ such that $f'(x_1)=4x_1^3+3x_1^2+8x_1+2=0$ and $f(x_1)$ is the global minimum. Then on the real axis,
\begin{align*}
f(z)&\geq f(x_1)=x_1^4+x_1^3+4x_1^2+2x_1+3\\
&=x_1^4+\dfrac{1}{4}(4x_1^3+16x_1^2+8x_1+12)\\
&=x_1^4+\dfrac{1}{4}(13x_1^2+10)>0.
\end{align*}
It is enough to consider the number of roots in the right half plane. We take a contour $C$ to be a right half circle with radius $R$,
$$\int_C \dfrac{f'}{f}dz=\int_{-\pi/2}^{\pi/2}+\int_{Ri}^{-Ri} \dfrac{f'}{f}dz.$$
For the first integral, $z=Re^{i\theta}$
\begin{align*}
\int_{-\pi/2}^{\pi/2} \dfrac{f'}{f}dz&=\int_{-\pi/2}^{\pi/2} \dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}Re^{i\theta}id\theta\\
&\to4\pi i \text{ as }R \to \infty
\end{align*}
For the second integral, $$\int_{Ri}^{-Ri} \dfrac{f'}{f}dz=\int_{Ri}^{-Ri} \dfrac{df}{f-0}.$$ It is the winding number of $f$ around 0. If we set $z=yi$, we have $f=y^4-4y^2+3+i(2y-y^3)$. We find all the zero of the real and imaginary parts, that is $y=\pm 1,\pm \sqrt{3}$ and $y=0,\pm\sqrt{2}$.
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
y & -\infty & -\sqrt{3} & -\sqrt{2} & -1 & 0 & 1 & \sqrt{2} & \sqrt{3} & \infty \\ \hline
Re(f) & + & 0 & - & 0 & + & 0 & - & 0 & + \\
\hline
Im(f) & 0 & + & 0 & - & 0 & + & 0 & - & 0 \\
\end{array}
Thus, $\dfrac{1}{2\pi i}\displaystyle\int_{Ri}^{-Ri} \dfrac{f'}{f}dz \to -2$ as $R \to \infty$. Combining our results, $\displaystyle\int_C \dfrac{f'}{f}dz \to 0$ as $R \to \infty$, which implies that there is no root in the right half plane. Thus there are 0, 2, 2, 0 roots in quadrant I, II, III, IV respectively.
| {
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} |
Study the Irreducibility of polynomial How could I study the irreducibility of the polynomial below
$((x-1)(x-2) \cdots (x-n))+1$ in $\mathbb{Q}[x]$
| Proposition:
a) The polynomial $P_n(x)=(x-1)(x-2) \cdots (x-n)+1 \in \mathbb{Q}[x]$ is irreducible for $n\neq 4$
b) $P_4(x)=(x^2-5x+5)^2$
Proof:
Suppose that $P_n(x)$ is reducible and write $P_n(x)=f(x)g(x)$ with $\operatorname {deg}f(x), \operatorname {deg}g(x)\lt n$ and assume (Gauss) that $f(x),g(x)\in \mathbb Z[x]$.
Then evaluating at the integers $1\leq k \leq n$ we see that $P_n(k)=1=f(k)g(k)$, forcing $$f(k)=g(k)=\epsilon_k=\pm1$$ But then the polynomial $f(x)-g(x)$ has $n$ zeros (namely $1,\cdots,n$) and degree $\lt n$.
This implies $f(x)-g(x)=0$ so that $f(x)=g(x)$ and $$P_n(x)=(x-1)(x-2) \cdots (x-n)+1 =(f(x))^2 \quad (\bigstar)$$ This shows that the degree $n$ of $P_n$ must be even: $n=2m$.
Finally evaluating at $x=\frac 32$, we get from $(\bigstar)$:$$-\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdots \frac{4m-3}{2}+1=f(\frac 32)^2$$
The right-hand side $(f(x))^2$ is always $\geq 0$ but the left-hand side is $\geq 0$ only for $m=1$ or $2$, so that we already see that $P_n(x)$ can be factored as $f(x)g(x)$ only if $n=2m$ equals $2$ or $4$.
$\bullet $ For $m=1$ we get $P_2(x)=(x-1)(x-2)+1=x^2-3x+3$, which is irreducible.
$\bullet \bullet$ For $m=2$ we get $$P_4(x)=(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2$$ which is thus (strangely!) the only reducible polynomial of the family $P_n(x)$ .
| {
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Proving $\sum\limits_{k=0}^{\infty} \left\lfloor\frac{n+2^k}{2^{k+1}}\right\rfloor=n$
Find $f(n)$ for each natural number $n$, where
$$f(n)=\sum_{k=0}^{\infty} \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor.$$
I tried with a few different small $n$, $f(n)$ seems to be $n$. But is there a way to formally prove it?
| An easier theorem might be: for any natural number $n$ and any $0\le\alpha<1$,
$$\sum_{k=0}^{\infty}\left\lfloor\frac{n+\alpha+2^k}{2^{k+1}}\right\rfloor=n$$
It's true for $n=0$ because $\alpha+2^k<2^{k+1}$ and for $n=1$ because $2\le1+\alpha+1<4$ but $1+\alpha+2^k<2^{k+1}$ for $k>0$. Then if true for all natural numbers up to $n$,
$$\begin{align}\sum_{k=0}^{\infty}\left\lfloor\frac{2n+\alpha+2^k}{2^{k+1}}\right\rfloor&=\left\lfloor\frac{2n+\alpha+1}{2}\right\rfloor+\sum_{k=1}^{\infty}\left\lfloor\frac{2n+\alpha+2^k}{2^{k+1}}\right\rfloor\\
&=n+\sum_{k=0}^{\infty}\left\lfloor\frac{2n+\alpha+2^{k+1}}{2^{k+2}}\right\rfloor\\
&=n+\sum_{k=0}^{\infty}\left\lfloor\frac{n+\frac{\alpha}2+2^{k}}{2^{k+1}}\right\rfloor=n+n=2n\end{align}$$
And also
$$\begin{align}\sum_{k=0}^{\infty}\left\lfloor\frac{2n+1+\alpha+2^k}{2^{k+1}}\right\rfloor&=\left\lfloor\frac{2n+1+\alpha+1}{2}\right\rfloor+\sum_{k=1}^{\infty}\left\lfloor\frac{2n+1+\alpha+2^k}{2^{k+1}}\right\rfloor\\
&=n+1+\sum_{k=0}^{\infty}\left\lfloor\frac{2n+1+\alpha+2^{k+1}}{2^{k+2}}\right\rfloor\\
&=n+1+\sum_{k=0}^{\infty}\left\lfloor\frac{n+\frac{1+\alpha}2+2^{k}}{2^{k+1}}\right\rfloor=n+1+n=2n+1\end{align}$$
So it follows that it's true for all natural numbers $n$ and all $0\le\alpha<1$.
Hopefully it's clear why I can make this assertion: build the binary representation of $n$ starting from the high bit to low bit and at each stage it can be seen that the growing number satisfies the conclusion of the theorem, starting from $1$ and ending at $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding Jordan-normal form of a special $n\times n$ matrix Let $$A := \begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_n & -a_{n - 1} & -a_{n - 2} & \cdots & -a_1
\end{pmatrix}$$
I have to find $B$, $J$ such that $A = BJB^{-1}$ such that $J$ is a Jordan matrix. I already proved that $$\det(\lambda E - A) = \lambda^n + a_1\lambda^{n - 1} + ... + a_n$$ (just simple induction). I know that this gives the eigenvalues, and if $a_n, ..., a_1$ are known, I can find the eigenvalues and generalised eigenvectors, and write down the Jordan normal form, but I don't see how there is a general solution to this problem, as $(A - \lambda E)^n$ does not have a nice closed form. This is homework for a differential equation course, so maybe we can use some of that theory here?
| If you can get the companion matrix into a diagonal block decomposition of smaller companion matrices, one for each eigenvalue, you get a recipe for each block. For example $(x-c)^4 = x^4 - 4cx^3 + 6 c^2 x^2 - 4 c^3 x + c^4.$ We have
$$
A =
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-c^4 & 4 c^3 & - 6 c^2 & 4 c \\
\end{array}
\right)
$$
Take
$$
R =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
c & 1 & 0 & 0 \\
c^2 & 2c & 1 & 0 \\
c^3 & 3 c^2 & 3 c & 1 \\
\end{array}
\right)
$$
then
$$
R^{-1} =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
-c & 1 & 0 & 0 \\
c^2 & -2c & 1 & 0 \\
-c^3 & 3 c^2 & -3 c & 1 \\
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
-c & 1 & 0 & 0 \\
c^2 & -2c & 1 & 0 \\
-c^3 & 3 c^2 & -3 c & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-c^4 & 4 c^3 & - 6 c^2 & 4 c \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
c & 1 & 0 & 0 \\
c^2 & 2c & 1 & 0 \\
c^3 & 3 c^2 & 3 c & 1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrr}
c & 1 & 0 & 0 \\
0 & c & 1 & 0 \\
0 & 0 & c & 1 \\
0 & 0 & 0 & c \\
\end{array}
\right)
$$
| {
"language": "en",
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"answer_id": 2
} |
Prove that for a given point on an ellipse, the sum of the distances from each focal point is constant
An ellipse has equation $\frac{x^2}{25}+\frac{y^2}{9} = 1$ and $P(p,q)$ is a point on the ellipse. Points $F_1$ and $F_2$ have coordinates $(-4,0)$ and $(4,0)$. Show that the sum of the distances $|PF_1|$ +$|PF_2|$ does not depend on the value of p.
First, to find distance $|F_1P|$
$|F_1P| = \sqrt{(p+4)^2 + q^2}$
as $\frac{p^2}{25}+\frac{q^2}{9} = 1$
$\frac{9p^2}{25}+q^2 = 9$
$q^2 = 9-\frac{9p^2}{25}$
$|F_1P| = \sqrt{p^2+8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2+8p-25}$
$= \sqrt{\frac{1}{25}(16p^2+200p-625)}$
$= \frac{1}{5}\sqrt{(4p+25)^2}$
$= \frac{1}{5}(4p+25)$
Finding the distance $|F_2P|$
$|F_2P| = \sqrt{(4-p)^2 + q^2}$
$= \sqrt{p^2-8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2-8p-25}$
$= \sqrt{\frac{1}{25}(16p^2-200p-625)}$
$= \frac{1}{5}\sqrt{(4p-25)^2}$
$= \frac{1}{5}(4p-25)$
Therefore
$|F_1P|+|F_2P|= \frac{1}{5}(4p-25)+\frac{1}{5}(4p+25) = \frac{8}{5}p$
This is exactly what we're trying to disprove. I must have made a mistake somewhere. Please can someone explain where I went wrong. My answer appears to be the negative of what it should be. I expect the correct answer will be 10 units but for some reason it does not seem to work.
| The key mistake you made here is your simplification of square roots.
Actually, since a square root is positive always, you should get $\vert F_1P\vert = \frac{|4p+25|}{5} $ and $ \vert F_2P\vert = \frac{|4p-25|}{5}$.
However, note $ -5 \leq p \leq 5$, so $ -20 \leq 4p \leq 20$ and $\frac{|4p+25|}{5} = \frac{4p+25}{5}, \frac{|4p-25|}{5} = \frac{-4p+25}{5}$ Summing the two distances, $p$ cancels out.
| {
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Schwarzian derivative as the derivative of the logarithm. There is one interesting fact in Wikipedia article on Schwarzian derivative, namely, if $F(z, w)=\log{\frac{f(z)-f(w)}{z-w}}$ then (sign error fixed) $6\frac{\partial^2 F(z, w)}{\partial z \partial w}|_{z=w} = (Sf)(w).$ I cannot see it. We have $\frac{\partial^2 F(z, w)}{\partial z \partial w} = \frac{f'(w)f'(z)}{(f(z)-f(w))^2}-\frac{1}{(z-w)^2}.$ When $|z-w|$ is small both of summands are large so none of them has no limit but their difference has. I brought them to a common denominator, but nothing happened. How to compute this limit?
| This is “just” an exercise in developing Taylor series.
Consider a fixed point $w$ where $f(w) \ne \infty$ and $f'(w) \ne 0$.
Then
$$
f(z) = a_0 + a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3 + \ldots
$$
where $a_n = \frac{f^{(n)}(w)}{n!}$, and $\ldots$ indicates “higher-order
terms of $(z-w)$” to shorten the notation.
Now
$$
\frac{1}{f(z)-f(w)} = \frac{1}{a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3 + \ldots} \\
= \frac{1}{a_1(z-w)(1 + \frac{a_2}{a_1}(z-w)+ \frac{a_3}{a_1}(z-w)^2 + \ldots)} \\
= \frac{1}{a_1(z-w)} \left( 1 - \frac{a_2}{a_1}(z-w) + \bigl( \frac{a_2^2}{a_1^2} - \frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
$$
and therefore
$$
\frac{1}{(f(z)-f(w))^2} = \frac{1}{a_1^2(z-w)^2} \left( 1 - 2\frac{a_2}{a_1}(z-w) + \bigl( 3\frac{a_2^2}{a_1^2} - 2\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
$$
On the other hand,
$$
f'(w)f'(z) = a_1 (a_1 + 2a_2(z-w) + 3a_3(z-w)^2 + \ldots) \\
= a_1^2 (1 + 2\frac{a_2}{a_1}(z-w) + 3\frac{a_3}{a_1}(z-w)^2 + \ldots)
$$
so that
$$
\frac{f'(w)f'(z)}{(f(z)-f(w))^2} = \frac{1}{(z-w)^2} \left( 1 + \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
= \frac{1}{(z-w)^2} + \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr) + \ldots
$$
which gives the desired
$$
\lim_{z \to w} \left( \frac{f'(w)f'(z)}{(f(z)-f(w))^2} - \frac{1}{(z-w)^2} \right)
= \frac{a_3}{a_1} - \frac{a_2^2}{a_1^2}
= \frac 16 \frac{f'''(w)}{f'(w) } - \frac 14 \left(\frac{f''(w)}{f'(w)}\right)^2 = \frac 16 (Sf)(w)
$$
Finally, due to the identity theorem, this relation holds everywhere
and not only at points where $f'(w) \ne 0$.
| {
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Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$ I am in search of a correct and shortcut techniques to prove this.
Otherwise I have calculated each remainder according to the power of $20$ to prove this :-
$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.
Therefore
$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$
For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$
Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$
If possible just show any short-cut correct way to prove this. Any help is appreciated.
| Sum is equal to $\frac{20^{22}-1}{20-1}=\frac{20^{22}-1}{19}$,
but $20^{22}\cong 1 \mod 19,23$.
| {
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How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$ Let $x,y,z$ rationals Show that if $(x+y+z)^3=9(x^2y+y^2z+z^2x)$ then $x=y=z$
I tried this :
Let $x$ be the smallest variable
Write $y=a+x$ and $z=b+x$ Prove $a=b=0$
by factoring the equation as the sum of three squares.
any suggestions?
| First of all $x=y=z$ is a solution to the equality, so we need to prove that all solutions other than this are not valid.
Let $y=a+x$ and $z=b+x$ where $b,a \in Q$
$(x+y+z)^{3} = 9(x^{2}y + y^{2}z + z^{2}x)$
$\Rightarrow (a+b+3x)^{3} = 9(x^{2}(a+x) + (a+x)^{2}(b+x) + (b+x)^{2}x)$
By equating coefficients, we get,
$\Rightarrow (a+b)^{3} = 9(a^{2}b)$
Now let $m= \frac{b}{a}$
$\Rightarrow m^{3} + 3m^{2} - 6m + 1=0$
Let $f(m)= m^{3} + 3m^{2} - 6m + 1$
Now, here is some food for thought,
If we can prove that $b$ or $a$ is irrational , we would arrive at a contradiction because $x,y,z$ are rational numbers.
If $\frac {b}{a}$ is irrational then either $b$ or $a$ will be irrational and we will arrive at a contradiction, which would give us no solution other than $x=y=z$.
So, all we need to do is prove that the cubic $f(m)=0$ has no rational roots.
$m^{3} + 3m^{2} - 6m + 1 = 0$
Let's assume that $f(m)=0$ has rational roots.
Substitute $m=\frac {p}{q}$ , where $gcd(p,q)=1$ and $p,q \in I$
$(\frac {p}{q})^{3} + 3(\frac {p}{q})^{2} - 6(\frac {p}{q}) + 1=0$
$\Rightarrow p(p^{2} + 3pq - 6q^{2}) = -q^{3}$
We know that $gcd(p,q)=1 \Rightarrow gcd(p,q^{3})=1$.
So $p$ must divide $-1$
Therefore the possible values of $p$ are $\pm 1$
Similarly we can write ,
$\Rightarrow q(q^{2} - 6pq + 3p^{2}) = -p^{3}$
So $q$ must divide $-1$
Therefore the possible values of $q$ are $\pm 1$
The possible roots for $f(m)=0$ are $\frac{p}{q} = \pm 1$.
But $f(1)=-1$ and $f(-1)=9$ .
Therefore, $f(m)=0$ has no rational roots.
Therefore, $\frac {b}{a}$ is irrational which is a contradiction.
Hence , $x=y=z$ is the only solution.
| {
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How can I solve this find the digits problem more rigorously?
