Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How does $\cos x=\frac12(e^{ix}+e^{-ix})$? I have seen the following definition many times:
$$\cos x=\frac12(e^{ix}+e^{-ix})$$ However, it makes little sense to me as it appears far from obvious. Please help me understand this definition, either a derivation or explanation of how the values on the right equal $\cos x$.
| Taylor series of $\cos(x)$:
$$\begin{align}
\cos(x)&=\color{green}{1-\frac{x^2}{2!}+\frac{x4}{4!}-\cdots}
\end{align}$$
Taylor series of $e^x$:
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$
Taylor series of $e^{ix}$:
$$\begin{align}
e^{ix}&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}\cdots\\
&=\color{green}{1}+ix\color{green}{-\frac{x^2}{2!}}-\frac{ix^3}{3!}\color{green}{+\frac{x^4}{4!}}\cdots
\end{align}$$
Taylor series of $e^{-ix}$:
$$\begin{align}
e^{-ix}&=1-ix+\frac{(-ix)^2}{2!}+\frac{(-ix)^3}{3!}+\frac{(-ix)^4}{4!}\cdots\\
&=\color{green}{1}-ix\color{green}{-\frac{x^2}{2!}}+\frac{ix^3}{3!}\color{green}{+\frac{x^4}{4!}}\cdots
\end{align}$$
So adding the above two together, the imaginary parts cancel out each other and the real parts are twice those in $\cos(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Determine the elements of a triangle knowing relationships between lengths and angles A triangle's sides' lengths are three successive numbers (meaning b=a+1, c=a+2). The smallest angle is a half of the triangle's biggest angle.
Find the area of this triangle and its angles.
Solution:
Sides are 4, 5 and 6.
Area: $15\sqrt{7}/4$
Angles : $41° 24' 34"$ and $82°49'8"$
| Hint:
Solve
$$\frac{\sin\theta}{a}=\frac{\sin3\theta}{a+1}=\frac{\sin2\theta}{a+2}.$$
$$\frac{\sin\theta}a=\frac{4\sin\theta\cos^2\theta-\sin\theta}{a+1}=\frac{2\sin\theta\cos\theta}{a+2}\implies\frac1a=\frac{4\cos^2\theta-1}{a+1}=\frac{2\cos\theta}{a+2}\\\implies a^2-3a-4=0\implies a=4$$ by eliminating $\cos\theta$. Next, $\cos\theta=3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| Put $u=x+\frac 1x$
and let
$u^\boxed{\tiny n}=x^n+\frac 1{x^n}$.
By expansion it is clear that
$$u^3=u^\boxed{\tiny{3}}+3u$$
and
$$u^5=u^\boxed{\tiny5}+5u^\boxed{\tiny3}+10u$$.
Eliminating $u^\boxed{\tiny3}$ gives
$$u^{\boxed{\tiny5}}=u^5-5u^3+5u=1-5+5=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.
Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
| You've got the right idea about getting an inequality. Since exponential functions grow much faster than polynomials, we should expect $3^b$ to dominate $2b^2+1$ at some point. We can show that it happens at $b=2$ using induction.
Induction hypothesis: Suppose that $2b^2+1<3^b$ for some particular value $b>2$.
Induction step: We want to prove $2(b+1)^2<3^{b+1}$ holds as well.
Let's work on the left side. We have
$$ \begin{array}{ll} 2(b+1)^2+1 & = 2b^2+4b+3 \\ & = [2b^2+1]+[2(2b)+1]+1 \\ & < [2b^2+1]+[2b^2+1]+1 \\ & <3^b+3^b+3^b \\ & =3^{b+1}. \end{array} $$
That is, we've proved $2(b+1)^2<3^{b+1}$, and so the claim follows.
If you're not very familiar with induction, what we've just shown is that if the inequality holds for a particular value of $b$, then it holds for the next value of $b$ as well. This leads to a chain
$$ \begin{array}{lc} & 2(3^2)+1<3^3 \\ \implies & 2(4^2)+1<3^4 \\ \implies & 2(5^2)+1<3^5 \\ \implies & 2(6^2)+1<3^6 \\ \implies & \vdots \end{array}$$
Thus, $2b^2+1<3^b$ holds for all $b>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
| I would emphasize that
$$ \sin (x + A) = \sin x \cos A + \cos x \sin A $$
You have
$$ \sin x \; \; \frac{1}{3} + \cos x \; \; \frac{\sqrt 8}{3} = 1 $$
We can take $A = \arctan \sqrt 8,$ so that $\cos A = 1/3$ and $\sin A = \sqrt 8 / 3. $ So, once again, you have
$$ \sin (x + \arctan \sqrt 8) = 1. $$
On e of the answers is
$$ x = \frac{\pi}{2} - \arctan \sqrt 8 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the value of the fraction
What common fraction is equivalent to $$\frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \text{ ?}$$
I didn't see how to relate the numerator to the denominator. We can't find the value of the numerator and denominator easily it seems, so what else should we try?
| $$\sum_{n=1}^{\infty }\frac{1}{n^3}=\sum_{n=1}^{\infty }\frac{1}{(2n)^3}+\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)$$
so
$$\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)-\frac{1}{8}\zeta(3)=\frac{7}{8}\zeta(3)$$
$$\dfrac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots}=\frac{\frac{7}{8}\zeta(3)}{\zeta(3)}=???$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Approximation of the Gamma function for small value It is well known that $\Gamma(u) \stackrel{u \to 0}{\sim} \frac{1}{u}$.
I am looking for more precise information on the behavior of $\Gamma(x)$ when $x$ is small, ie: $x\to 0$.
My question is then, are there accurate (for small value) inequalities for the Gamma function ?
Any other information on the behavior is also very welcome.
| $\Gamma(z)$ has a pole at zero, with Laurent series
$$
\frac{1}{z}-\gamma+ \left( {\frac {{\pi }^{2}}{12}}+{\frac {{\gamma}^{2}
}{2}} \right) z+ \left( -{\frac {\zeta \left( 3 \right) }{3}}-{\frac
{{\pi }^{2}\gamma}{12}}-{\frac {{\gamma}^{3}}{6}} \right) {z}^{2}+
\left( {\frac {{\pi }^{4}}{160}}+{\frac {\zeta \left( 3 \right)
\gamma}{3}}+{\frac {{\pi }^{2}{\gamma}^{2}}{24}}+{\frac {{\gamma}^{4}
}{24}} \right) {z}^{3}+ \left( -{\frac {\zeta \left( 5 \right) }{5}}-
{\frac {{\pi }^{4}\gamma}{160}}-{\frac {\zeta \left( 3 \right) {\pi }
^{2}}{36}}-{\frac {\zeta \left( 3 \right) {\gamma}^{2}}{6}}-{\frac {{
\pi }^{2}{\gamma}^{3}}{72}}-{\frac {{\gamma}^{5}}{120}} \right) {z}^{4
}+ \left( {\frac {61\,{\pi }^{6}}{120960}}+{\frac {\zeta \left( 5
\right) \gamma}{5}}+{\frac {{\pi }^{4}{\gamma}^{2}}{320}}+{\frac {
\left( \zeta \left( 3 \right) \right) ^{2}}{18}}+{\frac {\zeta
\left( 3 \right) {\pi }^{2}\gamma}{36}}+{\frac {\zeta \left( 3
\right) {\gamma}^{3}}{18}}+{\frac {{\pi }^{2}{\gamma}^{4}}{288}}+{
\frac {{\gamma}^{6}}{720}} \right) {z}^{5}+O \left( {z}^{6} \right)
$$
Of course $\gamma$ is Euler's constant and $\zeta$ is Riemann's zeta function.
I wonder if it looks better if those $\pi^2$ and $\pi^4$ terms are written in terms of $\zeta(2)$ and $\zeta(4)$ and so on?
$$
\frac{1}{z}-\gamma+ \left( {\frac {\zeta \left( 2 \right) }{2}}+{\frac {
{\gamma}^{2}}{2}} \right) z+ \left( -{\frac {\zeta \left( 3 \right)
}{3}}-{\frac {\zeta \left( 2 \right) \gamma}{2}}-{\frac {{\gamma}^{3}
}{6}} \right) {z}^{2}+ \left( {\frac {9\,\zeta \left( 4 \right) }{16}}
+{\frac {\zeta \left( 3 \right) \gamma}{3}}+{\frac {\zeta \left( 2
\right) {\gamma}^{2}}{4}}+{\frac {{\gamma}^{4}}{24}} \right) {z}^{3}+
\left( -{\frac {\zeta \left( 5 \right) }{5}}-{\frac {9\,\zeta
\left( 4 \right) \gamma}{16}}-{\frac {\zeta \left( 3 \right) \zeta
\left( 2 \right) }{6}}-{\frac {\zeta \left( 3 \right) {\gamma}^{2}}{
6}}-{\frac {\zeta \left( 2 \right) {\gamma}^{3}}{12}}-{\frac {{\gamma}
^{5}}{120}} \right) {z}^{4}+ \left( {\frac {61\,\zeta \left( 6
\right) }{128}}+{\frac {\zeta \left( 5 \right) \gamma}{5}}+{\frac {9
\,\zeta \left( 4 \right) {\gamma}^{2}}{32}}+{\frac { \left( \zeta
\left( 3 \right) \right) ^{2}}{18}}+{\frac {\zeta \left( 3 \right)
\zeta \left( 2 \right) \gamma}{6}}+{\frac {\zeta \left( 3 \right) {
\gamma}^{3}}{18}}+{\frac {\zeta \left( 2 \right) {\gamma}^{4}}{48}}+{
\frac {{\gamma}^{6}}{720}} \right) {z}^{5}+O \left( {z}^{6} \right)
$$
Maybe part of the $\zeta(4)$ should be a $\zeta(2)^2$ and similarly for the $\zeta(6)$ to make the pattern recognizable?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$
And I'm not sure how to go on from here. Help would be appreciated.
| $$\sqrt{n^2+n}-n = n(\sqrt{1+1/n}-1)$$
Which by the binomial theorem is equal to
$$n\left(-1 + \sum_{k=0}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right) = n\left(\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right)$$
(With the product equal to 1 for $k = 0$.) So now move the n inside the sum to get
$$\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}} = \frac{1}{2} + \sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$$
So $\epsilon$ would be the absolute value of $\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}$, and all that remains is to show this approaches zero for sufficiently large n. Well, we'll start by saying...
$$\left|\sum_{k=2}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^{k-1}}\right| = \left|\sum_{k=2}^\infty\frac{(-1)^{k-1}\cdot\frac{1}{2}\cdot\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}}\right| < \sum_{k=2}^\infty\frac{\frac{1}{2}\prod_{l=1}^{k-1}\left(\frac{2l-1}{2}\right)}{k! n^{k-1}} < \sum_{k=2}^\infty\frac{(k-1)!}{2\cdot k!n^{k-1}} = \sum_{k=2}^\infty\frac{1}{2kn^{k-1}}$$
And I should hope you can take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
equation of lines which intersect another line at given angle.
Find the equation of $2$ lines through the origin which intersect the line
$\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}$ at an angle of $\displaystyle \frac{\pi}{3}$
$\bf{My\; Try::}$ Let equation of line be $\displaystyle \frac{x-0}{a} = \frac{y-0}{b} = \frac{z-0}{c}$
Where $<a,b,c>$ be the direction cosine of line parallell to that line.
and Given Line is $\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}$
Where $<2,1,1>$ be the direction cosine of line parallell to that line
and Given $\displaystyle \frac{\pi}{3}$ be the angle between $<a,b,c>$ and $<2,1,1>$
So $\displaystyle \cos \frac{\pi}{3} = \frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}\Rightarrow \frac{1}{2}=\frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}$
Now how can i solve it, Help required, Thanks
| you must set
$$a=3+2t$$
$$b=3+t$$
$$c=t$$ with a real number $t$ since the second line intersect the line above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Need help with a trigonometry problem (w/ picture) I have this trigonometry problem I got when programming a code library for cameras in games. I made a picture in Paint to explain the problem as simple as possible. Here's a link:
The known values are random but it shouldn't be a problem in this case. However, I want to know how, and if, I can get the unkown values in the picture.
Thanks in advance!
|
The law of cosines:
$$
\begin{cases}
\text{a}^2=\text{b}^2+\text{c}^2-2\text{b}\text{c}\cos\left(\alpha\right)\\
\text{b}^2=\text{a}^2+\text{c}^2-2\text{a}\text{c}\cos\left(\beta\right)\\
\text{c}^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(\gamma\right)
\end{cases}
$$
And the law of sines:
$$\frac{\text{a}}{\sin\left(\alpha\right)}=\frac{\text{b}}{\sin\left(\beta\right)}=\frac{\text{c}}{\sin\left(\gamma\right)}$$
Where:
So, we can set up the equations for your problem:
$$\begin{cases}
\text{a}^2=\text{b}^2+8^2-2\text{b}\cdot8\cdot\cos\left(\alpha\right)\\
\text{b}^2=\text{a}^2+8^2-2\text{a}\cdot8\cdot\cos\left(\beta\right)\\
8^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(50^\circ\right)\\
\frac{\text{a}}{\sin\left(\alpha\right)}=\frac{\text{b}}{\sin\left(\beta\right)}=\frac{8}{\sin\left(50^\circ\right)}\\
\alpha+\beta=130^\circ
\end{cases}\Longleftrightarrow
\begin{cases}
\text{a}^2=\text{b}^2+64-16\text{b}\cos\left(130^\circ-\beta\right)\\
\text{b}^2=\text{a}^2+64-16\text{a}\cos\left(\beta\right)\\
64=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(50^\circ\right)\\
\sin\left(\beta\right)=\frac{\text{b}\sin\left(50^\circ\right)}{8}\\
\alpha=130^\circ-\beta
\end{cases}
$$
So, we get:
$$128-16\cdot\frac{8\sin\left(\beta\right)}{\sin\left(50^\circ\right)}\cdot\cos\left(130^\circ-\beta\right)-16\text{a}\cos\left(\beta\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) - \int \ln\sin(u) du
\tag{1}
\label{eq:20161030-1}
\end{align}
\begin{align}
\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2}
\tag{2}
\label{eq:20161030-2}
\end{align}
To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$
\begin{equation}
\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy
= -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u}
\tag{3}
\label{eq:20161030-3}
\end{equation}
Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits,
\begin{align}
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\
&= \frac{\pi}{2}\ln2
\end{align}
I would be interested in seeing other solutions.
| Transform our integral by letting $\theta=\arcsin x$, then
$$
\begin{aligned}
\int_{0}^{1} \frac{\arcsin x}{x} d x &=\int_{0}^{\frac{\pi}{2}} \frac{\theta \cos \theta d \theta}{\sin \theta} \\
&=\int_{0}^{\frac{\pi}{2}} \theta d(\ln (\sin \theta)) \\
&=-\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta
\end{aligned}
$$
By my post, $\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta=-\frac{\pi}{2}\ln 2$, we can now conclude that
$$\boxed{\int_{0}^{1} \frac{\arcsin x}{x} d x= \frac{\pi}{2}\ln 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = 2 \\
5^2 + 5 + 4 = 1 \\
6^2 + 6 + 4 = 2 \\
7^2 + 7 + 4 = 5 \\
8^2 + 8 + 4 = 10 \\
9^2 + 9 + 4 = 6 \\
10^2 + 10 + 4 = 4 $
But this process is very long one. I am seeking a short method by which I can solve such problem.
Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.
| As we can use the formula for the roots of a quadratic over any field with characteristic $\;\neq2\;$, calculating this quadratic's discriminant we get
$$\Delta=1-4\cdot4=-15=-4=7\pmod{11}$$
and now check the squares modulo $\;11\;$ :
$$0^2=0\;,\;\;1^2=1\;,\;\;2^2=4\;,\;\;3^2=9\;,\;\;4^2=5\;,\;\;5^2=3\;,\;\;6^2=3$$
and that is enough (why?), and thus there exists no $\;x\in\Bbb F_{11}\;$ with $\;x^2=7\;$ so the quadratic has no roots in $\;\Bbb F_{11}\;$ and is thus irreducible.
We could, of course, also use quadratic reciprocity instead of calculating the squares modulo $\;11\;$ (this is highly advisable for big primes) :
$$\left(\frac7{11}\right)\stackrel{Q.R}=-\left(\frac47\right)=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges I am trying to show if this series converges or diverges and I know it converges since for very large values of n,
$$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$
which is convergent from the alternating series test since lim 1/n = 0 and 1/n is a decreasing sequence. I am going back and forth using the limit comparison test and alternating series test and can't seem to come up with anything. So right now i have $$\frac{(-1)^n(n^2+3n-7)}{n^3+1} \lt \frac{(-1)^n(n^2+3n-7)}{n^3}$$
and whenever I try to show the right handed side to be convergent I come back to the same problem as with the left hand side.
| I'll take the opportunity for showing some useful techniques for the manipulation of similar series.
Let $\omega=\frac{1+i\sqrt{3}}{2}$ and $\overline{\omega}=\omega^{-1}=\frac{1-i\sqrt{3}}{2}$. By the residue theorem
$$ \frac{x^2+3x-7}{x^3+1} = \frac{-3}{x+1}+\frac{2+\frac{2i}{\sqrt{3}}}{x-\omega}+\frac{2-\frac{2i}{\sqrt{3}}}{x-\overline{\omega}} \tag{1}$$
where:
$$ \sum_{n\geq 0}\frac{-3(-1)^n}{n+1} = -3\log(2)\tag{2} $$
so the original series equals
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} &=& -3\log(2)-4-\sum_{n\geq 0}\frac{(-1)^n 4n}{n^2+n+1}\\&\stackrel{ILP}{=}&-3\log(2)-4-2\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(\frac{s\sqrt{3}}{2}\right)-\sin\left(\frac{s\sqrt{3}}{2}\right)}{\sqrt{3}\cosh\left(\frac{s}{2}\right)}\,ds\\&=&-3\log(2)-4-\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(s\sqrt{3}\right)-\sin\left(s\sqrt{3}\right)}{\cosh\left(s\right)}\tag{3}\end{eqnarray*}$$
where ILP stands for Inverse Laplace Transform. By the Cauchy-Schwarz inequality, the absolute value of the last integral appearing in the RHS of $(3)$ is not larger than
$$ \frac{8}{\sqrt{3}}\int_{0}^{+\infty}\frac{ds}{\cosh s} = \frac{4 \pi}{\sqrt{3}}.\tag{4}$$
We may also simplify the RHS of $(3)$ a little more and state that:
$$ \sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} = -\log(8)-4-\frac{2\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}+\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sin(s\sqrt{3})}{\cosh(s)}\,ds.\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right)=\frac{\log 2-1}{2}$ - looking for an elementary solution As stated in the title, I am looking for the most elementary proof of the following identity:
$$ \sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right) =
\frac{\log 2-1}{2}\tag{1}$$
I have a proof that exploits $2\,\text{arctanh}\frac{1}{2k}= \log(2k+1)-\log(2k-1)$, summation by parts and Stirling's inequality, but I have the strong feeling I am missing something quite trivial, maybe related with some Riemann sum or with
$$ \sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right) = -\sum_{m\geq 1}\frac{\zeta(2m)}{4^m(2m+1)}. \tag{2}$$
Any help is appreciated, thanks in advance.
