Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Determine the general term of a Recurrent Rational Sequence How to determine the general term of the $(u_n)$ sequence with $u_0 \in \mathbb R_+$ and $u_{n+1}=\frac{u_n^2+1}{1+2u_n}$ ?
Source : les dattes à Dattier
| 1/The trick :
Let $t,u,v$ functions with $t\circ v=v\circ u$, then : $t^n\circ v=v\circ u^n$.
Here, we have choosen, $t,u,v$ of the type : $$t(x)=\frac{x^2+1}{2x+1},
u(x)=x^2 \text{ and } v(x)=\frac{ax+b}{cx+d}$$
2/For the solution proposed by Francesco, the form is :
$$h_n=f\circ g^n\circ f^{-1}$$
with $$f(x)=\frac{\sqrt 5}{2}\coth(x)-\frac{1}{2}$$
$$g(x)=2x$$
We have : $h_n\circ h_m=h_{n+m}$
and a conjecture: $$h_1(x)=\frac{1+x^2}{2x+1}$$
3/
we go to show : $h_1(x)=\frac{1+x^2}{2x+1}$
we have : $$(1):\coth(2x)=\frac{\coth(x)^2+1}{2\coth(x)}$$
More : $$f=w \circ \coth$$
with $$w(x)=\frac{\sqrt{5}}{2}x-\frac{1}{2}$$
and $$w^{-1}(x)=\frac{(2x+1)}{\sqrt{5}}$$
so
$$h_1=w \circ \coth \circ g \circ \coth^{-1} \circ w^{-1} $$
$$\text{ with } w(x)=\frac{\sqrt{5}}{2}x-\frac{1}{2} \text{ and } g(x)=2x$$
with using of $(1)$ then :
$$ \coth \circ g \circ \coth^{-1}(x)=\frac{x^2+1}{2x} $$
so
$$h_1(x)=w\left(\frac{w^{-1}(x)^2+1}{2w^{-1}(x)}\right)=\frac{\sqrt{5}}{2}\left(\frac{(\frac{2x+1}{\sqrt{5}})^2+1}{2(\frac{2x+1}{\sqrt{5}})}\right)-\frac{1}{2}$$
$$h_1(x)=\frac{\sqrt{5}}{2}\left(\sqrt{5}\frac{(\frac{4x^2+4x+1}{5})+1}{2(2x+1)}\right)-\frac{1}{2}=\frac{\sqrt{5}}{2}\left(\frac{\sqrt{5}}{5}\frac{4x^2+4x+1+5}{2(2x+1)}\right)-\frac{1}{2}$$
$$h_1(x)=\frac{\sqrt{5}}{2}\left(\frac{1}{2\sqrt{5}}\frac{4x^2+4x+6}{(2x+1)}\right)-\frac{1}{2}$$
$$h_1(x)=\frac{1}{4}\left(\frac{4x^2+4x+6}{2x+1}-\frac{4x+2}{2x+1}\right)$$
$$h_1(x)=\frac{1}{4}\left(\frac{4x^2+4}{2x+1}\right)$$
$$h_1(x)=\frac{x^2+1}{2x+1}$$
4/
about the case $a_{n+1}=a_n^2-2$
This case is interested because, for the previous case, resolved by Francesco and Jean-Marie, we could make changes of invertible variables, here not.
We use, a function $v$, no invertible, so $u(x)=x^2$ and $t(x)=x^2-2$ are not conjugate, but the trick continues to work and here
$$v(x)=x+\frac{1}{x}$$
we have :
$$\text{ if } a_{n+1}=a_n^2-2, \text{ and }b \in \mathbb C \text{ with } a_0=b+\frac{1}{b} \text{ then } a_n=b^{2^{n}}+\frac{1}{b^{2^n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What number should be subtracted from What number should be subtracted from $4x^3+5x+3$
so that the resulting polynomial leaves remainder $-80$ when divided by $2x+5$?.
Let the required number to be subtracted be $K$.
let: $$P(x)=4x^3+5x+3-k$$
$$g(x)=2x+5=2(x+5/2)$$
Comparing $g(x)$ with $x-a $ , $a =\frac {-5}{2}$
Solving then gives $K=8$
but the answer in my book is $-152$. how?
| Let $f (x)=2x+5$ and $p (x)=4x^3+5x-3$. We have that: $$p (-\frac {5}{2}) -k = -80$$ $$\Rightarrow 4 (-\frac {5}{2})^3 +5 (-\frac {5}{2}) +3 - k = -80$$ $$-75+3-k = -80$$ $$\boxed {k = 8} $$ Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find y from given information.
Find $y$ given that $$x^{2}+\frac{1}{x^{2}} = 98;$$ and
$$x^{3}+\frac{1}{x^{3}} = y.$$
I've got $970$, but that's not the full answer, since I got half of the points. What else could I get here?
| We know,
$(x + \frac 1x)^2 = (x^2 + \frac 1{x^2} + 2.x.\frac 1x)$
$(x + \frac 1x)^2 = (98 + 2)$
$(x + \frac 1x)^2 = 100$
$(x + \frac 1x) = \pm10$
Case 1 - $(x + \frac 1x) > 0$
$(x + \frac 1x) = 10$
Now cube of above equation,
$(x + \frac 1x)^3 = 1000$
$x^3 + \frac 1{x^3} + 3.x.\frac 1x(x + \frac 1x) = 1000$
$x^3 + \frac 1{x^3} + 3(10) = 1000$
$x^3 + \frac 1{x^3} = 1000 - 30$
$x^3 + \frac 1{x^3} = 970$
Also,
$x^3 + \frac 1{x^3} = y$
So y = 970.
Case 2 - $(x + \frac 1x) < 0$
$(x + \frac 1x) = -10$
And we get y = -970.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:
$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$
Now I elevate everything to the square:
$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$
All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$
And here, I don't know where to go, and it doesn't seem as I simplified the problem.
Any hints of help will be appreciated
| $$
\frac{2-\sqrt 3}{2 \sqrt3} = \frac{1}{\sqrt 3} - \frac{1}{2}
$$
Now, suppose we assume that the opposite of what is given is true. We get a contradiction:
$$
\frac{1}{\sqrt 3} - \frac 12 \geq \frac 1{10} \implies\frac{1}{\sqrt 3} \geq \frac{3}{5} \implies \frac{1}{3} \geq \frac{9}{25} \implies 25 \geq 27
$$
which is not true. Hence, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Recursion - series examples Given:
$$p(n+1) = 2^n \cdot \sqrt{2\cdot {\left( 1 - \sqrt{1 - \left(\frac{p(n)}{2^n}\right)^2}\right)}}$$
And:
$$p(2) = 2\cdot \sqrt{2}$$
Find $p(n)$. I an unable to come up with a generalization for $p(n)$. Please help.
| \begin{align*}
P_{2} &= 2\sqrt{2} \\
P_{3} &= 4\sqrt{2-\sqrt{2}} \\
P_{4} &= 8\sqrt{2-\sqrt{2+\sqrt{2}}} \\
P_{5} &= 16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}
\end{align*}
$P_{n}$ looks like the perimeter of regular $2^{n+1}$-sided polygon of "diameter" $1$:
$$P_{n}=2^{n} \sin \frac{\pi}{2^{n}}$$
It's not difficult to verify by considering:
\begin{align*}
2^{n}\sqrt{2
\left(
1-\sqrt{1-\sin^{2} \frac{\pi}{2^{n}}}
\right)} &=
2^{n}\sqrt{2
\left(
1-\cos \frac{\pi}{2^{n}}
\right)} \\ &=
2^{n}\sqrt{4\sin^{2} \frac{\pi}{2^{n+1}}} \\ &=
2^{n+1}\sin \frac{\pi}{2^{n+1}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to integrate $\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$ without involving complex numbers? The integral is $$\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$$
Consider
$$\int \frac{\mathrm{d}x}{ax^2+bx+c}$$
There are a total of three cases, depending on the discriminant of $ax^2+bx+c$. Two of which are shown here (#1).
The third one, i.e. when $\Delta=0$, simply means evaluating $$\frac{1}{a}\int \frac{\mathrm{d}x}{\big(x+\frac{b}{2a}\big)^2} $$
For the first two cases, you can see different substitutions are used so as to prevent $i$ from appearing in the answer, hence resulting in an $\arctan$ function and a $\ln$ function respectively. The discrepancy arises when we change $+(4ac-b^2)$ to $-(b^2-4ac)$.
In general, is it possible to evaluate
$$\int \frac{\mathrm{d}x}{a\sin^2 x+b\sin x+c}$$ such that the result does not contain complex numbers when $\Delta_{\sin x}<0$?
The only approach I can think of when $\Delta_{\sin x}>0$ is by partial fraction decomposition, which differs from the method of substitution used above.
| Let $\sin x =\frac{1-t^2}{1+t^2}$, or $t= \tan(\frac\pi4-\frac x2)$
\begin{align}
&\int \frac{1}{\sin^2 x+\sin x+1}\ dx
=\int\frac{2+2{t^2}}{3+t^4} \ dt\\
=&\ (1+\frac1{\sqrt3}) \int \frac{1+\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt +(1-\frac1{\sqrt3}) \int \frac{1-\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt\\
=&\ \frac{\sqrt3+1}{\sqrt{6\sqrt3}}\tan^{-1}\frac{t-\frac{\sqrt3}t}{\sqrt{2\sqrt3}}
-\frac{\sqrt3-1}{\sqrt{6\sqrt3}}\coth^{-1}\frac{t+\frac{\sqrt3}t}{\sqrt{2\sqrt3}}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the sum of a function series with indefinite integral We have the series
$$
f(x) = \sum\limits_{n=0}^{\infty}\frac{n}{n+2}(x^2-4)^{2n}
$$
and it's asked to find the result of the series.
I've tried this approach
$$
\frac{n}{n+2} = \left(\frac{n+2}{n+2}-\frac{2}{n+2}\right) = 1 - \frac{2}{n+2}
$$$$
y = (x^2-4)^{2}\Rightarrow \sum\limits_{n=0}^{\infty}\frac{n}{n+2}(x^2-4)^{2n} = \sum\limits_{n=0}^{\infty}\left(1 - \frac{2}{n+2}\right)y^{n}
$$
then
$$
\sum\limits_{n=0}^{\infty}y^{n} - \frac{2}{n+2}y^{n} = \sum\limits_{n=0}^{\infty}y^{n} - \frac{2}{y^2}\sum\limits_{n=0}^{\infty}\frac{1}{n+2}y^{n+2}
$$$$
\frac{1}{1-y} - \frac{2}{y^2}\sum\limits_{n=0}^{\infty}\int y^{n+1}dy
$$
which leads to
$$
\frac{1}{1-y} - \frac{2}{y}\sum\limits_{n=0}^{\infty}\int y^{n}dy = \frac{1}{1-y} - \frac{2}{y}\int\left(\sum\limits_{n=0}^{\infty} y^{n}\right)dy
$$
and finally to
$$
\frac{1}{1-y} - \frac{2}{y}\int\frac{dy}{1-y} = \frac{1}{1-y}+\frac{2\ln|1-y|}{y}
$$$$
f(x) = \frac{1}{1-(x^2-4)^{2}}+\frac{2\ln(1-(x^2-4)^{2})}{(x^2-4)^{2}}
$$
where we consider values $|(x^2-4)| < 1$ and $|(x^2-4)| \neq 0$.
Is this process right? I'm not sure about taking the $y$ out of the sum, and splitting the sum in two.
| $$
\begin{align}
-\sum_{n=0}^{\infty}\frac{y^n}{n+2} &= -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\frac{y^{n+2}}{n+2} = -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\,\int_{\color{red}{0}}^{\color{red}{y}}t^{n+1}\,dt = -\frac{1}{y^2}\,\int_{0}^{y}\,\sum_{n=0}^{\infty}\,t^{n+1}\,dt \\[2mm]
&= -\frac{1}{y^2}\,\int_{0}^{y}\frac{t}{1-t}\,dt = -\frac{1}{y^2}\,\int_{0}^{y}\frac{t\color{red}{-1+1}}{1-t}\,dt = \frac{1}{y^2}\,\int_{0}^{y}\left(1-\frac{1}{1-t}\right)dt \\[2mm]
&= \frac{1}{y^2}\left[\color{white}{\frac{}{}}t+\log(1-t)\color{white}{\frac{}{}}\right]_{0}^{y}\, = \color{red}{\frac{y+\log(1-y)}{y^2}} \\[6mm]
\implies f(x) &= \frac{1}{1-(x^2-4)^2}+2\,\frac{(x^2-4)^2+\log\left(1-(x^2-4)^2\right)}{(x^2-4)^4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the rank of the following matrix depending on $\lambda\in\Bbb R$ Find the rank of the following matrix depending on $\lambda\in\Bbb R$.
$$A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$$
My attempt:
$$\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}\sim\begin{pmatrix}
1&2&3&4\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6
\end{pmatrix}\sim\begin{pmatrix}
1&0&-3&-8\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6\\
\end{pmatrix}$$
For $\lambda=4$ we have:
$$\begin{pmatrix}
1 &0&-3&-8\\
0&0&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6
\end{pmatrix}\sim\begin{pmatrix}
1&0&0&1\\
0&0&0&-1\\
0&0&-1&-3\\
0&-1&0&3\end{pmatrix}\sim\begin{pmatrix}
1&0&0&0\\
0&0&0&-1\\
0&0&-1&0\\
0&-1&0&0\\
\end{pmatrix}$$
$\Rightarrow r(A)=4$
For $\lambda\neq 4$ we have:
$$\begin{pmatrix}
1&0&0&1\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&0&3\\
\end{pmatrix}\sim\begin{pmatrix}
1&0&0&1\\
0&0&0&3\lambda-13\\
0&0&-1&-3\\
0&-1&0&3
\end{pmatrix}$$
For$\lambda=\frac{13}{3}\Rightarrow r(A)=3$ and for $\lambda\neq \frac{13}{3} \Rightarrow r(A)=4$
Is this correct? Thanks!
| It looks fine to me!! Or, we have $A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$ with $\lambda \in \Bbb{R}$ and we know: $$r(A) \leq 4 \text{ and } \big[r(A)= 4 \leftrightarrow \det(A)\neq 0\big]$$
We have $\displaystyle\det(A)= 13 - 3 \lambda$ therefore: $$ r(A) =4\leftrightarrow \bigg(\det(A) \neq 0 \leftrightarrow 13 - 3 \lambda \neq 0 \leftrightarrow 13 \neq 3 \lambda \leftrightarrow \lambda \neq \frac{13}{3}\bigg)$$ $$4 \neq r(A) < 4\leftrightarrow\bigg(\det(A)=0 \leftrightarrow 13 - 3 \lambda = 0 \leftrightarrow 13 = 3 \leftrightarrow \lambda=\frac{13}{3} \bigg) $$
But if $\det(A)=0$ and exists a $T \in M(A)_{3}$ with $\det(T) \neq 0$ and such that for all $R\in M(A)_{3+1=4}$ which contain $T$ we have $\det(R)=0$ then $r(A)=3$
$(M(A)_{n}:=\{X| X \text{ is minor of order }n \text{ for }A\})$
And, with $\lambda=\frac{13}{3}$, we have a $T \in M(A)_{3}$ with $\det(T) \neq 0$: $$\det(T)=\det\begin{pmatrix}
2&\frac{13}{3}&6\\
3&6&8\\
4&7&9
\end{pmatrix}=\frac{-1}{3}\neq 0$$ and $M(A)_{3+1=4}=\{A\}$ and $\det(A)=0$ because $\lambda=\frac{13}{3}$ then $r(A)=3$
Summary:
$r(A)=4 \leftrightarrow \Bbb{R}\ni\lambda\neq \frac{13}{3}$
$r(A)=3 \leftrightarrow \Bbb{R}\ni\lambda =\frac{13}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Determine convergence $\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)} $ I tried Ratio test for this series and got 1.
$$\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)}$$
Any suggestions?
I believe the limit of the sequence goes to $\infty$ , but I don't know how to prove.
Thanks a lot!
| Since
\begin{align}\frac{n!2^n}{1\cdot3\cdot5\cdots(2n+1)}=&\frac{2\cdot4\cdot6\cdots2n}{1\cdot3\cdot5\cdots(2n+1)}\\=&2\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots\frac{2n}{2n-1}\cdot\frac{1}{2n+1}\\ \geq&\frac{2}{2n+1}\geq\frac{2}{2n+2}=\frac{1}{n+1}\end{align}
we have that
$$
\sum_{n=0}^\infty\frac{n!2^n}{1\cdot3\cdot5\cdots(2n+1)}\geq\sum_{n=0}^\infty\frac{1}{n+1}=\infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| From second equation we get $y=-z^2-\frac{x}{z}-\frac{1}{z^2}$ (it's obvious that $z\neq0$).