Adjoin to the digits $739$ three more digits so that the resulting number $739 \text{_ _ _}$ is divisible by $6, 7, 8$, and $9$.
I can do some quick guess and check as well as some little tricks with $9$, $7$, $6$, and $8$ to arrive to an answer, but how could I show this with more rigor (like more concrete number theory)?
Answers: $739368$ and $739872$
| HINT
Let indicate
$$n=739xyz=7\cdot10^5+3\cdot10^4+9\cdot10^3+x\cdot10^2+y\cdot10+z$$
and consider
*
*$n\equiv 0 \pmod 6 \implies 4+4x+4y+z\equiv 0 \pmod 6\implies 4x+4y+z=2+6k$
*$n\equiv 0 \pmod 7 \implies 3+2x+3y+z \equiv0 \pmod 7 \implies 2x+3y+z=4+7h$
*$n\equiv 0 \pmod 8\implies 4x+2y+z\equiv 0 \pmod 8 \implies 4x+2y+z=8s$
*$n\equiv 0 \pmod 9 \implies 1+x+y+z \equiv 0 \pmod 9\implies x+y+z=8+9t$
| {
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If $1,a_1,a_2,...,a_{n-1}$ are the $n$ roots of $1$,then $(1-a_1)(1-a_2)...(1-a_{n-1}) \space ?$ If $1,a_1,a_2,...,a_{n-1}$ are the $n$ roots of unity,
then how can we find the value of $$(1-a_1)(1-a_2)...(1-a_{n-1}) \space ?$$
My Approach:
If $n=2$ , then $1,-1$ are the roots of unity
$\therefore (1-a_1)=(1-(-1))=2$
for $n=3 \space :$ $1,\omega,\omega^2$ are the roots of unity
$\therefore (1-a_1)(1-a_2)=(1-\omega)(1-\omega^2)$
$\quad \quad \quad =1-\omega^2 -\omega +1=3$
so we conclude for $n$ the value $(1-a_1)(1-a_2)...(1-a_{n-1}) \space = n$
but i want a direct process to evaluate $(1-a_1)(1-a_2)...(1-a_{n-1}) $ (without generalisation),so how can i do so?
| Consider the polynomial $x^n-1=(x-1)(x-a_1)...(x-a_{n-1})$. Divide by $x-1$ to get the polynomial $x^{n-1}+x^{n-2}+...+x+1$. Evaluate at $x=1$ to get the answer $n$.
| {
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2016 ARML Individual Problem #8. Tricky Substitutions to find coefficients of a polynomial Let $P(x)$ be the polynomial $x^3+Ax^2+Bx+C$ for some constants $A$, $B$, and $C$. There exists constants $D$ and $E$ such that for all $x$, $P(x+1)=x^3+Dx^2+54x+37$ and $P(x+2)=x^3+26x^2+Ex+115$. Compute the ordered triple $(A,B,C)$.
Try to substitute values of $x$ into $P(x+1)$ and $P(x+2)$ to make your life easier.
| $P(x+1)=x^3+Dx^2+54x+37$
$P(x+1)=(x+1)^3+A(x+1)^2+B(x+1)+C=x^3+3x^2+3x+1+Ax^2+2Ax+A+Bx+B+C=x^3+x^2(A+3)+x(3+A+B)+(1+A+B+C)$
Comparing this to our values:
$$1+A+B+C=37\implies A+B+C=36$$
$$A+3=D$$
$$3+A+B=54 \implies A+B=51$$
Substituting $A+B$, $51+C=36$, and thus $C=-15$
Continue with the same process, incorporating:
$P(x+2)=x^3+26x^2+Ex+115$ and
$P(x+2)=(x+2)^3+A(x+2)^2+B(x+2)+C$, then grouping.
| {
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What is the value of $\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots }}}}$? How can we find the value of
$$\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots}}}}=?$$
My Approach:
Let $$f(n)=\sqrt[n]{1+ \sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}}} \tag{1},$$
then $f(2)$ is our solution.
So, doing $n$th power in both sides of $(1)$, we get:
$${ \{ f(n) \} }^n =1+f(n+1)$$
$$\implies { \{ f(n) \} }^n - f(n+1) = 1 \tag{2}$$
Now how can I solve $(2)$ ? Any help please…
| Estimation of upper bound.
L signs the value of the expression, then we can write:
\begin{align}L^2=1+\left(1+(1+\cdots+(1+(1+s)^{{1/n}})^{{1/{n-1}}}\cdots)^{1/4}\right)^{1/3}\end{align}
where
\begin{align}s=\sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}} \tag{1}\end{align}
Accordance with Bernoulli$^,$s inequality if $0\le{r}\le1$
and $x\ge-1$ then $(1+x)^r\le(1+xr)$
In our case $r=\frac{1}{k}, k=3,4, ....n $ and $s\gt1 $.
First apply to $(1+s)^{1/n}$ we have that $\le1+\frac{s}{n}$
then outward apply to all power factor we get the following:
\begin{align}L^2\le1+1+\left(1+(1+\cdots+(1+(1+\frac{s}{n}))\frac{1}{n-1}\cdots)\frac{1}{4}\right)\frac{1}{3}\end{align}
Perform the multiplications:
\begin{align}L^2\le2 +2(\frac{1}{3!} +\frac{1}{4!}+......\frac{1}{(n-1)!}+\frac{s}{n!})\end{align}
Namely
\begin{align}L^2\le2 +2\sum_{k=3}^n\frac{1}{k!}\end{align}
We could apply to $s$ the above procedure then we have:
\begin{align}L^2\le2 +2\sum_{k=0}^{\infty}\frac{1}{k!}-5\end{align}
\begin{align}L\le \sqrt{2e-3}=1,56094\end{align}
| {
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Show that $\sqrt{ab} \leq \sqrt[n]{\frac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \frac{a+b}{2}$ Let $a,b,\lambda \in \mathbb R,$ with $a,b > 0, \lambda ≥1,$ and $ n \in \mathbb N^*.$
Show that $\sqrt{ab} \leq \sqrt[n]{\frac{a^n+b^n+\lambda((a+b)^n-a^n-b^n)}{2+\lambda(2^n-2)}} \leq \frac{a+b}{2}$
What is the good method to solve such a thing? I do not really have any idea for the moment ... It was given at an oral exam.
$(\sqrt{a}-\sqrt{b})^2 ≥0 \implies \sqrt{ab} \leq \frac{a+b}{2}$
| the right-hand side of your inequality is equivalent to $$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
ok, lets prove it,raise to the power $n$ gives:
$$a^n+b^n+\lambda(a+b)^n-a^n\lambda-b^n\lambda\le \left(\frac{a+b}{2}\right)^n(2+2^n\lambda-2\lambda)$$
after simplifying and rearranging we obtain:
$$(a^n+b^n)(1-\lambda)\le \left(\frac{a+b}{2}\right)^n2+(a+b)^n+(a+b)^n\lambda-2\lambda\left(\frac{a+b}{2}\right)^n$$
and so we get
$$(a^n+b^n)(1-\lambda)\le 2(1-\lambda)\left(\frac{a+b}{2}\right)^n$$
for $\lambda=1$ is the inequality true, for $$\lambda>1$$ we have
$$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
| {
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Why does conversion from base $b_1$ to $b_2$ give this strange result in Wolfram Alpha? I was trying to convert the $5$-ary number $44444444_5$ to base $25$ using Wolfram Alpha:
44444444 from base 5 to base 25
Curiously, I got the result $0000_{25}$.
Clearly, $0\times25^{3}+0\times25^{2}+0\times25^{1}+0\times25^{0}$ does not equal $44444444_5$. So why am I getting this result? Is this a bug?
| It's not a bug, and it's not $0000$, it's $oooo$. $o$ being the symbol $o$ (the one between $n$ and $p$ in the alphabet).
The letter $o$ in base $25$ stands for the twenty-fourth "digit" ( the "digits" in base $25$ are $1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,g,h,i,j,k,l,m,n,o$), which means it equals $24$ in base $10$, which means $$oooo_{25} = (24 + 24\cdot 25 + 24\cdot 25^2 + 24\cdot 25^3)_{10} = 390624_{10}$$
which fits well with the fact that
$$44444444_5 = (4+4\cdot 5 + 4\cdot 5^2+4\cdot 5^3+4\cdot 5^4+4\cdot 5^5+4\cdot 5^6+4\cdot 5^7)_{10} = 390624_{10}$$
Another way of seeing this (without calculating the number) is
$$\begin{align}44444444_5 &= 4\cdot(1+5+5^2+5^3+5^4+5^5+5^6+5^7)
\\& = 4\cdot((1+5)+(5^2+5^3)+(5^4+5^5)+(5^6+5^7))
\\& = 4\cdot(1+5+5^2(1+5) + 5^4(1+5)+5^6(1+5))
\\&= 4\cdot(6+6\cdot25+6\cdot25^2+6\cdot25^3)
\\&= 24 + 24\cdot 25 + 24\cdot 25^2 + 24\cdot 25^3
\\&= oooo_{25}\end{align}$$
| {
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Proving Inequality Using Graph
Given: 0 < a < b, Prove: $\sqrt{b} - \sqrt{a} < \frac{1}{3}(\frac{b}{\sqrt{a}}-\frac{a}{\sqrt{b}})$
As usual, we need to start from one side and reach the other by deduction. However, I am not sure I am going to get there. The solution may require Calculus but not to a highly advanced level.
The graph of $y=\frac{1}{2\sqrt{x}}$ is:
This shows us that the mean value of the function between a and b is less than the mean of f(a) and f(b) as the graph is concave.
In other words:
For the range $[a,b]$,
$f_{mean} \ne \frac{f(a)+f(b)}{2}$
| Rewrite like this (divide by $ab$): $${1\over \sqrt{b}} - {1\over \sqrt{a}} < \frac{1}{3}(\frac{1}{a\sqrt{a}}-\frac{1}{b\sqrt{b}})$$
and then
$${1\over \sqrt{b}} +\frac{1}{3}\frac{1}{b\sqrt{b}} < \frac{1}{3}\frac{1}{a\sqrt{a}}+{1\over \sqrt{a}}$$
If we put $a'={1\over \sqrt{a}}$ and $b'={1\over \sqrt{b}}$
then we get $0<b'<a'$ and
$$3b' + b'^3 <3a'+a'^3$$
Now observe
$$f(x) = 3x+x^3 $$
Since $f'(x) =3+3x^2>0$ the $f$ is strictly increasing and thus conclusion.
| {
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Prove that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is contained in $[3,5]$.
Prove that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is contained in $[3,5]$.
By contradiction: Assume that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is not contained in $[3,5]$. So, there exists $a$ such that $a$ is an element of $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$, but $a$ is not an element of $[3,5]$, so $a < 3$ or $a > 5$. Using $a < 3$, we know that there is a rational number $b > 0$ such that $a = 3-b$.
I'm not sure where to go from here to get a contradiction.
| So if $a < 3$, prove that there is an $x$ so that $a < 3-x^2$. That would prove that $a \not \in \cap [3-x^2, 5+x^2]$.
If $a > 5$, prove that there is an $x$ so that $5+x^2 < a$. That would prove that $a \not \in \cap [3-x^2, 5+x^2]$
To prove the first. If $a < 3$, then $3- a > 0$. Have you proven that square positive square roots exist for every positive real number yet? If so just let $x <\sqrt{3 -a}$ so $3 - x^2 > 3 - (\sqrt{3 -a })^2 = a$. But if you haven't proven that let $0 < x < \min(3-a, 1)$. Since $x < 1$, $x^2 < x$. so $x^2 < 3-a$ and $3 - x^2 > 3-(3-a) = a$.
| {
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How would I find the standard matrix of this linear transformation? Let $T$ be a linear transformation on the plane with
$T\begin{pmatrix}-5\\ -4\end{pmatrix}=\begin{pmatrix}-5\\ -5\end{pmatrix}$
and
$T\begin{pmatrix}-4\\ -3\end{pmatrix}=\begin{pmatrix}-1\\ 2\end{pmatrix}$
Find a standard matrix for $T(x)$.
Since the answer should be a $2$x$2$ matrix, I thought that I would solve for $T$ in the first equation and then solve for $T$ in the second equation. Then combine the two results into a standard matrix. But I am not sure how I would solve for $T$ as the inverse of a $2$x$1$ matrix is not possible.
So I am assuming there must be a different approach.
If so, any help or guidance will be highly appreciated!
| Express the the standard basis in terms of $\begin{pmatrix}-5\\ -4\end{pmatrix}$ and $\begin{pmatrix}-4\\ -3\end{pmatrix}$ ,i.e., find $x_1$ and $x_2$ such that $\begin{pmatrix} 1\\ 0\end{pmatrix} = x_1\begin{pmatrix}-5\\ -5\end{pmatrix} + x_2\begin{pmatrix}-4\\ 3\end{pmatrix} $. Do the same for the basis $\begin{pmatrix} 0\\ 1\end{pmatrix} $. Now, compute $T(\begin{pmatrix} 0\\ 1\end{pmatrix})$ and $T(\begin{pmatrix} 1\\ 0\end{pmatrix})$ using the fact that $T$ is a linear transformation. From the values of $T(\begin{pmatrix} 0\\ 1\end{pmatrix})$ and $T(\begin{pmatrix} 1\\ 0\end{pmatrix})$, you can construct your matrix.
| {
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Inverse of Power Series I am reading something that considers the following power series in $\mathbb{C}$:
$$ f(z) = z + a_2z^2 + a_3z^3 + a^4z^4 + ... \\ |z| < 1 $$
They define:
$$g(z) = z + b_0 + b_1z^{-1} + b_2z^{-2} + b_3z^{-3} + ... \\ |z| > 1 $$
And assert: $g(z) = \frac{1}{f(z^{-1})}$ and $f(z) = \frac{1}{g(z^{-1})}$. I'm having trouble seeing this though. Ideas?
| By direct calculation, we see that
\begin{align}
f(z^{-1}) =&\ \frac{1}{z}+a_2\frac{1}{z^2}+a_3\frac{1}{z^3}+\ldots=\ \frac{1}{z}\left(1+a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right)
\end{align}
which means
\begin{align}
\frac{1}{f(z^{-1})} =&\ \frac{z}{1+\left(a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right)} = z\sum^\infty_{n=0}(-1)^n\left( a_2\frac{1}{z}+a_3\frac{1}{z^2}+\ldots \right)^n\\
=&\ z\left( 1+\sum^\infty_{k=1} b_{k-1}\frac{1}{z^k}\right).
\end{align}
| {
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numbers with digits untill 7 and divisible by 7 I have these problems from Brilliant.org:
a) How many 7-digit positive integers are there, satisfying that it is divisible by 7 and each of its digits is 1, 2, 3, 4, 5, 6, or 7?
b) The same question as a), but digit 0 is also allowed.
For a) I got $7^6$ and I think I can prove it, but what I really like to get help or solution for b).
| The answer to (a) is $7^6$ like you claim. This is because for every choice of the first $6$ digits of the number, there is exactly one units digit that will make the number divisible by $7$.
While typing the solution for part (b) that is at the bottom of this answer, I realised that there is a much simpler solution. I've kept the write-up, but will include the simpler solution here.
We reason in a similar way to part (a). There aree $8^6$ options for the last $6$ digits of the number. For each of these, there is exactly one choice for the first digit of the number that will make the number divisible by $7$. We only have one choice even when the final $6$ digits of the number form a multiple of $7$, because we exclude the possibility of leading zeros, and so in this case the leading digit must be a $7$.
Original solution for part (b):
Part (b) is slightly more complicated. Let $a_n$ be the number of $n$-digit numbers which are divisible by $7$ and which consist only of the digits from $0$ to $7$. Note that $a_1 = 1$. (We don't include $0$ as a one-digit number. This will give us the correct answer later because we don't want leading zeros.)
To find $a_{n+1}$, we again look at the first $n$ digits of the number, and then see how many choices there are for the last digit to make the number divisible by $7$. If the number formed by the first $n$ digits is divisible by $7$, then the last digit must be $0$ or $7$. For each of the $a_n$ possibilities for the first $n$ digits, there are thus $2$ options for the last digit.
On the other hand, suppose that the number formed by the first $n$ digits is not divisible by $7$. There is then $7 \cdot 8^{n - 1} - a_n$ options for the first $n$ digits. (The first $n$ digits can be any of the $7 \cdot 8^{n - 1}$ $n$-digit numbers without a leading zero, except for the $a_n$ of them that are divisible by $7$.) For each choice of the first $n$ digits, there is exactly one choice for the final digit that makes the whole number divisible by $7$, so we get an additional $7 \cdot 8^{n - 1}$ numbers is this case.
We find that
$$
a_{n + 1} = 2 a_n + 7 \cdot 8^{n - 1} - a_n = a_n + 7 \cdot 8^{n - 1}.
$$
We thus have that
$$
a_n = a_{n - 1} + 7 \cdot 8^{n - 2} = a_{n - 2} + 7 \cdot 8^{n - 3} + 7 \cdot 8^{n - 2} = \dots = a_1 + 7 \cdot 8^0 + \dots + 7 \cdot 8^{n - 3} + 7 \cdot 8^{n - 2} = 1 + 7 \cdot \frac{8^{n - 1} - 1}{8 - 1} = 8^{n - 1}.
$$
It follows that the answer for part (b) is $8^6$.
| {
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Limit: $\lim_{x\to 0}\frac{x^3 - 3x^2+1}{x^3}$ I want find the limit of $\frac{x^3 - 3x^2+1}{x^3}$ as $x\to 0$.
