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$1/(x^6 + 1)$ partial fraction decomposition, with computer? This is a re-post from StackOverflow, I was advised to post it here.
https://stackoverflow.com/questions/64101194/partial-fraction-decomposition
How do I find the constants A,B,C,D,K,S such that
$$
\frac{1}{x^6+1} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2-\sqrt 3 x+1} + \frac{Kx+S}{x^2+\sqrt 3 x+1}
$$
is true for every real x.
I need some sympy code maybe, not sure. Or... any other Python lib which could help here.
I tried by hand but it's not easy at all... and after 1 hour of calculating, I found that I have probably made some mistake.
I tried partial fraction decomposition in SymPy but it does not go that far.
I tried Wolfram Alpha too, but it also does not decompose to that level of detail, it seems.
WA attempt
See the alternate forms which WA gives below.
EDIT:
I did a second try entirely by hand and I got these:
\begin{align}A &= 0,\\
B &= \frac13,\\
C &= -\frac1{2\sqrt3},\\
D &= \frac13,\\
K &= \frac1{2\sqrt3},\\
S &= \frac13.
\end{align}
Could someone verify if these are correct?
And in general... how can I automate this task via SymPy or WA?
| What you got is indeed correct since we have $$
\frac{1}{3}\frac{1}{x^2+1} + \frac{-\frac{x}{2\sqrt{3}}+\frac{1}{3}}{x^2-\sqrt 3 x+1} + \frac{\frac{x}{2\sqrt{3}}+\frac{1}{3}}{x^2+\sqrt 3 x+1}
$$
$$=\frac{1}{3}\frac{1}{x^2+1}+\frac{2-x^2}{3(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}+1)}$$
$$=\frac{1}{3}\frac{1}{x^2+1}+\frac{2-x^2}{3(x^4-x^2+1)}$$
$$=\frac{1}{3}\big[\frac{x^4-x^2+1+(2-x^2)(x^2+1)}{(x^2+1)(x^4-x^2+1)}\big]$$
$$=\frac{1}{3}\frac{3}{x^6+1}=\frac{1}{x^6+1}.$$
WA also agrees.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x$ In this question,
$$\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x=\lim_{x\to -\infty}\left(\sqrt{x^2+9x+1}-x\right)\cdot \frac{\sqrt{x^2+9x+1}+x}{\sqrt{x^2+9x+1}+x}=$$
$$=\lim_{x\to -\infty}\frac{9x+1}{\sqrt{x^2+9x+1}+x}=\lim_{x\to -\infty}\frac{\frac{9x+1}{\sqrt{x^2}}}{\frac{\sqrt{x^2+9x+1}}{\sqrt{x^2}}+\frac x{\sqrt{x^2}}}=\frac{-9}{\sqrt{1}-1}$$
I got an answer that is $-9/0$ where that $0$ is just indefinitely approaching $0$. Therefore, I got the answer $-\infty$. However, the correct answer is $+\infty$.
May I know what mistakes have I made?
| Answer :
$\lim_{x \to - \infty } \sqrt{x^2+9x+1 }-x$ = $\lim_{x \to - \infty } - x(\sqrt{1+\frac{9}{x}+\frac{1}{x^2 }}+1)$=$-\infty$
Because
$\lim_{x \to - \infty } \frac{9}{x}$=$0$
And
$\lim_{x \to - \infty } \frac{1}{x^2 }=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $X_n \sim \text{Beta}(n, n)$, show that $[X_n - \text{E}(X_n)]/\sqrt{\text{Var}(X_n)} \stackrel{D}{\longrightarrow} N(0,1)$ Let $X_n \sim \mathbf{B}(n,n)$ (Beta distribution), with pdf
$$
f_n(x) = \frac{1}{\text{B}(n,n)}x^{n-1}(1 - x)^{n-1},~~ x \in (0,1).
$$
Knowing that $\text{E}(X_n) = 1/2$ and that $\text{Var}(X_n) = 1/[4(2n+1)]$, prove that
$$
2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \stackrel{D}{\longrightarrow} N(0,1).
$$
I thought about doing it by the definition of convergence in distribution, but the cdf of $ 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$ is obscene. I wouldn't know how to calculate the limit $\text{lim}_{n \to \infty} F_{Y_n}(x) $ where $Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$.
Then I thought about proving convergence in probability, since converge in probability $\Rightarrow$
convergence in distribution. The problem is that it may not even converge in probability so it would be wasted work.
Edit:
I did some work and this is where I'm at:
Definition. A sequence of random variables $X_1, X_2, ...$, converges in distribution to a random variable X if
$$
\text{lim}_{n \to \infty} F_{X_n}(x) = F_X(x)
$$
So we have to prove that
$$
\text{lim}_{n \to \infty} F_{Y_n}(x) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy
$$
Where $Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) $.
Now,
$$
\begin{align}
P(Y_n \leq x) & = P(2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \leq x) \\
& = P(X_n - 1/2 \leq \frac{x}{2\sqrt{2n+1}} \\
& = P(X_n \leq \frac{x}{2\sqrt{2n+1}} + 1/2) \\
& = F_{X_n} \Bigl( \frac{x}{2\sqrt{2n+1}} + \frac{1}{2} \Bigr) \\
& = \frac{1}{B(n,n)}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt
\end{align}
$$
We use Stirling's approximation to $\text{B}(n,n)$:
$$
B(a, b) \approx \sqrt{2\pi} \frac{a^{a - 1/2}b^{b - 1/2}}{(a + b)^{a + b - 1/2}}
$$
So $\text{B}(n, n) \approx \frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}} $, after simplification.
Substituting the Stirling approximation (we do this because it converges asymptotically and we're taking the limit), we obtain
$$
\frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt.
$$
So what's left to do is prove that
$$
\text{lim}_{n \to \infty} \frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy.
$$
Edit 2: I asked my professor for guidance on how to finish the last step. All he said was "apply the limit theorem to solve directly".
| The answer is in portuguese because I'm a native portuguese speaker.
O último cálculo na demonstração acima é um problema computacional excessivamente difícil. Aqui a ideia é apresentarmos uma demonstração alternativa, que se dá pelos seguintes passos:\
1º: Mostramos que a densidade de $ Y_n $ converge para a densidade de $ Z $, onde $ Z \sim N(0, 1). $\
2º: Invocamos o \textit{Lema de Scheffé} para terminar a demonstração. O Lema de Scheffé é um resultado em Teoria da Medida que, no nosso caso, implica que se $ f_{Y_n}(x) \longrightarrow f_Z(x) $, então $ F_{Y_n}(x) \longrightarrow F_Z(x) $, provando a definição Convergência em Distribuição. Em resumo, temos um trabalho facilitado por causa de um resultado mais forte e sofisticado.\
Muito bem, ao diferenciar as equações (1) e (4), obtemos
$$
f_{Y_n}(x) = f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}}.
$$
Agora temos que demonstrar que
$$
\text{lim}_{n \to \infty} f_{Y_n}(x) = \text{lim}_{n \to \infty} f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.
$$
De fato,
\begin{align*}
f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} & = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)} (\frac{x}{2\sqrt{2n+1}} + \frac{1}{2})^{n-1} (\frac{1}{2} - \frac{x}{2\sqrt{2n+1}})^{n-1} \\
& = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)}(\frac{1}{4} - \frac{x^2}{4(2n+1)})^{n-1} \\
& = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4} - \frac{x^2}{4(2n+1)})^{n-1} \\
& = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4})^{n-1}(1 - \frac{x^2}{2n-1})^{n-1} \\
& = \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1}.
\end{align*}\
Aplicando o limite $\text{lim}_{n \to \infty}$ na úlima expressão acima, obtemos
\begin{align*}
\text{lim}_{n \to \infty} \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1} &= \frac{1}{\sqrt{\pi}} \frac{1}{\sqrt{2}} \text{lim}_{n \to \infty} (1 + \frac{(-x^2/2)}{n - \frac{1}{2}})^{n-1} \\
&= \frac{1}{\sqrt{2\pi}}e^{-x/2}.
\end{align*}
Mostramos que $ \text{lim}_{n \to \infty} f_{Y_n}(x) = f_Z(x) $. Agora, pelo \textit{Lema de Schéffe}, temos que $ \text{lim}_{n \to \infty} F_{Y_n}(x) = F_Z(x) $, o que prova a convergência em distribuição desejada e termina a demonstração.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
| For $x\geq1$ we obtain:
$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>0.$$
For $0<x<1$ we have:
$$x^{12}-x^9+x^4-x+1=(1-x)+x^4(1-x^5)+x^{12}>0.$$
For $x\leq0$ it's obvious that $$x^{12}-x^9+x^4-x+1>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
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Series of rational fractions I read in Gradshteyn I , Ryzhik I Table Of Integrals, Series And Products (7Ed , Elsevier, 2007)(Isbn 0123736374)(1220S)the following series
$$-\frac{x}{x-1}=\sum _{k=0}^{\infty } -\frac{x^{2^k}}{x^{2^{k+1}}-1}=\sum _{k=1}^{\infty } \frac{2^k x^{2^k-1}}{x^{2^k-1}+1}$$
the serie $$\frac{2^k x^{2^k}}{x^{2^k}+1}$$ it is easy calculating using Van Wijngaarden's trick positive series in alternate serie but when come from the second series??
| We show the following is valid for |x|<1:
\begin{align*}
\frac{x}{1-x}&=\sum_{k=0}^\infty\frac{2^kx^{2^{k}}}{1+x^{2^{k}}}\tag{1}\\
&=\sum_{k=0}^{\infty}\frac{x^{2^{k}}}{1-x^{2^{k+1}}}\tag{2}\\
\end{align*}
Since we have
\begin{align*}
\frac{2^{k+1}x^{2^{k+1}}}{1-x^{2^{k+1}}}=\frac{2^kx^{2^k}}{1-x^{2^k}}-\frac{2^kx^{2^k}}{1+x^{2^k}}
\end{align*}
we obtain by telescoping
\begin{align*}
\color{blue}{\sum_{k=0}^\infty}&\color{blue}{\frac{2^kx^{2^k}}{1+x^{2^k}}}\\
&=\lim_{N\to\infty}\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1+x^{2^{k}}}\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}-\sum_{k=0}^N\frac{2^{k+1}x^{2^{k+1}}}{1-x^{2^{k+1}}}\right)\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}-\sum_{k=1}^{N+1}\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}\right)\\
&=\lim_{N\to\infty}\left(\frac{x}{1-x}-\frac{2^{N+1}x^{2^{N+1}}}{1-x^{2^{N+1}}}\right)\tag{$2^{N+1}x^{2^{N+1}}\to0$}\\
&\,\,\color{blue}{=\frac{x}{1-x}}
\end{align*}
and the claim (1) follows.
Since we have
\begin{align*}
\frac{x^{2^{k}}}{1-x^{2^{k+1}}}&=\frac{\left(1+x^{2^{k}}\right)-1}{1-x^{2^{k+1}}}=\frac{1}{1-x^{2^{k}}}-\frac{1}{1-x^{2^{k+1}}}
\end{align*}
we obtain again by telescoping
\begin{align*}
\color{blue}{\sum_{k=0}^\infty}&\color{blue}{\frac{x^{2^k}}{1+x^{2^{k+1}}}}\\
&=\lim_{N\to\infty}\sum_{k=0}^N\frac{x^{2^{k}}}{1+x^{2^{k+1}}}\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{1}{1-x^{2^{k}}}-\sum_{k=0}^N\frac{1}{1-x^{2^{k+1}}}\right)\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{1}{1-x^{2^{k}}}-\sum_{k=1}^{N+1}\frac{1}{1-x^{2^{k}}}\right)\\
&=\lim_{N\to\infty}\left(\frac{1}{1-x}-\frac{1}{1-x^{2^{N+1}}}\right)\tag{$x^{2^{N+1}}\to0$}\\
&=\frac{1}{1-x}-1\\
&\,\,\color{blue}{=\frac{x}{1-x}}
\end{align*}
and the claim (2) follows.
Note: We find in edition 07 of Table of integrals, series, and products by I.S. Gradshteyn and I.M. Ryzhik formula 1.121(1.):
\begin{align*}
\frac{x}{1-x}=\sum_{k=1}^\infty\frac{2^{k-1}x^{2^{k-1}}}{1+x^{2^{k-1}}}=\sum_{k=1}^\infty\frac{x^{2^{k-1}}}{1-x^{2^{k}}}\qquad\qquad[x^2<1]
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?
In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both binary and decimal?
I feel like there probably isn't, but I can't think of a simple counterargument.
(A computer search by one of the commenters suggests there is no such integer up to $10^{100000}$.)
| For a prime $p$, let $\nu_p(n)$ denote the exponent of $p$ in the prime factorization of $n$. We will use the Lifting the Exponent lemma.
Firstly, note that
$$a=\nu_2(2^a+2^b)=\nu_2(10^c+10^d)=c,$$
so we have $2^a+2^b=10^a+10^d$. Let $b=a+x$ and $d=a+y$, and write
$$1+2^x=5^a(1+10^y).\tag{1}\label{eq1}$$
We observe that
$$2^b<2^a+2^b<2\cdot 10^d\implies a+x< 1+(a+y)\log_2(10)<1+\frac 72a+\frac 72y,$$
so $x<1+\frac 52a+\frac 72y$, and
$$2^a+2^{a+x}=10^a+10^{a+y}\implies 2^{a+x}>10^{a+y}\implies a+x>a+y\implies x>y.$$
We see that, from (*),
$$5^a|1+2^x|2^{4x}-1\implies a\leq \nu_5(2^{4x}-1)=1+\nu_5(x).$$
On the other hand, looking at powers of $2$, we see that, modding out by $2^y$,
$$1\equiv 5^a\bmod 2^y,$$
which implies that
$$y\leq \nu_2(5^a-1)=1+\nu_2(a).$$
We now have the inequalities
$$y\leq 1+\nu_2(a),\ \ a\leq 1+\nu_5(x),\ \ x<1+\frac 52a+\frac 72y.$$
Since the functions on the right sides of the first two inequalities are very small, this is basically enough to finish -- we just need to actually carry out the bounding. We note that
$$\nu_p(x)\leq \log_p(x)=1+\log_p\left(\frac xp\right)<1+\frac{x}{p}.$$
As a result,
$$y<2+\frac{a}{2},\ a<2+\frac{x}{5},\ x<1+\frac 52a+\frac 72y.$$
This gives
$$a<2+\frac{2+5a+7y}{10}<2+\frac{2+5a}{10}+\frac{7}{10}\left(2+\frac a2\right)=\frac{18}5+\frac{17a}{20},$$
which gives that $a<24$. In addition, we have $y<2+a/2=14$ and $x<110$, so this is just a finite case check (which it seems like one of the commenters has performed to a satisfactory bound).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the zeros of $f(x)=x^3−4x^2+x−4$ I am to find the zeros and multiplicities of $f(x)=x^3−4x^2+x−4$.
The solution provided in the answers section of my book is 4 with multiplicity 1. I arrived at $2\pm\sqrt(8)$.
My working:
$$x^3-4x^2+x-4$$
$$x(x^2-4x+1)-4$$
Then, focusing on the quadratic in the middle, I used 3BlueOneBrownVideo (start from ~23 minutes) to find the zeros:
$$m = \frac{-b}{2}=\frac{4}{2}=2$$
$$d^2=m^2-p=2^2+4=8$$
$$r,s=m\pm\sqrt{d^2}$$
$$r,s=2\pm\sqrt{8}$$
The zeros I arrive at are therefore $2\pm\sqrt{8}$
How can I arrive at 4 per the solution? Granular baby steps much appreciated.
| I think the following at least is right: $$x^3-4x^2+x-4=x^2(x-4)+1(x-4)=(x-4)(x^2+1),$$ which gives $$\{4,\pm i\}$$
| {
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Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule?
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}\right)=-\frac{2y}{\left(x+y^2\right)^2}$$
Then I multiply each partial derivative with its respective equal and add it together:
$$(fx)(32r\cos(\theta)) + (fy)(3r(\sin(\theta))$$
Then I sub in X and Y in the partial derivatives. And then plug in $r$ and $\theta$:
$$-\left(\frac{1}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi }{4}\right)\right)^2\right)^2}\right)\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)\right)+-\left(\frac{2\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:\:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)^2\right)^2}\right)\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)$$
And I show my answer in the screenshot above. But for some reason the software is telling me it's wrong. I've also tried not multiplying the partial derivatives by the $32rcos(\theta)$ thing but it's still giving me the wrong answer.
What am I doing wrong? Thank you.
| It is easier if you substitute $x$ and $y$ directly $$g(r,\theta)=\frac{1}{x+y^2}=\frac{1}{32r\cos(\theta)+9r^2\sin^2(\theta)}$$
So $$\frac{\partial g}{\partial \theta}=-\frac{-32r\sin(\theta)+18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$
Or $$\frac{\partial g}{\partial \theta}=\frac{\partial g }{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g }{\partial y}\frac{\partial y}{\partial \theta}$$
$$=-\frac{1}{(x+y^2)^2}(-32r\sin(\theta)+2y(3r\cos(\theta)))$$
$$=\frac{32r\sin(\theta)-18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$
Then evaluating at $(r,\theta)=(2\sqrt{2},\frac{\pi}{4})$ you should get $-\frac{1}{1250}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A cyclic inequality of degree 10 Suppose that $x,y,z\geq 0$. I would like to prove that
$$(x^5+y^5+z^5)^2\geq (x+y+z)(x^3y^6+y^3z^6+z^3x^6).$$
I can prove this inequality using some standard methods. For example, I can let $x=1, y=1+p$, and $z=1+p+q$. Then Mathematica can simplify the difference of the two sides of the inequality to show that all of the coefficients are positive. I don't like this solution, since it relies heavily on a complicated algebraic simplification. Do you have a better proof?
| Another way:
After homogenization we need to prove that:
$$\sum_{cyc}(x^{10}+2x^5y^5-x^7z^3-x^6z^4-x^6z^3y)\geq0$$ or
$$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)\geq0,$$ which is true by AM-GM and SOS:
$$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)=$$
$$=\frac{1}{37}\sum_{cyc}(30x^7z^3+4y^7x^3+3z^7y^3-37x^6z^3y)+\sum_{cyc}(x^{10}+2x^5y^5-x^4y^6-2x^3y^7)\geq$$
$$\geq\sum_{cyc}\left(\sqrt[37]{\left(x^7z^3\right)^{30}\left(y^7x^3\right)^4\left(z^7y^3\right)^3}-x^6z^3y\right)+$$
$$=\sum_{cyc}x^3(x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+3xy^5+2y^6)=$$
$$=\sum_{cyc}\left(x^3(x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+3xy^5+2y^6)-(x^{10}-y^{10})\right)=$$
$$=\sum_{cyc}y^5(x-y)^2(2x^3+3x^2y+2xy^2+y^3)\geq0.$$
The following inequality is also true.
Let $x$, $y$ and $z$ be non-negative numbers. Prove that:
$$(x^5+y^5+z^5)^2\geq(x^2+y^2+z^2)(x^7y+y^7z+z^7x).$$
| {
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"url": "https://math.stackexchange.com/questions/3851194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answer one question.
PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough
Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.
| I do not think it would take a long time to simplify these expression by hand. Rewrite it as
$$\frac {5}{2}\cdot\frac {10}{3}\cdot\frac {17}{4}\cdot\frac {26}{5}\text{ vs }\frac {11}{5}\cdot \frac {13}{4}\cdot\frac {13}{3}\cdot \frac {11}{2}$$
Denominators go away, and the factor $13$ in the numerator as well:
$$1700\text{ vs }13\cdot 121,$$
and the latter seems to be easy to estimate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does recurrence $a_n=\frac12(a_{n-1}+\frac\beta{a_{n-1}})$ become $b_n=\frac12\frac{b_{n-1}^2+\beta-\alpha^2}{b_{n-1}+\alpha}$ when $b_n=a_n-a$?