Thus, $$x^2+\left(z^2+\frac{1}{z^2}+1+\frac{x}{z}\right)^2=4$$ or
$$\left(1+\frac{1}{z^2}\right)x^2+\frac{2}{z}\left(z^2+\frac{1}{z^2}+1\right)x+\left(z^2+\frac{1}{z^2}+1\right)^2-4=0,$$
which gives
$$\frac{1}{z^2}\left(z^2+\frac{1}{z^2}+1\right)^2-\left(1+\frac{1}{z^2}\right)\left(\left(z^2+\frac{1}{z^2}+1\right)^2-4\right)\geq0$$ or
$$(z^4+z^2-1)^2\leq0,$$
which gives $z^2=\frac{\sqrt5-1}{2}$, $x=-\frac{\frac{1}{z}\left(z^2+\frac{1}{z^2}+1\right)}{1+\frac{1}{z^2}}$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Fourier Series Proof where $f(x)$ satisfies the symmetry condition $f(L-x)=f(x)$ Let $f(x)$, (where $−L < x < L$), be an odd function that satisfies the symmetry condition $f (L − x) = f (x)$, for all $x$.
Show that $A_n =0$, $\forall n$ while $Bn =0$ for all even $n$.
Some definitions:
Fourier Series: $A_0+\sum_{n=1}^{\infty}(A_ncos(\frac{n\pi x}{L})+B_nsin(\frac{n\pi x}{L}))$
where,
$A_0=\frac{1}{2L}\int_{-L}^{L}f(x)dx$
$A_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx$
$B_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L})dx$.
Since $f(x)$ is odd it follows that $A_o=A_n=0$. However, I'm not sure how to use the symmetry condition to prove $Bn =0$ for all even $n$.Any comments are welcome.
| (Hoping no calculation mistakes will happen)
For all $n$ you have
$$
B_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi x}{L} \right) dx
$$
Specifically for $2n$ you have
$$
B_{2n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \left( \frac{2 \pi n x}{L} \right) dx
$$
Defining $g(x) = f\left(x - \frac{L}{2} \right)$ you have
$$
g(x) = f \left( x - \frac{L}{2} \right) = f \left(-L + x + \frac{L}{2} \right) = -f \left(L - x - \frac{L}{2} \right)$$ $$= \color{red}{-f \left(L - (x + \frac{L}{2}) \right)} = - f \left(x + \frac{L}{2} \right) = f \left(-x - \frac{L}{2} \right) = g(-x)
$$
i.e. $g(x) = g(-x)$, $g$ is even
$$
B_{2n} = \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n (x-L/2)}{L} \right) dx
$$
We specifically have
$$
\sin \left( \frac{2 \pi n (x-L/2)}{L} \right) = \sin \left( \frac{2 \pi n x }{L} \right) \cos \left( n \pi \right) - \cos \left( \frac{2 \pi n x }{L} \right) \sin \left( n \pi \right) = (-1)^n \sin \left( \frac{2 \pi n x }{L} \right)
$$
Which allows to state
$$
B_{2n} = \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n (x-L/2)}{L} \right) dx = (-1)^n \frac{2}{L} \int_{-L/2}^{L/2} g(x) \sin \left( \frac{2 \pi n x}{L} \right) dx = 0
$$
Because $g$ is even and the sin is odd in a symmetric interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show this inequality $\sum\sqrt{\frac{x}{x+2y+z}}\le 2$ Let $x,y,z,w>0$ show that
$$\sqrt{\dfrac{x}{x+2y+z}}+\sqrt{\dfrac{y}{y+2z+w}}+\sqrt{\dfrac{z}{z+2w+x}}+\sqrt{\dfrac{w}{w+2x+y}}\le 2$$
I tried C-S, but without success.
| By C-S:
$(LHS)^2\le \sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}$
$\sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}\le 4 \ \ \iff \ \ (a-c)^2(b-d)^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can we show that $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$
$$4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}\tag1$$
$\phi$;Golden ratio
I understand that we can use
$$\arctan{1\over a}+\arctan{1\over b}=\arctan{a+b\over ab-1}$$
that would take quite a long time and simplifying algebraic expressions involving surds is also difficult task to do.
How else can we show that $(1)={\pi\over 2}$?
| according to "How else can we show that"
Take $f(x)=4\arctan\left({1\over \sqrt{x^3}}\right)-\arctan\left({1\over \sqrt{x^6-1}}\right)$
now you can show $f(1.618)=\dfrac{\pi}{2}$
This question is very interesting to me .I was working on $\phi $ properties ,like $\phi^2=\phi+1 \to \phi^3=\phi^2+\phi$ or $\psi=\dfrac{1}{\phi}$
we can see this $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$ as $4a-b=\dfrac{\pi}{2} $ so $4a=\dfrac{\pi}{2} +b \to \tan(4a)=-\cot(b)$ (if you prove this,prove the first relation)
$$\phi=\dfrac{1+\sqrt5}{2} \\
\phi^3=\phi^2+\phi=(\dfrac{1+\sqrt5}{2})^2+(\dfrac{1+\sqrt5}{2})=\dfrac{1+5+2\sqrt5+2+2\sqrt5}{4}=\dfrac{8+4\sqrt5}{4}=2+\sqrt5 \to \\
\phi^6-1=(2+\sqrt5)^2-1=4+5+4\sqrt5-1=8+4\sqrt5$$ now redude to
$$4\arctan\left({1\over \sqrt{2+\sqrt5}}\right)-\arctan\left({1\over \sqrt{8+4\sqrt5}}\right)=\dfrac{\pi}{2} \to 4a-b=\dfrac{\pi}{2} \to
\tan(4a)= ? -\cot(b)$$
$$\tan a=\dfrac{1}{\sqrt{2+\sqrt5}} \to
tan 2a=\dfrac{2\tan a}{1-\tan^2a}=\\\frac{2\dfrac{1}{\sqrt{2+\sqrt5}}}{1-(\dfrac{1}{\sqrt{2+\sqrt5}})^2}=\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5} \\
\tan(4a)=\dfrac{2\tan 2a}{1-\tan^22a}=\\\dfrac{2\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5}}{1-(\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5})^2}=\\
\dfrac{4\sqrt{2+\sqrt5}(1+\sqrt5)}{(1+\sqrt5)^2-4(2+\sqrt5)}=\\-2\sqrt{2+\sqrt5}$$ Other side we have $$-\cot b =-(\sqrt{8+4\sqrt5})=-2\sqrt{2+\sqrt5} \checkmark$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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converting $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$ I have a question. It might seem silly.
Suppose I have $(x^3-2)$.
I know that $ 2^{1/3}$ is a root. $\frac{-1 \pm \sqrt{3}i}{2^{2/3}}$ are also roots of $(x^3-2)$.
How do you convert $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$?
Thanks.
| The principal value of the third root of $-2\;$ is $$(-2)^{1/3}= |-2|^{1/3}\left(\cos(\frac{2}{3}\pi) + i \sin(\frac{2}{3}\pi)\right)
=\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Multiply with $-2$ to get
$$-(-2)^{1/3}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Now multiply your expression with $2^{1/3}$ to get
$$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Therefore
$$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-(-2)^{1/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use induction to prove that the inequality, $n! < \left(\frac{n}{2}\right)^n$ continues to hold after $n=6$. I am confused how you would continue with the inductive step. Thanks
| Basis step,
$P_6 : 6!<(\frac{6}{2})^6\rightarrow 720<3^6=729$. True
Inductive step, suppose
$P_n$ is true, prove that $P_{n+1}$ is also true.
Note $P_n:n!<(\frac{n}{2})^n$ and $P_{n+1}:(n+1)!<(\frac{n+1}{2})^{n+1}$
Observe that,
$$\begin{align}(\frac{n+1}{2})^{n+1}&=(\frac{n}{2}+\frac{1}{2})^{n+1} \\&>(\frac{n}{2})^{n+1}+(n+1)\cdot\frac{n^n}{2^{n+1}}+\frac{n(n+1)}{2}\cdot(\frac{n^{n-1}}{2^{n+1}})\tag{binomial}\\&=\frac{(5n+3)}{4}\cdot(\frac{n}{2})^n\\&=(n+\frac{n+3}{4})\cdot(\frac{n}{2})^n\\&>(n+1)\cdot(\frac{n}{2})^n,\forall n\ge6\end{align}$$
So, start with $P_n$,
$$\begin{align}n!(n+1)&<(\frac{n}{2})^n\cdot(n+1)\\&<(\frac{n+1}{2})^{n+1}\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/2134256",
"timestamp": "2023-03-29T00:00:00",
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Probability to find 'X' number of Balls. A bag contains 5 Red, 4 Blue and x green balls. Two balls are drawn at random from the bag. If the probability of both being green is 1/7, find x.
I am clear with the proceedings but stuck at solving equation of factorials, so please show detailed steps.
| Total balls = x + 9
Green balls = x
Probability of 2 balls both green $$= \frac{\binom x2}{\binom{x + 9}{2}}$$
$$\frac 17= \frac{\frac{x!}{2! × (x - 2)!}}{\frac{(x+9)!}{2! × (x + 7)!}}$$
$\frac 17= \frac{x!}{2! × (x - 2)!} × \frac{2! × (x + 7)!}{(x+9)!}$
$\frac 17= \frac{x!}{(x - 2)!} × \frac{(x+7)!}{(x+9)!}$
$\frac 17= \frac{x \cdot (x-1) \cdot (x-2)!}{(x - 2)!} × \frac{(x + 7)!}{(x+9)(x+8)(x+7)!}$
$\frac 17= \frac {x \cdot (x-1)}{(x+9) \cdot (x+8)}$
$x^2 + 17x + 72 = 7x^2 - 7x$
$6x^2 - 24x -72 = 0$
$x^2 - 4x -12 = 0$
$x^2 -6x +2x - 12 = 0$
$x(x-6) +2(x-6) = 0$
$(x + 2 )(x - 6) = 0$
Either x = -2 or x = 6.
But x can't be negative.
So Green balls x = 6
| {
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"timestamp": "2023-03-29T00:00:00",
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Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$ Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$
| Let $a\geq b\geq c$.
Since $f(x)=x^3$ is a convex function on $[0,3]$ and $(2,1,0)\succ(a,b,c)$,
by Karamata we obtain $$9=2^3+1^3+0^2\geq a^3+b^3+c^3$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmatrix}$ and $B^tA^t=(AB)^t=\begin{pmatrix}
5 &17 &1\\
2 &7 &1\\
4 &13 &0
\end{pmatrix}$
So we have $\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1 &0
\end{pmatrix}\cdot X\cdot\begin{pmatrix}
5 &17 &1\\
2&7&1\\
4&13&0
\end{pmatrix}=I$
Now how do I get $X$ from here?
| From $ABXB^tA^t=I$, we see that $det(A) \ne 0 \ne det(B)$. Hence $A$ and $B$ are invertible.
Again , from $ABXB^tA^t=I$, we see
$X=(AB)^{-1}((AB)^{-1})^t$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given is $\frac{100}{101}$, but I am not sure how.
So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$
The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator.
| $$a_1+a_2+a_3+...+a_{100}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}$$
Note that $-\frac{1}{2}$ cancel $\frac{1}{2}$, $-\frac{1}{3}$ cancel $\frac{1}{3}$ and go on until $-\frac{1}{100}$ cancel $\frac{1}{100}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin\frac{\pi}n·\sin\frac{2\pi}n···\sin\frac{(n-1)\pi}n=\frac{n}{2^{n-1}}$ How to prove that
$$
\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}
$$
using the roots of $(z+1)^n-1=0$?
My rough idea is to solve $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roots to prove the equality.
| Given the equation
$$(z+1)^n = 1 \tag{1}$$
Its roots are, $$z = e^{2\pi i \frac{k}{n}} - 1, \;\; 0 \leq k < n$$
Now,
$$\prod_{k = 1}^{n-1} (e^{2\pi i \frac{k}{n}} - 1)$$
is the product of all roots except $z = 0$. If you open up (1) and cancel one out of from both sides, you can take z common and the remaining polynomial will have the other roots. From Vieta's formulae, we know that,
$$\prod_{k = 1}^{n-1} (e^{2\pi i \frac{k}{n}} - 1) = (-1)^{n-1}n$$
The above can be written as,
$$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}(e^{\pi i \frac{k}{n}} - e^{-\pi i \frac{k}{n}}) = (-1)^{n-1}n$$
$$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}2i \sin \frac{\pi k}{n} = (-1)^{n-1}n$$
$$e^{\pi i \frac{n-1}{2}}2^{n-1}i^{n-1}\prod_{k = 1}^{n-1} \sin \frac{\pi k}{n} = (-1)^{n-1}n$$
$$\prod_{k = 1}^{n-1}\sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}$$
| {
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"url": "https://math.stackexchange.com/questions/2138997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| $$ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}= $$
$$ =\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}. $$
Let $\cos\frac{\pi}{7}=x$.
Hence,
$$2x^2-(4x^3-3x)-x=-\frac{1}{2},$$
which gives your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| In fact $$(x^2 - x+1)(x^2 + x+1)$$ $$ = (x^2+1 -x)(x^2+1 +x)$$ $$= (x^2+1)^2 -x^2$$ (remember the $\alpha^2-\beta^2 =(\alpha - \beta)(\alpha + \beta)$ formula? here $\alpha = x^2+1$ and $\beta = -x $) $$ = x^4 + x^2 +1$$
Well, if you really want to factor $x^4+1$, see here. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Volume of tetrahedron given $6$ sides. I'm given $6$ side lengths and asked to find the volume. I feel lost
Tetrahedron $ABCD$ has $AB=1.5$, $AC=1$, $BC=2$, $BD=2.5$, $AD=1.5$, and $CD=3$.
I got the area of triangle ABC by Heron's formula to be $\sqrt{2.25(2.25-1)(2.25-2)(2.25-1.5)}$.
How do I find the height of the tetrahedron?
| By the Cayley-Menger determinant
$$ 288 V^2=\det\begin{pmatrix}0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^3 & d_{14}^2\\ 1 & d_{21}^2 & 0 & d_{23}^2 & d_{24}^2\\1 & d_{31}^2 & d_{32}^2 & 0 & d_{34}^2 \\1 & d_{41}^2 & d_{42}^2 & d_{43}^2 & 0\end{pmatrix} $$
and in our case $d_{12}=d_{21}=\frac{3}{2}$, $d_{13}=d_{31}=1$, $d_{14}=d_{41}=\frac{3}{2}$, $d_{23}=d_{32}=2$, $d_{24}=d_{42}=\frac{5}{2}$, $d_{34}=d_{43}=3$. Since the involved matrix is symmetric but not positive definite, that tells us Houston has a problem: the sides of $ACD$ do not fulfill the triangular inequality, since $AC+AD\color{red}{<}CD$, hence there is no tetrahedron with such side lengths.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140214",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Various methods to find value of $\sin 18^\circ$ To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.
Are there any other elegant ways to find its value?
Thanks
| Let $x = \sin(18^{\circ})\implies x^2 = \dfrac{1-y}{2}$, whereas $y = \cos(36^{\circ})$. And square again $y^2 = \dfrac{1+\cos(72^{\circ})}{2}= \dfrac{1+x}{2}\implies 2x^2+y = 1 = 2y^2-x\implies 2(x^2-y^2) + (x+y) = 0\implies (x+y)(2x-2y+1) = 0\implies 2x= 2y-1\implies 4x^2=2-2y = 1-2x\implies 4x^2+2x-1=0$. You can now solve for $x$ to get the answer in terms of radical which is the expected form of the answer.
| {
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Splitting the function $ \frac{x}{(1+x^2)^2}$ into partial fractions. I have a function that I am trying to split into partial fractions in order to integrate the function.
The function is:
$$\int \frac{x}{(1+x^2)^2}dx$$
I am trying to split $\dfrac{x}{(1+x^2)^2}$ into partial fractions.
While trying I am splitting it into these fractions:
$\dfrac{Ax+D}{1+x^2}+\dfrac{Cx+D}{(1+x^2)^2}$
I get $C=1$ which ends up again where I started from. I am confused on what to do here.
| Since
$1+x^2
=(1+ix)(1-ix)
$,
if we ask Wolfy for
$\dfrac{1}{(1+ax)^2(1+bx)^2}
$
we get
$-\dfrac{2 a^2 b}{(a - b)^3 (a x + 1)}
+ \dfrac{a^2}{(a - b)^2 (a x + 1)^2}
+ \dfrac{2 a b^2}{(a - b)^3 (b x + 1)}
+ \dfrac{b^2}{(a - b)^2 (b x + 1)^2}
$.