I forgot to mention that I would like to avoid proving
$$\forall t\in(0,1),\qquad \sum_{k\geq 1}\frac{4t^2}{4k^2-t^2}=2-\pi t \cot\frac{\pi t}{2} \tag{3}$$
for first. That clearly allows us to compute the LHS of $(1)$ as an integral, but requires Herglotz trick or something similar (Weierstrass products, digamma function, whatever).
| $$
\begin{align}
& \zeta^{\ast}(2n)=\zeta(2n)-\frac{2}{2^{2n}}\zeta(2n) \Rightarrow \frac{\zeta(2n)}{2^{2n}}=\frac{\zeta(2n)-\zeta^{\ast}(2n)}{2} \Rightarrow \\[4mm]
& \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta^{\ast}(2n)}{2n+1} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\Gamma(2n)\zeta(2n)-\Gamma(2n)\zeta^{\ast}(2n)}{\Gamma(2n)(2n+1)} = \\[4mm]
& \frac{1}{2} \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \left(\frac{x^{2n-1}}{e^x-1}-\frac{x^{2n-1}}{e^x+1}\right) \,dx = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2x}-1} \,dx = \\[4mm]
& \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{(2n\color{red}{+1-1}) \space x^{2n-1}}{(2n+1)!} \,dx = \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \left( \frac{x^{2n-1}}{(2n)!}-\frac{x^{2n-1}}{(2n+1)!} \right) \,dx = \\[4mm]
& \int_{0}^{\infty} \left( \frac{x^{-1}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)!} - \frac{x^{-2}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right) \,dx \normalsize = \\[4mm]
& \int_{0}^{\infty} \left( \frac{\cosh(x)-1}{x (e^{2x}-1)} - \frac{\sinh(x)-x}{x^2 (e^{2x}-1)}\right) \,dx \\[4mm]
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\
\end{align}
$$
Both integral are convergent, directly calculate:
$$
\int_{0}^{\infty} \frac{\cosh(x)-1}{x^1 (e^{2x}-1)} \,dx = \frac{1}{2}\left(\log{\pi}-\log{2}\right) \quad\&\quad \int_{0}^{\infty} \frac{\sinh(x)-x}{x^2 (e^{2x}-1)} \,dx = \frac{1}{2}\left(\log{\pi}-\log{e}\right) \\[4mm]
\Rightarrow \sum_{k=1}^{\infty} \left(1-2k\space\text{arctanh}\frac{1}{2k}\right) = - \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{\log{2}-1}{2} \\[4mm]
$$
Or simplify first, and then calculate:
\begin{align}
& \int_{0}^{\infty} \left( \frac{\cosh(x)-1}{x (e^{2x}-1)} - \frac{\sinh(x)-x}{x^2 (e^{2x}-1)}\right) \,dx = \frac{1}{2} \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} - \frac{x}{e^{x}} - \frac{1}{e^{x}} \right) \,dx \\[4mm]
& = \frac{1}{2} \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} \color{red}{-1} - \frac{x}{e^{x}} - \frac{1}{e^{x}} \color{red}{+1} \right) \,dx \qquad \small \{\text{both convergent}\} \\[4mm]
& \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} - 1 \right) \,dx = -\log{2} \\
& \{\small \lim_{x\rightarrow 0} [\Gamma(x)\zeta^{\ast}(x)+\Gamma(x)\zeta(x)] = \lim_{x\rightarrow 0} (2 - 2^{1-x}) \Gamma(x) \zeta(x) = -\log{2} \normalsize\} \\[4mm]
& \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}} + \frac{1}{e^{x}} - 1 \right) \,dx = -1 \\
& \{\small \lim_{x\rightarrow 0} [\Gamma(x) + \Gamma(x-1)] = \lim_{x\rightarrow 0} (1 - 1/(1-x)) \Gamma(x) = -1 \normalsize\} \\[4mm]
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$
$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$
Now I do the nested radicals formula and I get $254\sqrt{41}$ which is none of those where did I mistaked?
| When I see expression where both $\alpha = a+b\sqrt{n}$ and $\beta =a-b\sqrt n$ occur, I immediately calculate $\alpha + \beta = 2a$ and $\alpha\beta = a^2-nb^2$ since they are guaranteed to be integers (more precisely, the minimal polynomial of both of them is $x^2 - (\alpha+\beta)x+\alpha\beta$ which might be helpful in some cases; if you are not familiar with the term, just ignore this remark). So, we have $$x = \sqrt{\alpha^3}-\sqrt{\beta^3}\implies x^2 = \alpha^3 -2\sqrt{(\alpha\beta)^3}+\beta^3\\
\implies x^2 = (\alpha + \beta)(\alpha^2-\alpha\beta+\beta^2)-2\sqrt{(\alpha\beta)^3}\\
\implies x^2 = (\alpha + \beta)((\alpha-\beta)^2+\alpha\beta)-2\sqrt{(\alpha\beta)^3}$$
Now, in your case $\alpha\beta = 1369 = 37^2$, so we have $$x^2 = 90((2\cdot 4\sqrt{41})^2+1369)-2\cdot 37^3 = 258064\implies x= 508$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Convergence of the series: $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$
Does the series $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$ converge or diverge?
Could you give me some hints? Thanks for helping.
| Easier than it looks. Let $m$ be the closest integer to $n \sqrt 2.$ We have
$$ | 2 n^2 - m^2 | \geq 1. $$
they are integers and $\sqrt 2$ is irrational. $2 n^2 - m^2 \neq 0$ is an integer.
$$ | n \pi \sqrt 2 - m \pi | = \left| \frac{\pi (2 n^2 - m^2)}{n \sqrt 2 + m} \right| \geq \frac{\pi }{n \sqrt 2 + m} > \frac{\pi}{3n\sqrt 2} > \frac{1}{2n} $$
This is as $n$ gets large... Therefore
$$ \left| \sin \left(n \pi \sqrt 2 \right) \right| > \frac{1}{3n} $$
and the thing you are summing is smaller than
$$ 3 \; \; \left( \frac{1}{n \log^2 n} \right) $$
which sum converges by the integral test
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
| Continuing from Momo's answer where was proposed the interesting change of variable $x=\tan(\theta)-1$, the integral reduces to $$I=\int \frac{d\theta}{\cos^2(\theta)\,(\sin(\theta)-\cos(\theta))}$$ Using the tangent half-angle substitution (just as Momo proposed), it reduces to $$I=\int\frac{2 \left(t^2+1\right)^2}{\left(t^2-1\right)^2 (t^2+2t-1)}\,dt$$ Using partial fraction decomposition $$\frac{2 \left(t^2+1\right)^2}{\left(t^2-1\right)^2 (t^2+2t-1)}=\frac{4}{t^2+2 t-1}+\frac{1}{t+1}-\frac{1}{(t+1)^2}-\frac{1}{t-1}+\frac{1}{(t-1)^2}$$ which makes $$I=\frac 2{1-t^2}+\log\left(\frac{1+t}{1-t}\right)+\sqrt 2\log\left(\frac{\sqrt 2-1-t}{\sqrt 2+1+t}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the smallest n integer number such that 24^n does not divide 50! What is the smallest n integer number such that 24^n does not divide 50! (factorial) ?
This is what I've done so far, but it seems a bit complicated
| $24=\color\red{2^3}\cdot\color\green{3^1}$
The multiplicity of $\color\red{2}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}50}\Big\lfloor\frac{50}{\color\red{2}^n}\Big\rfloor=25+12+6+3+1=\color\red{47}$
The multiplicity of $\color\green{3}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}50}\Big\lfloor\frac{50}{\color\green{3}^n}\Big\rfloor=16+5+1=\color\green{22}$
Therefore:
*
*The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $50!$ is $\Big\lfloor\frac{\color\red{47}}{\color\red{3}}\Big\rfloor=15$
*The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $50!$ is $\Big\lfloor\frac{\color\green{22}}{\color\green{1}}\Big\rfloor=22$
Therefore:
*
*The maximum value of $n$ such that $24^n$ divides $50!$ is $\min(15,22)=15$
*The minimum value of $n$ such that $24^n$ does not divide $50!$ is $15+1=16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find and prove the limit $\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$ I'm trying to find and prove the limit $$\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$$
which according to WolframAlpha happens to be $27$.
The furthest I could get is
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n} = 4\left(1 + \frac{3^{n}}{2^n+3^n}\right)$$
| $$2^{n+2}+3^{n+3}\sim_\infty 3^{n+3}, \qquad 2^n+3^n\sim_\infty 3^n,$$
hence
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n}\sim_\infty\frac{3^{n+3}}{3^n}=3^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2^{3k+1}2^4 + 5(3^{k})(3^2) $
$p(k+2) = 2^{3k+1+(1-1)}2^4 + 5(3^{k})3^2 $
$p(k+2) = 2^{3k-1}2^6 +5(3^{k})3^2$
I am new to induction and I don't know how to continue.
A point in the right direction would be greatly appreciated, thank you.
| Hint $\ {\rm mod}\ 11\!:\,\ 2x \equiv 8^n - 3^n\equiv (-3)^n-3^n\equiv 0\ $ by $\,n\,$ even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use mathematical induction to prove an assertion The assertion: $n^3 + 5n$ is divisible by $6$
I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.
This is what I have so far:
For $n=k$: $k^3 + 5k = 6t$
For $n= k+1$:
$(k+1)^3 + 5(k+1)$
$= k^3 + 3k^2 + 8k + 6$
$= (k^3 + 5k) + 3k + 3k^2 + 6$
$= 3k^2 + 3k + 6t + 6$
$= ???$
I cannot pull out a $6$ because that would leave halves that cannot be counted as integers. How should I proceed?
| Proof by induction
First, show that this is true for $n=1$:
$1^3+5\cdot1=6$
Second, assume that this is true for $n$:
$n^3+5n=6k$
Third, prove that this is true for $n+1$:
$(n+1)^3+5(n+1)=$
$n^3+3n^2+3n+1+5n+5=$
$\color\red{n^3+5n}+3n^2+3n+1+5=$
$\color\red{6k}+3n^2+3n+1+5=$
$6k+3n^2+3n+6=$
$6k+6+3n^2+3n=$
$6(k+1)+3\cdot\color\green{n(n+1)}=$
$6(k+1)+3\cdot\color\green{2m}=$
$6(k+1)+6m=$
$6(k+1+m)$
Please note that the assumption is used only in the part marked red.
Proof by modular arithmetic
Consider the following cases:
*
*$n\equiv0\pmod6 \implies n^3+5n\equiv0^3+5\cdot0\equiv6\cdot 0\equiv0\pmod6$
*$n\equiv1\pmod6 \implies n^3+5n\equiv1^3+5\cdot1\equiv6\cdot 1\equiv0\pmod6$
*$n\equiv2\pmod6 \implies n^3+5n\equiv2^3+5\cdot2\equiv6\cdot 3\equiv0\pmod6$
*$n\equiv3\pmod6 \implies n^3+5n\equiv3^3+5\cdot3\equiv6\cdot 7\equiv0\pmod6$
*$n\equiv4\pmod6 \implies n^3+5n\equiv4^3+5\cdot4\equiv6\cdot14\equiv0\pmod6$
*$n\equiv5\pmod6 \implies n^3+5n\equiv5^3+5\cdot5\equiv6\cdot25\equiv0\pmod6$
Please note that this method is handy only for a relatively small divisor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $b = ln(\sqrt[3]{\frac{18}{27-k}})$ for $9 = \frac{k}{3-2e^{-b3}}$ I did:
$$9 = \frac{k}{3-2e^{-b3}} \Leftrightarrow \frac{k}{9} = 3-2e^{b3} \Leftrightarrow \frac{k}{9}-\frac{3}{1} = -2e^{-3b} \Leftrightarrow \frac{\frac{k-27}{9}}{\frac{-2}{1}} \Leftrightarrow -\frac{k-27}{18} = \frac{1}{e^{2b}} \Leftrightarrow -\frac{18}{k-27} = e^{3b} \Leftrightarrow \sqrt[3]{-\frac{18}{k-27}} = e^b \Leftrightarrow b = ln{(\sqrt[3]{-\frac{18}{k-27}})}$$
But my book says that the solution if $b = ln(\sqrt[3]{\frac{18}{27-k}})$
It seems I changed the sign somewhere, but I cannot see where.
What did I do wrong?
| Please note that
$$ -\frac{18}{k-27} = -\frac{18}{-(27-k)} = \not-\frac{18}{\not-(27-k)} = \frac{18}{27-k}$$
Therefore
$$b = \ln\left(\sqrt[3]{-\frac{18}{k-27}}\right) = \ln\left(\sqrt[3]{\frac{18}{27-k}}\right)$$
And hence your solution is the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$ We have
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{2^{7/3}}+\frac{\sqrt{3}}{2^{4/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag1$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-2}=\frac{3}{2}-\frac{\ln(\sqrt[3]{2}-1)}{2^{5/3}}+\frac{\sqrt{3}}{2^{2/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag2$$
with the first discussed in this post. (Update) Courtesy of the answers below, we also have,
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}=\frac{4}{3}+\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{5/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{5/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag3$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-3}=\frac{4}{3}-\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{3/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{3/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag4$$Walpha gives a closed-form for $(3)$, but uses two log functions and two arctans. The two answers below show it can be simplified further similar to $(1),(2)$.
| What you need is $\enspace\displaystyle \arctan x+\arctan y=\arctan\frac{x+y}{1-xy}\enspace$ because the values in front of the two $\arctan$ are equal of the results of wolframalpha. Also the values in front of the two $\ln$ are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$ - my answer is wrong (why?) Need to find
$$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$$
One thing I use is
$$\lim_{x \to 1}\frac{x^{1/m}-1}{x^{1/n} - 1} = n/m$$
$$ \lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}} = 3\lim_{x \to 1}\frac{x^{1/2} - x^{1/3}}{(1-x^{1/2})((1-x^{1/3})} = 3\lim_{x \to 1}\frac{x^{2/3} (x^{1/6}-1)^2 - 2x^{1/3}+2x^{5/12}}{(1-x^{1/2})((1-x^{1/3})}= 3\lim_{x \to 1}\frac{x^{1/3} (x^{1/6}-1)^2 }{(1-x^{1/2})((1-x^{1/3})} + \frac{2x^{8/12}(x^{1/12} - 1)^2+4x^{5/24}-4x^{4/24}}{(1-x^{1/2})((1-x^{1/3})} + ... = 3*(1/6 + 2*(3/12*2/12) + 4*(3/24*2/24)... = 3(1/6 + 1/12 + 1/24+...) = 3(\frac{1/6}{1-1/2}) = 1 $$
Answer in my book is 1/2.
Important note I'm in the part of book where integration and differntiation is not covered.
| Put $$t=x^{\frac{1}{6}}.$$
we compute
$$\lim_{t\to 1}\frac{3}{1-t}\left(\frac{1}{1+t+t^2}-\frac{1}{1+t}\right)$$
$$=\lim_{t\to 1}\left(\frac{1}{1-t}\right)\frac{-1}{2}$$
$=+\infty$ at $1^+$ and $-\infty$ at $1^-$.
| {
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"url": "https://math.stackexchange.com/questions/2009665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that the determinant of a special $4 \times 4$ matrix is nonzero? Show that if $a,b,c,d \in \mathbb R$ and at least one is different from $0$, then
$$\begin{vmatrix}
a & b & c & d\\
b & -a & d &-c \\
c & -d & -a &b \\
d & c & -b &-a
\end{vmatrix} \neq 0$$
| I like the Artem methode but this is a direct methode :
We can assume that $a\neq0$ :
\begin{eqnarray}
\begin{vmatrix}
a & b & c & d\\
b & -a & d &-c \\
c & -d & -a &b \\
d & c & -b &-a
\end{vmatrix} &=&\frac{1}{a}\begin{vmatrix}
a^2 & b & c & d\\
ab & -a & d &-c \\
ac & -d & -a &b \\
ad & c & -b &-a
\end{vmatrix} \\
&=& \frac{1}{a}\begin{vmatrix}
a^2 +b^2+c^2+d^2 & b & c & d\\
(ab-ab+dc-cd) & -a & d &-c \\
(ac-bd-ac+bd) & -d & -a &b \\
(ad+cb-bc-ad) & c & -b &-a
\end{vmatrix}((C_1)+b(C_2)+c(C_3)+d(C_4)\to (C_1) ) \\
&=& \frac{1}{a}\begin{vmatrix}
a^2 +b^2+c^2+d^2 & b & c & d\\
0 & -a & d &-c \\
0 & -d & -a &b \\
0 & c & -b &-a
\end{vmatrix} \\
&=& \frac{a^2 +b^2+c^2+d^2}{a}\begin{vmatrix}
-a & d &-c \\
-d & -a &b \\
c & -b &-a
\end{vmatrix} \\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-d(ad-bc)-c(db+ac) \right)\\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-ad^2+dbc-cdb-ac^2 \right)\\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2+b^2+c^2) \right)\\
&=& -(a^2+b^2+c^2+d^2)^2
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$
I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral.