$a_n = \frac{1}{2}(a_{n-1} + \frac{\beta}{a_{n-1}})$ for $ n > 0$ with $a_0 = 1$. Changing variables in this recurrence, and letting $b_n = a_n - a$, we find by simple algebra that $b_n = \frac{1}{2}\frac{b_{n-1} ^2 + \beta - \alpha^2}{b_{n-1} + \alpha}$.
I've tried making the substitutions and doing algebra but can't come up with it. Would somebody please help show how to get from the expression for $a_n$ to the one for $b_n$ using simple algebra?
| $b_n = a_n - a$, $b_{n-1} = a_{n-1} -a$. Hence
$$\begin{align}b_n = a_n - a &= \frac12(a_{n-1} + \frac \beta{a_{n-1}})-a
\\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a})-a
\\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a}-2a)
\\&=\frac12(b_{n-1}-a + \frac \beta{b_{n-1}+a})
\\&=\frac1{2(b_{n-1}+a)}((b_{n-1}+a)(b_{n-1}-a)+\beta)
\\&=\frac{b_{n-1}^2-a^2+\beta}{2(b_{n-1}+a)}
\\&\left(=\frac12\frac{b_{n-1}^2+\beta-a^2}{b_{n-1}+a}\right)
\end{align}$$
| {
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y - \frac{2}{y} \implies x = \frac{y^{2}-2}{y}$$
Rearrange, $xy=y^{2}-2 \implies y^{2}-xy-2 = 0$
Then, I tried to use the equation $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, and I got:
$y = \frac{x \pm \sqrt{x^{2}+8}}{2}=f^{-1}$
Next, I tried to find $(f^{-1}){'}$, for both $+$ and $-$
$\frac{1}{2} \cdot (x' + ((x^{2}+8)^{\frac{1}{2}})')$
Apply chain rule:
$((x^{2}+8)^{\frac{1}{2}})'$ = $\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x$
Then do the same for -
And I got $(f^{-1})'$ = $\begin{cases} \frac{1}{2}\cdot(1+\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \\ \frac{1}{2}\cdot(1-\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \end{cases}$
When a = 1,
$(f^{-1})'=\begin{cases} \frac{\sqrt{9}+1}{2\sqrt{9}} = \frac{2}{3} \\ \frac{\sqrt{9}-1}{2\sqrt{9}} = \frac{1}{3} \end{cases}$
I need some help. It says when $x < 0$, but both of my answers are positive. How do I interpret it?
| So the issue is you have not used $x<0$. By the way be careful when you use a variable like x. in one part you have used it for the domain and in one for the range set.
By using the quadratic formula you have obtained two solutions, so your question is which is the right one at $a=1$. When $a=1$
$$ 1= x-\frac{2}{x} $$
Hence,
$$x^2-x-2=0 $$
using the quadratic formula you will get two solution +1, -2, so you have only use the negative one. And hence your answer will be 1/3.
Another interesting way to solve this is, as you $f(-2)=1=a$, differentiate $f^{-1}(f(x))=x$ at x=-2 using the chain rule and you should get your required.
Hope this helped.
| {
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"source": "stackexchange",
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Intersection of median and segment between two sides of a triangle
In triangle $ABC,$ $M$ is the midpoint of $\overline{BC},$ $AB=12,$ and $AC=16.$ Points $E$ and $F$ are taken on $\overline{AC}$ and $\overline{AB},$ respectively, and $\overline{EF}$ and $\overline{AM}$ intersect at $G.$ If $AE=2AF,$ then what is $EG/GF?$
This seemingly-easy problem (at least for my standards) is driving me crazy. I tried an analytical approach:
WLOG, assume that $\triangle{ABC}$ is right. (There is no specific angle measures.) Fix the points on the cartesian plane such that $A = (0, 0), B = (0, 12), \text{ and } C = (0, 16).$ Then $M$ is at $(8, 6)$, and the equation of line $AM$ is $y=\frac{3}{4}x.$ Next, let $E = (4, 0) \text{ and } F = (0, 8).$ The equation of line $EF$ is $y=-2x+8.$ Therefore, we have the system of equations $$y=\frac{3}{4}x$$$$y=-2x+8$$
Solving gets $$x=\frac{32}{11} \text{ and } y=\frac{24}{11}.$$
Therefore, the ratio of $EG$ to $GF$ is just $\frac{\frac{32}{11}}{4-\frac{32}{11}} = \frac{8}{3}.$
However, my approach is incorrect. Can anyone point out any flaws and present a solution to the correct answer? I also tried using mass points to no avail.
TIA!
| This is an application to the law of sines.
In $\triangle ABC$:
$$\frac{\sin\angle ABC}{AC}=\frac{\sin\angle ACB}{AB}$$
In $\triangle ABM$:
$$\frac{\sin\angle BAM}{BM}=\frac{\sin\angle ABC}{AM}$$
In $\triangle CAM$:
$$\frac{\sin\angle CAM}{CM}=\frac{\sin\angle ACB}{AM}$$
In $\triangle FAG$:
$$\frac{\sin\angle BAM}{FG}=\frac{\sin\angle AGF}{AF}$$
In $\triangle EAG$:
$$\frac{\sin\angle CAM}{GE}=\frac{\sin\angle AGE}{AE}$$
Since $\angle AGF+\angle AGE=180^\circ$, you have $$\sin\angle AGF=\sin\angle AGE$$
Putting everything together, and if I did not do any mistakes, I get $$\frac{AC}{AB}=\frac{\frac{FG}{AF}}{\frac{GE}{AE}}$$
As for your error, $AE=4$, $AF=8$, so you have $AE=AF/2\ne 2AF$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| $$\eqalign{1 - 2 \sin^2 x \cos^2 x &= 1 - (1-\cos^2 x) \cos^2 x - \sin^2 x (1 - \sin^2 x)\cr
&= 1 - \cos^2 x + \cos^4 x - \sin^2 x + \sin^4 x\cr
&= \cos^4 x + \sin^4 x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}? How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}?
I have drawn a diagram with a circle centre $(4+3i)$ and radius = $2$. I think to find the min/max values of $Arg(z)$ I need to find the points of tangency? I don't understand how to do it only with the diagram and any hints would be much appreciated!
The min possible value of $|z|$ is $3$, and max value of $|z|$ is $7$.
Also, I am on a chapter has more simple questions like this, so I'm looking for a not too complicated way of solving this :)
| Let $y=mx$ be the equation of a line tangent to the circle $(x-4)^2 + (y-3)^2 =4$. Then substituting the first into the second, we get
$$
\begin{align}
(x-4)^2 + (mx-3)^2 &= 4 \\
(m^2 + 1) \, x^2 - (6m+8) \, x +21 &= 0
\end{align}
$$
Since the line is tangent to the circle, the discriminant of the above must be zero.
$$
\begin{align}
(6m+8)^2 - 4 (m^2 + 1)(21) &= 0 \\
12m^2 - 24m + 5 &= 0 \\
m &= \frac{6 \pm \sqrt{21}}{6}
\end{align}
$$
Finally, we have $\tan \theta = m$, so
$$\theta_1 = \tan^{-1} \left( \frac{6 - \sqrt{21}}{6} \right) = 0.232$$
and
$$\theta_2 = \tan^{-1} \left( \frac{6 + \sqrt{21}}{6} \right) = 1.055$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
| Answer :
$\frac{x^3-4x^2-4x+16}{\sqrt{x^2 - 5x+4}}= \frac{x(x^2 - 4)-4(x^2 - 4)}{\sqrt{x^2 - 5x+4}}$=$\frac{(x^2 - 4)(x-4)}{\sqrt{x^2 - 5x+4}}$
$\sqrt{x^2 - 5x+4} = 0 $ if $ x =(1, 4) $
Suppose $x≠(1,4)$
$(x^2 - 4)(x-4)$ =0
$\Rightarrow$ $ (x - 2)(x+2)(x-4)=0$
$\Rightarrow $ the solution is ($x=2$ or $ x=- 2)$
Because if $x =4 $the denominator equal $0$
So $x=4$ not a solution
Finally :
$S=(2,-2) $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Stuck on Mathematical Induction Proof I have the following question: Prove with mathematical induction that $3^n+4^n\le 5^n$ for all $n\ge 2$.
$$\text{Assume true: }3^n+4^n \le 5^n \text{. Prove that $3^{n+1}+4^{n+1} \le 5^{n+1}$} \\
= 3\cdot3^n+4\cdot4^n \\ =3\cdot3^n+4^n(3+1)\\
=3\cdot3^n+3\cdot4^n+4^n \\
=3(3^n+4^n)+4^n \\
\le 3(5^n)+4^n \\
(5-2)(5^n)+4^n \\
5^{n+1} - 2\cdot5^n+4^n\\
5^{n+1} - 2\cdot(3^n+4^n) +4^n$$
I am stuck on. I see now that I double the $5^n$, and this leads me nowhere. Where can I go from here to solve this? I have tried playing with the algebra, but I am not getting anywhere with this last step.
| How about this, with the same first step:
$$\begin{align*}
3^{n+1} + 4^{n+1} &\le 3 \cdot 3^n + 4 \cdot 4^n \\
&\le 5 \cdot 3^n + 5 \cdot 4^n \\
&\le 5(3^n + 4^n) \\
&\le 5 \cdot 5^n \\ &= 5^{n+1}
\end{align*}
$$
| {
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"source": "stackexchange",
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Let $f(x)$ be a polynomial satisfying $\lim_{x\to \infty} \frac {x^4 f(x)}{x^8+1} =3$, $f(2)=5$, $f(3)=10$,$f(-1)=2$,$f(-6)=37$. Find $f(0)$ It’s clear that $f(x)$ is a 4th degree polynomial.
If $f(x)=ax^4+bx^3+cx^2+dx+e$, then $a=3$
From the rest of the given data, I can form four linear equations, which should give me the value of $a,b,c,d$, but that’s far too tedious and time consuming, and I don’t think the question is meant to be solved that way. Is there any alternative method?
| It is clear that $f(x)$ is of deegree 4 and
$$\begin{align} f(2) &= 1 + 2^2
\\ f(-1) &= 1 + (-1)^2
\\ f(3) &= 1+ 3 ^ 2
\\ f(-6) &= 1+ (-6)^2 \end{align} $$
so $f(x) - (1+x^2)$ has roots $2,3,-1,-6$
$$ \Rightarrow f(x) - (1+ x^2) = A(x+1)(x-2)(x+6)(x-3) $$
now you can continue
| {
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Prove by deduction $\displaystyle \frac{a^2 + b^2}{2} \geq ab$ I am currently working in the following question:
Prove by deduction $$\frac{a^2 + b^2}{2} \geq ab$$ for all $a,b \in \mathbb{R}$
My question is: can prove instead the following?
$$\frac{a^2 + b^2}{2} - ab \geq 0$$
My solution:
$$\frac{a^2 + b^2}{2} - ab = \frac{a^2 + b^2 -2ab}{2} = \frac{(a-b)^2}{2} \geq 0$$
| You are correct since $$\frac{a^2+b^2}{2}-ab\geq 0\iff\frac{a^2+b^2}{2}\geq ab.$$
Alternatively since $(a-b)^2\geq0$ for all $a,b\in\mathbb R$ we have $$a^2+b^2\geq2ab$$ or $$\frac{a^2+b^2}{2}\geq ab$$
as required.
| {
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.
| Rather than carry the notation we write $s=\sin(x)$ and $c=\cos(x)$. Subtracting the equation from $s^2+c^2=1$ yields $s^3-s^2 + c^3-c^2=0$ or $s(s^2-1)+c(c^2-1) = s(-c^2) + c(-s^2) = 0$ and so either $s=0$ or $c=0$.
| {
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Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find the sum of the series - however, I have to find the limit. It resembles for me:
$$\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n^2-kn }{(k+n)^3-2kn^2-2k^2n}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)^3-2kn(n+k)}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)((k+n)^2-2kn)}$$
but I don't know what to do next and how to solve it. I would appreciate your help.
Edit: Does it has something in common with Riemann sum?
| First, we can omit the $\sqrt{n}$ part and only sum up to $n$. To see this,
$$
\sum_{k=n}^{n+\sqrt{n}} x_n \leq
\sum_{k=n}^{n+\sqrt{n}} \frac{4(n+\sqrt{n})^2}{4n^3} \leq \sum_{k=n}^{n+\sqrt{n}} \frac{(n+n)^2}{n^3} \leq
\sum_{k=n}^{n+\sqrt{n}} \frac{4}{n} \leq 4\frac{\sqrt{n}}{n} \mapsto 0
$$
The rest is the Riemannian sum of the integral
$$
\int_0^1 \frac{x^2 + x + 2}{x^3 + x^2 + x + 1} \mathrm{d}x = \int_0^1 \frac{1}{1+x} \mathrm{d}x + \int_0^1 \frac{1}{1+x^2}\mathrm{d}x.
$$
This can be easily computed using calculus.
| {
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For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}.]$(manually) Here is the question I want to answer letter $(e)$ of it manually:
Let $B = \mathbb{Q}[x_{1}, ... , x_{n}] \cong \mathbb{Q}^{[n]}$ and $A = \mathbb{Q}[e_{1}, ... , e_{n}]$ where $e_{i} \in B$ is the elementary symmetric polynomial of degree $i.$ The discriminant is the polynomial $\Delta^2,$ where $\Delta \in B$ is defined by:\
$$\Delta = \prod_{i > j}(x_i -x_j)$$
$(a)$ (Vandermonde determinant) Show that: $$ \Delta = \det
\begin{pmatrix}
1 & x_{1} & x_{1}^2 & \dots & x_{1}^{n-1}\\
1 & x_{2} & x_{2}^2 & \dots & x_{2}^{n-1}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x_{n} & x_{n}^2 & \dots & x_{n}^{n-1}
\end{pmatrix}$$
$(b)$ Show that $\Delta^2 \in B^{S_{n}}.$ Deduce that $\Delta^2 \in A.$
$(c)$ Show that $\Delta \notin B^{S_{n}}.$ Determine which subgroup of $S_{n}$ fixes $\Delta.$
$(d)$ For $n =2,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}].$
$(e)$ For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}].$
Here is the solution using Mathematica:
Here is the hint given to me to solve it manually:
Still I can not see how this hint will make the solution easier manually, could anyone help me in that please?
| What a strange hint. There's a much better way to do it, and I have no idea why the author didn't set $a_3 = 1$ from the start.
I learned this argument from David Speyer. Computing the discriminant as a polynomial in the $e_i$ is equivalent to computing the discriminant of the polynomial $f(x) = x^3 - e_1 x^2 + e_2 x - e_3$. Up to a translation, which does not affect the discriminant, we can remove the quadratic term from this polynomial: if we substitute $x = y + \frac{e_1}{3}$, then we get a new polynomial (brace yourself)
$$\begin{align} g(y) &= \left( y^3 + e_1 y^2 + \frac{e_1^2}{3} y + \frac{e_1^3}{27} \right) - e_1 \left( y^2 + \frac{2e_1}{3} y + \frac{e_1^2}{9} \right) + e_2 \left( y + \frac{e_1}{3} \right) - e_3 \\
&= y^3 + \left( - \frac{e_1^2}{3} + e_2 \right) y + \left( -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3 \right). \end{align}$$
Set $p = - \frac{e_1^2}{3} + e_2, q = -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3$, so that this new polynomial can be written $g(y) = y^3 + py + q$. Now it suffices to compute the discriminant of this new polynomial as a function of $p$ and $q$, then plug in these expressions for $p$ and $q$ in terms of the $e_i$.
The point of doing this is that we've now removed most of the terms from the discriminant: $p$ has degree $2$, $q$ has degree $3$, and $\Delta^2$ has degree $6$, which implies that it must be some linear combination $ap^3 + bq^2$ of $p^3$ and $q^2$. So we only have two coefficients to compute. We can actually compute these coefficients by computing the discriminant of two specific polynomials:
*
*Set $p = -1, q = 0$. Then $g(y) = y^3 - y = y(y + 1)(y - 1)$ has roots $0, \pm 1$, so its discriminant is $\left( (0 - 1)(1 - (-1))( (-1) - 0) \right)^2 = 4$, which gives $a = -4$.
*Set $p = 0, q = -1$. Then $g(y) = y^3 - 1 = (y - 1)(y - \omega)(y - \omega^2)$ has roots $1, \omega, \omega^2$ where $\omega$ is a primitive third root of unity (satisfying $1 + \omega + \omega^2 = 0$), so its discriminant is $\left( (1 - \omega)(\omega - \omega^2)(\omega^2 - 1) \right)^2 = (1 - \omega)^6 = (- 3 \omega)^3 = -27$, which gives $b = -27$.
So we get $\boxed{ \Delta^2 = -4p^3 - 27q^2 }$ (already a useful formula in practice), and substituting $e_1, e_2, e_3$ in gives the full cubic discriminant (of more dubious value; I have never used it).
| {
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Combinatorics / Inclusion-Exclusion Principle I have this problem:
A bakery sells $4$ varieties of donuts but there are only $6$ chocolate and $7$ glazed.
How many ways can we buy $12$ donuts if the selection must include at least $3$ glazed donuts (order is irrelevant)?
In my attempt, I first subtracted the $3$ glazed we have to include then used stars and bars to find the number of ways assuming unlimited amounts of everything: $12-3 = 9$ spots to fill, with $4$ varieties $= {12\choose9} = 220$ ways.
From here I was thinking we can subtract the number of ways that include more than the amounts we have, e.g., the ways that have $7$, $8$, or $9$ chocolate donuts and the ways that have $5$, $6$, $7$, $8$, or $9$ glazed donuts.
However, I'm not sure if this is the best way to approach this? I don't really understand where the inclusion-exclusion principle would be involved.
| The question implies that there are at least $12$ donuts of each of the other varieties available.
If we select three glazed donuts to begin with, then we must select nine more donuts from the six chocolate, four glazed, and unlimited (for our purposes) number of the other two varieties available. The number of ways we can do this is the number of solutions of the equation
$$a + b + c + g = 9 \tag{1}$$
in the nonnegative integers subject to the restrictions that $c \leq 6$ and $g \leq 4$.
A particular solution of equation 1 corresponds to the placement of $4 - 1 = 3$ addition signs in a row of nine ones. For instance,
$$1 1 1 + 1 1 1 1 + 1 1 +$$
corresponds to the solution $a = 3$, $b = 4$, $c = 2$, $g = 0$. The number of such solutions is
$$\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3}$$
since we must select which $4 - 1 = 3$ of the $9 + 4 - 1 = 12$ positions required for $9$ ones and $4 - 1 = 3$ addition signs will be filled with addition signs, as you found.
From these, we must subtract the number of solutions in which $c > 6$ or $g > 4$. Notice that these restrictions cannot be violated simultaneously since $7 + 5 = 12 > 9$.
Suppose the restriction that $c \leq 6$ is violated. Then at least seven chocolate donuts must be placed in the box. Let $c' = c - 7$. Then $c'$ is a nonnegative integer. Substituting $c' + 7$ for $c$ in equation 1 yields
\begin{align*}
a + b + c' + 7 + g & = 9\\
a + b + c' + g & = 2 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{2 + 4 - 1}{4 - 1} = \binom{5}{3}$$
solutions.
Suppose the restriction that $g \leq 4$ is violated. Then at least five additional glazed donuts must be placed in the box. Let $g' = g - 5$. Then $g'$ is a nonnegative integer. Substituting $g' + 5$ for $g$ in equation 1 yields
\begin{align*}
a + b + c + g' + 5 & = 9\\
a + b + c + g' & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{4 + 4 - 1}{4 - 1} = \binom{7}{3}$$
solutions.