Putting $a=i, b=-i$,
we get both
$\dfrac{i}{4 (x + i)}
- \dfrac{1}{4 (x + i)^2}
- \dfrac{i}{4 (x - i)}
- \dfrac{1}{4 (x - i)^2}
$
and
$\dfrac{2 x^2}{(x^2 + 1)^4}
+ \dfrac{1}{(x^2 + 1)^4}
+ \dfrac{x^4}{(x^2 + 1)^4}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141382",
"timestamp": "2023-03-29T00:00:00",
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Substitutions for the trigonometric integral $\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$ I evaluated
$$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$$
and got to a result that is very similar to what I can check with Mathematica but I am not sure if it is equivalent, what I did is the following,
$$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$$
$$= \int \cos^4\left(\sin \theta\right)\cos\left(\sin\theta\right)\cos \theta d\theta$$
$$= \frac 14\int \left(1+2\cos\left(2\sin\theta\right)+\cos^2\left(2\sin\theta\right)\right)\cos\theta d\theta$$
$$=\frac 14 \int \cos\left(\sin\theta\right)\cos \theta d\theta\;+\;\frac 12 \int \cos\left(2\sin\theta\right)\cos\left(\sin\theta\right)\cos \theta d\theta\;$$
$$+\;\frac 18 \int \left(1+\cos\left(4\sin \theta\right)\right)\cos\left(\sin\theta\right)\cos\theta d\theta$$
I use $\cos A \cos B = \frac 12 \left(\cos\left(A-B\right) + \cos\left(A+B\right)\right)$ and I get,
$$= \frac 14 \int \cos\left(\sin\theta\right)\cos\theta d\theta\;+\;\frac 14 \int \cos\left(\sin \theta\right)\cos\theta\,d\theta\,+\, \frac 14 \int\cos\left(3\sin\theta\right)\cos\theta d\theta \;$$
$$+\; \frac 1{8} \int \cos\left(\sin\theta\right)\cos \theta\,d\theta \;+\; \frac 1{16} \int \cos\left(3\sin\theta\right)\cos \theta d\theta \;+\;\frac 1{16}\int \cos\left(5\sin\theta\right)\cos \theta d\theta$$
wihch is equal to
$$=\frac 5{8} \int \cos\left(\sin\theta\right)\cos \theta\,d\theta \;+\;\frac 5{16} \int \cos\left(3\sin\theta\right)\cos \theta d\theta \;+\;\frac 1{16}\int \cos\left(5\sin\theta\right)\cos \theta d\theta$$
Edit (correct solution)
Applying the correct substitution, letting $u = \sin \theta$ such that $du = \cos \theta\,d\theta$ to get,
$$\frac 58 \int \cos u\,du\;+\;\frac 5{16} \int \cos 3u\, du\;+\; \frac 1{16} \int \cos 5u \,du$$
yields the same result as $Mathematica$, that is,
$$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta \;=\; \frac 58 \sin\left(\sin\theta\right)\;+\; \frac 5{48} \sin\left(3\sin\theta\right) \;+\; \frac 1{80} \sin\left(5\sin\theta\right)$$
Of course the substitution could and should be applied right at the beginning, for the sake of simplicity. I wanted to correct the last step while leaving the question as it was for others that might make the same mistake I did and for whomever might find that information useful.
| Try this method once -
Put $\sin\theta =t$
$\cos\theta d\theta = dt$
So we have,
$\int \cos^5t$ dt
Now solve it.
Any doubt feel free to ask.
| {
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Let F be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$.
Let $F$ be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$, where ,b,c,d $\in F$.
I tried using the formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm \sqrt{b^2-4c}}{2}$, and I am stumped by the fact that $2\equiv0 \mod 2$, so there will be division by $0$.
I also considered another method: To let $x+i$ and $x+j$ be the factor of $x^2+bx+c$. Then, $(x+i)(x+j)=x^2+(i+j)x+ij$, which implies $i+j=b$ and $ij=c$. But completely stumped as well?
How should I attempt this question? I have basic elementary group theory algebra knowledge.
| If you multiply by $b^{-2}$ you get that $x^2+bx+c=0$ is equivalent to $(b^{-1}x)^2+b^{-1}x+d(d+1)=0$, which is equivalent to $b^{-1}x(b^{-1}x+1)=d(d+1)$. It therefore reduces to solving $y(y+1)=d(d+1)$ with $y=b^{-1}x$. Those solutions are $y=d$ and $y=d+1$, hence the original solutions are $x=bd$ and $x=b(d+1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge of the answer?
Solution to a problem
| Given an expression like $\sqrt{a\pm\sqrt{b}}$, can we simplify it to an expression of the form $\left|c\pm\sqrt{d}\right|$?
Well, suppose we could. Then we would find that
$$
(c\pm\sqrt{d})^2 = a\pm\sqrt{b}
$$
Carrying out the square, we get
$$
c^2+d\pm2\sqrt{4c^2d} = a\pm\sqrt{b}
$$
We might therefore try to equate $c^2+d$ with $a$, and $c^2d$ with $b/4$. In this case, we have $a = 4$ and $b = 12$. Can we find two numbers whose sum is $4$, and whose product is $12/4 = 3$? The obvious answer is $1$ and $3$, and in this case, $1$ is a perfect square, so it makes a good choice for $c^2$. Then $d = 3$, and we have
$$
\sqrt{4-\sqrt{12}} = \left|1-\sqrt{3}\right| = \sqrt{3}-1
$$
Since the value desired is not restricted to the positive value, then $1-\sqrt{3}$ will also work. (We could also have had $c = -1$, but that just yields $-1+\sqrt{3}$ again.)
If you want an algorithmic way of finding two values whose sum is $s$ and whose product is $p$, note that these two values are the roots to the quadratic equation
$$
x^2-sx+p = 0
$$
Solve for the roots, and $c^2$ can be any positive root, if one exists, and $d$ is then the other root.
| {
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Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)\cdot \left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
Now $\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
So need to prove $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>10^{2009}$.
Please help with this, and I'm very enthusiast to know your method to solve above problem.
| You have $x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$, but $x^2-2xy+2y^2 = (x-y)^2+y^2\geq y^2$, where $y = 2^{\frac{5^6-1}{2}}$. Therefore, $$y^2 = (2^{10})^{\tfrac{5^6-1}{10}}>10^{\tfrac{3(5^6-1)}{10}} = 10^{2009},$$
because $2^{10} = 1024>1000 = 10^3.$
| {
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Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$? When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$?
Examples:
$$7^2+4=53$$
$$11^2+4= 5^3$$
$$31^2+4= 5 \cdot 193$$
$$47^2+4=2213$$
$$89^2+4=5^2 \cdot 317$$
All of the prime factors above are $\equiv 1 \mod 4$.
Is there no counter example and if so, how to prove this?
| $x^2 + 4 \equiv 3 \pmod 4$ is the same as proving $x^2 + 0 \equiv 3 \pmod 4$ since $4 \equiv 0\pmod 4$.
If $x$ is even is easy to see that $x^2 \equiv 0 \pmod 4$ and when $x$ is odd it has the form $2n+1$ then:
$x=(2n+1)^2= 4n^2 + 4n + 1 = 4n(n+1) + 1$ thus $4n(n+1) \equiv 0 \pmod 4$ and $4n(n+1) + 1\equiv 1 \pmod 4$
| {
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Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$?
Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$ ?
I wrote $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$ as
$$2\Re\left(\left({-1 + i\sqrt{3}\over 2}\right)^n\right) = 2 \implies \Re\left(\left({-1 + i\sqrt{3}\over 2}\right)^n\right) = 1$$
But I don't know how to proceed now. I can't use polar representation of complex numbers. Please provide me some direction.
| First of all, note that $\left| \frac{-1 \pm i\sqrt 3}{2} \right| = 1$, hence this holds for all $n$th powers of these quantities also.
The triangle inequality states that $|a| + |b| \geq |a+b|$, with equality holding if and only if $a$ is a scalar multiple of $b$.
In our case, if we assume that the final condition holds, we have that for $a,b = \left(\frac{-1 \pm i\sqrt 3}{2}\right)^n$ (it doesn't matter which is which), $2 = |a| + |b| \geq |a+b| = 2$, which means equality is being obtained in the triangle inequality.
This means that $\left(\frac{-1 + i\sqrt 3}{2}\right)^n$ is a scalar multiple of $\left(\frac{-1 - i\sqrt 3}{2}\right)^n$. However, since both have modulus $1$, we can check that this happens if and only if $\left(\frac{-1 + i\sqrt 3}{2}\right)^n = \left(\frac{-1 - i\sqrt 3}{2}\right)^n$.
Hence, dividing both sides by $\left(\frac{-1 - i\sqrt 3}{2}\right)^n$ (it can't be zero, as it has modulus $1$), we get that (I leave you to see this) $\left(\frac{-1 + i\sqrt{3}}{2}\right)^n = 1$.
If you know your roots of unity well, this gives $n = 3k$, where $k$ is an integer.
| {
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If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with the same condition,
which we can prove by AM-GM and C-S:
$$\sum\limits_{cyc}\frac{1}{1-ab}=3+\sum\limits_{cyc}\left(\frac{1}{1-ab}-1\right)=3+\sum\limits_{cyc}\frac{ab}{1-ab}\leq$$
$$\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(2a^2+2b^2+2c^2-2ab)}\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(a^2+b^2+2c^2)}\leq$$
$$\leq3+\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2},$$
but for the starting inequality this idea does not work.
By the way, I have a proof of the following inequality.
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=3$. Prove that:
$$\sum\limits_{cyc}\frac{1}{(4-ab)^2}\leq\frac{1}{3}$$
(we can prove it by SOS and uvw).
This inequality is weaker and it not comforting.
We can assume of course that all variables are non-negatives.
Thank you!
| Let $x = \dfrac{1}{1-ab}$, $y = \dfrac{1}{1-bc}$, $z = \dfrac{1}{1-ac}$.
We have proven that
$$x + y + z \leq \dfrac{9}{2}.$$
Therefore, $x + y + z = \dfrac{9}{2} - \varepsilon$ for some $0 \leq \varepsilon \leq \dfrac{9}{2}$. If we assume that $x,y,z \geq 0$, which we can, we can square both sides and obtain
$$(x^2 + y^2 + z^2) + (2xy + 2yz + 2xz) = \bigg{(} \dfrac{9}{2} - \varepsilon \bigg{)}^2 \leq \dfrac{81}{4}$$
We can say that $x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2(xy + yz + xz)$. We now seek to minimize $xy + yz + xz$ in order to maximize the upper bound.
We are given that $a^2 + b^2 + c^2 = 1$, and since $xy + yz + xz$ is a symmetric polynomial in $x,y,z$, we should seek $x=y=z>0$ as our minimum, which can be shown the be the case using very straightforward techniques from calculus (Second derivative test should suffice, proving the minimum for $a,b,c$ suffices). We find that $x=y=z$ when $a=b=c=\dfrac{1}{\sqrt{3}}$, so that $x=y=z=\dfrac{1}{1-1/3}=\dfrac{3}{2}$. We can then say that
$$xy + yz + xz \geq 3*(3/2)^2 = 3*(9/4) = 27/4$$
Therefore,
$$x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2*\dfrac{27}{4} = \dfrac{81}{4} - \dfrac{54}{4} = \dfrac{27}{4}$$
| {
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Prove: $\exists M\in\mathbb{R}$ such that $\forall x\in (1,3)$, $\left|\frac{5x^2 - 2x - 4}{5(x^2 + 1)}\right|\leq M$ I am not sure how to find the correct upper and lower limits for this problem. I did find the breakdown of the inequality to be $\frac{5x^2+|2x|+4}{|5(x^2 + 1)|}$. Thanks for any help you can give!
| $|\frac {5x^2 - 2x -4}{5(x^2 + 1)}|\le M \iff$
$-M \le \frac {5x^2 - 2x -4}{5(x^2 + 1)} \le M \iff$
$-M*5(x^2 + 1) \le 5x^2 - 2x - 4 \le M*5(x^2 + 1)\Leftarrow$
$10M \le 5x^2 - 2x - 4 \le 10M$.
Now $5 < 5x^2 < 45$ for $1 < x < 3$.
And $-10 < -2x - 4 < -6$ for $1 < x < 3$.
So $-5 < 5x^2 - 2x - 4 < 39$.
So if $M = 3.9$ then
$-10M < -5 < 5x^2 - 2x - 4 < 39 = 10M \implies$
$- M*5(x^2 + 1) \le -10M < 5x^2 - 2x - 4 < 10M \le M*5(x^2 + 1)\implies$
$|\frac {5x^2 - 2x -4}{5(x^2 + 1)}|\le M$.
| {
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Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit?
$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$
I dont know how to simplify $\frac{2^n }{ n}$.
I cannot just do $\frac{2^n}{n}<\epsilon$ right ?
| For $\; n \geq 3 \;$ we have
$$\frac{2^n}{n!} \;\; = \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} \cdot \frac{2}{6} \cdot \;\; \cdots \;\; \cdot \frac{2}{n} \;\; \leq \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \left(\frac{2}{3}\right)^{n-2} \;\; \longrightarrow \;\; 0$$
| {
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Prove using induction the inequality.
$\forall$:n∈${N}$
$\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$
I tried with no any success...
| For $n=1$ we get $2\geq\frac{4}{3}$, which is true.
If $\binom{2n}{n}\geq\frac{4^n}{2n+1}$ we obtain:
$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+2)(2n+1)}{(n+1)^2}>\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}$$ and it remains to prove that
$$\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}>\frac{4^{n+1}}{2n+3}$$ or
$$2(2n+3)>4(n+1),$$
which is obvious.
| {
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Evaluating $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$ I'm trying to evaluate the following limit:
$$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$$
I've tried multiplying by the conjugate and variable substitution. I had a look at wolfram alpha and it said that $\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}=\sqrt{2}$, though I'm interested in the process to achieve that.
Any help would be much appreciated / actually finding the limit.
Thanks
| Note: I am using the limit $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ and the identity $1-\cos 2A=2\sin^2 A$.
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x} & = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 \left(x^2/2\right)}}{2 \sin^2 \left(x/2\right)}\\
& = \lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{\sqrt{2}\sin^2 \left(x/2\right)}\\
& = \frac{1}{\sqrt{2}}\lim_{x \to 0} \frac{\sin \left(x^2/2\right)}{x^2/2}\frac{(x/2)^2}{\sin^2 \left(x/2\right)}.2\\
& =\sqrt{2}.
\end{align*}
| {
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How to calculate the series $\sum\limits_{n=1}^{\infty} \arctan(\frac{2}{n^{2}})$? I encountered the series
$$
\sum_{n=1}^{\infty} \arctan\frac{2}{n^{2}}.
$$
I know it converges (by ratio test), but if I need to calculate its limit explicitly, how do I do that? Any hint would be helpful..
| \begin{align*}
\sum_{n=1}^\infty\arctan\left ( \frac{2}{n^2} \right ) &=-arg \prod_{n=1}^\infty\left (1-\frac{2i}{n^2} \right ) \\
&=-arg \prod_{n=1}^\infty\left (1-\frac{(\sqrt{2i})^2}{n^2} \right ) \\
&=-arg\left(\frac{\sin(\pi\sqrt{2i})}{\pi\sqrt{2i}} \right ) \\
&=-arg\left(-\frac{(1/2+i/2)\sinh\left(\pi \right )}{\pi} \right ) \\
&= \frac{3\pi}{4}
\end{align*}
| {
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Validating the inequality. $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}$, $x>0$
Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.
We have done the Lagrange's mean value theorem and intermediate value theorem, do these have anything to do with the inequality.
Kindly provide some hint.
| I expand my comment into an answer.
We have for $t > 0$ $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and hence if $0 < t < x$ then $$1 - t + \frac{t^{2}}{1 + x} < \frac{1}{1 + t} < 1 - t + t^{2}$$ and integrating the above with respect to $t$ on interval $[0, x]$ we get $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3(1 + x)} < \log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$
| {
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How to integrate this $\int \frac{\sqrt{x}}{1+x^4}dx$? How to integrate this ? $$\int \frac{\sqrt{x}}{1+x^4}dx$$
Any hint or idea on how to proceed?
Edit: Here is the final answer using Wolfram Alpha
| Let $x= \frac1{t^2}$. Then
$$\int \frac{\sqrt{x}}{1+x^4}dx=
-\int \frac{2t^4}{1+t^8}dt
=\frac1{\sqrt2}\int \left(
\frac{t^2}{t^4+\sqrt2t^2+1}
-\frac{t^2}{t^4-\sqrt2t^2+1} \right)dt $$
Integrate
$$\begin{align}
I(a)& =\int\frac{t^2}{t^4+at^2+1}dt\\
&=\frac12\int \frac{t^2+1}{t^4+at^2+1}dt -
\frac12\int \frac{1-t^2}{t^4+at^2+1}dt \\
&=\frac1{2\sqrt{2+a}}\tan^{-1}\frac{t-\frac1t}{\sqrt{2+a}}
- \frac1{2\sqrt{2-a}}\coth^{-1}\frac{t+\frac1t}{\sqrt{2-a}} \\
\end{align}$$
Thus,
$$\begin{align}
& \int \frac{\sqrt{x}}{1+x^4}dx
=\frac1{\sqrt2}\left[I(\sqrt2)-I(-\sqrt2)\right]+C
\end{align}$$
| {
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The following equation to solve :$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$ The following equation to solve :
$$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$$
My try:
$$\frac{2}{\sin 2x}=\sqrt{2}(\cos x+\sin x)$$
$$\left(\frac{2}{\sin 2x}\right)^2=(\sqrt{2}(\cos x+\sin x))^2$$
$$\left(\frac{2}{\sin 2x}\right)^2=2(1+\sin 2x)$$
$$2\sin^2 2x +2\sin ^3 2x=4$$
$$2\sin^2 2x +2\sin ^3 2x-4=0$$
$t=\sin 2x$
$$2t^3+2t^2-4=(t-1)(t^2+2t+4)$$
$$\sin 2x =1\\$$
is it right ?
| Avoid squaring whenever possible as it immediately introduces extraneous roots.
Method $\#1:$ Put $u=\dfrac{\cos x+\sin x}{\sqrt2}\implies2u^2=1+\sin2x$
$$\implies\dfrac2{2u^2-1}=2u\iff2u^3-u-1=0$$
Clearly, $u=1$ is a root. Please check for the other roots.
