Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why the orbit is of dimension $12$? Let $SL_3$ acts on the variety consisting of all nilpotent $3$ by $3$ matrices over $\mathbb{C}$ by conjugation. Let $S_p$ be the orbit of the matrix $$ a=\left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right). $$ Why the orbit $S_p$ over $\mathbb{R}$ is of dimension $12$?
I think that $S_p$ consisting of all matrices $g a g^{-1} $, $g \in SL_3$. But how to show that $S_p$ over $\mathbb{R}$ is of dimension $12$? Thank you very much.
| By orbit-stabilizer, the dimension of the orbit should be the dimension of the group minus the dimension of the stabilizer, so we should compute the dimension of the stabilizer. Suppose
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}.$$
Multiplying the matrices, this tells us
$$\begin{pmatrix} 0 & a & b \\ 0 & d & e \\ 0 & g & h \end{pmatrix} = \begin{pmatrix} d & e & f \\ g & h & i \\ 0 & 0 & 0 \end{pmatrix}.$$
Therefore $d,g,h=0$ and $a=e=i$ and $b=f$. So a stabilizing matrix looks like
$$\begin{pmatrix}a & b & c \\ 0 & a & b \\ 0 & 0 & a\end{pmatrix}.$$
In order to be in ${\rm SL}_3$ it must satisfy $a^3=1$, so $a$ is a cube root of unity. This leaves two degrees of freedom ($b$ and $c$), so the stabilizer has complex dimension $2$. The group ${\rm SL}_3(\Bbb C)$ has complex dimension equal to $3^2-1=8$. Therefore the orbit has complex dimension $8-2=6$, which means it has real dimension equal to $2\cdot 6=12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the limit is zero. But it is supposed to be $1/3$. Where is my mistake?
| Use the fact
$$e^{\sin(x)} \sim_{x\sim 0} 1+x+x^2/2-x^4/8. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating a rational function. How to integrate $$\int_1^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt3x^2+1}{\rm d}x$$
The bottom is not factorizable hence no partial fractions. There seems no other way.
| $$\begin{align}\int_{1}^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt 3x^2+1}dx&=\int_{1}^{\infty}\frac{2x^3-1}{(x^3+1)^2+\sqrt 3x^2}dx\\&=\int_{1}^{\infty}\frac{(2x^3-1)/(x^2)}{\left((x^3+1)^2+\sqrt 3x^2\right)/(x^2)}dx\\&=\int_{1}^{\infty}\frac{2x-\frac{1}{x^2}}{\left(x^2+\frac 1x\right)^2+\sqrt 3}dx\\&=\int_{2}^{\infty}\frac{du}{u^2+\sqrt 3}\ \ \ \ (\text{set $x^2+\frac 1x=u$})\\&=\int_{\arctan\frac{2}{\sqrt[4]{3}}}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt[4]{3}}\ \ \ \ (\text{set $u=\sqrt[4]{3}\tan\theta$})\\&=\frac{1}{2\sqrt[4]{3}}\left(\pi-2\arctan\frac{2}{\sqrt[4]{3}}\right).\end{align}$$
| {
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Prove a relationship involving floor functions I am trying to prove that a particular expression is a lower bound for a very unusually-behaved function. The whole proof will be complete if I can just nail down the details of one technical lemma involving the floor function. I am completely confident that the lemma is true, but have had a devil of a time proving it.
The Lemma is as follows:
Let $n$ and $b$ be natural numbers, with $1 \leq b \leq n$. Suppose that there exists an integer $k$ such that
$$\frac{n + \sqrt{n^2 + b + 2}}{b+2} < k < \frac{n+\sqrt{n^2+b+1}}{b+1}$$
Then there exists an integer lying strictly between
$$ \left( \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor +1 \right)\sqrt{n^2+b+1}$$
and
$$ \left( \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor +1 \right)\sqrt{n^2+b+2}$$
(where $\lfloor x \rfloor$ denotes the floor function).
The hypothesis can easily be restated in the form:
Suppose $ \left\lfloor \frac{n+\sqrt{n^2+b+2}}{b+2} \right\rfloor < \left\lfloor \frac{n+\sqrt{n^2+b+1}}{b+1} \right\rfloor$
and it's clear that the two inequalities are closely related, but I have not been able to get from one to the other, and I would appreciate any assistance. If there's interest, I can provide more background to the problem.
Edited to add: In fact, I am now almost certain that the Lemma above is actually if-and-only-if, and would also be happy to have help with the proof of the converse.
| Fix $n,b$ with $1\le b\le n$.
Let $k\in\mathbb Z$. For $i\in\{1,2\}$ define
$$\tag1\delta_i:=k- \frac{n+\sqrt{n^2+b+i}}{b+i}.$$
Then $k$ is strictly between the bounds iff $\delta_1<0<\delta_2$.
Then for $i\in\{1,2\}$
$$\begin{align}k\sqrt{n^2+b+i}&=\left( \frac{n+\sqrt{n^2+b+i}}{b+i}+\delta_i\right)\sqrt{n^2+b+i}\\
&=\frac{n^2+b+i+n\sqrt{n^2+b+i}}{b+i}+\delta_i\sqrt{n^2+b+i}\\
&=1+n\cdot \frac{n+\sqrt{n^2+b+i}}{b+i}+\delta_i\sqrt{n^2+b+i}\\
&\tag2=1+nk+\delta_i(\sqrt{n^2+b+i}-n).\end{align}$$
Note that
$$n<\sqrt{n^2+b+i}\le\sqrt{n^2+n+2}\le\sqrt{n^2+2n+1}=n+1 $$
so that $$\tag30<\sqrt{n^2+b+i}-n\le 1$$
with equality on the right if and only if $n=b=1$ and $i=2$. In other words, we have
$$\tag{4}0<\sqrt{n^2+b+i}-n< 1\rlap{\qquad\text{if $n>1$}.}$$
Lemma. Let $k\in\mathbb Z$. Then $k$ is strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$ if and only if $1+nk$ is strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Proof.
Because of $(3)$ and $(2)$, the conclusion holds if and only if $\delta_1<0<\delta_2$. But that is equivalent to the premise. $_\square$
Proposition. Let $k=\left\lfloor\frac{n+\sqrt{n^2+b+2}}{b+2}+1\right\rfloor$. There exists an integer strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$
if and only if there exists an integer strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Proof.
Assume there exists an integer strictly between $\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$.
Then $k$ is certainly such an integer because it is the smallest integer strictly larger than $\frac{n+\sqrt{n^2+b+2}}{b+2}$.
Then by the lemma, $1+nk$ is strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
Conversely assume that $m$ is an integer strictly between $k\sqrt{n^2+b+1}$ and $k\sqrt{n^2+b+2}$.
For $n=b=1$ this means that $2\sqrt 3<m<4 $, which is impossible. Therefore $n>1$. Define $\delta_{1,2}$ per $(1)$.
Then by definition of $k$, we have $0<\delta_2\le 1$.
Then from $(2)$ and $(4)$
$$m<k\sqrt{n^2+b+2}=1+nk+\underbrace{\delta_2(\sqrt{n^2+b+2}-n)}_{\in(0,1)}$$
so that $m\le 1+nk$. Using $(2)$ again we get
$$ 1+nk\ge m>k\sqrt{n^2+b+1}=1+nk+\delta_1(\sqrt{n^2+b+1}- n)$$
so that $\delta_1(\sqrt{n^2+b+1}- n)<0$ and by $(3)$ ultimately $\delta_1<0$. But $\delta_1<0<\delta_2$ is equivalent to $k$ being strictly between
$\frac{n+\sqrt{n^2+b+2}}{b+2}$ and $\frac{n+\sqrt{n^2+b+1}}{b+1}$.$_\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation Proof Dealing With 3s Multiples So the problem is as follows:
Prove that if the sum of digits of a decimal $n$ is three's multiple, then n is three's multiple by direct proof. For example, $11234567$ is 3's multiple because $1+1+2+3+4+5+6+7=24$, and in fact, $11234567$ is $3744856 * 3$. Use $10^n \equiv 1 \pmod3$ if necessary.
I am having a lot of problems with this question seeing as it is a rather difficult problem. Could someone please help me solve this? Thank you!
| Maybe you could first prove that the sum of the digits subtracted from the number is a multiple of 9.
For example, a 4 digit decimal number such as $1234=1\times1000+2\times100+3\times10+4$
But $1000=999+1$, $100=99+1$ and $10=9+1$.
So $1234=1\times\left(999+1\right)+2\times\left(99+1\right)+3\times\left(9+1\right)+4$.
And finally, $1234=1\times999+2\times99+3\times9+\left(1+2+3+4\right)$, or $1234-\left(1+2+3+4\right)=999+2\times99+3\times9$.
I'm sure you can come up with the rest of the proof. All you have to do is show that every power of $10$ is $1$ more than a multiple of $9$. And remember, $9$ is a power of $3$ so the difference between the digital sum and the number is also a multiple of 3.
Okay - here is a proof that any power of 10 is 1 more than a multiple of 9, by induction:
$10^0=1=0\times9+1$
$10^1=10=9\times1+1$
$10^2=100=9\times11+1$
Clearly our hypothesis is true for $n=0,1,2$.
Assume it is true for some arbitrary $n$, so $10^n$ is $1$ more than a multiple of $9$.
More formally: $10^n=9\times m+1$, for some integer $m$.
Then $10^{n+1}=10^n\times10\\
=\left(9\times m + 1\right)\times10\\
=9\times10\times m+10\\
=9\times10\times m+9+1\\
=9\times\left(10\times m+1\right)+1$
We are given $m$ is an integer, so $10\times m$ is an integer and so is $10\times m+1$ (multiplication and addition are closed on the integers). Then $9\times\left(10\times m+1\right)$ is $9$ times an integer or in other words, a multiple of $9$.
So clearly, when $10^n$ is $1$ more than a multiple of $9$ so is $10^{n+1}$.
| {
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Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$ Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is:
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$
| $$
\begin{align}
\int_0^1\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=2\int_{1/2}^1\sqrt{\frac{1-x}{x}}\,\mathrm{d}x\tag{1}\\
&=4\int_{1/\sqrt2}^1\sqrt{1-x^2}\,\mathrm{d}x\tag{2}\\
&=4\int_0^{\pi/4}\sin^2(x)\,\mathrm{d}x\tag{3}\\
&=2\int_0^{\pi/4}1-\cos(2x)\,\mathrm{d}x\tag{4}\\
&=\left[2x-\sin(2x)\vphantom{\int}\right]_0^{\pi/4}\tag{5}\\[6pt]
&=\frac\pi2-1\tag{6}
\end{align}
$$
Explanation:
$(1)$: $x\mapsto2x-1$
$(2)$: $x\mapsto x^2$
$(3)$: $x\mapsto\cos(x)$
$(4)$: $\cos(2x)=1-2\sin^2(x)$
$(5)$: integrate
$(6)$: evaluate
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this integral by a simple way? I'm given $$\int \frac{x^3}{\sqrt{x^4+x^2+1}}dx$$
My attempt,
Let $u=x^2$, $du=2xdx$
$$=\frac{1}{2}\int \frac{u}{\sqrt{u^2+u+1}}du = \frac{1}{2}\int \frac{u}{\sqrt{(u+\frac{1}{2})^2+\frac{3}{4}}}du$$
Let $s=u+\frac{1}{2}$, $ds=du$
$$=\frac{1}{2}\int\frac{s-\frac{1}{2}}{\sqrt{s^2+\frac{3}{4}}} ds$$
Let $s=\frac{\sqrt{3}}{2}\tan p$, $ds=\frac{\sqrt{3}}{2}\sec^2 p\,dp$
So, $$\sqrt{s^2+\frac{3}{4}}=\sqrt{\frac{3\tan^2 p}{4}+\frac{3}{4}} =\frac{\sqrt{3}}{2}\sec p$$
and
$$p=\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)$$
$$=\frac{\sqrt{3}}{4}\int \frac{2\left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p}{\sqrt{3}}dp$$
$$=\frac{1}{2}\int \left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p\,dp$$
$$=\frac{\sqrt{3}}{4}\int \tan p\sec p\,dp-\frac{1}{4}\int \sec p\,dp$$
Let $w=\sec p$, $dw=\tan p\sec p\,dp$
$$=\frac{\sqrt{3}}{4}\int 1 dw-\frac{1}{4}\int \sec p\,dp$$
$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\int \sec p\,dp$$
$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\ln (\tan p+\sec p)+c$$
$$=\frac{1}{4}\sqrt{3}\sec p-\frac{1}{4}\ln (\tan p+\sec p)+c$$
$$=\frac{1}{4}\sqrt{3}\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)-\frac{1}{4}\ln \left[\tan\left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)+\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)\right]+c$$
$$=\frac{1}{4}\sqrt{4s^2+3}+\frac{1}{8}\left[\ln 3 -2\ln \left(\sqrt{4s^2+3}+2s\right)\right]+c$$
$$=\frac{1}{2}\sqrt{u^2+u+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2\sqrt{u^2+u+1}+2u+1\right)\right]+c$$
$$=\frac{1}{2}\sqrt{x^4+x^2+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)\right]+c$$
$$=\frac{1}{8}\left[4\sqrt{x^4+x^2+1}-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)+\ln 3\right]+c$$
| Substituting $\frac{2x^2+1}{\sqrt{3}}=u\implies x=\frac{\sqrt{\sqrt{3}u-1}}{\sqrt{2}}$, the integral becomes:
$$\begin{align}
\int\frac{x^3}{\sqrt{x^4+x^2+1}}\,\mathrm{d}x
&=\int\frac{\left(\sqrt{3}u-1\right)^{3/2}}{\sqrt{6}\sqrt{u^2+1}}\cdot\frac{\sqrt{6}\,\mathrm{d}u}{4\sqrt{\sqrt{3}u-1}}\\
&=\frac14\int\frac{\sqrt{3}u-1}{\sqrt{u^2+1}}\,\mathrm{d}u\\
&=\frac{\sqrt{3}}{8}\int\frac{2u\,\mathrm{d}u}{\sqrt{u^2+1}}-\frac14\int\frac{\mathrm{d}u}{\sqrt{u^2+1}}\\
&=...\\
\end{align}$$
This seems to me to be the most direct route to the antiderivative.
| {
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Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
&=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
\end{align}$
$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\
&=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\
\end{align}$
$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\
&=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\
\end{align}$
is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$
I have studied maths upto $12$th grade.
| This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic behaviour of sequences Could anyone explain in details how these approximations as $n \to \infty$ are found? ($a$ is a positive real number)
*
*${x_n} = \frac{1}{n}\left( {\frac{a}{3} - \frac{3}{2}} \right) + O\left( {\frac{1}{{{n^3}}}} \right)$ for ${x_n} = \sqrt[3]{{{n^3} + an}} - \sqrt {{n^2} + 3} $
*${y_n} = \frac{{{{( - 1)}^n}}}{{{n^{\frac{a}{2}}}}} - \frac{1}{{2{n^{\frac{{3a}}{2}}}}} + o\left( {\frac{1}{{{n^{\frac{{3a}}{2}}}}}} \right)$ for ${y_n} = \frac{{{{( - 1)}^n}}}{{\sqrt {{n^a} + {{( - 1)}^n}} }}$
The idea is to study the behaviour of a series whose $n-$th term is given by the above expressions in terms of the positive real $a$ and these asymptotic developments which I don't understand are part of the solution.
| We have $\sqrt[3]{n^3+an}=n\sqrt[3]{1+\frac{a}{n^2}}$. We expand the latter using Newton's Binomial Theorem to get $$\sqrt[3]{n^3+an}=n\sum_{k\ge 0} {1/3 \choose k} \left(\frac{a}{n^2}\right)^k=n(1+\frac{a}{3n^2}-\frac{a^2}{9n^4}+O(n^{-6}))$$
Repeating, we have $\sqrt{n^2+3}=n\sqrt{1+\frac{3}{n^2}}$, so
$$\sqrt{n^2+3}=n\sum_{k\ge 0}{1/2 \choose k}\left(\frac{3}{n^2}\right)^k=n(1+\frac{3}{2n^2}-\frac{3^2}{8n^4}+O(n^{-6}))$$
Subtracting, we get $$\sqrt[3]{n^3+an}-\sqrt{n^2+3}=n\left(\frac{a}{3n^2}-\frac{3}{2n^2}-\frac{a^2}{9n^4}+\frac{9}{8n^4}+O(n^{-6})\right)$$
| {
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Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$ b)$\frac{\pi}{6}\quad$ c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$
| Recall $1+\tan^2(x) = \sec^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$. Hence,
$$1+\tan^2(x) = 8 \sin^2(x) \implies \sec^2(x) = 8\sin^2(x) \implies 8\sin^2(x)\cos^2(x) = 1$$
This gives us
$$2\sin^2(2x) = 1 \implies \sin(2x) = \pm \dfrac1{\sqrt2} \implies 2x = \dfrac{n\pi}2 + \dfrac{\pi}4 \implies x = \dfrac{n\pi}4 + \dfrac{\pi}8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $.
What is the integral of: $\int \frac{1}{5+3\sin x}dx$
My attempt:
Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have:
$\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $
I'll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So:
$2\int \frac 1 {5t^2+6t+5}dt = \frac 2 5 \int \frac 1{(t+\frac 3 5)^2+0.64}dt=\frac 2 5(\frac 5 4\arctan((t+\frac 3 5)\frac 5 4))=\frac 1 2 \arctan(\frac{5t+3}{4}) $
But if I'll place $\tan \frac x 2=t$ I won't be able to simplify it further and since the online calculator's answers don't have $\tan \frac x 2$ there, I believe I made a mistake. What is wrong with what I did and is there a better way to do it?
| Taking from
$5t^2$+6t+5, instead of expanding, complete the square.