Hence, the number of ways to select $12$ donuts, of which at least three are glazed, from the four varieties given the restrictions that only six chocolate and seven glazed donuts are available is
$$\binom{12}{3} - \binom{5}{3} - \binom{7}{3}$$
| {
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If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I concluded that $n^2+1=2a^2$ and $n+1=2b^2$ for some $a,b\in\mathbb{N}$ and it is here where I need some help.
Please only provide hints.
| I. If $1+n+n^2+n^3=z^2$, and odd $n=2y+1$, then$$1+n+n^2+n^3=8y^3+16y^2+12y+4=z^2$$
Since $4$ divides $z^2$, then if $z^2=4w^2$ $$2y^3+4y^2+3y+1=w^2$$ and factoring $$(y+1)(2y^2+2y+1)=w^2$$
Now $y+1$ and $2y^2+2y+1$ are relatively prime, since if integer $k$ divides $y+1$, then $k$ divides $2y(y+1)=2y^2+2y$, and hence $k$ does not divide $2y^2+2y+1$. If their product is square, then both $y+1$ and $2y^2+2y+1$ are squares.
II. Let $2y^2+2y+1$, which is odd, $=(2t+1)^2=4t^2+4t+1$. Then$$y^2+y=2t^2+2t$$and$$\frac{y(y+1)}{2}=\frac{2t(t+1)}{2}$$Hence the LHS is a triangle number which is double a triangle number, and it lies in the sequence $$6, 210, 7140, 242556, 8239770, 279909630, 9508687656, ...$$and $y$ belongs in the sequence $$3, 20, 119, 696, 4059, 23660, 137903,...$$III. With a little more argument, drawing upon the sequence of solutions to the Pell equation$$b^2=2a^2 {^+_-} 1$$ it is possible to rule out all members of the $y$-sequence except $y=3$, and so prove that $n=7$.
I don't know if this approach is better than one by way of Pythagorean triples, but identifying $y$ as a factor in this relatively small sub-set of the triangle numbers seems a good way to go.
| {
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How to prove the equation $\int_{a}^{b} f(x) \mathrm{d} x = \frac{b-a}{2}[f(a)+f(b)]-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta)$? In numerical analysis, we know that the following equation is true:
$$\int_{a}^{b} f(x) \mathrm{d} x = \frac{b-a}{2}[f(a)+f(b)]-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta), \quad \eta \in(a, b)$$
I wonder if there is any profound and concise way to prove this conclusion. It is better to use Taylor series to prove this conclusion.
| The link Trapezoid rule shows the error term is bounded but is a little unsatisfactory since it does not prove equality in this particular case.
Simplifying to the interval [0,1], If we assume $f$ is twice continuously differentiable on $[0,1]$ then, using integration by parts we can proceed thus,
\begin{aligned}
\int_0^1 f(x)~dx &= \Big[ \big(x-\tfrac{1}{2}\big) f(x) \Big]_0^1 - \int_0^1\big(x-\tfrac{1}{2}\big) f'(x) ~ dx \\
&=\frac{f(1)+f(0)}{2}-\Big[\big(\tfrac{1}{2}x^2-\tfrac{1}{2}x\big)f'(x)\Big]_0^1+\int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x\big)f''(x)~dx \\
&= \frac{f(1)+f(0)}{2}+\int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x\big)f''(x)~dx
\end{aligned}
Now, $f''(x)$ is continious and therefore bounded, say $m \leqslant f’’(x) \leqslant M$. The expression $\tfrac{1}{2}x^2-\tfrac{1}{2}x \leqslant 0 $ throughout $[0,1]$, and therefore
\begin{aligned}
M\int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x \big)~dx \leqslant \int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x \big)f’’(x) ~dx \leqslant m \int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x \big)~dx.
\end{aligned}
where the inequalities can be taken as strict unless $f’’$ is constant. This becomes,
\begin{aligned}
-\frac{M}{12} \leqslant \int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x\big)f’’(x)~dx \leqslant -\frac{m}{12}
\end{aligned}
Moreover, being continuous, $f’’$ takes every value between $m$ and $M$ so there exists $\xi\in(0,1) ^1$ such that,
\begin{aligned}
\int_0^1 \big(\tfrac{1}{2}x^2-\tfrac{1}{2}x\big)f’’(x)~dx = -\frac{1}{12}f’’(\xi)
\end{aligned}
which completes the proof, that there exists $\xi\in(0,1)$ such that:
\begin{aligned}
\int_0^1 f(x) ~dx = \frac{f’’(0)+f’’(1)}{2} - \frac{1}{12} f''(\xi).
\end{aligned}
The result for the case $[a,b]$ can be derived through a change of variable.
Corrected to refer to $f’’$ consistently
Footnote: if $f’’$ is not constant we can choose $\xi$ in the interior of $[0,1]$ because the inequalities are then strict; if $f’’$ is constant any point $\xi$ will serve.
| {
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Prove $0 < x < \pi /2 \implies \sin x > x/\sqrt{x^2+1}$ using Mean Value Theorem I'm solving the following problem:
Show that if $0 < x < \pi /2$ then $\sin x > \dfrac{x}{\sqrt{x^2+1}}$.
One of the hints given is to apply mean value theorem for $\sin (x)$ on the interval $[0,x]$
This is my attempt so far:
Let $f(x) = \sin(x)$
Since all trigonometric functions are continuous and $\sin (x)$ is differentiable, mean value theorem can be applied.
$$\frac{\sin x - \sin 0}{x - 0} = \cos c$$
$$\frac{\sin x}{x} = \cos c$$
We know that $0 < c < x$
So, $0 < c < x \leq \pi / 2$
So, $0 \leq \cos c < 1$
Also, $\cos c > \cos x$
$$\cos c > \sqrt{1 - \sin^2 x}$$
$$\frac{\sin x}{x} > \sqrt{1 - \sin^2 x}$$
$$\sin x > x\sqrt{1 - \sin^2 x}$$
Now after this I'm stuck. I'm not sure how to bring $\sqrt{x^2 + 1}$ into the proof! I did think over it and was able to find some relations involving it like:
$$\sqrt{x^2 + 1} > 1$$
But I think I'm going the wrong path. How should I complete the proof ?
| You are on the right path: square both sides of $\sin(x)>x (1-\sin^2(x))^{1/2}$ and get $\sin^2(x)>x^2(1-\sin^2(x))=x^2-x^2\sin^2(x)$, then add $x^2\sin^2(x)$ to both sides to obtain $\sin^2(x)(1+x^2)>x^2$. Now divide both sides by $(1+x^2)$ and take the square root.
| {
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5x + q $$
$$ \text{Since } x + 1 \text{ and } x - 3 \text{ are factors of } f(x), \text{ then } f(-1) = 0 \text{ and } f(3) = 0, \text{ i.e.} $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \color{red}{\leftarrow (1)} \\
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \color{blue}{\leftarrow (2)}
\end{align}
$$ \text{From } \color{red}{(1)}: $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \\
1 + p(-1) + 5(1) + (-5) + q &= 0 \\
1 - p + 5 - 5 + q &= 0 \\
1 - p + q &= 0 \\
q &= p - 1 \color{limegreen}{\leftarrow (3)}
\end{align}
$$ \text{From } \color{blue}{(2)}: $$
\begin{align}
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \\
81 + 27p + 45 + 15 + q &= 0 \\
27p + q + 60 + 81 &= 0 \\
27p + q + 141 &= 0 \\
q &= -27p - 144 \color{orange}{\leftarrow (4)}
\end{align}
$$ \color{orange}{(4)} + \color{limegreen}{(3)}: $$
\begin{align}
-27p - 144 &= p - 1 \\
-27p - p &= 144 - 1 \\
-28p &= 143 \\
28p &= -143 \\
p &= -\frac{143}{28} \\
\therefore p &= -5\frac{3}{28} \color{mediumpurple}{\leftarrow (5)}
\end{align}
$$ \text{Substitute } \color{mediumpurple}{(5)} \text{ into } \color{limegreen}{(3)}: $$
\begin{align}
q &= -5\frac{3}{28} - 1 \\
\therefore q &= -6\frac{3}{28}
\end{align}
The answers were $ p = -5, q = -6 $.
Where did I go wrong?
| $$q = -27p - \color{red}{141} {\leftarrow (4)}$$
| {
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Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$
Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$
Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq
\frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$
My Solution.
We write the inequality as
$$\big[2(b+c)-a\big]^2 \big[4\left( {{\mkern 1mu} b + {\mkern 1mu} c} \right){a^2} + 5{\mkern 1mu} \left( {2{\mkern 1mu} c + b} \right)\left( {c + 2{\mkern 1mu} b} \right)a + 3{\mkern 1mu} \left( {b + c} \right)\left( {6{\mkern 1mu} {b^2} + 19{\mkern 1mu} bc + 6{\mkern 1mu} {c^2}} \right)\big]$$
$$+\big[a-2(b+c)\big]\left( {b}^{2}+3bc+{c}^{2} \right) \left( 36{b}^{2}+77bc+36{ c}^{2} \right)+2bc \left( b+c \right) \left( b-c \right) ^{2}\geq 0,$$
which is true.
You can see the text to check here. I'm hoping for alternative proof..
Thank you!
| Let $c=\min\{a,b,c\}$, $b=c+u$ and $a-2(b+c)=v$.
Thus, $u\geq0$, $v\geq0$, $a=4c+2u+v$ and
$$(ab+ac+bc)\prod_{cyc}(a+b)\left(5\sum_{cyc}\frac{a}{b+c}-\frac{a^2+b^2+c^2}{ab+ac+bc}-10\right)=$$
$$=745vc^4+2(2u^2+745uv+247v^2)c^3+(6u^3+1074u^2v+741uv^2+109v^3)c^2+$$
$$+(2u^4+329u^3v+355u^2v^2+109uv^3+8v^4)c+$$
$$+2uv(18u^3+27u^2v+13uv^2+2v^3)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}$ I have two inequalities today :
*
*$$\color{red}{a.~~~~~ e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}}$$
*$$\color{purple}{b.~~~~~ e ^{-\frac{x^{2}}{2}}\leq\sqrt{e}e ^{-|x|}}$$
I tried to relate these inequalities to this famous inequality
$$e^{x}\geq 1+x$$
but without success so I considere the function :
$$f(x)=e ^{-\frac{x^{2}}{2}}-\frac{2}{\sqrt{e}}\frac{1}{1+x^{2}}$$
But the calculations were too much.
Thanks!
| For $a$, you can instead consider $y = x^2$ for positive $y$ to get $$e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}} \frac{1}{1+y} \to (1+y)e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}}$$
You can then find the maximum of the left hand side by differentiating and finding the roots. It would end up being at $y = 1$, and $(1+1) e^{-\frac{1}{2}} = \frac{2}{\sqrt{e}} \le \frac{2}{\sqrt{e}}$.
For $b$, you can multiply by $e^{\frac{x^2}{2}}$ on both sides to get $$\sqrt{e}e^{-\left|x\right|}e^{\frac{x^{2}}{2}} = e^{\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2}} \ge 1$$
Taking logs, $$\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2} \ge 0$$
This is easy to show since $$\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2} = \frac{\left(\left|x\right|-1\right)^{2}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| Solve for $c$:
$$c=\frac{-b\pm\sqrt{12-3b^2}}{2}$$
To prove the inequality, we need to prove it for the larger of the two values of $c$ (the one with the sign "plus"):
$$b+c=\frac{b+\sqrt{12-3b^2}}{2}\le 2$$
i.e.
$$\sqrt{12-3b^2}\le 4-b$$
which is defined for $|b|\le 2$ - and in that interval the RHS is positive so we can square both sides:
$$12-3b^2\le 16-8b+b^2$$
or: $4b^2-8b+4=4(b-1)^2\ge 0$, which is obviously true. The equality is satisfied for $b=1$, which gives $c=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equation is $E=\frac{1}{4}$. Now I thought that for a non-homogeneous equation, where the term $d(n)$ does not depend on the underlying recursion has the form $d(n) = k\mu^n$ and $\mu$ is not a root of the homogeneous equation, the solution is given by
$$a_n = \frac{k\mu^n}{\Phi(\mu)}$$
where $\Phi$ is the characteristic equation of the homogeneous one. In my case $d(n) = \frac{1}{4}\frac{2}{3}^{n}$, so $k=\frac{1}{4}$ and $\mu = \frac{2}{3}$. Thus the solution should be given by
$$a_n = \frac{\frac{1}{4}\frac{2}{3}^n}{\frac{2}{3}-\frac{1}{4}}=\frac{\frac{1}{4}\frac{2}{3}^n}{\frac{5}{12}}=\frac{3}{5}\frac{2}{3}^n$$
However, the solution should be $$\frac{3}{5}\frac{2}{3}^n-\frac{3}{5}\frac{1}{4}^n$$. What did I wrong?
Note: the question arises from another problem, see here
| Note that $$4^na_n-4^{n-1}a_{n-1}=\left(\dfrac{8}{3}\right)^{n-1}$$ now telescope.
Add: Let me compete the computation to get a closed form. After taking the summation $$4^na_n-a_0=\sum_{k=1}^n\left(\dfrac{8}{3}\right)^{k-1}=\dfrac{1-\left(\dfrac{8}{3}\right)^{n}}{1-\left(\dfrac{8}{3}\right)}$$ and hence $$4^na_n=a_0+\dfrac{3}{5}\left(\left(\dfrac{8}{3}\right)^n-1\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \cdot 1} = 9^2 + 7^2 = 130
$$
Therefore, $10|(9^{n+1}+7^{2n})$, is true.
Inductive Step: There exists an integer $k$ such that $(9^{n+1}+7^{2n}) = 10k$. Let $P(k) = 9^{k+1} + 7^{2k}$. Therefore,
$$
P(k+1) = 9^{(k+1)+1} + 7^{2(k+1)} = 9^{k+2} + 7^{2k+2}
$$
$$
P(k+1) = (9^1 \cdot 9^{k + 1}) + (7^2 \cdot 7^{2k})
$$
$$P(k+1) = (2+7)(9^{k+1}) + (7^2 \cdot 7^{2k}) = (2 \cdot 9^{k+1}) + (7 \cdot 9^{k+1}) + (7 \cdot 7 \cdot 7^{2k})
$$
$$
P(k+1) = (2 \cdot 9^{k+1}) + 7(9^{k+1} + 7 \cdot 7^{2k})
$$
| Instead of trying to factor out numbers to make it divisible by 10, I was able to solve the equation by rearranging the formula that equals 10k to
$$
9^{+1} = 10 - 7^{2}
$$
If you substitue it into P(k + 1) you get:
$$
P(k+1) = 9^1 \cdot (10m - 7^{2k}) + (49 \cdot 7^{2k})
$$
Which reduces to
$$
P(k+1) = 90m + (40 \cdot 7^{2k}) = 10(9m + (4 \cdot 7^{2k})
$$
which makes it divisible by 10
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int [e^{\tan x} (1-\tan x)^2\cos^2 x]dx$ Let $\tan x=t$
$$\int [e^t (1-\tan x)^2 \cos^2 x \frac{1}{\sec^2x} ]dt$$
$$\int [e^t (\cos x-\sin x)^2 \cos^2 x ]dt$$
$$\int [e^t (1-\sin 2x)\cos^2 x] dt$$
How do I solve further?
| $(1-\tan x)^2 = 1+\tan^2x-2\tan x $
So, if you let $\tan x = t \Rightarrow dx = \dfrac{dt}{\sec^2x} = \dfrac{dt}{1+t^2}$ and $\cos^2x = \dfrac{1}{1+t^2}$
$\begin{align}\Rightarrow I = \int e^{t}(1+t^2-2t)\dfrac{dt}{(1+t^2)^2} & = \int e^t\left[\dfrac{1}{1+t^2 }-\dfrac{2t}{(1+t^2)^2}\right]dt\end{align}$
Now, $\dfrac{d}{dt}\left(\dfrac{1}{1+t^2}\right) = \dfrac{-2t}{(1+t^2)^2}$
We have
$\begin{align}\int e^x(f(x)+f'(x))dx = e^xf(x)\end{align}$
So, you have $I = \begin{align}e^t\dfrac{1}{1+t^2} +c = e^{\tan x}(\cos^2 x) +c\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} +
\frac{21 + 2r}{3 + q} +
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} +
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.
| Titu's lemma gives us
$$ \sum \frac{q}{ 3 + p } = \sum \frac{ q^2 } { 3q + pq} \geq \frac{ (p+q+r)^2 } { \sum 3q + pq } = \frac{ 9}{9 + pq+qr+rs}.$$
Since $(p+q+r) ^2 \geq 3 (pq+qr+rs)$, so $3 \geq pq+qr+rs $ and thus
$$ \sum \frac{q}{ 3 + p } \geq \frac{9}{9 + pq+qr+rs} \geq \frac{3}{4}.$$
The original problem could be approach in as similar manner.
$$\sum \frac{7+2b}{1+a} = \sum \frac{ (7+2b)^2}{(7+2b)(1+a)} \geq \frac{ (21 + 2a+2b+2c)^2}{ \sum 7 + 2b + 7a + 2ab} = \frac{23^2}{30 + 2ab+2bc+2ca}. $$
Then, since $ \frac{1}{3} \geq ab+bc+ca$, thus
$$\sum \frac{7+2b}{1+a} \geq \frac{23^2}{30 + 2ab+2bc+2ca} \geq \frac{69}{4}.$$
| {
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"source": "stackexchange",
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Simplifying the derivative of $\sin(\cos^2 x)\cos(\sin^2 x)$ Differentiating the term
$$\sin(\cos^2 x)\cos(\sin^2 x)$$
leads me through the chain and product rule to
$$-\sin(2x)\cos(\cos^2 x)\cos(\sin^2 x)+(-\sin(\sin^2 x)\sin(2x)\sin(\cos^2 x))$$
where the derivative of $\sin^2 x$ equals to
$$\frac{d}{dx} \sin^2 x = 2\sin x \frac{d}{dx} \sin x = 2\sin x \cos x = \sin 2x$$
and $\frac{d}{dx} cos^2 x$ to $-\sin 2x$ respectively.
Through factorization, I can then simplify the term to
$$-\sin 2x\ (\cos(\cos^2 x)\cos(\sin^2 x) + \sin(\sin^2 x)\sin(\cos^2 x))$$
The problem starts here where I fail to find a simplification for the second factor which, according to Wolfram Mathematica, should lead to $\cos(\cos 2x)$ and ultimately to
$$-\sin(2x)\cos(\cos 2x)$$
How and which trigonometric identities could I apply to get to that? Is my approach right?
| If we start with $$y=\sin \left(f(x)^2\right) \cos \left(g(x)^2\right)$$
$$y'=2 f(x) f'(x) \cos \left(f(x)^2\right) \cos \left(g(x)^2\right)-2 g(x) g'(x)\sin
\left(f(x)^2\right) \sin \left(g(x)^2\right)$$ If $f(x)=\cos(x)$ and $g(x)=\sin(x)$, then
$$2 f(x) f'(x)=-2 \sin (x) \cos (x)=-\sin(2x)$$
$$2 g(x) g'(x)=+2 \sin (x) \cos (x)=+\sin(2x)$$ So
$$y'=-\sin(2x)\Big[\cos \left(f(x)^2\right) \cos \left(g(x)^2\right)+\sin
\left(f(x)^2\right) \sin \left(g(x)^2\right) \Big]$$
$$\cos \left(f(x)^2\right) \cos \left(g(x)^2\right)+\sin
\left(f(x)^2\right) \sin \left(g(x)^2\right)= \cos\left(f(x)^2-g(x)^2 \right)$$ Continue and get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability that the cards are in AP
Box $1$ contains three cards bearing numbers $1, 2, 3$ ; box $2$ contains five cards bearing numbers $1, 2, 3, 4, 5$ ; and box $3$ contains seven cards bearing numbers $1, 2, 3, 4, 5, 6, 7$. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i$th box, $i = 1, 2, 3.$ What's the probability that $x_1 , x_2 , x_3$ are in an arithmetic progression?