$\implies1=\dfrac{\cos x+\sin x}{\sqrt2}=\cos\left(x-\dfrac\pi4\right)$
$\implies x-\dfrac\pi4=2m\pi$ where $m$ is any integer.
Method $\#2:$
$\cos x+\sin x=\sqrt2\cos\left(x-\dfrac\pi4\right)$
$\sin2x=\cos2\left(x-\dfrac\pi4\right)=\left(\sqrt2\cos\left(x-\dfrac\pi4\right)\right)^2-1$
| {
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How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calculating the single real solution for the cubic equation $$ z^3 - 33 z - 112 = 0 $$
Using that cubic equation, I can show that $ z = 7 $ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7. But what if I didn't know about the cubic equation? Is there a more direct way of reducing the complicated expression to a simpler one?
| I'll outline the basic method to simplify$$\sqrt[3]{56+19\sqrt{5}}+\dfrac {11}{\sqrt[3]{56+19\sqrt{5}}}\tag{1}$$
To simplify the nested radical, note that we have this general outline:$$\sqrt[n]{A+B\sqrt[m]{C}}=a+b\sqrt[m]{C}\tag{2}$$
So therefore, we have$$\begin{align*}\sqrt[3]{56+19\sqrt{5}} & =a+b\sqrt{5}\\56+19\sqrt{5} & =\underbrace{(a^3+15ab^2)}_{56}+\underbrace{(3a^2b+5b^3)}_{19}\sqrt{5}\end{align*}$$
So we get this system of equations:$$\begin{align*} & a^3+15ab^2=56\\ & 3a^2b+5b^3=19\end{align*}\tag{3}$$
$(3)$ has real solutions as $(a,b)=\left(\dfrac 72,\dfrac 12\right)$ so$$\sqrt[3]{56+19\sqrt{5}}=\dfrac {7+\sqrt5}2\tag4$$
Using $(4)$, and through some algebraic manipulations, we have$$\begin{align*}\color{blue}{\sqrt[3]{56+19\sqrt5}}+\dfrac {11}{\color{red}{\sqrt[3]{56+19\sqrt{5}}}} & =\color{blue}{\dfrac {7+\sqrt{5}}2}+\dfrac {11}{\color{red}{\frac {7+\sqrt{5}}2}}\\ & =\dfrac {7+\sqrt{5}}{2}+\dfrac {22}{7+\sqrt{5}}\\ & =\dfrac {7+\sqrt{5}}2+\dfrac {22(7-\sqrt5)}{44}\\ & =\dfrac {7+\sqrt5}2+\dfrac {7-\sqrt5}2\\ & =\boxed 7\end{align*}$$
Just like what you got!
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Find a $3\times3$ orthogonal matrix $A$ . How can we find a $3\times3$ orthogonal matrix $A$ such that $$A\,\left[\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right]^T=\left[\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\right]^T?$$
My try:I don't know how to proceed in order to get that matrix.Thank you
| step 1, Find an otho-normal matrix that maps one of your principle component vectors to $(\frac {1}{\sqrt 3},\frac {1}{\sqrt 3},\frac {1}{\sqrt 3})$
$(\frac {1}{\sqrt 3},\frac {1}{\sqrt 3},\frac {1}{\sqrt 3})$ can be our first vector.
a vector othogonal to that (but not necessarily normal) is (1,-1,0).
and vector to the otehr two (but not necessarily normal) is (1,1,-2).
Now that we have the orthogonal set of vector, we can divide each by their norms to get an ortho-normal set.
$\begin {bmatrix}
\frac 1{\sqrt 3}& \frac 1{\sqrt 2} & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 3}& -\frac 1{\sqrt 2} & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 3}& 0 & -\frac 2{\sqrt 6}\end{bmatrix}$
This transform takes $(1,0,0)$ to $(\frac {1}{\sqrt 3},\frac {1}{\sqrt 3},\frac {1}{\sqrt 3})$
So now we transform our vector $(\frac 1{\sqrt2},\frac 1{\sqrt2}, 0)$ to $(1,0,0)$ with an ortho-normal transformation. And the product of the two, will do what we need it to do.
$\begin {bmatrix}
\frac 1{\sqrt 3}& \frac 1{\sqrt 2} & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 3}& -\frac 1{\sqrt 2} & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 3}& 0 & -\frac 2{\sqrt 6}\end{bmatrix}\begin {bmatrix}
\frac 1{\sqrt 2}& \frac 1{\sqrt 2} & 0\\
-\frac 1{\sqrt 2}& \frac 1{\sqrt 2} & 0\\
0& 0 & 1\end{bmatrix}=\begin {bmatrix}
\frac 1{\sqrt 6}-\frac 12& \frac 1{\sqrt 6}+\frac 12 & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 6}+\frac 12& \frac 1{\sqrt 6}-\frac 12 & \frac 1{\sqrt 6}\\
\frac 1{\sqrt 6}& \frac 1{\sqrt 6} & -\frac {2}{\sqrt 6}\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. Prove $(a_n)$ is Cauchy $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.
let $|a_2 - a_1 | =x$.
$|a_3-a_2| \leq \frac{1}{2}x$
$|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$
In general, for every $n$:
$|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$
Let $\epsilon >0$
for every $m,n > 0 $, say $m>n$ the following holds:
$$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$
$$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x
}{2^{n-1}}$$.
I can say that for big enough $n,m$, $|a_m-a_n|<\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
| Your first steps are correct. To finish
$$|a_m - a_n| \leqslant \frac{1}{2^{m-2}} + \frac{1}{2^{m-3}} + \dots + \frac{1}{2^{n-1}} \\ = \frac{1}{2^{n-1}} \sum_{k=0}^{m-n-1}\frac{1}{2^k} \ < \frac{1}{2^{n-1}} \sum_{k=0}^{\infty}\frac{1}{2^k} = \frac{1}{2^{n-1}}\frac{1}{1- 1/2} = \frac{1}{2^{n-2}} .$$
Since $1/2^{n-2} \to 0$ as $n \to \infty$ it follows that for any $\epsilon > 0$ there exists $N$ such that if $n \geqslant N$ and for any $m > n$, then $|a_m - a_n| < \epsilon$ and the sequence is Cauchy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\angle BDC$ Quadrilateral $ABCD$ , $\angle ABD = 17^{\circ}, \angle DBC = 34^{\circ}, \angle ACB = 43^{\circ}, \angle ADB = 13^{\circ}$, Find $\angle BDC$.
| Nice problem!
Let $X$ be a point symmetric to $A$ with respect to $BD$. Let $Y$ be a point symmetric to $A$ with respect to $BX$. Let $Z$ be a point symmetric to $A$ with respect to $DX$.
Then $DA=DX=DZ$, $BA=BX=BY$, and $AX=XY=XZ$. Angle chasing gives
$$\angle ZXY=360^\circ - \angle AXZ - \angle YXA = 360^\circ - 2\angle AXD - 2\angle BXA = \angle XDA + \angle ABX = 2\angle BDA + 2 \angle ABD = 2\cdot 13^\circ+2\cdot 17^\circ = 60^\circ$$
which along with $XZ=XY$ implies that $XYZ$ is equilateral. Thus
$$\angle BZA = \angle BZX + \angle XZA = 30^\circ + 13^\circ = 43^\circ.$$
We also have
$$\angle DBZ = \angle DBX + \angle XBZ = 2\angle ABD = 34^\circ.$$
This means that $Z=C$. Therefore
$$\angle CDB = \angle ZDB = 3\angle BDA = 39^\circ.$$
Below is a trigonometric solution. Let $\angle CDB = x$. We use Snellius' theorem thrice:
\begin{align}
\frac{AC}{DC} & = \frac{\sin(13^\circ + x)}{\sin 64^\circ}, \\
\frac{DC}{CB} & = \frac{\sin 34^\circ}{\sin x}, \\
\frac{CB}{AC} & = \frac{\sin 86^\circ}{\sin 51^\circ}.
\end{align}
Multiplying yields
$$1=\frac{\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ}{\sin 64^\circ \sin x \sin 51^\circ}$$
so $$\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ = \sin 64^\circ \sin x \sin 51^\circ.$$
Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$ twice we get
$$\sin(13^\circ + x)\left(\cos 52^\circ - \cos 120^\circ \right) = \sin x \left(\cos 13^\circ + \cos 65^\circ\right).$$
Since $\cos 120^\circ = -\frac 12$, we have
$$\sin(13^\circ + x) \cos 52^\circ + \frac 12 \sin(13^\circ + x) = \sin x \cos 13^\circ + \sin x \cos 65^\circ.$$
Multiplying by two and using $2\sin A \cos B = \sin(A+B) + \sin(A-B)$ we infer
$$\sin(65^\circ + x) + \sin(x-39^\circ) + \sin(13^\circ + x) = \sin(x+13^\circ) + \sin(x - 13^\circ) + \sin(x+65^\circ) + \sin(x-65^\circ).$$
Therefore
$$\sin(x - 39^\circ) = \sin(x-13^\circ) + \sin(x-65^\circ).$$
We use now $\sin A + \sin B = 2 \sin \frac{A+B}2 \cos\frac{A-B}2$:
$$\sin(x-39^\circ) = 2\sin(x-39^\circ)\cos 26^\circ$$
or
$$\sin(x-39^\circ)(1-2\cos 26^\circ)=0.$$
Since $\cos 26^\circ \neq \frac 12$, we have $\sin(x-39^\circ)=0$ and so $x=39^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Number of triangles with integer sides Total number of right angle triangles whose inradius is $2013$ and sides are integer
Attempt: assuming that $a,b,c>0$ are the sides of a triangle.
so form a right angle triangle with sides $a,b,c$ and right angle at $C$
$\displaystyle r=(s-c)\tan \frac{C}{2}=(s-c)=\frac{a+b-c}{2}$ and $c^2=a^2+b^2$
$\displaystyle 2r = a+b-\sqrt{a^2+b^2}$ so $(\sqrt{a^2+b^2})^2 = (a+b-2r)$
$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$
$\displaystyle ab-2r(a+b)=0$
Could someone help me how to solve it, thanks.
|
$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$
This should be
$$a^2+b^2=(a+b)^2+4r^2-4r(a+b)=a^2+b^2+2ab\color{red}{+4r^2}-4r(a+b)$$
from which we have
$$(a-2r)(b-2r)=2r^2=2^1\times 3^2\times 11^2\times 61^2\tag1$$
where $r=2013=3\times 11\times 61$.
We may suppose that $$-2r\lt a-2r\le b-2r\tag2$$
Note that a necessary and sufficient condition is $(1)$ and
$$(c=)\ a+b-2r\gt 0\iff (a-2r)+(b-2r)\gt -2r\tag3$$
*
*If both $a-2r$ and $b-2r$ are negative, then, from $(1)$ and $(2)$, there is only one case $(a-2r,b-2r)=(-3721,-2178)$. However, this is not sufficient since this does not satisfy $(3)$.
*If both $a-2r$ and $b-2r$ are positive, then, from $(1)$ and $(2)$, the number of cases is the half of the number of the positive divisors of $2^1\times 3^2\times 11^2\times 61^2$, and every case is sufficient since every case satisfies $(3)$.
Therefore, the answer is $$\frac{(1+1)\times (2+1)\times (2+1)\times (2+1)}{2}=\color{red}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $h(85)$ if $h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$ let : $h: \mathbb{R}\to \mathbb{R}$ ane for any real number
$$h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$$
then :
$$h(85)=?$$
My Try: $$x=0:h(3)+2h(5)=17\\x=1:h(5)+2h(3)=13\\+\\3h(3)+3h(5)=30\\h(3)+h(5)=10$$
now ?
| Let $h(x) \equiv Ax+B$, then $$A(x^2+x+3)+B+2A(x^2-3x+5)+2B\equiv 6x^2-10x+17$$
$$
\left \{
\begin{align*}
3A &= 6 \\
-5A &= -10 \\
13A+2B &= 17 \\
\end{align*}
\right.$$
On solving, $(A,B)=\left( 2, -\dfrac{9}{2} \right)$
Alternatively,
*
*$x=\dfrac{-1 \pm \sqrt{329}}{2} \implies x^2+x+3=85 \quad \text{and} \quad x^2-3x+5=89 \mp \sqrt{329}$
*$x=\dfrac{3 \pm \sqrt{329}}{2} \implies x^2-3x+5=85 \quad \text{and} \quad x^2+x+3=89 \pm \sqrt{329}$
*If you substitute $x=\dfrac{-1+\sqrt{329}}{2}$ and $x=\dfrac{3-\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89-\sqrt{329})$.
*If you substitute $x=\dfrac{-1-\sqrt{329}}{2}$ and $x=\dfrac{3+\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89+\sqrt{329})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Checking if two matrices are similar I have two matrices
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
and $$ \begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
They are not diagonalizable. Share the same characteristic polynomial, the same trace, same determinant, eigenvalues, rank. What could I use more to say if they are similar or not?
| The first of your matrices is in Jordan canonical form, which must also be the Jordan canonical form of the second one (because the eigenvalue $2$ has algebraic multiplicity $2$ but geometric multiplicity $1$). So they are similar.
More concretely, we have
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} =
\begin{pmatrix}1 \\ &2 \\ &&1 \end{pmatrix}
\begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix}
\begin{pmatrix}1 \\ &2 \\ &&1 \end{pmatrix}^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$
The hint says to consider $(n+2)(n+1)S$.
Doing so, I get:
$(n+2)(n+1)S = \frac{1}{2}{n+2 \choose 0} + \frac{1}{3}{n+2 \choose 1} + \cdots + \frac{1}{n+2}{n+2 \choose n}$,
but I don't see how this is any easier to solve.
I wonder if somebody might help by expanding on the given hint, and maybe offering their suggestions as to how to use it properly.
Thanks.
| Integration Approach
Since
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n\frac1{k+2}\binom{n}{k}x^{k+2}
&=\sum_{k=0}^n\binom{n}{k}x^{k+1}\\
&=x(1+x)^n
\end{align}
$$
we have
$$
\begin{align}
\sum_{k=0}^n\frac1{k+2}\binom{n}{k}
&=\int_0^1x(1+x)^n\,\mathrm{d}x\\
&=\int_1^2(x-1)x^n\,\mathrm{d}x\\
&=\frac{2^{n+2}-1}{n+2}-\frac{2^{n+1}-1}{n+1}
\end{align}
$$
Pre-calculus Approach
lab bhattacharjee has already given a hint for this approach, but I was working on adding it so I will include it.
Since
$$
\binom{n}{k}=\frac{(k+2)(k+1)}{(n+2)(n+1)}\binom{n+2}{k+2}
$$
and
$$
\binom{n+1}{k+1}=\frac{k+2}{n+2}\binom{n+2}{k+2}
$$
we have
$$
\begin{align}
\sum_{k=0}^n\frac1{k+2}\binom{n}{k}
&=\frac1{(n+1)(n+2)}\sum_{k=0}^n(k+1)\binom{n+2}{k+2}\\
&=\frac1{(n+1)(n+2)}\left[\sum_{k=0}^n(k+2)\binom{n+2}{k+2}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\
&=\frac1{(n+1)(n+2)}\left[(n+2)\sum_{k=0}^n\binom{n+1}{k+1}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\
&=\frac1{(n+1)(n+2)}\left[(n+2)\left(2^{n+1}-1\right)-\left(2^{n+2}-(n+2)-1\right)\right]\\
&=\frac{2^{n+1}}{n+1}-\frac{2^{n+2}-1}{(n+1)(n+2)}
\end{align}
$$
Noting that $\frac1{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}$ and $2^{n+2}=2\cdot2^{n+1}$, we see that the two approaches give the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\int^4_2 xln(x) \:dx$ using integration by parts. Let $f'(x) = x$ and $g(x) = \ln(x) \implies f(x) = \dfrac{x^2}{2}$ and $g'(x) = \dfrac{1}{x}$.
$$\int^4_2 x\ln(x) \:dx = \left[ \dfrac{x^2}{2}\ln(x) \right]^4_2 - \int^4_2 \dfrac{1}{x}\dfrac{x^2}{2} \: dx = 8\ln(4) - 2\ln(2) - 1$$
Apparently, this answer is incorrect. I would greatly appreciate it if people could please explain what I'm doing wrong.
| $$\int _{ 2 }^{ 4 } xln(x)\: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\int _{ 2 }^{ 4 }{ \frac { { x } }{ 2 } } \: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\left[ { \frac { x^{ 2 } }{ 4 } } \right] ^{ 4 }_{ 2 }=\frac { 16 }{ 2 } \ln { 4 } -\frac { { 2 }^{ 2 } }{ 2 } \ln { 2 } -\left[ \frac { 16 }{ 4 } -\frac { 4 }{ 4 } \right] =\\ =14\ln { 2 } -3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$ Help with this limit:
$$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$$
Using L'Hopital rule gives $2m$ but I need it without using L' Hopital.
| In the same spirit as Yves Daoust's answer, start developing $\sqrt{1+x^2}$ using the generalized binomial theorem or Taylor series. This would give
$$\sqrt{1+x^2}=1+\frac{x^2}{2}+O\left(x^3\right)$$ $$\sqrt{1+x^2}+x=1+x+\frac{x^2}{2}+O\left(x^3\right)$$
$$\sqrt{1+x^2}-x=1-x+\frac{x^2}{2}+O\left(x^3\right)$$ Now, the binomial theorem again
$$(\sqrt{1+x^2}+x)^m=1+m x+\frac{m^2 x^2}{2}+O\left(x^3\right)$$
$$(\sqrt{1+x^2}-x)^m=1-m x+\frac{m^2 x^2}{2}+O\left(x^3\right)$$ from which the result easily follows.