=5(t+$\frac{6}{10})^2$-$\frac{16}{5}$
Rewrite in the form
=$\frac{-1}{5}$$\int$$\frac{1}{16/25-(t+6/10)^2}$
=$\frac{-1}{5}$[$\frac{5}{4}$$tanh^{-1}$($\frac{t+6/10}{4/5})]$+C
=$\frac{-1}{4}$[$tanh^{-1}$($\frac{5t}{4}$+$\frac{3}{2}$)]+C
Still getting answer in terms of tan(x/2).
| {
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Evaluation of $\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx$
Evaluate the integral
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$
My Attempt:
Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to
$$
-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt
$$
Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to
$$
\begin{align}
-\frac{1}{4}\int t^{-3/4}\,dt &= -u^{1/4}+\mathcal{C}\\
&= -\left(1+t^4\right)^{1/4}+\mathcal{C}
\end{align}
$$
So we arrive at the solution
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$
Question: Is there any other method for solving this problem?
| Alternate Solution:
$$
\begin{align}
\int\frac{1}{x^2\cdot \left(x^4+1\right)^{3/4}}\,dx &= \int\frac{1}{x^2\cdot x^3\cdot \left(1+x^{-4}\right)^{3/4}}\,dx\\
&= \int\frac{1}{x^{5}\cdot \left(1+x^{-4}\right)^{3/4}}\,dx
\end{align}
$$
Now Let $(1+x^{-4}) = t$. Then $x^{-5}\,dx = -\frac{1}{4}\,dt$, so we have
$$
-\frac{1}{4}\int \frac{1}{t^{3/4}}dt = -\frac{1}{4}\cdot 4\cdot t^{1/4}+\mathcal{C} = -\left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C}
$$
| {
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In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has the form $2^a \cdot 5^b$, and a decomposition of 1000000 into a product of five factors has the form
$$
1000000 = (2^{a_1} \cdot 5^{b_1})(2^{a_2} \cdot 5^{b_2})(2^{a_3} \cdot 5^{b_3})(2^{a_4} \cdot 5^{b_4})(2^{a_5} \cdot 5^{b_5})
$$
where $a_i$ and $b_i$ are nonnegative integers which satisfy the
conditions
$$
a_1 + a_2 + a_3 + a_4 + a_5 = 6, b_1 + b_2 + b_3 + b_4 + b_5 = 6
$$
The total number of systems of $a_i$ which satisfy the first equation is $210$ and the same number is for $b_i$. Thus the total number of decompositions is $210 \cdot 210 = 44100$.
However, in this enumeration, factorizations which differ only in the brder of the factors have been counted separately; that is, some factorizations are counted several times each.
To get the number of distinct unordered decompositions I must, at first, substract the number of ordered decompositions with at least two identical factors and, at second, divide resulting number by $5!$ to leave only unordered ones.
And I'm stuck at the step of counting the number of ordered decompositions with at least two identical factors.
The number of decompositions with $k$ identical factors is $(\lfloor\dfrac{a}{k}\rfloor + 1)(\lfloor\dfrac{b}{k}\rfloor+1){5 \choose k}$, that is, the number of decompositions with two identical factors is $16 \cdot {5 \choose 2} = 160$, three identical factors - $9 \cdot {5 \choose 3} = 90$, four ones - $4 \cdot {5 \choose 4} = 20$ and five ones - $4 \cdot {5 \choose 5} = 4$. Thus the number of distinct decompositions must be equal $\dfrac{44100-160-90-20-4}{5!}$, but this number is not integral. I suppose that the numbers of decompositions with $k$ identical factors overlap and I misuse inclusion-exclusion principle. But I have no idea how I can count that overlapped decompositions.
Please, help!
Thanks!
| There are 5 types of possibilities for the unordered exponents $a_1,\cdots,a_5$ for 2:
$\textbf{1)}$ $6+0+0+0+0,\;\; 2+1+1+1+1$ $\;\;$(4 of a kind)
$\textbf{2)}$ $5+1+0+0+0,\;\; 3+0+1+1+1,\;\;4+2+0+0+0$ $\;\;$(3 of a kind)
$\textbf{3)}$ $4+1+1+0+0,\;\; 0+2+2+1+1$ $\;\;$(2 pairs)
$\textbf{4)}$ $3+3+0+0+0,\;\; 0+0+2+2+2$ $\;\;$(full house)
$\textbf{5)}$ $3+2+1+0+0$ $\;\;$ (1 pair)
Since we have the same 5 possibilities for the unordered exponents $b_1,\cdots,b_5$ for 5,
we can count the number of ways to pair up these types to get distinct divisors, using the fact that digits matched to repeated digits must be distinct, and their order doesn't matter:
$\textbf{1)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 1): $\;\;\;2(2\cdot1)(1)={\color{red}4}$ ways
$\textbf{2)}$ and 2): $\hspace{1.03 in}\;\;\;(3\cdot3)(1)={\color{red}9}$ ways
$\textbf{2)}$ and 3) $\;$and$\;$ $\textbf{3)}$ and 2): $\;\;\;2(3\cdot2)(2)=\color{red}{24}$ ways
$\textbf{2)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 2): $\;\;\;2(3\cdot1)(1+2\cdot3)=\color{red}{42}$ ways
$\textbf{3)}$ and 3): $\hspace{1.03 in}\;\;\;(2\cdot2)(1+2\cdot2)=\color{red}{20}$ ways
$\textbf{3)}$ and 4) $\;$and$\;$ $\textbf{4)}$ and 3): $\;\;\;2(2\cdot2)(1)=\color{red}{8}$ ways
$\textbf{3)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 3): $\;\;\;2(2\cdot1)(2\cdot3+3\cdot2)=\color{red}{48}$ ways
$\textbf{4)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 4): $\;\;\;2(2\cdot1)(3)=\color{red}{12}$ ways
$\textbf{5)}$ and 5): $\hspace{1.03 in}\;\;\;3\cdot3+3\cdot3\cdot2=\color{red}{27}$ ways
This gives a total of $\color{red}{194}$ ways of expressing $10^6$ as a product of 5 distinct positive integers.
| {
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Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that
*
*$\color{red}{2\le} a\le b\le c$
*$\sqrt{abc}\in\mathbb N$
*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$
Examples : The followings are positive integers.
$$\frac{(2-1)(8-1)(49-1)}{\sqrt{2\cdot 8\cdot 49}},\ \frac{(6-1)(24-1)(529-1)}{\sqrt{6\cdot 24\cdot 529}},\frac{(7-1)(63-1)(3844-1)}{\sqrt{7\cdot 63\cdot 3844}}$$
Then, I began to try to find every such triplet. Then, I found
$$(a,b,c)=(k,km^2,(km^2-1)^2)$$
where $k,m$ are positive integers such that $k\ge 2$ and $km^2\ge 3$, so I knew that there are infinitely many such triplets. However, I can neither find the other triplets nor prove that there are no other triplets. So, here is my question.
Question : How can we find every such triplet $(a,b,c)$?
Added : There are other triplets : $(a,b,c)=(k,k,(k-1)^4)\ (k\ge 3)$ by a user user84413, $(6,24,25),(15,15,16)$ by a user Théophile. Also, from the first example by Théophile, I got $(2k,8k,(2k-1)^2)\ (k\ge 3)$.
Added : $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)\ (k\ge 2)$ found by a user coffeemath. From this example, I got $(k^2,(k+1)^2,(k-1)^2(k+2)^2)\ (k\ge 2)$.
Added : I got $(a,b,c)=(2(2k-1),32(2k-1),(4k-3)^2)\ (k\ge 5)$.
Added : I got $(a,b,c)=(k,(k-1)^2,k(k-2)^2)\ (k\ge 4)$.
Added : A squarefree triplet $(6,10,15)$ and $(4,k^2,(k+1)^2)\ (k\ge 2)$ found by a user martin.
Added : user52733 shows that $(6,10,15)$ is the only squarefree solution.
| (Too long for a comment.)
The two solutions,
$$a,\,b,\,c = k^2,\;(k+1)^2,\;(k+2)^2$$
$$a,\,b,\,c = 2^2,\;k^2,\;(k+1)^2$$
by users coffeemath and martin, respectively, are special cases of the more general solution,
$$a,\,b,\,c = k^2,\;(km\pm1)^2,\;(km\pm2)^2$$
where coffeemath's had $m=1$, while martin's had $k=2,\, m = \frac{n}{2}$.
| {
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Show that $\sin\left(\frac\pi3(x-2)\right)$ is equal to $\cos\left(\frac\pi3(x-7/2)\right)$ Show that
$$\sin\left(\frac\pi3(x-2)\right)$$
is equal to
$$\cos\left(\frac\pi3(x-7/2)\right)$$
I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.
| The two angles $\frac{\pi}{3}(x-2)$ and $\frac{\pi}{3}(x-7/2)$ differs by $\pi/2$.
In fact,
$$ \frac{\pi}{3}(x-2) - \frac{\pi}{3}(x-\frac{7}{2}) = \frac{\pi}{3}x- \frac{2}{3}\pi - \frac{\pi}{3}x + \frac{7}{6}\pi = - \frac{7-4}{6}\pi = -\frac{\pi}{2}\, , $$
thus $\sin(\frac{\pi}{3}(x-2)) = \sin(\frac{\pi}{3}(x-7/2) - \frac{\pi}{2}) = \cos(\frac{\pi}{3}(x-7/2))$.
| {
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limit of sequnce in real numbers For every tow real numbers $a$ and $b$,with this condition that $0<a<b$ ,define sequence ${x_{n}}$ to the following:
$x_{1}=a$ $x_{2}=b$
${x_{n}}=\frac{ x_{n-1}+ x_{n-2} }{2}$ (for $n>2$)
then which of the following options is limit of ${x_{n}}$ ?
1)b
2)$\frac{a+2b}{3}$
3)$\frac{a+b}{2}$
4)$\frac{3a+5b}{8}$
I think that option 1 is true because if we draw ${x_{n}}$ on the axiom of real numbers then limit of ${x_{n}}$ is b but it is not true.
| The characteristic equation for the recurrence is
$$y^2 - \dfrac{y}2 - \dfrac12 = 0 \implies (y-1)(y+1/2) = 0$$
This gives us
$$x_n = c_0 1^n + c_1 \left(-\dfrac12\right)^n$$
Hence, note that $\lim_{n \to \infty} x_n = c_0$. Setting $n=1$, we have $c_0 - \dfrac{c_1}2 = a$ and setting $n=2$, we have $c_0 + \dfrac{c_1}4 = b$. This gives us $c_0 = \dfrac{a+2b}3$.
Hence, we have that
$$\color{blue}{\boxed{\lim_{n \to \infty} x_n = \dfrac{a+2b}3}}$$
EDIT:
It cannot converge to $b$ because we have $x_3 = \dfrac{a+b}2$ and $x_4 = \dfrac{a+3b}4$. This means any further $x_n$ has to be in the interval $\left[\dfrac{a+b}2,\dfrac{a+3b}4\right]$ and clearly since $a<b$, we have $$a < \dfrac{a+b}2 < \dfrac{a+3b}4 < b$$
| {
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Show that there is a large value
Suppose not all 4 integers, $a, b, c, d$ are equal. Start with $(a, b, c, d)$ and repeatedly replace $(a, b, c, d)$ by $(a - b, b - c, c - d, d - a)$. Then show that at least one number of the quadruple will become arbitrarily large.
Let $x_k$ be the pair with:
$x_k = (a_k, b_k, c_k, d_k)$ and $x_{k+1} = (a_k - b_k, b_k - c_k, c_k - d_k, d_k - a_k)$ then it follows that,
$a_k - b_k + b_k - c_k + c_k - d_k + d_k - a_k = 0 = L$
But I cant show any further. The idea is to find an invariant. I did find one but I can't do anything with it.
Actually using
$x_k = (a_k, b_k, c_k, d_k)$
It is seen that,
$a_k = a_{k-1} - b_{k-1}$, then
$a_k + b_k + c_k + d_k = a_{k-1} - b_{k-1} + b_{k-1} - c_{k-1} + ... = 0$
So I have found the invariant.
$(a_k + b_k + c_k + d_k)^2 = a_k^2 + 2a_kb_k + 2a_kc_k + 2a_kd_k + b_k^2 + 2b_kc_k + 2b_kd_k + c_k^2 + 2c_kd_k + d_k^2 $
| This is problem $E5$ under The invariance principle in Arthur Engel's book titled "Problem Solving Strtegies"
We have
$$(a_{n+1},b_{n+1},c_{n+1},d_{n+1}) = (a_n-b_n,b_n-c_n,c_n-d_n,d_n-a_n)$$
As you have rightly observed, the invariant is $a_n+b_n+c_n+d_n = 0$ for all $n \geq 1$, where $a_0 = a$, $b_0=b$, $c_0 = c$ and $d_0 = d$. Let us now look at $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$. We have
\begin{align}
a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 & = (a_n-b_n)^2+(b_n-c_n)^2+(c_n-d_n)^2 + (d_n-a_n)^2\\
& = 2(a_n^2+b_n^2+c_n^2+d_n^2) - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n) & (\spadesuit)
\end{align}
We shall now show that $ - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n)$ is non-negative for $n \geq 1$. Since $a_n+b_n+c_n+d_n=0$, we have
\begin{align}
0 & = (a_n+b_n+c_n+d_n)^2 = (a_n+c_n)^2 + (b_n+c_n)^2 + 2(a_n+c_n)(b_n+d_n)\\
& = (a_n+c_n)^2 + (b_n+c_n)^2 + 2a_nb_n + 2b_nc_n + 2c_nd_n + 2d_na_n
\end{align}
This gives us
$$-2(a_nb_n+b_nc_n+c_nd_n+d_na_n) = (a_n+c_n)^2 + (b_n+c_n)^2 \geq 0 \,\,\, (\clubsuit)$$
Making use of $(\clubsuit)$ in $(\spadesuit)$, we obtain that
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2(a_n^2+b_n^2+c_n^2+d_n^2)$$
This gives us
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2^n(a_1^2+b_1^2+c_1^2+d_1^2)$$
Since $a, b,c,d$ are distinct, one of the terms $a_1,b_1,c_1$ or $d_1$ is non-zero. Hence, $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$ grows unbounded, which means the largest of the terms must keep growing unbounded.
| {
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| Notice that $(n^2 - 1)(n^2 + 1) = n^4 - 1$. Fermat's little theorem tells us that $n^4 \equiv 1 \pmod 5$. This means that $n^4 - 1$ is a multiple of $5$ if $n$ is not, and therefore $n^2 (n^4 - 1)$ is also a multiple of $5$. For example, if $n = 2$, then $n^4 - 1 = 15$ and $n^2 (n^2 - 1)(n^2 + 1) = 60$.
This leaves us the case where $n$ is a multiple of $5$ to consider. Obviously $n^4 - 1$ is not a multiple of $5$. But $n^2$ is. For example, $n = 5$ gives us $15600$.
| {
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Different Law of Cosines using Sine instead: $c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$ Playing around with Trig and the Law of Cosines (LoC), I came up with this formula given a triangle with sides $a$, $b$, $c$ where we are given $a$, $b$ and angle $\theta$ between them:
$$c^2 = a^2 + b^2 - 2ab\sqrt{1-\sin^2(\theta)}$$
Far from me the idea that I could've stumbled onto something no one's ever derived before, but I've never seen this formula and was just curious whether it has a name or is never considered because it offers no advantage over the LoC (needing the same amount of initial information) and is slightly more complicated.
Also, is my proof correct?
Here's my work; here I use $C$ for the angle:
$$c^2 = x^2 + h^2$$
$$h = a \sin(C)$$
$$h^2 = a^2 \sin^2(C)$$
$$x = b - (b-x)$$
$$(b-x) = \sqrt{a^2 - h^2} = \sqrt{a^2 - a^2 \sin^2(C)} = \sqrt{a^2 (1-\sin^2(C))}$$
$$x = b-a\sqrt{1-\sin^2(C)}$$
$$x^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C))$$
Therefore:
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C)) + a^2 \sin^2(C)$$
$$c^2 = b^2 - 2ab \sqrt{1-\sin^2(C)} + a^2 (1-\sin^2(C) + \sin^2(C))$$
$$c^2 = a^2 + b^2 - 2ab \sqrt{1-\sin^2(C)}$$
| Really this just boils down to the identity $$\cos^2{x} = 1 - \sin^2{x}$$
So no, your formula is really no different from the Law of cosines. Your proof looks fine however.
Having believed I found a new trigonometric law once (and being incorrect), my advice would be to always exhaust your result for any known identities to see if you simply have another representation of a known law or expression. Resources like Wolfram Alpha can help for easier cases.
| {
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Deriving a contradiction How can I derive a contradiction from the following nasty statement:
Assume $\sqrt{5} = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8},$ with $a,b,c,d \in \mathbb{Q}$?
This is the last piece of an effort to prove that the polynomial $x^4-2$ is irreducible over $\mathbb{Q}(\sqrt{5}).$
| Starting from
$$\tag1m\sqrt 5=a+b\alpha+c\alpha^2+d\alpha^3 $$
(where $\alpha^4=22$) with $a,b,c,d\in\mathbb Z$ you find
$$\begin{align}5m^2&=(a+b\alpha+c\alpha^2+d\alpha^3)^2\\&=(a^2+2c^2+4bd)+(2ab+4cd)\alpha+(b^2+2d^2+2ac)\alpha^2+(2ad+2bc)\alpha^3\end{align}$$
If you alredy know that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$, this implies
$$ \begin{align}a^2+2c^2+4bd-5m^2&=0\\
ab+2cd&=0\\
b^2+2d^2+2ac&=0\\
ad+bc&=0\end{align}$$
From the thirs we have immediatelythat $b$ must be even.
Assume $m$ is odd.
Then from the first, $a$ must be odd. Then the first reads $1+2c^2+0-5\equiv 0\pmod 8$, i.e., $c^2\equiv 2\pmod 4$, which is absurd.
Therefore $m$ is even. Then from the first $a$ must be even. Also from the first, $c$ must be even. And from the thord $d$ must be even.