My attempt: Total cases are: $3\cdot5\cdot7=105$. Favorable instances are $10$.
$ \{ (1,1,1), (1,2,3), (1,3,5), (1,4,7), (2,2,2), (2,3,4), (2,4,6), (3,3,3), (3,4,5), (3,5,7)\}$.
So, my answer is $\frac{10}{105}$, but the answer is given as $\frac{11}{105}$. Which case am I missing?
| As others have commented, $(3,2,1)$ is missing. Instead of listing the choices, observe that $x_1,x_2,x_3$ are in AP iff $x_1+x_3=2x_2$. This implies either both $x_1,x_3$ are odd or both are even since $x_1+x_3$ is even. We have $2$ odd values of $x_1$ and $4$ odd values for $x_3$, giving $2\times 4$ pairs. We also have $1$ even value of $x_1$ and $3$ even values of $x_3$, giving $1\times 3$ pairs.
For all pairs $(1+1)/2=1\le x_2=(x_1+x_3)/2\le(3+7)/2=5$ i.e. $x_2\in\{1,2,...,5\}$. Total number of pairs is $8+3=11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Unifying a real-valued function over two branches I have a function that I want to evaluate numerically. These three forms listed below are all identical algebraically for nonnegative real $u$ (aside from removable discontinuities at $u=0$ and $u=1$), but none avoids complex numbers for its entire domain.
$$\begin{aligned}
f_1(u) &= \frac{1}{\sqrt{1-u^2}}\tan^{-1}\frac{\sqrt{1-u^2}}{u} \\
f_2(u) &= \frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{2}-\tan^{-1}\frac{u}{\sqrt{1-u^2}}\right) \\
f_3(u) &= \frac{1}{\sqrt{u^2-1}}\ln\left(u+\sqrt{u^2-1}\right)
\end{aligned}$$
The first two have imaginary values of the square root for $u>1$ and the third has imaginary values of the square root for $u<1$. None are well-conditioned numerically in close proximity to $u=1$.
Any suggestions on a different form for numeric evaluation?
| Using the series expansion
$$\ln(x)=\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}$$
gives, according to the third representation,
\begin{align}f(x)&=\frac1{\sqrt{x^2-1}}\ln\left(x+\sqrt{x^2-1}\right)\\&=\frac1{\sqrt{x^2-1}}\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x-1+\sqrt{x^2-1}}{x+1+\sqrt{x^2-1}}\right)^{2n+1}\\&=\frac1{\sqrt{x^2-1}}\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{\sqrt{x^2-1}}{x+1}\right)^{2n+1}\tag{Rationalize}\\&=\frac1{x+1}\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x^2-1}{(x+1)^2}\right)^n\\&=\frac1{x+1}\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x-1}{x+1}\right)^n\end{align}
which is convergent for all $x\ge0$ and is well-conditioned near $x=1$. It doesn't seem to be the case that there is a nice representation of $f$ over $\mathbb R_{\ge0}$, but for $x>1$ it is related to the inverse hyperbolic tangent function.
| {
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Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$ I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan normal form).
I could find that $\lambda=3$ is $A$'s only eigenvalue and its algebraic multiplicity is $3$.
Afterwards, I solved the system $(A-3I_3)X=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and I got that $x=\begin{pmatrix} -a-b \\ a \\ b \end{pmatrix}$ for some parametres $a, b\in \mathbb{C}$.
I picked the eigenvectors $X_1=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $X_2=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ and then I got stuck. I tried solving both the equations $(A-3I_3)X'=X_1$ and $(A-3I_3)X''=X_2$, but these have no solutions, so I can't find any generalised eigenvectors. What should I do?
| Since you have two linearly independent eigenvectors, you know that there are two blocks for $\lambda=3$. That is already enough to tell you what the Jordan form looks like.
So it is not really necessary to look for generalized eigenvectors, if you only want Jordan form. But still, it might be useful to try this - especially since you've mentioned problems with this part in your question.
We want to have a solution for $(A-3I)X'=X$, where $X$ is an eigenvector. We do not know which eigenvector to chose for $X$, so let us try to do this with arbitrary eigenvector.
$\left(\begin{array}{ccc|c}
1 & 1 & 1 &-a-b \\
-2 &-2 &-2 & a \\
1 & 1 & 1 & b
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
0 & 0 & 0 &-a-2b \\
0 & 0 & 0 & a+2b \\
1 & 1 & 1 & b
\end{array}\right)$
Already after these two steps we see, that this is only possible if $a+2b=0$. For example, if we choose $b=1$ and $a=-2$, then the above system is
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{array}\right)$$
and one of the solutions is $(1,0,0)^T$. This is what we can choose for our generalized eigenvector.
Another thing worth noticing is that if we want to have $(A-3I)X'=X$ then $X$ needs to be in the column space of the matrix $(A-3I)$. So we have $X$ in the span of $(1,-2,1)^T$. For example, we can choose $X'=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $X=\begin{pmatrix}1\\-2\\1\end{pmatrix}$.
Once we found all eigenvectors and generalized eigenvectors, we can use them to get a matrix such that $A=PJP^{-1}$. So for
$$P=
\begin{pmatrix}
1 & 1 &-1 \\
-2 & 0 & 1 \\
1 & 0 & 0 \\
\end{pmatrix}
\qquad\text{and}\qquad
J=
\begin{pmatrix}
3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{pmatrix}
$$
we have $$A=PJP^{-1}.$$
One can check this by direct computation. Here is also a link to WolframAlpha.
| {
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$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.
Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.
Also, just some hints would be appreciated, thanks
| Because
$$\sum \frac{a^2}{c+a} = \sum \frac{c^2}{c+a}.$$
Therefore, we need to prove
$$2\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqslant \frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}.$$
Suppose $a \geqslant b \geqslant c,$ then
$$a^2 \geqslant b^2 \geqslant c^2 \quad \text{and} \quad \frac{1}{b+c} \geqslant \frac{1}{c+a} \geqslant \frac{1}{a+b}.$$
Using the Rearrangement inequality
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geqslant \frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}.$$
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geqslant \frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}.$$
The proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove $\ e^{n^2} \ge n^n $ for every n by induction Prove for any $\ n \ge 1 $
$$\ e^{n^2} \ge n^n $$
my attempt:
true for $\ n = 1 $ then for $\ n+ 1 $ :
$\ e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} \cdot e^{2n} \cdot e $
$\ ( n+1)^{n+1} = (n+1)^n\cdot (n+1) \le (n+n)^n\cdot(n+1) = (2n)^n \cdot(n+1) = 2^n \cdot n^n \cdot(n+1) \le 2^n \cdot e^{n^2} \cdot(n+1) $
from here it's obvious $\ e^{2n} \ge 2^n $ but I can't find a way to get rid of $\ n+1 $ and $\ e $?
| You can continue your work by taking $\ln$ of both sides and we have $2n+1 \geq n\ln2 + \ln(n+1) \geq n/2 + n+1=3n/2 + 1$. Then we get that $n/2 \geq 0$ with reversible steps.
| {
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos\alpha = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.
But the answers are $\frac{\pi}{3}$ and $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?
| $$
f(\cos\alpha)=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg(\frac{1}2\cos\alpha+\frac{\sqrt3}2\sin\alpha\bigg)\\
= \cos^{-1}(\cos\alpha) +\cos^{-1}\bigg(\cos\frac \pi 3\cos\alpha+\sin\frac{\pi}3\sin\alpha\bigg)\\
=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg[\cos\bigg(\frac\pi3-\alpha\bigg)\bigg]\\
= \pm\alpha+2\pi k_1 \pm\ (\frac\pi3-\alpha)+2\pi k_2 \\
=\pm\alpha \pm\ (\frac\pi3-\alpha)+2\pi k
$$
where $k,k_1,k_2 $ are integers.
For the first case we get the prescribed solution by taking the $+$ signs in both and $k=0$. In the second case, if we take the $+$ sign on the first and the $-$ sign on the second, and $k=0$ we get:
$$
2\alpha-\frac\pi3=2\cos^{-1}\big(\frac13 \big)-\frac\pi3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble Understanding Proof of Lagrange's Identity Using Binomial Formula I've been reading a book on elementary mathematics and am having trouble understanding a proof that they give for Lagrange's identity using the binomial formula.
Lagrange's identity is stated as :
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}^{2} = \binom{2n}{n}
\end{equation}
In the proof they say :
"Writing $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ and applying Newton's binomial formula to both sides we get :
\begin{equation}
\binom{2n}{n} = \sum_{i+j=n} \binom{n}{i} \binom{n}{j} = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \sum_{i=0}^{n} \binom{n}{i}^{2}
\end{equation}
$\square$."
A lot of detail seems to be missing here. Could someone provide the missing details here ? I cannot understand the proof due to the elision of details.
| The binomial coefficient $\binom{2n}{n}$ is the coefficient of $x^n$ in the left-hand side.
In the right hand side you get $x^n$ from terms $x^i$ and $x^j$ where $i+j=n$. The term $x^i$ has coefficient $\binom{n}{i}$ and similarly for $x^j$, so you have
$$
\binom{2n}{n}=\sum_{i+j=n}\binom{n}{i}\binom{n}{j}
$$
Now observe that if $i+j=n$, then $j=n-i$ and that
$$
\binom{n}{n-i}=\binom{n}{i}
$$
and that $i$ can run from $0$ to $n$, so we finally get
$$
\binom{2n}{n}=\sum_{i=0}^n\binom{n}{i}^{\!2}
$$
As an example with $n=3$: in the right hand side we have
$$
\binom{3}{0}+\binom{3}{1}x+\binom{3}{2}x^2+\binom{3}{3}x^3
$$
multiplied by itself, but we just need to consider the terms with $x^3$, so
$$
\binom{3}{0}\cdot\binom{3}{3}x^3+\binom{3}{1}x\cdot\binom{3}{2}x^2
+\binom{3}{2}x^2\cdot\binom{3}{1}x+\binom{3}{3}x^3\cdot\binom{3}{0}
$$
| {
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how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded? I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$
| Claim. If $a_n$ is increasing, then $a_n<3$.
Otherwise, if $a_n\ge 3$, for some $n\in\mathbb N$, then
Case A. $a_n=3$, then $a_{n+1}=\sqrt{a_n+6}=\sqrt{3+6}=\sqrt{9}=3$. Hence $a_n$ is not increasing.
Case B. $a_n>3$. Then $a_{n+1}>a_n>3$ and
$$
a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{\sqrt{(a_n+6}-a_n)(\sqrt{a_n+6}-a_n)}{\sqrt{a_n+6}+a_n}\\=\frac{a_n+6 -a_n^2}{\sqrt{a_n+6}+a_n}=-\frac{(a_n+2)(a_n-3)}{\sqrt{a_n+6}+a_n}<0.
$$
Contradiction.
Hence, $a_n$ is upper bounded by $3$ and lower bounded by $a_1$.
| {
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Solution of |x| + 3x = 2 + 6i. Where is an error in my thinking? So I solved the equation: $|x| + 3x = 2 + 6i$ but I don't know where my error is. And I know there is an error because Wolfram Alpha shows that the only solution is $ x = 2i $ . In my calculations I have 2 solutions.
$$|x| + 3x = 2 + 6i$$
$$\sqrt{a^2 + b^2} +3(a + bi) = 2 +6i$$
$$\sqrt{a^2 + b^2} = 2 +6i - 3a -3bi$$
$$\sqrt{a^2 + b^2} = 2-3a +(6 -3b)i$$
$$6 -3b=0$$
$$b=2$$
$$\sqrt{a^2 + 4} = 2-3a $$
$$a^2 + 4 = 4-12a+9a^2 $$
$$8a^2-12a=0 $$
$$4a(2a-3)=0 $$
$$a = 0 || a=3/2 $$
Why $x = 3/2 + 2i$ is not an answer?
| When you square both sides of $\sqrt{a^2+4} = 2-3a$, an extra solution is generated.
In particular, the solution $a = 3/2$ is the solution to $\sqrt {a^2+4} = 3a-2$, which is a perfectly valid solution for the squared equation $a^2+4 = (2-3a)^2$.
However for this $a$, $\sqrt{a^2+4} > 0$ but $2-3a < 0$, rendering it invalid for the original equation. For this reason, we must check whether all our solutions satisfy the original equation. There are no invalid steps in your calculations otherwise.
| {
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated
| The set of solutions is the segment $\{\alpha i\mid-1\le\alpha\le1\}$.
This follows from the following geometrical argument. $|z-i|$ is the distance from $z$ to $i$; $|z+i|$ is the distance from $z$ to $-i$. Which point $z\in\mathbb C$ can have the sum of distances to $i$ and to $-i$ equal to $2$, knowing that the distance between $i$ and $-i$ is already $2$? It must be colinear with $i$ and $-i$ (or otherwise the LHS would be $\gt 2$ due to triangle inequality) and it must lie in the actual segment with endpoints $i,-i$, or otherwise one of the moduli $|z-i|,|z+i|$ would alone be $\gt 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
I got this question on my exam today and had no idea how to do it.
There must be a connection to complex numbers since its the topic its connected to. I only learnt how to use the De Moivre theorem with complex numbers in polar form so I was completely lost here.
Sorry if its a dumb question :<.
| So you begin by writing
$$
\tan 6\theta \equiv {\sin 6\theta \over \cos 6\theta}
$$
From there you apply De Moivre's Theorem which states:
$$
(\cos \theta +\iota\sin\theta)^n \equiv (\cos n\theta + \iota\sin n\theta)
$$
In this case we have $n = 6$, so:
$$
(\cos \theta +\iota\sin\theta)^6 \equiv (\cos 6\theta + \iota\sin 6\theta)
$$
Expanding the left side with the binomial theorem gives:
$$
\cos6\theta + \iota \sin6\theta = C^6 + 6 i C^5 S - 15 C^4 S^2 - 20 i C^3 S^3 + 15 C^2 S^4 + 6 i C S^5 - S^6
$$
Where $C = \cos\theta$ & $S = \sin\theta$
Separating the real and imaginary parts, we get:
$$
\cos6\theta = C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6 \\
\sin6\theta = 6 C^5 S - 20 C^3 S^3 + 6 C S^5
$$
From here
$$
\tan6\theta = {6 C^5 S - 20 C^3 S^3 + 6 C S^5 \over C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6}
$$
Dividing both the numerator and the denominator by $\cos^6\theta$ or $C^6$
$$
\tan6\theta = {{[6 C^5 S - 20 C^3 S^3 + 6 C S^5]\over C^6} \over {[C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6]\over C^6}}
$$
Which simplifies to:
$$
\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$? As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:
$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$
The lower estimate
$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$
which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate
$$R_k<\ln\left(1+\frac{1}{k}\right)$$
is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?
Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.
| In the same spirit as @Robert Israel in his answer, we could use
$$\sum_{n=N+1}^\infty \frac{1}{n^2}=\psi ^{(1)}(N+1)$$ and use the series expansion of the rhs
$$\psi ^{(1)}(N+1)=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{30
N^5}+\frac{1}{42
N^7}-\frac{1}{30 N^9}+O\left(\frac{1}{N^{11}}\right)$$ which, since alternating, allows to propose as sharp bounds as required.
What could be interesting is to look at the expansion of @Robert Israel's results
$$\frac{1}{N+\frac 12} - \log \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{8
N^4}+O\left(\frac{1}{N^5}\right)$$
$$\frac{1}{N+\frac 12} - \log \left(\frac{2N+3}{2N+1}\right) + \frac{1}{N+1}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}+\frac{1}{8
N^4}+O\left(\frac{1}{N^5}\right)$$
Much less accurate : in my former group, we used for numerical purposes the simple double inequality
$$\sinh \left(\frac{1}{N+1}\right)<\psi ^{(1)}(N+1)<\frac{1}{2} \sinh \left(\frac{2}{N}\right)$$
Later, playing with Padé-like approximants, I found better bounds (much better for the lower than for the upper)
$$\color{blue}{\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}<\psi ^{(1)}(N+1)}<\frac{2 N^2+7 N+7}{2 (N+1)^2 (N+2)}$$
$$\Delta=\psi ^{(1)}(N+1)-\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}=\frac{1}{45 N^5}+O\left(\frac{1}{N^6}\right)$$
Continuing for answering this question, a better one
$$\color{blue}{\frac{5N(1302 N^2+573 N+697) } {6(1085 N^4+1020 N^3+910 N^2+285 N+27 ) }<\psi ^{(1)}(N+1)}$$ and for this one $$\Delta=\frac{207}{15190 N^8}+O\left(\frac{1}{N^9}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding positive integer(s) $n$ for which $3^{n-1} + 5^{n-1} | 3^{n} + 5^{n}$: Is my solution correct? The problem is the following
Find all positive integers $n$ such that $3^{n-1} + 5^{n-1}$ divides $3^{n} + 5^{n}$.
And I attempted as the following:
As $3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n}$,
$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 5(3^{n-1} + 5^{n-1}) = -2\cdot 3 ^ {n-1}$$
And
$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 3(3^{n-1} + 5^{n-1}) = 2\cdot 5 ^ {n-1}$$
These expressions say that
$$2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \ p \in \mathbb{Z}^{+}$$
And
$$2\cdot 5 ^ {n-1} = q(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \ q \in \mathbb{Z}^{+}$$
By addition, it becomes clear that $q = 2 - p$.
Hence, the last expression is equivalent to $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$.
Solving for $p$, it is equal to $\frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$.
Here, $p$ is odd only for $n = 1$, hence it is the only solution.
Is this attempt a solution for the problem? And for the effort, Thanks!
| What you did is mostly correct, but your argument that $p = \frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$ is only odd for $n = 1$ is not true since for all odd $n$, the result is either not an integer or it's an odd integer. Instead, $p$ is odd, as you stated, but with $2 - p \gt 0$ from $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$ since the left side is positive, the right side must be as well, means only $p = 1$ works. Thus, $2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1})$ gives $3^{n-1} = 5^{n-1} \implies n = 1$, so your solution is correct.