So I write
$$\frac{x^3 - 3x^2+1}{x^3} = 1 - \frac{3x^2 +1}{x^3}$$ and
$$\lim_{x \to 0^- }(\frac{x^3 - 3x^2+1}{x^3})= \lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3})$$ Here, the limit must be $\pm \infty$ and I think when $x \to 0^-$ as $x^2$ is something positive, $x^3$ is something negative and since we have $- \frac{3x^2 +1}{x^3}$ we should have $+\infty$. However, it must be $-\infty$. Similarly I think $\displaystyle\lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3}) =-\infty$ but it's $+ \infty$.
What do I miss, how should I think?
| Write $$f(x)=g(x)h(x),$$ where $$g(x)=x^3-3x^2+1,\quad h(x)=\frac{1}{x^3}.$$
Now, approaching from the right, we have $$\lim_{x\to 0^{+}}g(x)=1,\quad \lim_{x\to 0^{+}}h(x)=\infty.$$ Similarly, approaching from the left, we have $$\lim_{x\to 0^{-}}g(x)=1,\quad\lim_{x\to 0^-}h(x)=-\infty.$$ The limits regarding $h(x)$ can be seen by looking at the sketch for the corresponding graph.
Since the right hand/left hand limits aren't equal; $\lim_{x\to 0}f(x)$ does not exist.
| {
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Solving for $x$ from given equation Sorry for the trivial question. I was wondering if there was an easy way to solve this equation.
The equation is:
$$(x-1)(x-2)(x+3)(x-6)+36=0$$
I know this can be solved the traditional way, where after multiplying all terms, we could factorize the entire thing, but I ended up with a really messy equation. Is there a simpler method I'm missing?
Thanks for any help :-)
| \begin{align*}
(x-1)(x-2)(x+3)(x-6)+36&=0\\
(x^2-3x+2)(x^2-3x-18)+36&=0\\
(x^2-3x)^2-16(x^2-3x)&=0\\
(x^2-3x)(x^2-3x-16)&=0\\
x&=0, 3\textrm{ or }\frac{3\pm\sqrt{73}}{2}
\end{align*}
| {
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Show that $\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$
Show that $$\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$$
I am struggling with this one. I have tried substitutions and the residue theorem but haven't got anywhere. It would be nice to see an 'elementary' solution if there is one, but any will do.
| First integrate by parts to get:
\begin{align}
\int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx= \frac{\pi}{3}\log(8)-\frac{1}{3}\int^\pi_0 \log(5-3\cos(x))\,dx
\end{align}
The integral on the right, can be done with many methods, for example, notice that we have with substituting $u=\pi-x$ that:
\begin{align}
\int^\pi_0 \log(5-3\cos(x))\,dx&=\frac{1}{2}\int^\pi_0 \log(25-9\cos^2(x))\,dx\\
&=\frac{\pi}{2}\log(25)+\frac{1}{2}\int^\pi_0\log\left(1-\frac{9}{25}\cos^2(x)\right)\,dx\\
&=\frac{\pi}{2}\log(25)+\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx\\
\end{align}
We know from this link that: $$\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx=\pi\log(9/10)$$
Putting everything together yields:
\begin{align}
\int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx=\frac{\pi}{3}\log(8)-\frac{1}{3}\left( \frac{\pi}{2}\log(25)+\pi\log(9/10)\right)=\frac{2\pi\log(4/3)}{3}\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2716687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to write the polar form of $x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2$? $$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\
r=(x^2+y^2)^{1/2}\\
x^2/r^4-y^2/r^4$$
Convert $x^2-y^2$ to polar form
$$x=r\cos(\theta)$$
$$y=r\sin(\theta)$$
$$(r\cos(\theta))^2-(r\sin(\theta))^2=r^2 \cos(2 θ)$$
$$r^2 \cos(2 θ)/r^4= \cos(2\theta)/r^2$$
Is that correct?
| Yes, you are correct.
$$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\
\\
x^2/r^4-y^2/r^4 = $$
$$\frac{\cos 2\theta}{r^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof By Induction: $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ Here's what I have so far: $$\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$$ where $n\ge2$.
Test for base case
$$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} > \sqrt 2$$
Simplify
$$ 1 + .707... > 1.414... $$
$$ 1.707... > 1.414... $$
This statement is true.
Thus we can assume $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ is true for $\forall n$
Now prove $n + 1$
Expand
$$\sum_{i=1}^n \frac{1}{\sqrt i} = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} \gt \sqrt n $$
Add the next n in the series, and add it to both sides of the inequality.
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \gt \sqrt n + \frac{1}{\sqrt{n + 1}} $$
Rewrite
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} = \sum_{i=1}^{n + 1} \frac{1}{\sqrt i}$$
LHS is complete now we have to try to make RHS = $\sqrt {n + 1}$
$$\sqrt n + \frac{1}{\sqrt{n + 1}} = \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}}$$
Given that n > 0
$$\frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} = \frac{\sqrt{n(n+1)} + 1} {\sqrt{n+1}}$$
Now since this is my first induction involving inequalities or summation, I was hoping you guys could show me what the next step would be, since I'm really not sure. Thanks in advance!
| You can prove this without induction. In fact, when $n=1$ it's ok; and when $n>1$, for each $i=1,..,n-1$ we have $$\frac{1}{\sqrt{i}}>\frac{1}{\sqrt{n}}$$
Now, $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}}>n\cdot\frac{1}{\sqrt{n}}=\sqrt{n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix - charpoly, minpoly and diagonalization
Let $a\in K$. Consider the matrix
$$
M = \begin{bmatrix}
1-2a & -a & 0 & 0 \\
4a & 1+2a & 0 & 0 \\
0 & 0 & 2 & 3 \\
0 & 0 & 0 & 1 \\
\end{bmatrix} \in M_{4\times4}(K)
$$
a) Deduce the eigenvalues of $M$ from the characteristic polynomial of $M$.
b) Compute the minimal polynomial of $M$.
a) Using $\det(x\operatorname{Id}-M)$, I got a fairly long computation so I shall post only the final lines:
$$(x^2-2x-8a^2+1)(x^2-3x+2)=0.$$
If we develop, we get: $x^4-5x^3+9x^2-7x-8a^2x^2+24a^2x-16a^2+2$, and the real zeros are either $\pm1,2$ after we check, we get $1,2$.
Therefore, the eigenvalues are $x_1=1$ and $x_2=2$.
Regarding b): is minpoly of $M$ in this case the same as the charpoly of $M$? I do not know how to compute it with both $x$ and $a$ in the equations.
c) Prove $M$ is diagonalizable if and only if $a=0$.
| First, check this property:
$$
A = \begin{bmatrix} A_{11} & A_{12} \\O & A_{22}\end{bmatrix},\\
B = \begin{bmatrix} B_{11} & B_{12} \\O & B_{22}\end{bmatrix},\\
AB = \begin{bmatrix} A_{11}B_{11} & A_{11}B_{12}+A_{12}B_{22} \\O & A_{22}B_{22}\end{bmatrix},\\\\
det(A) = det(A_{11})\,det(A_{22})
$$
when $A_{11}$, $B_{11}$ are $m \times m$, $A_{12}$, $B_{12}$ are $m \times n$, $A_{22}$,$B_{22}$ are $n \times n$. (It can be proved by dirty calculations)
In a), your characteristic polynomial is actually wrong, and you don't need to develop.
You can get and solve $(x^2-2x+1)(x^2-3x+2)=0 : \\(x-1)^3(x-2)=0 \Leftrightarrow x=1 \: or\: x=2$
Now b) is clear : A's minimal polynomial should be a form of $(x-1)^i (x-2)$ for some $i\geq1$, and you can find out $i = \cases{1 \:\:\: {if} \:\:\:a=0 \\2 \:\:\:otherwise}$ using property above.
c) is trivial : $A$ is diagonalizable iff its minimal polynomial's roots are those of its characteristic polynomial with multiplicity 1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all solutions to the congruence $x^3\equiv 1 {\pmod{77}}$ Find all solutions to the congruence $x^3\equiv 1 \pmod{77}$
I was able to do this for $x^2 \equiv 1\pmod{77}$ since $x$ would either be $1$ or $-1$, in each of $\bmod 7$ and $\bmod 11$. I don't know what $x$ would be in this case.
I can do the Chinese remainder theorem to find the final answers.
| As a solution to $x^2 \equiv 1 \mod 77$ you should have gotten
$x \equiv \pm 1 \mod 7$ and $x \equiv \pm 1 \mod 11$. As this is FOUR systems of equations each system of equations should have yielded one unique solution.
System 1: $x \equiv 1\mod 7$ and $x \equiv 1 \mod 11$ yields $x \equiv 1 \mod 77$.
System 2: $x \equiv 1 \mod 7$ and $x \equiv -1 \mod 11$ yields $x \equiv 43 \mod 77$. And $43^2 = 24*77 + 1$.
System 3: $x \equiv -1 \mod 7$ and $x \equiv 1 \mod 11$ yields $x \equiv 34 \mod 77$. And $34^2 = 15*77 + 1$.
System 4: $x \equiv -1 \mod 7$ and $x \equiv - 1 \mod 11$ yields $x \equiv 76\equiv -1 \mod 77$.
====
For $x^3 = 1 \mod 77$.
To solve $x^3 \equiv 1 \mod 7$. Note that if $x\not \equiv 0$ then $x^6 \equiv 1 \mod 7$ so $(x^2)^3 \equiv 1 \mod 7$ and so if $x \equiv (\pm 1)^2,(\pm 2)^2, (\pm 3)^2 \equiv 1, 4, 2 \mod 7$ then $x^3 \equiv 1 \mod 7$.
To solve $x^3 \equiv 1 \mod 11$, note that $\phi(11) = 10$. $\gcd(3,10) = 1$. For $x^k \equiv 1 \mod 11$ then $x\equiv y^j$ for some $j$ and $10|j*k$. That can only happen if $k = 10$ or $k < 10$ and $j=10$. So $x \equiv 1 \mod 11$ is the only solution to $x^3 \equiv 1 \mod 11$.
So we have $3$ systems of equations.
$x \equiv 1 \mod 7$ and $x\equiv 1 \mod 11$ yields $x \equiv 1 \mod 77$.
$x \equiv 2 \mod 7$ and $x \equiv 1 \mod 11$ yields $x\equiv 23 \mod 77$. And $23^3 = 12167 = 1 + 158*77$.
$x \equiv 4 \mod 7$ and $x \equiv 1 \mod 11$ yeilds $x \equiv 67 \mod 77$. And $67^3 \equiv (-10)^3 \equiv -1000 + 770 \equiv -230 + 385 \equiv 155 - 154 \equiv 1 \mod 77$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Residue Complex Function I've applied the formula $$\lim_{z \to i} \frac{d}{dz}{(z-z_0)^2.f(z)}$$ and I keep getting a value of 0. Could someone prove my sanity by starting this off?
| Here a solution with less calculations.
*
*$f(z) = \frac{z^2-1}{(z^2+1)^2} = \frac{z^2+1-2}{(z^2+1)^2}= \frac{1}{z^2+1} - \frac{2}{(z^2+1)^2} $
*$Res_{z=i}\frac{1}{z^2+1} = \lim_{z\rightarrow i}\frac{z-i}{(z-i)(z+i)}=\frac{1}{2i}$
*For $Res_{z=i}\left(- \frac{2}{(z^2+1)^2}\right) = -2 Res_{z=i} \frac{1}{(z^2+1)^2}$ you need to calculate $$\frac{d}{dz}\frac{(z-i)^2}{(z-i)^2(z+i)^2} = \frac{d}{dz}\frac{1}{(z+i)^2}= \frac{-2}{(z+i)^3} \stackrel{z\rightarrow 0}{\longrightarrow}\frac{1}{4i}$$
*All together: $$Res_{z=i}f(z) = \frac{1}{2i} - 2\cdot\frac{1}{4i} = 0$$
p.s.: You can check your residues with wolfram alpha. Here the : output for above residue
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A curious property of an acute triangle Many years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:
$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.
The property is that $AF = AG$
Proof:
Let $H$ be the midpoint of $AB$ (and the centre of the respective semicircle)
$$AG^2 = AD^2 + GD^2 = \left(AC\cdot \cos\angle A\right)^2 + GD^2$$
Since $HG = AH = \frac{AB}{2}$ is the radius of the semicircle
$$GD^2 = HG^2 - HD^2 = \left(\frac{AB}{2}\right)^2 - \left(\frac{AB}{2} - AC\cdot\cos\angle A\right)^2 = \\ = AB\cdot AC\cdot \cos\angle A - \left(AC\cdot\cos\angle A\right)^2$$
which gives
$$AG^2 = AB\cdot AC\cdot \cos\angle A$$
Analogously ($I$ is the midpoint of $AC$)
$$AF^2 = AE^2 + FE^2 = \left(AB\cdot \cos\angle A\right)^2 + FE^2$$
$$FE^2 = FI^2 - EI^2 = \left(\frac{AC}{2}\right)^2 - \left(AB\cdot \cos\angle A - \frac{AC}{2}\right)^2 = \\ = AC\cdot AB\cdot \cos\angle A - \left(AB\cdot \cos\angle A\right)^2$$
which finally gives
$$AF^2 = AC\cdot AB\cdot \cos\angle A$$
Questions:
*
*Is this a known property?
*Is there a better more elegant proof?
| HINT:
$$AF^2 = AE \cdot AC\\
AG^2 = AD \cdot AB \\
AE \cdot AC = AD \cdot AB $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720518",
"timestamp": "2023-03-29T00:00:00",
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Which normal to $y=x^2$ forms the shortest chord? The line between the pair of points common to the normal and the parabola is the chord formed by that normal.
I tried solving this problem using calculus but I was unable to solve the equation I got from the derivative. Is there a better solution, perhaps using a geometric argument?
My attempt:
I came up with a function for the length of the chord passing through the point $x$ by finding the equation of the normal at $x$, then finding its points of intersection with $y=x^2$ and finally finding the distance between those two points, which came out to be
$$L(x) = \sqrt {\left( 2x + \frac{1}{2x} \right )^2 + \left ( 1 + \frac{1}{4x^2}\right )^2}$$
The minimum of $L(x)$ conincides with the minimum of $L(x)^2$. Hence, I tried to minimize $L(x)^2$. I attempted to solve $\frac{d}{dx} L(x)^2 = 0$ for $x$ however this problem reduced to solving $32x^6 - 6x^2 - 1 = 0$, which I was unable to solve even after substituting $t=x^2$. I plugged the equation into Wolfram to get $x=\frac{1}{\sqrt{2}}$. So the shortest chord must be the line formed by the points $\left(\frac{1}{\sqrt{2}},\frac{1}{2} \right)$ and $(-\sqrt{2}, 2)$.
| The slope between the points $(a,a^2)$ and $(b,b^2)$ can be calculated two ways,
$$\frac{-1}{2a} = \frac{a^2-b^2}{a-b}$$
The distance squared between the points is
$$(a-b)^2 + (a^2-b^2)^2$$
I rearranged the first formula to plug in to the right part of that second equation since I figured this is easy to plug in and might end up lowering the degree of the polynomial later,
$$-\frac{a-b}{2a} = a^2-b^2$$
$$(a-b)^2 + \frac{(a-b)^2}{4a^2}$$
I then decide to factor,
$$(a-b)^2\left(1+\frac{1}{4a^2} \right)$$
At this point I'm satisfied that I should probably use the first equation to solve for b and plug in and differentiate,
$$\frac{-1}{2a} = a+b$$
$$-b = a + \frac{1}{2a}$$
$$\left( 2a + \frac{1}{2a} \right)^2 \left(1+\frac{1}{4a^2} \right)$$
Clean up the fractions a bit,
$$\frac{(4a^2+1)^2}{4a^2} \frac{(4a^2+1)}{4a^2}$$
$$\frac{(4a^2+1)^3}{16a^4}$$
Now we can differentiate and set this to 0 to obtain our minimum,
$$\frac{3(4a^2+1)^2 4*2a*(16a^4) - (4a^2+1)^3 * 16 * 4 a^3 }{(16a^4)^2} = 0$$
At this point nearly everything divides out,
$$3(4a^2+1)^2 4*2a*(16a^4) - (4a^2+1)^3 * 16 * 4 a^3=0$$
$$6a^2 - (4a^2+1) =0$$
$$2a^2 - 1 =0$$
$$a = \pm \frac{\sqrt{2}}{2}$$
At this point the problem is downhill, but no cubic shows up which seemed to be the problem you were coming to that I never saw.
Furthermore the question is still unanswered because you wanted a purely geometric solution not using calculus. I think that's the real question since it seems possible just like Heron's Problem
| {
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Prove that the sequence converges to using the $\varepsilon$-$N$ definition of a limit Question: Prove that the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_n=\frac{2n}{3n+1}$ converges to $\frac{2}{3}$ using the $\varepsilon$-$N$ definition of a limit.
I'm still a beginner when it comes to these types of questions. I've had a go at it but I'm pretty sure this is nonsense. Would appreciate some feedback and suggestions. Thanks!