Just for your curiosity, if you were using the initial expansions up to $O\left(x^4\right)$, you would get
$$(\sqrt{1+x^2}+x)^m=1+m x+\frac{m^2 x^2}{2}+\frac{1}{6} m \left(m^2-1\right) x^3+O\left(x^4\right)$$
$$(\sqrt{1+x^2}-x)^m=1-m x+\frac{m^2 x^2}{2}-\frac{1}{6} m \left(m^2-1\right) x^3+O\left(x^4\right)$$ which would make
$$\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}=2 m+\frac{1}{3} m \left(m^2-1\right) x^2+O\left(x^3\right)$$ which will show the limit and also how it is approached.
| {
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"url": "https://math.stackexchange.com/questions/2173674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$
Find the value of $abc$.
Please check if my answer is correct or not.
$a+b+c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32 +a+b+c$
$(a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) = 32 +a+b+c$
$a+b+c = 4$
$a^2+b^2+c^2 = 6$
$ab+bc+ca = 5$
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + 3= 12$
$(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 12$
$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 3$
$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{2abc+\sum{a^2b} + \sum{ab^2}} = 3$
$7 = (ab+bc+ca)(a+b+c)-abc$
$abc = 13$
| $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=9$, eqI
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=32$, eqII
$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}=122$ eqIII
Let $a+b+c=k \Rightarrow$
$ a=k-(b+c), b=k-(a+c), c=k-(a+b)$. (eqIV)
Using (eqI): $\frac{k-(b+c)}{b+c}+\frac{k-(a+c)}{a+c}+\frac{k-(a+b)}{a+b}=9 \Rightarrow \frac{k}{b+c}-1+\frac{k}{a+c}-1+\frac{k}{a+b}-1=9 \Rightarrow$
$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $ (eqV).
Lets apply (eqIV) in (eqII):
$\frac{(k-(b+c))²}{b+c}+\frac{(k-(a+c))²}{a+c}+\frac{(k-(a+b))²}{a+b}=32$
$\frac{k²-2k(b+c)+(b+c)²}{b+c}+\frac{k²-2k(a+c)+(a+c)²}{a+c}+\frac{k²-2k(a+b)+(a+b)²}{a+b}=32$
$k²(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-6k+(b+c)+(a+c)+(a+b)=32$
$k.[k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})]-4k=32$
Using (eqV):
$ k.12-4k=32 \Rightarrow k=4$
Lets do the same with (eqIV) in (eqIII):
$\frac{(k-(b+c))³}{b+c}+\frac{(k-(a+c))³}{a+c}+\frac{(k-(a+b))³}{a+b}=122$
$\frac{k³-3k²(b+c)+3k(b+c)²+(b+c)³}{b+c}+\frac{k³-3k²(a+c)+3k(a+c)²+(a+c)³}{a+c}+\frac{k³-3k²(a+b)+3k(a+b)²+(a+b)³}{a+b}=122$
$k²(\frac{k}{b+c}+\frac{k}{a+c}+\frac{k}{a+b})
-3k²(1+1+1)+3k((b+c)+(a+c)+(a+b))-((b+c)²+(a+c)²+(a+b)²)=122$
$k²(12)-9k²+3k(2k)-2(a²+b²+c²+ab+bc+ac)=122$
Using that k=4, $144-2(a²+b²+c²+ab+bc+ac)=122$
$ a²+b²+c²+ab+bc+ac = 11$
But we know that $k=a+b+c=4 \Rightarrow (a+b+c)²=(a²+b²+c²)+2(ab+bc+ac)$
Lets call $S_2=a²+b²+c²$ and $\sigma_2=ab+bc+cd$. So
$ S_2+\sigma_2= 11$
$ S_2+2\sigma_2= 16$
Solving this, we have $S_2=6$ and $\sigma_2=5$.
But $(a+b+c)^3=a³+b³+c³+3(a+b+c)(\sigma_2)-3abc \Rightarrow S_3=a³+b³+c³=4^3-3*4*5+3abc$
$\Rightarrow S_3=4+3abc$
$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $
$ (\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=3 $
$ \frac{(b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b)}{(b+c)(a+c)(a+b)}= 3$
$ \frac{(ab+ac+bc+c²)+(ba+b²+ac+bc)+(a²+ab+ac+bc)}{a²(b+c)+b²(a+c)+c²(a+b)+2abc}=3 $
$ \frac{S_2+3\sigma_2}{a²(4-a)+b²(4-b)+c²(4-c)+2abc}=3 $
$ \frac{S_2+3\sigma_2}{4S_2-S_3+2abc}=3 $
$ \frac{6+3*5}{4*6-(4+3abc)+2abc}=3 $
$ abc=13 $
| {
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"url": "https://math.stackexchange.com/questions/2174348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$ If $a,b,c$ ${\in}$ $\mathbb{R} $ such that
$a+2b+c=4$, then find the $max$ value of
$ab+bc+ca$.
I always get stuck with max, min questions. We cannot apply AM:GM here, I have not studied much calculus yet.
Can you do this with graphs or by plane algebra?
I don't really know what tag to put for this!
| $$a+b+c=4-b$$
$$a^2+b^2+c^2+2(ab+ac+bc)=16-8b+b^2$$
$$ab+bc+ca=\frac{16-8b-a^2-c^2}2$$
Since $b=\frac{4-a-c}2$,
$$ab+bc+ca=\frac{16-16+4a+4c-a^2-c^2}2=\frac{a(4-a)+c(4-c)}2\le\frac{4+4}2=4$$
| {
"language": "en",
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"source": "stackexchange",
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Trig Integration $\displaystyle\int\sec^3(x)\tan^2(x)\,\mathrm{d}x$ $$\int \sec^3\left(x\right)\tan^2\left(x\right)\,\mathrm{d}x$$
Hi, for the question above, I think that substituting $\tan(x)$ would be ok but I couldn't figure out the final step.
$$\int\sec^2(x)\sec(x)\tan^2(x)\,\mathrm{d}x$$
$$\int \sec\left(x\right)u^2(x)\,\mathrm{d}u$$
The answer to this solution is $\sec(x)\tan^3(x)/3$ but it is not correct.
| The popular approach is to convert it into integrals of odd powers of secant and then use reduction formulas derived from integration by parts, but there is an alternative.
You can make the substitution $u = sec(x) + tan(x)$.
Note that $1/u = sec(x) - tan(x)$,
because $(sec(x)+tan(x))((sec(x)-tan(x)) = sec^2(x)-tan^2(x) = 1$.
We have:
$$
\begin{align}
\dfrac{u^2 + 1}{2u} &= \dfrac{[(\sec(x) + \tan(x))^2] + [1]}{2[\sec(x) + \tan(x)]} \\
&= \dfrac{[\sec^2(x) + 2\sec(x)\tan(x) + \tan^2(x)] + [\sec^2(x) - \tan^2(x)]}{2[\sec(x) + \tan(x)]}\\
&= \dfrac{[2\sec^2(x) + 2\sec(x)\tan(x)}{2[\sec(x) + \tan(x)]}\\
&= \dfrac{2\sec(x)[\sec(x) + \tan(x)]}{2[\sec(x) + \tan(x)]}\\
&= \sec(x)\\
\end{align}
$$
And also:
$$
\begin{align}
\dfrac{u^2 - 1}{2u} &= \dfrac{[(\sec(x) + \tan(x))^2] - [1]}{2[\sec(x) + \tan(x)]} \\
&= \dfrac{[\sec^2(x) + 2\sec(x)\tan(x) + \tan^2(x)] - [\sec^2(x) - \tan^2(x)]}{2[\sec(x) + \tan(x)]}\\
&= \dfrac{[2\tan^2(x) + 2\sec(x)\tan(x)}{2[\sec(x) + \tan(x)]}\\
&= \dfrac{2\tan(x)[\sec(x) + \tan(x)]}{2[\sec(x) + \tan(x)]}\\
&= \tan(x)\\
\end{align}
$$
And lastly:
$$
\begin{align}
du &= (\sec(x) + \tan(x))' dx \\
&= (\sec(x)\tan(x) + \sec^2(x)) dx \\
&= \sec(x)(\sec(x) + \tan(x)) dx \\
&= \left(\dfrac {u^2 + 1}{2u}u\right) dx\\
&= \left(\dfrac {u^2 + 1}{2}\right) dx\\
\end{align}
$$
$$\implies dx =\left(\dfrac { 2 }{u^2 + 1}\right)du$$
So our integral becomes:
$$
\begin{align}
\int \tan^2(x)\sec^3(x) dx &= \int \left(\dfrac{u^2 - 1}{2u}\right)^2 \left(\dfrac{u^2 + 1}{2u}\right)^3 \left(\dfrac { 2 }{u^2 + 1}\right)du \\
&= \frac{1}{16} \int \dfrac{(u^2 - 1)^2(u^2 + 1)^2}{u^5}du \\
&= \frac{1}{16} \int \dfrac{(u^4 - 1)^2}{u^5}du \\
&= \frac{1}{16} \int \dfrac{u^8 - 2u^4 + 1}{u^5}du \\
&= \frac{1}{16} \int \left(u^3 - \dfrac{2}{u} + u^{-5}\right) du \\
&= \frac{1}{16} \left(\frac{1}{4}u^4 - 2\ln |u| - \frac{1}{4}u^{-4}\right) + C\\
&= \frac{1}{64}(u^4 - u^{-4}) - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{64}(u - u^{-1})(u + u^{-1})(u^2 + u^{-2}) - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{16}\left(\frac{u^2 - 1}{2u}\right)\left(\frac{u^2 + 1}{2u}\right)\left(u^2 + \left(\frac{1}{u}\right)^2\right) - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{16}[\tan(x)\sec(x)]\left[(\sec(x)+\tan(x))^2 + (\sec(x)-\tan(x))^2\right] - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{16}[\tan(x)\sec(x)][2\tan^2(x)+2\sec^2(x))] - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{16}[\tan(x)\sec(x)][4\tan^2(x)+2] - \frac{1}{8}\ln |u| + C\\
&= \frac{1}{4}tan^3(x)sec(x)+\frac{1}{8}tan(x)sec(x) - \frac{1}{8}\ln |sec(x)+tan(x)| + C\\
\end{align}
$$
If you're satisfied with an answer in terms of $u$, that's fine.
But I can see why people use the reduction formula.
| {
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Factorise the expression Factorise of the expression $$ a^4+b^4-c^4-2a^2b^2+4abc^2~ \\(S:(a + b - c) (a + b + c) (a^2+ b^2 + c^2 - 2 a b))$$
I tried to solve this problem:
$$ a^4+b^4-c^4-2a^2b^2+4abc^2\rightarrow\\\
(a^2-b^2)^2-c^4+4abc^2\rightarrow\\\
(a^2-b^2-c^2)(a^2-b^2+c^2)+4abc^2\\\
((a-b)(a+b)-c^2)((a-b)(a+b)+c^2)+4abc^2$$ but I stopped here. Can anyone help me?
| As a quadratic in $c^2\,$, the reduced discriminant of $-(c^4-4ab \cdot c^2-a^4-b^4+2a^2b^2)$ is:
$$
\frac{1}{4}\Delta = 4a^2b^2 +a^4+b^4-2a^2b^2=(a^2+b^2)^2
$$
Therefore the roots are $c^2=2ab \pm (a^2+b^2) = \{(a+b)^2, -(a-b)^2\}\,$ giving the factorization:
$$
\begin{align}
-\left(c^2-(a+b)^2\right)\left(c^2+(a-b)^2\right) & = -(c-a-b)(c+a+b)(c^2+a^2-2ab+b^2) \\[3px]
& = (a+b-c)(a+b+c)(a^2+b^2+c^2-2ab)
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$ We have to find the number of solution for the given equation:
$$\sqrt{3x^2+x+5} = x-3.$$
There are two solution one is
By using graph we get one solution
By squaring both sides we get no solution
I want to know which solution is correct
| we have $$\sqrt{3x^2+x+5}=x-3$$ it must be $$x\geq 3$$ and after squaring we get
$$3x^2+x+5=x^2-6x+9$$ this is equivalent to $$2x^2+7x-4=0$$ solving this quadratic equation we obtain
$$x_{1,2}=-\frac{7}{4}\pm\sqrt{\frac{49}{16}+\frac{32}{16}}$$
Can you finish this?
| {
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"url": "https://math.stackexchange.com/questions/2181765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle t^3+at^2+(b-3)t+1=0,$ where $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $|t|\geq 2$
Could some help me to solve it, thanks
| The minimum number of real roots is $2$ counted with multiplicity, or $1$ counted without multiplicity.
Since $p(x)$ is a reciprocal polynomial (that is, $x^6p(\frac1x)=p(x)$), its roots come in pairs $r,\frac1r$. In particular, if $1$ is a root of $p(x)$ then it is a double root. It is possible for these to be the only roots: take for example, $a=-\frac{19}{12}$ and $b=\frac1{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving for $p$ and $k$. For the question,
$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$
I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.
What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square of the two quadratics of $F(x)$ separately then added the two terms while also correcting for coefficients. I doubt this operation however went for it anyway due to any alternatives seem wrong.
I reached the term $-\frac {39}4k + 11$ then promptly equated it to $12 - 2k$.
I reached the value of $-\frac{4}{31}$ for $K$.
My textbook has answers for $k = \frac{2}{7}$ and $p=\frac{80}{7}$.
| You can try completing the square directly. Notice that the minimum of $F(x)$ lies at $x = -2$, and $F(-2) = p$. This means that
$$F(x) = a(x + 2)^2 + p$$
for some constant $a$. We can't assume $a = 1$ because any non-zero value of $a$ can work, so we leave it there as it is.
Now we go ahead and solve as usual by comparing coefficients:
$$a(x + 2)^2 + p = x^2 + 6x + 20 + k(x^2 - 3x - 12)$$
$$ax^2 + 4ax + (4a + p) = (k + 1)x^2 + (6 - 3k)x + (20 - 12k)$$
Comparing coefficients, we have
$$k + 1 = a~~~~,~~~~4a = 6 - 3k~~~~,~~~~4a + p = 20 - 12k$$
Firstly,
$$4a = 6 - 3(a - 1)$$
$$4a = 9 - 3a$$
$$7a = 9$$
$$a = \frac{9}{7}$$
Secondly,
$$\begin{align}k &= a - 1 \\&= \frac{9}{7} - 1 \\&= \frac{2}{7}\end{align}$$
And lastly,
$$4a + p = 20 - 12k \implies 4\cdot\frac{9}{7} + p = 20 - 12 \cdot\frac{2}{7}$$
$$p = \frac{80}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$ Consider the integrals $(1)$ and $(2)$, how does on show that
(1): $I=J$
(2): and $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$
$$\int_{0}^{\pi/2}{\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=I\tag1$$
$$\int_{0}^{\pi/2}{\tan^3 x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=J\tag2$$
An attempt:
$u=\tan x$ $\implies du=\sec^2 x dx$
$(1)$ becomes
$$\int_{0}^{\infty}{u\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^2}\tag3$$
$v=1+u^2$ then
$${1\over 2}\int_{1}^{\infty}{\mathrm dv\over \sqrt{v[1+(v-1)^3]}}\tag4$$
$u=\tan^3 x$ $\implies du=3\tan^2 x\sec^2 x dx$
$(2)$ becomes
$${1\over 3}\int_{0}^{\infty}{u^{1/3}\over\sqrt{(1+u^2)(1+u^6)}}\cdot{\mathrm du\over 1+u^{2/3}}\tag5$$
We are not sure how to continue from here...
| On the path of Lab Bhattacharjee,
$J=\displaystyle \int_0^{+\infty} \dfrac{1}{\sqrt{(1+x)(1+x^3)}}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}\displaystyle J&=\int_{-1}^{1} \dfrac{1}{\sqrt{1+3x^2}}dx\\
\end{align}$
Perform the change of variable $y=\sqrt{3}x$,
$\begin{align}
J&=\dfrac{1}{\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{1}{\sqrt{1+x^2}}dx\\
&=\dfrac{1}{\sqrt{3}}\Big[\text{arcsinh}(x)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\Big[\ln\left(x+\sqrt{1+x^2}\right)\Big]_{-\sqrt{3}}^{+\sqrt{3}}\\
&=\dfrac{1}{\sqrt{3}}\ln\left(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\right)\\
&=\boxed{\dfrac{1}{\sqrt{3}}\ln\left(7+4\sqrt{3}\right)}
\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Improper integral $\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)} $ I know that this integral is convergent, but I don't know how to prove it.