Thus: For any solution $(m,a,b,c,d)\in\mathbb Z^5$ of $(1)$, all variables are even. But if there exists a nontrvial solution at all, then certainly there exists one with at least one variable odd, obtained by dividing out the larges common power of $2$. We conclude that $m=a=b=c=d=0$ is the only integer solution to $(1)$. Therefore, there is no rational solution for
$$\sqrt 5=a+b\alpha+c\alpha^2+d\alpha^3 $$
| {
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Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ Can we possibly compute the following integral in terms of known constants?
$$\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$$
Some progress was already done here http://integralsandseries.prophpbb.com/topic279.html but still we have a hypergeometric function. What's your thoughts on it?
UPDATE: The question was also posted on mathoverflow here https://mathoverflow.net/questions/206253/calculate-in-closed-form-int-01-int-01-fracdx-dy1-xy1-x1-y#
| An incomplete answer.
Tackling the inner integral with elementary methods,
\begin{align}
\int^1_0\frac{{\rm d}x}{1-ax+ax^2}
&=\int^1_0\frac{{\rm d}x}{a\left(x-\frac{1}{2}\right)^2+1-\frac{a}{4}}\\&=\int^\frac{1}{2}_{-\frac{1}{2}}\frac{{\rm d}x}{ax^2+1-\frac{a}{4}}\\
&=\frac{1}{\sqrt{a\left(1-\frac{a}{4}\right)}}\int^\frac{\sqrt{a}}{2\sqrt{(1-\frac{a}{4})}}_{-\frac{\sqrt{a}}{2\sqrt{(1-\frac{a}{4})}}}\frac{{\rm d}x}{1+x^2}\\&=\frac{4}{\sqrt{a(4-a)}}\arctan\left(\sqrt{\frac{a}{4-a}}\right)
\end{align}
Thus
\begin{align}
\iint\limits_{[0,1]^2}\frac{{\rm d}A}{1-xy(1-x)(1-y)}
&=4\int^1_0\arctan\left(\sqrt{\frac{x(1-x)}{x^2-x+4}}\right)\frac{{\rm d}x}{\sqrt{x(1-x)(x^2-x+4)}}\\
&=16\int^1_0\arctan\left(\sqrt{\frac{1-x^2}{15+x^2}}\right)\frac{{\rm d}x}{\sqrt{(1-x^2)(15+x^2)}}\\
&=\frac{16}{\sqrt{15}}\int^1_0\operatorname{F}\left(\arcsin{x}\right)\frac{x\ {\rm d}x}{\sqrt{(1-x^2)(15+x^2)}}\\
&=\frac{8}{15}\int^{\operatorname{K}}_{-\operatorname{K}}u\operatorname{sn}(u)\ {\rm d}u
\end{align}
where $\operatorname{sn}(u)$ is a Jacobi elliptic function with modulus $k=15^{-1/2}i$.
Let $R$ be the parallelogram with vertices $-\operatorname{K}-2i\operatorname{K}'$, $\operatorname{K}-2i\operatorname{K}'$, $\operatorname{K}+2i\operatorname{K}'$, $-\operatorname{K}+2i\operatorname{K}'$. Parameterising along the contour,
\begin{align}
\int_{\partial R}z^{\color{red}{2}}\operatorname{sn}(z)\ {\rm d}z
&=-8i\operatorname{K}\left(\frac{4}{\sqrt{15}}\right)\int^{\operatorname{K}}_{-\operatorname{K}}u\operatorname{sn}(u)\ {\rm d}u+2i\int^{2\operatorname{K}'}_{-2\operatorname{K}'}(\operatorname{K}^2-t^2)\operatorname{cd}(it)\ {\rm d}t\\
&=2\pi i\sum_{\pm}\operatorname*{Res}_{z=\pm i\operatorname{K}'}z^2\operatorname{sn}(z)=-4\pi\sqrt{15}\operatorname{K}\left(\frac{4}{\sqrt{15}}\right)^2
\end{align}
$\operatorname{cd}(t)$ has an anti-derivative, namely
$$2i\operatorname{K}^2\int^{2\operatorname{K}'}_{-2\operatorname{K}'}\operatorname{cd}(it)\ {\rm d}t=-4i\sqrt{15}\operatorname{K}^2\left.\ln\left(\operatorname{nd}(t)+\frac{i}{\sqrt{15}}\operatorname{sd}(t)\right)\right|^{2i\operatorname{K}'}_0$$
Otherwise, this seems to be as far as I can go for now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
} |
Numbers of the form $5 \cdot 2^{n}-1$ divisible by $3^k$ for large values of $k$ Let $n_k$ be the smallest integer such that $5 \cdot 2^{n_k}-1$ is divisble by $3^k$ where $k$ is a positive integer. Can one say something about the growth of $n_k$ with respect to $k$ ? Is it exponential ?
The first values are $n_k=1,1,13,31,139,\ldots$ for $k=1,2,3,4,5,\ldots$.
| Here is some progress I made ...
Looking closely at the values given by Lucian, we remark that $$n_k - n_{k-1} = 2 \cdot a_{k} \cdot 3^{k-2} \text{ where } a_{k}=0,1,2 \text{ for } k>1.$$
Example: $n_6 - n_5= 324 = 4 \cdot 3^4$.
My sketch of proof below:
First, by induction, we have
$$ 2^{2 \cdot 3^{k-2}} \equiv 1 + 3^{k-1} \pmod{3^k}.$$
Then, assuming that $$5 \cdot 2^{n_{k-1}} \not\equiv 1 \pmod{3^k}$$ we can write
$$5 \cdot 2^{n_{k-1}} \equiv 1 + i \cdot 3^{k-1} \pmod{3^k} \text{ where } i=1,2,$$ and we obtain
$$5 \cdot 2^{n_{k-1} + 2(3-i)3^{k-2}} \equiv (1 + i \cdot 3^{k-1})(1 + 3^{k-1})^{3-i} \equiv 1 \pmod{3^k}.$$
So we can take $a_{k} = 3-i$.
Finally we take $a_k = 0$ in the simple case $5 \cdot 2^{n_{k-1}} \equiv 1 \pmod{3^k}$.
$\square$
Thus we have the general formula
$$n_k = 1 + 2\sum_{j=2}^k a_{j} \cdot 3^{j-2}.$$
The sequence of $a_k$ values ($k\geq2$) is $$0, 2, 1, 2, 2, 0, 1, 2, 2, 0, 1, 1, 0, 0, 2, 0, 1, 0, 2, \ldots$$
and $n_{17} = n_{16} = n_{15} + 4 \cdot 3^{14} = 19641181$.
An interesting result is:
$$\text{if } a_k \neq 0 \text{ then } 2 \cdot 3^{k-2} < n_k < 2 \cdot 3^{k-1}.$$
But the problem remains if $a_k=0$ for many consecutive $k$ integers. And it is difficult to say much more, since $a_k$ sequence appears to be pseudorandom. Any idea ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Dynamical System transformation How can the system $$\frac{dx}{dt}=-y+\epsilon x(x^2+y^2)$$$$\frac{dy}{ dt}=x+\epsilon y(x^2+y^2)$$ be transformed into $$\frac{dr}{dt}=\epsilon r^3$$ $$\frac{d\theta}{dt}=1$$ via polar coordinates?
I sub in $x=r\cos(\theta)$ and $y=r\sin(\theta)$ into both of the original equations, but there are $\frac{dr}{dt}$ and $\frac{d\theta}{dt}$ terms in both equations then. And adding/subtracting the equations doesn't seem to produce anything 'nice'.
| The given equations are
$\dfrac{dx}{dt} = - y + \epsilon x (x^2 + y^2), \tag{0A}$
$\dfrac{dx}{dt} = x + \epsilon y (x^2 + y^2); \tag{0B}$
we have
$r^2 = x^2 + y^2; \tag{1}$
thus,
$2r\dfrac{dr}{dt} = 2x\dfrac{dx}{dt} + 2y \dfrac{dy}{dt}, \tag{2}$
whence
$r \dfrac{dr}{dt} = x\dfrac{dx}{dt} + y \dfrac{dy}{dt}; \tag{3}$
using (0A)-(0B) in (3) yields
$r \dfrac{dr}{dt} = x(-y + \epsilon x(x^2 + y^2)) + y(x + \epsilon y(x^2 + y^2))$
$= -xy + \epsilon x^2 (x^2 + y^2) + xy + \epsilon y^2 (x^2 + y^2) = \epsilon (x^2 + y^2)(x^2 + y^2)$
$= \epsilon r^2 r^2 = \epsilon r^4; \tag{4}$
assuming $r \ne 0$ it follows that
$\dfrac{dr}{dt} = \epsilon r^3. \tag{5}$
We further have
$x = r\cos \theta, \tag{6}$
$y = r\sin \theta; \tag{7}$
thus
$\dfrac{dx}{dt} = \dfrac{dr}{dt} \cos \theta - r\dfrac{d\theta}{dt} \sin \theta; \tag{8}$
$\dfrac{dy}{dt} = \dfrac{dr}{dt} \sin \theta + r\dfrac{d\theta}{dt} \cos \theta; \tag{9}$
using (0A), (1), (5), (6) and (7) in (8),
$-r \sin \theta + \epsilon r^3 \cos \theta = \dfrac{dr}{dt} \cos \theta - r\dfrac{d\theta}{dt} \sin \theta = \epsilon r^3 \cos \theta - r\dfrac{d\theta}{dt} \sin \theta, \tag{10}$
whence
$-r\sin \theta = - r\dfrac{d\theta}{dt} \sin \theta, \tag{11}$
and whence
$\sin \theta \ne 0 \Rightarrow \dfrac{d\theta}{dt} = 1; \tag{12}$
additionally, using (0B), (1), (5), (6), (7) in (9):
$r \cos \theta + \epsilon r^3 \sin \theta = \dfrac{dr}{dt} \sin \theta + r\dfrac{d\theta}{dt} \cos \theta = \epsilon r^3 \sin \theta + r\dfrac{d\theta}{dt} \cos \theta, \tag{13}$
or
$r\cos \theta = r\dfrac{d\theta}{dt} \cos \theta ; \tag{14}$
finally,
$\cos \theta \ne 0 \Rightarrow \dfrac{d \theta}{dt} = 1. \tag{15}$
Since $\cos \theta$, $\sin \theta$ cannot both vanish, we have shown that
$\dfrac{d \theta}{dt} = 1 \tag{16}$
in all cases. Together with (5), this completes the transformation to polars. Note that this derivation covers the cases $x = 0$, $y = 0$, in which the transformations
$\tan \theta = \dfrac{y}{x}; \;\; \cot \theta = \dfrac{x}{y} \tag{17}$
may become problematic.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Express a quadratic form as a sum of squares using Schur complements I was able to figure out the first part of this problem, but I have no concept of how it relates to Schur complements, so I'm not sure (no pun intended) how to proceed. The question is as follows:
Consider $2x^2 + 2xy + 2y^2 + z^2 + 2xz$. Write the symmetric matrix representing this quadratic form. Now, express this as a sum of squares by using this symmetric matrix and Schur complements.
I determined the symmetric matrix representation as:
$$\begin{bmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 0 & 1 \\
\end{bmatrix} $$
And that's as far as I've gotten. Any help would be much appreciated.
| I think what the question means is to apply the $L^TDL$ decomposition that produces Schur complements, rather than to use Schur complements themselves. In other words, you are expected to apply recursively the decomposition
$$
\pmatrix{A&B\\ B^T&C}=
\pmatrix{I&BC^{-1}\\ 0&I}
\pmatrix{A-BC^{-1}B^T&0\\ 0&C}
\pmatrix{I&0\\ C^{-1}B^T&I}
$$
to diagonalise a symmetric matrix by matrix congruence. In your case, apply $L^TDL$ decomposition once, we get
$$
\left(\begin{array}{cc|c}
2&1&1 \\
1&2&0 \\
\hline
1&0&1
\end{array}\right)
=\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1}
\pmatrix{1&1&0\\ 1&2&0\\ 0&0&1}
\pmatrix{1&0&0\\ 0&1&0\\ 1&0&1}.
$$
Apply the decomposition once more to the submatrix
$
\left[\begin{array}{c|c}
1&1\\
\hline
1&2
\end{array}\right],
$
we get
\begin{align}
\pmatrix{
2&1&1\\
1&2&0\\
1&0&1}
=\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1}
\pmatrix{1&\frac12&0\\ 0&1&0\\ 0&0&1}
\pmatrix{\frac12&0&0\\ 0&2&0\\ 0&0&1}
\pmatrix{1&0&0\\ \frac12&1&0\\ 0&0&1}
\pmatrix{1&0&0\\ 0&1&0\\ 1&0&1}.
\end{align}
Therefore, your matrix can be written as $L^TDL$ for some invertible matrix $L$, where $D=\operatorname{diag}(\frac12,2,1)$. Put $(u,v,w)=(x,y,z)L^T$, your quadratic form becomes $\frac12u^2+2v^2+w^2$, which is a sum of squares.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the roots of the polynomial? (Cardano's Method) $y^3-\frac7{12}y-\frac7{216}$
This is part of Cardano's method, so I've gotten my first root to be:
$y_1=\sqrt[3]{\frac7{432}+i\sqrt{\frac{49}{6912}}}+\sqrt[3]{\frac7{432}-i\sqrt{\frac{49}{6912}}}$
I am at a loss on how to evaluate this. I know this is one of three real roots, but I don't know what to do now.
| One can proceed as described above and find the three roots:
$\left\{\frac{1}{6} \left(\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i
\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i
\sqrt{3}\right)}\right),-\frac{7^{2/3} \left(1+i \sqrt{3}\right)}{6\ 2^{2/3}
\sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2}
\left(1+3 i \sqrt{3}\right)},-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{6\ 2^{2/3}
\sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{12} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2}
\left(1+3 i \sqrt{3}\right)}\right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Compute $A^{2013}$.
QUESTION 6
Let $$A=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}$$
Compute $A^{2013}$ (10 marks)
Solution First, observe that
$$A^{2013}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}^{2013}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}$$
Further, let $T:\mathbb R^3\to\mathbb R^3$ be the linear operator of the left multiplication with the matrix $$\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}$$
Notice that $T$ acts as follows: it sends $x_1$ to $-x_1$ and rotates the plane $(x_2,x_3)$ by the angle $\pi/3$ in the counter-clockwise direction. It means that $T^6$ is the identity transformation and hence $T^{2010}$ is also the identity (because $2010$ is a multiple of $6$). Thus $T^{2013}=T^3$, which, according to its geometric meaning, does the following: sends $x_1$ to $-x_1$ and rotates the plane $(x_2,x_3)$ by the angle $\pi$. The rotation by the angle $\pi$ is minus identity and hence the whole transformation $T^3$ is minus identity. Thus, $$A^{2013}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}^{2013}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}\\=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\cdot\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-1&0\\0&1&-2\\0&0&1\end{bmatrix}$$
I don't really understand the part where the solution talks about the way $T$ acts (second half of the solution). Can anyone explain to me why is that so?
| Notice that we can write $A$ on this form
$$A=P\operatorname{diag}(-1,R(\pi/3))P^{-1}$$
where
$$R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
is the matrix of rotation of angle $\theta$. Now using the bloc multiplication we get
$$A^{2013}=P\operatorname{diag}(-1,R^{2013}(\pi/3))P^{-1}=P\operatorname{diag}(-1,R(2013\times\pi/3))P^{-1}\\=P\operatorname{diag}(-1,R(\pi))P^{-1}=P\operatorname{diag}(-1,-1,-1)P^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Common tangent to a circle and ellipse Hey guys i am noy able to solve this problem.So please do help me in solving this.The equation of common tangent to ellipse
\begin{equation*}
x^2 +2y^2=1
\end{equation*}
and circle
\begin{equation*}
x^2 +y^2=\frac{2}{3}
\end{equation*}
is?
| A method that would also work for other pairs of conics: $$x^2 +2y^2-1=0 , x^2 +y^2-\frac23=0$$ have dual conics $$X^2+\frac12Y^2-1=0, \frac23X^2+\frac23Y^2-1=0$$ which intersect in $$(X,Y)=(-\frac1{\sqrt2},-1),(-\frac1{\sqrt2},1),(\frac1{\sqrt2},-1),(\frac1{\sqrt2},1)$$
and by substituting into the equation of the universal line $xX+yY+1=0$ we get the four common tangent lines $$-\frac1{\sqrt2}x-y+1=0,-\frac1{\sqrt2}x+y+1=0,\frac1{\sqrt2}x,-y+1=0,\frac1{\sqrt2}x+y+1=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving Recurrence Relations using Iteration $$a_0 = 2; \qquad a_k=4a_{k-1}+5 ~ \forall\ k\ge 1$$
I have already tried solving for $a_1$ through $a_5$.
| $$\begin{align*}
a_k - a_{k-1} & = 4a_{k-1}+5 -(4a_{k-2}+5) \\
&= 4(a_{k-1} - a_{k-2}) \\
&= 4^2(a_{k-2}-a_{k-3}) \\
& =4^{k-1}(a_1-a_0) \\
& =11\cdot 4^{k-1} \\
\Rightarrow a_k & = (a_k-a_{k-1})+(a_{k-1} - a_{k-2})+\cdots +(a_1-a_0)+a_0 \\
& = 11\cdot 4^{k-1} +11\cdot 4^{k-2} +\cdots + 11\cdot 4 + 11 + 2 \\
& = \ldots
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $\left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$ Question:
Simplify
$$ \left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$$
Attempted solution:
Rewriting it to look a bit better:
$$\left(ab \sqrt[4]{\frac{a^{3}}{\sqrt{b\sqrt{b}}}}\right)^{2}$$
Rewriting the denominator by replacing square roots with powers of fractions:
$$\left(ab \sqrt[4]{\frac{a^{3}}{(b (b^{\frac{1}{2}}))^{\frac{1}{2}}}}\right)^{2}$$
Combining b:s in the denominator:
$$\left(ab \sqrt[4]{\frac{a^{3}}{b^{\frac{3}{4}}}}\right)^{2}$$
Distributing the 4th root:
$$\left(\frac{ab \cdot a^{\frac{12}{16}}}{b^{\frac{3}{16}}}\right)^{2}$$
Combining a:s with a:s and b:s with b:s:
$$\left(b^{\frac{13}{16}} a^{\frac{28}{16}}\right)^{2}$$
Squaring gives the final result:
$$b^{26} a^{56}$$
The answer turns out to be:
$$a^{\frac{7}{2}} b^{\frac{13}{8}}$$
This is quite far away from the result I reached. Where did I go wrong, and are there are key insights that are useful for solving questions with lots and lots of square, cube and higher roots?
| we have $a^2b^2\sqrt{\frac{a^3}{b^{3/4}}}=\sqrt{\frac{a^4b^4a^3}{b^{3/4}}}=a^{7/2}(b^{13/4})^{1/2}=a^{7/2}b^{13/8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to find the eigenvectors. $$\begin{vmatrix}
1&1&1&1\\
1&1&0&0\\
1&0&1&0\\
1&0&0&1\\
\end{vmatrix}$$
for this matrix the eigenvalues are $(1,1,1-\sqrt3,1+\sqrt 3)$
I if I try find the eigenvectors in the cases where the eigenvalue is $1$ I end up with two equations and $4$ unknowns.