Solution: Fix $\varepsilon>0$. We need to find $N\in\mathbb{N}$ such that $n>N \implies|a_n-\frac{2}{3}|<\varepsilon.$
We have $|a_n-\frac{2}{3}| < \varepsilon.$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have
$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2}{3}(\frac{1}{n}-1)| < \varepsilon$
The sequence is positive for all $n$ so the absolute values are redundant
$\iff \frac{2}{3}(\frac{1}{n}-1) < \varepsilon$
$\iff \frac{1}{n}-1 < \frac{3\varepsilon}{2}$
$\iff \frac{1}{n} < \frac{3\varepsilon}{2}+1$
$\iff \frac{1}{n} < \frac{3\varepsilon+2}{2}$
$\iff n> \frac{2}{3\varepsilon+2}$
So choose any $N>\frac{2}{3\varepsilon+2}$ and the definition is satisfied. Q.E.D.
| Complete the $\epsilon-n_0$ argument:
We have
$|\dfrac{2n}{3n+1} -\dfrac{2}{3}|=\dfrac{2}{3(3n+1)} \lt$
$\dfrac{2}{3(3n)}\lt \dfrac {2}{n}.$
Let $\epsilon >0$ be given.
Archimedes :
There is a $n_0 \in\mathbb{Z^+}$ with
$n_0 \gt 2/\epsilon.$
For $n \ge n_0:$
$|\dfrac{2n}{3n+1} -\dfrac{2}{3}|=$
$\lt \dfrac{2}{n} \le \dfrac{2}{n_0} \lt \epsilon$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Total number of triangles in a triangular grid I once tried to find how to quickly find the total number of squares in a bigger, $n\times n$ grid. I’m sure you’ve seen one of these puzzles, and the bigger the grid, the more embedded squares there are and the more chance you have of miscounting. So I finally came up with the answer
$$ \sum^n_{k=1}k^2$$ which can be simplified to $$\frac{n(n+1)(2n+1)}{6}$$ Then later on, I wanted to try the same with a triangular (equilateral) grid, where the size $n$ corresponds to the side of the triangle.
So I eventually found a formula that has two summation terms, one inside the other, with some even/odd evaluator ($n$ is the size of a side): $$T(n)=\sum^n_{k=1}\bigg(\frac{3-(-1)^{n-k}}{2}\sum^k_{j=1}j\bigg)$$
The fraction in between the two sums is the even/odd evaluator, and determines if $n-k$ is even or odd. Now this looks long and tedious, and for large grid sizes, only a computer could do it without a chance of mistaking. Now for the important part, my question is this: Is there a way to simplify the formula in a similar way than the square grid one? Yes, I did try Wolfram Alpha, and it just gave me back the summation form of the formula.
| First, lets start with what we know. We know that
$$ A_m:= \sum_{k=1}^mk=\frac{m(m+1)}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; B_m:=\sum_{k=1}^mk^2=\frac{m(m+1)(2m+1)}{6}$$
Now lets try to work out $$ C_n :=\sum_{k=1}^n(-1)^kk \; \; \; \; \; \; \; \; \; \; \; \; \; \; D_n:=\sum_{k=1}^n(-1)^kk^2$$
assuming tha $n$ is even. Observe that $$C_n=-A_n+4A_{n/2}=-\frac{n(n+1)}{2}+4\frac{\frac n2 (\frac n2+1)}{2}=\frac n2$$
$$D_n=-B_n+8B_{n/2}=-\frac{n(n+1)(2n+1)}{6}+8\frac{\frac n2(\frac n2+1)(n+1)}{6}=\frac{n(n+1)}{2}$$
For $n$ being odd, we simply break the sum into two parts and use the above, i.e.
$$C_n=\sum_{k=1}^n(-1)^kk=\sum_{k=1}^{n-1}(-1)^kk-n=\frac {n-1}{2} -n=-\frac{n+1}{2}$$
Similarly,
$$D_n=\frac{n(n-1)}{2}-n^2=-\frac{n(n+1)}{2}$$
Now, I assume that your formula for $T(n)$ is correct, so using that, we get
\begin{align}
\ T(n) & = \sum_{k=1}^n \biggl(\frac{3-(-1)^{n-k}}{2}\sum_{j=1}^kj\biggr) \\
\ & = \sum_{k=1}^n \biggl(\frac{3-(-1)^{n-k}}{2} \frac{k(k+1)}{2} \biggr) \\
\ & = \sum_{k=1}^n \biggl(\frac 34 k^2 + \frac 34 k - \frac{(-1)^{n-k}}{4}k^2-\frac{(-1)^{n-k}}{4}k \biggr) \\
\end{align}
Now supposing that $n$ is even
\begin{align}
\ & = \frac 34 \frac{n(n+1)(2n+1)}{6}+ \frac 34 \frac{n(n+1)}{2}-\frac{(-1)^n}{4}\frac{n(n+1)}{2} -\frac{(-1)^n}{4} \frac n2 \\
\ & = \frac{n(n+1)(n+2)}{4} - \frac{(-1)^nn(n+2)}{8} \\
\ & = \frac{n(n+2)(2n+2-(-1)^n)}{8} \\
\ & = \frac{n(n+2)(2n+1)}{8}
\end{align}
Similarly, for $n$ odd, we plug in the odd expression for the sums
\begin{align}
\ & = \frac 34 \frac{n(n+1)(2n+1)}{6}+ \frac 34 \frac{n(n+1)}{2}+\frac{(-1)^n}{4}\frac{n(n+1)}{2} +\frac{(-1)^n}{4} \frac {n+1}{2} \\
\ & = \frac{n(n+1)(n+2)}{4} + \frac{(-1)^n(n+1)^2}{8} \\
\ & = \frac{(n+1)(2n^2+3n-1)}{8}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Can't solve non linear system I'm trying to solve this system:
$\begin{cases}(1+y)(1+z) = 2\lambda x \\ (1+x)(1+z) = 2\lambda y \\ (1+x)(1+y)= 2\lambda z \\ x^2+y^2+z^2 - 1 = 0\end{cases}$
We can assume that $\lambda \neq 0$
How would you solve this? No matter in which direction I go it seems unpleasant.
| We have
$$
\left\{
\begin{array}{rcl}
(1+x)(1+y)(1+z) &=&2\lambda x(1+x)\\
(1+x)(1+y)(1+z) &=&2\lambda y(1+y)\\
(1+x)(1+y)(1+z) &=&2\lambda z(1+z)\\
x^2+y^2+z^2&=&1
\end{array}
\right.
$$
Note in this case that if we switch as unknowns $x$, $y$ and $z$ with each other the system does not change. It is suggested that one of the solutions is $x=y=z$. Then $x^2+y^2+z^2=1$ and $x=y=z$ implies $3x^2=1$, $3y^2=1$, $3z^2=1$, that is, $x=\pm\frac{1}{\sqrt{3}}$, $y=\pm\frac{1}{\sqrt{3}}$ and $z=\pm\frac{1}{\sqrt{3}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does one prove that $(x^4 - 5)$ and $(x^4 + 5)$ are irreducible over $\mathbb{Q}$? I have factorised $x^8 - 25$ into $(x^4 - 5)(x^4 + 5)$ but I'm unsure about how to go about showing that it is fully factorised.
Thanks very much in advance.
| A very hands on way since your polynomials are of degree $4$ (not very high degree). Suppose that $x^4-5$ is reducible over $\mathbb{Q}$ (the other polynomial the same argument). Then there exist nonconstant polynomials $p(x),q(x)$ with degree smaller than $4$ and with rational coefficients such that $x^4-5=p(x)q(x)$. There are two possibilities, either $p(x)$ is of degree $1$ and $q(x)$ is cubic or both are quadratic. Say $p(x)=x+a, q(x)=x^3+bx^2+cx+d$ with $a,b,c,d\in\mathbb{Q}$. Then upon multiplication we get
$$p(x)q(x)=(x+a)(x^3+bx^2+cx+d)=x^4+(a+b)x^3+(c+ab)x^2+(d+ac)x+ad$$
On the other hand
$$p(x)q(x)=x^4-5$$
This implies $a+b=0,c+ab=0,d+ac=0,ad=-5$ which yield $a=-b$ and consequently $c=b^2, d=b^3$ giving $ad=-bb^3=-b^4=-5$ or $b=\sqrt[4]{5}\notin\mathbb{Q}$. Hence a contradiction to our assumption $b\in\mathbb{Q}$. In the other case
$$p(x)q(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d)x^2+(ad+bc)x+bd$$
Again by looking at the coefficients we obtain $a+c=0, b+d=0,ad+bc=0$ and $bd=-5$. These imply $bd=-b^2=-5$ or $b=\sqrt{5}\notin\mathbb{Q}$. Again a contradiciton. For cases of higher degrees of course this method is laborious and you should refer to other ways like Eisenstein's theorems for polynomial reductions.
| {
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"url": "https://math.stackexchange.com/questions/2727605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does passing to ideals solve the problem of unique factorization? $A:=\mathbb{Z}[\sqrt{-5}]$ is not a UFD, because for instance
$$21 = 3 \cdot 7 = \left( 1+2\sqrt{-5}\right) \cdot \left(1-2\sqrt{-5}\right).$$
But since $A$ is a Dedekind domain, we should have unique factorization of ideals into prime ideals (this is, after all, the whole point of introducing Dedekind domains). But is it not the case that the ideals $(3)$,$(7)$,$\left( 1+2\sqrt{-5}\right)$,$\left( 1-2\sqrt{-5}\right)$ are all different prime ideals and that
$$(21) = (3) \cdot (7) = \left( 1+2\sqrt{-5}\right) \cdot \left(1-2\sqrt{-5}\right)?$$
What am I missing?
| $3, 7, 1 + 2\sqrt{-5}$, and $1 - 2\sqrt{-5}$ are irreducible elements of $A$, but not prime elements. Hence the principal ideals generated by these aren't prime ideals. The prime ideals dividing these principal ideals are not principal, they are generated by two of these irreducible elements. We have
$$(21) = (3, 1 + 2\sqrt{-5})\cdot (3, 1 - 2\sqrt{-5})\cdot (7, 1 + 2\sqrt{-5})\cdot (7, 1 - 2\sqrt{-5})\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle Given that roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle, find condition on $b$.
If $a$ is a root then $a(\frac{1}{2}+i\frac{\sqrt 3 }{2})$ i also a root. Sum of root is $-1 = a(\frac{3}{2}+i\frac{\sqrt3}{2})$ and product is $a^2(\frac{1}{2}+i\frac{\sqrt 3 }{2}) = \frac{b}{3}$
I am asking is this correct and also does veeta's formula work in complex numbers? What is your method and how to proceed using this, find $a$ and put in second equation to get $b$ is complicated.
Thanks a lot!!
| As you say, if $a$ is a root, then $a\varepsilon$, where $\varepsilon = {1\over 2}+i{\sqrt{3}\over 2}$ is also a root. It is easy to see that $\varepsilon^2 = \varepsilon-1$. By Vieta we have $$a \varepsilon +a = -1\implies a = {-1\over \varepsilon+1}$$ so by second Vieta formula we have $$ b= 3a ^2\varepsilon = {3\varepsilon\over (\varepsilon+1)^2}={3\varepsilon\over \varepsilon^2+2\varepsilon+1} = {3\varepsilon\over \varepsilon-1+2\varepsilon+1}= 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional equation: $ f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$
$f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17
\forall x \in \mathbb R $, then find the function $f(x)$
I have:
$f(15/4)= 16/3$ (both quadratics intersect at $x=1/2$)
$f(3)= 3$ (by substituting $x=0$ and $x=1$ and then solving the simultaneous equations obtained)
$f(5)= 7$
I am not getting anything fruitful from these.
Could anyone provide me a hint on how to proceed?
Edit:
Someone had commented (now it's deleted) that we can assume $f(x) = ax +b$, how can we do that? I got the right answer using that. Is it always okay to assume that way? When is it a reliable assumption?
| Replace $x$ by $1-x$,
\begin{align*}
f((1-x)^2+(1-x)+3)+2f((1-x)^2−3(1-x)+5)&=6(1-x)^2−10(1-x)+17\\
f(x^2-3x+5)+2f(x^2+x+3)&=6x^2-2x+13\\
\end{align*}
Solving with $f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$, we have $f(x^2+x+3)=2x^2+2x+3$.
$f(x)=2x-3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Rotation area of a function Im faced with the following problem.
Calculate the area of the "cone" that is created when the function
$$y = \frac{x^2}{32}$$
$$0≤y≤24$$ rotates around the y-axis.
If we draw the cone that is created we can see that the hight will be 24. However we have to change our function.
$$y = \frac{x^2}{32} \implies \sqrt{32y} = x = f(y)$$
The formula for the rotation area is the following:
$$Area = 2\pi\cdot\int f(x)\cdot \sqrt{1+f'(x)}dx $$
I've tried to go continue from this however, i cant manage the integral part. Can any1 help me?
/John
| Your formula is missing a square in the derivative.
The rotational area is given by
$$ 2\pi \int x ds = 2\pi \int x \sqrt{dx^2 + dy^2} $$
where $ds$ is the arc length element of the curve $y = \dfrac{x^2}{32}$
If you choose to integrate w.r.t $y$, the integral looks like this
\begin{align}
2\pi \int x \sqrt{dx^2 + dy^2} &= 2\pi \int_0^{24} x(y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy \\
&= 2\pi \int_0^{24}4\sqrt{2y} \sqrt{1 + \left(\frac{2\sqrt{2}}{\sqrt{y}}\right)^2} dy \\
&= 8\pi\sqrt{2} \int_0^{24} \sqrt{y} \sqrt{1+\frac{8}{y}} dy \\
&= 8\pi\sqrt{2} \int_0^{24} \sqrt{y+8} \ dy \\
&= 8\pi\sqrt{2}\cdot \frac{2}{3} (y+8)^{3/2}\Bigg|_0^{24}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find Roots of $x^3-7$ Relative to Roots of Unity and $\sqrt{7}$ find roots of $$x^3-7$$ relative to roots of unity and $\sqrt[3]7$
one root is $\sqrt[3]7$, remaining roots might be something like
$$\sqrt[3]7,(\sqrt[3]7)^2w,(\sqrt[3]7)^3w $$
where $w=e^{i 2 \pi/3 }$ [and wolram alpha is sayin $x_1=-\sqrt[3]{7},x_2 = \sqrt[3]{7},x_3=(-1)^{2/3} \sqrt[3]{7}$ ]
Appreciate confirmation, whether this is wrong. or more elaboration. Trying to work out a problem in abstract algebra but my complex analysis is a bit rusted. Thanks.
| The roots of unity are $1, \omega, \omega^2$ where $\omega = e^{i\frac 23 \pi}$
So $(1)^3, (\omega)^3, (\omega^2)^3$ all equal $1$.
So $(\sqrt[3]{7})^3,(\omega*\sqrt[3]{7})^3,(\omega^2*\sqrt[3]{7})^3$ will all equal $7$.
So $x = \sqrt[3]{7},\omega*\sqrt[3]{7},\omega^2*\sqrt[3]{7}$ are the three roots of $x^3 - 7 =0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ a+b+c=0,\ a^2+b^2+c^2=1$ implies $ a^4+b^4+c^4=\frac{1}{2}$ This is a high school problem. It is solved by algebraic calculation
(not difficult) - $\ast$.
But I have a problem in interpretation the result.
Here intersection of two surfaces $a+b+c=0,\ a^2+b^2+c^2=1$ is a
circle $C$. That is, by $\ast$, $a^4+b^4+c^4$ is constant on it. What does make
it so ?
Proof : Note that $S:=\{
(a,b,c)|a^4+b^4+c^4=\frac{1}{2}\}$ is a convex surface. In further,
there is $x,\ y\in S$ s.t. $|x|<1<|y|$.
And I guess that $C$ is not a geodesic : Consider a norm on $\mathbb{R}^3$. For instance $\| (a,b,c) \|_1 := |a|+|b|+|c|$ In this case unit ball wrt this norm is octahedron. In further, unit balls of standard Euclidean norm, infinite norm are unit sphere $S_2$, a cube $S_3$, respectively. Clearly, intersection of $S_2$ and plane $\{ (a,b,c)| a+b+c=0\}$ is a geodesic (So is $S_3$). Here considering $S$ and $S_3$, we can assume that $S$ is like $S_3$. That is, $S$ has largest Gaussian curvature at eight points. Hence geodesic is far from these points, but $C$ is not.
I enumerate some these exercises, but how can we interpret the answer ?
[add] Consider $ \Delta':=\{ (x,y,z)\in S_2 | x,\ y,\ z\geq 0\} $, which
is an equilateral geodesic triangle of side length $\frac{\pi}{2}$.
In further $\Delta \subset \Delta'$ is also an equilateral geodesic
triangle of side length $\frac{\pi}{3}$, which touches mid points of
sides in $\Delta'$. Clearly it is not in a plane.
Then we have a claim that if $ f (x,y,z)=(x^4,y^4 ,z^4)$, then
$f(\partial \Delta )$ is in a plane.