A somewhat easier way to see this is that, from what you wrote, $3^{n-1} + 5^{n-1} \mid -2\cdot 3^{n-1} \implies 3^{n-1} + 5^{n-1} \le \left|-2\cdot 3^{n-1}\right| = 2\cdot 3^{n-1}$. For $n \gt 1$, we have $5^{n-1} \gt 3^{n-1} \implies 3^{n-1} + 5^{n-1} \gt 2\cdot 3^{n-1}$, so only $n = 1$ is a possible solution. Since $n = 1$ gives $2 \mid 8$, this means only $n = 1$ works.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can from here find all x such that $x \in \mathbb{Z}$ ??
i did use the hint for $y = 2x+1$ so that $x =\frac{y-1}{2}$
made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$
$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$
I can not understand with the help of the clues, can some 1 some formal proof ?
| Hint:
If $2x+1=y$
$$x^3-3x+2=(y-1)^3/8-3(y-1)/2+2=\dfrac{y^3-3y^2-9y+27}8$$ which will be divisible by $y$
$\iff y$ divides $y^3-3y^2-9y+27$ as $y$ is odd
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there an elegant way to find the equation of a plane given its line and angle of intersection with another plane? This is a problem I give sometimes to my calculus students:
Let $\mathcal{Q}$ be the plane defined by $x+2y+3z=6.$ Find a plane $\mathcal{P}$ that meets $\mathcal{Q}$ at an angle of $\theta = \frac{\pi}{3}$ such that $\mathcal{P},\mathcal{Q}$ intersect along the line $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle.$ (Hint: Write $\mathcal{P}$ as $ax+by+cz=6$, and use the conditions to solve for $a,b,c.$)
The intended solution: We know that $\mathcal{P}$ contains the point $\langle 3,0,1\rangle$ and also $\langle 3,0,1 \rangle+t \langle -3,1,\frac{1}{3} \rangle$ for any $t$, in particular for $t=1.$ Then we have:
$$
\langle 3,0,1\rangle\cdot \langle a,b,c\rangle = 6,\qquad \langle 0,1,4/3\rangle\cdot \langle a,b,c\rangle = 6
$$
We also know the angle of intersection:
$$
\frac{\langle a,b,c\rangle \cdot \langle 1,2,3 \rangle}{\sqrt{a^2+b^2+c^2}\;\sqrt{14}}=\cos(\pi/3)=\frac{1}{2}
$$Putting these together, we are solving:
$$
\begin{cases}
3a+c=6\\
b+\frac{4}{3}c = 6\\
\frac{a+2b+3c}{\sqrt{14(a^2+b^2+c^2)}}=\frac{1}{2}
\end{cases}
$$Solving the first two equations for $a$ gives $c=6-3a$, $b=4a-2.$ Plugging this into the third equation gives:
$$
56=a^2+(4a-2)^2+(6-3a)^2
$$
This is just a quadratic equation in $a.$ Solving it produces
$$
\{a,b,c\}=\left\{1\mp\frac{\sqrt{273}}{13},2\mp\frac{4\sqrt{273}}{13},3\pm\frac{3\sqrt{273}}{13} \right\}
$$
Not terrible but not entirely straightforward. I'm wondering if there is a simpler way to reach this conclusion, maybe by exploiting the normal vectors of the planes?
| a direction within the plane $Q$ that is orthogonal to the line is
$(1,2,3)$ crossed with $(-9,3,1);$ this is $7(-1,-4,3).$ Let us name $u= (-1,-4,3).$
Length of $u$ is $\sqrt{26},$ so make a unit vector
$$ v = \frac{1}{\sqrt{26}}(-1,-4,3) $$
The unit normal to $Q$ is
$$ w = \frac{1}{\sqrt{14}} (1,2,3) $$
Next let $$ N = v \cos \theta + w \sin \theta $$
be a unit vector at angle $\theta$ from $v.$ You want $\theta = \frac{\pi}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating functions and a closed form for the Fibonacci sequence - the big picture I have spent the last few hours trying to understand one way of deriving a closed form for the Fibonacci sequence. As part of improving my mathematical maturity, I am trying to learn to see the "big picture" of what I'm doing as opposed to myopically following the details of a derivation or proof and then forgetting most of what I've done within a few weeks or months.
Below I will post my derivation so that it can be checked for errors and general clarity, but what I would like also ask what the proper intuition or high-level perspective is on what I've done.
As I look over my work, the key takeaways I see are that we put the Fibonacci sequence into the form of a generating function, and in particular we managed to put the generating function into a compact rational form. Once this was done, the rest of the work essentially involved analyzing the rational form of the generating function and extracting information from it until we were able to write the generating function (in its formal power series form) in two different ways, compare coefficients and thus derive a closed form for the Fibonacci sequence.
If these observations are apt, then perhaps the big takeaway is that generating functions can be useful when they make it possible to package a sequence into a form that allows algebraic and/or analytic techniques to be brought to bear on the sequence, when it is represented as a generating function.
I would appreciate if anyone can tell me whether my perspective is reasonable or not.
My exposition:
The Fibonacci numbers are a sequence $1, 1, 2, 3, 5, 8, 13, \dots$. The first two numbers are 1, and then every subsequent number is the sum of the prior two. Let $(a_n)$ be the sequence of Fibonacci numbers and $f(x) = a_0 + a_1 x + a_2 x^2 + \dots$ be the generating function. Consider that
\begin{align*}
x^2 f(x) + x f(x) &= a_0 x^2 + a_1 x^3 + a_2 x^4 + \dots + a_0 x + a_1 x^2 + a_2 x^3 + \dots\\
&= a_0 x + (a_0 + a_1) x^2 + (a_1 + a_2) x^3 + \dots\\
&= a_0 x + a_2 x^2 + a_3 x^3 + \dots\\
&= f(x) - a_0 - a_1 x + a_0 x\\
&= f(x) - 1 - x + x \text{ (using the known values for } a_0 \text{ and } a_1)\\
&= f(x) - 1.
\end{align*}
Thus $f(x) = \frac{1}{1 - x - x^2}$. Now factor $1 - x - x^2$ as $(1 - \alpha x)(1 - \beta x)$, so that $- \alpha - \beta = -1$ (or more naturally, $\alpha + \beta = 1$) and $\alpha \beta = -1$. If you solve this system of equations you will end up with $\alpha = \frac{1 + \sqrt 5}{2}$ and $\beta = \frac{1 - \sqrt 5}{2}$, or the reverse. Now consider the partial fraction decomposition
\begin{align*}
f(x) &= a_0 + a_1 x + a_2 x^2 + \dots\\
&= \frac{1}{1 - x - x^2}\\
&= \frac{1}{(1 - \alpha x)(1 - \beta x)}\\
&= \frac{a}{1 - \alpha x} + \frac{b}{1 - \beta x}\\
\Rightarrow 1 &= a(1 - \beta x) + b(1 - \alpha x).
\end{align*}
This gives us another system of equations such that $a + b = 1$ and $-a \beta - \alpha b = 0$ (or more naturally, $a \beta + \alpha b = 0$.) If you solve \textit{this} system of equations you get $a = \frac{\sqrt 5 + 1}{2 \sqrt 5}$ and $b = \frac{\sqrt 5 - 1}{2 \sqrt 5}$.\
Thus,
\begin{align*}
f(x) &= a_0 + a_1 x + a_2 x^2 + \dots\\
&= \frac{1}{1 - x - x^2}\\
&= \frac{1}{(1 - \alpha x)(1 - \beta x)}\\
&= \frac{a}{1 - \alpha x} + \frac{b}{1 - \beta x}\\
&= a(1 + \alpha x + \alpha^2 x^2 + \dots) + b(1 + \beta x + \beta ^2 x^2 + \dots)\\
&= (a + b) + (a \alpha + b \beta)x + (a \alpha ^2 + b \beta^2)x^2 + \dots
\end{align*}
which explicitly means that $a_k = (\frac{\sqrt 5 + 1}{2 \sqrt 5})(\frac{1 + \sqrt 5}{2})^k + (\frac{\sqrt 5 - 1}{2 \sqrt 5})(\frac{1 - \sqrt 5}{2})^k$, as desired.
| It's a good approach. One thing that can be simplified a little bit is:
$$
f(x)=\frac{1}{(1-\alpha x)(1-\beta x)} = \frac{1}{\alpha - \beta} \cdot \frac{\alpha (1-\beta x) - \beta(1-\alpha x)}{(1-\alpha x)(1-\beta x)} = \frac{\alpha/(\alpha - \beta)}{1-\alpha x} - \frac{\beta/(\alpha - \beta)}{1-\beta x}.
$$
(And this is not hindsight.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $x$ and $y$ are unequal real numbers
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$
For the above question, I got $x$ not equal to $-1,0,1$ and $y = 1/x$.
Then, I tried finding $x-y$ using $\big((x+y)^2-4xy)\big)^{\frac{1}{2}}$ and then solving for $x$ and $y$.
I think I over complicated it and there must be some easier way. Please help me out.
| Intuitively, one should let $a=x+y$ and $b=x-y$.
Then from the condition $\sqrt {xy}=1$ one obtains $a^2-b^2=4$.
The other condition becomes
$$a = \sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}$$
Cubing both sides, we have
\begin{align}a^3 &= \frac ab + 3\left(\sqrt[3]{\frac ab}\right)^2\sqrt[3]{\frac ba}+ 3\sqrt[3]{\frac ab}\left(\sqrt[3]{\frac ba}\right)^2+ \frac ba
\\&=\frac ab + \frac ba+ 3\left(\sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}\right)
\\&=a\left(\frac 1b + \frac b{a^2}\right)+ 3a\end{align}
Eliminating $a$ from both sides gives
$$a^2 = \frac 1b + \frac b{a^2} + 3$$
$$4+b^2=\frac1b + \frac b{4+b^2} + 3$$
$$1+b^2=\frac {4+b^2+b^2}{b(4+b^2)}$$
$$b(4+b^2)(1+b^2)-2b^2 =4$$
Expanding $b(4+b^2)(1+b^2)-2b^2$ gives $b^5+5b^3-2b^2+4b$, which seems very familiar...
| {
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Can fractional/decimal radicals/roots exist? For questions like "What is the 1/2th root of x would the answer be $x^2$?
My logic is that since $$
\sqrt[\cfrac{1}{2}]{x}=x^{1/{(\cfrac{1}{2}})}
$$
Which simplifies to $x^2$.
So as a general rule it could be $$
\sqrt[\cfrac{1}{a}]{x}=x^{1/{(\cfrac{1}{a}})}
=x^a
$$
And with a different denominator $$\sqrt[\cfrac{b}{a}]{x}=x^{1/{(\cfrac{b}{a}})}
=x^{\cfrac{a}{b}}$$
This corresponds to how decimal/fractional exponents denote radicals (their inverse) while fractional radicals are easier shown with exponents.
Example : (2/3rd root of 4)
$$\sqrt[\cfrac{2}{3}]{4}=4^{1/{(\cfrac{2}{3}})}
=4^{\cfrac{3}{2}}= 8$$
Example (22/7th root of π) :
$$\sqrt[\cfrac{22}{7}]{π}=π^{1/{(\cfrac{22}{7}})}
=π^{\cfrac{7}{22}}
\approx 1.439$$
Example (1/2th root of 1/4) :
$$\sqrt[\cfrac{1}{2}]{\cfrac{1}{4}}=\cfrac{1}{4}^{1/(\cfrac{2}{1})}
=\cfrac{1}{4}^{(\cfrac{2}{1})}
=\cfrac{1}{4}^{2}
=\cfrac{1}{16}
$$
| The definition is:
$$\sqrt[a]{x} = x^{\frac{1}{a}}.$$
Hence, if $a = \frac{1}{b}$, then:
$$\sqrt[\frac{1}{b}]{x} = x^{b}.$$
As a consequence:
$$\sqrt[\frac{1}{2}]{x} = x^2.$$
Notice that $a = \frac{1}{b}$ implies that, if for example $b = \frac{22}{91}$, then $a = \frac{91}{22}$. If for example $b = \sqrt{3}$, then $a = \frac{1}{\sqrt{3}}$. And so on.
| {
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
| Let $$S=1^1+3^2+\cdots+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2=\sum_{k=1}^{n}[4 k^2-4k+1]$$
$$S=4\sum_{k=1}^n k^2- 4\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$
$$S=4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2}+n=\frac{n(4n^2-1)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function I am trying to derive the Hardy-Ramanujan asymptotic formula
$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$
from Radmacher's formula for the partition function $p(n)$ given by
$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
where
$$A_{k}(n)=\sum_{h=0, (h,k)=1}^{k-1}e^{\pi i(s(h,k)-2n\frac{h}{k})}$$ and
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}\right\rfloor-\frac{1}{2}\right)$$
G.E. Andrews, and any other literature on this topic, says that we can obtain the H-R asymptotic expression from the first term of the Rademacher series, i.e. for $k=1$. I don't know how to approach this as simply calculating for $k=1$ does not give the desired result. Could we perhaps try and use Lapalce's method, or the method of steepest descent?
| Rough sketch: Take the first term, and you'll get:
$$\begin{align}p\left(n\right)&\sim\frac{1}{\pi\sqrt{2}}A_{1}\left(n\right)\cdot\frac{d}{dx}\left(\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right)\\&\sim\frac{1}{\pi\sqrt{2}}\frac{d}{dx}\left(\frac{\sinh\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{x}}\right)\\&\sim\frac{1}{2\pi\sqrt{2}}\frac{d}{dx}\left(\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{x}}\right)\\&=\frac{1}{2\pi\sqrt{2}}\left(\frac{\pi\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{\sqrt{6}x}-\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{2x\sqrt{x}}\right)\\&=\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{4x\sqrt{3}}\left(1-\frac{\sqrt{6}}{2\pi\sqrt{x}}\right)\\&\sim\frac{\exp\left(\pi\sqrt{\frac{2x}{3}}\right)}{4x\sqrt{3}}\end{align}$$
In the first line, we take the first term (as it is the leading term). We can see $A_1(n)=1$ and we write $x-1/24$ as $x$, since we're looking for as $x \rightarrow \infty$.
In the third-line, we take the leading term of the $\sinh(x)$ function.
$$\sinh\left(x\right)=\frac{e^{x}-e^{-x}}{2}\sim\frac{e^{x}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How does $(abc) = (ac)(ab)$? I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$
However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.
Thank you in advance!
| We can also conveniently calculate the product of transpositions in cycle notation by switching to the two-line notation of permutations.
Considering the first identity $(abc)=(ac)(ab)$ we have the representation
\begin{align*}
(abc)\equiv\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\qquad\qquad&
(ac)\equiv\begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\\
&(ab)\equiv\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
\end{align*}
We obtain
\begin{align*}
\color{blue}{(ac)(ab)}&\equiv \begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\circ\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
&=\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\tag{1}\\
&\,\,\color{blue}{\equiv(abc)}\tag{2}
\end{align*}
Comment:
*
*In (1) we use the convention to multiply out from right to left:
\begin{align*}
&a\to b\to b\\
&b\to a\to c\\&
c\to c\to a
\end{align*}
*In (2) we switch back to the cycle notation: $a\to b\to c\to a$.
In the same way we can calculate e.g. the next identity with four elements:
\begin{align*}
\color{blue}{(ad)(ac)(ab)}&
\equiv \begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\left(\begin{pmatrix}a&b&c&d\\c&b&a&d\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&a&c&d\end{pmatrix}\right)\\
&=\begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&c&a&d\end{pmatrix}\\
&=\begin{pmatrix}a&b&c&d\\b&c&d&a\end{pmatrix}\\
&\,\,\color{blue}{\equiv (abcd)}
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Triangle related problem
In the figure below if $tan \alpha=\frac{m}{n}$, where $m$ and $n$ are relatively prime, find the value of $m$ and $n$.
My approach is as follow
$\frac{{SinA}}{{14}} = \frac{{SinB}}{{15}} = \frac{{SinC}}{{13}}$
$\Delta APB$
$\angle PBA = B - \alpha $
$\angle APB = 180 - \left( {\alpha + B - \alpha } \right) = 180 - B$
$\frac{{Sin\left( {180 - B} \right)}}{{13}} = \frac{{Sin\alpha }}{{BP}} \Rightarrow SinB = \frac{{13Sin\alpha }}{{BP}}$
$\Delta APC$
$\angle PAC = A - \alpha $
$\angle APC = 180 - \left( {\alpha + A - \alpha } \right) = 180 - A$
$\frac{{Sin\left( {180 - A} \right)}}{{15}} = \frac{{Sin\alpha }}{{AP}} \Rightarrow SinA = \frac{{15Sin\alpha }}{{AP}}$
$\Delta BPC$
$\angle PCB = C - \alpha $
$\angle BPC = 180 - \left( {\alpha + C - \alpha } \right) = 180 - C$
$\frac{{Sin\left( {180 - C} \right)}}{{13}} = \frac{{Sin\alpha }}{{PC}} \Rightarrow SinC = \frac{{13Sin\alpha }}{{PC}}$
$\frac{{SinA}}{{SinB}} = \frac{{\frac{{15Sin\alpha }}{{AP}}}}{{\frac{{13Sin\alpha }}{{BP}}}} = \frac{{14}}{{15}} \Rightarrow \frac{{{{15}^2}}}{{13 \times 14}} = \frac{{AP}}{{BP}}$
How do I approach from here?
| P is first Brocard point of the triangle.
We have the relation $$\cot \alpha = \cot A + \cot B + \cot C$$
To quickly find these, we partition the triangle as below
so $$ \cot B = \dfrac{5}{12}, \cot C = \dfrac{3}{4} $$
and in any $\triangle$, $$ \cot A\cot B + \cot B\cot C +\cot C\cot A =1$$
Update :
Following @JeanMarie's link, we see $$ \cot \alpha = \cot A + \cot B + \cot C = \dfrac{a^2+b^2+c^2}{4\triangle} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ideas how i could continue or whats wrong?
| Assuming you want to prove that for each $n\geq 5$ one has $n! > 4n^2+3$ just note that it easily holds for $n=5$. Now assume that it holds up to $n \geq n-1 \geq \ldots 5.$ Then
$$(n+1)! = (n+1)n! > (n+1)(4n^2+3).$$
It is only left to prove that $(n+1)(4n^2+3) > 4(n+1)^2+3$ for each $n\geq 5.$ In fact,
$$(n+1)(4n^2+3)-4(n+1)^2-3 = 4n^3-5n-4.$$
So let $p(n) = 4n^3-5n-4.$ Note that $p(5) > 0.$ Moreover, $p'(n) = 12n-5 > 0,~\forall n\geq 5.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| By binomial theorem$$5^{6n+6}-2^{2n+2}$$
$$={(3+2)}^{6n+6}-2^{2n+2}$$
$$=3k+2^{6n+6}-2^{2n+2}$$
$$=3k+2^{2n+2}({(3-1)}^{4n+4}-1)=3k+2^{2n+2}(3m)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Three inequalities for number $e$: $\frac{e}{2n+2}I have to prove the following inequalities for number $e$:
$$\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}<\frac3{2n}$$
Here's what I have tried:
$\textbf{First inequality:}$ $\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n$
I use that $\log{\left(\frac{n+1}n\right)}=\int_n^{n+1}\frac1xdx\leq \frac12\left(\frac1n+\frac1{n+1}\right)$.
Operating we have: $n\log{\left(\frac{n+1}n\right)}\leq 1-\frac1{2n+2}$
Exponentiating and using $\frac1{1+x}\geq e^{-x}$, we have:
$$\frac{e}{2n+3}\leq e-\left(1+\frac1n\right)^n$$
This is similar to what i need to prove, but not the same...
$\textbf{Second inequality:}$ $e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}$
I don't know even how to start for this inequality.
$\textbf{Third inequality:}$ $\frac{e}{2n+1}<\frac3{2n}$
I think this is trivial as $e<3$ and $2n<2n+1$ (correct me if I am wrong).
| Update: The book I mentioned below provides more than what you need. Here I give only the things you need to prove the two inequalities.
Step 1: We prove $a_n=\left(1+\frac 1n \right)^{n+\frac 12}$ is decreasing. You can check out this post Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? but the book's method is different and simpler.
Note that $$\log \frac{1+x}{1-x} = 2 \left( \frac x1 + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right)$$
Let $x=\frac{1}{2n+1}$ then
$$
\log a_n = \left(n+\frac 12\right)\log \left(1+\frac 1n \right) = (2n+1)\left(\frac{1}{2n+1}+\frac{1}{3(2n+1)^3}+\frac{1}{5(2n+1)^5}+ \cdots \right)\\
=1+\frac{1}{3(2n+1)^2}+\frac{1}{5(2n+1)^4}+ \cdots
$$
Therefore $a_n$ is (strictly) decreasing.
Step 2: We prove $b_n=\left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right)$ is monotone decreasing.