I'm not sure not to take the equivalent of the cross product for $4\times1$ matrices (I'm not sure it exists?) so how can I find the other two orthogonal eigenvectors?
| For $\lambda =1$
$$\begin{align} (A - I)v = 0 &\implies \begin{pmatrix}0&1&1&1\\1&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\\v_4\end{pmatrix}= 0\\&\implies \begin{cases}v_1 = 0\\v_2 + v_3 + v_4 =0\end{cases}\end{align}$$
Then the eigenvectors associated to $\lambda =1 $ are of the form $$\begin{pmatrix}0\\-v_3-v_4\\v_3\\v_4\end{pmatrix}$$
Taking for example, $v_3 = 1$ and $v_4 = -1$, and then $v_3 = 1$ and $v_4 =2$ we have $$\begin{pmatrix}0\\0\\1\\-1\end{pmatrix} , \begin{pmatrix}0\\-3\\1\\2\end{pmatrix}$$
Notice that you may choose different values for $v_3$ and $v_4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does my solutions manual take away or add 180 when finding $\theta$? These are the provided notes:
These are the provided questions:
I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)
| Consider the following diagram:
The range of the arccosine function is $[0, \pi]$. Since you are working in degrees, this corresponds to $[0^\circ, 180^\circ]$. Thus, you can calculate the measure of the obtuse angle $\theta$ whose cosine is $-3/5$ directly since
$$\cos\theta = -\frac{3}{5} \Longrightarrow \theta = \arccos\left(-\frac{3}{5}\right)$$
However, this will not work for sine or tangent since the range of the arcsine function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (or, in degrees, $[-90^\circ, 90^\circ]$) and the range of the arctangent function is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (in degrees, $(-90^\circ, 90^\circ))$.
For an obtuse angle $\theta$ such that $\sin\theta = \frac{1}{4}$, we draw a right triangle in the second quadrant with opposite side of length $|y| = |1| = 1$, hypotenuse of length $r = 4$, and adjacent side of length $|x| = |-\sqrt{15}| = \sqrt{15}$. The reference triangle we draw in the first-quadrant, has adjacent side of length $\sqrt{15}$, opposite side of length $1$, and hypotenuse of length $4$. The hypotenuse of the reference triangle forms an angle of $180^\circ - \theta$ with the positive $x$-axis, as shown in the diagram. Since the sine function satisfies the property
$$\sin(180^\circ - \theta) = \sin\theta$$
we obtain
$$\sin(180^\circ - \theta) = \frac{1}{4}$$
Moreover, since $\theta$ is an obtuse angle, $180^\circ -\theta$ is an acute angle, so it falls within the range of the arcsine function. Thus,
$$180^\circ - \theta = \arcsin\left(\frac{1}{4}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arcsin\left(\frac{1}{4}\right)$$
It would be inconvenient to draw a right triangle for an obtuse angle $\theta$ such that $\sin\theta = 0.7890$. However, we can still use the property that $\sin(180^\circ - \theta) = \sin\theta$ to conclude that
$$\sin(180^\circ - \theta) = 0.7890$$
Since $180^\circ - \theta$ is an acute angle, we obtain
$$180^\circ - \theta = \arcsin(0.7890)$$
Now, solve for $\theta$.
For an obtuse angle $\theta$ such that $\tan\theta = -\frac{6}{11}$, we draw a second-quadrant angle with adjacent side of length $|x| = |-6| = 6$ and opposite side of length $|y| = |11| = 11$. The hypotenuse has length $\sqrt{157}$. The reference triangle in the first-quadrant has adjacent side of length $6$, opposite side of length $11$, and hypotenuse of length $\sqrt{157}$. The angle the hypotenuse of the reference triangle forms with the positive $x$-axis is $180^\circ - \theta$. Since $\theta$ is obtuse, $180^\circ - \theta$ is acute. Since the tangent function has the property that
$$\tan(180^\circ - \theta) = -\tan\theta$$
we obtain
$$\tan(180^\circ - \theta) = -\left(-\frac{6}{11}\right) = \frac{6}{11}$$
Since $180^\circ - \theta$ is an acute angle, it falls within the range of the arctangent function. Thus,
$$180^\circ - \theta = \arctan\left(\frac{6}{11}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arctan\left(\frac{6}{11}\right)$$
For the obtuse angle $\theta$ such that $\tan\theta = -3.8522$, we use the property that $\tan(180^\circ - \theta) = -\tan\theta$ to conclude that
$$\tan(180^\circ - \theta) = -(-3.8522) = 3.8522$$
Since $180^\circ - \theta$ is an acute angle,
$$180^\circ - \theta = \arctan(3.8522)$$
Now, solve for $\theta$.
| {
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Prove/disprove an inequality $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ Is $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ TRUE in $(0, \frac{π}{2})$ ?
I proceed by taking two functions $f(x) = \sqrt{1 + x}-1 $ and $g(x) = \frac{x}{2} - \frac{x^2}{8} $.
Then $f(0) = 0 = g(0),$ while $f(\frac{π}{2}) > g (\frac{π}{2})$.
Also both functions are strictly increasing in $(0, \frac{π}{2})$.
I found nothing else. Does it help?
| Integration by parts gives
$$ \sqrt{1+x} = 1 +\frac{x}{2} - \frac{x^2}{8} + \int_0^x \frac{(x-t)^2}{2} \frac{3}{8(1+t)^{5/2}} \, dt, $$
and it is easy to see that the integral on the right-hand side is positive for any $x>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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partial fraction decomposition help now very quick way to solve 2x/x^2-4 as a partial fraction. I have tried the long way and it took over 30 minutes, I got it right but is it easier another way?
| Write $$\frac{2x}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$
Then note that if we multiply both sides by $x-2$ and let $x \to 2$ we obtain
Write $$\left.\frac{2x}{x+2}\right|_{x=2}=B+\left.\frac{A(x-2)}{x+2}\right|_{x=2}$$
which gives $B=1$.
Similarly, note that if we multiply both sides by $x+2$ and let $x \to -2$ we obtain
Write $$\left.\frac{2x}{x-2}\right|_{x=-2}=A+\left.\frac{B(x+2)}{x-2}\right|_{x=-2}$$
which gives $A=1$.
Putting it all together reveals that
$$\frac{2x}{x^2-4}=\frac{1}{x+2}+\frac{1}{x-2}$$
| {
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"url": "https://math.stackexchange.com/questions/1279525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When is $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ solvable?
Exercise: For which prime numbers does the equation $x^2 - 75 y^2 = 0$ have non-trivial solution in the $p$-adic integers $\mathbb{Z}_p$?
For $p\neq 5$, the non-trivial solvability of the equation $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ is equivalent to the solvability of the equation $f(z):=z^2 - 75=0$ in the $p$-adic numbers.
But if $z = u \cdot p^{-n}$ with $u \in \mathbb{Z}^\times_p$ and $n\in\{1, 2, 3, \ldots\}$, then
$$u^2 \cdot p^{-2n} - 75 = 0 \Longleftrightarrow u^2 - 75 p^{2n} = 0$$
and we would have to conclude that $u^2 \equiv 0\bmod{p}$, which is not possible, so $z\in\mathbb{Z}_p$.
Now let $p > 5$. Find the $p$ prime for which the Legendre-symbol $$\left(\frac{75}{p}\right) = \left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}} \left(\frac{p}{3}\right) = +1\, .$$
$\left(\frac{p}{3}\right) = +1$ exactly if $p\equiv 1 \bmod{3}$ and $(-1)^{\frac{p-1}{2}} \equiv +1 \bmod{p}$ exactly if $p = 1 \bmod{4}$.
So for $\left( \frac{75}{p} \right) = +1$, either $p \equiv 1 \bmod{3}$ and $p\equiv 1\bmod{4}$ or $p \equiv 2 \bmod{3}$ and $p\equiv 3\bmod{4}$. By the Chinese remainder theorem this is equivalent to $p \equiv \pm 1 \bmod{12}$, e.g.:
$$\{11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, \ldots \}\, .$$
By Hensel's lemma in these $\mathbb{Z}_p[X]$ the solutions are also liftable because $f'(z) =2 z \not \equiv 0 \bmod{p}$.
Now the remaining cases:
$p = 2$: $z^2 - 75 \equiv 0 \bmod{4} \Longleftrightarrow z^2 \equiv 3 \bmod{4}$, which is not possible.
$p = 3$: $z^2 - 75 \equiv 0 \bmod{9} \Longleftrightarrow z^2 \equiv 3 \bmod{9}$, which is not possible.
$p = 5$: $z^2 - 75 = 0$ in $\mathbb{Q}_5$, which means $z^2 - 75 \equiv 0 \bmod{125}$ must be solvable in $\mathbb{Z}_p$. So $\bigl(\frac{z}{5}\bigr)^2 \equiv 3 \bmod{5}$ must be solvable, but since $\left( \frac{3}{5} \right) = -1$ this is not possible.
To sum up the results: $$x^2 - 75 y^2 = 0 \text{ has non-trivial solutions in $\mathbb{Z}_p$} \Longleftrightarrow p \equiv \pm 1 \bmod{12}\, .$$
Is the proof ok? How could one make it cleaner? Thanks!
| Proof looks OK but I would streamline it quite a bit.
$x^2-75y^2=0$ is just the same as $x^2=3(5y)^2$ so for $x,y <>0$ and $p<>5$ (which makes sure that 5 has an inverse) we get $3=[x(5y)^{-1}]^2$ which means that 3 is a quadratic residue mod p or, in Legendre symbol notation,
$$(\frac{3}{p})=+1$$
Using the quadratic reciprocity theorem, hence
$$(\frac{p}{3})(\frac{3}{p})=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}$$, we derive that
$$(\frac{p}{3})=(-1)^{\frac{p-1}{2}}$$
Which excludes $p=2$ and $p=3$, and else indeed leads to $p\equiv \pm 1 \bmod{12}$, as you describe; quite simply because either $p\equiv 1 \bmod{4}$ and then we need $(\frac{p}{3})=+1$ thus $p\equiv 1 \bmod{3}$, or $p\equiv 3 \bmod{4}$ and then we need $(\frac{p}{3})=-1$ thus $p\equiv 2 \bmod{3}$.
The first leads to $p\equiv +1 \bmod{12}$; the second to $p\equiv -1 \bmod{12}$.
$p=2$ means $x^2=y^2$ which gives only $x=y=1$ as a near-trivial solution;
$p=3$ or $p=5$ means $x^2=0$ which gives only trivial solutions $x=0$, $y=$any.
| {
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How to establish the identity How do I establish the identity in this problem? Struggling with this one at the moment.
$$\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\tan\theta}{1-\tan^2\theta}$$
| We have
$$\frac{\tan \theta}{1-\tan^2 \theta}=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{\sin\theta(\cos^2\theta)}{\cos\theta(\cos^2\theta-\sin^2\theta)}=\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
and this is where I get stuck... How to continue?
When I put $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{n}{6}$
When I put $\frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{1}{6} * (n^{3} - n +1)$
| Oh, lol!
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n-1)! + (n-2)!}{3!(n-2)!} = \frac{(n-2)!((n-1) + 1)}{3!(n-2)!}=\frac{n}{3!} = \frac{n}{6}$
Correct? :)
How come Wolfram Alpha gives two different results?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the following formula is true for $n \geq 1$ by induction Prove that the following formula is true for $n \geq 1$ by induction.
$a_{n} = a_{n-1} + 4n - 3 \\ a_{n} = 2n^{2} - n + 1 \\ a_{1} = 2$
My attempt follows below. I almost succeed in proving the formula for $n \geq 2$, but I fail (more specifically, I fail when I try to prove the formula for $n = k + 1$ and also, I do not know how to prove it for $n \geq 2$. Here is my attempt:
*
*Let $n = 2 \\ 2 + 2*4 - 3 = 7 (a) \\ 2*2^{2} - 2 + 1 = 7 (b) \\ (a) = (b) \space QED$
*Suppose the formula is true for $n = k$, which gives us:
$ \\ a_{k-1} + 4k - 3 = 2k^{2} - k +1 \\$ This leads to the formula being true for $n = k + 1$: $a_{k+1-1} + 4(k+1) -3 = 2(k+1)^{2} - (k+1) + 2$
$ RHS = a_k + 4k -2 \\ LHS = 2(k^2 + 2k + 1) - k - 1 + 1 = 2k^2 +4k+2-k=(2k^2 - k +1) + 4k +1 =^{*} a_{k-1} + 4k -3 +4k+1 = a_{k-1} + 8k -2 $
From the above you see that RHS is not equal to LHS... What am I doing wrong? And when it has been shown that RHS = LHS, how does one continue to show that the formula is true not only for $n \geq 2$ but also for $n \geq 1$?
*is where I use the assumption from (2)
| I find working with $n$ is better with $k$. So: $$a_n = a_{n-1} + 4n-3=2(n-1)^2-(n-1)+1+4n-3=2n^2-4n+2-n+2+4n-3=2n^2-n+1$$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ramanujan's sum related to $\tan^{-1}(e^{-\pi x/2})$ While reading Ramanujan's Collected Papers I came across a nice formula which he mentions without proof $$\tan^{-1}(e^{-\pi x/2}) = \frac{\pi}{4} - \left(\tan^{-1}\frac{x}{1} - \tan^{-1}\frac{x}{3} + \tan^{-1}\frac{x}{5} - \cdots\right)$$ where $x$ is any real number. Ramanujan proves similar formulas with terms involving $\tan^{-1}(x^{2})$ but leaves the above one as if it is obvious.
I tried to simplify RHS term by term starting with $$\frac{\pi}{4} - \tan^{-1}x = \tan^{-1}\frac{1 - x}{1 + x}$$ and then
\begin{align}
RHS &= \frac{\pi}{4} - \tan^{-1}x + \tan^{-1}\frac{x}{3}\notag\\
&= \tan^{-1}\frac{1 - x}{1 + x} + \tan^{-1}\frac{x}{3}\notag\\
&= \tan^{-1}\left(\dfrac{\dfrac{1 - x}{1 + x} + \dfrac{x}{3}}{1- \dfrac{1 - x}{1 + x}\cdot\dfrac{x}{3}}\right)\notag\\
&= \tan^{-1}\left(\dfrac{3 - 2x + x^{2}}{3 + 2x + x^{2}}\right)\notag\\
\end{align}
My guess is that the argument of $\tan^{-1}$ probably looks like some convergent to a suitable continued fraction expansion of $e^{-\pi x / 2}$ but I am just not able to figure it out. An obvious difficulty is that any expansion of $e^{-\pi x / 2}$ as a continued fraction would involve $\pi$ as well as.
Please suggest any other approach to prove the Ramanujan's formula.
Update: If I differentiate with respect to $x$ I get the LHS as $$\frac{-\pi e^{-\pi x / 2}}{2(1 + e^{-\pi x})}$$ and the RHS comes out to be $$-\left(\frac{1}{1 + x^{2}} - \frac{3}{3^{2} + x^{2}} + \frac{5}{5^{2} + x^{2}} - \cdots\right)$$ This is similar to the partial fraction expansion of $\tanh(x)$ given by $$\frac{\tanh x}{8x} = \sum_{k = 1}^{\infty}\frac{1}{(2k - 1)^{2}\pi^{2} + 4x^{2}}$$ but not exactly as desired. For the proof of formula for $\tanh x$ see this.
| If we differentiate both terms, we get an identity that can be easily proved by considering the logarithmic derivatives of the Weierstrass products for the $\sinh$ and $\cosh$ functions.
We just need to prove:
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)+\frac{z^2}{2n+1}} = \frac{\pi}{4\cosh\frac{\pi z}{2}}.\tag{1}$$
From:
$$\cot z = \sum_{k\in\mathbb{Z}}\frac{1}{z-k\pi},\qquad \frac{1}{\sin z}=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z-k\pi}\tag{2}$$
it follows that:
$$ \frac{\pi}{2\sin \frac{\pi z}{2}}=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{ z-2k}\tag{3}$$
and by replacing $z$ with $z-1$ we also have:
$$ -\frac{\pi}{2\cos \frac{\pi z}{2}}=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{ z-(2k+1)}\tag{4}$$
so we just need to replace $z$ with $iz$ to get $(1)$.
By integrating both sides of $(1)$ between $0$ and $x$ Ramanujan's identity follows, since:
$$ \int\frac{\pi\,dz}{4\cosh\frac{\pi z}{2}} = \arctan\left(\tanh\frac{\pi z}{4}\right)=\frac{\pi}{4}-\arctan(e^{-\pi z/2}).\tag{5}$$
Another way to prove $(1)$ is to notice that, through the Laplace transform of $\cos x$:
$$\int_{0}^{+\infty}\frac{\cos (\alpha x)}{e^x+e^{-x}}\,dx =\sum_{k\geq 0}(-1)^k\int_{0}^{+\infty}e^{-(2k+1)x}\cos(\alpha x)\,dx = \sum_{k\geq 0}\frac{(-1)^k (2k+1)}{\alpha^2+(2k+1)^2}$$
but the LHS can be also computed through the residue theorem.