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.
| My approach is a little different. I would sub $x=u^{1/n}$ in the integral and get
$$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 du \frac{u^{\frac1n-1}}{1+u} = \frac1n \int_0^1 du \, u^{1/n} \left (\frac1u - \frac1{1+u} \right ) \\= 1-\frac1n \int_0^1 du \frac{u^{1/n}}{1+u}$$
We then note that $u^{1/n} = e^{\log{u}/n}$ and that $n$ is large enough for the following series expansion to be valid:
$$I(n) = 1- \sum_{j=0}^{\infty} \frac1{n^{j+1}} \int_0^1 du \frac{\log^j{u}}{1+u} $$
Due to the nature of the limit we are posed, we go out to second order; thus
$$I(n) = 1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right ) $$
Then
$$\begin{align}I(n)^n &= e^{n \log{\left [1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right )\right ]} } \\ &= e^{n\left [-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} - \frac{\log^2{2}}{2 n^2}+O \left ( \frac1{n^3} \right ) \right ] } \\ &= \frac12 \left [1+\left (\frac{\pi^2}{12} - \frac12 \log^2{2} \right ) \frac1n +O \left ( \frac1{n^2} \right ) \right ]\end{align}$$
The sought after limit is then
$$\lim_{n \to \infty} n \left [I(n)^n-\frac12 \right ] = \frac{\pi^2}{24} - \frac14 \log^2{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fraction Decomposition I have the following problem:
Suppose $x + y + z = 0$. Show that $$\frac{x^5 + y^5 + z^5}{5}= \frac{x^3 + y^3 + z^3}{3} \times \frac{x^2 + y^2 + z^2}{2}$$ and $$\frac{x^7 + y^7 + z^7}{7}= \frac{x^2 + y^2 + z^2}{2} \times \frac{x^5 + y^5 + z^5}{5}$$
I thought this was a fun problem to tackle, but I've been stuck on this for hours, and I can't seem to get anywhere. I've tried expanding, but no matter what I do, I just get a huge mess.
I also cannot find how to incorporate the initial assumption ($x + y + z = 0$) anywhere. Have I missed something?
| You've surely been trying to express one of the unknowns using the others and plugging in. Try to preserve the symmetry between them instead:
$$(x+y+z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz$$
But we know that the left hand side is a zero and, moreover, wherever something like $x+z$ appears we can replace that by a $-y$ (again, strictly keeping exchange symmetry). So the same equation can be written as
$$0 = x^3 + y^3 + z^3 + 3x^2(-x) + 3y^2(-y) + 3z^2(-z) + 6xyz$$
and even better than that!
$$\begin{aligned}
0 &= -2(x^3 + y^3 + z^3) + 6xyz \\
xyz &= \frac{x^3 + y^3 + z^3}3.
\end{aligned}$$
Surely replacing the right hand side by the left in your problem will make the multiplications simpler! Try to proceed from here :-)
Hint: apart from $S_{111} = xyz$ and $S_k = x^k+y^k+z^k$, among which you have a first relation here, you'll need other symmetric forms like $S_{11} = xy+xz+yz$.
Update: I have been able to derive both your equations this way so it's guaranteed to work. It just takes some time. There's a whole theory about monomial and power-sum symmetric polynomials but I didn't want to go into that.
Starting from $S_1 = 0$, you can derive things like
$$\begin{aligned}
S_2 &= -2S_{11} \\
S_3 &= 3S_{111} \\
S_4 &= \frac12 S_2^2
\end{aligned}$$
and also some useful observations like
$$\begin{aligned}
S_{kl} &= S_k S_l - S_{k+l} \\
S_{(a+c)(b+c)c} &= S_{ccc} S_{ab} = S_{111}^c S_{ab}
\end{aligned}$$
With these you're one little theorem away from $S_5/5 = S_3S_2/6$ and $S_7/7 = S_5S_2/10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given ${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$, find $\sin 4 x$. If
$${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$$
then find the value of $\sin(4x)$.
The options given are:
a) $x$
b) $4x$
c) $2x$
I tried applying many trigo identities but none of them is working and the radicals are posing a big problem...need help...Thanks!!
| For convenience, write
$$ s := \sin x \qquad c := \cos x \qquad m := 1 +\sqrt{1+x} \qquad n := 1 + \sqrt{1-x}$$
The initial equation can then be rewritten as
$$m s = n c \qquad\to\qquad m^2 s^2 = n^2 c^2 \qquad \to\qquad c^2 = \frac{m^2}{m^2+n^2} \qquad s^2 = \frac{n^2}{m^2 + n^2}$$
Note that $m$ and $n$ are both strictly positive. Since (presumably real) $x$ must lie between $1$ and $-1$, we know that $\cos x \geq 0$; since $\tan x$ must be positive in the original equation, $\sin x$ is, too. Thus, we have ...
$$c = \frac{m}{\sqrt{m^2+n^2}} \qquad s = \frac{n}{\sqrt{m^2+n^2}}$$
Now ...
$$\begin{align}
\sin 4x &= 2 \sin 2x \cos 2x \\[4pt]
&= 4 sc ( 2 c^2 - 1 ) \\[4pt]
&= \frac{4mn}{m^2 + n^2} \left(\frac{2 m^2}{m^2 + n^2} - 1\right) \\[4pt]
&= \frac{4mn(m^2-n^2)}{(m^2 + n^2)^2} \\[4pt]
&= \frac{8 x (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})}{8 (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})} \\[4pt]
&= x
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $TC = 7+2x+x^2$ and $TR = 10x$ Was wondering whether you might be able to take a look over my working and see if I'm doing this right! I was using the box method before, but have tried to do this one this way!
Find the break-even points in the case where total cost function $TC=7+2x+x^2$ and total revenue function is $TR=10x$.
Therefore, we need to solve: $$TC = TR$$
$$7+2x+x^2=10x\\
\therefore x^2+2x+7-10x=0\\
\therefore x^2-8x+7 = 0
$$
Now having the equation in the quadratic form $ax^2+bx+c$, we can solve using the quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
$$
Plugging in the values $a=1$, $b=-8$ and $c=7$, we get:
$$\begin{align}
x & = \frac{-(-8)\pm \sqrt{(-8)^2-(4\cdot1\cdot7)}}{2\cdot1}\\\\
& = \frac{8\pm \sqrt{64-28}}{2}\\\\
& = \frac{8\pm \sqrt{36}}{2}\\\\
& = \frac{8\pm 6}{2}\\
\end{align}
$$
Therefore $x$ can equal:
$$\begin{align}\\
x & =\frac{8+6}{2}\\\\
& = \frac{14}{2}\\\\
& = 7
\end{align}
$$
OR:
$$\begin{align}\\
x & =\frac{8-6}{2}\\\\
& = \frac{2}{2}\\\\
& = 1
\end{align}
$$
Therefore, the break-even points could equal either $7$ or $1$.
| Yes, your follow through is correct. Although I would like to point out one thing: You didn't have to use the quadratic formula to find the roots of the polynomial.
Whenever you get a polynomial, factoring usually is the quickest way to find the roots.
For your polynomial, $x^2-8x+7=0$, we see that we need two numbers to have a product of $7$ and sum to $8$. $1,7$ fit quite nicely. Thus, your quadratic can be factored into$$x^2-8x+7=(x-1)(x-7)=0\tag1$$
From which we see $x=1,7$ the two solutions.
Personally, I think it's quicker to factor than use the quadratic formula. Although the quadratic formula does have the nice feature of working on $any$ polynomial.
| {
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Evaluate $ \int_{0}^{\infty} \frac{1}{x^3+x+1}dx$ I want to evaluate the following integral
$$ \int_{0}^{\infty} \frac{1}{x^3+x+1}\>dx$$
via Residue theorem. Or, any other methods are welcome!
Actually i just compute this via mathematica, but it seems
the command
Integrate[1/(x^3 + x + 1), {x, 0, Infinity}]
gives some terrible output not a compact or simple form.
My first trial is factor $x^3+x+1$ into linear terms but it does not seem easy,
I have difficulties for finding poles, $i.e$, finding zeros for $x^3+x+1=0$. And having trouble for finding proper contour.
| Note that $x^3+x+1$ has only one real root below per the Cardano’s formula
$$r= \sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{\sqrt{93}+9}{18}}= -0.6823
$$
Then, decompose the integrand accordingly to integrate as follows
\begin{align}
&\int_0^\infty \frac{1}{x^3+x+1}dx\\
=&\ \int_0^\infty \frac1{(x-r)(x^2+rx-1/r)}dx\\
=&\ \int_0^\infty \frac r{2r +3}\left(-\frac1{x-r}+\frac{x+2r}{x^2+rx-1/r}\right)dx\\
=&\ \frac r{2r +3}\bigg(-\ln\frac{x-r}{\sqrt{ x^2+rx-1/r}}
+\frac{ 3r \tan^{-1}\frac{2x+r}{\sqrt{1-3/r} } }{\sqrt{1-3/r}}\bigg)_{0}^{\infty}\\
=&\ \frac {3r}{2r +3}\bigg(\ln(-r)^{\frac12}
-\frac{(-r)^{\frac32}}{\sqrt{3-r} } \bigg(\frac\pi2 + \tan^{-1}\frac{(-r)^{\frac32}}{\sqrt{3-r} }\bigg)\bigg) \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?
| Divide both sides by $2^{n+1}$.
$$9^{n+1}=27$$
$$n+1=\log_9(27)=\frac 32$$
$$n=\frac 1 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2 + y^2 - 3 = 0$ has no rational solutions, then $x^2 + y^2 - 3^k = 0$ has no rational solutions.
*Explain why $x^2 + y^2 – 3 = 0$ not having any rational solutions (Exercise 20) implies $x^2 + y^2 – 3^k = 0$ has no rational solutions for $k$ an odd, positive integer. (Book of Proof by Hammack)
Please critique my proof:
For the sake of contradiction, suppose that $x^2 + y^2 – 3 = 0$ has no rational solutions but $x^2 + y^2 – 3^k = 0$ has rational solutions for $k = 2j + 1$,
$j\in\mathbb{Z}$. It follows that there are integers $a$, $b$, $c$ and $d$, with $c ≠ 0$
and $d ≠ 0$, such that $x = a/c$ and $y = b/d$ are solutions of $x^2 + y^2 – 3^k = 0$.
Now, we have that $a^2/c^2 + b^2/d^2 = 3^{2k} \cdot 3^1$, it follows that $a^2/(3^{2k}c^2) + b^2/(3^{2k}d^2) = 3$. This in turn leads to that there are some integers $e = 3kc$ and $f =3kd$ such that $x = a/e$ and $y = b/f$ are rational solutions of $x^2 + y^2 = 3$, a contradiction.
| Essentially correct, but more complicated than required. You make some confusion about $k$ and $j$, though.
Suppose $x$ and $y$ are rational solutions of $x^2+y^2=3^k$, with $k$ an odd integer. Then $k=2j+1$ and so
$$
\left(\frac{x}{3^j}\right)^{\!2}+\left(\frac{y}{3^j}\right)^{\!2}=3
$$
where $x/3^j$ and $y/3^j$ are rational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving a set of 2x2 matrices are linearly independent. I'm trying to prove the following matrices are linearly independent.
$ \beta = \left\{\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \right\} $.
My lecture notes say that because the matrix below is row equivalent to the identity matrix, the vectors in $\beta$ must be linearly independent.
$ \begin{bmatrix} 1 & -1 & 0 & 2 \\ 0 & 1 & 3 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}$
Could somebody please explain why this is?
Thanks,
Jack
| Hint
The basis of $\mathbb{M}_{2\times 2}(\mathbb{R})$ is as follow
$$B=\left\{e_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},e_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, e_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, e_4=\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}\right\}$$
and
$$\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=e_1-e_2+0\times e_3+2e_4\\
\qquad\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}=0\times e_1+e_2+3e_3+0\times e_4\\
\qquad\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}=e_1+0\times e_2+e_3+0\times e_4\\
\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=e_1+e_2+e_3+e_4$$
Edit
let
$$\alpha\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}+\beta\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}+\gamma\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}+\theta\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=0_{2\times 2}$$
therefore
$$(\alpha+\gamma+\theta)e_1+(-\alpha+\beta+\theta)e_2+(3\beta+\gamma+\theta)e_3+(2\alpha+\theta)e_4=0_{2\times 2}$$
in other words
$$ \begin{bmatrix} 1 & 0 & 1 & 1 \\ -1 & 1 & 0 & 1 \\ 0 & 3& 1 & 1 \\ 2 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}\alpha\\\beta\\\gamma\\\theta\end{bmatrix}=0$$
| {
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maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| Since $z=4-x-2y$,
$$\begin{align}xy+yz+zx&=xy+y(4-x-2y)+(4-x-2y)x\\&=xy+4y-xy-2y^2+4x-x^2-2xy\\&=-x^2+(4-2y)x-2y^2+4y\\&=-(x^2+(2y-4)x)-2y^2+4y\\&=-\left((x+y-2)^2-(y-2)^2\right)-2y^2+4y\\&=-(x+y-2)^2+(y-2)^2-2y^2+4y\\&=-(x+y-2)^2-y^2+4\\&\le 4\end{align}$$
The equality is attained when $x+y-2=y=0$, i.e. $x=2,y=0,z=2$.
| {
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Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
| HINT
I would say - simply divide the number $1$ manually expression $(2+2x+x^2)$:
More attention and gradually comes out:
$1:(2+2x+x^2) = 1/2 - x/2 + x^2/4 - x^4/8 + x^5/8 - x^6/16 + \cdots$
$\Rightarrow f(x)=1/2 \cdot(1 - x + x^2/2 - x^4/4 + x^5/4 - x^6/8 + \cdots)$
| {
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"timestamp": "2023-03-29T00:00:00",
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About the diophantine equation $y^3=8x^6+2x^3y-y^2$
How I can solve the equation $y^3=8x^6+2x^3y-y^2$ in integers?
I made the substition $x^3=z$ and got the equation $8z^2+2yz-y^3-y^2=0$. So I decided to apply the general formula for quadratic equations and thus I got $$z=\frac{-2y\pm \sqrt{32y^3+36y^2}}{16}=\frac{-y\pm y\sqrt{8y+9}}{8}.$$
So in order to have $z\in \mathbb{Z}$, $8y+9$ must be a perfect square, let's say $8y+9=k^2$, with $k\in \mathbb{Z}$. From this we get $8y=(k+3)(k-3)$, but what could I do next? Also, is there another approach, like an appropriate factorization, to solve this equation? Thanks in advance.
| What you have so far means that $k=2m+3$, so $8y+9 = 4m^2+12m+9\Rightarrow y =\dfrac{m(m+3)}{2}\in\mathbb{Z}, \forall m$. Then, $x^3 = z = \dfrac{m(m+3)}{16}(-1\pm (2m+3) ) = \dfrac{m(m+1)(m+3)}{8}$ or $x^3 = -\dfrac{m(m+2)(m+3)}{8}$.
Now, the only options for the first case are $2x = m+1,m+2 $ and you can do the same for the second equation as well. That should determine what $m$ should be and therefore give you all the solutions.
| {
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Find all solutions to $xyz=1,000,000;$ $x,y,z \in \mathbb Z$ I had previously solved $xy=1,000,000;$ $x,y \in \mathbb Z$, I believe: 1,000,000 has 49 factors so there are 49 pairs since $x$ and $y$ could be both positive or both negative.
Please would you help me find as efficient and systematic a way as possible.
EDIT: since there are many solutions, finding how many there are would be fine. However, unfortunately, (1,1,1000000) is considered distinct from (1000000,1,1).
Thank you
| If $x$, $y$, and $z$ are positive integers with $xyz=2^6\cdot5^6$, then $x=2^a\cdot5^{a'}$, $y=2^b\cdot5^{b'}$, and $z=2^c\cdot5^{c'}$ with $0\le a,a',b,b',c,c'$ and $a+b+c=a'+b'+c'=6$. There are $1+2+3+4+5+6+7=28$ non-negative solutions to $a+b+c=6$, and likewise for $a'+b'+c'=6$, so there is a total of $28\cdot28$ positive triples with product $1{,}000{,}000$. Each of these triples can be modified in three different ways to give a triple with two negative signs, so the total number of integer solutions to $xyz=1{,}000{,}000$ is $4\cdot28\cdot28=3136$.
To explain the $1+2+3+4+5+6+7$, if $a=6$, then there's only $1$ choice for $b$ and $c$, namely $b=c=0$, while if $a=5$, there are $2$ choices, down to $a=0$, for which there are $7$ solutions to $b+c=6$.
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How does $\angle ACB$ change as the point $C$ changes? Let $A = (-4,0)$, $B = (4,0)$, $C = (x,y)$.
How would you determine how angle $\angle ACB$ changes in relation to $x$ and $y$? Must a further variable be introduced? If so, will knowing the distance between $C$ and the origin help? I've been puzzling at this for about a week now but cannot come to anything, so I assume there is a gap in my knowledge of such analytical geometry. Assuming that this is done through differential calculus in relation to geometry, once an equation is derived, it should be easy to find local minima and maxima.
| If $\theta = \angle ACB$, then $AC = (x + 4, y)$ and $BC = (x - 4, y)$, so
\begin{align*}
\cos\theta
&= \frac{AC \cdot BC}{\|AC\|\, \|BC\|} \\
&= \frac{x^{2} - 16 + y^{2}}{\sqrt{(x + 4)^{2} + y^{2}}\, \sqrt{(x - 4)^{2} + y^{2}}} \\[12pt]
&= \sqrt{\frac{(x^{2} - 16 + y^{2})^{2}}{[(x + 4)^{2} + y^{2}][(x - 4)^{2} + y^{2}]}} \\[12pt]
&= \sqrt{\frac{(x^{2} - 16)^{2} + 2y^{2}(x^{2} - 16) + y^{4}}{(x^{2} - 16)^{2} + 2y^{2}(x^{2} + 16) + y^{4}}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prime Factorization of $15^8 + 16$ I have created a problem and I am not able to find the answer. What is the prime factorization of $15^8 + 16$? The thing is I can find an answer with a calculator easily, but what is the step by step approach to the problem.
| \begin{align*}
n^8+n+1 &= (n^2+n+1)(n^6-n^5+n^3-n^2+1) \\
15^8+15+1 &= (15^2+15+1)(15^6-15^5+15^3-15^2+1) \\
&= 241 \times 10634401 \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find Power Series representation of the function $f(x) = {x\over 2x^2 + 1}$?
Find Power Series representation of the function $f(x) = \dfrac x{2x^2 + 1}$?