$$\int_0^\infty\frac{{\rm d}x}{1+x^4\sin^2(x)}$$
| Split the integral up into portions where $\sin(x)$ is small (i.e., where the best bound we can hope for is $\frac{1}{1+x^4\sin^2(x)} \le 1$), and portions where $\sin(x)$ is relatively large so that we can use the growth of $x^4$ to our advantage. For the former, we use intervals $I_n =(n\pi - \frac{1}{n^\alpha}, n\pi + \frac 1 {n^\alpha})$ for some $\alpha > 0$. We see $$\sum_{n\in \mathbb N}\int_{I_n} \frac{dx}{1+x^4\sin^2(x)} \le \sum_{n\in\mathbb N}\int_{I_n} dx = 2 \sum_{n\in\mathbb N} \frac 1 {n^\alpha}.$$
Outside these intervals, we are in the intervals $$J_n = \left[ \left(n-\tfrac 1 2\right)\pi, n\pi - \tfrac{1}{n^\alpha} \right] \cup \left[ n\pi + \tfrac 1 {n^\alpha}, \left(n+\tfrac 1 2 \right)\pi\right].$$ In these intervals, we can use $$\lvert \sin(x) \rvert \ge \lvert x-n\pi\rvert / 2$$ for $x \in \left(\left(n-\tfrac 1 2\right)\pi, \left(n+\tfrac 1 2\right)\pi\right)$. Indeed, this inequality can be proven easily for the interval around $x=0$ and then extended periodically to any $n \in \mathbb Z.$ Using this, we see $$\frac 1 {1+x^4 \sin^2(x)}\le \frac{1}{1+\tfrac 1 4x^4(x-n\pi)^2}, \,\,\,\,\,\, x \in J_n.$$ However, on $J_n$, the distance from $x$ to $n\pi$ is bigger than $1/n^\alpha$ and $x \ge \left(n-\frac{1}{2}\right)\pi \ge \frac{n\pi}{2}$, so we have $$\frac 1 {1+x^4 \sin^2(x)}\le \frac{1}{1+\tfrac 1 4x^4(x-n\pi)^2} \le \frac{1}{1 + \tfrac 1 {32}\pi^4 n^{4-2\alpha}} , \,\,\,\,\,\, x \in J_n.$$ Thus these intervals contribute at most $$\sum_{n\in \mathbb N}\int_{J_n} \frac{dx}{1+x^4\sin^2(x)} \le \sum_{n\in\mathbb N}\int_{J_n} dx = 1 \sum_{n\in\mathbb N} \frac 1 {1 + \tfrac 1 {32}\pi^4 n^{4-2\alpha}}.$$ Hence, we have \begin{align*} \int^\infty_0 \frac{dx}{1+x^4\sin^2(x)} &\le \sum_{n \in \mathbb N}\int_{I_n} \frac{dx}{1+x^4\sin^2(x)} + \sum_{n\in \mathbb N}\int_{J_n} \frac{dx}{1+x^4\sin^2(x)} \\ &\le 2\sum_{n\in \mathbb Z} \frac{1}{n^\alpha} + \sum_{n\in\mathbb N} \frac 1 {1 + \tfrac 1 {32}\pi^4n^{4-2\alpha}}\end{align*} Finally, if we can choose an $\alpha >0$ so that both sums converge, then the integral will also converge. It turns out that $1 < \alpha < 3/2$ will make both sums converge (just use the comparison test for both; we just need the exponents on $n$ to be bigger than $1$).
| {
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"source": "stackexchange",
"question_score": "2",
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Find $1^2+3^2+...99^2$ given $1^1+2^2+...+100^2$ and $1^1+2^2+...+50^2$ We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?
| This should help: $$2^2+4^2+6^2+\dots+100^2=2^2(1^2+2^2+3^2+\dots+50^2)=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*$x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
*$x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
*$x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
*$x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$
Therefore, $x = e$.
| Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to
$$e=x^9=x(x^2)^4=xe^4=x$$and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192881",
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"question_score": "7",
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$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{6\pi}{7})}$$
$$L=\frac{1}{1+2\cos (\frac{6\pi}{7})}+\frac{1}{1+2\cos (\frac{4\pi}{7})}+\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
what now ?
| Use factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
or if $7x=2m\pi$ where $m$ is any integer
$\sin4x=\sin(2m\pi-3x)=-\sin3x$
$\implies4\sin x\cos x\cos2x=\sin x(4\sin^2x-3)$
$\implies4\sin x\cos x(2\cos^2x-1)=\sin x\{4(1-\cos^2x)-1\}$
So, the roots of $4\cos x(2\cos^2x-1)=4(1-\cos^2x)-1\iff8\cos^3x+4\cos^2x-4\cos x-3=0\ \ \ \ (1)$
are $7x=2m\pi$ where $m\equiv\pm1,\pm2,\pm3\pmod7$
Now if $u=\dfrac1{1+2\cos x}\iff \cos x=?$
Replace the values of $\cos x$ in $(1)$
| {
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Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization obtained?
| This solution was suggested by dxiv.
Solution
The given quartic polynomial factors into a product of two monic, quadratic polynomials:
\begin{equation*}
x^{4} - 144x^{2} + 1728x - 5184 = (x^{2} + ax + b)(x^{2} + cx + d) ;
\end{equation*}
\begin{equation*}
(x^{2} + ax + b)(x^{2} + cx + d) = x^{4} + (a + c)x^{3} + (ac + b + d)x^{2} + (ad + bc)x + bd .
\end{equation*}
Since $c = -a$,
\begin{equation*}
d - b = \dfrac{12^{3}}{a}
\qquad \text{and} \qquad
b + d = a^{2} - 12^{2}
.
\end{equation*}
So,
\begin{equation*}
b = \frac{1}{2a} \Bigl(-12^{3} - 12^{2}a + a^{3} \Bigr)
\qquad \text{and} \qquad
d = \frac{1}{2a} \Bigl( 12^{3} - 12^{2}a + a^{3} \Bigr) .
\end{equation*}
Moreover, since $bd = -5184 = -3(12^{3})$, and
\begin{equation*}
bd = \frac{1}{4a^{2}} \Bigl(a^{6} - 2(12^{2}) a^{4} + 12^{4} a^{2} - 12^{6}\Bigr) ,
\end{equation*}
\begin{equation*}
a^{6} - 2(12^{2}) a^{4} + 2(12^{4}) a^{2} - 12^{6} = 0 .
\end{equation*}
$\pm12$ are roots of this polynomial equation in the variable $a$. If $a = 12$, $b = -72$, $c = -12$, and $d = 72$; if $a = -12$, $b = 72$, $c = 12$, and $d = -72$. In either case,
\begin{align*}
&x^{4} - 144x^{2} + 1728x - 5184 \\
&\qquad = (x^{2} + 12x - 72)(x^{2} - 12x + 72) \\
&\qquad = \bigl(x - 6\sqrt{3} + 6\bigr)\bigl(x + 6\sqrt{3} + 6\bigr)(x^{2} - 12x + 72) .
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^6 -3r^4 +3r^2 - 1 = (r-1)^3(r+1)^3$. How do you get this?
| Observe you have
\begin{align}
r^6-3r^4+3r^2-1 = (r^2-1)^3
\end{align}
by the binomial theorem. Then it follows $(r^2-1) = (r+1)(r-1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thanks
| $\dfrac{2x - \frac{1}{x^2}}{x^4 -x^2 +2x -2 -\frac{1}{x} + \frac{1}{x^2}}$
We want to write the denominator in $x^2 + \frac{1}{x}$ since it is derivative in the nominator
Reordering
$$x^4 +2x +\frac{1}{x^2}-2 -x^2 - \frac{1}{x} = \left(x^2 +\frac{1}{x}\right)^2 - 2 -(x^2 + \frac{1}{x}) $$
Let $ u = x^2 + \frac{1}{x^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| Compare $a^2+b^2+c^2-ab-bc-ca$ with $(a+b+c)^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Integrating $\frac{1}{u}$ Say I have $\int_{}{}\frac{\frac{1}{2}}{u + 1}du$
Why is $\frac{1}{2}\ln(u + 1) \neq \frac{1}{2}\ln(2u + 2)$?
Assuming I did this:
$$\int_{}{}\frac{\frac{1}{2}}{u + 1}\ du = \int_{}{}\frac{1}{2u + 2}\ du$$
| You are forgetting the constant for an indefinite integral.
Note that
$$\frac{1}{2} \int \frac{du}{u+1} = \frac{1}{2} \ln (u+1) + C,$$
and
$$\frac{1}{2} \ln(2u+2) = \frac{1}{2} \ln (u+1 ) + \frac{\ln 2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is
(a)-1
(b)9
(c)6
(d)1
Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.
Please keep in mind that this is an objective question in one of the competitive exam and you get around 5 mins to solve it.
Also this is asked in elementary section, so only knowledge of basic calculus and polynomials is assumed.
| $f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$
$= x^4(x^4 - 8x^2 + 16) +19x^4 - 16x^4 - 12 x^3 + 14x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x^2 - 4x + 4) + 14x^2 -12x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x - 2)^2 + 2(x^2 - 4x + 4) + 1$
$= (x - 2)^2(x^4(x+2) + 3x^2 +2) + 1 \ge 1$ with equality holding iff $x = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If $B= ((X-1)^2, X-1, 1)$ is a base of $\mathbb{R}_2[X]$, how do I the coordinates of a polynomial in this base? Suppose you have a base: $B_2=((X-1)^2, (X-1), 1)$ and a polynomial of the form $a + bX + cX^2$ in base $B_1(X^2, X, 1)$, how do I find the its coordinates in $B_2$?
I figured maybe trying to find the matrix of the base change, and as $B_2=((X^2 -2X +1), (X-1),1))$ this would give: $$ \begin{bmatrix}
1 & 0 &0 \\
-2& 1& 0\\
1& -1& 1
\end{bmatrix}$$
And the coordinates of the polynomial in $B_1$ are $(a,b,c)$, then I would simply multiply the vector of the coordinates by the $B_2$, which yields this $$[a-2b+c, b-c, c ] $$
Is this correct?
| \begin{align}
x^2 &= 1\cdot(x-1)^2 + 2\cdot(x-1) + 1\cdot1 \\
x &= 1\cdot(x-1) + 1\cdot1 \\
1 &= 1\cdot1 \\ \\
\Rightarrow ax^2 + bx + c &= a((x-1)^2 + 2(x-1) + 1) + b((x-1) + 1) + c \\
&= a\cdot(x-1)^2 + (2a + b)\cdot(x-1) + (a+b+c)\cdot 1
\end{align}
That is we need something that maps $(a,b,c)$ to $(a, 2a+b, a+b+c)$.
The matrix that belongs to this transformation is the one on the left side of the following equation.
$$\pmatrix{1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 1 & 1} \pmatrix{a\\b\\c} = \pmatrix{a\\ 2a + b \\ a+ b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How many different knight's moves are there on an $n \times n$ chessboard? I have found there to be $48$ total moves on a $4 \times 4$ board... and $96$ on a $5 \times5$... but I can not see the relevance to each other in terms of a "$n \times n$" board.
By "moves" I am referring to from each space the amount of moves available from it.
Example
| Extending your example ...
For the top row you get 2,3,4,4,....,4,4,3,2
For the second row you get 3,4,6,6,....,6,6,4,3
For the third 4,6,8,8,...,8,8,6,4
For the fourth also 4,6,8,8,...,8,8,6,4
...
[and then the mirror of that for the bottom 4 rows]
So:
With $n\ge 4$ you get:
$2+3+(n-4)*4+3+2=10 +4*(n-4)$ for rows 1 and $n$
$3+4+(n-4)*6+4+3=14+6*(n-4)$ for rows 2 and $n$
$(n-4)*(4+6+(n-4)*8+6+4=(n-4)*(20+8*(n-4))$ for all other rows
For a total of
$$2*(10+14)+ (2*(4+6)+20)*(n-4)+ 8*(n-4)^2 = $$
$$48+40*(n-4)+8*(n-4)^2$$
Check: for $n=4$ we indeed get 48. For $n=5$, we get $48+40+8=96$ (So your 98 was a bit off)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\log\left(\left(\frac{b}{a}\right)^{\frac{b}{3}}\right) + \log\left(\left(\sqrt[3]{\frac{a}{b}}\right)^{9a}\right)=1$, then what is $a^2+b^2$? Let $a$ and $b$ be two positive integers where $b$ is a multiple of $a$. If $\log\left(\left(\frac{b}{a}\right)^{\frac{b}{3}}\right) + \log\left(\left(\sqrt[3]{\frac{a}{b}}\right)^{9a}\right)=1$then what is $a^2+b^2$?
| $b = na$
$b/3 (\log b - \log a) + 3a (\log a - \log b) = 1$
$(na/3 - 3a)(\log n + \log a - \log a) = 1$
$a (n/3 - 3) \log n = 1$
$n = 10, a = 3, b = 30$ is a solution
you can see that you won't find more solution in integers (e.g. n = 10^k)
| {
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"timestamp": "2023-03-29T00:00:00",
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prove an inequality with conditions Let $a,b, c \in \mathbb{R}$ such that $0\le a\le1,0\le b\le1 , 0\le c\le1.$
If $$a+b\leq c+1,
\\ a+c \leq b+1,
\\b+c\leq a +1$$
can we prove that
$a^2+b^2+c^2\le 1+2abc$ ?
| Let $a+1-b-c=x$, $b+1-a-c=y$ and $c+1-a-b=z$.
Hence, $x$, $y$ and $z$ are non-negatives such that $x+y+z=3-a-b-c\leq3$,
$a=\frac{2-y-z}{2}$, $b=\frac{2-x-z}{2}$, $c=\frac{2-x-y}{2}$
and we need to prove that
$$\frac{1}{4}\sum_{cyc}(2-x-y)^2\leq1+\frac{1}{4}\prod_{cyc}(2-x-y)$$ or
$$\sum_{cyc}(4+2x^2-8x+2xy)\leq4+8-\prod_{cyc}(x+y)+\sum_{cyc}(-8x+2x^2+6xy)$$ or
$$4(xy+xz+yz)\geq(x+y)(x+z)(y+z)$$
and since $1\geq\frac{x+y+z}{3}$, it's enough to prove that
$$\frac{4}{3}(x+y+z)(xy+xz+yz)\geq(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(x^2y+x^2z+2xyz)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}$ let $f(x)= (x^3+2x)[\frac{x}{2}]$
[x]:floor function
then :
$$\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}=?$$
My try :
$$\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{h}=f'(2)$$
$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-f(2)}{x-2}=f'(2)$$
$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-12}{x-2}=f'(2)$$
Now what ?
| Observe that as $h\rightarrow 0$, $h^2\rightarrow 0^+$ since $h^2$ is always nonnegative. Therefore, we can write:
$$
\lim_{h\rightarrow 0}\frac{f(2-h^2)-f(2)}{h^2}=\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}
$$
by setting $k=h^2$. Then, by substituting the formula for $f$, we get
\begin{align*}
\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}&=\lim_{k\rightarrow 0^+}\frac{((2-k)^3+2(2-k))\left\lfloor\frac{2-k}{2}\right\rfloor-(2^3+4)\left\lfloor\frac{2}{2}\right\rfloor}{k}\\
&=\lim_{k\rightarrow 0^+}\frac{(-k^3+6k^2-14k+12)\left\lfloor\frac{2-k}{2}\right\rfloor-12}{k}
\end{align*}
Since, when $k$ is slightly larger than $0$, $2-k$ is slightly less than $2$, $\frac{2-k}{2}$ is slightly less than $1$, so its floor is $0$. This implies that we're looking at
$$
\lim_{k\rightarrow0^+}\frac{-12}{k}
$$
which does not exist.
On the other hand,
$$
\lim_{h\rightarrow 0}\frac{f(2+h^2)-f(2)}{h^2}
$$
does exist and the limit is $14$.
| {
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"url": "https://math.stackexchange.com/questions/2206971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$ If $f:R\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$
My Attempt :
$$f(x)=x^2-3$$
$$y=x^2-3$$
Then how to proceed further?
| $$
f\left( x \right) = x^{2} - 3
$$
The function has two $x$ values that map to $y = -1$.
$$
\color{blue}{f^{-1}(y) = \sqrt{x+3}} \qquad \Rightarrow \qquad f^{-1}\left( -1 \right) = \sqrt{2}
$$
$$
\color{red}{f^{-1}(y) = -\sqrt{x+3}}\qquad \Rightarrow \qquad f^{-1}\left(-1 \right) = -\sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $Col(A)$ and $Col(B)$ are linearly independent, then $AB = 0$ What does $Col(A)$ and $Col(B)$ are linearly independent mean?
It is from an old linear algebra book and I don't see this used very often, does it mean a basis of $Col(A)$ and a basis of $Col(B)$ are linearly independent?
Assume the above definiton, then it means $Col(A)$ and $Col(B)$ are linearly independent iff $Col(A) \cap Col(B) = \{0\}$.
Just from this how could I show $AB = 0$?
Edit: I guess we need $A$ and $B$ to be symmetric, from the counter example $\begin{bmatrix} 1 & 2 \\ 1 &2 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 &1 \end{bmatrix}$.
| First, an example:
$$
\begin{align}
\mathbf{A} \mathbf{B} &= \mathbf{0} \\[5pt]
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)\,
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
&=
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\end{align}
$$
Certainly $\mathbf{A} \mathbf{B} = 0$.