That was exactly the point of a currently deleted question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Show that $ \int_{-\infty}^{\infty} \frac{x^3}{(x^2+4)(x^2+1)}\, dx$ does not converge I noticed that $\displaystyle \int_{-a}^{b} \frac{x^3}{(x^2+4)(x^2+1)}$ will converge to $0$ whenever $a=b$ and will converge to some value whenever $a,b$ are in the reals (excluding infinity). How would you show that this integral does not converge over the interval $(-\infty, \infty)$ to someone who is not convinced?
When I evaluated this integral using complex residues for the interval $(-\infty, \infty)$ $(Res[z=2i] + Res[z=i])$,I got a complex value as an answer.
| Beside the simple point from Aloizio Macedo which would apply to many cases, may be, going further, you could use the fact that $$\frac{x^3}{(x^2+4)(x^2+1)}=\frac{4 x}{3 \left(x^2+4\right)}-\frac{x}{3 \left(x^2+1\right)}=\frac 23 \frac{2 x}{x^2+4}-\frac 16 \frac{2 x}{x^2+1}$$ So $$\int\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{2}{3} \log \left(x^2+4\right)-\frac{1}{6} \log \left(x^2+1\right)$$ Already, the doefficients show what would be the problem $\frac 23 -\frac 16=+\frac 12$.
Since you already noticed that, if $a=b$, the result is $0$ then $$\int_{-a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\int_{a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{1}{6} \log \Big(\frac{(b^2+4)^4\,(a^2+1)}{(a^2+4)^4\,(b^2+1)}\Big)$$ which, forgetting a constant which depend on $a$, is basically $\log(b)$ for large values of $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Estimate from below $\int_0^\pi e^{-t}\cos nt dt$ without calculate it. Estimate from below the following integral
$$\int_0^\pi e^{-t}\cos nt dt$$
without calculate it. Here $n\in\mathbb N$. Any suggestions please?
| A simpler way.
Since
$\cos(z) = \Re(e^{iz})$,
$\begin{array}\\
I(n)
&=\int_0^\pi e^{-t}\cos nt dt\\
&=\Re \int_0^\pi e^{-t}e^{int} dt\\
&=\Re \int_0^\pi e^{-t+int} dt\\
&=\Re \frac{e^{t(-1+in)}}{-1+in}\big|_0^\pi\\
\end{array}
$.
$\begin{array}\\
\frac{e^{t(-1+in)}}{-1+in}
=\frac{e^{-t}+e^{int}}{-1+in}\\
=\frac{e^{-t}+\cos(nt)+i\sin(nt)}{-1+in}\frac{-1-in}{-1-in}\\
=\frac{(e^{-t}+\cos(nt)+i\sin(nt))(-1-in)}{-1+n^2}\\
=\frac{-(e^{-t}+\cos(nt)+i\sin(nt))-in(e^{-t}+\cos(nt)+i\sin(nt))}{-1+n^2}\\
=\frac{-e^{-t}-\cos(nt)-i\sin(nt))-ine^{-t}-in\cos(nt)+n\sin(nt)}{-1+n^2}\\
=\frac{-e^{-t}-\cos(nt)+n\sin(nt)-i(\sin(nt)+ne^{-t}+n\cos(nt))}{-1+n^2}\\
\end{array}
$
Therefore
$\begin{array}\\
I(n)
&=\frac{-e^{-t}-\cos(nt)+n\sin(nt)}{-1+n^2}\big|_0^\pi\\
&=\frac{(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))-(-e^{0}-\cos(0)+n\sin(0))}{-1+n^2}\\
&=\frac{(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))-(-1-1)}{-1+n^2}\\
&=\frac{2+(-e^{-\pi}-\cos(n\pi)+n\sin(n\pi))}{-1+n^2}\\
\end{array}
$
If $n$ is even,
$I(n)
=\frac{2+(-e^{-\pi}-1)}{-1+n^2}
=\frac{1-e^{-\pi}}{-1+n^2}
$.
If $n$ is odd,
$n=2m+1$,
$\begin{array}\\
I(n)
&=\frac{2+(-e^{-\pi}-(\cos((2m+1)\pi)+n\sin((2n+1)\pi))}{-1+n^2}\\
&=\frac{2+(-e^{-\pi}-n\sin((2m+1)\pi))}{-1+n^2}\\
&=\frac{2+(-e^{-\pi}-n(-1)^m))}{-1+n^2}\\
&=\frac{2-e^{-\pi}-n(-1)^{\lfloor n/2\rfloor)}}{-1+n^2}\\
\end{array}
$.
If $n=4m+1$,
$I(n)
=\frac{2-e^{-\pi}-n}{-1+n^2}
=\frac{2-n-e^{-\pi}}{-1+n^2}
$.
If $n=4m+3$,
$I(n)
=\frac{2-e^{-\pi}+n}{-1+n^2}
=\frac{2+n-e^{-\pi}}{-1+n^2}
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?
| Perhaps a little bit less tedious then a binomial expansion to the power of $6$, you could use the triple and double angle identities $\cos 3x=4\cos^3x-3\cos x$ and $\cos 2x=2\cos^2x-1$, so that we have
$$\begin{align}\cos (6x)&=\cos(3\cdot 2x)\\&=4\cos^3(2x)-3\cos(2x)\\&=4(2\cos^2x-1)^3-3(2\cos^2x-1)\\&=4(8\cos^6x-12\cos^4x+6\cos^2x-1)-6\cos^2x+3\\&=32\cos^6x-48\cos^4x+18\cos^2x-1\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to proceed with Euclidean algorithm for finding greatest common divisor of two polynomials. I am trying to find
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2).
\end{equation*}
I have done the first step of long division and found.
\begin{equation*}
x^4-x^3-4x^2-x+5=(x^2-2x)(x^2+x-2)-5x+5
\end{equation*}
so we have
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2)=gcd(x^2+x-2,-5x+5)
\end{equation*}
now is where I am stuck. For the next step do we need to divide $x^2+x-2$ by $-5x+5$ or can we simply divide $x^2+x-2$ by $-x+1$ since the $gcd$ needs to a monomial?
Also if I did divide by $-x+1$ instead of $-5x+5$ would this change my procedure at all when reversing the algorithm to find the polynomials that multiply $(x^4-x^3-4x^2-x+5,x^2+x-2)$ to give $gcd(x^4-x^3-4x^2-x+5,x^2+x-2)$. (Bezout's lemma)
| You are working with polynomials with coefficients in $\Bbb Q$ or in a larger field. $\Bbb Z[x]$ is not a principal ideal domain ($\langle x,2\rangle$ is not principal, for example), and the existence of $\gcd$ is not guaranteed.
That said, $5$ is an unit, and hence $-x+1$ and $-5x+5$ are associated. Therefore, you can change one by another when you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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the series: compute $ \sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4} $ Compute
$$
\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4}
$$
the result is $\frac{\pi^4+30\pi^2-384}{768}$, so I'm sure the sums $\sum\frac{1}{n^2}$ and $\sum\frac{1}{n^4}$ should appear in the solution.
The standard method $\frac{1}{4n^2-1}=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})$ allows to compute $\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\frac{1}{2}$ and $\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}=\frac{\pi^2-8}{16}$, what about higher powers?
| Although it is tedious, you can use partial fractions to get:
$\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(4n^2-1)^4}$ $= \dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right)$ $+\dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^2}+\dfrac{1}{(2n-1)^2}\right)$ $+\dfrac{1}{8}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^3}-\dfrac{1}{(2n-1)^3}\right)$ $+\dfrac{1}{16}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^4}+\dfrac{1}{(2n-1)^4}\right)$.
The first and third summations telescope to $-1$.
Since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \zeta(2) = \dfrac{\pi^2}{6}$, we have $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n)^2} = \sum_{n = 1}^{\infty}\dfrac{1}{4n^2} = \dfrac{1}{4}\zeta(2) = \dfrac{\pi^2}{24}$.
Hence $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n-1)^2} = \dfrac{\pi^2}{6}-\dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$, and $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n+1)^2} = \dfrac{\pi^2}{8} - 1$.
You can do a similar thing for the fourth sum.
| {
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A tough limit problem involving $1/(\sin x - \sin a)$ and its generalization Long back I had encountered the following problem in Hardy's Pure Mathematics (originally from the infamous Mathematical Tripos 1896):
If $$f(x) = \frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a}$$ then show that $$\frac{d}{da}\{\lim_{x \to a}f(x)\} - \lim_{x \to a}f'(x) = \frac{3}{4}\sec^{3}a - \frac{5}{12}\sec a$$
I had solved it using Taylor series expansions and even then it involved good amount of calculation. I suppose applying L'Hospital would be even more arduous.
On the other hand both the Taylor series and L'Hospital Rule are discussed later in Hardy's book suggesting that it could be solved via elementary techniques (i.e. using algebra of limits and Squeeze theorem and if needed one can use mean value theorem). Please let me know if such a solution is possible.
Also I believe there might be a suitable generalization applicable to functions of type $$g(x) = \frac{1}{\phi(x) - \phi(a)} - \frac{1}{(x - a)\phi'(a)}$$ and perhaps the expression $$\frac{d}{da}\{\lim_{x \to a}g(x)\} - \lim_{x \to a}g'(x)$$ has some significance. Any ideas in this direction would be helpful.
| $$\begin{align} \lim_{x \to a} f(x) &= \lim_{x\to a}\frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a} \\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)}\\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)} \end{align}$$
Let $ y = x - a$; Since $\sin x - \sin a = 2\sin \left(\dfrac{x-a}{2}\right) \cos \left(\dfrac{x+a}{2}\right)$
$$\begin{align}\sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)} & = \dfrac{\sec^2 a}{2}\lim_{y \to 0} \dfrac{(y - \sin y) \cos a + \sin a (1 - \cos y)}{y \sin(y/2)} \\=\ & \sec^2 a\lim_{y \to 0} \dfrac{(y - \sin y) \cos a}{y^2} + \sec^2 a\lim_{y\to0 }\dfrac{\sin a (1 - \cos y)}{y^2 }. \end{align}$$
It is common knowledge that $\lim_{y \to 0} \dfrac{y -\sin y}{y^2} = 0$ and $\lim_{y\to0 }\dfrac{ 1 - \cos y}{y^2 } = \dfrac12$ can be proved only using standard limits(I omitted the proof for the sake of brevity).
Therefore $$\bbox[5px,border-style: solid; border-color:black; border-width: 2px]{\lim_{x \to a} f(x) = \dfrac12 \sec a \tan a.} \tag 1$$
$$f'(x) = \dfrac{1}{(x-a)\cos a} - \dfrac{\cos x}{(\sin x - \sin a)^2}$$
$$\begin{align}\lim_{x \to a} f'(x) = & \sec a \lim_{x\to a}\dfrac{(\sin x -\sin a)^2 - (x-a)^2\cos x \cos a }{(x-a)^2 (\sin x - \sin a)^2} \\= \ & \dfrac{\sec^3 a}{4} \lim_{x\to a}\dfrac{(\sin x -\sin a)^2 - (x-a)^2\cos x \cos a }{(x-a)^2 \sin^2\left(\dfrac{x-a}{2}\right)} \end{align}$$
Letting $ y = x - a$ the above limit becomes,
$$\begin{align}\sec^3 a\lim_{y \to 0} \dfrac{(\sin(y + a ) - \sin a)^2 - y^2 \cos (y +a )\cos a}{y^4}. \end{align}$$
Expanding the numerator completely in terms of $\sin a, \cos a, \sin y $ and $\cos y$, we get
$$\begin{align}\sec^3 a \lim_{y \to 0} \dfrac{\sin^2 y \cos^2 a + \cos^2 y \sin^2 a + \sin^2 a + 2\sin a \cos a \sin y \cos y \\- 2 \sin a \cos a \sin y - 2\sin^2 a \cos y - y^2 \cos y \cos a^2 + y^2 \sin y \sin a \cos a}{y^4}\end{align}$$
Collecting terms simplifies this to
$$\sec^3 a\lim_{y\to 0} \dfrac{\cos^2 a(\sin^2 y - y^2 \cos y) + \sin a \cos a \sin y (2 \cos y - 2+ y^2) + \sin^2 a (\cos y - 1)^2}{y^4}$$
*
*Solving $\lim_{y \to 0} \dfrac{\sin^2 y - y^2 \cos y}{y^4}$ :
$$\begin{align} \lim_{y \to 0} \dfrac{\sin^2 y - y^2 \cos y}{y^4} =& \lim_{y \to 0} \dfrac{\sin^2 y - y^2 }{y^4}+\lim_{y \to 0}\dfrac{1- \cos y}{y^2} \\= \ &\lim_{y \to 0} \left(\dfrac{\sin y - y }{y^3}\right)\left(\dfrac{\sin y + y }{y}\right)+\lim_{y \to 0}\dfrac{1- \cos y}{y^2}\end{align}$$
Using $\lim_{y \to 0} \dfrac{\sin y - y}{y^3} = \dfrac{-1}{6}$(proved here using only standard limits) and couple other familiar limits, we get the limit as $\dfrac 16$.
*
*Solving $\lim_{y \to 0} \sin y\dfrac{(2 \cos y - 2 + y^2)}{y^4}$:
$$\begin{align}\lim_{y \to 0} \sin y\dfrac{(2 \cos y - 2 + y^2)}{y^4}
= &\lim_{y \to 0} \dfrac{\sin y}{y}\left(\dfrac{ y^2 - 4\sin^2(y/2)}{y^3}\right) \\=\ &\lim_{y \to 0} \dfrac14\left(\dfrac{ y/2 - 2\sin(y/2)}{(y/2)^3}\right)(y + 2 \sin(y/2)) \\ = \ & \dfrac16 \cdot 0 = 0.
\end{align}$$
*
*Lastly we use our "common knowledge" to show to conclude that $\lim_{y \to 0} \dfrac{(\cos y - 1)^2}{y^4} = \dfrac14$.
Using above three bullet points we get $$\lim_{x\to a} f'(x) = \dfrac{\sin^2 a\sec^3 a}{4} + \dfrac{\sec a}{6} = \dfrac{\sec^3 a}{4} - \dfrac{\sec a}{12}.$$
Hence $$\bbox[5px,border-style: solid; border-color:black; border-width: 2px]{\lim_{x \to a} f'(x) = \dfrac{\sec^3 a}{4} - \dfrac{\sec a}{12}.} \tag 2$$
Using $(1)$ and $(2)$
$$\begin{align}\dfrac{d}{ da}\{\lim_{x \to a}f(x)\} - \lim_{x \to a}f'(x)=& \sec^3 a - \dfrac12\sec a - \dfrac{\sec^3 a}{4} + \dfrac{\sec a}{12}\\ = \ & \dfrac{3\sec^3 a}{4} - \dfrac{5 \sec a}{12}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Scale the distance of lattice points I know that for a hexagonal lattice generated by (0,1) and ($\sqrt{3}/2$,1/2) (i.e., when the distance between lattice points is 1), the number of lattice points in a circle of given radius $r$ can be calculated by the equation below:
$$n(r) = \sum_{x=-\lfloor \frac{r}{\sqrt{3}}\rfloor}^{\lfloor \frac{r}{\sqrt{3}}\rfloor} (2\lfloor \sqrt{r^2-3x^2} \rfloor+1) + \sum_{x=\frac{1}{2}-\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\rfloor}^{\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\rfloor-\frac{1}{2}} 2\lfloor \sqrt{r^2-3x^2} +\frac{1}{2} \rfloor$$
I would like to transform this sum for the case when the distance between lattice points is $d$ where $d$ is an integer. Is the scaling $r$ by $d$ is sufficient? I mean is the following equation for counting the number of lattice points when the distance between lattice points is $d$ correct?
$$n(r) = \sum_{x=-\lfloor \frac{r}{d\sqrt{3}}\rfloor}^{\lfloor \frac{r}{d\sqrt{3}}\rfloor} (2\lfloor \sqrt{\frac{r^2}{d^2}-3x^2} \rfloor+1) + \sum_{x=\frac{1}{2}-\lfloor \frac{r}{d\sqrt{3}}+\frac{1}{2}\rfloor}^{\lfloor \frac{r}{d\sqrt{3}}+\frac{1}{2}\rfloor-\frac{1}{2}} 2\lfloor \sqrt{\frac{r^2}{d^2}-3x^2} +\frac{1}{2}\rfloor$$
| Yes, if the first one is correct (which I will take your word for!) because you are simply saying how many points there are in a circle of radius $\frac rd$ instead of radius $r$, which is equivalent to what you are looking for.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Product of two integers of the form $x^2+my^2$ is of the same form. Let $x,y,a,b\in \mathbb Z$. Prove that there are integers $c$ and $d$
so that
\begin{equation*}
(x^2+y^2m)(a^2+b^2m)=c^2+d^2m.
\end{equation*}
I'm stuck, I took the product and got $x^2a^2+b^2y^2m^2+(a^2y^2+x^2b^2)m$ but the numbers in the parenthesis need not be squares. Thank you in advance.
| Try the following
$$\begin{align}(ax+mby)^2+m(ay-bx)^2
&=a^2x^2+2maxby+m^2b^2y^2+ma^2y^2-2maxby+mb^2x^2\\
&=(a^2+mb^2)(x^2+my^2)\end{align}$$
The easy way to remember it is the identity with $m=1$
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How can I determine the value of $a_1 + \displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2}$ For reals $x \ge 3$, let $f(x)$ denote the function $f(x) = \frac{-x + x\sqrt{4x - 3}}{2}$. Now suppose that $a_1, a_2, \ldots, a_{2013}$ is a sequence of real numbers such that $a_1 > 3, a_{2013} = 2013$, and for $n = 1, 2, \ldots, 2012$, $a_{n + 1} = f(a_n)$.