I'm not sure how to tackle this...I'm supposed to find interval of convergence.
| $\sum x^n = \frac {1}{1-x}$ when $x$ is in the radius of convergence
$\sum (-x)^n = \frac {1}{1+x}\\
\sum (-x^2)^n =\sum (-1)^nx^{2n} = \frac {1}{1+x^2}\\
\sum (-1)^n(\sqrt 2 x)^{2n} =\sum (-1)^n(2^n) x^{2n} \frac {1}{1+2x^2}\\
x\sum (-1)^n(2^n)x^{2n} = \frac {x}{1+2x^2}\\
\sum (-1)^n(2^n)x^{2n+1} = \frac {x}{1+2x^2}$
The series conveges by the root test if:
$\lim_\limits{n\to \infty}|\sqrt[n]{a_n} x|<1$
$|\sqrt[n]{a_n}| = |2^{\frac {n}{2n+1}} x|<1$
$|x|<\frac {\sqrt 2}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
maximum value of the expression : $2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1$ I have to find out the maximum value of the expression :
$2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1.$
| By Cauchy Schwarz inequality $$(2x+3y+z)^2\le (4+9+1)(x^2+y^2+z^2)$$
$\therefore$ the maxium value that can be attained is when $2x=3y=z$ , and the value attained is $\sqrt{14}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral calculus sine functions: $\frac{1}{2\pi }\int_{-\pi }^{\pi }\frac{\sin\left((n+1/2)\,x\right)}{\sin\left(x/2\right)}\,dx = 1$ For an integer, $n$, how do I show the following?
$$
\frac{1}{2\pi }\int_{-\pi }^{\pi }\frac{\sin\left((n+1/2)\,x\right)}{\sin\left(x/2\right)}\,dx = 1.
$$
Can I use induction?
| To use induction, first establish a base case. If $n=0$, then we see trivially that
$$\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((n+1/2)x\right)}{\sin(x/2)}\,dx=\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left(x/2\right)}{\sin(x/2)}\,dx=1$$
Next, we assume that for some integer $N\ge 1$ we have
$$\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((N+1/2)x\right)}{\sin(x/2)}\,dx=1$$
We now examine the integral for $n=N+1$. Proceeding, we have
$$\begin{align}
\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((N+1+1/2)x\right)}{\sin(x/2)}\,dx&=\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((N+1/2)x\right)+2\cos((N+1)x)\sin(x/2)}{\sin(x/2)}\,dx\\\\
&=\color{blue}{\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((N+1/2)x\right)}{\sin(x/2)}\,dx}+\color{red}{\frac1\pi \int_{-\pi}^\pi \cos((N+1)x)\,dx}\\\\
&=\color{blue}{1}+\color{red}{0}\\\\
&=1
\end{align}$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| You should be able to do the third term in your decomposition.
For the second term, follow Mark Bennet's hint: write the fraction as $$\frac{x+2}{9(1+x+x^2)} = \frac{(x+2)(1-x)}{9(1-x^3)} = \frac{2-x-x^2}{9} \cdot \frac{1}{1-x^3},$$ then write the second factor as an infinite geometric series with common ratio $x^3$.
For the first term, you need to observe that $$f(x) = \frac{1}{1-x} = \sum_{k=0}^\infty x^k$$ implies $$f'(x) = \frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1},$$ and $$f''(x) = \frac{2}{(1-x)^3} = \sum_{k=2}^\infty k(k-1) x^{k-2}.$$ Then you have to put this all together and collect like powers in $x$. Tedious, but computationally feasible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Edit: Found another solution, removed old answer (it was incorrect anyway)
You have $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$. By simple algebraic manipulation you can get the
$$x^4-x^3+x^2-x+1 = 0$$
Now notice that $x^4 = x^3-x^2+x-1$ and multiplying both sides by $x$ you get $x^5 = x^4-x^3+x^2-x=-1$.
Therefore
$$x^{2000}+\frac{1}{x^{2000}} = ({x^{5}})^{400}+\frac{1}{(x^{5})^{400}} = (-1)^{400}++\frac{1}{(-1)^{400}} = 1+1 = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$ - not able to evaluate residue Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$,
Here is my attempt:
Replace cos (ax) with $e^{iaz}$, then take the real part such that:
$\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx = R.P[\int_{c}^{} \frac{e^{iaz}}{z^2+z+1} dz]$ where c is the upper half circle.
$\int_{c}^{} \frac{e^{iaz}}{z^2+z+1} dz = 2\pi i Res[\frac{e^{iaz}}{z^2+z+1}, -1+\frac{i\sqrt{3}}{2}]$ (considering only the poles inside my contour )..
then I got stuck , how can I follow from here?
| In this case, you can take the integral to be
$$\operatorname{Re}{\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^2+x+1}} $$
Assume $a \gt 0$ for now. The roots of the denominator are at $x_{\pm} = -\frac12 \pm i \frac{\sqrt{3}}{2} = e^{\pm i 2 \pi/3}$.
Because we assumed $a \gt 0$, we consider the contour integral
$$\oint_C dz \frac{e^{i a z}}{z^2+z+1} $$
where $C$ is a semicircle in the upper half plane. By the residue theorem, we need only consider the pole at $x_+$. As the radius of the semicircle goes to infinity, we find that
$$\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^2+x+1} = i 2 \pi \frac{e^{-\sqrt{3} a/2} e^{-i a/2}}{i \sqrt{3}} $$
Thus, when $a \gt 0$,
$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{x^2+x+1} = \frac{2 \pi}{\sqrt{3}} e^{-\sqrt{3} a/2} \cos{\left ( \frac{a}{2} \right )} $$
When $a \lt 0$, we must use a contour in the lower-half plane. In this case, we use the residue at the pole at $x_-$. Thus, for all values of $a$:
$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{x^2+x+1} = \frac{2 \pi}{\sqrt{3}} e^{-\sqrt{3} |a|/2} \cos{\left ( \frac{a}{2} \right )} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $x$ : $(2-3\mathbb i)x^6 + 1 + 5\mathbb i = 0$ $x^6 = \frac{ 1 + 5\mathbb i}{-2+3\mathbb i}$
How do I convert the right hand side into polar form to find 6th roots?
For numerator we have argument $= \arctan(5) + \pi$
And for denominator argument $= \arctan(-3/2) + \pi$
But $\arctan(5)$ and $\arctan(-3/2)$ are not in standard $\arctan$ ratios
| I prefer to put z instead of x in the equation.
any way
$$Z^6=1-i\\z^6=\sqrt{((-1)^2+(1^2)}.exp(arctan(\frac{-1}{1}))=\\ z^6=\sqrt 2.e^{-\frac{\pi}{4}i}\\
z^6=\sqrt 2.e^{(-\large\frac{\pi}{4}i+2k\pi)}\\$$ now go to the power of $\frac{1}{6}$
$$\sqrt[6]{z^6}=(\sqrt 2.e^{\large(-\frac{\pi}{4}i+2k\pi)})^{\frac{1}{6}}\\z=
\sqrt[12]{2}.e^{\frac{\large(-\frac{\pi}{4}i+2k\pi)}{6}}$$ now put
$$k=0,1,2,3,4,5$$ to find six root of this equation
$$k=0 \to z_0=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i)}{6}}\\
k=1 \to z_1=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i+2\pi)}{6}}\\
k=2 \to z_2=\sqrt[12]{2}.e^{\large\frac{(-\frac{\pi}{4}i+4\pi)}{6}}\\...\\k=5 \to
\sqrt[12]{2}.e^{\frac{(\large-\frac{\pi}{4}i+10\pi)}{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Lagrange multipliers: how could this be outside the restriction? I have to get the extrema of function $f(x,y)=\cos^2(x)+\cos^2(y)$ restricted to $x+y=\frac{\pi}{2}$. So this is what I did:
$$g(x,y)=x+y=\frac{\pi}{2}$$
$$\nabla f(x,y)=\lambda.\nabla g(x,y)$$
So I got the gradients of both functions:
$$\nabla f(x,y)=<-2\cos(x)\sin(x); -2\cos(y)\sin(y)>$$
$$\nabla g(x,y)=<1;1>$$
From where I got the equation system where:
$$
\begin{cases}
-2\cos(x)\sin(x)=1 \\
-2\cos(y)\sin(y)=1 \\
x+y=\frac{\pi}{2}
\end{cases}
$$
I solved the first one as:
$$\sin(2x)=-1 \\
x=\frac{arcsin(-1)}{2} \\
x=-\frac{\pi}{4}
$$
So, since the second equation is the same for $y$, then also $y=-\frac{\pi}{4}$
From all this, I get the extrema must be at $(-\frac{\pi}{4}; -\frac{\pi}{4})$. However, this contradicts my restriction of $x+y=\frac{\pi}{2}$ because of $x$ and $y$ signs.
So, where have I gone wrong?
| Hint
From $\nabla f=\lambda \nabla g$ we get
\begin{cases}\sin 2x=-\lambda\\ \sin 2y=-\lambda\end{cases} Since $\sin t$ is $2\pi$-periodic we have that $\sin 2t$ is $\pi$-periodic. Thus we get that $2y=2x+k\pi,$ for some integer $k.$ That is, $y=x+\frac {k\pi}2.$ Using the constraint $x+y=\frac \pi2$ we get that $2x+\frac {k\pi}2=\frac \pi2.$ That is, $x=\frac{(1-k)\pi}{4}.$ Thus $y=\frac {(k+1)\pi}4.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-0.0064$ (Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$
I did it in two ways:
*
*Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is decreasing and also $\lim_\limits{n\to +\infty}{\frac{n^3}{4^n}}=0$.
*Using the n-th root test (Cauchy's criterion) and absolute convergence, proving that $\lim_\limits{n\to +\infty}{\frac{(\sqrt[n]n)^3}{4}}=\frac{1}{4}<1$.
However, Wolfram Alpha states another interesting result: That this series sums exactly to $-0.0064$. I would like to see how that result is obtained, so a proof for it.
| As in other answers consider $$\sum_{n=1}^{+\infty}n^3 x^n$$ and rewrite $$n^3=n(n-1)(n-2)+a n(n-1)+bn$$ Expanding and identifying coefficients, you should get $a=3$ and $b=1$. So
$$\sum_{n=1}^{+\infty}n^3 x^n=\sum_{n=1}^{+\infty}n(n-1)(n-2) x^n+3\sum_{n=1}^{+\infty}n(n-1) x^n+\sum_{n=1}^{+\infty}n x^n$$
$$\sum_{n=1}^{+\infty}n^3 x^n=x^3\sum_{n=1}^{+\infty}n(n-1)(n-2) x^{n-3}+3x^2\sum_{n=1}^{+\infty}n(n-1) x^{n-2}+x\sum_{n=1}^{+\infty}n x^{n-1}$$
$$\sum_{n=1}^{+\infty}n^3 x^n=x^3\left(\sum_{n=1}^{+\infty} x^{n}\right)'''+3x^2\left(\sum_{n=1}^{+\infty} x^{n}\right)''+x\left(\sum_{n=1}^{+\infty} x^{n}\right)'$$ Now, for $|x|<1$ $$\sum_{n=1}^{+\infty} x^{n}=\frac x {1-x}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'=\frac{1}{(1-x)^2}$$ $$\left(\sum_{n=1}^{+\infty} x^{n}\right)''=\frac{2}{(1-x)^3}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'''=\frac{6}{(1-x)^4}$$ which make
$$\sum_{n=1}^{+\infty}n^3 x^n=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac{x}{(1-x)^2}={x(x^2+4x+1)\over (1-x)^4}$$ Using now $x=-\frac 14$ leads to $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-\frac{4}{625}=-0.0064$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution to $\mathbf{x}'=A\mathbf{x}$, with $A$ anti-symmetric implies $|\mathbf{x}|=|\mathbf{x}_0|$ Let $A$ be anti-symmetric. That is, $A^T=-A$. Show that the solution to the $\mathbf{x}'=A\mathbf{x}$ satisfies $|\mathbf{x}|=|\mathbf{x}_0|$. Hint: Compute $d/dt|\mathbf{x}|^2$.
The given solution:
$d/dt|\mathbf{x}|^2=2\mathbf{x}'\cdot\mathbf{x}=2A\mathbf{x}\cdot\mathbf{x}=2A|\mathbf{x}|^2=0,$ since $A\mathbf{x}\cdot\mathbf{x}=\mathbf{x}A^T\mathbf{x}=-\mathbf{x}A\mathbf{x}.$ Thus, the norm is preserved.
So, I get up until the "since" part. Is that just true because $A$ is anti-symmetric? That man Also, how does that imply that the given quantity is $0?$ I've tried to see it. This is as far as I can go: We have that $A|\mathbf{x}|^2=-\mathbf{x}A\mathbf{x}=-\mathbf{x}\cdot\mathbf{x}'=-e^{tA}\mathbf{x_0}A\mathbf{x}$.
| The expression $2A|\mathbf{x}|^2$ makes no sense as $A$ is a matrix while $|\mathbf{x}|^2$ is a number so let's stop one step before it. The product rule gives us the equation $\frac{d}{dt} |\mathbf{x}|^2 = 2(A\mathbf{x}) \cdot \mathbf{x}$. Assuming that $\mathbf{x}$ is a column vector and calculating the dot product using the definition and $A^T = -A$, we have
$$ (A\mathbf{x}) \cdot \mathbf{x} = (A \mathbf{x})^T \cdot \mathbf{x} = \mathbf{x}^T A^T \mathbf{x} = -\mathbf{x}^T A \mathbf{x} = - \mathbf{x} \cdot (A \mathbf{x}) = -(A \mathbf{x}) \cdot \mathbf{x}$$
which shows that $(A\mathbf{x}) \cdot \mathbf{x} = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've worked these problems before, I usually end up performing a u-substitution after the trig sub, but I'm not sure where to do that or even if this is the right strategy for this integral.
$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$
$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$
| Hint: Let $u=36x^2+1$, then $du=72xdx$ and thus
$$\frac{x^5}{(36x^2+1)^{3/2}}dx=\frac{(u-1)^2}{36^2u^{3/2}}\frac{du}{72}.$$
Now we get something easier to handle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\
\sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 - x_3} + \sqrt{1 - x_4} = 2\sqrt{3}
\end{array}
\right.
$$
| HINT:-
Since the roots have to be positive, we see that each radical must reduce to the form $a\sqrt5$ for the first equation , and $b\sqrt3$ for the second equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An interesting result in ratio and proportions If
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d}=k
$$
Also
$$
\frac{a^2}{b^2}=\frac{c^2}{d^2}=k^2
$$
And
$$
\left( \frac{a+c}{b+d} \right)^2=k^2
$$
Also
$$
\frac{a^2 + c ^2}{b^2+d^2}=k^2
$$
Hence
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
My question is... if the reverse is true. That is...
if
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
$$
Then can we assume that :
$$
\frac{a}{b}=\frac{c}{d}
$$
??
| if we factorizing $$\frac{a^2+c^2}{b^2+d^2}-\left(\frac{a+c}{b+d}\right)^2=\frac{2 (a d-b c) (a b-c d)}{(b+d)^2 \left(b^2+d^2\right)}=0$$ and we get $$ad-bc=0$$ or $$ab-cd=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How should I evaluate the integral $\lim_{n\rightarrow\infty}n\int\frac{\cos(nx)}{1+x^2}\,dx$? Q1: How should I evaluate the integral $$\lim_{n\rightarrow\infty}n\int_{\mathbb{R}}\frac{\cos(nx)}{1+x^2}\,dx$$ (Hint: dominated convergent theorem and integration by parts) ?
Possible Solution:
\begin{align}
n\int\frac{\cos(nx)}{1+x^2}\,dx&=n\left(\left[\frac{1}{1+x^2}\frac{\sin(nx)}{n}\right]_{-\infty}^\infty+\int\frac{\sin(nx)2x}{n(1+x^2)^2}\,dx\right)\\
&=2\int\frac{x\sin(nx)}{n(1+x^2)^2}\,dx=\frac{2}{n}\int \frac{-x}{(1+x^2)^2}\,d(\cos(nx))\\
&=\frac{2}{n}\left(\left[-\frac{x \cos(nx)}{(1+x^2)^2}\right]_{-\infty}^\infty+\int \cos(nx)\frac{(1+x^2)^2-2x(1+x^2)}{(1+x^2)^4}\,dx\right)\\
&=\frac{2}{n}\int \cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx.
\end{align}
Hence
$$\lim\frac{2}{n}\int \cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx=\int\lim\frac{2}{n}\cos(nx)\frac{1+x^2-2x}{(1+x^2)^3}\,dx=0$$
by Dominated Convergent Theorem.
I am still working on this hence I will update this post if there are more steps coming out.
| By $nx=t$ and using integration by parts $\int_{-\infty}^{\infty}n\frac{cos(nx)}{1+x^2}dx = \frac{sin(t)}{1+(t/n)^2}|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{sin(t)}{(1+(t/n)^2)}\frac{2t}{n^2}dt \\ $. Here the first term goes to zero and second term is bounded by $\int_{-\infty}^{\infty}\frac{1}{(1+(t/n)^2)}\frac{2t}{n^2}dt$. The last term is an odd function hence the integral is 0. Hence the answer should be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving polynomial equation with calculus I saw a question stating that
Let $f(x)$ is a polynomial of degree $5 $.
$ x-1|f(x)+1 $ and $x+1|f(x)-1$ . Find $f(x)$ without using calculus.
And because I am not good at calculus , I was not able to solve the question with or without using calculus.
Please help me in solving it.
A hint for how to start could work.
| Followup on previous comment:
The two conditions are equivalent to $f(1)=−1$ and $f(−1)=1$ but that's not enough to determine the polynomial univocally.
Let the quintic be $f(x) = ax^5+bx^4+cx^3+dx^2+ex+f$, then the two conditions give:
$$
\begin{align}
\begin{cases}
a+b+c+d+e+f & = -1 \\
-a+b-c+d-e+f &= 1
\end{cases}
\end{align}
$$
Adding and subtracting the two equations:
$$
\begin{align}
\begin{cases}
a+c+e &= -1 \\
b+d+f & = 0
\end{cases}
\end{align}
$$
The above give the necessary and sufficient relations that the coefficients of $f(x)$ must satisfy, but do not determine $f(x)$ univocally. For example, all of the following satisfy the given conditions:
*
*$f(x) = -x^5$
*$f(x) = x^5 - x^3 - x$
*$f(x) = x^5 + 3 x^4 - x^3 - 5 x^2 - x + 2$
*$f(x) = a x^5 + b x^4 + cx^3 +d x^2 - (a+c+1)x - b-d \quad \text{for} \;\forall a,b,c,d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question
| Call $k$ the common value of
$$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$$
and then
$$\log(a^ab^bc^c)=a\log a+b\log b+c\log c=\left(a(b-c)+b(c-a)+c(a-b)\right)\cdot k=0$$
which implies that $a^ab^bc^c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve $\sqrt{3}^n=3^3$ I got answer but i could not understand, can anyone explain me ?