Note that $$b_n = a_n \frac{1+\frac{1}{2n}}{\left(1+\frac 1n\right)^{\frac 12}} = a_n \left( \frac{(1+\frac{1}{2n})^2}{1+\frac 1n}\right)^{\frac 12} = a_n \left(1+\frac{1}{4n(n+1)} \right)^{\frac 12}$$
Since $a_n$ and $1+\frac{1}{4n(n+1)}$ are both strictly decreasing, so is $b_n$.
Step 3: Now we go back to prove $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}$$
The first inequality $ \iff \left( 1+\frac 1n\right)^{n+1} < e\left( 1+\frac{1}{2n}\right)\tag 1$
Denote $f(x)=x+x \log \left( 1+\frac x2 \right)-(1+x)\log(1+x), 0 < x \leqslant \frac 1n.$
$$f'(x)=\frac{x}{x+2} - \log \frac{1+x}{1+\frac x2}>\frac{x}{x+2}-\frac{1+x}{1+\frac x2}+1=0, f(0)=0.$$
Therefore $$f\left(\frac 1n\right) >0 \implies (1).$$
The second inequality $$\iff e < \left(1+\frac 1n\right)^n \left(1+\frac{1}{2n}\right) \tag 2$$
From step 2, $\forall m>n$
$$
\left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2n+2}\right) \geqslant \left(1+\frac 1m\right)^m\left(1+\frac{1}{2m}\right)\\
\implies \left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2{n+1}}\right) \geqslant e, m\to \infty. \blacksquare
$$
These inequalities are not trivial. They are a series of problems from Problems and Theorems in Analysis I by Polya and Szego.
167. We define $$x_n = y_n e^{-\frac{1}{12n}}, y_n = n! n^{-n-\frac 12}e^n, n=1,2,3,\ldots$$
Then each interval $(x_k, y_k), k=1, 2, 3, \ldots, $ contains the interval $(x_{k+1}, y_{k+1})$ as a subinterval.
168. The sequence $$a_n=\left(1+\frac 1n \right)^{n+p}, n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $p\geqslant \frac 12$.
169. The sequence $$a_n=\left(1+\frac 1n\right)^n\left(1+\frac xn\right), n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $x\geqslant \frac 12.$
170. Let $n$ be a positive integer. Then we have $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Elementary algebraic system with integer and positive rational unknowns The problem is to determine the solutions of the system $x+y+z=3$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=5$, given that $x$ is integer and $y,z$ are positive rationals.
From the hypotheses I have obtained that $y$ and $z$ are rational solutions of the quadratic equation
$$(5x-1)t^2-(5x-1)(3-x)t+(3-x)x=0$$
with integer coefficients, therefore its discriminant $\left( 5\,x-1 \right) \left( x-3 \right) \left( 5\,{x}^{2}-12\,x+3
\right)$ has to be a perfect square.
But I cannot see a clear continuation from here.
| (There was an old problem which looks similar.)
Solution:
First note that x<3, so if $x>0$ we can easily find $x=1$ is not good; and $x=2$ yields $(y,z)=(\frac 13, \frac 23)$ or $(\frac 23, \frac 13)$.
Now suppose $x<0, w=-x, w \in \mathbb N$. Your discriminant becomes $$(5w+1)(w+3)(5w^2+12w+3)=(5w^2+16w+3)(5w^2+12w+3)\\=(5w^2+14w+3)^2-(2w)^2$$
So we have a Pythagorean triple. There are only two possibilities:
Case 1: $5w^2+14w+3=k(m^2+n^2), 2w=2kmn,m>n, (m,n)=1, $ and not both odd.
But
$$k(m^2+n^2)>5w^2+14w=5k^2m^2n^2+14kmn>5k^2m^2+14kn^2>km^2+kn^2, \Rightarrow \Leftarrow.$$
Case 2: $5w^2+14w+3=k(m^2+n^2), 2w=k(m^2-n^2),m>n, (m,n)=1, $ and not both odd.
Again
$$k(m^2+n^2) > 5w^2 = 5 \frac{k^2}{4} (m+n)^2(m-n)^2 > k^2(m+n)^2 > k(m^2+n^2), \Rightarrow \Leftarrow.$$
Therefore the only solutions are $(2,\frac 13, \frac 23), (2, \frac 23, \frac 13).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The 'ratio' of a 2x2 matrix Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$
to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.
My instinct is to go with a proof by induction, but I really can't see a way to prove this.
| Theorem
There are $x_n$ such that
$$ A_n :=
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
= A^n
$$
Proof by induction
For $n=1$ it is trivial: $x_1 := 1$.
Let the theorem be true for a $n$.
Then on the one hand
$$ A^{n+1} = A^n
\cdot
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
\cdot
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
aa_n + cbx_n & b(a_n + dx_n) \\
c (d_n + ax_n) & dd_n + bcx_n
\end{pmatrix}.
$$
On the other hand
$$ A^{n+1} =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\cdot
A^n
=
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\cdot
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
=
\begin{pmatrix}
aa_n + cbx_n & b(d_n + ax_n) \\
c (a_n + dx_n) & dd_n + bcx_n
\end{pmatrix}.
$$
Thus $a_n + dx_n = d_n + ax_n =: x_{n+1}$. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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I cannot solve these separable differential equations There is no explanation in the book on how to solve these and I can't find any help online. Step-wise calculators also don't make sense.
(1) $y'=(y^2-1)x, \; y(0)=0$,
(2) $xy'=y^2-2y, \; y(1)=1 \; x\geq0$.
I'll post the question and write my attempts underneath straight after
| For the first one, note that you can re-write
\begin{align}
y' &= (y+1)(y-1)x \\
\implies \displaystyle \int \frac{dy}{(y+1)(y-1)} &= \int x dx
\end{align}
Whence
\begin{align}
\frac{1}{2}\displaystyle \int \frac{1}{y-1} - \frac{1}{y+1}&= \int x dx
\end{align}
Thus
$$\ln\frac{y-1}{y+1} = x^2+C$$
Then high school algebra can assist in re-arranging to make $y$ the subject.
For the second DE this can be written on the right as $y(y-2)$ and upon finding partial fractions
\begin{align}
\displaystyle \int \frac{dy}{y(y-2)} &= \int \frac{1}{x}dx \\
\implies \frac{1}{2}\displaystyle \int \frac{1}{y-2}-\frac{1}{y} &= \int \frac{1}{x}dx \\
\implies \ln(y-2)-\ln y &= \ln x + C \\
\implies \ln \frac {y-2}{y} &= \ln x+C\\
\implies \ln \left( 1-\frac{2}{y} \right) &= \ln x +C
\end{align}
and the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.
| Using the spherical coordinates (with the convention you used) we have, $$yz= r^2\sin\theta\cos\theta\sin\varphi,$$ $$x^2+y^2+z^2 = r^2$$ $$\,dx\,dy\,dz = r^2\sin\theta\,dr\,d\varphi\,d\theta.$$ Also, the region $E$ is bounded by $x^2+y^2+z^2-x = 0$, which is a sphere centered at $(\tfrac{1}{2},0,0)$ with radius $\tfrac{1}{2}$.
The inside of this sphere is given by $x^2+y^2+z^2-x \le 0$, i.e. $r^2-r\sin\theta\cos\varphi \le 0$. Since $r \ge 0$, this simplifies to $0 \le r \le \sin\theta \cos\varphi$ or $r = 0$ (which is just one point).
In order for $0 \le r \le \sin\theta\cos\varphi$ to be a nontrivial range, we need $\sin\theta\cos\varphi \ge 0$. Since $\sin\theta \ge 0$ for all $\theta \in [0,\pi]$, we only need to restrict the bounds of $\varphi$ to be over the range where $\cos\varphi \ge 0$. We can do this either by $\varphi \in [0,\tfrac{\pi}{2}] \cup [\tfrac{3\pi}{2},2\pi]$ or $\varphi \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. The second is simpler, so we will use that instead.
So the integral becomes $$\iiint\limits_{E}\dfrac{yz \,dx\,dy\,dz}{x^2+y^2+z^2} = \int_{-\pi/2}^{\pi/2}\int_{0}^{\pi}\int_{0}^{\sin\theta\cos\varphi}r^2\sin^2\theta\cos\theta\sin\varphi\,dr\,d\theta\,d\varphi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the determinant of a matrix with $0$s on the diagonal, $1$s in the first row and column, and $x$ elsewhere. How do you attempt to solve the deterimnant of
$$D_n=\begin{vmatrix}
0 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & x & x & \cdots & x \\
1 & x & 0 & x & \cdots & x \\
1 & x & x & 0 & \ddots & \vdots \\
\vdots & \vdots & \vdots & x & \ddots & x \\
1 & x & x & \cdots & x & 0
\end{vmatrix}$$
My approach is to try and create a similar matrix with finite dimensions (something like a $4\times 4$) and then try and simplify it, but I always get stuck in the process of simplifying...
In general, how do you approach these types of questions?
| We have $$x^2D_n = \begin{vmatrix}
0 & x & x & \cdots & x \\
x & 0 & x & \cdots & x \\
x & x & 0 & \cdots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x & x & x & \cdots & 0
\end{vmatrix} = x^n\begin{vmatrix}
0 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 0
\end{vmatrix} = x^n\det(J-I)$$
where $J$ is the matrix with all entries equal to $1$. We know that $J$ has eigenvalues $0,0, \ldots, 0, n$ with multiplicity. Indeed, the respective basis of eigenvectors are $$e_2-e_1, e_3-e_2,\ldots, e_n-e_{n-1}, e_1+e_2+\cdots+e_n.$$The matrix $J-I$ therefore has eigenvalues $-1,-1, \ldots, -1, n-1$ with multiplicity. Determinant is the product of eigenvalues so $$\det(J-I)=(-1)^{n-1}(n-1)$$
We conclude $D_n = (-1)^{n-1}(n-1)x^{n-2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &= 1 \\
a^2 + \frac{1}{2a^2} &= 1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &= 1 \\
1 \pm i + 4b^2 &= 2 \\
\pm i + 4b^2 &= 1 \\
4b^2 &= 1 \pm i \\
2b &= \sqrt{1 \pm i} \\
b &= \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}
Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?
| Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.
You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.
In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a root of the polynomial $p \in \mathbb Q(\sqrt 2) [x]$
$$p(x) = x^2 - (1 +\sqrt 2).$$
And $p$ is irreducible over $\mathbb Q(\sqrt 2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Steps for Velocity given vector I am asked to find velocity for the space curve
$$r(t) = \langle \cos(3t), 1-\sin(3t), 3\sin(3t)+\cos(3t)\rangle$$
at $t=\pi / 4$
I have differentiated each component and substituted $t$ in said components. Are those the correct steps to find the answers?
I got this as my answer:
$$\langle -3 \sqrt{2}, 3 \sqrt{2} , -9 \sqrt{2} - 3 \sqrt{2} \rangle.$$
| The steps you stated you did are the correct ones. In particular, differentiating each component leads to
$$\frac{dr(t)}{dt} = \langle -3\sin(3t), -3\cos(3t), 9\cos(3t) - 3\sin(3t) \rangle \tag{1}\label{eq1A}$$
Next, $t = \frac{\pi}{4}$ means $3t = \frac{3\pi}{4}$, with $\sin(3t) = \frac{1}{\sqrt{2}}$ and $\cos(3t) = -\frac{1}{\sqrt{2}}$. Thus, \eqref{eq1A} gives
$$\left\langle -\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}, -\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} \right\rangle \tag{2}\label{eq2A}$$
However, your answer has the result in \eqref{eq2A} multiplied by $2$. Also, note that the $z$ component can be simplified to
$$-\frac{9}{\sqrt{2}} - \frac{3}{\sqrt{2}} = -6\sqrt{2} \tag{3}\label{eq3A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding specific interval The question is to find $[a,b]$ such that $$\forall x,y : x\in [a,\frac {a+b}{2}],y\in[\frac{a+b}{2},b]\Rightarrow x+y,xy \in [a,b]$$
I tried like below
$$a \leq x \leq\frac {a+b}{2}\\
\frac {a+b}{2} \leq y \leq b\\
\to \frac {3a+b}{2} \leq x+y \leq \frac {a+3b}{2}\\\to
a\leq \frac {3a+b}{2}\leq \frac {a+3b}{2}\leq b$$ this all get $a\leq b $
and $$a.\frac {a+b}{2} \leq xy \leq b.\frac {a+b}{2}\\
a\leq a.\frac {a+b}{2} \leq b.\frac {a+b}{2}\leq b$$ for a>o will be $$2a\leq a^2+ab \to 2\leq a+b\\
b.\frac {a+b}{2} \leq b \to a+b\leq 2$$
this mean $a+b=2$
first :Am I on right track?
second: I got stuck on this,because It revert to trivially $a<b$
Please guide me to find the solution? or give an Idea to take over
thanks in advance
| $$a \le \frac{3a+b}2 \iff 0 \le a+b$$
$$\frac{a+3b}{2}\le b \iff a + b \le 0$$
Hence we have $a+b=0$.
$$a=-b$$
$|x| \le b$, $|y| \le b$, we want $|xy| \le b$
$$b^2 \le b$$
$$b^2-b \le 0$$
$$b(b-1) \le 0$$
Hence,
$$0 \le b \le 1, a = -b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$
Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
{\rm d}t=\dfrac{\pi }{\sqrt{2}}$.
We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?
| I found a possible method here which is straightforward. I will just carry it here.
\begin{aligned}
\lim _{x \rightarrow 1} u(x) &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t} \cdot \frac{1+t}{(t-1)^{2}}}{\sqrt{\frac{1+x}{(x-1)^{2}}-\frac{1+t}{(t-1)^{2}}}} d t \\
&=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t}(1+t)(1-x)}{(1-t) \sqrt{(x-t)(3-t-x-x t)}} d t \\
&=\lim _{y \rightarrow 0} \int_{y}^{1} \frac{(2-s) \sqrt{1-s} y}{s \sqrt{(s-y)(2 s+2 y-s y)}} d s \quad(\text { Let } s=1-t, y=1-x)
\\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y} \cdot y \cdot y}{u y \sqrt{(u y-y)\left(2 u y+2 y-2 u y^{2}\right)}} d u \quad(\text { Let } s=u y) \\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y}}{u \sqrt{(u-1)(2 u+2-u y)}} d u\\
&=\sqrt{2} \int_{1}^{+\infty} \frac{1}{u \sqrt{u^{2}-1}} d u \\
&=\sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-m^{2}}} d m \quad\left(\text { Let } m=\frac{1}{u}\right) \\
&=\frac{\sqrt{2}}{2} \pi
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?
| Note
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} dx
= \int_{0}^{1} \frac{1+\frac1{x^{2}}}{x^{2}+\frac1{x^2}} dx
=\int_0^1 \frac {d(x-\frac1x)}{(x-\frac1x)^2+2}
= \frac\pi{2\sqrt2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can such a matrix be singular? Let $A$ be an $n\times n$ matrix that satisfies
i. All diagonal entries of $A$ are positive, even integers,
ii. All non-diagonal entries of $A$ are positive, odd integers,
iii. $A$ is symmetric: $A_{ij}=A_{ji}$.
iv. If $i\neq j$, then $A_{ij}<A_{ii}$ and $A_{ij}<A_{jj}$.
Question: can such $A$ be singular (i.e. $|A|=0$)?
I can prove the answer is NO, when $n$ is even, and without using condition iv: reduce $A$ mod $2$ we get a matrix with zero diagonal and all other entries $1$; then we can compute directly that the determinant of such matrix is odd, in particular not zero. When $n$ is odd this argument says $|A|$ is even, I don't know if it can be zero, and it seems condition iv will be important here.
I encountered it when I was trying to solve a question in this post.
| I found:
$$
\begin{pmatrix}
15 & 0 \\
0 & 15 \\
11 & 11
\end{pmatrix}
\cdot
\begin{pmatrix}
14 & 1 & 11 \\
1 & 14 & 11 \\
\end{pmatrix}
=
\begin{pmatrix}
210 & 15 & 165 \\
15 & 210 & 165 \\
165 & 165 & 242
\end{pmatrix}
$$
The general idea is to start with a product of rank-$(n-1)$ matrices and to get the configuration of odd and even numbers. This looks like the following, assuming that $1$ represents odd and $0$ represents even:
$$
\begin{pmatrix}
1 & 0 & \cdots & 0 & 0 \\
0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
1 & 1 & \cdots & 1 & 1 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
0 & 1 & \cdots & 1 & 1 & 1 \\
1 & 0 & \cdots & 1 & 1 & 1\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
1 & 1 & \cdots & 0 & 1 & 1\\
1 & 1 & \cdots & 1 & 0 & 1
\end{pmatrix}
$$
Then we observe that the row vectors of the first matrix should point approximately in the same direction as the column vectors of the second matrix, all the vectors should have approximately the same length, and two row vectors of the first matrix (two column vectors of the second matrix) should enclose an angle that is as close as possible to the right angle - in order to keep the non-diagonal elements as small as possible.
In the end, we have to ensure that the resulting matrix is symmetric.
So we can use the following
$$
\begin{pmatrix}
a+n-2 & 0 & \cdots & 0 & 0 \\
0 & a+n-2 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & a+n-2 & 0 \\
0 & 0 & \cdots & 0 & a+n-2 \\
b & b & \cdots & b & b \\
\end{pmatrix}
\cdot
\begin{pmatrix}
a & 1 & \cdots & 1 & 1 & b \\
1 & a & \cdots & 1 & 1 & b\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
1 & 1 & \cdots & a & 1 & b\\
1 & 1 & \cdots & 1 & a & b
\end{pmatrix}
$$
and adjust $a$ and $b$ with an even $a$ and an odd $b$ such that all conditions are fulfilled.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Why do diagonal entries of odd potencies of adjacency matrices of forests sum up to zero? I came across the statement that the trace (sum of diagonal entries) of an odd potency of an adjacency matrix of a forest is always zero. I have tried out a few options and it seems to be correct for forests or trees but not for other graphs, e.g.:
$A = \begin{bmatrix} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}$ and $A^3 = \begin{bmatrix} 0 & 4 & 0 & 3 & 3 \\ 4 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 1 & 1 \\ 3 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 0 & 0 \end{bmatrix}$
While if I add one edge to create a circle:
$B = \begin{bmatrix} 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ and $B^3 = \begin{bmatrix} 2 & 4 & 4 & 3 & 0 \\ 4 & 2 & 3 & 1 & 0 \\ 4 & 3 & 2 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$
Can someone explain to me why this is the case?
| If $A$ is the adjacency matrix of a graph, then $A^n_{ij}$ counts the number of paths of length $n$ from vertex $i$ to vertex $j$. If you are in a forest, because there are no cycles, every step along a path is either one step closer or one step father away from any particular vertex. Because of this, if we start at vertex $i$, we will alternate between an even distance away and an odd distance away. In particular, at odd times, we are an odd distance away, and therefore NOT back at our starting point. So all the diagonal entries of $A^n$ are $0$ when $n$ is odd.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other. Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.
I'm not sure how to solve this at all, maybe Vieta's formulas for quadratics can help.
| One can certainly use Vieta's formulas, but I'll give a self-contained solution. (My argument essentially reproves them though.)
Suppose $r_1, r_2$ are the two roots of $f$. By the Fundamental Theorem of Algebra (and using the fact that $f$ is monic--i.e., the leading coefficient of $x^2$ is $1$), we may write
$$f(x) = x^2+bx+9 = (x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2.$$
Since $-r_1$ and $-r_2$ are the roots of $g$ and $g$ is also monic, we have
$$g(x) = x^2+ax+c = (x+r_1)(x+r_2) = x^2+(r_1+r_2)x+r_1r_2.$$
Comparing these two equations, we see $c=9$ and $a=-b$. Thus
$$h(x) = f(x)+g(x) = (x^2+bx+9)+(x^2-bx+9) = 2x^2+18.$$
So $h$ has two imaginary roots: $\pm 3i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$ I don’t know how to begin solving. Can I get a hint?