[my answer] Consider $SO(3)$-action on $S^2(1)$. Here what is $smallest$ invariant set ? It is $S^2$. Consider a finite subgroup of $SO(3)$. For instance, a group $H$ which acts on cube whose center is origin. So smallest invariant set of $H$ is not $S^2$. It may be union of finite number of great circles, $T$. And note that $F(x,y,z)=x^4+y^4+z^4$ is invariant under $H$. Hence we can guess that $F$ is constant on $T$.
| The geometric approach that you have outlined...
The intersection of $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$
A plane cutting a sphere, producing a circle.
$a^4 + b^4 + c^4 = \frac 12$ is a convex surface that resembles a cube with rounded edges and corners.
As it happens every point on the circle $C$ lies on this "cubeoid." But, it is still not obvious that this would be the case.
You could parameterize the circle:
$\begin {bmatrix}
\frac {\sqrt 3}{3} & \frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\
\frac {\sqrt 3}{3} & -\frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\
\frac {\sqrt 3}{3} & 0 &-\frac {\sqrt 6}{3}\end{bmatrix}\begin {bmatrix} 0\\\cos\theta\\\sin\theta\end{bmatrix}= \begin {bmatrix} \frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 6}{3}\sin\theta\end{bmatrix}$
Would give you a parameterization of the circle.
And $a^4 +b^4 + c^4$ indeed does equal $\frac 12$
I don't think that this is easier than the more algebraic approaches...But it does work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find the indefinite integral $\frac{x\sin x}{1+\cos^2 x}$
Evaluate $$\int\frac{x\sin x}{1+\cos^2 x}dx$$
My attempt:
$$I=\int\frac{x\sin x}{1+\cos^2 x}dx=\int x\frac{\sin x}{1+\cos^2 x}dx=\\x\int \frac{\sin x}{1+\cos^2 x}dx-\int \left[\frac{d}{dx}x\int\frac{\sin x}{1+\cos^2 x}dx\right]dx$$
$I'=\displaystyle \int \frac{\sin x}{1+\cos^2 x}dx$
Let $u=\cos x$
$\therefore \dfrac{du}{dx}=-\sin x$
$\implies du=(-\sin x)dx$
$\therefore \displaystyle I'=-\int \frac{du}{1+u^2} $
$\implies I'=-\dfrac{1}{1}\tan^{-1}\left(\dfrac{u}{1}\right)+C \implies I'=-\tan^{-1}(u)+C$
$\implies I'=-\tan^{-1}(\cos x)+C$
$\therefore \displaystyle \int \frac{\sin x}{1+\cos^2 x}dx=-\tan^{-1} (\cos x)+C$
$\therefore \displaystyle I=x\cdot[-\tan^{-1} (\cos x)]-\int [-\tan^{-1} (\cos x)] dx$
$\implies \displaystyle I=-x \tan^{-1} (\cos x)+\int \tan^{-1} (\cos x)dx$
I cannot understand how to proceed further. Please help.
| I doubt you will be able to evaluate the integral without limits, since this link shows that the integral is very complicated, and has polylogarithms.
With the limits given and using your progress so far, $$\begin{align}\int_0^\pi\frac{x\sin x}{1+\cos^2 x}\,dx&=\left[-x\tan^{-1}(\cos x)\right]_0^\pi+\int_0^\pi\tan^{-1}(\cos x)\,dx\\&=\frac{\pi^2}4-\int_{-\pi/2}^{\pi/2}\tan^{-1}(\sin x)\,dx\end{align}$$The second term is an integral of an odd function on a symmetric interval about $0$. So it is zero. Therefore the answer is $\frac{\pi^2}4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Composition of $h(x)=f(|g(x)|)+|f(g(x))|$
$f(x)=\begin{cases}x-1,\ &-1\leq x\leq0\\ x^2,\ &0< x\leq1\end{cases}$ and $g(x)=\sin x$. Find $h(x)=f(|g(x)|)+|f(g(x))|$.
$$f(|g(x)|)=\begin{cases}|\sin x|-1,\ &-1\leq |\sin x|\leq0\\ \sin^2(x),\ &0< |\sin x|\leq1\end{cases}$$.$$\implies f(|g(x)|)=\begin{cases}0-1,\ &x=n\pi\\ \sin^2(x),\ &x\in R-{n\pi}\end{cases}$$
Also,$$|f(x)|=\begin{cases}1-x,\ &-1\leq x\leq0\\ x^2,\ &0< x\leq1\end{cases}$$ $$\implies|f(g(x))|=\begin{cases}1-\sin x,\ &-1\leq \sin x\leq0\\ \sin^2x,\ &0< \sin x\leq1\end{cases}$$ $$\implies|f(g(x))|=\begin{cases}1-\sin x,\ &(2n+1)\pi\leq x\leq2n\pi\\ \sin^2x,\ &2n\pi<x<(2n+1)\pi\end{cases}$$
Now how can I add these up, also answer is provided with $x$ in periods which don't incorporate $\pi's$, is any another elegant way exists to solve this problem? Please help.
| I will only focus on $x \in [0, 2 \pi]$.
$\begin{align}
f(\vert g(x) \vert) & = \begin{cases}
\vert \text{sin}(x) \vert - 1, & -1 \le \vert \text{sin}(x) \vert \le 0 \\
\vert \text{sin}(x) \vert^{2}, & 0 \lt \vert \text{sin}(x) \vert \le 1 \\
\end{cases} \\
& = \begin{cases}
\vert \text{sin}(x) \vert - 1, & \text{sin}(x) = 0 \\
\text{sin}^{2}(x), & 0 \lt \vert \text{sin}(x) \vert \le 1 \\
\end{cases} \\
& = \begin{cases}
- 1, & x = n \pi, n\in \mathbb N\\
\text{sin}^{2}(x), & x \not = n \pi \\
\end{cases} \\
\end{align}$
Basically just trying to simplify $f(\vert g(x) \vert)$.
$\begin{align}
f(g(x)) = & \begin{cases}
\text{sin}(x) - 1, & -1 \le \text{sin}(x) \le 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
\text{sin}(x) - 1, & \pi \le x \le 2 \pi \\
\text{sin}^{2}(x), & 0 \lt x \le \pi \\
\end{cases} \\
\end{align}$
Also just trying to simplify $f(g(x))$.
Now, a general rule to write down absolute value functions:
$$ \vert f(x) \vert = \begin{cases}
-f(x), & f(x) \lt 0 \\
f(x), & f(x) \ge 0 \\
\end{cases}$$
So:
$\begin{align}
\vert f(g(x)) \vert = & \begin{cases}
1 - \text{sin}(x), & -1 \le \text{sin}(x) \le 0 \\
- \text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\text{sin}(x) - 1, & -1 \le \text{sin}(x) \le 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
1 - \text{sin}(x), & -1 \le \text{sin}(x) \le 0 \\
- \text{sin}^{2}(x), & \text{for no } x \in \mathbb R \\
\text{sin}(x) - 1, & \text{for no } x \in \mathbb R \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
1 - \text{sin}(x), & -1 \le \text{sin}(x) \le 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
\end{align}$
Now, adding up both functions:
$\begin{align}
f(\vert g(x) \vert) + \vert f(g(x)) \vert = & \begin{cases}
- 1, & \text{sin}(x) = 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} + \begin{cases}
1 - \text{sin}(x), & -1 \le \text{sin}(x) \le 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
- 1, & \text{sin}(x) = 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} + \begin{cases}
1, & \text{sin}(x) = 0 \\
\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
0, & \text{sin}(x) = 0 \\
2\text{sin}^{2}(x), & 0 \lt \text{sin}(x) \le 1 \\
\end{cases} \\
& = \begin{cases}
0, & x \in \{0,\pi\} \\
\text{sin}^{2}(x), & 0 \lt x \lt \pi \\
\end{cases} \\
& = \text{sin}^{2}(x), \text{for } x \in [0,\pi]
\end{align}$
I hope this is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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What's the Jordan form of $J^2$? $$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.
So it could be
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or }
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?
| "But why can't the Jordan chain have the length of 4 and 1?" Because the nilpotency degree of your matrix $A=J^2$ is $3$, so $A^3=0$. But a $5\times 5$-matrix $B$ with full Jordan block does not satisfy $B^3=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?
$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$
I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system:
$$
\begin{cases}
4x^2 - 4y^2 = 15xy\\
x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y
\end{cases}
$$
Then I put $x$ from the second equation in the first one:
$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$
When I solved it I got these results:
*
*$$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$
*$$x_2 = -3,\ y_2 = -3.$$
But these results are incorrect.
These are the answers from my book:
*
*$$x_1 = 12,\ y_1 = 3.$$
*$$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$
| Hint:
$$2x-5y=9\iff2(x-2)=5(y+1),$$
$\implies\dfrac{x-2}5=\dfrac{y+1}2=z$(say)
$$\dfrac{15}4=\dfrac{5z+2}{2z-1}-\dfrac{2z-1}{5z+2}$$
$$\iff15(2z-1)(5z+2)=(5z+2)^2-(2z-1)^2$$
which is a Quadratic equation in $z$ on rearrangement
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to get the value of $a^{-3} - b^{-3}$ I have a statement that says:
If $a - b = 3$, and $a$ is the reciprocal of $b$, that is equal to $a = \frac{1}{b}$, get the value of $a^{-3} - b^{-3}$
So i have,
$a = 3 + b$, $b = a - 3$, $a = \frac{1}{b}$
$3 + b = \frac{1}{b}$, then ($b^2 + 3b - 1 = 0$ )
or
$a = \frac{1}{a-3}$, then ($a^2 -3a -1 = 0$ )
So, i need the value of $a^{-3} - b^{-3}$
Now i use the Quadratic formula for $a$:
$a = \frac{3 \pm\sqrt{13}}{2}$, then $a^-3 = \frac{8}{(3\pm\sqrt{13})^3}$
Now i use the Quadratic formula for $b$:
$b = \frac{-3 \pm\sqrt{13}}{2}$, then $b^-3 = \frac{8}{(-3\pm\sqrt{13})^3}$
Finally,
= $\frac{8}{(3\pm\sqrt{13})^3} - \frac{8}{(-3\pm\sqrt{13})^3}$, I will use $+$ instead of the $\pm$, so
= $\frac{8}{(3+\sqrt{13})^3} - \frac{8}{(-3+\sqrt{13})^3}$
= $\frac{8}{40\sqrt{13}+144} - \frac{8}{40\sqrt{13}-144}$
= $-36$
In fact, the result is correct but the development, has been very long, for a question that I am supposed to answer in less than 2 minutes, So, what other way do you see to solve this exercise? I've been looking for it on my own, but I do not clarify with another form.
| So $a=b^{-1}$ and therefore $c=a^{-3}-b^{-3}=a^{-3}-a^{3}$
Now $a-b=a-\frac 1a$ and $(a-\frac 1a)^3=a^3-a^{-3}-3(a-\frac 1a)$ so that $27=-c-9$ and $c=-36$
| {
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Is $\frac{\sqrt3+\sqrt5}{2}$ an algebraic integer?
Find the minimal polynomial of $\sqrt3+\sqrt5$ over $\mathbb Q$. Is $\Large \frac{\sqrt3+\sqrt5}{2}$ an algebraic integer?
The first part is fine and I found that the minimal polynomial of $\sqrt3+\sqrt5$ over $\mathbb Q$ is $x^4-16x^2+4$. I have some questions about the written answer.
Let $\alpha = \sqrt3+\sqrt5 $ so will try to find if $\frac{\alpha}{2}$ is algebraic integer.
The real answer:
$\alpha^4-16\alpha^2+4=0$
$\Large (\frac{\alpha}{2})^4-(\frac{16}{2^2})(\frac{\alpha}{2})^2+(\frac{4}{2^4})=0$
$\Large (\frac{\alpha}{2})^4-4(\frac{\alpha}{2})^2+(\frac{1}{4})=0$
Minimal polynomial of $\Large \frac{\alpha}{2}$
is $x^4-4x^2+\frac{1}{4}$, which is not defined over $\mathbb Z$ hence $\Large \frac{\alpha}{2}$ is not an algebraic integer.
I've highlighted the line which confuses me. Can anyone show me where this comes from?
I thought they may have substituated $\alpha$ with $\frac{\alpha}{2}$ in the minimal polynomial but then I don't understand where $\frac{16}{2^2}$ and $\frac{4}{2^4}$ came from?
| \begin{align}\alpha^4-16\alpha^2+4=0&\iff\frac{\alpha^4-16\alpha^2+4}{2^4}=0\\&\iff\left(\frac\alpha2\right)^2-4\left(\frac\alpha2\right)^2+\frac14=0.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate: $\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}$ Calculate: $$\lim\limits_{x\to 0}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}$$
In the numerator, by factoring out x I was able to get to: $$\sqrt{x}\sqrt{1 + \frac{1}{\sqrt{x}}}\sqrt[4]{1 + \frac{1}{\sqrt{x}}}$$
In the denumerator my success is:
$$\sqrt{x}(2 + 4\sqrt[-\frac{3}{8}]{x})$$
So I could cancel out $$\frac{\sqrt{x}}{\sqrt{x}}$$
But it will leave me with
$$\lim\limits_{x\to 0}{\frac{\sqrt{1 + \frac{1}{\sqrt{x}}}\sqrt[4]{1 + \frac{1}{\sqrt{x}}}}{2 + 4\sqrt[-\frac{3}{8}]{x}}}$$
What to do next?
| Note that the function is not real for negative $x$.
$\sqrt[\frac{-3}{8}]{x}=\frac{1}{\sqrt[\frac{3}{8}]{x}}\to \infty$ as $x\to0^+$. So, it is not the way to do.
\begin{align*}
\lim\limits_{x\to 0^+}{\frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}&=\lim\limits_{x\to 0^+}{\frac{ \sqrt{x+\sqrt[4]{x}\sqrt{\sqrt{x}+1}} }{ 2\sqrt{x}+4\sqrt[8]{x}}}\\
&=\lim\limits_{x\to 0^+}{\frac{ \sqrt[8]{x}\sqrt{\sqrt[4]{x^3}+\sqrt{\sqrt{x}+1}} }{ 2\sqrt[8]{x}(\sqrt[8]{x^3}+2)}}\\
&=\lim\limits_{x\to 0^+}{\frac{ \sqrt{\sqrt[4]{x^3}+\sqrt{\sqrt{x}+1}} }{ 2(\sqrt[8]{x^3}+2)}}\\
&=\frac{1}{4}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$
If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$
I can solve for two terms only but for additional $c\sqrt5$ term, I got stuck. I also want to know how to solve this for general $n$ terms i.e. $a\sqrt2+b\sqrt3+c\sqrt5+\ldots+n \text{ terms}=0$ will imply $a=b=c=\ldots=0$
| Assume $a\sqrt 2+b\sqrt3+c\sqrt 5=0$. Then $5c^2=2a^2+2ab\sqrt 6+3b^2$. From the irrationality of $\sqrt 6$, we conclude that $ab=0$. If $b\ne 0$, this makes $a= 0$ and $5c^2=3b^2$, so $\sqrt{15}=\frac{5c}b$ rational, contradiction.
If $a\ne 0$, this makes $b=0$ and $5c^2=2a^2$, ao $\sqrt{10}=\frac{5c}{a}$ rational, contradiction. We conclude that $a=b=0$, but then also $c=0$.
The general statement is:
Let $a_1,\ldots, a_n$ be different square-free positive integers. Then $\sqrt{a_1},\ldots, \sqrt{a_n}$ are linearly independent over the rationals.
Its proof involves but a bit deeper techniques. Edit: Found it, see my answer here
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Compute the line integral of $f (x, y) = x^2 + y^2$ over the parabola $y = x^2$ from the point $x = 0$ to $x = 1$. I ended up with
$$\int_1^0 t^2(f^2+1)(1+4t^2)^{\frac{1}{2}}\,dt.$$ I don't know if I'm right from there and I wouldn't even know how to solve it from there anyway. My lecturer didn't bother with giving answers so I don't know.
| $\int_C f(x,y)\|dr\|$
$x = t\\y = t^2\\ \frac {dx}{dt} = 1\\\frac {dy}{dt} = 2t\\\|dr\| = \sqrt {(\frac {dx}{dt})^2+(\frac {dy}{dt})^2}\ dt = \sqrt {1+4t^2}\ dt$
$\int_0^1 (t^2 + t^4)\sqrt {1+4t^2} \ dt$
Do you want to use a trig substitution or hyperbolics?
$t = \frac 12 \tan \theta\\dt = \frac 12 \sec^2\theta \ d\theta$
$\int_0^{\frac{\pi}{4}} (\frac 14 \tan^2\theta + \frac 1{16} \tan^4\theta)\frac 12\sec^3\theta\ d\theta\\
\int_0^{\frac {\pi}{4}} \frac 18 \tan^2\theta\sec^3\theta + \frac 1{32} \tan^4\theta\sec^3\theta d\theta$
vs.
$t = \frac 12 \sinh u\ du\\ dt = \frac 12 \cosh u\ du$
$\int_0^{\sinh^{-1} 2} \frac 18 \sin^2 u\cosh^2 u + \frac 1{32} \sinh^4 u\cosh^2 u du\\
\int_0^{\sinh^{-1} 2} \frac 18 \left(\frac {e^u - e^{-u}}{2}\right)^2\left(\frac {e^u + e^{-u}}{2}\right)^2 + \frac 1{32} \left(\frac {e^u - e^{-u}}{2}\right)^4\left(\frac {e^u + e^{-u}}{2}\right)^2 du\\
$
I think the second one looks easier.