$\sqrt{3}^n=3^3$
$\implies{3}^{n/2}=3^3$
$\implies \frac{n}{2}=3\implies n=6$
How we get ${3}^{n/2}$ in the second line and how we get $\frac{n}{2}$ in third line.
| $$\sqrt{x} = x^{\frac{1}{2}}$$
Since if $\sqrt{x}\sqrt{x} = x$ by definition then it follows that $x^kx^k = x^1$ where $2k = 1$ so $k = \frac{1}{2}$. If you choose to agree with this, then
$$\sqrt{3^n} = \sqrt{3}^\frac{n}{2} = 3^3 \implies \frac{n}{2} = 3 \implies n = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but giving it to wolfram alpha gives five complex set of answers. Another idea that looks to be nice is this:
$\sum\limits_{}^{cyc}\frac{1}{xy+z-1}=\sum\limits_{}^{cyc}\frac{x}{x^2-x+4}$
Maybe it gives the answer but it is to hard to calculate. Any hints?
| If we transform the whole thing into a single fraction, then both numerator and denominator will be symmetric polynomials on three variables.
The good thing about them is that one can then use the Fundamental Theorem of Symmetric Polynomials to write $P(x,y,z)$ in a unique way as a polynomial $P'(e_1(x,y,z),e_2(x,y,z),e_3(x,y,z))$ where
$$e_1(x,y,z) = x + y + z \qquad e_2(x,y,z) = xy + xz + yz \qquad e_3(x,y,z) = xyz.$$
We already know that $e_1(x,y,z) = 2$ and $e_3(x,y,z) = 4$. We can compute $e_2$ since $e_1^2 - (x^2 + y^2 + z^2) = 2e_2$:
$$e_2 = \dfrac{1}{2}.$$
Now the only thing we have to do is to decompose the numerator and denominator as a polynomial on $e_1,e_2,e_3$ to find the value of the initial problem. This, however, may not be very elegant.
The whole thing gives $P/Q$ where:
$$P = (x^2yz+xy^2z+xyz^2) + (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - (xy+xz+yz) - (2x+2y+2z) + 3,$$
$$Q = (x^3yz+xy^3z+xyz^3) + (x^2y^2z^2) + (x^2y^2+x^2z^2+y^2z^2) -( x^2yz+xy^2z+xyz^2) - (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) + (x+y+z) + (xyz)- 1.$$
Therefore
$$x^2yz+xy^2z+xyz^2 = xyz(x+y+z) = 8$$
$$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$
$$2x+2y+2z = 2(x+y+z) = 4$$
So $P = 8 - 11 - \dfrac{1}{2} - 4 + 3 = -\dfrac{9}{2}$.
Conversely,
$$x^3yz + xy^3z + xyz^3 = (xyz)(x^2 + y^2 + z^2) = 12$$
$$x^2y^2z^2 = (xyz)^2 = 16$$
$$x^2y^2+x^2z^2+y^2z^2 - x^2yz-xy^2z-xyz^2 = (xy+xz+yz)^2 - 3(xyz)(x+y+z) = -\dfrac{95}{4}$$
$$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$
so $Q = 12 + 16 -\dfrac{95}{4} + 11 + 2 + 4 - 1 = \dfrac{81}{4}$.
Finally, $P/Q = -\dfrac{9\cdot 4}{2\cdot 81} = -\dfrac{2}{9}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
QR algorithm for "general" square matrices
*
*Can QR algorithm find repeat eigenvalues (https://en.wikipedia.org/wiki/QR_algorithm) ?
I.e. does it support the case when not all N eigenvalues for real matrix N x N are distinct?
*How to extend QR algorithm to support finding complex eigenvalues ?
*Is it possible to extend QR algorithm to work with not full rank matrices ?
| Summary
The QR Decomposition resolves all of your matrix types.
a) Repeated eigenvalues
The three eigenvalues of $\mathbf{A}$ are
$$
\lambda \left( \mathbf{A} \right) = \left\{ 1, 1, 1 \right\}
$$
The decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{Q\,R} \\
% A
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right]
&=
% Q
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right]
% R
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right]
%
\end{align}
$$
b) Complex eigenvalues
The three eigenvalues of $\mathbf{A}$ are
$$
\lambda \left( \mathbf{A} \right) = \left\{ 1 + 2 i, 1 - 2 i \right\}
$$
The decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{Q\,R} \\
% A
\left[
\begin{array}{cr}
3 & -2 \\
4 & -1 \\
\end{array}
\right]
&=
% Q
\frac{1}{5}
\left[
\begin{array}{rc}
3 & 4 \\
-4 & 3 \\
\end{array}
\right]
% R
\left[
\begin{array}{cr}
5 & -2 \\
0 & 1 \\
\end{array}
\right]
%
\end{align}
$$
c) Matrices with rank defect
The following matrix has rank $\rho = 1$, therefore the matrix has both a row and a column rank defect.
The decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{Q\,R} \\
% A
\left[
\begin{array}{rrr}
1 & -1 & 1 \\
-1 & 1 & -1 \\
\end{array}
\right]
&=
% Q
\frac{1}{\sqrt{2}}
\left[
\begin{array}{r}
1 \\ -1 \\
\end{array}
\right]
% R
\left[
\begin{array}{crc}
\sqrt{2} & -\sqrt{2} & \sqrt{2} \\
\end{array}
\right]
%
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.
I'm not sure how to use Lagrange multipliers for it... any ideas?
| Let $f(x,y,z) = \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} $
$$\begin{eqnarray*}
\vec \nabla f \cdot \hat x &=& +\frac{1}{(y+z)} - \frac{y}{(x+z)^2}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat y &=& -\frac{x}{(y+z)^2} + \frac{y}{(x+z)}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat z &=& -\frac{x}{(y+z)^2} - \frac{y}{(x+z)^2}+ \frac{z}{(x+y)}
\end{eqnarray*}$$
You should be able to demonstrate that $$\vec\nabla f=0 \implies x=y=z$$
and since $f(x,x,x)=\frac 32$ you are pretty well done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$ I need to prove this equation
$$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$$
where $x$, $y$ and $n$ are nonnegative integers satisfying $y \geq n$.
The sign $(-1)^k$ suggests using binomial expansion, I've tried this, but without success. Is there a better way?
| Standing the conditions you gave, the $y \choose n$ is not null, and we can multiply by it both sides, giving
$$
\left( \begin{gathered}
y - x \\
n \\
\end{gathered} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
y \\
n \\
\end{gathered} \right)\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)/\left( \begin{gathered}
y \\
k \\
\end{gathered} \right)}
$$
The RHS can be developed as
$$
\begin{gathered}
\sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
y \\
n \\
\end{gathered} \right)\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)/\left( \begin{gathered}
y \\
k \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
y \\
k \\
\end{gathered} \right)\left( \begin{gathered}
y - k \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)/\left( \begin{gathered}
y \\
k \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
y - k \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( \begin{gathered}
y - k \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
k - x - 1 \\
k \\
\end{gathered} \right)} = \hfill \\
= \left( \begin{gathered}
y - x \\
n \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
which is valid for $x$ and $y$ even complex, with the only condition that ${y \choose n }\ne 0$.
*
*---Note ----*
1st step) Trinomial Revision :
$\left( \begin{gathered}
y \\
n \\
\end{gathered} \right)\left( \begin{gathered}
n \\
k \\
\end{gathered} \right) = \left( \begin{gathered}
y \\
k \\
\end{gathered} \right)\left( \begin{gathered}
y - k \\
n - k \\
\end{gathered} \right)$
2nd step) simplifying $ {y \choose k }$
3rd step) VanderMonde Convolution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
show that; strange sum yields triangular numbers $\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$ $1,3,6,10,15,...$ for $n=1,2,3,...$ it is the n-th Triangular numbers.
I find it unsual that this sum yields even triangular numbers;
$$\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$$
How can I show that? Any hints into this strange sum?
I can't figure it out where to start!
| As recommended in the comment section of OPs question we can follow the nice answer of Fallager.
We obtain
\begin{align*}
\left(\cos\frac{k\pi}{2n+1}+i\sin\frac{k\pi}{2n+1}\right)^{2n+1}&=(-1)^k\tag{1}\\
\\
\sum_{j=0}^{2n+1}\binom{2n+1}{j}\left(\cos\frac{k\pi}{2n+1}\right)^j\left(i\sin \frac{k\pi}{2n+1}\right)^{2n+1-j}&=(-1)^k\tag{2}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(\cos\frac{k\pi}{2n+1}\right)^{2j}\left(i\sin \frac{k\pi}{2n+1}\right)^{2n+1-2j}&=0\tag{3}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(i\tan \frac{k\pi}{2n+1}\right)^{2n+1-2j}&=0\tag{4}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(i\tan \frac{k\pi}{2n+1}\right)^{2n-2j}&=0\tag{5}\\
\end{align*}
Comment:
*
*In (1) we use De Moivre's formula
*In (2) we apply the Binomial theorem
*In (3) we take the imaginary part of (2) i.e. with index $j$ even
*In (4) we divide by $\left(\cos \frac{k\pi}{2n+1}\right)^{2n+1}$
*In (5) we divide by $\tan \frac{k\pi}{2n+1}$
We conclude $\left(\tan\frac{k\pi}{2n+1}\right)^2$ with $0\leq k\leq 2n$ are the zeros of the following polynomial in $z$
\begin{align*}
\sum_{j=0}^n\binom{2n+1}{2j}\left(-z\right)^{n-j}
\end{align*}
According to Vieta's formulas the sum of the zeros is
\begin{align*}
\sum_{k=0}^{n}\left(\tan\frac{k\pi}{2n+1}\right)^2=\frac{\binom{2n+1}{2}}{\binom{2n+1}{0}}=n(2n+1)=T_{2n}
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the sum of the roots of a fourth-degree polynomial is equal to the coefficient of $x^3$
but I didn't get that.
| Consider instead the polynomial in $x$
$P(x)=\det\begin{pmatrix}1&x&x^2&x^3&x^4\\1&a&a^2&a^3&a^4\\1&b&b^2&b^3&b^4\\1&c&c^2&c^3&c^4\\1&d&d^2&d^3&d^4\end{pmatrix}$.
If you use Laplace's expansion in the first row, you'll notice that $P(x)$ has degree $4$.
Also, $P(a)=P(b)=P(c)=P(d)=0$ because plugging $x=a,b,c,d$ creates two equal rows and the determinant vanishes.
Now, by Vieta's relations, and recalling that the terms in the Laplace's expansion alternate signs, the term in $x^3$ is the sum of the roots, so
$a+b+c+d=\frac{\det\begin{pmatrix}1& a& a^2& a^4\\1& b& b^2& b^4\\1& c& c^2& c^4\\1& d& d^2& d^4\end{pmatrix}}{\det\begin{pmatrix}1& a& a^2& a^3\\1& b& b^2& b^3\\1& c& c^2& c^3\\1& d& d^2& d^3\end{pmatrix}}$
Now we can use Vandermonde's determinant and finish the problem:
$\det\begin{pmatrix}1& a& a^2& a^4\\1& b& b^2& b^4\\1& c& c^2& c^4\\1& d& d^2& d^4\end{pmatrix}=(a+b+c+d)(d-a)(d-c)(d-b)(c-a)(c-b)(b-a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 0
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Find the least possible integer for which $\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$ Question
Find the least possible no for which $$\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$$
How do I reach a stage where I can deduce definitely the smallest integer value of $n$. I keep getting stuck after I cube on both sides and I still get cube root terms which I can't seem to simplify . Any help is appreciated.Thanks :)
| Let $n = x^3$
\begin{array}{c}
\sqrt[3]{n+1} - \sqrt[3] n < \frac 1{12} \\
\sqrt[3]{x^3+1} - x < \frac 1{12} \\
\sqrt[3]{x^3+1} < x + \frac 1{12} \\
x^3+1 < x^3 + \dfrac 14x^2 + \dfrac{1}{48}x + \dfrac{1}{1728} \\
\dfrac 14x^2 + \dfrac{1}{48}x - \dfrac{1727}{1728} > 0 \\
x^2 + \dfrac{1}{12}x - \dfrac{1727}{432} > 0 \\
\left( x + \dfrac{1}{24} \right)^2 - \dfrac{6911}{1728} > 0\\
x + \dfrac{1}{24} > \dfrac{\sqrt{6911}}{24\sqrt 3} \\
x > \dfrac{\sqrt{6911}}{24\sqrt 3} - \dfrac{1}{24} \\
x > \dfrac{1}{72} (\sqrt{20733} - 3) \\
x^3 > \left(\dfrac{1}{72} (\sqrt{20733} - 3)\right)^3 \\
n > \dfrac{865 \sqrt{20733} - 7776}{15552} \\
n > 7.5
\end{array}
The smallest value is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Do not use series expansion or L' Hospital's rule: $f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ In the following function
$$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},(x\neq 0)$$
is continuous at $x= 0$. Find $A$ and $B$. Also find $f(0)$.
I first thought of using L hospital. But my sir told me to
'Do not use series expansion or L' Hospital's rule.�'
Now how can I proceed?
| Hint. If one admits that
$$
\lim_{u \to 0}\frac{\sin u}u=1 \tag1
$$ then using
$$
\sin(2x)=2\sin x \cos x,\quad \sin(3x)=3\sin x-4\sin^3x,
$$ one may observe that, for $x \neq0$,
$$
\begin{align}
\frac{\sin 3x + A \sin 2x + B\sin x}{x^5}
&=\frac{\sin x}{x}\cdot\frac{4\left(\cos x+\frac{A}4 \right)^2+\left(B-1-\frac{A^2}4\right)}{x^4}
\\&=\frac{\sin x}{x}\cdot\left[\left(\frac{\sin^2(\frac{x}2)}{(\frac{x}2)^2}-\frac{A+4}{2x^2} \right)^2+\frac{B-1-\frac{A^2}4}{x^4}\right] \tag2
\end{align}
$$ implying for the existence of the limit as $x \to 0$ that
$$
A+4=0,\quad B-1-\frac{A^2}4=0
$$ that is
$$
A=-4,\quad B=5.
$$
From $(2)$ one sees that in this case $f(0)=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac{x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1$ for $a+b+c=m$
If $a+b+c=m$, prove that:
$$\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac {x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1.$$
My Attempt:
$a+b+c=m$
$a=m-b-c$
$b=m-a-c$
$c=m-a-b$,
L.H.S.$=\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}$.
Now, how should I move on?
| Note that
$$\frac{x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}} = \frac{x^a}{x^a} \frac{x^a}{x^a+x^b+x^c}.$$
Using a similar reasoning for the other two terms, we find that
$$\begin{align*} &\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}\\ &= \frac{x^a}{x^a+x^b+x^c} + \frac{x^b}{x^a+x^b+x^c} + \frac{x^c}{x^a+x^b+x^c} \\ &= 1. \end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple exercise on Galois theory Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.
| $x^6-2x^4-8x^2+16 = x^6+x^4+x^2+1$ in $\mathbb {F}_3[x]$.
$x^6+x^4+x^2+1 = \dfrac{x^8-1}{x^2-1} = (x^2 + 1) (x^4 + 1)$
Therefore, the splitting field of $x^6-2x^4-8x^2+16$ is the same as the splitting field of $x^4 + 1=(x^2 + x + 2) (x^2 + 2 x + 2)$, which is $\mathbb {F}_9$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a system of linear differential equations with repeated eigen values I have this problem where to solve the system,
$$x'=4x+y-z$$
$$y'=2x+5y-2z$$
$$z'=x+y+2z$$
using a linear algebraic solution. I have found the eigen values of the
$$\begin{bmatrix}
4 & 1 & -1 \\
2 & 5 & -2 \\
1 & 1 & 2 \\
\end{bmatrix}$$
as 3,3 and 5. When evaluating the corresponding eigen vectors for 3, the following occurs.
$$(A-3I)x=0$$
$$\begin{bmatrix}
1 & 1 & -1 \\
2 & 2 & -2 \\
1 & 1 & -1 \\
\end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = 0$$
We can say ok and $x_3=x_1+x_2$ and then a set of eigen vectors which are not multiples of each other are formed. As the next step, I have to find $\rho$ such that $(A-3I)\rho = \eta_{\lambda =3}$. There I'm getting nowhere because of the ambiguity of $\eta_{\lambda=3}$. I am new to this eigen things. Am I doing something terribly wrong or what?
| You have correctly calculated your eigenvalues. To avoid ambiguous notation, denote your eigenvalues as follows:
$\lambda_1=5$
$\lambda_2=3$
$\lambda_3=3$.
Using your eigenvalues, use the fact that:
$A\mathbf{x}=\lambda_1 \mathbf{x}$, $A\mathbf{x}=\lambda_2 \mathbf{x}$ and $A\mathbf{x}=\lambda_3 \mathbf{x}$, where $\mathbf{x}=\begin{pmatrix}x\\y\\z \end{pmatrix}$ and $A=\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}$ to evaluate your eigenvectors.
Hence for $\lambda_1$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix}=5 \begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix}$.
For $\lambda_2$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix}=3 \begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix}$.
For $\lambda_3$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_3\\y_3\\z_3 \end{pmatrix}=3 \begin{pmatrix}x_3\\y_3\\z_3 \end{pmatrix}$.
From this, we deduce for $\lambda_1$ the following system:
$4x_1+y_1-z_1=5x_1$
$2x_1+5y_1-2z_1=5y_1$
$x_1+y_1+2z_1=5z_1$
Note that these simultaneous equations are redundant – they are all essentially the same. Hence, we deduce from one of these equations that $y_1=x_1+z_1$.
We repeat the same process for $\lambda_2$ (no need for $\lambda_3$ since both have same eigenvalues) and obtain $z_2=x_2+y_2$, as you mentioned earlier. The next step to find the eigenvectors is to find the smallest possible nonzero integer values for $x_n, y_n, z_n$ for each of your corresponding eigenvalues such that it satisfies your equation. These are your eigenvectors.
For $\lambda_1$, we obtain $v_1=\begin{pmatrix}1\\2\\1 \end{pmatrix}$. For $\lambda_2$ we obtain two possible solutions $v_2=\begin{pmatrix}1\\0\\1 \end{pmatrix}$ and $v_3=\begin{pmatrix}-1\\1\\0 \end{pmatrix}$. The eigenvectors will be the same for $\lambda_3$.