Given this, how can I determine the value of $$a_1 + \displaystyle\sum_{i = 1}^{2012}\frac{a_{i + 1}^3}{a_i^2 + a_ia_{i + 1} + a_{i + 1}^2}$$
| Let $$\displaystyle A= a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2} = a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3-a_i^3+a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$
$$\displaystyle = a_1 +\sum_{i=1}^{2012} (a_{i+1} -a_i) +\sum_{i=1}^{2012} \frac{a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$
The first sum is solved using telescoping series property to get $\displaystyle -a_1+a_{2013}$.
Now we have $\displaystyle f(x)=\frac{-x+x\sqrt{4x-3}}{2}$ and $\displaystyle a_{i+1}=f(a_i)$
$$\displaystyle x^2 +xf(x) +f(x)^2 = x^2 +f(x) (x+f(x))= x^2 +f(x) \left (x+ \frac{-x+x\sqrt{4x-3}}{2}\right )$$
$$\displaystyle =x^2+ \frac{-x+x\sqrt{4x-3}}{2} \frac{x+x\sqrt{4x-3}}{2} $$
$$\displaystyle = x^2 + \frac{-4x^2 +4x^3}{4} = x^3$$ $$\displaystyle \Rightarrow a_i^2+a_ia_{i+1}+a_{i+1}^2 = a_i^3 \quad\text{(let } \displaystyle x=a_i\text{)}$$
$$\displaystyle \Rightarrow A= a_1-a_1+a_{2013} +\sum_{i=1}^{2012} 1 = 2013+2012=4025$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ We want to find an expression for $A^n = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)^n$ for an arbitrary "n".
I have tried writing out a few elements of the sequence as $n \to \infty$:
*
*$A^2 = \left( \begin{array}{cc} 9 & 16 \\ 8 & 17 \end{array} \right)$
*$A^3 = \left( \begin{array}{cc} 41 & 84 \\ 36 & 51 \end{array} \right)$
However, a pattern doesn't seem to appear.
This is where I want to ask my question: if we put this matrix into reduced-row echelon form, would an expression of the $(reduced matrix)^n$ work as an expression for the original matrix $A$?
i.e. reduced-row matrix $ = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)^n$. Then, we know that any diagonal matrix to the $n^{th}$ is just the diagonal entries to the $n^{th}$ and this would make an expression easy to come up with.
Thank you!
| In problems such as these one of the most efficient ways to calculate powers of a matrix is through diference equations. For the matrix of this problem the following is obtained.
Given
\begin{align}
A = \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)
\end{align}
it is found that
\begin{align}
A^{2} = \left( \begin{array}{cc} 9 & 16 \\ 8 & 17 \end{array} \right)
\hspace{10mm}
A^{3} = \left( \begin{array}{cc} 41 & 84 \\ 42 & 83 \end{array} \right).
\end{align}
The problem asks for the values of $A^{n}$. This is done by letting
\begin{align}
A^{n} = \left( \begin{array}{cc} a_{n} & b_{n} \\ c_{n} & d_{n} \end{array}
\right).
\end{align}
With this and $A^{n} = A \cdot A^{n-1}$ then
\begin{align}
\left( \begin{array}{cc} a_{n} & b_{n} \\ c_{n} & d_{n} \end{array} \right)
= \left( \begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array} \right)
\left( \begin{array}{cc} a_{n-1} & b_{n-1} \\ c_{n-1} & d_{n-1} \end{array} \right).
\end{align}
From this equation it is quickly determined that
\begin{align}
a_{n} &= a_{n-1} + 4 c_{n-1} \hspace{12mm} b_{n} = b_{n-1} + 4 d_{n-1} \\
c_{n} &= 2 a_{n-1} + 3 c_{n-1} \hspace{10mm} d_{n} = 2 b_{n-1} + 3 d_{n-1}.
\end{align}
These equations are reducible to the second order difference equation, $\phi_{n} \in \{ a_{n} , b_{n}, c_{n}, d_{n} \}$,
\begin{align}
\phi_{n+2} = 4 \phi_{n-1} + 5 \phi_{n}
\end{align}
which has the solution
\begin{align}
\phi_{n} = A \, 5^{n} + B \, (-1)^{n}.
\end{align}
Now returnig to the first few powers of $A$ the conditions
\begin{align}
a_{1} &= 1, a_{2} = 9, a_{3} = 41 \\
b_{1} &= 4, b_{2} = 16, b_{3} = 84 \\
c_{1} &= 2, c_{2} = 8, c_{3} = 42 \\
d_{1} &= 3, d_{2} = 17, d_{3} = 83
\end{align}
are obtained. The second order difference equation equation can be solved for each of the condition sets and is determined that
\begin{align}
a_{n} &= \frac{1}{3} \left( 5^{n} + 2 (-1)^{n} \right) \hspace{10mm}
b_{n} = \frac{2}{3} \left( 5^{n} - (-1)^{n} \right) \\
c_{n} &= \frac{1}{3} \left( 5^{n} - (-1)^{n} \right) \hspace{12mm}
d_{n} = \frac{1}{3} \left( 2 \cdot 5^{n} + (-1)^{n} \right).
\end{align}
Putting this together the $n^{th}$-power of $A$ is given by
\begin{align}
A^{n} = \frac{1}{3} \left( \begin{array}{cc} 5^{n} + 2 (-1)^{n} & 2(5^{n} - (-1)^{n}) \\ 5^{n} - (-1)^{n} & 2 \cdot 5^{n} + (-1)^{n} \end{array}
\right).
\end{align}
| {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Find the maximum and minimum of the function $f$ Find the maximum and minimum of $f(x, y)=xy-y+x-1$ at the set $x^2+y^2\leq 2$.
I have done the following:
Since the region $x^2+y^2\leq 2$ is closed, $f$ has a maximum and a minimum, which is either at the boundary or at the critical points of the function.
To find the critical points we do the following:
$$\nabla f=\overrightarrow{0} \Leftrightarrow (y+1, x-1)=(0, 0) \Rightarrow x=1, y=-1$$
So, we have that the only critical point is $(1, -1)$, which satisfies the contraint.
To find the extremas of $f$ at the boundary $x^2+y^2=2$, we will use Lagrange multipliers therem.
The constraint is $g(x, y)=x^2+y^2-2=0$.
We are looking for $x$, $y$ and $\lambda$ such that $$\nabla f(x, y)=\lambda \nabla g(x, y) \tag 1$$ and $$g(x, y)=0 \tag 2$$
$$(1) \Rightarrow (y+1, x-1)=\lambda (2x, 2y) \Rightarrow \lambda =\frac{1}{2}$$
$$\Rightarrow x=y+1$$
$$(2) \Rightarrow y=\frac{-1+\sqrt{3}}{2}, y=\frac{-1-\sqrt{3}}{2}$$
SO, the extremas are the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$.
Calculating the values of $f$, we have: $$f\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )=\frac{1}{2} \\ f\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )=\frac{1}{2} \\ f(1, -1)=0$$
SO, $f$ has a local maximum at the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$ and an absolute minimum at $(1, -1)$.
Is this correct?? Could I improve something??
| Looks good, I have just a few comments:
"So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema.
"$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum and minimum"
| {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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Is the argument I used to evaluate the convergence of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$ right?
If $a,b,c$ be real constants, analyze the convergence of $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n+a}{(n+b)(n+c)}$$
What I tried to:
I compared the general term of my series to $\frac{1}{n}$: $$\lim \limits_{n \to \infty} \frac{\frac{n+a}{(n+b)(n+c)}}{\frac{1}{n}}= $$ $$\lim\limits_{n \to \infty} \frac{n^2 (1 + \frac{a}{n})}{n^2 \left ( 1 + \frac{b+c}{n} + \frac{bc}{n^2} \right ) } = 1 $$
So, as $\frac{n+a}{(n+b)(n+c)} \sim \frac{1}{n}$ when $n \to \infty$, the series (conditionally) converges because the alternating harmonic series converges.
| Since if $b\neq c$ then there exists $A,B\in\mathbb{R}$ such that $$\frac{n+a}{(n+b) (n+c)}=\frac{A}{n+b} +\frac{B}{n+c} $$
the series is convergent by Leibniz test.
If $b=c$ then $$\frac{n+a}{(n+b) (n+c)}=\frac{1}{n+b} +\frac{a-b}{(n+b)^2}$$
and the series is also convergent by Leibniz test.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Joint probability distribution (over unit circle) A couple of two continuous random variables $(X,Y)$ is distributed uniformly over the closed unity circle (so $-1\leq x \leq 1$ , $y$ analog). $U$ is defined as the distance from $O$ to the point $(X,Y)$. Calculate U:
My calculations so far:
$$
f_{XY}(x,y) = \frac{1}{\pi} ; (x,y) \in S
$$
Where S is the unit circle.
$$
\begin{align}
F_{U}(u) &= Pr((x,y): \sqrt{X^2+Y^2}\leq u)\\
&= \int_{S}\int f_{X}(x)f_{Y}(y)dxdy\\
&= \int_{-u}^{u}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{4}{\pi^2}dydx\\
&= \frac{8}{\pi^2}\left(u\sqrt{1-u^2}+\arcsin{u}\right)
\end{align}
$$
When $0<u\leq1$. However the answer is $F_{U}(u) = u^2$ when $0<u\leq1$.
Where am I going the wrong way in my argumentation?
| Answer: On the second line of your derivation.
Let's do it correctly:
$$
\begin{eqnarray}
F_U\left(u\right) &=& \Pr\left(\sqrt{X^2+Y^2} \leqslant u \right) \\
&=& \int_S f_{X,Y}\left(x,y\right) I\left(\sqrt{X^2+Y^2} \leqslant u\right) \mathrm{d}x\mathrm{d}y \\
&=& \left.\int_{-1}^{1} \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} I\left(\sqrt{X^2+Y^2} \leqslant u\right) \mathrm{d} y \right)\mathrm{d} x \right|_{\text{assuming } 0<u \leqslant u} \\
&=& \int_{-u}^{u} \left(\int_{-\sqrt{u^2-x^2}}^{\sqrt{u^2-x^2}} \frac{1}{\pi} \mathrm{d} y \right)\mathrm{d} x \\
&=& \left.\int_{-u}^{u} \frac{2}{\pi} \sqrt{u^2-x^2} \mathrm{d}x \right|_{x = z \cdot u} \\
&=& u^2 \int_{-1}^{1} \frac{2}{\pi} \sqrt{1-z^2} \mathrm{d}z = u^2
\end{eqnarray}
$$
As a side note, it would have been easier to work in polar coordinates here.
| {
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A recursive sequence is defined by... A sequence is defined recursively by $a_1=1$ and $a_{n+1} = 1 + \frac{1}{1+a_{n}}$.
Find the first eight terms of the sequence $a_n$. What do you notice about the odd terms and the even terms? By considering the odd and even terms separately, show that $a_n$ is convergent and deduce that its limit is $\sqrt{2}$.
EDIT: Ok, so I was being an idiot and forgot how to basic math for a little. The first 8 terms are as follows:
EVENS:
a2 = 1.5
a4 = 1.46666...
a6 = 1.415731...
a8 = 1.41426...
ODDS:
a1 = 1
a3 = 1.4
a5 = 1.4054...
a7 = 1.41395...
From these I see that the even terms are decreasing while the odd terms are increasing. How can I use these to prove that $\{a_n\}$ converges? Does it have something to do with alternating series or something similar?
| We have
$a_{n+1}
= 1 + \frac{1}{1+a_{n}}
= \frac{2+a_n}{1+a_{n}}
$.
Therefore,
$a_n > 1$
for all $n$.
Also,
$\begin{array}\\
a_{n+1}-\sqrt{2}
&= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\
&= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\
&= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\
&= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\\
\end{array}
$
so
$\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n}
=\dfrac{\sqrt{2}-1}{1+a_{n}}
$.
Therefore
(1)$a_n-\sqrt{2}$
alternates in sign
and
(2)$\big|\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n}\big|
<\sqrt{2}-1
$.
This implies that
$a_n-\sqrt{2}
\to 0$
so that
$\lim_{n \to \infty} a_n
=\sqrt{2}
$.
| {
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"source": "stackexchange",
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prime number and order problem Does anyone can solve problem stated at
Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$?
?
It should have the additional constraint, k < p.
| Let $o$ be the order of $n$ to base $2$.
In other words, $o$ is the smallest positive natural number with $2^o\equiv 1\ \ (\ mod\ n\ )$
Because of $2^{n-1}\equiv 1\ (\ mod\ n\ )$, $o$ must be a divisor of
$n-1=kp^2$.
$o$ cannot be a divisor of $k$, because then we would have $2^k\equiv 1\ (\ mod\ n\ )$
So, $o$ must be a multiple of $p$. So, if $n$ is not prime, it must have
the form $(ap+1)(bp+1)=abp^2+(a+b)p+1=kp^2+1$ , with $a,b>0$
It follows $abp+a+b=kp$
Because of $k<p$, we have $ab<p$. Furthermore, we have $a+b\equiv 0\ (\ mod\ p\ )$.If we assume $a<b$, the only way to fulfill these equations is $a=1,b=p-1$.
So, $n=(p+1)((p-1)p+1)=p^3+1$
Which contradicts $n=kp^2+1$ with $k<p$.
So, $n$ must indeed be prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ I want to find the power series expansion of $f(z)=\frac{1}{3-z}$ about the point $4i$ and to find the radius of convergence, what does this take?
Is this just a taylor series with $z=4i$ subbed in? What about finding radius of convergence?
| This is an expansion in powers of $(z-4i)$, so the first step is to locate $z-4i$ in this expression.
$$
\frac 1 {3-((z-4i)+4i)} = \frac 1 {(3-4i) - (z-4i)}.
$$
To get it into the form $\dfrac 1 {1-r}$ we need to divide the numerator and denominator by the constant term in the denominator, which is $3-4i$:
\begin{align}
& \frac 1 {(3-4i) - (z-4i)} = \frac {\frac 1 {3-4i}} {\frac{3-4i}{3-4i} - \frac{z-4i}{3-4i}} = \frac {\frac 1 {3-4i}} {1 - \frac{z-4i}{3-4i}} = \frac {3+4i} {25} \cdot\frac 1 {1-\frac{z-4i}{3-4i}} \\[10pt]
= {} & \frac{3+4i} {25} \cdot \frac 1 {1-r} = \frac{3+4i} {25} \left( 1+r + r^2 + r^3 + \cdots \right) \\[10pt]
= {} & \frac{3+4i} {25} \left( 1 + \frac1{3-4i}(z-4i) + \frac1{(3-4i)^2}(z-4i)^2 + \frac1{(3-4i)^3}(z-4i)^3 + \cdots \right) \\[10pt]
= {} & \frac{3+4i} {25} \left( 1 + \frac{3+4i} {25} (z-4i) + \left(\frac{3+4i} {25} \right)^2 (z-4i)^2 + \left(\frac{3+4i} {25} \right)^3 (z-4i)^3 + \cdots \right)
\end{align}
This converges when $|r|<1$, i.e. when $\left|\dfrac{z-4i}{3-4i}\right|<1$, i.e. when $|z-4i|<|3-4i|=5$. Thus the radius of convergence is $5$, and the center of the circle of convergence is $4i$. The number $3$ is exactly on the boundary of the disk of convergence. So are $-i$, $-3$, and $9i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
| The reciprocal is
$$\frac{x^2+xy+y^2}{(x+y)^2}=1-\frac{xy}{(x+y)^2} $$
and by the AMGM inequality, $\sqrt{xy}\le \frac{x+y}2$ with equality iff $x=y$, hence $\frac{xy}{(x+y)^2}\le \frac14$ with equality iff $x=y$ and from this $\frac{x^2+xy+y^2}{(x+y)^2}\ge \frac 34$ and finally $\frac{(x+y)^2}{x^2+xy+y^2}\le \frac 43$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
How to distribute $x \times \sqrt{x^2 + 1}$ How do I distribute the x in this problem?
How do I "gain access" so to say. Does it become $x^\frac{1}{2}(x^2 + 1)^\frac{1}{2}$ or $(x^3+x)^\frac{1}{2}$ ?
Or do I need to even do that in order to integrate $\int x(x^2+1)^\frac{1}{2}$ ?
| $x\sqrt{x^2+1} = \dfrac{x(x^2+1)}{\sqrt{x^2+1}}=(x^2+1)\sqrt{\dfrac{x^2}{x^2+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Boolean Algebra fundementals A disjunction A OR B truth table has A , B , and A OR B but mine has A ,B C, with A or B or C could some please explain this
| It's really all the same thing.
$$\begin{array}{c|c|cc}
A & B & A\vee B \\ \hline
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 & \star\end{array}
\qquad
\begin{array}{c|c|c|cc}
A & B & C & A\vee B\vee C \\ \hline
1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & \star\end{array}$$
etc. The truth value of $A\vee B\vee C$ is $1$ whenever any of the $A$, $B$ or $C$ is $1$. It is only zero if all of them are zero (in the rows marked $\star$). Replace $1$ with $T$ and $0$ with $F$ to your liking.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, the limit is
$$2\lim\limits_{x\to+\infty}\frac{x\sqrt{1-\frac{1}{x}}\times t}{x^n}$$
What to do next, if this is a good approach?
| Use this properties: $$\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\tag{1}$$
$$\lim_{x\to\infty}\left[f(x)\pm g(x)\right]=\lim_{x\to\infty}f(x)\pm \lim_{x\to\infty}g(x)\tag{2}$$
if and only if the limit of $f(x)$ and $g(x)$ exists
Therefore:
$$\begin{align}\lim_{x\to\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}&=\lim_{x\to\infty}\left(\frac{x+\sqrt{x^2-1}}{x}\right)^n+\lim_{x\to\infty}\left(\frac{x-\sqrt{x^2-1}}{x}\right)^n\\&=\lim_{x\to\infty}\left[\frac{x(1+\sqrt{1-\frac{1}{x^2}})}{x}\right]^n+\lim_{x\to\infty}\left[\frac{x(1-\sqrt{1-\frac{1}{x^2}}}{x}\right]^n\\&=\underbrace{\lim_{x\to\infty}\left(1+\sqrt{1-\frac{1}{x^2}}\right)^n}_{2^n}+\underbrace{\lim_{x\to\infty}\left(1-\sqrt{1-\frac{1}{x^2}}\right)^n}_{0}\\&=2^n\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1 = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red 4$ and $46 = 6\times 7 + \color \red 4$.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c \geq x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. $x$ is supposed to be the largest possible value that holds the assertion. Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?
| Answer for the second question.