The coeficients are going to get messy though
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Simple algebra simplification I have
$ \frac{-8rS^{2} \pm \sqrt{64r^{2}S^{4} + 4}}{2}$
Which simplified to
$ -4rS^{2} \pm \sqrt{16r^{2}S^{4} +1}$
But I can’t seem to get this
| Note that
$$
64r^2S^4+4=4(16r^2S^4+1)
$$
so
$$
\sqrt{64r^2S^4+4}=\sqrt{4(16r^2S^4+1)}=2\sqrt{16r^2S^4+1}
$$
Finally,
\begin{align}
\frac{-8rS^2\pm\sqrt{64r^2S^4+4}}{2}
&=\frac{-8rS^2\pm2\sqrt{16r^2S^4+1}}{2}\\[6px]
&=\frac{2(-4rS^2\pm\sqrt{16r^2S^4+1}\,)}{2}\\[6px]
&=-4rS^2\pm\sqrt{16r^2S^4+1}
\end{align}
Your original equation apparently was
$$
x^2+8rS^2x-1=0
$$
Whenever you have a factor $2$ that can be extracted from the coefficient of the degree $1$ term, the simplification above can be performed: if the equation is $ax^2+2\beta x+c=0$, the quadratic formula yields
\begin{align}
\frac{-2\beta\pm\sqrt{4\beta^2-4ac}}{2a}
&=\frac{-2\beta\pm\sqrt{4(\beta^2-ac)}\,}{2a}\\[6px]
&=\frac{-2\beta\pm2\sqrt{\beta^2-ac}}{2a}\\[6px]
&=\frac{-\beta\pm\sqrt{\beta^2-ac}}{a}
\end{align}
In your case $a=1$, $\beta=-4rS^2$ and $c=-1$.
It's not necessary to remember one more formula, just to recall that the simplification can be done.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the integral$ \int_{-\infty}^{\infty}\frac{b\tan^{-1}\big(\frac{\sqrt{x^2+a^2}}{b}\big)}{(x^2+b^2)(\sqrt{x^2+a^2})}\,dx$. I am attempting to evaluate
$$\int_{-\infty}^{\infty}\dfrac{b\tan^{-1}\Big(\dfrac{\sqrt{x^2+a^2}}{b}\Big)}{(x^2+b^2)(\sqrt{x^2+a^2})}\,dx. $$
I have tried using the residue formula to calculate the residues at $\pm ib,\pm ia,$ but it got messy very quickly. Then I tried to use a trigonometric substitution $x=a\tan(\theta)$; $dx=a\sec^{2}(\theta)\,d\theta$ which led me to the integral $$\int_{-\infty}^{\infty}\dfrac{b\tan^{-1}\Big(\dfrac{a\sec(\theta)}{b}\Big)\sec(\theta)}{(a^2\tan^{2}(\theta)+b^2)}\,d\theta.$$ The bounds for this integral seem incorrect, but I am more worried about the actual expression before I deal with the bounds, which may have to be changed into a double integral where $0\leq\theta\leq2\pi$ and the second bound would range from $-\infty$ to $\infty$. I am wondering if there is some kind of substitution I have missed, but I have hit the wall. The OP of this problem said there were cases that would come into play, but when I asked him whether or not those cases arose from $b<0$ and $b>0$ he told me they did not. The cases most likely arise from whether $a$ and $b$ are positive or negative, because the case where $b=0$ is trivial, and in the case where $a=0$ I used wolframalpha and the integral evaluates to $\dfrac{\pi\ln(2)\lvert b \rvert}{b^2}$ for $\Im(b)=0 \land \Re(b)\neq0.$ Contour integration may be necessary. I am stuck on this problem and I would greatly appreciate the help. Thank you for your time.
| We will use Mark Viola’s derivations to account for the case when $\lvert\,a\,\rvert< \lvert\,b\,\rvert$.
Integrating $(2)$ (See Mark Viola's answer above) and using $\lim_{c\to \infty}F(c)=0$, we find that
$$\begin{align}
F(c)&=\pi\,\left(\frac{\arctan\left(\frac{bc}{\sqrt{a^2-b^2}\sqrt{a^2+c^2}}\right)-\arctan\left(\frac{b}{\sqrt{a^2-b^2}}\right)+\pi/2-\arctan\left(\frac{c}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}\right)
\end{align}.$$
Now we will account for our case’s assumption and arrive at
$$\begin{align}
F(c)&=\pi\,\left(\frac{-i\tanh^{-1}\left(\frac{bc}{\sqrt{b^2-a^2}\sqrt{a^2+c^2}}\right)+i\tanh^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)+\pi/2+i\tanh^{-1}\left(\frac{c}{\sqrt{b^2-a^2}}\right)}{i\sqrt{b^2-a^2}}\right) \\
&=\pi\,\left(\frac{-\tanh^{-1}\left(\frac{bc}{\sqrt{b^2-a^2}\sqrt{a^2+c^2}}\right)+\tanh^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)-i\pi/2+\tanh^{-1}\left(\frac{c}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}\right)
\end{align}.$$
We replace $c=b$ to obtain our result
$$\int_{-\infty}^\infty \frac{b\arctan\left(\frac{\sqrt{x^2+a^2}}{b}\right)}{(x^2+b^2)\sqrt{x^2+a^2}}\,dx=\\\pi\,\left(\frac{-\tanh^{-1}\left(\frac{b^2}{\sqrt{b^2-a^2}\sqrt{a^2+b^2}}\right)-i\pi/2+2\tanh^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Interesting property of sum of powers of integers from 1 to 114. When I have this list of specific X values:
$X: 1, 2, 3, 4, \ldots, 112, 113, 114.$
$$\sum_{n=1}^{114}n = 6555$$
$$6555/19 = 345$$
The sum of these $X$ values divided by $19$ is an integer.
Then I square each $X$ value:
$X^2: 1, 4, 9, 16, \ldots, 12544, 12769, 12996.$
The sum of these $X^2$ values divided by $19$ is also an integer.
Then I cube each $X$ value, and the pattern continues.
Is there any reason for this property mathematically? Or does it just exist?
Many thanks.
| This is because $19|114$.
$\sum_{k=1}^n k = \frac {n(n+1)}{2}$
And $\sum_{k=1}^n k^2 = \frac {n(n+1)(2k + 1)}6$.
And as $19$ is a prime number that divides $114$, $19$ will divide both $\frac {19*20}{10}$ and $\frac {19*20*39}6$.
In general if $n = 19*m$ then
$\sum_{k=1}^n k = \frac {n(n+1)}2 = \frac {19m(19m + 1)}2 = 19\frac {m(19m+1)}2$ and since either $m$ is even or $19m + 1$ is even, $2|m(19m+1)$.
Likewise $\sum_{k=1}^n k^2 =\frac {19m(19m + 1)(38m + 1)}6 = 19 \frac {m(19m + 1)(38m + 1)}6$. Again either $m$ or $19m + 1$ is even so $2|m(19m+1)(38m + 1)$. And if $m$ has remainder $0$ when divided by $3$ then $3|m$. If $m$ has remainder $1$ when divided by $3$ then $3|38m + 1$. And if $m$ has remainder $2$ when divided by $3$ then $3|19m+1$ So $3|m(19m+1)(38m + 1)$. So $6|m(19m+1)(38m+1)$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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Subgroups and invariants in a unitary group $U(2)$ mixing multiplets This is related to the post, but a simplified version of the problem. However, the form of $P_x,P_y,P_z$ are different and richer.
Let $$G=U(2),$$ be the unitary group. Here we consider $G$ in terms of the fundamental representation of U(2). Namely, all of $g \in G$ can be written as a rank-2 (2 by 2) matrices.
Can we find some subgroup of Lie group, $$k \in K \subset G= U(2) $$ such that
$$
k^T \{P_x, P_y, P_z, -P_x, - P_y, - P_z\} k =\{P_x, P_y, P_z, -P_x, - P_y, - P_z\}.
$$
This means that set $\{P_x, P_y, P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$.
Here $k^T$ is the transpose of $k$.
What is the full subset (or subgroup) of $K$?
Here we define:
$$
P_x =
\left(
\begin{array}{ccc}
0 & 1 \\
1 & 0 \\
\end{array}
\right),\;\;\;\; P_y =
\left(
\begin{array}{ccc}
-1 & 0 \\
0 & 1 \\
\end{array}
\right),\;\;\;\; P_z =
\left(
\begin{array}{ccc}
-i & 0 \\
0 & -i \\
\end{array}
\right).$$
This means that $k^T P_a k= \pm P_b$ which may transform $a$ to a different value $b$, where $a,b \in \{x,y,z \}$. But overall the full set $ \{P_x, P_y, P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$.
There must be a trivial element $k=$ the rank-2 identity matrix. But what else can it allow?
How could we determine the complete $K$?
|
Answer. The subgroup $K$ of invariant matrices in $U(2)$ is isomorphic to the finite group
$$
\operatorname{SL}(2,3)\to G\to\mathbb{Z}_2) \rtimes_\varphi \mathbb{Z}_2
\quad\simeq\quad
\mathbb{Z}_2\to G\to S_4) \rtimes_\varphi \mathbb{Z}_2
$$
where $\operatorname{SL}(2,3)$ is the special linear group of degree 2 over a field of 3 elements (order 24), which is isomorphic to the binary von Dyck group with the parameters (2,3,3); it has the presentation: $\langle a,b,c \mid a^3=b^3=c^2=abc\rangle$.
In other words, $K\cong G\rtimes_\varphi\mathbb{Z}_2$ for some subgroup $G\subset K$ and $G/\operatorname{SL}(2,3)\cong\mathbb{Z}_2$ or $G/\mathbb{Z}_2\simeq S_4$.
For whom is interested, I list the all $96=24\times 4$ invariant matricies of $K$:
$$
\begin{gather*}
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
\frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix},
\frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \\
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},
\frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ -i & 0 \end{pmatrix},
\frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & -1 \\ i & i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & -1 \\ -i & -i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ i & -1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & -i \\ -1 & -i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & -1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & i \\ -1 & i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix}
\end{gather*}
$$
product with $\langle i\rangle=\{\pm I,\pm iI\}$.
For given orders, the numbers of elements in $K$ and $\mathbb{Z}_4\times S_4$ are as follows:
order | K | Z4 x S4
--------------------------
1 | 1 | 1
2 | 19 | 19
3 | 8 | 8
4 | 20 | 44
6 | 8 | 8
8 | 24 | 0
12 | 16 | 16
--------------------------
total | 96 | 96
This information shows that $K$ (GAP ID [96, 192]) in $U(2)$ is not isomorphic to the previous answer $\mathbb{Z}_4\times S_4$ (GAP ID [96,186]) in $U(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Equation involving the logarithm Find the set of values of $k$ satisfying this equation for only one real root of $x.$
$$ \log(kx) = 2 \log(x+2)$$
I think that for the sake of satisfying the domain restriction:
*
*$ k \cdot x > 0 $
*$x+2 > 0$
*$ k \cdot x = (x+2)^2$
*$ b^2-2ac = 0$.
I am not sure what to do from here.
|
Find the set of values of $k$ satisfying
\begin{align} \log(kx) &= 2
\log(x+2) \tag{1}\label{1}
\end{align}
for only one real root of $x$.
Equation
\begin{align}
k \cdot x &= (x+2)^2
\tag{2}\label{2}
\end{align}
follows from \eqref{1}
and can be transformed as
\begin{align}
(x+2)^2 -kx&=0
\\
(x+2)^2 -k(x+2)+2k&=0
\end{align}
with the roots
\begin{align}
x+2&=\tfrac12\,k\pm\tfrac12\sqrt{k^2-8k}
\tag{3}\label{3}
.
\end{align}
The domain restrictions
\begin{align}
kx&>0
\tag{4}\label{4}
,\\
x+2&>0
\tag{5}\label{5}
\end{align}
suggest that we need to consider two cases:
\begin{align}
\text{Case 1. }\quad
k&>0,\quad x>0
\tag{6}\label{6}
,\\
\text{Case 2. }\quad
k&<0,\quad x\in(-2,0)
\tag{7}\label{7}
.
\end{align}
In the first case
there is only one value ($k=8$) that is useful for \eqref{1}.
In the second case,
when $k<0,\quad x\in(-2,0)$,
since the left-hand side of \eqref{3}
is positive,
we need to check,
for which values of negative $k$
\begin{align}
x+2&=\tfrac12\,k+\tfrac12\sqrt{k^2-8k}
\tag{8}\label{8}
\end{align}
holds.
From equation \eqref{8} and condition \eqref{7},
\begin{align}
0&<\tfrac12\,k+\tfrac12\sqrt{k^2-8k}
<2
\tag{9}\label{9}
\end{align}
Let $\kappa=-k$, $\kappa>0$. Then
condition
\begin{align}
0&<-\tfrac12\,\kappa+\tfrac12\sqrt{\kappa^2+8\kappa}
\end{align}
always holds for any $\kappa>0$, that is,
for any $k<0$.
Next, consider
\begin{align}
-\tfrac12\,\kappa+\tfrac12\,\kappa\,\sqrt{1+\frac8\kappa}
<2
,\\
-\kappa+\kappa\,\sqrt{1+\frac8\kappa}
<4
,\\
\kappa\,\sqrt{1+\frac8\kappa}
<4+\kappa
,\\
\kappa^2+8\kappa
<16+8\kappa+\kappa^2
,
\end{align}
which also holds for all $\kappa>0$, that is, for all $k<0$
.
Summarizing,
the answer is:
\begin{align}
k\in(-\infty,0)\cup\{8\}
.
\end{align}
Illustration for the Case 1:
Illustration for the Case 2:
| {
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Find the value of $xyz$
$x$, $y$, $z$ are real numbers such that
$①$ $ x+y+z=\sqrt{3}$
$②$ $xy+yz+zx=1$
In this case, find the value of $xyz$.
$$$$
Here's my attempt.
$$$$
$3=(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = x^2 + y^2 + z^2 + 2$
$\therefore \ \ x^2 + y^2 + z^2 = 1 $
$\\$
$ x^3 +y^3 + z^3 - 3xyz=(x+y+z)(x^2 + y^2 + z^2 - xy-yz-zx) = 0$
$\therefore \ \ x^3 + y^3 + z^3 = 3xyz $
$\\$
What can I do... after these?
$$$$
I guessed the answer like this.
If $ x=y=z = \frac{\sqrt{3}}{3}$ , then these satisfy $①, ②$.
So, $\ $ $xyz=\frac{\sqrt{3}}{9}$....
Thanks in advance.
| Another property of AM-GM is that if the arithmetic means and geometric means are equal, all the variables are equal.
Since $\frac{x^3+y^3+z^3}{3} ≥ ({x^3 \cdot y^3 \cdot z^3})^{1/3}$, we get $x^3+y^3+z^3 ≥ xyz$, and as a consequence $x = y = z$. Therefore, you only need to solve the equation $u^3 + u^3 + u^3 = u \cdot u \cdot u$, and then substitute into equation $1$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression, then prove that:
$$\cos (\dfrac {B-C}{2})=2\sin (\dfrac {A}{2})$$
My Attempt:
Since, $a,b,c$ are in AP
$$2b=a+c$$
$$\sin A+\sin C=2\sin B$$
$$2\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2.\sin (\dfrac {A+C}{2}).\cos (\dfrac {A+C}{2})$$
$$2\cos (\dfrac {A+C}{2})=\cos (\dfrac {A-C}{2})$$
| You can't, because it is not true. Consider a right-angle triangle with the sides $3,4,5$. Then:
$$\cos \left(\frac{B-C}{2}\right)=\sqrt{\frac{1+\cos{(B-C)}}{2}}=\sqrt{\frac{1+\cos B\cos C+\sin B\sin C}{2}}=\frac{3}{\sqrt{10}},\\
2\sin \frac A2=2\sqrt{\frac{1-\cos A}{2}}=\frac{2}{\sqrt{10}}.$$
| {
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} |
British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9 $ (A corollary of C-S I believe)
Is there an elementary proof without this fact (or C-S in general)?
| By the RMS-AM-GM inequalities (root-mean square vs. arithmetic vs. geometric means):
$$
\frac{1}{\sqrt{3}} = \sqrt{\frac{x^2+y^2+z^2}{3}} \;\ge\; \frac{x+y+z}{3} \;\ge\; \sqrt[3]{xyz} \quad\implies\quad \begin{cases}\begin{align}x+y+z \,&\le\, \sqrt{3} \\[5px] xyz \,&\le\, \dfrac{1}{3\sqrt{3}}\end{align}\end{cases}
$$
Multiplying the latter gives $\,xyz(x+y+z) \le \dfrac{1}{3}\,$, which is the inequality to prove. As with all means inequalities, the equality holds iff $\,x=y=z\,$.
| {
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if $|az^2+bz+c|\le 1$, find the maximum of $|a|+|b|$ Let $a,b,c$ be complex numbers, and $f(z)=az^2+bz+c$, such that if the complex number $|z|\le 1$, then we have $|f(z)|\le 1$. Find the maximum of $|a|+|b|$.
if we $a,b,c$ be real and $z$ be real, I can find the maximum is $2$,because $f(z)=2z^2-1$ such it. and
$$a=\dfrac{1}{2}[f(1)+f(-1)]-f(0),b=\dfrac{1}{2}[f(1)-f(-1)]$$
so
$$|a|+|b|=\dfrac{1}{2}|f(1)+f(-1)-2f(0)|+\dfrac{1}{2}|f(1)-f(-1)|\le 2$$But for complex, maybe this answer is also $2?$,I'm not sure.
| A comment on the uniqueness of the solution of achile hui: and @Martin R: and a generalization
Take $f(z) = az^2 + bz + c$ with $a$, $b\ge 0$, $|f(z)|\le 1$ for $|z|\le 1$ ( enough for $|z|=1$) and such that $a+b=\frac{2}{\sqrt{3}}$, the maximum,
With $\omega=e^{i \pi/3}$ we have
$$a+b = \frac{f(\omega)-f(\omega^{-1})}{i \sqrt{3}}$$
so its absolute value does not exceed $\frac{2}{\sqrt{3}}$. Since we have equality we conclude
$f(\omega)=i $ and $f(\omega^{-1})=-i$.