Therefore, from the eigenvalues and eigenvectors, we may substitute and obtain the general solutions:
$\begin{pmatrix}x(t)\\y(t)\\z(t) \end{pmatrix}=c_1\begin{pmatrix}1\\2\\1 \end{pmatrix}e^{5t}+c_2\begin{pmatrix}1\\0\\1 \end{pmatrix}e^{3t}+c_3\begin{pmatrix}-1\\1\\0 \end{pmatrix}e^{3t}$.
Please do not hesitate to ask if you have any doubts or questions.
| {
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The sum of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. What can $p$ be?
The summation of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. Find all possible solutions for $p$, with proof. Here $p$ is a prime number.
This is a preparation problem for the upcoming BdMO. I can't find a way to start. Any hint will be helpful.
| Let $p_n$ denote the $n$-th prime. Note that $1+2+\cdots+p_n = p_n\cdot\dfrac{p_n+1}{2}$.
By Bertrand's postulate, $p_n < 2p_{n-1}$. Hence, $0 < \dfrac{p_n+1}{2} < \dfrac{2p_{n-1}+1}{2} < 2p_{n-1}$.
Therefore, $\dfrac{p_n+1}{2}$ is not divisible by $2p_{n-1}$.
If $n \ge 3$, then $2$ and $p_{n-1}$ are distinct primes, and so $\dfrac{p_n+1}{2}$ is not divisible by both $2$ and $p_{n-1}$.
Then, since $p_n$ is distinct from $2$ and $p_{n-1}$, we have $p_n\cdot\dfrac{p_n+1}{2}$ is not divisible by both $2$ and $p_{n-1}$.
Hence, $1+2+\cdots+p_n$ is not divisible by all of $p_1 = 2, \ldots, p_{n-1}, p_n$ for $n \ge 3$.
For $n = 1$, we have $p_1 = 2$, and $1+2 = 3$ is not divisible by $p_1 = 2$.
For $n = 2$, we have $p_2 = 3$, and $1+2+3 = 6$ is divisible by $p_1 = 2$ and $p_2 = 3$.
Therefore, the only prime $p$ such that $1+2+\cdots+p$ is divisible by all primes less than or equal to $p$ is $p = 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why $x^5y\equiv y^5x\pmod {240}$ if $x,y$ have similar parity? Let $x,y$ be any 2 integers of similar parity. Why do we have :$$x^5y\equiv y^5x\pmod {240}$$
Any hint anyone?
| $$x^5y-xy^5=(x^5-x)y-(y^5-y)x$$
Now $p|(a^p-a)$ by Fermat's little theorem for prime $p$
$a^5-a$ is divisible by $a^3-a$
If $x,y$ are even, we are done.
For odd $a,a=2c+1$(say)
$(2c+1)^2=8\dfrac{c(c+1)}2+1=8d+1$ for some integer $d$
$(2c+1)^4=(8d+1)^2=16(d+4d^2)+1$
$\implies(2c+1)^{4m+1}\equiv2c+1\pmod{16}$
$\implies a^5\equiv a\pmod{16}$ for odd $a$
| {
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Multiplicative inverse using irreducible polynomial I am trying to compute the multiplicative inverse of:
$x^4 + x^2 + x$
Using the irreducible polynomial:
$x^8 + x^4 + x^3 + x + 1$
Over $GF(2)=\mathbb{F}_2$.
I tried applying the extended euclidean algorithm but to no avail. At some point during the algorithm I end up with the remainder $169/144$ and am unsure how to proceed from there on. Any help would be greatly appreciated.
Thanks in advance!
| $\mathbb{F}_2[x]/(x^8+x^4+x^3+x+1) \cong \mathbb{F}_{2^8}$.
If $\alpha \in \mathbb{F}_{2^8}$ and $\alpha\ne0$, then $\alpha^{2^8-1}=1$, and so $\alpha^{-1}=\alpha^{2^8-2}$.
Therefore, the inverse of $x^4 + x^2 + x$ in $\mathbb{F}_{2^8}$ is $(x^4 + x^2 + x)^{2^8-2}$.
This is an answer, even if not a very practical one.
The extended Euclidean algorithm (*) gives
$$
9= (12 x^7 - 15 x^6 + 9 x^5 - 6 x^4 + 18 x^3 - 6 x^2 - 6 x + 12)(x^4+x^2+x)+(-12 x^3 + 15 x^2 - 21 x + 9)(x^8+x^4+x^3+x+1)
$$
from which we get that the inverse of $x^4+x^2+x$ is $12 x^7 - 15 x^6 + 9 x^5 - 6 x^4 + 18 x^3 - 6 x^2 - 6 x + 12=x^6 + x^5$.
(*) computed by WA.
| {
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How to quickly solve $y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx$? I'm currently trying to solve $$y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx,$$ which is a GRE math subject test problem. I was able to get the answer ($\sqrt{2}$) by breaking up the integral $$\int_{-\pi/4 }^{\pi/4 } \cos x \ dx+\int_{-\pi/4 }^{\pi/4 } \sqrt{1+x^2}\sin^3x\cos^3x \ dx.$$ Then I used $$1+x^2=(z-x)^2$$ to get $$t=\frac{z^2-1}{2z}$$ as well as $$\sin x = \frac{2z}{1+z^2}, \ \ \cos x = \frac{1-z^2}{1+z^2}, \ \ dx = \frac{2 \ dz}{1+z^2}.$$ Note that $$z=\tan(x/2).$$ I then plugged these into $$\int_{-\pi/4 }^{\pi/4 }\sqrt{1+x^2}\sin^3x\cos^3x \ dx$$ to get something rather ugly
\begin{align*}
&\int_{-\tan{\pi/8}}^{\tan{\pi/8}}\sqrt{1+ \frac{z^2-1}{2z}^2 } \cdot\left( \frac{2z}{1+z^2}\right)^3\cdot\left(\frac{1-z^2}{1+z^2}\right)^3\cdot\left(\frac{2}{1+z^2}\right)dz \\
&= \int_{-\tan{\pi/8}}^{\tan{\pi/8}} \frac{16 z^3 (1 - z^2)^3 \sqrt{\frac{(z^2 - 1)^2}{4 z^2} + 1}}{(z^2 + 1)^7}dz.
\end{align*}
Since this is an odd function, the solution is $0$ and thus $$y=\int_{-\pi/4 }^{\pi/4 } \cos x \ dx + 0 =\sqrt{2}.$$ Even though I was able to get to the answer, I'd imagine on the math GRE subject test, I would not have had enough time to figure all this out. Are there any tricks for quickly solving definite integrals like this?
| Hint. The function
$$
x \mapsto \sqrt{1+x^2}\sin^3x\cos^3x, \qquad x \in \left[-\frac{\pi}4, \frac{\pi}4\right]
$$ is odd as is $x \mapsto \sin (x)$.
| {
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Showing that $f(x) = x^3$ is injective? This is my attempt. Is it right? Is there a simpler way? Is there a way which relies on less background knowledge?
A function is injective iff $f(x) = f(y) \implies x = y$. This is equivalent to $x \not = y \implies f(x) \not = f(y)$. We will prove this latter statement by contradiction:
Suppose there are two numbers $x$, $x+a$ with $a >0$ and $x \not = x+a$ but $x^3 = (x+a)^3$. Expanding the RHS, subtracting $x^3$, and dividing both sides by $a>0$ yields $0 = 3x^2 + 3xa + a^2$. Taking this last equation as a quadratic in $x$, it has a real solution iff the discriminant is non-negative. But the discriminant is $9a^2 - 12a^2 = -3a^2$, so there are no real $x$ which satisfy $x^3 = (x+a)^3$.
| Suppose $$x^3=y^3$$ then $$(x-y)(y^2+xy+x^2)=0$$ so either $x=y$ or $$x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac {3y^2}4=0$$
But this is strictly non-negative and is zero only if $x=y=0$
| {
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Proving the inequality I need to prove this:
$\binom{n}{2}a^2 + \binom{m}{2}b^2
\leq
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2$
I tried this (WLOG $n\leq m$):
$\binom{n}{2}a^2 + \binom{m}{2}b^2 = \frac{n(n-1)}{2}a^2 + \frac{m(m-1)}{2}b^2
\\
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2
= \frac{n+m-1}{2}\frac{(na+mb)^2}{n+m}$
How would you do it?
| I have to rush off now; if I remember, I'll come back and finish this answer.
Combinatorially: I'll interpret everything as areas.
Say we have $n$ sticks of length $a$, and $m$ sticks of length $b$. Then $a^2$ is the area of the square we get by placing two $a$-length sticks orthogonally; there are $\binom{n}{2}$ ways to do this, so $\binom{n}{2} a^2$ is the sum of the areas of all the squares we can make this way.
Similarly, $\binom{m}{2} b^2$ is the sum of the areas of all the $b$-squares we can make.
On the other hand, the right-hand side is $\binom{n+m}{2}$ the number of ways we can pick two sticks, multiplied by the square of $\frac{na+mb}{n+m}$ the average length of a stick.
The left-hand side has counted all the area we can possibly make by picking two of the same kind of stick; while the right-hand side has counted every possible rectangle but has counted each as being of area "the square of the average side length" instead of each rectangle's true area.
Probability beckons!
Divide through by $\binom{n+m}{2}$, so that the right-hand side becomes the square of the average length of a stick, while the left-hand side becomes "the total area we can make by picking two sticks of the same size, divided by the number of ways to pick two sticks of any size".
That's a quantity less than or equal to "the total area we can make by picking two sticks of the same size, divided by the number of ways to pick two sticks of the same size", which is just the mean area of a rectangle made of two sticks of the same size.
So it would be enough to show that the mean area of a rectangle made of two sticks of the same size is less than or equal to the square of the mean length of a stick.
| {
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If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
| We have $ a^2+b^2-ab=c^2$, then $ (a-b)^2=c^2-ab\ge 0$. So $ c^2\ge ab $. This implies that at least one of $ c\ge a $ or $ c\ge b $ is true.
First case: $ c\ge a $. Then $ b^2-ab=b (b-a)=c^2-a^2\ge 0$, so $ b\ge a $. Thus we have $b^2-c^2=ab-a^2=a (b-a)\ge 0$, so $ b\ge c $. Therefore $(a-b)(b-c)\le 0$.
Second case: $ c\ge b$. Then $ a^2-ab=a (a-b)=c^2-b^2\ge 0$, so $ a\ge b $. Hence $(a-b)(b-c)\le 0$.
| {
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Another way of proving :$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$ Prove that
$$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$$
My try
$x-x^2+x^3+x^5-x^7=x(1-x^2)+x^5(1-x^2)=x(1+x^4)(1-x^2)$
$$\int_{0}^{1}{x(1+x^4)(1-x^2)\over (1+x^4)\ln{x}}dx$$
Applying Frullani theorem
$$\int_{0}^{1}{x-x^3\over \ln{x}}dx$$
$$\int_{0}^{1}{x-x^3\over \ln{x}}dx=-\ln{2}$$
| Maybe some one will be interested in that
We can show
$$ \int^1_0 \frac{x^a-1 }{\log x} dx =\log \left ( a+1\right)$$
Using differentiation under the integral sign.
Generally
$$ \int^1_0 \frac{x^a-x^b }{\log x} dx =\int^1_0 \frac{x^a-1-(x^b-1)}{\log x} dx =\log \left ( \frac {a+1}{b+1}\right)$$
Hence
$$ \int^1_0 \frac{x-x^3 }{\log x} dx =\log \left ( \frac {2}{4}\right)=-\log 2$$
| {
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Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly
Say a rational number is $kingly$ if it is the square of another rational number. Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly.
If $\dfrac{n+7}{25n+7}$ is kingly, then we can write $\dfrac{n+7}{25n+7} = \dfrac{p^2}{q^2}$ for some $p,q \in \mathbb{Z}^+$ and $\gcd(p,q) = 1$. Rearranging the equation gives $$q^2(n+7) = p^2(25n+7).$$
How can I continue?
| Option 1: $n+7$ and $25n+7$ are both perfect squares
Not possible:
All perfect squares are equivalent to $0,1$ or $4$ modulo $5$
$25n+7\equiv 2\pmod 5$
Option 2: $(n+7)$ divides $(n+25)$ and the ratio is a perfect square.
$\lim_\limits {n\to \infty} \frac {n+7}{25n+7}= \frac 1{25}$
There are 3 possible values that $\frac {n+7}{25n+7}$ could take on where the ratio is a perfect square.
$\frac {n+7}{25n+7} = \frac 14 \implies n =1$
$\frac {n+7}{25n+7} = \frac 1{9} \implies 16 n = 56$ and $n$ is not an integer.
$\frac {n+7}{25n+7} = \frac 1{16} \implies 9 n = 105$ and $n$ is not an integer.
Indeed I have overlooked something.
$n = 20 \implies \frac {27}{507} = \frac {9}{169}$
$p^2(25n+7) = q^2(n+7)\\
(25p^2 - q^2) n = 7(q^2-p^2)$
$(5p+q) | 7(q+p)(q-p)$ and $5p-q |7(q+p)(q-p)$
| {
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Show that $x^3+2y^3+4z^3=0 $ has no non trivial solutions without using infinite descent The question is to show that the equation $$x^3+2y^3+4z^3=0 $$ has no (non trivial) integer solutions. I know it can be done using infinite descent, but how do you do it using only the modular arithmetic argument?
| Mod $9$ also solves it.
By Binomial Theorem $(3k\pm 1)^3\equiv \pm 1\pmod{9}$,
so $x^3\equiv\{0,\pm 1\}\pmod{9}$
(also $x^3\equiv \{0,\pm 1\}\pmod{8}$, so mod $8$ solves it similarly, as a comment has suggested).
$x^3+2y^3+4z^3\equiv 0\pmod{9}$
is equivalent to $x^3+2y^3\equiv -4z^3\pmod{9}$
$x^3+2y^2\equiv \{\pm 3,\pm 2, \pm 1, 0\}\pmod 9$
while $-4z^3\equiv \{0,\pm 4\}\pmod{9}$.
Therefore $x^3+2y^3\equiv -4z^3\equiv 0\pmod{9}$.
So $3\mid x,y,z$. Therefore $(x,y,z)$ is a solution if and only if $\left(\frac{x}{3},\frac{y}{3},\frac{z}{3}\right)$ is also a solution, so $(x,y,z)=(0,0,0)$ (see, e.g., proof by infinite descent). As a comment has said, you must use infinite descent because $x\equiv y\equiv z\equiv 0$ is a solution mod anything.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$
Prove that $$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
To do this I unified three terms on the left side with a common denominator and then factored the numerator (with the aid of Wolfram Alpha... as the numerator is a 5th order expression).
$$\frac{b²c²(c-b)+c²a²(a-c)+a²b²(b-a)}{abc(a-b)(b-c)(c-a)}=\dots=\frac{(ab+bc+ca)(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}$$
My question is would there be a better or easier way to lead from the left side to the right side? Actually, the original question was to 'simplify' the left side without showing the right side. I am not sure if I could do so without Wolfram Alpha. So that's why I'm asking an easier way.
What I've thought of was, for example,
$$f(x):=(x-a)(x-b)(x-c)$$
$$\text{LHS}=abc\left(\frac1{a^2f'(a)}+\frac1{b^2f'(b)}+\frac1{c^2f'(c)}\right)$$
which didn't help.
| Claim:
$$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
Proof:
Note that the Lagrange interpolating polynomial through $(a,p),(b,q),(c,r)$ is given by:
$$f(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}p+\frac{(x-a)(x-c)}{(b-a)(b-c)}q+\frac{(x-a)(x-b)}{(c-a)(c-b)}r$$
We seek the value of $f(0)$ when $p=\dfrac{1}{a},q=\dfrac{1}{b},r=\dfrac{1}{c}$, i.e. when $y=f(x)$ interpolates $y=\dfrac{1}{x}$ at $x=a,b,c$.
We then notice that $1-xf(x)$ is a degree $3$ polynomial interpolant passing through $(0,1),(a,0),(b,0),(c,0)$, and is in fact the unique such polynomial.
By inspection, one must have
$$1-xf(x)=\frac{(x-a)(x-b)(x-c)}{(-a)(-b)(-c)}=\left(1-\frac{x}{a}\right)\left(1-\frac{x}{b}\right)\left(1-\frac{x}{c}\right)$$
Expanding to leading order,
$$1-xf(x)=1-x\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\text{ higher order terms}$$
which, upon a moment's thought, gives that $f(0)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ as desired.
Note that this method gives a natural generalisation to when we have more than just $a,b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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A Viéte-like infinite product for $\sin (\pi/7)$ In the article "The unruly $\sin (\pi/7)$ of Samuel Moreno, the following infinite product is given:
Also, the same article shows that $\sin (\pi/7)$ is equal to the infinite nested radical
$ \sin (\pi/7) = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2+...}}}}}}}}}}$
The question is: are these two expressions fundamentally different for $\sin(\pi/7)$, or is it possible to rewritte the infinite nested radical into that infinite product by simple algebraic manipulations?
| With the help of Half angle cosine formula it is possible to express many sine and cosine angles as finite or infinite nested square roots of 2 here.
In this perspective $2\sin\frac{\pi}{7}$ can be solved as follows
$2\sin\frac{\pi}{7}$ = $\sqrt{2-2\cos\frac{2\pi}{7}}$ = $\sqrt{2-\sqrt{2+2\cos\frac{4\pi}{7}}}$ = $\sqrt{2-\sqrt{2-2\cos\frac{3\pi}{7}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{6\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-2\cos\frac{\pi}{7}}}}$ = $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{2\pi}{7}}}}}$
One can observe repetition of $2\cos\frac{2\pi}{7}$ subsequently and $\therefore$ $2\sin\frac{\pi}{7}$ can be represented simply as
$$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+...2\cos\frac{2\pi}{7}}}}}}}}$$ with repeating pattern $--+$ infinitely and first $-$ as single nonrepeating sign
But getting infinite product has to be worked out
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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finding the geometry of complex numbers when you have terms in cube When I solve for the following $$\left|{\frac{1+z}{1-z}}\right|=2$$
I get $$ 3x^2 +3y^2-10x +3=0 $$ and this isnt the equation of a circle nor a line so what geometry is that?
the correct possibilities point to either a circle of radius 4/3 or 5/4 or a circle centered in (0,5/3) or a circle centred in (1,0) or radius 4/3 but with what I ended up with I can't figure which one it is.
| For this case:
$$3x^2 +3y^2-10x +3=0 \rightarrow 3\left(x^2 -\frac{10}{3}x\right)+3y^2+3=0$$
$$3\left[\left(x -\frac{5}{3}\right)^2-\frac{25}{9}\right]+3y^2+3=0$$
$$3\left(x -\frac{5}{3}\right)^2-\frac{25}{3}+3y^2+3=0$$
$$\left(x -\frac{5}{3}\right)^2+(y-0)^2=\left(\frac{4}{3}\right)^2$$
It is a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that given fraction is power of two I do not really know where to start with the following:
Prove that
$$\frac{\displaystyle{ \prod_{k = 1}^{2n - 1} k^{\min(k, 2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}$$ is a power of two.