Look at the matrix-vector multiplication
$$
\mathbf{A} \mathbf{B} x \overset{?}{=} \mathbf{0}
$$
Define
$$
y = \mathbf{B} x
$$
When does
$$
\mathbf{A} y = \mathbf{0}?
$$
When the vector $y$ is not in the column space of $\mathbf{A}$. Yet the vector $y$ is in the column space of $\mathbf{B}$ by construction:
$$
y = \mathbf{B} x = x_{1} \mathbf{B}_{1} + x_{2} \mathbf{B}_{2} + \dots
$$
It is assembled using the column vectors of $\mathbf{B}$.
The proof demonstrates that if the matrix product is $\mathbf{0}$, the multiplicand matrices must be linearly independent.
As noted by @Nan Li, it does not show the reverse direction, that the product of regular matrices is necessarily $\mathbf{0}$.
| {
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"url": "https://math.stackexchange.com/questions/2213429",
"timestamp": "2023-03-29T00:00:00",
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Solve in $\mathbb Z^3$.
Let $x$, $y$ and $z$ be integer numbers. Solve the following equation.
$$x^2+y^2+z^2=45(xy+xz+yz)$$
My trying.
It's a quadratic equation of $z$ and we need $\Delta=n^2$ for an integer $n$,
but it gives a very ugly expression.
Thank you!
| For Will Jagy, I am sorry!
Let $x-y=a$, $y-z=b$ and $z-x=c$.
Hence, $a^2+b^2+c^2\vdots11$ and $a+b+c=0$.
Thus, $a^2+ab+b^2\vdots11$, which says that $a\vdots11$ and $b\vdots11$ and $x\equiv y\equiv z(\mod11)$,
which gives $x\vdots11$, $y\vdots11$ and $z\vdots11$ (if $x\equiv y\equiv z\equiv r(\mod11)$ then $r^2\vdots11$).
Id est, an infinite descent ends this problem.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta=\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=2\frac{(1+2\cos\theta)^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=0
$$
so, 0 area is clearly wrong. Where was my mistake?
| The outer curve of the limacon lies between $\theta=\pm\frac{2\pi}{3}$ but using symmetry it is twice the area between $\theta=0$ and $\theta=\frac{2\pi}{3}$
\begin{eqnarray}
A&=&2\int_0^{2\pi/3}\frac{1}{2}r^2d\theta\\
&=&\int_0^{2\pi/3}(1+2\cos\theta)^2d\theta\\
&=&\int_0^{2\pi/3}3+4\cos\theta+2\cos(2\theta)\,d\theta
=[3\theta+4\sin\theta+\sin(2\theta)]_0^{2\pi/3}\\
&=&2\pi+\frac{3\sqrt{3}}{2}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that each of the following is an isomorphism Verify that each of the following is an isomorphism:
1) $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x,y,z)=(x+y,y+z,z+x)$.
2) $T: M_{2,2} \to \mathbb{R}^4$ defined by $T\left(\begin{bmatrix}a & b\\ c & d \end{bmatrix}\right) = (a+b, d, c, a-b)$.
For the 1), should we find $S: \mathbb{R}^3 \to \mathbb{R}^3$ such that $S(x,y,z)=(x, y, z)$ first? And in general do we have to show that each of them is linear.
| I will leave you to check that these are actually linear transformations. However, the fact that I am about to express them as matrices implies that they are linear transformations.
1) We can represent $T$ by the matrix $$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}_.$$
By this I mean that $T(v) = Av$.
Then $T$ is invertible if and only if $\det(A) \neq 0$.
Since $\det(A) = 2$, we have that $T$ is invertible.
2) Let $$e_{1,1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \ e_{1,2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \ e_{2,1} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \ e_{2,2} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}_.$$
These form a basis of $M_{2,2}$. If we use the standard basis for $\mathbb{R}^4$, the matrix of this transformation becomes:
$$A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0\end{pmatrix}_.$$
Since $\det(A) = 2 \neq 0$, we have that $T$ is invertible.
| {
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Evaluation of two Euler type sums We know that the harmonic number sum (also called Euler type sum) enter link description here
$$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^2}{2^n}}}} = {\rm{L}}{{\rm{i}}_4}\left( {\frac{1}{2}} \right)\; + \frac{1}{{16}}\zeta (4) + \frac{1}{4}\zeta (3)\log 2 - \frac{1}{4}\zeta \left( 2 \right){\log ^2}2 + \frac{1}{{24}}{\log ^4}2,$$
How to calculate the closed form of the following Euler type Sums
$$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^3}{2^n}}}} ,\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^4}{2^n}}}}.$$
Here the harmonic numbers are defined by
$$H^{(k)}_n:=\sum\limits_{j=1}^n\frac {1}{j^k}\quad {\rm and}\quad H^{(k)}_0:=0.$$
| By Cauchy product we have
$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
divide both sides by $x$ then integrate from $x=0$ to $1/2$ and use the fact that $\int_0^{1/2}x^{n-1}=\frac1{n2^n}$ we have
$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\sum_{n=1}^\infty \frac{1}{n2^n}\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
rearrange to get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=3\operatorname{Li}_5\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\frac12\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx$$
Substitute the first sum
\begin{align}
\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2)
+\frac12\ln^22\zeta(3)\\
&\quad-\frac18\ln2\zeta(4)-
\frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52
\end{align}
along with the result from Song's solution
$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\ dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$
we get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If P congruent to $-1 \bmod$ 8 then $2^{(p-1)/2} - 1$ divisible by $p$ I am trying to prove the following statement:
If $p$ congruent to $-1 \bmod 8$, then $$2^{\frac{p - 1}{2}} - 1$$ divisible by $p$.
I can assume $p$ is congruent to $-1 \bmod 8$, then I must prove
$2^{\frac{p - 1}{2}} - 1$ is divisible by $p$.
So $p + 1 = 8n$, but I'm confused how I can use this assumption to prove $2^{\frac{p - 1}{2}} - 1$ divisible by $p$.
| Let $p=8k-1$ and let $n=(p-1)/2 = 4k-1$. Then note that
$$1\cdot 2\cdot 3\cdots n \equiv (-(p-1))\cdot 2 (-(p-3)) \cdot 4 (-(p-5)) \cdots (n-1)(-(p-n)) \pmod{p},$$
where we replaced each odd number $x$ by a congruent even number $-(p-x)$.
Now count the minus signs in the last product. Because $n$ is odd, there are $(n+1)/2 = 2k$ of them, so their product equals $1$. So the congruence above becomes
$$n! \equiv (p-1)\cdot 2 (p-3) \cdot 4 (p-5) \cdots (n-1)(p-n) \pmod{p}.$$
Next, note that $2n = p-1$, $2(n-1) = p-3$, $2(n-2) = p-5, \ldots, n+1=4k =p-n,$ so we have
$$n! \equiv (2n)\cdot 2\cdot(2(n-1))\cdot 4 \cdots (n-1)\left(2\frac{n+1}{2}\right) \pmod{p}. $$
Rearrange the right side to get
$$n! \equiv 2\cdot 4 \cdot 6 \cdots (n-1)(n+1)\cdots (2n) \equiv 2^n n! \pmod{p}.$$
Cancel the $n!$ and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Does $\lim_{m \to \infty}\sum_{n=1}^m (-1)^n (\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2) $ exist? This question is based on
an answer and comment
to this question:
convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$
Does
$\displaystyle \lim_{m \to \infty} \sum_{n=1}^m (-1)^n
\left[ \sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2 \right]
$
exist?
The answers there show that
$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}
\to 2
$,
but are not precise enough
to show that the difference is monotonic,
so the alternating series theorem
can not be applied.
| We can express $2$ as the telescoping sum $$\sum_{k=n^2}^{(n+1)^2-1} (2\sqrt{k+1}-2\sqrt{k}),$$
which lets us rewrite the sum over $n$ as $$\sum_{n=1}^\infty (-1)^n \sum_{k=n^2}^{(n+1)^2-1} \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$
I claim that this series is actually absolutely convergent, for which we just need to prove the convergence of $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$
Consider the inside term: $$\frac1{\sqrt k} - 2\sqrt{k+1} + 2\sqrt k = \frac{2k+1 - 2\sqrt{k(k+1)}}{\sqrt k} = \frac{(\sqrt{k+1} - \sqrt k)^2}{\sqrt k}.$$
We have $$\left(\sqrt k + \frac1{2\sqrt k}\right)^2 = k + 1 + \frac1{4k} > k+1 \implies \sqrt k + \frac1{2\sqrt k} > \sqrt{k+1},$$
so $\sqrt{k+1} - \sqrt k < \frac1{2\sqrt k}$, and therefore $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right) < \sum_{k=1}^\infty \frac1{4k \sqrt k}$$ which converges by the $p$-series test.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) & e_2'=(-1,1). \\
\end{array}
$$
I find the eigenvalues and vectors for $AA'$:
$$
\begin{array}{ccc}
\lambda_1=10 & \lambda_2=8 & \lambda_3=0 \\
e_1=(1,0,2) & e_2=(0,1,0) & e_3=(-2,0,1), \\
\end{array}$$
and so my SVD should be:
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
However, this gives
$$\left(
\begin{array}{cc}
1 & 1 \\
-2 & 2 \\
2 & 2 \\
\end{array}
\right)$$ instead of $A$.
To get $A$ I need to decompose in the following way
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
This last decomposition is as if I had multiplied the first $e_2$ by $-1$.
These eigenvalues and vectors were derived from Mathematica, just to be sure I was using the correct elements.
Any help in explaining why my first decomposition doesn't work would be appreciated.
Edit: The book I'm using doesn't tell which of the orthonormal eigenvectors I have to use. For each eigenvalue, I have two orthonormal eigenvectors, $e_i$ and $-e_i$.
| You need to match the left singular vectors to the right ones, or vice versa. E.g. after you have computed $e_1'$ and $e_2'$, you could get the two corresponding left singular vectors as $e_1=Ae_1'/\|Ae_1'\|=\frac1{\sqrt{5}}(1,0,2)^T$ and $e_2=Ae_2'/\|Ae_2'\|=(0,\color{red}{-1},0)^T$ (note: the sign of $e_2$ here is different from yours). The remaining left singular vector can be any unit vector orthogonal to the previous two singular vectors (in this case, it must be $\pm e_1\times e_2$, where the sign is unimportant). Then
\begin{align*}
A&=
\pmatrix{e_1&e_2&\pm e_1\times e_2}
\pmatrix{\frac{\|Ae_1'\|}{\|e_1'\|}&0\\ 0&\frac{\|Ae_2'\|}{\|e_2'\|}\\ 0&0}
\pmatrix{\frac{(e_1')^T}{\|e_1'\|}\\ \frac{(e_2')^T}{\|e_2'\|}}\\
&=\pmatrix{\frac{1}{\sqrt{5}}&0&\frac{\pm2}{\sqrt{5}}\\
0&-1&0\\ \frac{2}{\sqrt{5}}&0&\frac{\mp1}{\sqrt{5}}}
\pmatrix{\sqrt{10}&0\\ 0&\sqrt{8}\\ 0&0}
\pmatrix{\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}&\frac{1}{\sqrt{2}}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Rationalise the Denominator Rationalise the denominator and simplify fully:
$$\dfrac{6}{\sqrt{7} + 2}$$
I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?
I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sqrt{7}}{9} = \frac{2 \sqrt{7}}{ 3}$ .
| I'll assume your expression was $\frac{6}{\sqrt{7}+2}$, since it's the only way you obtain that denominator.
$$\frac{6}{\sqrt{7}+2}\times\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{6\sqrt{7}-12}{\sqrt{7}^2-2^2}=\frac{6\sqrt{7}-12}{3}={2\sqrt{7}-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| Using Taylor's formula for polynomials:
*
*$f(x+1)=f(x)+f'(x)\cdot 1+f''(x)\cdot\dfrac 12$,
*$f(x-1)=f(x)-f'(x)\cdot 1+f''(x)\cdot\dfrac 12$.
Therefore $\;f(x+1)+f(x-1)=2f(x)+f''(x)=4x^2-2x+10$. If we set $f(x)=ax^2+bx+c$, this equation simplifies to
$$ax^2+bx+c+a=2x^2-x+5,\enspace\text{whence}\quad a=2, b=-1, c=3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$ Show that: $$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\sqrt{\frac{p+1}{2}}\sqrt{\frac{q+1}{2}}}{2\sqrt{\frac{p+q+2}{2}}},\; p,q > -1$$
Here's the question.
| if $p=q=1$, we get
$$\int_0^\frac {\pi}{2} \frac {\sin (2\theta)d\theta}{2} $$
$$=\frac {1}{4}[-\cos (2\theta)]_0^\frac {\pi}{2} $$
$$=\frac {1}{2} \neq \frac {1}{2\sqrt {2}} $$
Your formula doesn't seem to be correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
| Let $p$ be a prime, write $x=p^au, y=p^bv$ where $u,v$ are relatively prime with $p$. $xy=p^{a+b}uv$. Suppose $a<b$ If $xy$ divides $x^2+y^2=p^{2a}u^2+p^{2b}v^2$, we have ${{p^{2a}u^2+p^{2b}v^2}\over{p^{a+b}}}$ is an integer. This implies that $p^{a+b}$ divides $p^{2a}u^2$. This is impossible. We deduce that $a\geq b$ and similarly$b\geq a$. This implies that $x=y$ or $x=-y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
| If you know the value of $n$ then $$3+7+12+18+25+\ldots=\\3n-3n+3+7+12+18+25+\ldots$$ As $3=1+2$ we can write as $$\\1+2+3+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+\ldots-3n=\\\left(\sum_{n=3}^n{\frac{n(n+1)}{2}}\right)-3n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find a,b and c. Any method? Given that $$\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}=\frac{16}{38}$$, find $a,b,c$.
I've been figuring for this quite some time? Is it possible to solve?
| Sure it is, but you'll end up with infinite possibilities:
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{38-16a}{16}\\
b+\frac{1}{c+1}&=\frac{16}{38-16a}\\
\frac{1}{c+1}&=\frac{16-(38-16a)b}{38-16a}\\
c&=\frac{38-16a}{16-(38-16a)b}-1
\end{align*}$$
You can chose $a$ and $b$ freely, and then $c$ will be determined by the two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need help to prove the last inequality of this mathematical induction Here is the original question:
Let $$x_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n - 1}{2n}$$
Then show that $$x_n\leq\frac{1}{\sqrt{3n+1}} \text{, for all } n = 1, 2, 3, \ldots \text{ .}$$
It is trivial to show that the statement is true for $n=1$.
I assumed it is true for some arbitrary positive integer $k$. Hence, we have
$$x_k\leq\frac{1}{\sqrt{3k+1}}$$
Now, after proper grouping, we arrive at the expression for $x_{k+1}$:
$$
x_{k+1}=x_k\cdot\frac{2k+1}{2k+2}
$$
which implies that
$$
x_{k+1}\leq \frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}
$$
From here, ho do I arrive at the fact that
$$
x_{k+1}\leq \frac{1}{\sqrt{3(k+1)+1}}=\frac{1}{\sqrt{3k+4}}?
$$
| Say this inequality is correct for $x_n$. So we can say that: $x_n \leq \dfrac{1}{\sqrt{3n+1}}$ $x_{n+1} = x_n * \dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2}$ So know if we prove the statement below, the inequality is proved: $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}}$ In order to do that, we assume that this statement is correct and then use it to derive an obvious statement. Then we can claim that the initial statement that we had, is correct. $\dfrac{1}{\sqrt{3n+1}}*\dfrac{2n+1}{2n+2} \leq \dfrac{1}{\sqrt{3n+4}} => \dfrac{2n+1}{2n+2} \leq \sqrt{\dfrac{3n+1}{3n+4}} => \dfrac{4n^2+4n+1}{4n^2+8n+4}\leq\dfrac{3n+1}{3n+4}=>$
$12n^3+28n^2+19n+4\leq12n^3+28n^2+20n+4=>n\geq0$
Which is an obvious statement which means for any $n\geq 0$ this statement is correct. So the inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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inertia of a lamina I am given a lamina and the area that it covers is rectangular. Its density, $d$, varies with position: $d=x^2+y^2$. If I want to find the moment of inertia about the $x$ and $y$ axis, can I do the integrals in polar form? I say this because I am aware that I need to (double) integrate $y^2(x^2+y^2)$ to find inertia about $x$-axis, and this isn't easy to do in Cartesian! My lamina is rectangular so polar surely wouldn't work?