My failed attempt
Let $t =\tan (x/2)$
Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve
| Continuing from your method is messy but possible:
Let $\frac{(3-t^2)}{(3+t^2)^2 } = \frac{A}{(3+t^2)} + \frac{B}{(3+t^2)^2}$. Then $3-t^2 = (3+t^2)A + B$. When $t = 0, 3 = 3A + B$, and when $t = 1, 2=4A+B$, so $A = -1, B= 6$. Thus:
$$I = -2 \int_0^1 \frac{1}{3+t^2} \ dt + 12 \int_0^1 \frac{1}{(3+t^2)^2} \ dt $$
For the second integral, substitute $t = \sqrt{3} \tan u, dt = \sqrt{3} \sec^2 u$. Then $(3+t^2)^2$ $ =(3(1 + \tan^2 u))^2 = 9 \sec^4 u$, and so:
$$I_2 = \frac{\sqrt{3}}{9} \int_0^1 \cos^2u \ du = \frac{\sqrt3}{18} \int_0^1 1+\cos2u \ du= \left[\frac{\sqrt3}{18} (u+\frac{1}{2} \sin 2u)\right]_0^1 = \left[\frac{\sqrt3}{18} (\tan^{-1} \frac{t}{\sqrt3}+\frac{1}{2} \sin(2(\tan^{-1}\frac{t}{\sqrt3}))\right]_0^1$$
and therefore:
$$I = -2 \left[\frac{1}{\sqrt 3} \tan^{-1} \frac{t}{\sqrt3}\right]_0^1+12\left[\frac{\sqrt3}{18} (\tan^{-1} \frac{t}{\sqrt3}+\frac{1}{2} \sin(2(\tan^{-1}\frac{t}{\sqrt3}))\right]_0^1$$
$$=-2(\frac{1}{6\sqrt3} \pi - 0) + 12 \cdot \frac{\sqrt{3}}{18}(\frac{\pi}{6}+\frac{\sqrt3}{4}-0) = \frac{1}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the Expectation and Variance of 4 Independent Dice We were given this seatwork:
With four independent dice:
a) the expected value of the sum of the rolls,
b) the expected value of the product of the rolls, and
c) the variance of the sum of the rolls .
I was able to answer a and b, but I don't know how to get the variance. Here's my attempt:
\begin{align*}
S = \dfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5
\end{align*}
This is the expectation of one die. For the expectation of four dice, we could assume the expectation of the sum four dice is equal to the sum of the expectations of a die:
\begin{align*}
&= S + S + S + S\\
&= 4S\\
&= 4(3.5)\\
&= 14
\end{align*}
Similarly, we could also do this for the products. The expected product of four dice rolls is:
\begin{align*}
&= S \cdot S \cdot S \cdot S\\
&= S^4\\
&= 3.5^4\\
&= 150.06
\end{align*}
Are these assumptions correct?
| You have found the expected value of a dice roll to be $E(X)=3.5$, thus the variance of one roll is equal to:
\begin{align}
\operatorname{Var}(X) &= \sum_{k=1}^6 \frac{1}{6}\left(k - 3.5\right)^2 \\
&= \frac{35}{12} \approx 2.92.
\end{align}
Since the rolls of the 4 dies are independent the variance of their sum is equal to $4 \cdot \frac{35}{12}\approx 11.66$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.
Case 2: If $b^2 - 4ac$ is not a perfect square, $b = \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$
Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)
So the equation is not quadratic and $c = 0$.
$a + bp^\frac{1}{3} = 0$
$-\dfrac{a}{b} = p^\frac{1}{3}$
This is a contradiction and hence $b = 0$ and $a = 0$
Is there any other way to solve this?
| Let $K= \mathbb{Q}(p^{\frac{1}{3}}) \cong \mathbb{Q}[x]/(x^3-p).$ We have the following:
*
*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{1}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{1}{3}}$ is $x^3-p.$
*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{2}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{2}{3}}$ is $x^3-p^2.$
*$\text{Tr}_{K/\mathbb{Q}} (a) = 3a$, for all $a \in \mathbb{Q}.$
Now applying the trace to your equation we get
\begin{align*}
\text{Tr}_{K/\mathbb{Q}}(a + bp^\frac{1}{3} + cp^\frac{2}{3}) &= \text{Tr}_{K/\mathbb{Q}}(a)+ b \text{Tr}_{K/\mathbb{Q}}(p^\frac{1}{3} ) + c \text{Tr}_{K/\mathbb{Q}}(p^\frac{2}{3})\\
&= \text{Tr}_{K/\mathbb{Q}}(a)+0+0 = 3a = 0= \text{Tr}_{K/\mathbb{Q}}(0),\\
\end{align*}
thus $a=0.$ Next multiply your equation by $p^{\frac{1}{3}}$, apply the trace, and conclude that $c=0.$ Repeat.
Edit:
As Paramanand Singh correctly points out the problem reduces to showing that $f(x)= x^3-p$ is the minimal polynomial of $p^{\frac{1}{3}},$ which I assumed. However, this follows directly from Eisenstein's Criterion and the fact that $f(p^{\frac{1}{3}})=0.$ In light of this information, the polynomial given by the OP is of degree $2$, thus must be the zero polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
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Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.
I am stuck on page 7, particularly the following steps:
$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!
| Step 1 (in the question): $\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
Step 2: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
In step 1, $-\frac{1}{2}, -\frac{3}{2}... -\frac{2n-3}{2}$ has $(n-1)$ terms with negative signs as $2n - 3 = 1 + 2 (n-2)$ and there is $-\frac{1}{2}$ outside curly brackets too. So the negative sign is taken out in step 2 as $(-1)^n \times (-1)^n = (-1)^{2n} = 1$.
Also note there are $n$ $2's$ in the denominator of $\frac{1}{2} (-\frac{1}{2}), (-\frac{3}{2})... (-\frac{2n-3}{2})$
Step 3: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
i) Writing $4^n$ as $2^{2n}$, ii) In step 2, we multiplied numerator and denominator by $(n-1)!$. We now expand $(n-1)!$ in the numerator as $1 \times 2 \times 3 ...\times (n-1)$ and write in order.
Step 4: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
In this step we cancel $2^{2n-2}$ in the numerator with $2^{n-1}$ in denominator and we also use $2^{n-1}$ to double each term of $(n-1)!$ from $1 \times 2 \times 3 ..(n-1)$ to $2 \times 4 \times 6 ... \times (2n-2)$ which will lead to numerator as $(2n-2)!$
| {
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Integral of $1 / \sqrt {(a x + b) (p x + q)}$ I am working on this:
$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x$
valid where $(a x + b)(p x + q) > 0$.
I am specifically interested in the case where $\dfrac {b p - a q} p < 0$.
I do this:
Let $u = \sqrt {a x + b}$
$\leadsto x = \dfrac {u^2 - b} a$
$\leadsto \sqrt {p x + q} = \sqrt {p \left({\dfrac {u^2 - b} a}\right) + q}$
$= \sqrt {\dfrac {p \left({u^2 - b}\right) + a q} a}$
$= \sqrt {\dfrac {p u^2 - b p + a q} a}$
$= \sqrt {\dfrac p a} \sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) }$
Then by a standard result:
$\displaystyle \int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } = \int \frac {2 u \mathrm d u} {a \sqrt {\frac p a} \sqrt {u^2 - \left({\frac {b p - a q} p}\right) } u}$
$= \dfrac 2 {\sqrt {a p} } \int \dfrac {\mathrm d u} {\sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) } }$
Now let $\dfrac {b p - a q} p < 0$.
Let $c^2 = -\dfrac {b p - a q} p$.
This gives us:
$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x = \frac 2 {\sqrt {a p} } \int \frac {\mathrm d u} {\sqrt {u^2 + c^2} }$
$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} {\dfrac u c} + C$ (standard result)
$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \dfrac {\sqrt {a x + b} } {\sqrt {\dfrac {a q - b p} p} } + C$
(because $\dfrac {b p - a q} p = -\dfrac {a q - b p} p$)
$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p (a x + b) } {a q - b p} } + C$
But when I look in the book (in this case Murray Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), I see this as result $14.120$:
$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x = \dfrac 2 {\sqrt {-a p} } \arctan \sqrt {\dfrac {-p (a x + b) } {b p - a q} } + C$
Can't see where I'm doing it wrong.
| Amplified comment: "Convert an $\arcsin$ to an $\arctan$."
I learned this long ago because certain computer languages had only $\arctan$ as built-in function, and you had to use formulas if you wanted $\arcsin, \arccos$, etc. (BASIC had ATN arctangent, but none of the other "arc" trig functions)
So here it is:
$$
\arcsin t = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right)
$$
Here are the graphs.
Identical
| {
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Proving $\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$ for positive values such that $ab+bc+ca=3$ This problem is from my teacher.
Know that $a,b,c>0$, $ab+bc+ca=3$.
Prove that: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$$
I tried to change the number '$6$' into $2(ab+bc+ca)$.
Then get
$$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{2ab+2bc+2ca}{abc}=\frac{2}{c}+\frac{2}{b}+\frac{2}{a}$$
but i have no idea what to do next.
| By holder: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+3b+c}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1) \cdot (1+1+1)}$$ It hence suffices to prove $$\sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1)(1+1+1)}\le \frac{6}{abc}$$ $$\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})\le \frac{24}{a^3b^3c^3}\tag 1$$ Now let $p=a+b+c,q=ab+bc+ca=3,r=abc$ then we rewrite (1) as $$\frac{q}{r}+\frac{2p}{r}+5p\le \frac{24}{r^3}$$ $$\iff qr^2+2pr^2+5pr^3\le 24$$ Now as $q^2\ge 3pr \iff p\le \frac{3}{r}$ and $q=3$
it suffices to prove $$3r^2+6r+15r^2\le 24$$ which is true as $r\le 1$(by AM-GM)
Note in your original question you had $...a+2b+c$ i think its a typo and should be $a+3b+c$
all the same since $$\sum_{cyc}\sqrt[3]{({\frac{1}{a}+\frac{2}{bc}+a+2b+c})}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})}$$ the same proof applies here here so we are done in either case
| {
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Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.
Solution
We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$
Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$
Hence, the answer is $31a^4$.
The original exam
| Yes, there is a shorter solution that relies on the identity $$(x+y)(x^n + y^n) = (x^{n+1} + y^{n+1}) + xy(x^{n-1} + y^{n-1}),$$ which is easily verified by multiplication. Then if we let $f_n = x^n + y^n$, this may be written $$f_{n+1} = (x+y)f_n - xy f_{n-1} = f_1 f_n - xy f_{n-1}.$$
Next, observe that $f_0 = 2$ for any nonzero choice of $x, y$. In addition, we are given $f_1 = a$. Then $$f_2 = f_1^2 - xy f_0 = a^2 - 2xy,$$ and $$f_3 = f_1 f_2 - xy f_1 = a(a^2 - 2xy - xy) = a(a^2 - 3xy).$$ Since we are also given $f_3 = 10a^3$, it follows that $10a^3 = a(a^2 - 3xy)$ and if $a \ne 0$, we obtain $$xy = -3a^2.$$ This gives us the information we need to compute $$f_4 = f_1 f_3 - xy f_2 = a (10a^3) - (-3a^2) (a^2 - 2(-3a^2)) = 31a^4$$ and we are done. If $a = 0$, then $f_n$ is trivially $0$ for all $n \ge 1$.
Note this method furnishes the more general recursion $$f_{n+1} = a f_n + 3a^2 f_{n-1}, \quad f_0 = 2, \quad f_1 = a,$$ for which the non-recursive solution is $$f_n = \left(\frac{a}{2}\right)^n \left( (1 - \sqrt{13})^n + (1 + \sqrt{13})^n \right).$$
| {
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Simplify $\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}$ The double square roots can be removed by $8+2\sqrt7 = (\sqrt7+1)^2$ and $$4-\sqrt7 = \frac{1}{4}(16-4\sqrt7) = \frac{1}{4}(14+2-2\sqrt2\cdot\sqrt14) = \frac{1}{4}(\sqrt14-\sqrt2)^2$$The expression simplifies to $\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2}$. This problem was given to me as an MCQ, the four options were:
A) $1$,
B) $2$,
C)$32\sqrt3$
D) $36\sqrt3$.
I am not getting any of these options, $$\begin{eqnarray*}\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} &=& \frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} \cdot \frac{\sqrt7+1+\sqrt2}{\sqrt7+1+\sqrt2} \\ &=& \frac{1}{2}\cdot \frac{6\sqrt2+2\sqrt7-2}{6+2\sqrt7} \\ &=& \frac{1}{4}(9\sqrt2-3\sqrt14+4\sqrt7-10) \end{eqnarray*}$$ Is my calculation incorrect ?
| Let $\sqrt{8+2\sqrt{7}}=\sqrt{x}+\sqrt{y} \implies x+y+2\sqrt{xy}, \implies x+y=8, xy=7 \implies x=1,y=7$ and let $\sqrt{4-\sqrt{7}}=\sqrt{p}-\sqrt{q} \implies p+q=4, pq=7/4 \implies p=7/2, q=1/2.$
Then
$$F=\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}= \frac{\sqrt{7/2}-\sqrt{1/2}}{1+\sqrt{7}-\sqrt{2}}$$
Something is amiss!
| {
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Complete factorisation of $x^8-2x^4\cos (4\theta)+1$ with roots of unity
Factorize completely $x^8-2x^4\cos (4\theta)+1$ using complex numbers and $n$-th root of unity.
The answer given is
$$\prod_{r=0}^3 \left(x^2-2x\cos\left(\theta+\frac{r\pi}{2}\right)+1\right)\,.$$
Can anyone please help . I just want to know how to tackle this . I just want the conceptual trick to solve such sum.
| With the 4th root of unity $e^{\pm i\frac{a+2\pi r}4} $ for $x^4 - e^{\pm ia}=0$, factorize as follows
\begin{align}
x^8-2x^4\cos (4\theta)+1
=&\> (x^4-e^{i 4\theta})(x^4-e^{-i 4\theta}) \\
= &\prod_{r=0}^{3}(x-e^{i\frac{4\theta+2\pi r}4})
\prod_{r=0}^{3}(x-e^{-i\frac{4\theta+2\pi r}4}) \\
= &\prod_{r=0}^{3}(x-e^{i(\theta+ \frac{\pi r}2)})
(x-e^{-i(\theta+ \frac{i\pi r}2)}) \\
= &\prod_{r=0}^{3}(x²-2x\cos(\theta+\frac{\pi r}{2})+1)
\end{align}
| {
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"source": "stackexchange",
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Complex and real matrix for eigenvalues This is a short one but: Consider the real matrix $\alpha = \begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}$ and let $\beta \in M_n (\mathbb{C})$ be the same matrix but considered as a complex matrix.
Problem
with use of a computer program calculate the eigenvalues of the two matrices. i.e $\alpha$ over the real numbers and $\beta$ over the complex numbers.
Question
What does this actually mean. They are the same but over different numbers? Is it just the same matrix and thus the same result in this problem but got new rules in different problems where the fact that it is over complex numbers makes a difference? Does it make a difference here?
What does the matrix actually look like - I am thinking it is the same but the fact it is over complex numbers do not matter here.
| The characteristic polynomial of your matrix is
$$-x^3+3x^2-x+3 = -(x-3)(x^2+1)$$
The real root of this polynomial is $3$ which is an eigenvalue for $\alpha$ and $\beta$:
$$\begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}
\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix} = 3\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix}$$
Non-real roots of the polynomial are $\pm i$, which are eigenvalues only for $\beta$:
$$\begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}
\begin{pmatrix}
1+i \\ -2i \\ 2
\end{pmatrix} = i\begin{pmatrix}
1+i \\ -2i \\ 2
\end{pmatrix}, \quad \begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}
\begin{pmatrix}
1-i \\ 2i \\ 2
\end{pmatrix} = -i\begin{pmatrix}
1-i \\ 2i \\ 2
\end{pmatrix}$$
Note that there are no real nonzero vectors $v$ such that $\alpha v = \pm i v$ so they are not eigenvalues of $\alpha$ in this sense.
| {
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How many variables in the multinomial expansion have an exponent that's different from 2? Multinominal: $(a + b + c + d)^{10}$
Question: How many variables in the multinomial expansion have an exponent that's different from 2?
I want to solve it using generating functions.
So we can write: $t_1 + t_2 + t_3 + t_4 = 10$, When $t_i$ cannot use $x^2$.
That equals to the generating function:
$$
(1+x+x^3+x^4...)^4 = (1+x+\frac{x^3}{1-x})^4
$$
But now what? How can I keep expending this till I find the coefficient of $x^{10}$?
| $(1 + x + x^3 + x^4 + ...)$ can be simplified as $[(1 - x)^{-1} - x^2]$.
Now, we had to find coefficient of $x^{10}$ in ${[(1 - x)^{-1} - x^2]}^4$
${[(1 - x)^{-1} - x^2]}^4$ = $\binom{4}{0}$ $[({1 - x}) ^ {-4}]$ - $\binom{4}{1}$ $[({1 - x}) ^ {-3}]$ $x^2$ + $\binom{4}{2}$ $[({1 - x}) ^ {-2}]$ $x^4$ - $\binom{4}{3}$ $[({1 - x}) ^ {-1}]$ $x^6$ + $\binom{4}{4}$ $x^8$
Hence the answer will be $\binom{13}{3}$ - 4 $\binom{10}{2}$ + 6 $\binom{7}{1}$ - 4 $\binom{4}{0}$ i.e. 144
| {
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Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$
Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.
I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.
This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?
| EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" argument. Any feedback on the new proof will be much appreciated.
Note: In your post, you claimed the following:
$x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}.$
I do not see how the last equality follows. If it were true, then we would have $a_k=\dfrac{1}{a_1a_2...a_n}=1$, which is not necessarily the case. But anyways, it is possible to solve the problem using AM-GM and Cauchy-Schwarz.
Firstly, the case where $x_1=x_2=...=x_n=1$ is trivial. We only consider the case where $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$. Note that:
$$\dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_2} + ... + \dfrac{1}{n-1+x_n} \leq 1$$
\begin{align}
& \iff \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_1}\right) + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_2}\right) + ... + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_n}\right) \geq 0 \\
& \iff \dfrac{x_1-1}{n(n-1+x_1)} + \dfrac{x_2-1}{n(n-1+x_2)} + ... + \dfrac{x_n-1}{n(n-1+x_n)} \geq 0 \\
& \iff \dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \geq 0 \ (\because \dfrac{1}{n} > 0 ) \\
\end{align}
By Cauchy-Schwarz,
$$\left(\dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \right) \left[(x_1-1)(n-1+x_1) + ... + (x_n-1)(n-1+x_n) \right]$$
$$\geq \left[(x_1-1)+(x_2-1) + ... + (x_n-1) \right]^2$$
The R.H.S. of the preceding inequality is clearly non-negative. It thus suffices to prove that:
\begin{align}
& (x_1-1)(n-1+x_1) + (x_2-1)(n-1+x_2) + ... + (x_n-1)(n-1+x_n) > 0 \\
& \iff n(x_1-1)+(x_1-1)^2 + n(x_2-1)+(x_2-1)^2 + ... + \ n(x_n-1)+(x_n-1)^2 > 0 \\
& \iff (x_1-1)^2 + ... + (x_n-1)^2 + n\left[(x_1-1) + ... + (x_n-1) \right] > 0. \\
\end{align}
Since $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$, it is obvious that $(x_1-1)^2 + ... + (x_n-1)^2 > 0$. Now, we claim that $n\left[(x_1-1) + ... + (x_n-1) \right] \geq 0 $, which concludes our proof. To see why this is true, note that:
\begin{align}
n\left[(x_1-1) + ... + (x_n-1) \right] & = n(x_1 + ... + x_n -n) \\
& \geq n(n \sqrt[^n]{x_1...x_n} - n) \ (\text{By applying AM-GM inequality})\\
& = n(n-n) \ (\because x_1...x_n=1) \\
&=0.