Could you give me a hint which helps to solve this problem?
| Let $f(n)$ be the given expression. Then $f(1) = 1$ and $f(n+1)/f(n) = 2^{2n}$:
$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^{n+1}(n+2)^n \dotsb (2n-1)^3(2n)^2(2n+1)}{3^{2n-1} \dotsb (2n-1)^3(2n+1)} \cdot \frac{3^{2n-3} \dotsb (2n-1)}{(n+1)^{n-1}(n+2)^{n-2} \dotsb (2n-1)}$$
which simplifies to
$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^2(n+2)^2 \dotsb (2n-1)^2(2n)^2}{3^2 \dotsb (2n-1)^2} = \frac{(n+1)^2(n+2)^2 \dotsb (2n-1)^2(2n)^2 2^2 4^26^2\dotsb (2n)^2} {3^2 \dotsb (2n-1)^2 2^2 4^26^2\dotsb (2n)^2} = \frac{2^{2n} (2n)!^2}{(2n)!^2} = 2^{2n}$$
Then $$f(n) = f(1)\cdot \frac{f(2)}{f(1)} \dotsb \frac{f(n)}{f(n-1)} = 2^{2\cdot 0} \cdot 2^{2\cdot 1} \dotsb 2^{2\cdot (n-1)} = 2^{2(1+2+\dotsb (n-1))} = 2^{n(n-1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$ and one of the asymptotes of the hyperbola at $(1,-2)$. Find the centre.
A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$. It also intersects one of the asymptotes of the hyperbola at the point $(1,-2)$.
Find the centre of the hyperbola.
My approach:
The mid-point of the two points of contact $(-2,-6)$ and $(4,2)$ is the point $(1,-2)$ and we are given that the line cuts the asymptote at this point. Hence the answer is $(1,-2)$. But this seems intuitive and I am not sure whether my reasoning is correct. I am looking for another approach to the problem.
| Your intuition can be stated as the proposition below:
If $A$ and $B$ are two different points on a hyperbola and the midpoint of $AB$ lies on the asymptotes, then this midpoint is the center of the hyperbola.
Proof: Note that this proposition is irrelevant to coordinate systems, thus without loss of generality, the hyperbola can be assumed to be $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Suppose the coordinates of $A$ and $B$ are $(x_1, y_1)$ and $(x_2, y_2)$, respectively, then $x_1, x_2 ≠ 0$ and$$
\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1, \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1. \tag{1}
$$
Because the asymptotes are $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$ and $\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)$ lies on the asymptotes, then$$
\frac{(x_1 + x_2)^2}{a^2} - \frac{(y_1 + y_2)^2}{b^2} = 0. \tag{2}
$$
Now, suppose $x_1 + x_2 ≠ 0$, then $y_1 + y_2 ≠ 0$ by (2). From (1) there is$$
\frac{x_1^2 - x_2^2}{a^2} = \frac{y_1^2 - y_2^2}{b^2},
$$
and from (2) there is$$
\frac{(x_1 + x_2)^2}{a^2} = \frac{(y_1 + y_2)^2}{b^2},
$$
thus\begin{align*}
&\mathrel{\phantom{\Longrightarrow}}{} \frac{x_1 - x_2}{x_1 + x_2} = \frac{\dfrac{x_1^2 - x_2^2}{a^2}}{\dfrac{(x_1 + x_2)^2}{a^2}} = \frac{\dfrac{y_1^2 - y_2^2}{b^2}}{\dfrac{(y_1 + y_2)^2}{b^2}} = \frac{y_1 - y_2}{y_1 + y_2}\\
&\Longrightarrow \frac{2x_1}{x_1 + x_2} = \frac{x_1 - x_2}{x_1 + x_2} + 1 = \frac{y_1 - y_2}{y_1 + y_2} + 1 = \frac{2y_1}{y_1 + y_2}\\
&\Longrightarrow \frac{y_1}{x_1} = \frac{y_1 + y_2}{x_1 + x_2} \Longrightarrow \frac{y_1}{x_1} = \frac{y_2}{x_2} := c.
\end{align*}
Note that $x_1 + x_2 ≠ 0$, plugging $y_1 = cx_1$ and $y_2 = cx_2$ into (2) to get $\dfrac{1}{a^2} - \dfrac{c^2}{b^2} = 0$, then by (1) there is$$
1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = \left( \frac{1}{a^2} - \frac{c^2}{b^2} \right) x_1^2 = 0,
$$
a contradiction. Therefore, $x_1 + x_2 = 0$, which implies $y_1 + y_2 = 0$ by (2). Hence the midpoint of $AB$ is the center $(0, 0)$ of the hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Incorrect solution in limits $$
\lim_{n\to \infty}\left(\frac{1}{n^4}{+3^{\frac{2}{2+n}}}\right)^{n}
$$
So i re-write it like:
$\lim_{n\to \infty}e^{n\ln{\frac{1}{n^4}\ln3^{\frac{2}{2+n}}}}$ $=$ $e^{\frac{2n}{2+n}\ln{\frac{1}{n^4}\ln3}}=e^{{2}\ln{\frac{1}{n^4}\ln3}}$
So here, $\ln{\frac{1}{n^4}}$ give us minus infinity, but i think that somewhere i lost a way, so i need some hints
| Note that
$$
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a}
{n}} \right)^{\,n} = e^{\,a} \quad \quad \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{f(n)}}
{n}} \right)^{\,n} \ne \mathop {\lim }\limits_{n\; \to \;\infty } e^{\,f(n)}
$$
that's the fault in your derivation. Consider instead that we have:
$$
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a}
{{\left( {n + \alpha } \right)}} + \frac{b}
{{\left( {n + \beta } \right)^{\,2} }} + \cdots } \right)^{\,n + \gamma } = e^{\,a}
$$
Therefore we shall try and expand $3^{2/(2+n)}$ in powers of the exponent, i.e. Taylor of $3^x \quad | \, x=2/(2+n) \approx 0$
$$
\begin{gathered}
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {\frac{1}
{{n^{\,4} }} + 3^{\,\frac{2}
{{2 + n}}} } \right)^{\,n} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {e^{\frac{{2\ln 3}}
{{2 + n}}} + \frac{1}
{{n^{\,4} }}} \right)^{\,n} = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}}
{{2 + n}} + \frac{{\left( {2\ln 3} \right)^2 }}
{2}\frac{1}
{{\left( {2 + n} \right)^2 }} + \cdots + \frac{1}
{{n^{\,4} }}} \right)^{\,n} = \hfill \\
= \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}}
{{2 + n}}} \right)^{\,n} = e^{2\ln 3} = 9 \hfill \\
\end{gathered}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index? Let $\sigma(x)$ denote the sum of the divisors of $x$, and denote the abundancy index of $x$ as
$$I(x) = \dfrac{\sigma(x)}{x},$$
and the deficiency of $x$ as
$$D(x) = 2x - \sigma(x).$$
If the equation $I(a)=b/c$ has no solution $a \in \mathbb{N}$, then $b/c$ is said to be an abundancy outlaw.
Statement of the Problem
When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index?
Preliminary Results
The following lemmas are easy to show:
Lemma 1. If $p$ is an odd prime, and $(p+2)/p$ is the abundancy index of some integer $n$, then $n$ is deficient.
Lemma 2. If $p$ is odd, then $\gcd(p,p+2)=1$.
Lemma 3. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $D(n) = 2n - \sigma(n) \neq 1$.
Lemma 4. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $p < n$.
Lemma 5. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $\gcd(n,\sigma(n)) \neq 1$.
Remarks
In fact, one can show that, if $p$ is an odd prime and $I(n) = (p+2)/p$, then
$$\sigma(n) = \bigg(\dfrac{n}{p}\bigg)\cdot(p+2)$$
and
$$n = \bigg(\dfrac{\sigma(n)}{p+2}\bigg)\cdot(p).$$
(Note that $n/p$ and $\sigma(n)/(p+2)$ are (equal) integers because of Lemma 2.) Consequently, we obtain
$$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2}.$$
(Note further that both $\gcd(n,\sigma(n)) \leq n/3$ and $\gcd(n,\sigma(n)) \leq \sigma(n)/5$ hold.)
Added September 16 2017
Given that $X = A/B = C/D$ ($B \neq 0$, $D \neq 0$, and $B \neq D$), we can make use of the algebraic identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}$$
to get another expression for
$$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2}.$$
Indeed,
$$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2} = \frac{\sigma(n) - n}{2}.$$
This last finding implies that
$$\bigg(\frac{\sigma(n) - n}{2}\bigg) \mid n \iff (\sigma(n) - n) \mid (2n) \iff 2n = (\sigma(n) - n){d_1}$$
and
$$\bigg(\frac{\sigma(n) - n}{2}\bigg) \mid \sigma(n) \iff (\sigma(n) - n) \mid (2\sigma(n)) \iff 2\sigma(n) = (\sigma(n) - n){d_2}.$$
Note that $2 \mid (\sigma(n) - n)$. Additionally, notice that
$$2\gcd\left(n,\sigma(n)\right) = \gcd\left(2n, 2\sigma(n)\right) = \gcd\left((\sigma(n) - n){d_1},(\sigma(n) - n){d_2}\right)$$
$$= \left(\sigma(n) - n\right)\gcd({d_1},{d_2}) \iff \frac{2\gcd\left(n,\sigma(n)\right)}{\left(\sigma(n) - n\right)}=1=\gcd({d_1},{d_2}).$$
In fact,
$$d_1 = \frac{2n}{\sigma(n) - n} = p$$
and
$$d_2 = \frac{2\sigma(n)}{\sigma(n) - n} = p+2.$$
Double-checking if it is indeed the case that ${d_1}+2={d_2}$:
$$d_1 = \frac{2n}{\sigma(n) - n} + 2 = \frac{2n + 2(\sigma(n) - n)}{\sigma(n) - n} = \frac{2\sigma(n)}{\sigma(n) - n} = d_2.$$
So far so good!
More is actually true. One can also show that
$$p(2n - \sigma(n)) = (p - 2)n$$
so that
$$D(n) = (p - 2)\cdot\bigg(\dfrac{n}{p}\bigg) = (p - 2)\cdot\bigg(\dfrac{\sigma(n)}{p + 2}\bigg) = (p - 2)\cdot\gcd(n,\sigma(n)).$$
We therefore conclude that
$$\dfrac{D(n)}{n} = \dfrac{p - 2}{p} = \bigg(\dfrac{p - 2}{p + 2}\bigg)\cdot{I(n)}.$$
Added October 8 2017
We deduce that
$$\dfrac{p-2}{p}=\dfrac{D(n)}{n}<\dfrac{\phi(n)}{n}<\dfrac{n}{\sigma(n)}=\dfrac{p}{p+2},$$
whence there is still no contradiction.
Motivation
It is conjectured that $(p+2)/p$ is an outlaw, since if it were an index, then we would be able to produce an odd perfect number for $p=3$.
Here is my question:
To what extent can the following theorem be improved to hopefully produce some results towards proving the aforementioned conjecture?
Theorem If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy of $n$ in terms of the deficiency of $n$:
$$\dfrac{2n}{n + D(n)} < I(n) < \dfrac{2n + D(n)}{n + D(n)}.$$
| Too long to comment.
One can prove that
$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n)\tag1$$
$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag2$$
hold for any $a>0,b\gt -1$. Note here that $a,b$ are not necessarily integers.
However, it seems that we cannot prove the conjecture using $(1)(2)$.
About $(1)$ :
Let $x:=\frac{\sigma(n)}{n}$. Then, we get $$1\lt x\lt 2\tag3$$
Trying to find $a,b,c$ such that $-x+c\gt 0$ and
$$\frac{2}{3-x}\lt\frac{ax+b}{-x+c}\lt x$$
which is equivalent to
$$ax^2+(b-3a-2)x+2c-3b\lt 0\quad\text{and}\quad x^2+(a-c)x+b\lt 0\tag4$$
every $x$ such that $(4)$ has to satisfy $(3)$.
So, trying to find $a,b,c$ such that
$$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(a-c)^2-4b}\le \min(4+a-c,-a+c-2)\\\sqrt{(3a+2-b)^2-4a(2c-3b)}\le \min(a+2-b,a-2+b)\end{cases}$$
and choosing $b=2$ give
$$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(c-a)^2-8}\le \min(4-(c-a),(c-a)-2)\\\sqrt{9a^2-8a(c-3)}\le a\end{cases}$$
So, we see that choosing $c=a+3$ works, and that
$$\frac{2}{3-x}\lt\frac{2+ax}{a+3-x}\lt x,$$
i.e.
$$\frac{2n}{n+D}\lt\frac{(2a+2)n-aD}{(a+1)n+D}\lt \frac{\sigma(n)}{n}$$
holds for any $a\gt 0$.
About $(2)$ :
Trying to find $a,b,c$ such that $-x+c\gt 0$ and
$$x\lt\frac{a+bx}{-x+c}\lt\frac{4-x}{3-x}$$
which is equivalent to
$$x^2+(b-c)x+a\gt 0\quad\text{and}\quad (b+1)x^2+(a-3b-c-4)x+4c-3a\gt 0\tag6$$
every $x$ such that $(6)$ has to satisfy $x\not=2$.
So, trying to find $a,b,c$ such that
$$\begin{cases}b+1\gt 0\\
(b-c)^2-4a\le 0\\
(a-3b-c-4)^2-4(b+1)(4c-3a)\le 0\\
2^2+(b-c)\times 2+a=0\\
(b+1)\times 2^2+(a-3b-c-4)\times 2+4c-3a=0\end{cases}$$
and choosing $a=4$ give
$$\begin{cases}b\gt -1\\
c=b+4\end{cases}$$
So, we see that
$$x\lt\frac{4+bx}{-x+b+4}\lt\frac{4-x}{3-x},$$
i.e.
$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}$$
holds for any $b\gt -1$.
| {
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"url": "https://math.stackexchange.com/questions/2090850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Determinant composed from polynomials $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ Let $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ be two polynomials with real
coefficients, and $x_1$ and $x_2$ be two arbitrary real numbers. Consider the
following determinant $$D(x) = \begin{vmatrix}
1 & p_1(x_1) & p_2(x_1)\\
1 & p_1(x_2) & p_2(x_2)\\
1 & p_1(x) & p_2(x)\
\end{vmatrix}
$$
Show that $D(x) = m(x-x_1)(x-x_2)$
| Note that since $\det$ is multi linear and alternating, we have
\begin{eqnarray}
D(x) &=& \det \begin{bmatrix}
1 & x_1+a & x_1^2+bx_1+c \\
1 & x_2+a & x_2^2+bx_2+c \\
1 & x+a & x^2+bx+c \
\end{bmatrix} \\
&=& \det \begin{bmatrix}
1 & x_1 & x_1^2+bx_1+c \\
1 & x_2 & x_2^2+bx_2+c \\
1 & x & x^2+bx+c \
\end{bmatrix} \\
&=& \det \begin{bmatrix}
1 & x_1 & x_1^2+bx_1 \\
1 & x_2 & x_2^2+bx_2 \\
1 & x & x^2+bx \
\end{bmatrix} \\
&=& \det \begin{bmatrix}
1 & x_1 & x_1^2 \\
1 & x_2 & x_2^2 \\
1 & x & x^2 \
\end{bmatrix} \\
\end{eqnarray}
We see from this that $D$ is a second degree polynomial in $x$ and
$D(x_1) = D(x_2) = 0$. We can read off the coefficient $m$ as the coefficient of $x^2$ above, $m=x_2-x_1$.
(As Joonas notes below, the above is a Vandermonde matrix which has a well
known formula for the determinant.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$ I want to find the limit of this function by simply using algebraic manipulation. Though I have computed the limit through L' Hospital's method but still I want to compute the limit purely by function's manipulation to yield a form where limit can be applied
$$\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x} $$
Till now we have been taught basic limits such as $\lim_{x\to 0} \frac{e^x-1}{x}=1$ and that's why I have been trying to bring such form in this expression.
P.S. I got the current answer by L'Hospital's rule i.e. $0$
| $$\begin{align*}
& \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x}
+ \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x}
+ \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot
\lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}}
+ \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot
\lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\
& = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\
& = \frac{1}{2} - \frac{1}{2}\\
& = 0 \\
\end{align*}$$
| {
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"url": "https://math.stackexchange.com/questions/2094362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Compute the integral $\int_0^{\pi/2}\sin(1+\cos^2x)\mathrm{d}x $ My attempt.
$$\int\limits_0^{\pi/2}\sin(1+\cos^2x)\mathrm{d}x $$
Let $t=1+\cos^2x \Rightarrow \mathrm{d}t -2\sin x\cos x~ \mathrm{d}x. $
We have $\cos^2x=t-1 \Rightarrow \cos x=\sqrt{t-1}$ and $\sin x = \sqrt{2-t}.$ So $$\mathrm{d}x = \frac{-\mathrm{d}t}{2\sqrt{2-t}\sqrt{t-1}}$$
Now,
$$\frac{1}{2}\int\limits_1^2\frac{\sin t}{\sqrt{2-t}\sqrt{t-1}}\mathrm{d}t$$
I tried substituting by parts letting $u=\sin t$, and $dv = \int \mathrm{d}t/(\sqrt{(2-t)(t-1)})$.