We have
\begin{align}
a = k q_{1} + r_{1} \\
a = k q_{2} + r_{2} \\
a = k q_{3} + r_{3} \\
\end{align}
Where $a > b > c \geq k$.
Let suppose that $r_{1} = r_{2} = r_{3}$
We get $a - b = k (q_{1} - q_{2})$; $a - c = k (q_{1} - q_{3})$; and $b - c = k (q_{2} - q_{3})$.
We deduce that $k$ is a devider for $(a-b)$, $(a-c)$, and $(b-c)$. And the highest value for $k$ is the H.C.F of these three values.
Let prove now if $k$ is the H.C.F of $(a-b)$, $(a-c)$, and $(b-c)$, we would get $r_{1} = r_{2} = r_{3}$.
We have:
\begin{align}
a - b = k (q_{1} - q_{2}) + (r_{1} - r_{2}) \\
a - c = k (q_{1} - q_{3}) + (r_{2} - r_{3}) \\
b - c = k (q_{2} - q_{3}) + (r_{2} - r_{3}) \\
\end{align}
$k$ is a devider of $(a-b)$, $(a-c)$, and $(b-c)$, so $r_{1} - r_{2} = r_{2} - r_{3} = r_{1} - r_{3} = 0$ and consequently $r_{1} = r_{2} = r_{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 9
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Simple Absolute Value Inequality
$$|\frac{x-1}{x+3}|\leq 2$$
I solve it as follow:
$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{3}=-\frac{5}{3}\leq x\leq -7$
I now see that I did not take into consideration $x\neq -3$ which can not be
The book on the other side says that in answer is $x\leq -7$ or $x\geq -\frac{5}{3}$
Am I wrong?
| $\left|\dfrac{x-1}{x+3}\right| \leq 2 \iff |x-1| \leq 2|x+3| \iff (x-1)^2 \leq 4(x+3)^2 \iff x^2-2x+1 \leq 4(x^2+6x+9) \iff 3x^2 + 26x+35 \geq 0 \iff (x+7)(3x+5) \geq 0 \iff x \leq -7$ or $x \geq -\dfrac{5}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Closed form for the sum $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$ I want a closed form for the sum $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$
where: $a\ne 1<p<n;\quad p\in\mathbb Z$
I know a related identity,
$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1-\frac{p}{2}\right)\left\lfloor\frac{n}{p}\right\rfloor-\frac{p}{2}\left\lfloor\frac{n}{p}\right\rfloor^2\quad\;,$$
but I can't apply it to this problem!
Result for this sum:
$$\quad\displaystyle S=\dfrac{a^{n+1}}{a-1}\left\lfloor\frac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}$$
Can you prove it?
| Write $n=cp+r$, with $c,r\in\mathbb N\cup\{0\}$, and $r<p$. Then, for $k\leq n$,
$$
\left\lfloor\frac kp\right\rfloor=\begin{cases}0,&\ 0\leq k<p\\ 1,&\ p\leq k<2p,\\
\vdots\\
m,&\ mp\leq k<(m+1)p\\
\vdots\\
c,&\ cp\leq k\leq n
\end{cases}
$$
Then
$$
S=\sum_{k=0}^na^k\,\left\lfloor\frac kp\right\rfloor=\sum_{m=0}^{c-1}\sum_{k=mp}^{(m+1)p-1}ma^k\,+\sum_{k=cp}^{n}ca^k
=\sum_{m=1}^{c-1}m\,\sum_{k=mp}^{(m+1)p-1}a^k\,+\sum_{k=cp}^{n}ca^k\\
=\sum_{m=1}^{c-1}m\,\frac{a^{mp}-a^{(m+1)p}}{1-a}+c\,\frac{a^{cp}-a^{n+1}}{1-a}.
$$
Note that in the first sum, if we put together the negative term for index $m$ and the positive for $m+1$, we get
$$
-ma^{(m+1)p}+(m+1)a^{(m+1)p}=a^{(m+1)p}.
$$
So
$$
(1-a)S=\sum_{m=1}^ca^{mp}-ca^{n+1}=\frac{a^p-a^{(c+1)p}}{1-a^p}-ca^{n+1},
$$
or
$$
S=\frac{ca^{n+1}}{a-1}-\frac{a^{(c+1)p}-a^p}{(a^p-1)(a-1)}=\frac{ca^{n+1}}{a-1}-\frac{a^p(a^{pc}-1)}{(a^p-1)(a-1)}.
$$
Where
$$
c=\left\lfloor \frac np\right\rfloor.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?
| Considering that the answer is apparently
$$ \frac{7 \left(489857+280140 \sqrt{3}\right)}{8388608}, $$
I doubt there's an easy way.
You know $\sin{\frac{\pi}{6}}=\frac{1}{2}$, $\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}$, so using the half-angle formulae
$$ 2\sin^2{\frac{x}{2}} = 1-\cos{x},\\
2\cos^2{\frac{x}{2}} = 1+\cos{x} $$
will get us to $\sin{\frac{\pi}{24}}$ and $\cos{\frac{\pi}{24}}$, when applied twice.
In particular,
$$ 4\sin^4{\frac{x}{4}} = \left(1-\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}-2\cos{\frac{x}{2}} \\
4\cos^4{\frac{x}{4}} = \left(1+\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}+2\cos{\frac{x}{2}} $$
(At this point we see we have the nice identity
$$4(\sin^4{\theta}+\cos^4{\theta}) = 2(1+\cos^2{2\theta}) = 3+\cos{4\theta},$$
although sadly this is not of much help to us here.)
Therefore we need to raise everything to the power $6$, then divide by $4^6=2^{12}$. Set $a=\cos{(x/2)}$, then we have
$$ 4^6 ( \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} ) = (1-a)^{12}+(1+a)^{12} $$
At this point we break out the binomial theorem, and notice that quite a lot cancels: all the odd terms, in fact. Then
$$ \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} = \frac{1}{2^5} \left( 1 + a^6 + \binom{12}{2}(a^2 + a^{10}) + \binom{12}{4}(a^4+a^8) + \binom{12}{6}a^6 \right) = \frac{1}{2^{11}}(1+a^{12}+66(a^2+a^{10})+495(a^4+a^8)+924a^6) $$
Right, now the good news is that we don't need to know what $a$ is, just $a^2$. Indeed, the formula above gives
$$ \cos^2{\frac{\pi}{12}} = \frac{1}{2}\left(1+\cos{\frac{\pi}{6}}\right) = \frac{2+\sqrt{3}}{4}. $$
Therefore you just have to find $a^n$ for even $n$ up to $12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
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Recursive random draw Let $R(n)$ be a random draw of integers between $0$ and $n − 1$ (inclusive). I
repeatedly apply $R$, starting at $10^{100}$. What’s the expected number of repeated applications until I get zero?
| Just to fill in the algebra in the solution of BruceZ why $E(n) = E(n-1) + \frac{1}{n}, \:\:n > 1:$
Firstly, we re-arrange $E(n) = 1 + \frac{1}{n}\sum_{k=1}^{n-1}E(k)$ to $\sum_{k=1}^{n-1}E(k) = n \cdot [E(n)-1].$ (*)
We also have:
$E(n+1) = 1 + \frac{1}{n+1}\sum_{k=1}^{n}E(k) = 1 + \frac{1}{n+1} \sum_{k=1}^{n-1}E(k) + \frac{1}{n+1} E(n)$
After that we plug in (*) to get
$E(n+1) = 1 + \frac{1}{n+1}\sum_{k=1}^{n}E(k) = 1 + \frac{n}{n+1} [E(n)-1] + \frac{1}{n+1} E(n)$
and simplify:
$E(n+1) = 1 + \frac{(n+1) \cdot E(n) - n}{n+1} = 1 + E(n) - \frac{1}{n+1}$
to finally get
$E(n+1) = E(n) + \frac{1}{n+1}$
BTW, $E(n=10^{100}) = 100 \cdot \log_{10}{e} + \gamma = 230.2585 + 0.5772 = 230.8357$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317879",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Show that $\cos4x=8\sin^4x-8\sin^2x+1$ using De Moivre's theorem
Use De Moivre's theorem to show that $$\cos4x=8\sin^4x-8\sin^2x+1.$$
Hence show the one of the roots of the equation $8z^4-8z^2+1=0$ is $\sin\frac{\pi}{8}$ and express the other roots in polar form.
Deduce that $\sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and find an exact expression for $\sin \frac{11}{8}\pi$.
My attempt,
$\cos4x=Re(\cos4x+i\sin4x)$
$=Re(\cos x+i\sin x)^4$
$=Re(c^4+4c^3is+6c^2i^2s^2+4ci^3s^3+i^4s^4)$
$=c^4-6c^2s^2+s^4$
So, $\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$
$=(1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+\sin^4x$
$1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x$
$=8\sin^4x-8\sin^2x+1$
How to proceed then?
| let $\cos x =c $, $\sin x = s$
$\cos4x = (1-s^2)^2-6(1-ss)\cdot s^2+s^4$
$= 1 -2s^2 +s^4 - 6s^2 +6s^4 +s^4$
$= 8s^4 -8s^2 +1$
$\cos 4x = 0 = 8s^4 -8s^2 +1$
$\cos \frac{\pi}{2} =0$
$x= \frac{\pi}{8}$
so $\sin \frac{\pi}{8}$ is a solution
Look at the roots of $8s^4-8s^2 +1$ and use the quadratic equation to solve
we see that there is only one solution in range $(0,0.5)$ which is $\frac{\sqrt{(2-\sqrt{(2)})}}{2}$ so this has to be $\sin \frac{\pi}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\lim\limits_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}$
How to calculate
$$\lim_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}?$$
I know the result is $1/(6e)$.
| Re-write the logarithm of the original expression as follows
\begin{eqnarray*}
\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}} &=&\frac{1}{x^{2}}\ln \left(
1+\left( \frac{\sin x}{x}-1\right) \right) \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left(
\frac{\sin x}{x}-1\right) }\frac{\left( \frac{\sin x}{x}-1\right) }{x^{2}} \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left(
\frac{\sin x}{x}-1\right) }\frac{\left( \sin x-x\right) }{x^{3}}
\end{eqnarray*}
Now using the standard limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \ \lim_{u\rightarrow
0}\frac{\ln (1+u)}{u}=1,\ \ and\ \ \ \lim_{x\rightarrow 0}\frac{\sin x-x}{%
x^{3}}=-\frac{1}{6}.
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}}=1\cdot
\left( -\frac{1}{6}\right) =-\frac{1}{6}.
\end{equation*}
Taking back the exponential one gets
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right) ^{1/x^{2}}=e^{-1/6}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322993",
"timestamp": "2023-03-29T00:00:00",
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Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2=1}$ and origin $(1,-i)$? Is $|z+i| = |z-1|$ a circle with radius $\sqrt{1^2+1^2}=1$ and origin $(1,-i)$?
Because I know $|z+i| = 3$ is is a circle with radius $3$ and origin $(0,i)$.
| For finding out the locus of point $z$, let's assume, $z=x+iy$ then substitute it in the relation given as follows
$$|z+i|=|z-1| $$$$ \implies |x+iy+i|=|x+iy-1| $$$$ \implies \sqrt{x^2+(y+1)^2}=\sqrt{(x-1)^2+y^2}$$ $$\implies x^2+(y+1)^2=(x-1)^2+y^2$$
$$\implies x^2+y^2+2y+1=x^2-2x+1+y^2$$ $$\implies \color{blue}{y=-x} \quad \text{or} \quad \color{blue}{x+y=0}$$
It is obvious that the locus of the point $z$: $\color{blue}{|z+i|=|z-1|}$ is a straight line: $\color{blue}{x+y=0}$ passing through the origin having negative slope i.e. making an angle $135^o$ with positive real axis in the $\color{blue}{\text{complex plane}}$.
Hence, $\color{blue}{|z+i|=|z-1|}$ represents a $\color{blue}{\text{straight line}}$ $\color{red}{\text{not a circle}}$.
| {
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Finding the locus of midpoint of $AB$ The normal to the ellipse $b^2x^2+a^2y^2=a^2b^2$ is passing through the x-axis in point $A$ and through the Y-axis in point $B$.
Point $P$ is the midpoint of $AB$.
Need to find the locus of $P$.
My attempt :I tried to use the fact of the tangent to the ellipse $n^2=a^2m^2+b^2$ and to use slopes but I didn't succeed getting the locus.
Thanks.
| i will parametrize the ellipse by $$x = a \cos t, y = b \sin t $$ the slope of the tangent is $$\frac{dy}{dx} = -\frac{b\cos t}{a \sin t} $$ therefore the slope of the normal is $$\frac{a\sin t}{b \cos t}.$$ the equation of the normal line is $$(y-b\sin t)b \cos t = a \sin t(x -a \cos t) $$ the $x$-intercept is given by $$a \sin t(x - a \cos t) = -b^2\sin t \cos t \to x = \frac 1a(a^2-b^2)\cos t.$$ in the same way the $y$-intercept is $$y = \frac1b(b^2-a^2)\sin t.$$ therefore the midpoint is $$x = \frac1{2a}(a^2 - b^2)\cos t,\\y = \frac1{2b}(b^2 - a^2)\sin t $$ you can eliminate $t,$ and find that locus of the center of the two intercepts is the ellipse $$4a^2x^2 + 4b^2y^2 = (a^2-b^2)^2.$$
| {
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Summation of product of $m+1$ alternate numbers We know that
$$\begin{align}
\sum_{r=1}^n \prod_{j=0}^m (r+j)&=\sum_{r=1}^n r(r+1)(r+2)\cdots(r+m)\\
&=(m+1)!\sum_{r=1}^n\binom {r+m}{m+1}\\
&=(m+1)!\binom {n+m+1}{m+2}\\
&=\frac {(n+m+1)^\underline{m+2}}{m+2}\end{align}$$
but is there a similar closed form for
$$\sum_{r=1}^n \prod_{j=0}^m (r+2j)=\sum_{r=1}^n r(r+2)(r+4)\cdots (r+2m)$$
?
Edit 1
Here's a summary of results from wolframlpha for the first few values of $m$ (please excuse the messy alignment).
From the summary we observe some interesting patterns as follows:
*
*all results have factors $n,(n+1)$
*all results have a factor of $\frac 1{2(m+2)}$ or $\frac 1{m+2}$ where the product of the other factors results in a $2n^{m+2}$ term or $n^{m+2}$ term respectively
*for $m=2$: $(n+4),(n+5)$ are factors
*for $m=4$: $(n+8),(n+9)$ are factors
*for $m=6$: $(n+12),(n+13)$ are factors
*for $m=8$: $(n+16),(n+17)$ are factors
*this implies that for even $m$: $(n+2m), (n+2m+1)$ are factors
Perhaps the observations above might be helpful in working out the solution.
The solution could then possibly be of the form:
$$\begin{cases}
\dfrac {n(n+1)(n+2m)(n+2m+1)\cdot f(n,m)}{2(m+2)}\qquad \text{for even $m$}\\
\dfrac {n(n+1)\cdot g(n,m)}{2(m+2)}\qquad \qquad\qquad\qquad\qquad\text {for odd $m$}
\end{cases}$$
| Simply for my result, I have:
$$\sum_{r=1}^n\prod_{j=0}^m(r+2j)=S(n,m)=\frac{1}{2m+4}\left(\prod_{k=0}^m(2k+1)+\prod_{k=0}^{m+1}(n+2k)+\prod_{k=0}^{m+1}(n+2k-1)\right)$$
That's a polynomial of $n$ with $m+2$ degrees.