Thanks.
| It's easier to use Cartesian coordinates,
\begin{align*}
I_{xx} &= \int_{a}^{b} \int_{c}^{d} \rho y^2 \, dy \, dx \\
&= \int_{a}^{b} \int_{c}^{d} y^2(x^2+y^2) \, dy \, dx \\
&= \int_{a}^{b}
\left[
\frac{x^2(d^3-c^3)}{3}+\frac{d^5-c^5}{5}
\right] \, dx \\
&= \frac{(b^3-a^3)(d^3-c^3)}{9}+\frac{(b-a)(d^5-c^5)}{5} \\
I_{yy} &= \int_{a}^{b} \int_{c}^{d} \rho x^2 \, dy \, dx \\
&= \frac{(b^3-a^3)(d^3-c^3)}{9}+\frac{(b^5-a^5)(d-c)}{5} \\
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the partial fraction decomposition $\frac{x^3+5x^2+3x+6}{2x^2+3x}$. I have the following integral:
$$\int\frac{x^3+5x^2+3x+6}{2x^2+3x}dx$$
I'm trying to use partial fraction decomposition but I'm getting stuck at the following formula:
$$\int\frac{(x+6)(1+5x+x^2)}{x(2x+3)}-\frac{x+27}{2x+3}dx$$
I can't necessarily guarantee that this problem has a nice solution. But is there a way to circumvent the fact that $(1+5x+x^2)$ has a somewhat messy root? (I'm getting $\frac{5+-\sqrt{21}}{2}$). Or do I have to use that value?
| I will let you work out the detail, but the solution should be:
$ \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} =\cfrac{7}{4} + \cfrac{2}{x} + \cfrac{x}{2} - \cfrac{25}{4(2x+3)} $
Now, you can evaluate the integral:
\begin{equation} \int \cfrac{x^3 + 5x^2 + 3x + 6}{2x^2 + 3x} dx = \int \cfrac{7}{4} dx + \int \cfrac{2}{x} dx + \int \cfrac{x}{2} dx - \int \cfrac{25}{4(2x+3)} dx \end{equation}
\begin{equation} = \cfrac{7}{4}x + 2\ln(x) + \frac{x^2}{4} - \cfrac{25}{8}\ln(2x + 3) \end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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partial fractions for a function I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is:
$(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.
| As suggested by one user.
$\frac{(x^2 + x + 1)}{(2x^4+3x^2+1)} = \frac{ax+b}{2x^2+1} + \frac{cx+d}{x^2+1}$
$x^2+x+1 = (ax+b)(x^2+1) + (cx+d)(2x^2+1)$
Comparing terms containing $x^3$,
$0 = a + 2c$
Comparing terms containing $x^2$,
$1 = b + 2d$
Comparing terms containing $x$,
$1 = a + c$
Comparing terms containing constant terms,
$1 = b + d$
Hope you can solve above equations to find a, b, c and d.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find Jordan form of matrix 6x6 Find the Jordan form of matrix $$A = \begin{bmatrix}
2 & 0 & 1 & 0 & -1 & 0\\
0 & 2 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 5 & 0 & -9\\
2 & 0 & 1 & 0 & 3 & 0\\
0 & 0 & 0 & 1 & 0 & -1\\
\end{bmatrix}$$
I found $\lambda$ from $|A-\lambda I|= 0$ and $(\lambda -2)^6=0$
$$B=(A-2I) = \begin{bmatrix}
0 & 0 & 1 & 0 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 & 1 & 0\\
0 & 0 & 0 & 3 & 0 & -9\\
2 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & -3\\
\end{bmatrix}$$
$rankB=3 \Rightarrow$ solving $Bx=0$ I got:
$$x_1=[0,1,0,0,0,0]^T, \hspace{5pt} x_2=[0,0,0,3,0,1]^T, \hspace{5pt} x_3=[0,0,1,0,-1,0]^T$$
$$B^2 = \begin{bmatrix}
-2 & 0 & -2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
2 & 0 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
2 & 0 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}$$
Solving $B^2x=0$ I got: $$x'_1=[1,0,-1,0,0,0]^T, \hspace{3pt} x'_2=[0,1,0,0,0,0]^T, \hspace{3pt} x'_3=[0,0,0,1,0,0]^T, \hspace{3pt} x'_4=[0,0,0,0,1,0]^T, \hspace{3pt} x'_5=[0,0,0,0,0,1]^T$$
So I got only 5 vectors but I'm in $\mathbb{R}^6$ and $B^3$ is null matrix. So I don't know what to do next
| Since $\operatorname{rank} B = 3$, you have three linearly independent eigenvectors associated to the (only) eigenvalue $\lambda = 2$ so there are three Jordan blocks in the Jordan form of $A$. Since $B^3 = 0$ while $B^2 \neq 0$, the minimal polynomial of $A$ is $(x - 2)^3$ which means that the largest Jordan block in the Jordan form of $A$ must be a $3 \times 3$ block. This gives you only one option for the Jordan form of $A$:
$$ \begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 1 & 0 \\
0 & 0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 2\end{pmatrix}. $$
| {
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"url": "https://math.stackexchange.com/questions/2239186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Angle between vectors a and b if a + 3b is perpendicular to.... Determine the angle between vectors $a$ and $b( a \ne 0, b \ne0)$ if $a + 3b$ is perpendicular to $2a - b$ and $a + 7b$ is perpendicular to $2a + b$.
I've done this:
$(a + 3b) \cdot (2a - b) = 0$
$\dots 5a⋅b = 3b⋅b - 2a⋅a$
$(a + 7b) \cdot (2a + b) = 0$
$\cdots 15a⋅b = -2 a⋅a - 7b\cdot b$
which gives me:
$a⋅b = (-(b⋅b + a⋅a))/5$
$cos(\theta) = (-b - a)/5$
Now I'm not really sure how to proceed
| You have $$5a\cdot b=3b\cdot b-2a\cdot a$$ $$15a\cdot b=-7b\cdot b-2a\cdot a$$so subtract the 2nd equation from the first to get $$-10a\cdot b=10b\cdot b\Rightarrow b\cdot b=-a\cdot b$$
Also $$35a\cdot b=21b\cdot b-14a\cdot a$$ $$45a\cdot b=-21b\cdot b-6a\cdot a\Rightarrow80a\cdot b=-20a\cdot a$$ so similarly $$a\cdot a=-4a\cdot b$$
Then finally
$$\cos\theta=\frac{a\cdot b}{|a||b|}\\
\cos\theta=\frac{a\cdot b}{\sqrt{(a\cdot a) (b\cdot b)}}=\frac{a\cdot b}{\sqrt{(-4a\cdot b)(-a\cdot b)}}=\frac{1}{2}.$$
So $\theta=\pi/3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start.
Do I need contradiction?
| You may just ignore the hint and prove that $x^2+(x+1)^2=y^2$ has an infinite number of solutions in $\mathbb{N}\times\mathbb{N}$. Such identity is equivalent to
$$ (2x+1)^2-2 y^2 = -1 $$
hence it is enough to show that there are infinite natural solutions to $a^2-2b^2=-1$ with $a$ being odd. But, wait, if $a,b\in\mathbb{Z}$ and $a^2-2b^2=-1$ then $a$ has to be odd. And $a^2-2b^2$ is just the norm over $\mathbb{Z}[\sqrt{2}]$, where $3\pm2\sqrt{2}$ is an invertible element (unit norm). It follows that by raising $(3+2\sqrt{2})$ to some power we still get an invertible element, and a solution of $a^2-2b^2=1$. Since the norm of $1+\sqrt{2}$ is $-1$, in such a case $(a+2b)^2-2(a+b)^2=-1$ and we get a solution of $x^2+(x+1)^2=y^2$. For instance,
*
*$(3+2\sqrt{2})^3 = 99+70\sqrt{2}$
*$99^2-2\cdot 70^2 = 1$
*$239^2-2\cdot 169^2 = -1$
*$119^2+120^2=169^2$
or
*
*$(3+2\sqrt{2})^4 = 577+408\sqrt{2}$
*$577^2-2\cdot 408^2 = 1$
*$1393^2-2\cdot 985^2 = -1$
*$696^2+697^2=985^2$
In particular, if we define the Pell number $P_n$ as $\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$, we have that
$$ \left(\frac{P_{2n-1}+3 P_{2n}-1}{2}\right)^2+\left(\frac{P_{2n-1}+3 P_{2n}+1}{2}\right)^2 = \left(P_{2n-1}+2\,P_{2n}\right)^2.$$
Such well-known identity can be simplified a bit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$
a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.
b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$
So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to use this to hence factor the expression.
Also I was reading a solution which factorised the expression without using part a) but how does the second last line turn into the last line?
\begin{align}&\color{white}=(bc^3−cb^3)−(ac^3−ca^3)+(ab^3−ba^3)\\
&=bc(c^2−b^2)−ac(c^2−a^2)+ab(b^2−a^2)\\
&=(c−b)(c+b)bc−ac(c−a)(c+a)+ab(b−a)(b+a)\\
&=−(a−c)(b−c)(a−b)(a+b+c)\end{align}
Thanks in advance
| Hint (without using the first part): considering it as a $3^{rd}$ degree polynomial in $a\,$:
$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)\,a^3+(b^3-c^3)\,a+bc\,(b^2-c^2)$$
It can be easily verified that it has $a=b$ and $a=c$ as roots. Since the term in $a^2$ is missing, the sum of all three roots is $0$ by Vieta's relations, so the third root is $a=-(b+c)\,$. Therefore the polynomial is divisible by $(a-b)(a-c)(a+b+c)\,$. Comparing the leading coefficients, it follows that the scalar multiplier is $c-b\,$, so in the end:
$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)(a-b)(a-c)(a+b+c)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let a general term $T_n$ be defined as $T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$ Let a general term $T_n$ be defined as
$$T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$$
Then prove that
$\lim_{n\to\infty}(T_1 + T_2 +\cdots+T_n) \lt \frac{4}{27}.$
I tried finding pattern between terms ..
$T_1=\frac{1}{9} , \frac{T_2}{T_1}=(\frac{2}{5})^2, \frac{T_3}{T_2}=(\frac{3}{7})^2$
but could not think more of how to get a bound on the series.
Any help is appreciated.
| I do not think $\sum_{n\geq 1}T_n$ has a closed form, but such inequality can be improved a bit.
We have
$$ T_n = \left(\frac{n!}{(2n+1)!!}\right)^2 = \left(\frac{n!(2n)!!}{(2n+1)!}\right)^2 = \left(\frac{n!^2 2^n}{(2n+1)!}\right)^2 = \frac{4^n}{(2n+1)^2 \binom{2n}{n}^2}$$
and we may borrow a couple of useful lemmas from this answer:
$$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n x^{2n-1}}{n\binom{2n}{n}},\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}}\tag{2}$$
to derive:
$$ \sum_{n\geq 1}\frac{x^{2n+1}}{(2n+1)\binom{2n}{n}}=-x+\frac{4}{\sqrt{4-x^2}}\arcsin\frac{x}{2}\tag{3} $$
By $(1)$ and $(3)$ we get:
$$ \sum_{n\geq 1}T_n = \frac{1}{9}\cdot\phantom{}_3 F_2\left(1,2,2;\frac{5}{2},\frac{5}{2};\frac{1}{4}\right)=\int_{0}^{\pi/2}\left[-\sin(x)+\frac{4}{\sqrt{4-\sin^2 x}}\arcsin\frac{\sin x}{2}\right]\,dx $$
from which:
$$ \sum_{n\geq 1}T_n=-1+\int_{0}^{1}\frac{4\arcsin\frac{x}{2}}{\sqrt{(4-x^2)(1-x^2)}}\,dx=-1+\int_{0}^{\pi/6}\frac{4x}{\sqrt{1-4\sin^2 x}}\,dx \tag{4}$$
and by convexity the LHS of $(4)$ is bounded by
$$ -1+\int_{0}^{1}\frac{4\cdot\left[\frac{x}{2}+\left(\frac{\pi}{6}-\frac{1}{2}\right)x^3\right]}{\sqrt{(4-x^2)(1-x^2)}}\,dx = \frac{6-4\pi -9\log(3)+5\pi \log(3)}{6}<\color{red}{\frac{17}{127}}.\tag{5} $$
$(4)$ also provides a decent lower bound:
$$ \sum_{n\geq 1}T_n \geq -1+\int_{0}^{1}\frac{2x}{\sqrt{(1-x^2)(4-x^2)}}\,dx = \color{red}{-1+\log(3).}\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got:
$-2x^{3}-x^{2}-6x+9=0$
| So you got to the cubic equation $f(x)=-2x^{3}-x^{2}-6x+9=0$. When you come across a cubic like this, try evaluating $f(\pm1), f(\pm2)$, etc, to try and figure out some roots so you can factor it (you know $x_0$ a root if $f(x_0)=0$). Here you can see $1$ is a root, so factoring out $(x-1)$ gives $f(x)=(x-1)\underbrace{(2x^2+3x+9)}_{\text{no real roots}}$. So the only solution is $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\frac{a_k}{z^{n-k}},(0\leq k < n)$ less than $\frac{\lvert a_n \rvert}{2n}$ Suppose there is a polynomial:
$$P(z) = a_0 + a_1z +a_2z^2 +\dots +a_nz^n,\quad (a_n \ne 0)$$
Let $$w = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \frac{a_2}{z^{n-2}} + \dots + \frac{a_{n-1}}{z},$$
Why is it that for a sufficiently large positive number $R$, the modulus of each of the quotients in the expression $w$ is less than the number $\frac{\lvert a_n \rvert}{2n}$ when $\lvert z \rvert > R$
| Since, $a_n\not=0$ we have $|a_n|\not=0$. Now, we have to find a positive real number $R_0$ such that for any complex number $z$ with $|z|> R_0$ we have $$\left |\frac{a_0}{z^n}\right|=\frac{|a_0|}{|z^n|}=\frac{|a_0|}{|z|^n}<\frac{|a_n|}{2n},$$$$\left |\frac{a_1}{z^{n-1}}\right|=\frac{|a_1|}{|z^{n-1}|}=\frac{|a_1|}{|z|^{n-1}}<\frac{|a_n|}{2n},$$$$\vdots$$$$\left |\frac{a_{n-2}}{z^2}\right|=\frac{|a_{n-2}|}{|z^2|}=\frac{|a_{n-2}|}{|z|^2}<\frac{|a_n|}{2n},$$$$\left |\frac{a_{n-1}}{z}\right|=\frac{|a_{n-1}|}{|z|}<\frac{|a_n|}{2n}.$$
In other words, we have to find a positive real number $R_0$ such that for any complex number $z$ with $|z|> R_0$ we have $$\frac{2n|a_0|}{|a_n|}<|z|^n,\frac{2n|a_1|}{|a_n|}<|z|^{n-1},...,\frac{2n|a_{n-2}|}{|a_n|}<|z|^2,\frac{2n|a_{n-1}|}{|a_n|}<|z|.$$ But, the above inequalities hold if and only if $$\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}}<|z|,\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}}<|z|,...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}}<|z|,\frac{2n|a_{n-1}|}{|a_n|}<|z|.$$ So, we may take $$R_0:=1+\max\left\{\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}},\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}},...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}},\frac{2n|a_{n-1}|}{|a_n|}\right\}.$$ So, for any complex number $z$ with $$|z|>R_0>\max\left\{\left(\frac{2n|a_0|}{|a_n|}\right)^{\frac{1}{n}},\left(\frac{2n|a_1|}{|a_n|}\right)^{\frac{1}{n-1}},...,\left(\frac{2n|a_{n-2}|}{|a_n|}\right)^{\frac{1}{2}},\frac{2n|a_{n-1}|}{|a_n|}\right\}$$$$\implies|z|>\left(\frac{2n|a_k|}{|a_n|}\right)^{\frac{1}{n-k}}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies |z|^{n-k}>\frac{2n|a_k|}{|a_n|}\text{ for each }k=0,1,...,(n-2),(n-1)$$$$\implies\frac{|a_n|}{2n}>\frac{|a_k|}{|z|^{n-k}}=\frac{|a_k|}{|z^{n-k}|}=\left|\frac{a_k}{z^{n-k}}\right|\text{ for each }k=0,1,...,(n-2),(n-1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of a trigonometric function
Prove that the minimum value of $$a \sec \theta + b \csc \theta$$ is $$(a^{2/3} + b^{2/3})^{3/2}$$ where $0 < \theta < \frac{\pi}{2}$ and $a$ and $b$ are positive real numbers.
This is one part of a larger question whose first part I solved. I tried to express $a \sec \theta + b \csc \theta$ in a more concise manner, but couldn't do so without introducing more variables.
Even after equating the function to the minimum, it is not clear to me how to find the value of $\theta$ for which the function attains its minimum.
| $f(\theta)=a\sec\theta+b\csc\theta,\quad 0<\theta<\frac{\pi}{2}$
$f'(\theta) = a\cdot(-1)\cdot(-\sin\theta)\cdot\sec^2\theta+b\cdot(-1)\cdot(\cos\theta)\cdot\csc^2\theta$
$$f'(\theta)= \frac{a\sin\theta}{\cos^2\theta} -\frac{b\cos\theta}{\sin^2\theta}$$
Setting $f'(\theta)=0:$
$$f'(\theta) = \frac{a\sin^3\theta-b\cos^3\theta}{\sin^2\theta\cos^2\theta}=0$$
$$\implies a\sin^3\theta=b\cos^3\theta \implies \tan^3\theta=\frac{b}{a}\implies \tan\theta=\bigg(\frac{b}{a}\bigg)^{\frac{1}{3}}$$
Then $\cos\theta = \frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}\quad$ and $\sin\theta = \frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}$
Then $f(\theta) = a\cdot\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+b\cdot\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}} =(a^{2/3}+b^{2/3})\cdot \sqrt{a^{2/3}+b^{2/3}} = (a^{2/3}+b^{2/3})^{3/2}$
I leave it to you to show that the point is a minimum
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.