But the integral just gets more complicated.
| Use the integral representations of the Bessel function of the first kind at $0$,
$$\int_{0}^{\pi/2}\cos\left(z\cos \theta\right)\,\mathrm{d}\theta=\frac{\pi}{2}J_0\left(z\right).$$
So,
$$\begin{align*}
\int_0^{\frac{\pi}{2}}\sin\left(1+\cos^2x\right)\mathrm{d}x&=\int_0^{\frac{\pi}{2}}\sin\left(\frac{1}{2}\cos\left(2x\right)+\frac{3}{2}\right)\mathrm{d}x\\
&=\sin\left(\frac{3}{2}\right)\int_{0}^{\frac{\pi}{2}}\cos\left(\frac{1}{2}\cos\left(2x\right)\right)\mathrm{d}x+\cos\left(\frac{3}{2}\right)\underset{0}{\underbrace{\int^{\frac{\pi}{2}}_{0}\sin\left(\frac{1}{2}\cos\left(2x\right)\right)\mathrm{d}x}}\\
&=\frac{\pi}{2}\sin\left(\frac{3}{2}\right)J_0\left(\frac{1}{2}\right)
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions ? because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there is any quick way to find the decomposition of this rational function ?
| Rewriting it,
$$\left(\frac{1}{x(x-2)}\right)^3$$
Using Partial fractions,
$$\left(\frac{1}{x(x-2)}\right)^3 = \left(-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x-2}\right)\right)^3$$
Applying $(a-b)^3 = a^3-b^3+3a^2b-3ab^2$
$$\frac{1}{x^3(x-2)^3} = -\frac{1}{8}\left(\frac {1}{x^3}-\frac{1}{(x-2)^3}+\frac{3}{x^2(x-2)}-\frac{3}{x(x-2)^2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
| $$(x+y)^4=1 \\
x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$
Now, by AM-GM we have
$$x^3y \leq \frac{x^4+x^4+x^4+y^4}{4}\\
x^2y^2 \leq \frac{x^4+y^4}{2}\\
xy^3 \leq \frac{x^4+y^4+y^4+y^4}{4}\\
$$
Thus
$$1=x^4+4x^3y+6x^2y^2+4x^3+y^4 \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.
| One starts with
$$\begin{align}
\log(1+x)=&x-{x^2\over 2}+{x^3\over 3}+o(x^3)\\
\sin{X}=&X-{X^3\over 6}+o(X^3)
\end{align}$$
Now make $X=\log(1+x)$ in the development of the sine function to get
$$\begin{align}\sin(\log(1+x))=&x-{x^2\over 2}+{x^3\over 3}-{\left(x-{x^2\over 2}+{x^3\over 3}\right)^3\over 6}+o(x^4)\\
=&x-{x^2\over 2}+{x^3\over 6}+o(x^3)\end{align}$$
Now every Taylor series around $0$ is of the form $f(0)+f'(0)x+f''(0)x^2/2+f'''(0)x^3/6+o(x^3)$ and so $f'''(0)=1$. This is for the sine
Now if we raise it to the cube and do the same type of calculations, noting that we can limit the series of sine and logarithms to the first order, we get
$$\sin^3(\log(1+x))=x^3+o(x)$$
And this means $f'''(0)=6$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiability: What if Left-hand and Right-hand Limit are Equal in $x$ but differ from $f(x)$? Earlier today I asked this question: Derivative of Function with Cases: $f(x)=x^2\sin x^{-1}$ for $x\ne0$
Following the answers to my question I worked on my problem. Among other things I showed that the left hand limit of the derivative of the function from the question is equal to the left hand derivative. But that lead me to a new question:
A function is differentiable in $x$ iff $f'_+(x) = f'_-(x)$. But also a function isn't differentiable in $x$ if it is not continuous in $x$. What if both things are true?
The function $$f(x) = \begin{cases}x^2\sin x^{-1} & \text{if } x \neq 0\\ 10 & \text{else }\end{cases}$$ for instance. Here the right hand derivative is equal to the left hand derivative:
from the left:
\begin{align*}
\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} & = \lim_{x \to 0^-} \frac{x^2 \sin \left( \frac{1}{x} \right) - {0^2 \sin \left( \frac{1}{0} \right)}}{x - 0}\\
& = \lim_{x \to 0^-} \frac{{x} \left(x\sin \left( \frac{1}{x} \right) \right)}{{x}}\\
& = \lim_{x \to 0^-} \frac{{x\sin \left( \frac{1}{x} \right)}}{1}\\
& = 0
\end{align*}
from the right:
\begin{align*}
\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} & = \lim_{x \to 0^+} \frac{x^2 \sin \left( \frac{1}{x} \right) - {0^2 \sin \left( \frac{1}{0} \right)}}{x - 0}\\
& = \lim_{x \to 0^+} \frac{{x} \left(x\sin \left( \frac{1}{x} \right) \right)}{{x}}\\
& = \lim_{x \to 0^+} \frac{{x\sin \left( \frac{1}{x} \right)}}{1}\\
& = 0
\end{align*}
So, as the left and right hand limits are the same $f$ must be differentiable in $0$, right? But it is obviously not continuous at $0$, as $f(0) = 10$ but $\lim_{x \to 0} f(x) = 0$, so then it must not be differentiable... What am I missing here? Is the left-right-hand limit equality just a necessary rather than a sufficient precondition for differentiability?
| 1.There is no real number $\frac {1}{0}.$ Any computations using it are false or meaningless.
*If $f(x)=x^2 \sin 1/x$ for $x>0$ and $f(x)=10$ for $x\leq 0$ then for $x>0$ we have $$\frac {f(x)-f(0)}{x-0}=\frac {x^2\sin 1/x -10}{x}=x(\sin 1/x) -10/x $$which has no limit as $x\to 0.$ So $f$ has no "upper" derivative at $0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternative form of Eisenstein integers I just recently got into number theory and algebra so my knowledge is very limited. I stumbled upon the Eisenstein integers $\mathbb{Z}[\omega]$. They are of the form $\varepsilon = a + b \omega$, where $\omega = e^{2 \pi i/3} = -1/2 + i \sqrt{3}/2$.
Now my problem:
I want to show that you can write every Eisenstein integer in the form of $\varepsilon = \frac{a + bi \sqrt{3}}{2}$ where $a \equiv b \bmod 2.$
For now I got: $\omega = -1/2 + i \sqrt{3}/2 \Rightarrow \varepsilon = a + b \omega = a - b/2 + i \sqrt{3}b/2$ then $a = b + 2k \Rightarrow b = a - 2k \Rightarrow \varepsilon = a - a/2 + k + i \sqrt{3}b/2 = \frac{a + bi\sqrt{3}}{2} + k$.
Now I still have to deal with the superfluous $k \in \mathbb{Z}$. How is the argumentation to get rid of it?
| Your end number is $$\frac{(a+2k)+bi\sqrt{3}}{2}$$ which is of the right form.
Another way to argue is that if $x+y\sqrt{-3}$ is an integer where $x,y\in \mathbb{Q}$ then $x-y\sqrt{-3}$ is also an integer and thus
$2x$ and $x^2+3y^2$ are integers. This implies that $3(2y)^2$ is an integer and thus $2y$ is an integer since $\sqrt{3}$ is irrational.
So we have $x=\frac{a}{2}$ and $y=\frac{b}{2}$ and since $4|a^2+3b^2$ we see that $a$ and $b$ have the same parity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $a_{n+2}=3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$ is an integer Given the sequence
$$a_1=a_2=1;\ a_ {n+2} = 3a_n + 2\sqrt{2a_n^2 + 2a_{n+1}^2}$$
prove that $a_n$ is an integer for all $n\in\mathbb N$.
Attempt
It is enough to show that $2a_n^2 + 2a_{n+1}^2$ is a perfect square. That means it's an even perfect square and so divisible by 4; thus $a_n^2+a_{n+1}^2=2k^2$ for some integer $k$. I think can solve this diophantine equation but I can't relate the solution to the original problem. Can anyone help?
| We have that:
$$a_{n+2}-3a_n=2\sqrt{2a_n^2+2{a_{n+1}}^2}\Rightarrow (a_{n+2}-3a_n)^2=(2\sqrt{2a_n^2+2{a_{n+1}}^2})^2\Rightarrow\\ {a_{n+2}}^2-6a_na_{n+2}+9a_n^2=8a_n^2+8{a_{n+1}}^2\Rightarrow {a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_na_{n+2}+9{a_{n+1}}^2\Rightarrow \\{a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_n(3a_n+2\sqrt{2a_n^2+2{a_{n+1}}^2})+9{a_{n+1}}^2\Rightarrow\\ {a_{n+2}}^2+{a_{n+1}}^2=17a_n^2+9{a_{n+1}}^2+12a_n\sqrt{2a_n^2+2{a_{n+1}}^2}\Rightarrow\\ 2({a_{n+2}}^2+{a_{n+1}}^2)={(4a_n)}^2+\left(3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)^2+2\cdot (4a_n)\cdot \left(3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)\Rightarrow\\ 2({a_{n+2}}^2+{a_{n+1}}^2)=\left(4a_n+3\sqrt{2a_n^2+2{a_{n+1}}^2}\right)^2\ (1)$$
Now we will prove the claim with induction.
It is a matter of a few calculations to verify that $a_3$ and $a_4$ are integers.
Now we suppose that $a_n$ is an integer for every $4\leq n\leq k$. Then we immediately get that $\sqrt{2{a_{k-2}}^2+2{a_{k-1}}^2}$ is an integer as the difference of the integers $a_k$ and $3a_{k-2}$. From $(1)$ it follows that $\sqrt{2{a_{k-1}}^2+2{a_k}^2}$ is an integer too.
Hence $a_{k+1}=3a_{k-1}+\sqrt{2{a_{k-1}}^2+2{a_k}^2}\in \mathbb Z$ as a sum of integers and the inductive proof is complete. Thus we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?
How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?
My Working:
What I can think of is subtracting the no. of subsets that don't contain a single odd number from the total number of subsets as if we calculate it for single cases (like $1$ odd integer, $2$ odd integers, $\ldots$) it would be pretty long.
As there needs to be no odd integer, the maximum number of elements in the set is $5$ (only $5$ even integers are there in the superset).
Case 1: $0$ elements: $1$ set
Case 2: $1$ element: $5$ sets ($1$ even integer in each set)
Case 3: $2$ elements: $(5)(5)$ sets ($1$ element odd and $1$ even) $+\binom{5}{2}$ sets (both elements even)
which gives $35$ sets
Case 4: $3$ elements: $\cdots$
Problem:
This is getting complicated and I'm pretty sure I'll mess up if I proceed further. Is there any other way of solving this question?
| Others have already explained the easy solution, here is an alternative more similar to what you tried.
We want to know how many subsets contain exactly $k$ odd integers, from $k = 1$ to $5$, and sum.
*
*$k = 1$: $\binom{5}{1} = 5$ possible subsets of $\{1,3,5,7,9\}$
*$k = 2$: $\binom{5}{2} = 10$ possible subsets of $\{1,3,5,7,9\}$
*$k = 3$: $\binom{5}{3} = 10$ possible subsets of $\{1,3,5,7,9\}$
*$k = 4$: $\binom{5}{4} = 5$ possible subsets of $\{1,3,5,7,9\}$
*$k = 5$: $\binom{5}{5} = 1$ possible subsets of $\{1,3,5,7,9\}$
In each case, we can add some even integers, so we multiply by $2^5 = 32$.
Then,
$2^5 \sum_{k=1}^5 \binom{5}{k} = 32 (5+10+10+5+1) = 992$
But effectively this can be simpler:
$2^5 \sum_{k=1}^5 \binom{5}{k} = 2^5 \left( \sum_{k=0}^5 \binom{5}{k} - \binom{5}{0} \right) = 2^5 \left( 2^5 - 1 \right) = 2^{10} - 2^5 = 992$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder when $140^{67}+153^{51}$ is divided by $17$ Find the remainder when $$140^{67}+153^{51}$$ is divided by $17$.
$$140\equiv 4 \pmod {17}$$
$$67\equiv 16 \pmod{17}$$
$$153 \equiv 0 \pmod{17}$$
$$51\equiv 0 \pmod{17}$$
$$\Rightarrow 140^{67}+153^{51}\equiv 4^{16}+0 \equiv 1\pmod{17}$$
Solution should be $13$. What's wrong?
| One has $140\equiv 4\pmod{17}$. Now $4^4=256\equiv 1\pmod{17}$ and $67=16\cdot 4+3$ so
$$140^{67}\equiv (4^4)^{16}\cdot 4^3\equiv 4^3\equiv13 \pmod{17}$$
Similarly one has $153\equiv 0\pmod{17}$ and so $153^{51}\equiv 0\pmod{17}$.
So the answer is indeed $13$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, but I have no idea how to go about that)
$$\frac{(10x^4-18x^3-94x^2-8x+2)} {(10x+2)}$$
$$\frac{10x^4}{10x+2} + \frac{-18x^3}{10x+2} + \frac{-94x^2}{10x+2} + \frac{-8x}{10x+2} + \frac{2}{10x+2}$$
Simplifying some more (multiplying by $10x+2$)
$$10x^4-18x^3-94x^2-8x-2 = 0$$
But what do I do from here? Do I factor? Do do something else?
If I do factor How do I go about that? If not, what do I do instead?
| Polynomial long division:
\begin{array}{c|cc}
& &\color{red}{x^3} & \color{green}{-2x^2}& \color{blue}{-9x}& \color{orange}{+1}& \\ \hline
10x+2 & 10x^4 & -18x^3 & -94x^2& -8x& +2 \\
& \color{red}{10x^4} &\color{red}{+2x^3}\\
& & -20x^3 & -94x^2& \vdots & \vdots\\
& & \color{green}{-20x^3} & \color{green}{-4x^2}\\
& & & -90x^2& -8x & \vdots\\
& & & \color{blue}{-90x^2}& \color{blue}{-18x} \\
& & & & 10x & +2\\\
& & & & \color{orange}{10x} & \color{orange}{+2}\
\end{array}
$\therefore (10x^4-18x^3-94x^2-8x+2) =(10x+2)(x^3-2x^2-9x+1)$
| {
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Triangle inequality in C Show that $\left|Im(2+ z^{c} -4z^2) \right| \leq 9.5$ When $ \left| z \right| \leq \frac {3}{2}$
$z^c$= compliment of z
$\left|Im(2+ z^{c} -4z^2) \right| \leq \left| 2+z^{c} - 4z^2 \right|$
I have tried to split it up directly i have tried to force complete the square i always get a weird value or a number bigger than 9.5.
EDIT: Why cant i write
$\left| Im(2+ z^{c} -4z^2) \right| = \left| Im(z^{c} -4z^2)\right| \leq \left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\left| \frac {z^{c}}{2} - 2z^2 \right| \leq 2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left| 2z^2 \right|)$
and $2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left| 2z^2 \right|)= 2^{\frac {1}{2}}(\frac {3}{4} + \frac {9}{2}) = \frac {21}{2*2^{\frac {1}{2}}} \leq 9.5$
| You're trying to prove the unprovable. Let $z=re^{it}$, such that
$$Im(2+ z^{c} -4z^2) =Im(z^c)-4Im(z^2)=-r\sin (t)-4r^2sin(2t).$$
For $r=3/2$, this expression is $\approx-10.076$ at $t\approx 0.814$.
| {
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Integral $\int\frac{dy}{(y^2 + ay + b)^{3/2}}$ Need the indefinite integral with $a,b$ constants. This is not a homework question. Need it for a physics problem.
| $∫\frac{dy}{(y2+ay+b)^{3/2}} = ∫\frac{dy}{((y + \frac{a}{2})^2+\frac{1}{4}(4b-a^2))^{3/2}} = |\frac{a}{2} + y = u , dy = du| = ∫\frac{du}{(u^2+\frac{1}{4}(4b-a^2))^{3/2}} = | t = \frac{2u}{\sqrt{4b - a^2}} , dt = \frac{2du}{\sqrt{4b-a^2}} = \frac{4}{4b-a^2}∫\frac{dt}{(t^2+1)^{3/2}}= | t = tan(z) , dt = \frac{dz}{cos^2(z)}| = \frac{4}{4b-a^2}∫cos(z)dz = \frac{4sin(z)}{4b-a^2} + C= \frac{4sin(arctan(t))}{4b-a^2}+ C=-\frac{4t}{(a^2-4b)\sqrt{t^2+1}} + C= \frac{8u}{(4b-a^2)^{3/2}\sqrt{\frac{a^2-4b-4u^2}{a^2-4b}}} + C= \frac{2(a+2y)}{(4b-a^2)\sqrt{y^2 + ay+b}} + C $
| {
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} |
Calculate the following indefinite integral Calculate: $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx.$$
I have no idea how to start it.
| Let $u = x^2-x+1$, we have $u' = 2x-1$. Notice
$$\begin{align}
2x^2+x+\frac12 &= 2(x^2-x+1) + \frac32 (2x-1) = 2u + \frac32 u'\\
6x^2-7x+\frac{13}{2} &= 6(x^2-x+1) - \frac12(2x-1) = 6u - \frac12 u'
\end{align}
$$
The integrand of the integral can be rewritten as
$$\begin{align}
& \frac{(2u + \frac32 u')\cos(2x) + (6u - \frac12u')\sin(2x)}{u^{3/2}}\\
= & \left(-\frac12 \frac{u'}{u^{3/2}}\right)(\sin(2x)-3\cos(2x))
+ \frac{1}{u^{1/2}}(\sin(2x)-3\cos(2x))'\\
= & \left(\frac{\sin(2x)-3\cos(2x)}{u^{1/2}}\right)'
\end{align}$$
This implies the indefinite integral equals to
$$\frac{\sin(2x)-3\cos(2x)}{\sqrt{x^2-x+1}} + \text{ constant }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?
I want to find limit of this problem by a very specific method . I am uploading the pic of that method and my attempt.
Please guide me as to how to take this method ahead.
| Starting where you left: for $n\geq 1$,
$$
x_n = \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{k}{n^3}} \tag{1}
$$
While this furiously looks like a Riemann sum, it is not one due to the $\frac{k}{n^3}$ in the denominator. But then, we can use the squeeze theorem: as $1\leq k\leq n$,
$$
\frac{1}{1+\frac{1}{n^2}}\cdot \underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
= \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^2}}
\leq x_n \leq \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{1}{n^3}}
= \frac{1}{1+\frac{1}{n^3}}\cdot\underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k}{n}}_{}
$$
and since $\lim_{n\to\infty}\frac{1}{1+\frac{1}{n^2}} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n^3}} = 1$, by the squeeze theorem the limit will be that of
$$
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{k}{n} = \int_0^1 f(x)dx
$$
for some very simple function $f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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