So we must have
$$f(z) = \frac{2z^2 +4 z -1}{3\sqrt{3}} + \eta\cdot ( z^2-z+1) = f_0(z) + \eta\cdot ( z^2-z+1)$$ for some $\eta$ real.
The function $|f_0(e^{i \theta})|$ achieves a maximum at $\theta = \pi/3$ so the tangent to the curve $f_0(e^{i\theta)}$ is horizontal. Since $f$ is also an extreme, the tangent to $f(e^{i\theta})$ at $\theta = \pi/3$ has the same direction. Recall that $\frac{d}{d\theta} f(e^{i \theta})= i e^{i \theta} f'(e^{i \theta})$.
Therefore, we must have
$$\frac{ \eta (2 \omega -1) }{4 \omega + 4} \in \mathbb{R}$$ and that implies $\eta = 0$. Therefore $f=f_0$.
The plot of $\theta \mapsto \frac{2 e^{2 i \theta} + 4 e^{i \theta} -1}{3 \sqrt{3}}$ is a cardioid ( see pic )The fixed circle has center $(-\frac{1}{3\sqrt{3}},0)$ and radius $\frac{2}{3\sqrt{3}}$. The parametrization will match if we start rolling the circle from the right.
ADDED: After some calculations I found the maximum of $s|a| + |b|$, where $s\ge 0$ is a constant. There are two cases:
*
*$0\le s \le \sqrt{2}$, so $s = 2 \cos \theta$, where $\theta \in [\pi/4, \pi/2]$. Then $\max (2 \cos\theta |a| + |b|) =\frac{1}{\sin \theta}$.
*If $s\ge \sqrt{2}$ then $\max (s|a|+|b|)=s$.
The inequality $2 \cos \theta |a| + |b| \le \frac{1}{\sin \theta}$ is proved in the same way by evaluating the polynomial
$f$ at $\omega, \bar \omega = \cos \theta \pm i \sin \theta$.
If moreover $\theta \in [\pi/4, \pi/2]$, one checks that the value $\frac{1}{\sin \theta}$ is achieved for the polynomial
$$f_{\theta}(z) = \frac{\cos \theta\cdot z^2 - 2 \cos 2 \theta \cdot z + \cos \theta \cos 2 \theta }{2\sin^3 \theta}$$
$f_{\theta}$ was found so that $f(e^{\pm i \theta}) = \pm i$ and the curve $t \mapsto f(e^{\pm i t})$ has a horizontal tangent at $t = \phi$.
One checks that we have the equality
$$ 1- |f_{\theta}(e^{it})|^2 = \frac{- \cos (2 \theta) \cos^2 \theta\, (\cos t - \cos \theta)^2 }{\sin^6 \theta}$$
(so the need for $\theta \ge \pi/4$ ).
The uniqueness of the optimal polynomial (up to some phase changes) also holds.
Note that for $\theta = \pi/4$ we get $\max \sqrt{2} |a| + |b| = \sqrt{2}$ achieved for the (essentially) unique polynomial $z^2$.
If $s\ge \sqrt{2}$
we have
$$s|a| + |b| = \frac{s}{\sqrt{2}} \cdot ( \sqrt{2}|a| + \frac{\sqrt{2}}{s}|b|)\le \frac{s}{\sqrt{2}} (\sqrt{2}|a|+|b|)\le \frac{s}{\sqrt{2}} \cdot \sqrt{2} = s$$
Therefore, the maximum of $s|a| + |b|$ is $s$ ( achieved for $z^2$).
| {
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Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$
Deduce that $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\dfrac{1}{n+2}$
We know $\dfrac{1}{n+2}=\displaystyle \int_{0}^{1}t^{n+2-1}dt$
Now $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)(k+2)}\bigg)=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}-\dfrac{1}{k+2}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}\bigg)-\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+2)}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\int_{0}^1 t^{k+1-1}dt \space - \space \displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k \int_{0}^1 t^{k+2-1}dt$
$=\displaystyle \int_{0}^1 t^{1-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt- \int_{0}^1 t^{2-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt$
$=\displaystyle \int_{0}^1 t^{1-1}(1-t)^{n}dt \space - \space \int_{0}^1 t^{2-1}(1-t)^{n}dt$
$=\displaystyle \large \beta(1,n+1)$-$\displaystyle \large\beta(2,n+1)$
$=\dfrac{\Gamma(1)\Gamma(n+1)}{\Gamma(1+n+1)}-\dfrac{\Gamma(2)\Gamma(n+1)}{\Gamma(2+n+1)}= \underbrace{\dfrac{1}{n+2}}_{\text{which is exactly what I proved}}$
PS @Chappers Thankyou all users for correcting one nasty mistake.
| Hint:
Multiply and divide by $(n+1)(n+2)$ to turn the question into $$\frac {1}{(n+1)(n+2)} \sum_{k=0}^n (-1)^k \binom {n+2}{k+2}= \frac {1}{(n+1)(n+2)}\left[ \left(\sum_{k=-2}^n (-1)^k \binom {n+2}{k+2}\right)+(n+2)-1\right]=\frac {1}{n+2} $$
| {
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If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$?
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to?
Attempt:
$y= \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that?
The answer is:
$c= 4$
| Note:
$$y= \dfrac{x^2+x+c}{x^2+2x+c} \Rightarrow \\ (y-1)x^2+(2y-1)x+c(y-1)=0.$$
For $x$ to be real:
$$\Delta=(2y-1)^2-4c(y-1)^2=4(1-c)y^2+2(4c-2)y+1-4c\ge0.$$
The range $y\in\left[\frac65,\frac32\right]$ corresponds to the solution of the above inequality with $y_1=\frac56$, $y_2=\frac32$ and $y_1y_2=\frac54$:
$$y_1y_2=\frac{1-4c}{4(1-c)}=\frac54 \Rightarrow c=4.$$
| {
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"source": "stackexchange",
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About the relation $\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2$ It is well known that
$$\int\frac{1}{1+x^2}\,\mathrm{d}x=\tan^{-1}x+C \tag{1}$$
However, I integrated this differently and got an unusual result.
Suppose we make the substitution $x=\sinh\theta$ and $\mathrm{d}x=\cosh\theta\,\mathrm{d}\theta$ so the integral becomes $$\int\frac{\cosh\theta}{\cosh^2\theta}\,\mathrm{d}\theta=\int\frac{1}{\cosh\theta}\,\mathrm{d}\theta \tag{2}$$
By the definition of $\cosh\theta$, we can rewrite this as $$\int\frac{2e^\theta}{e^{2\theta}+1}\,\mathrm{d}\theta=2\tan^{-1}e^\theta+C \tag{3}$$
Using the fact that $e^\theta=\cosh\theta+\sinh\theta$, we get $e^\theta=x+\sqrt{1+x^2}$, so the answer is then $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C \tag{4}$$
Equating $(4)$ with $(1)$, we have $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C=\tan^{-1}x \tag{5}$$
Plugging in $x=0$, we find $C=-\frac\pi2$. We now have the following strange relationship
$$\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2 \tag{$\star$}$$
This leads me to wonder: Why is this true geometrically, and does this relationship extend into the complex plane?
| The geometric interpretation could be found in the figure below.
We have $\alpha = \frac{\pi}{2} - \arctan x , ~\beta = \frac{\pi}{2} - \arctan \left(x + \sqrt{1 + x^2}\right)$, and $\alpha = 2 \beta$.
Combining these three equations would lead to your results.
For a formal proof, let $\theta = \arctan \left(x + \sqrt{1 + x^2}\right)$. Since
$$ x + \sqrt{1 + x^2} \geq \sqrt{1 + x^2} - |x| > 0,$$
we have $\theta \in (0, \frac{\pi}{2})$.
Also, note that
$$
\begin{align}
\tan\left(2\theta - \frac{\pi}{2}\right)&= -\frac{1}{\tan(2\theta)}\\
&= -\frac{1 - \tan^2 \theta}{2 \tan \theta}\\
&= -\frac{1 - \left(x + \sqrt{1 + x^2}\right)^2}{2 \left(x + \sqrt{1 + x^2}\right)}\\
&= x.
\end{align}
$$
As $\left(2\theta - \frac{\pi}{2}\right) \in (-\frac{\pi}{2}, \frac{\pi}{2})$, i.e., $\left(2\theta - \frac{\pi}{2}\right)$ lies in the range of $\arctan(\cdot)$, we know that
$$ \arctan x = 2\theta - \frac{\pi}{2} = 2\arctan \left(x + \sqrt{1 + x^2}\right) - \frac{\pi}{2}. $$
| {
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Why is $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$ defined at -1, 0 and not at 1 Why does the function:
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
Is defined at $x = -1, x = 0$ (we will have $1/0$ in the fraction then) and not at $x = 1$ according to the following graph?
Live example
| Let consider
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
from here we need to set $x\neq1,-1,0$, morover we need that
$$\frac{2}{x+1} + \frac{1}{x}=\frac{3x+1}{x(x+1)}\neq 0\implies x\neq -\frac13$$
Simplifing the original expression we obtain
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}\to\bar f(x)=\frac{2x^2}{(3x+1)(x-1)}$$
which is well defined for $x\neq -\frac13,1$.
In conclusion the function $f(x)$ is not defined for $x=1,-1,0,-\frac13$ and $x=0$ and $x=-1$ are two point with removable discontinuity.
| {
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Express the sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ in terms of ${x_0, x_1, n}$ Given a sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ how can one express it in terms of ${x_0, x_1, n}$. Can this be generalized for ${x_n = \alpha x_{n-1} + \beta x_{n-2}}$
I've tried to use the following approach:
$$\eqalign{
& 1x_0 = x_01 \\
& zx_1 = x_1z \\
& z^2x_2 = (2x_1 + 3x_0)z^2 \\
& z^3x_3 = (2x_2 + 3x_1)z^3 \\
& z^4x_4 = (2x_3 + 3x_2)z^4 \\
& ... \\
& z^nx_n = (2x_{n-1} + 3x_{n-2})z^n
}$$
Then sum LHS with RHS which will produce:
$${
x_0 + \sum\limits_{k = 1 }^na_nz^n = x_0 +zx_1 + 2\sum\limits_{k = 2}^nx_{n-1}z^n + 3 \sum\limits_{k = 2}^n x_{n-2}z^n
}$$
Let
$${
G(z) = x_0 + \sum\limits_{k = 1 }^na_nz^n
}$$
Then RHS may be expressed in terms of ${G(z)}$. For example
$${
2\sum\limits_{k = 2}^na_{n-1}z^n = 2z\sum\limits_{k = 2}^na_{n-1}z^{n-1} = 2z(\sum\limits_{k = 1}^na_{n}z^{n} + x_0 - x_0) = 2z(G(z) - x_0)
}$$
Applying those transformations I eventually got ${G(z)}$ expressed in terms of z and ${x_1, x_0}$. But at this point I got stuck.
I got a sum of fractions:
$${
\frac{3x_0 - x_1}{4(1+z)}+\frac{x_0+x_1}{4(1-3z)}
}$$
I guess i could expand the fractions into series and find their sum, but i am not supposed to know about such expansions at the point of the book i took the problem from.
All of the above feels like a wrong approach. So the question is whether this can be done in a more elegant way.
| Use generating functions. Define $g(z) = \sum_{n \ge 0} x_n z^n$, take the recurrence shifted by 2, multiply by $z^n$ and sum over $n \ge 0$, recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} x_{n + 2} z^n
&= 2 \sum_{n \ge 0} x_{n + 1} z^n + 3 \sum_{n \ge 0} x_n z^n \\
\frac{g(z) - x_0 - x_1 z}{z^2}
&= 2 \frac{g(z) - x_0}{z} + 3 g(z)
\end{align*}$
Solve for $g(z)$, write as partial fractions; extract coefficient of $z^n$:
$\begin{align*}
g(z)
&= \frac{x_0 + (x_1 - 2 x_0) z}{1 - 2 z - 3 z^2} \\
&= \frac{x_0 + x_1}{4 (1 - 3 z)} + \frac{3 x_0 - x_1}{4 (1 + z)} \\
x_n
&= [z^n] g(z) \\
&= \frac{x_0 + x_1}{4} \cdot 3^n + \frac{3 x_0 - x_1}{4} \cdot (-1)^n
\end{align*}$
The last because the series is:
$\begin{align*}
\sum_{n \ge 0} a^n z^n
&= \frac{1}{1 - a z}
\end{align*}$
| {
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Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
| $$(2x^2-x)-(x-2)=2x^2-2x+2$$
Hence$$f(x)-g(x)=2\bigg(\frac{x^2-x+1}{x^2-x+1}\bigg)=2(1)=2$$
In effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values.
| {
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Find ${(ab)}^{-1}$ if $a^2=e, b^6=e, ab=b^4a$ where $a,b$ are elements of a group
If $a,b$ are elements of a group and $a^2=e, b^6=e, ab=b^4a$, then find the order of $ab$ and express ${(ab)}^{-1}$ in terms of $a^mb^n$ and $b^ma^n$
I could find the order of $ab$ to be 6 but struggling to find ${(ab)}^{-1}$ in terms of $a^mb^n$ and $b^ma^n$.
Please help me move forward.
| We have
$$b = a b^4 a = (a b a)^4= (b^4)^4=b^{16}= b^6 \cdot b^6 \cdot b^4 = b^4$$
so $b^3=e$ and $ab= b a$. From here it is easy to see that the order of $ab$ is $\le 6$ (it is in fact the product of the orders of $a$, $b$, these being relatively prime). Moreover, $(ab)^{-1}=a^{-1}b^{-1}=a b^2= b^2 a$.
| {
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Solving inverse function problems The following is my (likely completely incorrect) attempt.
$$f(x)= 3x + 5, \qquad \qquad g(x)= 1 - 2x$$
*
*Show that $f{^{-1}}(2) < g(-3)$
$$\frac{1}{3\cdot 2+5} < (1-2\cdot(-3))\implies\frac{1}{11} < 7$$
*Find the value of $a$ such that $f(2a) = g{^{-1}}(a)-2$
$$6a+5 = \frac{1}{-1-2a}$$
$$6a = \frac{1}{4-2a}$$
$$8a = \frac{1}{4}\implies a = \frac{1}{32}$$
*Solve $\frac{f(x)}{2} + \frac{g(x)}{3} = -23$
$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$
| *
*$f$: "Take $x$, multiply it by $3$ and add $5$".
*Inverse of $f$, $f^{-1}$ "Take a $y$, subtract 5, then divide by $3$"
| {
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$p^{q+1}+ q^{p+1}$ is a perfect square. Find all the prime numbers $p$ and $q$ such that $p^{q+1} + q^{p+1}$ is a perfect square. I tried this but I can't imagine where should I start fromCan some one give me a hint please.
| If $p,q$ are odd primes $\implies p,q \equiv ± 1\ mod\ 4$
$$p^{q+1} + q^{p+1} \equiv (±1)^{q+1}+(±1)^{p+1} \ mod\ 4 \equiv 2\ mod\ 4$$
And we know perfect squares are $\equiv 0,1 \ mod\ 4 \implies p\ or\ q$ must be even.
Let's take $p= 2$
$$2^{q+1} + q^{3} \equiv q \ mod\ 4 \implies q \equiv 1\ mod\ 4$$
Now,
$$2^{\cfrac{2(q+1)}{2}} + q^3 = k^2$$
Notice that $k$ is odd
$$k^2 - 2^{q+1} = q^3$$
$$\left(k+2^{\cfrac{q+1}2}\right)\left(k-2^{\cfrac{q+1}2}\right) = q^3$$
Case 1:
$$\left(k+2^{\cfrac{q+1}2}\right) = q^2 \ and\ \left(k-2^{\cfrac{q+1}2}\right) =q$$
Adding both,
$$2k = q+q^2 \implies k = \cfrac{q+q^2}{2}$$
$$2^{q+1} + q^3 = {\left(\cfrac{q+q^2}{2}\right)}^2 = \cfrac{q^2+q^4}{4} + q^3 \implies 2^{p+3} = q^2(1 + q^2)$$ a contradiction
Case 2:
$$\left(k+2^{\cfrac{q+1}2}\right) = q^3 \ and\ \left(k-2^{\cfrac{q+1}2}\right) =1$$
Using $k = 1 + 2^{\cfrac{q+1}2} \implies k^2 = 1 + 2^{q+1} + 2^{\cfrac{q+1}2 + 1}$
Putting $k^2$ in equation,
$$2^{q+1} + q^3 = 1 + 2^{q+1} + 2^{\cfrac{q+1}2 + 1} \implies q^3 - 1 = 2^{\cfrac{q+1}2 + 1}= (q-1)(q^2 + q + 1)$$
Again not possible
We are left with only option, $p,q$ must be even and only even prime is $2$
$$2^3 + 2^3 = 16 = 4^2$$
Only solution is $(p,q) = (2,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction.