\end{align}
| {
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"question_score": "8",
"answer_count": 5,
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} |
Evaluate:$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$ Evaluate:$$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$$
where $$\theta_{n}=\frac{\theta}{3^n}$$ and $0<\theta<\pi$
I did try to find relation between $\tan 3x$ and $\tan x$
$$\tan 3x-3\tan x=\frac{8\tan^3x}{1-3\tan^2x}$$
Also $$3\tan 3x-\tan x=\frac{8\tan x}{1-3\tan^2x}$$
so as to create some kind of telescopic action but couldn't split the expression given in the question
| Note that
$$ \frac{\tan x}{3-\tan^2 x} = \frac{\cot x - 3\cot(3x)}{8}. $$
So we have
$$ \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)}
= \frac{1}{8}\left( \frac{\cot \theta_n}{3^n} - \frac{\cot \theta_{n-1}}{3^{n-1}} \right). $$
Summing this from $n = 2$ to $\infty$ yields a telescoping series with the value
$$ \sum_{n=2}^{\infty} \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)}
= \frac{1}{8}\left( \lim_{N\to\infty} \frac{\cot \theta_N}{3^N} - \frac{\cot \theta_1}{3} \right)
= \frac{1}{8}\left( \frac{1}{\theta} - \frac{\cot (\theta/3)}{3} \right). $$
| {
"language": "en",
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Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}
Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!
| $$\frac{n}{n+1}-\frac{n+1}{n}=\frac{n2-(n^2+2n+1)}{n(n+1)}=-\frac{n+(n+1)}{n(n+1)}=-\bigg(\frac{1}{n+1}+\frac{1}{n}\bigg)$$
$$\therefore lim_{n\rightarrow\infty} \frac{n}{n+1}-\frac{n+1}{n}=0$$
| {
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"source": "stackexchange",
"question_score": "4",
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Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$
$$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$
$$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$
$$= (250\sqrt{2009} - 1259)^5$$
However, the answer given is $32$, so there could have been more simplifications.
As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?
| $a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$
That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe
$$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.
| The system
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
is a linear system of two (independent) equations in two variables, and it is readly checked that $x=\sin\theta, y=\cos\theta$ is a solution. Therefore it is the unique solution. Now $x^2+y^2=1$ follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $
So I can express $b=-a(\alpha+\beta)$
$c=a.\alpha\beta$
Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$
I want to know whether there is any different approach other than this method?
Any hint is higly valued. thank you!
| Solve the two equations to find
$$ay^2+by+c=0\implies y=\dfrac{-b\pm\sqrt{b^2-4ca}}{2a}\ \ \ \ (1)$$
$$cax^2-b(c+a)x+(c+a)^2=0\implies x=(c+a)\cdot\left(\dfrac{b\pm\sqrt{b^2-4ca}}{2ca}\right)\
\ \ \ (2)$$
Considering the opposite signs of the roots,
$$\dfrac yx=-\dfrac c{c+a}$$
Considering the same signs of the roots,
$$\dfrac yx=-\dfrac c{c+a}\cdot\dfrac{b+m\sqrt{b^2-4ca}}{b-m\sqrt{b^2-4ca}}\text{ where }m =\pm1$$
So, the ratios are not the same!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| Let $x=\sin{u}$. Substitute $\arcsin{x}=u$ to change variables, notice that $d(\cos^3{u})=-3\sin{u}\cos^2{u}\,du$ and get the integral by parts
$$
\int x\sqrt{1-x^2}\arcsin{x}\,dx =
\int{u}\,(\sin{u}\cos^2{u}\,du) =
-\frac{1}{3}\int{u}\,d(\cos^3{u}) =
\frac{1}{3}\Bigl(\int\cos^3{u}\,du - u\cos^3{u} \Bigr)
$$
After back substitution the last integral looks
$$
\frac{1}{3}\Bigl(\int(1-x^2)\,dx - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) =
\frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr)
$$
Don't forget a constant and get the answer
$$
\int{x\sqrt{1-x^2}\arcsin{x}}\,dx =
\frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) + C
$$
Check the math and good luck!
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$
I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:
$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$
So that:
$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$
Then i tried to separate this last integral by partial fractions, my result on this was:
$$\frac {1} {(x+a)(x^2 + 1)} = \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1}\right)$$
And the integral reduces to:
$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$
Then this last expression evaluates to:
$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$
Then integrating from 0 to 1 with respect to $a$ we will get:
$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$
(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)
But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.
Do i need some other approach? Or did i made any mistake?
Any help is appreciated.
| This would work:
$$I(a) = \int_0^1 \frac{\ln(ax+1)}{x^2+1} dx \\
I’(a) =\int_0^1 \frac{x}{(x^2+1)(ax+1)} dx \\ \overset{\text{partial fractions}}= \\ \frac{-2\ln |ax+1| +\ln(x^2+1)+2a\tan^{-1} x}{2(a^2+1)} \bigg |_0^1 \\ =-\frac{\ln(a+1)}{a^2+1}+\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}
$$
Integrating from $0$ to $1$, $$I(1)-I(0) = -I(1) +\int_0^1 \left(\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}\right) da$$
Hopefully you can finish.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The floored quadratic equation $a \lfloor x^2 \rfloor + b \lfloor x \rfloor + c = 0$ What are the roots of $f(x) = a\lfloor x^2 \rfloor +b\lfloor x \rfloor + c$?
(where $a \neq 0$ and $a,b,c \in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R}$ as usual)
My failed approach:
Let $\delta = x - \lfloor x \rfloor$ and get
\begin{align}
a \lfloor (\lfloor x \rfloor + \delta)^2 \rfloor + b\lfloor x \rfloor + c &= 0, \\
a \lfloor \lfloor x \rfloor^2 + 2\lfloor x \rfloor \delta + \delta^2 \rfloor + b\lfloor x \rfloor + c &= 0, \\
a \lfloor x \rfloor^2 + a\lfloor 2\lfloor x \rfloor \delta + \delta^2 \rfloor + b\lfloor x \rfloor + c &= 0.
\end{align}
From here there are 2 cases:
$$ a \lfloor x \rfloor^2 + a\lfloor 2\lfloor x \rfloor \delta \rfloor + b\lfloor x \rfloor + c = 0 \ \text{or}\ a \lfloor x^2 \rfloor + a\lfloor 2\lfloor x \rfloor \delta \rfloor + b\lfloor x \rfloor + (c + 1) = 0,$$
and then I'm stuck from here. This approach is probably not helpful since even if you get a quadratic equation in terms of $\lfloor x \rfloor$, you still have $\delta$ which is in terms of $x$, the variable of $f(x)$. Which is a difficult.
| Fix $a,b,c$ reals with $a\neq 0$, let us look at some necessary conditions that $$(\dagger)\quad\quad a \lfloor x^2\rfloor + b \lfloor x\rfloor +c =0$$ will have a root:
(1) It is necessary that $a,b,c$ satisfies an integral relation $$aN + b M +c =0$$ for some integers $N,M$. Otherwise we will never have a root. For instance if $a,b$ are both rational and $c$ is irrational, then there is no root to $(\dagger)$. This is a very weak necessary condition to meet. We know that $N$ needs to be roughly the square of $M$, so we will make this precise next on what integers $N,M$ can be.
(2) Suppose $(\dagger)$ has a solution for some $x$. Let's break it into cases.
If $x\ge 0 $, then $n\le x < n+1$ for some non-negative integer $n$. In this case $\lfloor x\rfloor =n$, while $n^2\le x^2 <(n+1)^2 = n^2 + 2n+1$. Hence $\lfloor x^2 \rfloor$ can take on any of $n^2, n^2+1 ,\ldots, n^2 +2n$. Hence for $(\dagger)$ to have non-negative $x$ solutions, we need to satisfy
$$
(\triangle)\quad\quad a (n^2 + k) + bn +c =0
$$
for some non-negative integer $n$ and $k=0,1,\ldots 2n$.
Once we have such $n$ and $k$, then we can work out what $x$ is.
We can analyze the scenario for a negative $x$ solution similarly.
If there is a negative $x$ solution, then $n\le x <n+1$ for some negative integer $n$. Note $(n+1)^2 < x^2 \le n^2$ as $x$ and $n$ are negative. So $\lfloor x \rfloor=n$ and $\lfloor n^2 \rfloor$ can take on any of $n^2 + 2n+1,\ldots ,n^2$ (again, note $n$ is negative).
Hence for $(\dagger)$ to have non-negative $x$ solutions, we need to satisfy
$$
(\triangle\triangle)\quad\quad a (n^2 - k) + bn +c =0
$$
for some negative integer $n$ and $k=0,1,\ldots -1-2n$.
Remark. Since we can actually solve for the real roots to $at^2+bt+c=0$, that will give you an approximate search range for such $n$, and you can brute force test out the relevant $(n,k)$ pairs that will satisfy $(\triangle)$ or $(\triangle\triangle)$. And then you can recover your $x$ values from $(n,k)$. (Note each $(n,k)$ pair will give you an interval of $x$ values.)
| {
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"timestamp": "2023-03-29T00:00:00",
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The common difference is equal to the common ratio. Four numbers are in A.P. The first, the second and the fourth are in G.P. Find the numbers if the common difference is equal to the common ratio.
Let the terms of the A.P. be $a_1,a_1+d,a_1+2d,a_1+3d$ and the terms of the G.P. will be $a_1,a_1+d,a_1+3d.$ We know that $d=q.$ I am not sure what to do from here.
| Let the four numbers be $a_1, a_2, a_3, a_4$. The common difference $d$ will make the terms equal to $a_1, a_1+d, a_1+2d, a_1+3d$, and the terms of the geometric sequence should be $a_1, a_1+d, a_1+3d$ as you correctly mentioned.
Since $d=r$, $a_1, a_1r, a_1r^2$ are also the terms of the geometric sequence.
So we have $a_1r=a_1+d$, $a_1r^2=a_1+3d$. Solving the first equation for $r$ gives $r=1+\frac{d}{a_1}$. Sustituting $r$ into the second equation gives $a_1(1+\frac{d}{a_1})^2=a_1+3d\Rightarrow a_1(1+\frac{2d}{a_1}+\frac{d^2}{a_1^2})=a_1+3d\Rightarrow a_1+2d+\frac{d^2}{a_1}=a_1+3d$
Simlifying gives $d^2=a_1d\Rightarrow d=0,a_1$.
A common difference of $0$ will give a common ratio of $1$ (since all $4$ terms are the same), so that's an extraneous solution.
The only other option is $d=a_1$.
In that case, substituting it back into $r=1+\frac{d}{a_1}$ gives $r=2$.
Therefore, since $d=r$, $d=r=a_1=2$, and so your only sequence is $2, 4, 6, 8$
| {
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Is there any way of calculating this long formula? $$\left\{\frac{-x^4+y^4-y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}\left\{1+\left\{\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}+\left\{\frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}^{\:}\left\{1+\left\{\frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}-2\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{y^2+x^2}\sqrt{y^2+x^2-1}}$$
I'm sorry if this question is not proper here, but is there any way(website) to calculate this formula? In principle, this should be zero, but when I do it by hand, it doesn't seem like it. And I want to verify it. Thanks!
| Your question is equivalent to asking whether the last term (without the minus sign) and the sum of the first two terms are equal. Expressing these two in the format of one big fraction, the denominators on both sides of the equation are the same: $(x^2+y^2)^{7/2} (x^2+y^2-1)^{7/2}$, therefore comparing the numerators suffices. Make the substitution $p=x^2$ and $q=y^2$ for simplification.
The numerator on the L.H.S. is
$
(-p^2+q^2-q)\big[ (p+q)^2 (p+q-1)^2 + q^2 \big] +(-q^2+p^2-p)\big[ (p+q)^2 (p+q-1)^2 + p^2 \big]
$
which can be simplified into
$-p^3-q^3-(p+q)^3(p+q-1)^2$.
The numerator on the R.H.S. is $2pq(p^2-q^2-p)$.
Notice how all terms when expanded on the R.H.S. must be in the form of $p^m q^n$ where $m,n \in \mathbb{Z} \neq 0$. But on the L.H.S., there are at least a $-p^5$ and $-q^5$ term.
| {
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Convergence of $\int_0^n f(x)\, dx - [\frac{1}{2} f(0) + f(1) +\dots + f(n-1) + \frac{1}{2} f(n)]$ If $f(x)$ is increasing, strictly concave, and twice continuously differentiable for $x > 0$, does the difference
$$
d_n = \int_0^n f(x)\, dx - [\tfrac{1}{2} f(0) + f(1) + f(2) +\dots + f(n-1) + \tfrac{1}{2} f(n)]
$$
converge to a finite positive limit when $n\to +\infty$?
The error form provided by the trapezoidal rule is not helpful since it just asserts that $d_n = -n f''(\xi_n)/12\geqslant 0$, for some $0 < \xi_n < n$, or that
$$
d_n\leqslant -\tfrac{1}{12}[f''(\xi_1) + f''(\xi_2) +\dots + f''(\xi_n)]
$$
with $k-1 < \xi_k < k$. Since $f(x)$ is increasing and concave, then $f''(x)$ is negative, bounded, and increasing for $x > 0$, but then again, how to assure the converge of $\sum_{k=1}^n f''(\xi_n)$ when $n\to +\infty$?
| Consider
$$
f(x)=-\frac{1}{(x+1)(x+2)}.
$$
Clearly $f$ is increasing and concave in $[0,\infty)$.
Then
$$
\frac{1}{2}f(0)+f(1)+\cdots+f(n-1)+\frac{1}{2}f(n)=\cdots=-\frac{3}{4}+\frac{1}{n+1}-\frac{1}{2(n+1)(n+2)}\\=-\frac{3}{4}+\frac{1}{n}+{\mathcal O}(n^{-2})
$$
while
$$
\int_0^n f(x)\,dx=\int_0^n \frac{dx}{x+2}-\int_0^n\frac{dx}{x+1}=\log(n+2)-\log 2-\log(n+1)\\=-\log 2-\log\Big(1+\frac{1}{n+1}\Big)\\=-\log 2-\frac{1}{n}+{\mathcal O}(n^{-2})
$$
Thus
$$
d_n\to -\frac{1}{2}+\log 2>0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.
Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$
I did this question a while ago and the answer is correct. However, when I started revising this topic again, I can't seem to figure out how I went from the first step $\left( \text{for this integral}\int \log _{3x}\left(x^2\right)dx\right)$ to the second step. Here's my attempt:
\begin{align}
\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx &= {\int{\frac{2}{\sin{\left(2 x \right)}} d x}} \\
&= {\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}} \\
&= {\int{\frac{\sec^{2}{\left(x \right)}}{\tan{\left(x \right)}} d x}} \\
&\stackrel{{u = \tan{\left(x\right)}}}{=} {\int{\frac{1}{u} d u}} \\
&= {\ln{\left(u \right)}} + c \\
&= \ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}+
\end{align}
\begin{align}
\int \log _{3x}\left(x^2\right)dx &= x\log _{3x}\left(x^2\right)-\int \frac{2\ln \left(3\right)}{\ln ^2\left(3x\right)}dx \\
&= x\log _{3x}\left(x^2\right)-\left(-\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}+\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)\right) \\
&= 2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right) + c
\end{align}
Therefore:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx = \left(\ln \left|\tan \left(x\right)\right|+c\right)\left(2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)+c\right)$$
| I think you are asking to verify that
$$\frac{\tan \left(x\right)}{\sin ^2\left(x\right)}.$$ RIGHT?
if so, $$\frac{\sin(x)/\cos(x)}{\sin ^2\left(x\right)}=\frac{1}{\sin(x)\cos(x)}=\frac{2}{2 \sin(x)\cos(x)}=\frac{2}{\sin(2x)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$ The following question I read in a book, but the book does not give proof. I doubt the correctness of the result
let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$
I think use binomial theorem
it maybe show $$\binom{2p^2}{k}\equiv 0,\pmod {p^2},k=1,2,\cdots,p^2-1,p^2+1,\cdots,2p^2-1\tag{2}$$
but other hand I think (2) is not right.becuase let $p=5,k=5$
we have $$\binom{2p^2}{k}=\binom{50}{5}=\dfrac{50\cdot 49\cdot 48\cdot 47\cdot 46}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ne 0\pmod {25}$$,so I think $(1)$ is not right?
| Let $p = 5$. For any integer $k$, we have
$$\begin{equation}\begin{aligned}
& ((5k + 1)(5k + 2))((5k + 3)(5k + 4)) \\
& \equiv (25k^2 + 15k + 2)(25k^2 + 35k + 12) \\
& \equiv (15k + 2)(10k + 12) \\
& \equiv 150k^2 + 200k + 24 \\
& \equiv -1 \pmod{25}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Using this for each group of a product of $4$ consecutive integers between multiples of $5$, we get
$$\begin{equation}\begin{aligned}
\binom{50}{25} & \equiv \frac{50(-1)45(-1)40(-1)35(-1)30(-1)}{25(-1)20(-1)15(-1)10(-1)5(-1)} \\
& \equiv \frac{2(45)(2)(35)(2)}{10(5)} \\
& \equiv 9(2)(7)(2) \\
& \equiv 2 \pmod{25}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $\varphi(25) = 20$, with $x = 1$, your congruence equation becomes
$$\begin{equation}\begin{aligned}
2^{50} \equiv 1 + 2 + 1 \pmod{25} \\
2^{10} \equiv 4 \pmod{25} \\
1024 \equiv 4 \pmod{25} \\
24 \equiv 4 \pmod{25}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Of course, this is not true, so this gives one specific counter-example to show the book's statement is not correct for all primes $p \gt 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$ Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$.
I get $x=\sqrt{11}$ or $x=10$ whereas my textbook says it's $x=\pm\frac{10}{3}$.
My working - initial attempt:
$$x^2-10+9=10$$
$$x^2-1=10$$
$$x^2=11$$
$$x=\sqrt{11}$$
My working - another attempt:
$$\ln(\frac{x^2-10}{9})=\ln(10)$$
$$\frac{x^2-10}{9}=10$$
$$x^2-10=90$$
$$x^2=100$$
$$x=10$$
Where am I going wrong? How can I arrive at $x=\pm\frac{10}{3}$? Don't both my solutions make sense? Why are they incorrect?
| Note that\begin{align*}\ln(x^2-10)+\ln(9)=\ln(10)&\iff\ln\bigl(9(x^2-10)\bigr)=\ln(10)\\&\iff9x^2-90=10\\&\iff x^2=\frac{100}9\\&\iff x=\pm\frac{10}3.\end{align*}
The error from your first attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln(10)\iff x^2-10+9=10,$$whereas the error in your second attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln\left(\frac{x^2-10}9\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$
To find the domain:
$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$
$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$
$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$
$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$
The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$
But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.
Thanks
| a)
The function $f$ is not injective, therefore not bijective and cannot have an inverse therefore.
But you can calculate the individual branches (i. e. partial inverses) of $f$.
b)
You didn't rearrange your equation correctly.
Let $y$ denote $f^{-1}(x)$.
$$\sin^2\left(\frac{2y+1}{3}\right)=x$$
$$\left(\sin\left(\frac{2y+1}{3}\right)\right)^2=x$$
$$\sin\left(\frac{2y+1}{3}\right)=\pm\sqrt{x}$$
$$\frac{2y+1}{3}=\arcsin(\pm\sqrt{x})$$
$$\frac{2y+1}{3}=\pm\arcsin(\sqrt{x})$$
$$y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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.