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$.
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus:
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.
| For a non-inductive solution, note that $3^2\equiv 2 \bmod 7$; then
$\begin{align}
3^{2n+1} + 2^{n+2} &\equiv 3\cdot 2^n + 4\cdot 2^n \\
&\equiv 7\cdot 2^n \bmod 7\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Lagrange multiplier for two constraints?
$$\begin{array}{ll} \text{maximize} & −x^2−y^2+4x+6y\\ \text{subject to} & x+y \le 6\\ & y \le 2\\ & x ,y \geq 0\end{array}$$
I tried using the Lagrange multiplier method and set:
$$L=(-x^2-y^2+4x-6y)+a(6-x-y)+b(2-y).$$
And differentiated to get:
$$\frac{dL}{dx}=-2x+4-a$$
$$\frac{dL}{dy}=-2y+6-a-b$$
$$\frac{dL}{da}=6-x-y$$
$$\frac{dL}{DB}=2-y$$
Then by complementary slackness:
$$-2x+4-a=-2y+6-a-b.$$
But I don't know what to do next?
| As $f(x,y)=-(x-2)^2-(y-3)^2+13$, there’s no stationary point in the interior of the function’s domain. Now consider the critical points on the boundary.
First take a look at the vertices.
For $x=0$ we have to find the critical points of $-y^2+6y$, that is $y=3$, but that’s not on the boundary.
Similarly:
For $y=0$ we find $x=2$ and $f(2,0)=4$.
For $y=2$ we find $x=2$ and $f(2,2)=12$.
For $y=6-x$ we find $x=2.5$ and $y=3.5$ which is out of bounds.
Now consider the edges, ie, the points where the boundary isn’t differentiable, namely $(0,0)$, $(0,2)$, $(4,2)$ and $(6,0)$. By comparison we find the minimum in $(6,0)$, its value is $-12$.
No need for Lagrange.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the rational root theorem (which uses unique factorization
in its proof).
| Suppose $\sqrt[3]{2} = \dfrac{a}{b} \in \mathbb{Q}$ where $a, b \in \mathbb{N}_+$, $(a, b) = 1$. Now suppose $n \in \mathbb{N}_+$ satisfies$$
(\sqrt[3]{2} - 1)^n < \frac{1}{b}.$$
Note that there exist $c_0, c_1, c_2 \in \mathbb{Z}$ such that$$
(\sqrt[3]{2} - 1)^n = c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0,
$$
thus\begin{align*}
\sqrt[3]{2} &= \frac{\sqrt[3]{2} (\sqrt[3]{2} - 1)^n}{(\sqrt[3]{2} - 1)^n} = \frac{\sqrt[3]{2} (c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0)}{c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0} = \frac{c_1 (\sqrt[3]{2})^2 + c_0 \sqrt[3]{2} + 2c_2}{c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0}\\
&= \frac{c_1 \left(\dfrac{a}{b}\right)^2 + c_0 \dfrac{a}{b} + 2c_2}{c_2 \left(\dfrac{a}{b}\right)^2 + c_1 \dfrac{a}{b} + c_0} = \frac{c_1 a^2 + c_0 ab + 2c_2 b^2}{c_2 a^2 + c_1 ab + c_0 b^2},
\end{align*}
and$$
c_2 a^2 + c_1 ab + c_0 b^2 = (c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0) b^2\\
\Longrightarrow 0 < c_2 a^2 + c_1 ab + c_0 b^2 = (\sqrt[3]{2} - 1)^n b^2 < b,
$$
contradictory to the minimality of $b$. Therefore, $\sqrt[3]{2} \not\in \mathbb{Q}$.
| {
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"url": "https://math.stackexchange.com/questions/2778191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Why this value is not the second root of the equation? $\sqrt{x+3}=x-2$
Why $\frac{5}{2} - \frac{\sqrt{21}}{2}$ is not root?
There is only one restriction: $\sqrt{x+3}$, but $\frac{5}{2} - \frac{\sqrt{21}}{2} > 0$.
$x^2-4x+4=x+3$, $x^2-5x+1=0$, $D=25-4=21>0$
$D>0$, =>, $x = \frac{ 5 \pm \sqrt{21} }{ 2 }$
| There is actually a second restriction to your equation, which is "hidden":
Since $\sqrt{x+3} \geq 0$ you must also have $x-2 \geq 0$, which your root fails.
Note that the second root actually satisfies $\sqrt{x+3}=-(x-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral with Substituting $u=e^x$ I have to find the indefinite integral: $\int \frac{1}{\sqrt{1+e^x}}\cdot dx$.
I tried substituting $u=e^x$ and then $v=1+u$, and I find that $\int \frac{1}{\sqrt{v}}=2\sqrt{v}+c=2\sqrt{u+1}+c=2\sqrt{1+e^x}+c$. However, when I take the derivative of this, I get $\frac{e^x}{\sqrt{1+e^x}}$, so I know I must have made a mistake somewhere. What have I done wrong?
Thanks for your time.
| You have to also transform $dx$ when you do a substitution. For example, if you substitute $u = e^{x}$, then you have $x = \log u$ and thus $\frac{dx}{du} = \frac{1}{u}\Rightarrow dx = \frac{du}{u}$.
With this subsitution, we get
$$\int\frac{dx}{\sqrt{1+e^{x}}} = \int\frac{du}{u\sqrt{1+u}}$$
which isn't exactly all that easy to handle either.
Below I present an alternative approach via $u$-substitution.
We multiply through with $e^{-x}$, then we have:
\begin{align}
\int\frac{dx}{\sqrt{1+e^{x}}} &= \int\frac{e^{-x}}{e^{-x}\sqrt{1+e^x}}\,dx\\
&=\int\frac{e^{-x}}{\sqrt{e^{-2x}+e^{-x}}}dx
\end{align}
We can now substitute $u=e^{-x}$, then we have $\frac{du}{dx} = -e^{-x}$ and therefore $e^{-x}\,dx = -du$.
Therefore, we have
\begin{align}
\int\frac{dx}{\sqrt{1+e^{x}}} &= -\int\frac{du}{\sqrt{u^2+u}}\\
&=-\int\frac{du}{\sqrt{\left(u+\frac{1}{2}\right)^2-\frac{1}{4}}}
\end{align}
Substituting $u+\frac{1}{2} = t$, then $\frac{du}{dt} = 1 \Rightarrow du = dt$ and we have
\begin{align}
\int\frac{dx}{\sqrt{1+e^{x}}} &= -\int\frac{dt}{\sqrt{t^2-\left(\frac{1}{2}\right)^2}}\\
&=-\operatorname{arcosh}\left(2t\right)+C\\
\end{align}
Reversing the substitutions, we have the final result
$$\int\frac{dx}{\sqrt{1+e^{x}}} = -\operatorname{arcosh}(2e^{-x}+1) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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expected win for an indexed objects game We have some $n$ objects indexed with $1$ to $n$.We pick randomly objects one by one without replacement. For every drawn object we get 1$, the game ends if the drawn object has a smaller index than the previous drawn one. (ex: if obj. labeled 3 was drawn after obj. labeled 5 game ends)
Let
$
X_k=
\begin{cases}
1, & \text{if at least $k$\$ are won} \\
0, & \text{otherwise}
\end{cases}
$
Q: what the PMF of $X_k$ and what is the expected amount of $ at the end ?
$X_k$ didn't seem easy to calculate at first I took another event.
Let
$
Y_k=
\begin{cases}
1, & \text{if exactly $k$\$ are won} \\
0, & \text{otherwise}
\end{cases}
$
I took a small case where $k=2,n=4$ to see the pattern:
$P(Y_3=1)=\large\frac{{4 \choose 2}}{\frac{4!}{(4-2)!}}$ generalized it to $P(Y_k=1)=\large\frac{{n \choose k}}{\frac{n!}{(n-k)!}}=\frac{1}{k!}$
with $\large{4 \choose 2}$ I just list all pos. comb. $obj_iobj_j$ where $i<j$
given that the above is correct then $X_k=\sum^n_{i=k}{Y_i}$ but when I calculate the expected win I get $e$ which is most likely wrong.
| The objects are $1,2,\dots , n$ in the sequel.
Note that it is combinatorically not enough to count the "increasing chains of $k$ symbols", since one has then to go at least once down after the last step. There may be for instance no chance to go down...
Or a small chance.
(From the context it is not clear what happens in the case of the extraction $1<2<\dots<n$, which never goes down, i suppose the game ends as a win for $Y_n$.)
So the probability of $A_k=\{Y_k=1\}$ for a $k<n$ is exactly the one to extract a path $0<s_1<s_2<\dots<s_k$, and after this, one more $s_{k+1}<s_k$, so let us use the notation $S$ for the role of $s_k$:
$$
\Bbb P(A_k)=
\frac 1{(n-0)\cdot(n-1)\dots(n-k)}
\sum_{S\ge k}\binom {S-1}{k-1}\cdot (S-k)\ .
$$
The condition $S\ge k$ can be replaced with $S>k$.
For $n=5$ we then compute:
$$
\begin{aligned}
\Bbb P(A_1) &= \frac 1{5\cdot 4}\left[
\binom 10\cdot 1+
\binom 20\cdot 2+
\binom 30\cdot 3+
\binom 40\cdot 4
\right]=\frac 1{5\cdot 4}(1+2+3+4)=\frac 12\ ,
\\
\Bbb P(A_2) &= \frac 1{5\cdot 4\cdot 3}\left[
\binom 21\cdot 1+
\binom 31\cdot 2+
\binom 41\cdot 3
\right]=\frac 1{5\cdot 4\cdot 3}(2+6+12)
=\frac {20}{5\cdot 4\cdot 3}\ ,
\\
\Bbb P(A_3) &=
\frac 1{5\cdot 4\cdot 3\cdot 2}\left[
\binom 32\cdot 1+
\binom 42\cdot 2
\right]=\frac 1{5\cdot 4\cdot 3\cdot 2}(3+12)=\frac {15}{5\cdot 4\cdot 3\cdot 2}\ ,\\
\\
\Bbb P(A_4) &=
\frac 1{5\cdot 4\cdot 3\cdot 2\cdot 1}\left[
\binom 43\cdot 1
\right]=\frac {4}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ ,
\\
\Bbb P(A_5) &=
\frac {1}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ .
\end{aligned}
$$
This strategy to count is complicated, but i inserted it since it is relatively close to the posted strategy. One other way to count is as follows. Let us consider again a path
$$
0<s_1<s_2<\dots <\boxed{s_k}>s_{k+1}\ .
$$
then the subset $\{s_1,s_2,\dots,s_k,s_{k+1}\}$ corresponds to $k$ possible choices of a path as above, here, $s_k$ is the maximal element of the set, $s_{k+1}$ is one of the remaining $k$ elements, and the rest are determined by their order to be placed on the first places of the path. So we can write again, $k<n$:
$$
\Bbb P(A_k)=\frac 1{(n-0)(n-1)\dots(n-k)}\cdot \frac 1k\binom n{k+1}\ .
$$
| {
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"url": "https://math.stackexchange.com/questions/2779308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Please explain how Richardson Extrapolation is used in this example I am having a hard time understanding what on earth Richardson Extrapolation is trying to do.
Consider the example of approximating $\pi$ by inscribing regular polygons in a unit circle.
The perimeter of a regular $n$-gon is $2n\sin \bigl(\dfrac{\pi}{n} \bigl)$, whereas the circumference of a unit circle is $2\pi$, so that $2n\sin \bigl(\dfrac{\pi}{n} \bigl) \approx 2\pi \implies n\sin \bigl(\dfrac{\pi}{n} \bigl) \approx \pi$. So let $c_n = n\sin \bigl(\dfrac{\pi}{n} \bigl)$.
If we put $n=\dfrac 1h$, then
$$c_n=\frac 1h \sin (\pi h)=\pi - \frac{\pi^3h^2}{6}+\frac{\pi^5h^4}{120}+\cdots$$
With a bit of trigonometric manipulations, we get the recurrence relation
$$c_{2n} = 2n\sin \biggl (\frac{\pi}{2n} \biggl)=2n\sqrt{\frac 12 \biggl(1-\sqrt{1-\Bigl(\frac {c_n}{n}\Bigl)^2}\biggl)}$$
so $c_2=2$, $c_4=2.8284$, $c_8 = 3.0615$, $c_{16} = 3.1214$ and so on.
I still understand what is happening up to here, but then it suddenly says, without any explanation,
Extrapolating between $c_4$ and $c_8$, we get $c_4^{(2)}=3.1391$; and between $c_8$ and $c_{16}$, we get $c_8^{(2)}=3.1214$.
Can anyone explain to me why we get these numbers?
Apparently, there is the formula
$$T_h^ {(2)} = \frac{4T_{h/2}-T_h}{3}$$
where $T_h$ is an approximation to $T$, and $T_h^{(2)}$ is a "more accurate formula".
But plugging in the numbers doesn't seem to work:
$$c_4^{(2)}=\frac{4c_2-c_4}{3}=\frac{4(2)-(2.8284)}{3}=1.7239 \neq 3.1391$$
| Richardson extrapolation is often phrased in terms of a stepsize $h$ which is small. Here I will phrase part of the theory in terms of an integer $n$ which is large.
In your case of the cirle, you have
$$ c_n = n \sin\left(\frac{\pi}{n}\right) = \pi - \frac{1}{3!} \frac{\pi^3}{n^2} + \frac{1}{5!}\frac{\pi^5}{n^4} + O(n^{-6}).$$
It follows that
$$ \pi - c_n = \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6}).$$
This is an example of an asymptotic error expansion. By replacing $n$ by $2n$ we discover that
$$ \pi - c_{2n} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{1}{16} \cdot \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6}).$$
We would like the ability to estimate the dominant error term, i.e., the term
$$ \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2}.$$
By subtracting $\pi-c_{2n}$ from $\pi-c_n$ we find
$$ c_{2n} - c_{n} = \frac{3}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{15}{16} \cdot \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6}).$$
It follows that
$$ \frac{c_{2n} - c_n}{3} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{5}{16} \cdot \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6})$$
We can suppress some information and write this as
$$ \frac{c_{2n} - c_n}{3} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} + O(n^{-4}).$$
Observe, that the number on the left hand side is an approximation of the dominant error term. Moreover, it can be explicitly computed.
By combining past expressions we have
$$ \pi - \left( c_{2n} + \frac{c_{2n} - c_n}{3} \right) = O(n^{-4}).$$
In short, if $c_{2n}$ is a good approximation, then (presumably) we can construct a better approximation using the formula
$$ c_{2n} + \frac{c_{2n} - c_n}{3}.$$
Do not subcumb to the temptation to write or compute this as
$$ \frac{4 c_{2n} - c_n}{3}. $$
The two formulae are equivalent in exact arithmetic, but the first formula is superior when using finite precision arithmetic.
| {
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"url": "https://math.stackexchange.com/questions/2780135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find value of $\tan\big(\frac{\pi}{25}\big)\cdot \tan\big(\frac{2\pi}{25}\big)\cdots\tan\big(\frac{12\pi}{25}\big)$ Find value of:
$$\displaystyle \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg)$$
The solution I tried:
Assume
$$P = \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg),$$
with the help of $\tan(\pi-\theta)=-\tan \theta$, then
$$P=\tan\bigg(\frac{13\pi}{25}\bigg)\cdot \tan\bigg(\frac{14\pi}{25}\bigg)\cdot \tan\bigg(\frac{15\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{24\pi}{25}\bigg)$$
which gives
$$P^2=\prod^{24}_{r=1}\tan\bigg(\frac{r\pi}{25}\bigg).$$
How do I proceed from here?
| Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\tan(2n+1)x=\dfrac{\sum_{r=0}^n\binom{2n+1}{2r+1}(-1)^r\tan^{2r+1}x}{\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer
So, the roots of $t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(2n+1)(-1)^nt=0$ are $\tan\dfrac{m\pi}{(2n+1)}$ where $0\le m\le 2n$
So, the roots of $t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(2n+1)(-1)^n=0$ are $\tan\dfrac{m\pi}{(2n+1)}$ where $1\le m\le 2n$
$$\implies\prod_{m=1}^{2n}\tan\dfrac{m\pi}{(2n+1)}=(-1)^n(2n+1)$$
$$\implies\prod_{m=1}^n\tan\dfrac{m\pi}{(2n+1)}=+\sqrt{2n+1}$$
as $0<\dfrac{m\pi}{(2n+1)}<\dfrac\pi2$ as $0<m\le n$
and as $\tan(\pi-y)=-\tan y,$
$\implies\tan\dfrac{m\pi}{(2n+1)}=-\tan\left(\pi-\dfrac{m\pi}{(2n+1)}\right)=-\tan\dfrac{(2n+1-m)\pi}{2n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.