The first few values of $m$
\begin{align*}S(n,1)&=\frac{1}{6}\left(\prod_{k=0}^1(2k+1)+\prod_{k=0}^2(n+2k)+\prod_{k=0}^2(n+2k-1)\right) \\ &=\frac{1}{6}\bigg(1.3+n(n+2)(n+4)+(n-1)(n+1)(n+3)\bigg)\\ &=\frac{1}{6}n(n+1)(2n+7)\end{align*}
\begin{align*}S(n,2)&=\frac{1}{8}\bigg(1.3.5+n(n+2)(n+4)(n+6)+(n-1)(n+1)(n+3)(n+5)\bigg) \\ &=\frac{1}{8}\cdot2n(n+1)(n+4)(n+5)\end{align*}
etc..,
I think so don't closed form for this sum, It is a similarity with the sum $\displaystyle \sum_{k=1}^n k^r$
| {
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Evaluating the Sum of $ \frac {1} {3}+\frac {1} {15}+\frac {1} {35}+\ldots +\frac {1} {4n^{2}-1} $ How would you evaluate the sum of this sequence ?
$$
\dfrac {1} {3}+\dfrac {1} {15}+\dfrac {1} {35}+\ldots +\dfrac {1} {4n^{2}-1}
$$
I realise the expression can be factorised but I can't really see what this can tell you.
| $$\sum_{n=1}^{\infty }\frac{1}{4n^2-1}=\sum_{n=1}^{\infty }\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$$
$$=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....\right]=\frac{1}{2}$$
| {
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Evaluating $\lim\limits_{x\to\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)$ I could transform it to
$$\lim_{x\to\infty}x\left(\dfrac{1}{e}-\left(1+\dfrac{1}{x}\right)^{-x}\right).$$
I think using sandwich theorem would help but I was unable to find the bounds.
| Another way based on Taylor expansions : consider $$A=\left(\dfrac{x}{x+1}\right)^x$$ Taking logarithms $$\log(A)=x\log\left(\dfrac{x}{x+1}\right)=-x\log\left(\dfrac{x+1}{x}\right)=-x\log\left(1+\dfrac{1}{x}\right)$$ Now, remember that, for small $y$, $$\log(1+y)=\frac {y}{1}-\frac {y^2}{2}+\frac {y^3}{3}+\cdots$$ Replace $y$ by $\frac {1}{x}$ which makes $$\log(A)=x\left(\frac {1}{x}-\frac {1}{2x^2}+\frac {1}{3x^3}+\cdots\right)=1-\frac {1}{2x}+\frac {1}{3x^2}+\cdots$$ So $$A=e \,e^{-\frac {1}{2x}+\frac {1}{3x^2}+\cdots}$$ Now, remember that, for small $y$, $$e^y=1+\frac{y}{1}+ \frac{y^2 }{2}+\cdots$$ Replace $y$ by $-(\frac {1}{2x}-\frac {1}{3x^2})$ and you should arrive to $$A=\frac{1}{e}+\frac{1}{2 e x}-\frac{5}{24 e
x^2}+\cdots$$
I am sure that you can take from here (and show not only the limit but also how it is approached).
| {
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$\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ For any natural number $m$,how to evaluate $\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ ?
| Let $$\displaystyle I = \int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx\;,$$ Where $x>0$
Now Let $x^{m}=t\;,$ Then $$\displaystyle mx^{m-1}dx = dt\Rightarrow dx = \frac{dt}{mx^{m-1}}=x\frac{dt}{mt} = \frac{t^{\frac{1}{m}}}{mt}dt$$
So $$\displaystyle I = \frac{1}{m}\int (t^3+t^2+t)\cdot (2t^2+3t+6)^{\frac{1}{m}}\cdot \frac{t^{\frac{1}{m}}}{t}dt$$
So $$\displaystyle I = \frac{1}{m}\int (t^2+t+1)\cdot (2t^3+3t^2+6t)^{\frac{1}{m}}dt$$
So Let $(2t^3+3t^2+6t)=u\;,$ Then $\displaystyle (t^2+t+1)dt = \frac{du}{6}$
So $$\displaystyle I = \frac{1}{6m}\int u^{\frac{1}{m}}du = \frac{1}{6m}\cdot \frac{u^{\frac{1}{m}+1}}{\frac{1}{m}+1}+\mathcal{C} = \frac{1}{6}\cdot \frac{(2t^3+3t^2+6t)^{\frac{m+1}{m}}}{m+1}+\mathcal{C}$$
So $$\displaystyle I = \frac{1}{6}\cdot \frac{(2x^{3m}+3x^{2m}+6x^{m})^{\frac{m+1}{m}}}{m+1}+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Resolve this system: Im tried to resolve this problem:
$$\max\quad f\left( x,y \right) =xy\quad \text{s.a}\quad \begin{cases} x^2 +y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
Well, i form the lagrangian and the respective gradient, so i had this system to resolve:
$$\begin{cases} y+2\lambda_1 x+\lambda_2=0 \\ x+2\lambda_1 y+ \lambda_2=0 \\ x^2+y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
And i can't find all the solutions, i need help-
| W/O using calculus,
Let $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$
$\implies \cos A\cos B+\cos A\sin B+\sin A=0\ \ \ \ (1)$
Method $\#1:$
$\iff\dfrac{-\sin A}{\cos B+\sin B}=\dfrac{\cos A}1=\pm\sqrt{\dfrac1{(\cos B+\sin B)^2+1}}$
$\implies\cos^2A=\dfrac1{2+2\cos B\sin B}$
Method $\#2:$
By $(1),\tan A=-(\cos A+\cos B)$
$\cos^2A=\dfrac1{1+\tan^2B}=\dfrac1{1+(\cos A+\cos B)^2}=\dfrac1{2+2\cos B\sin B}$
By anyone of the methods, we need to maximize $xy=\cos^2A\cos B\sin B=\dfrac{\cos B\sin B}{2+2\cos B\sin B}$
$2xy=\dfrac{\sin2B}{2+\sin2B}=1-\dfrac2{2+\sin2B}$
We need to maximize $\sin2B$ which is $\le1$
| {
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Prove the inequality $4+xy+yz+zx \ge 7xyz$ Let $x,y,z$ be non negative real numbers such that $x+y+z=3$, prove the following inequality:
$$4+xy+yz+zx \ge 7xyz$$
I tried MV and taking out one variable but I got nothing
| Note that because $x+y+z=3$ we have $1\geq xyz$ and $4\geq 4xyz$.
In addition, $xy+yz+xz \geq 3(xyz)^\frac{2}{3}$ but because $1\geq xyz$ we have $1\geq xyz^\frac{1}{3}$ and $3(xyz)^\frac{2}{3}\geq3xyz$. Hence, $xy+yz+xz \geq 3xyz$.
Adding the two inequalities we have $4+xy+yz+xz \geq 7xyz$.
| {
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Finding a (nonidentity) rational map of the plane with period $7$
Does there exist a nonidentity (which also is not a rotation) rational map $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with period $7$, i.e., for which the seventh iteration $f^7$ is the identity map?
I know such a function, $f:(x,y)\mapsto \left(\frac{1}{y},x(1+y)\right)$, with period $5$. Here is a verification done in Maple:
f := (x, y) ->( 1/y, x*(1+y));
(f@@5)(x, y);
$$\left( {\frac {1+x \left( 1+y \right) }{y} \left( 1+{\frac {1}{x \left( 1+y
\right) } \left( 1+{\frac {1+x \left( 1+y \right) }{y}} \right) }
\right) ^{-1}}
\right. ,$$ $$\left.{x \left( 1+y \right) \left( 1+{\frac {y}{1+x \left( 1+y \right) }
\left( 1+{\frac {1}{x \left( 1+y \right) } \left( 1+{\frac {1+x
\left( 1+y \right) }{y}} \right) } \right) } \right) \left( 1+{
\frac {1+x \left( 1+y \right) }{y}} \right) ^{-1}}\right)
$$
simplify((f@@5)(x, y)[1]),simplify((f@@5)(x, y)[2]);
$$ x,\,y$$
| The $2 \times 2$ rotation matrix for an (anticlockwise) rotation by $\frac{\pi}{7}$ is
$$\pmatrix{\cos \frac{\pi}{7} & -\sin \frac{\pi}{7} \\ \sin \frac{\pi}{7} & \cos \frac{\pi}{7}}.$$ Via the usual identification of that group with the group of Mobius transformations (and clearing the cosine factors for readability), the image of that matrix in $PSL(2, \Bbb R) = SL(2, \Bbb R) / \{\pm I\}$ under the canonical projection map can be identified with the rational function
$$f_7 : t \mapsto \frac{t - \alpha}{1 + \alpha t},$$
where $$\alpha := \tan \frac{\pi}{7}.$$
Clearly $f_7 \neq \text{id}$ and via the above identification (or by repeated application of the tangent sum identity) we see that $f_7^7 = \text{id}$.
So, the rational transformation
$$g_7 : (x, y) \mapsto (f_7(x), y)$$
is not the identity but satisfies $g_7^7 = \text{id}$.
We can produce a natural example that involves $y$ in a more essential way by replacing $t$ in the definition of $f_7$ with a complex variable $z = x + iy$ and decomposing $f_7$ into real and imaginary parts. This yields the rational map
$$h_7 : (x, y) \mapsto \left( \frac{\alpha x^2 + (1 - \alpha^2) x - \alpha + \alpha y^2}{(1 + \alpha x)^2 + \alpha^2 y^2}, \frac{(1 + \alpha^2) y}{(1 + \alpha x)^2 + \alpha^2 y^2} \right)$$
of period $7$, which can be rewritten in various ways.
Note that this construction is not specific to the number $7$. Replacing it, for example, with $3$ and using the special value $\tan \frac{\pi}{3} = \sqrt{3}$ gives the analogous maps
$$f_3 : x \mapsto \frac{x - \sqrt{3}}{1 + \sqrt{3} x}$$
and
$$h_3 : (x, y) \mapsto \left( \frac{\sqrt{3} x^2 - 2 x + \sqrt{3} y^2 - \sqrt{3}}{3x^2 + 2 \sqrt{3} x + 3 y^2 + 1} , \frac{4 y}{3x^2 + 2 \sqrt{3} x + 3 y^2 + 1} \right)$$
of period $3$.
| {
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Solve in positive integers: $5^x 7^y +4=3^z$ Solve in positive integers: $5^x 7^y +4=3^z$.
I tried to solve it with log but I couldn't complete.
| [Based on the work from @Adam Hughes]
We assume $x>1$ and rewrite it:
$$3^z-4=5^x7^y \Rightarrow
3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$
so $z=2j$ is even.
Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j+2)-(3^j-2)=4$--we have that they are coprime (any larger factor would then be a factor of $5^x7^y$ which is odd), i.e. one of the following two cases hold
$$\begin{cases} 3^j+2=5^x\quad ,\quad 3^j-2=7^y \\ 3^j+2=7^y\quad ,\quad3^j-2=5^x\end{cases}.$$
Case 2:
$3^j+2=7^y\quad ,\quad3^j-2=5^x \Rightarrow 7^y = 4 + 5^x$, which is impossible if you look at it $\mod 3$.
So we have Case 1:
$3^j-2=7^y\quad ,\quad3^j+2=5^x \Rightarrow 7^y+4 = 5^x$. If you look at it $\mod 4$, you get $2|y$, so we write $y=2 y_2$
We look at it $\mod 25$: this gives us
$7^y + 4 \equiv 49^{y_2} +4 \equiv (-1)^{y_2} + 4 \equiv 5^x \equiv 0$, which is impossible.
it remains to look at x=1, here we have the only solution $y=0,x=1,z=2$ (you can ignore this if you only want strictly positive solutions)
| {
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Series involving complex roots: $\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = \frac{(n-2)2^{n-1}+1}{2^n - 1}$ $$
\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} =
\frac{(n-2)2^{n-1}+1}{2^n - 1}
$$
Here $1,a_1,a_2,\dots,a_{n-1}$ are $n$-th roots of unity
I know the sum of roots is 0. I think the series should be some sort of telescopic.
| Let $y=\dfrac1{2-x}\iff x=\dfrac{2y-1}y$ where $x^n=1$
$\implies\left(\dfrac{2y-1}y\right)^n=1$
$\displaystyle\iff y^n(2^n-1)-\binom n1 2^{n-1}y^{n-1}+\cdots+(-1)^n=0$
Using Vieta's formula, $$\displaystyle\sum_{r=0}^{n-1}y_r=\dfrac{\binom n1 2^{n-1}}{2^n-1}=\dfrac{n 2^{n-1}}{2^n-1}$$
If $x=1,y=1=y_0$(say)
$\displaystyle\implies\sum_{r=1}^{n-1}y_r=\dfrac{n 2^{n-1}}{2^n-1}-y_0=\dfrac{n 2^{n-1}}{2^n-1}-1=\cdots$
| {
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Find the fourier series of the function Find the fourier series of the function
$g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)}$ for $x \not= k\pi$, and $g(k\pi) = \lim_{x\to k\pi} g(x)$, $(k \in \mathbb{Z})$
| $g(x)=\sum\limits_{n=1}^\infty \frac{\alpha^n}{sin(x)}(sin(n-2)x cos2x + cos(n-2)x sin2x)$
$ = \sum\limits_{n=1}^\infty \alpha^n\frac{sin(n-2)x}{sin(x)} + 2\sum\limits_{n=1}^\infty \alpha^n cos(n-1)x$
$= -\alpha + \alpha^2\sum\limits_{n=1}^\infty\alpha^n \frac{sin(nx)}{sinx} +
2\sum\limits_{n=1}^\infty\alpha^n cos(n-1)x$
where we can find $g(x)$ to be:
$g(x) = \frac{\alpha}{\alpha^2-1} + 2\sum\limits_{n=1}^\infty \frac{\alpha^n}{1-\alpha^2}cos(n-1)x$
$ = \frac{\alpha}{1-\alpha^2} + 2\alpha\sum\limits_{n=1}^\infty\frac{\alpha^n}{1-\alpha^2} cos(nx)$
now in our case $\alpha = \frac{1}{6}$
| {
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What is the value of $x*y$?
Given that $$\left(\frac{x}{y}\right)^{-2} + \left(\frac{y}{x}\right)^{-2} = \frac{10}{3}$$
find the value of $x*y$.
My question is, can we calculate the value of $x*y$ or not? If yes, then how? If not, then why?
| Let $u = (\frac{x}{y})^2$.
Then, your equation becomes: $\frac{1}{u} + u = \frac{10}{3}$.
Multiplying both sides by u, we get:
$1 + u^2 = \frac{10}{3} \cdot u$
Now, we have a quadratic equation to solve. Subtract $\frac{10}{3} \cdot u$ from both sides:
$1 + u^2 - \frac{10}{3} \cdot u = 0$
For look's sake, let's put our new polynomial in standard form:
$u^2 - \frac{10}{3} \cdot u + 1 = 0$
Now, solve for u. Notice that there are two solutions. So, because we don't have any indication of which u to use, we can only say that y^2 is either approximately $\frac{x}{1.7676} $ or approximately $\frac{x}{0.5657}.$ Therefore, we cannot calculate xy.
| {
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If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$? What I know is that for equations of type $x+y=8$, $xy$ attains its maximum value when $x=y$ and this can be proved by either solving the quadratic equation with completing the squares or finding the first and second derivatives of $xy$ w.r.t. $x$ to perform maxima/minima test. But both of these methods are applicable only when there is not more than one independent variable -- here $x$ is independent variable and $y=8-x$ is the dependent one.
I don't understand why $(a+1)(b+1)(c+1)(d+1)$ attains its maximum value at $a=b=c=d$ within the condition of $a+b+c+d=1$. How can we prove this fact? In some other similar posts I saw others using the Lagrange multiplier method. This method is too advanced for me to understand. I'm looking for some kind of algebraic proof similar to comleting with sqaure proof as in case of finding the maximum value of $xy$.
| You known AM-GM inequality?
$$(a+1)(b+1)(c+1)(d+1)\le\left(\dfrac{a+b+c+d+4}{4}\right)^4$$
if you don't known AM-GM inequality,you also can following
In fact ,if $x,y,z,w\in R^{+}$,then we have
$$x^4+y^4+z^4+w^4-4xyzw\ge 0$$
because
$$x^4+y^4+z^4+w^4-4xyzw=(x^2-y^2)^2+(z^2-w^2)+2(xy-zw)^2$$
so $=$ iff $x=y=z=w$
so
$$4xyzw\le \le x^4+y^4+z^4+w^4$$
let
$$\sqrt[4]{a+1}=x,\sqrt[4]{b+1}=y,\sqrt[4]{c+1}=z,\sqrt[4]{d+1}=w$$
so
$$4\sqrt[4]{(a+1)(b+1)(c+1)(d+1)}\le a+1+b+1+c+1+d+1=5$$
so
$$(a+1)(b+1)(c+1)(d+1)\le \left(\dfrac{5}{4}\right)^4$$
| {
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"url": "https://math.stackexchange.com/questions/1343435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3$ for $x,y,z>0$ By considering that $$\frac{x}{y}+\frac{y}{x} \geq 2$$I can show that $$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y} \geq 6$$
But how would one go from here to prove the required result? It feels like I'm almost there but I can't quite see how to finish it.
I can see that it can be killed off almost instantly using AM/GM of degree 3, but how would one acquire this result without the use of such a tool?
| By AM-GM inequality, since $x,y,z>0$:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3\sqrt[3]{\frac{x}{y}\frac{y}{z}\frac{z}{x}}=3$$
with equality iff $\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$, i.e. $x=y=z$.
Alternatively, rearrangement inequality. $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ is cyclic - two cases:
*
*wlog $x\ge y\ge z$. Then $\frac{1}{z}\ge\frac{1}{y}\ge \frac{1}{x}$, so (RHS oppositely sorted):
$$x\frac{1}{y}+y\frac{1}{z}+z\frac{1}{x}\ge x\frac{1}{x}+y\frac{1}{y}+z\frac{1}{z}=3$$
*wlog $x\ge z\ge y$. Then $\frac{1}{y}\ge\frac{1}{z}\ge\frac{1}{x}$, so (RHS oppositely sorted):
$$x\frac{1}{y}+z\frac{1}{x}+y\frac{1}{z}\ge x\frac{1}{x}+z\frac{1}{z}+y\frac{1}{y}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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system of modular equations. $x\equiv 2\pmod3$
$x\equiv 3\pmod 5$
$x\equiv 7 \pmod{11}$
How can I solve this system for $x$? I've tried all kinds of things using divisibility but no success. Any hints of solutions are greatly appreciated. What value of $x$ satisfies these three equations?
| When I create new variables, they're integers.
$x\equiv 2\,\pmod{\! 3}\iff x=3k+2$.
$x\equiv 3\,\pmod{\! 5}\iff 3k+2\equiv 3\,\pmod{\! 5}$
$\iff 3k\equiv 1\equiv 6\stackrel{:3}\iff k\equiv 2\,\pmod{\! 5}$.
I.e. $k=5m+2$, then $x=3(5m+2)+2=15m+8$.
Here we could've solved $3k\equiv 1\pmod{\! 5}$ by using Extended Euclidean algorithm (as below) to find $k_1,b\in\Bbb Z$ such that $3k_1+5b=1$, in which case $3k_1\equiv 1\pmod{\! 5}$.
EEA can be used in a table (as explained here), where e.g. $5=5(1)+3(0)$. We subtract consecutive rows:
$$\begin{array}{l|c|r}5& 5(1)& 3(0)\\\hline 3& 5(0)& 3(1)\\\hline 2& 5(1)& 3(-1)\\\hline 1& 5(-1)& 3(2)\end{array}$$
$5(-1)+3(2)=1$ implies $3(2)\equiv 1\pmod{\! 5}$, i.e. $k\equiv 2\pmod{\! 5}$.
Gauss' algorithm can also be used:
$3k\equiv 1\stackrel{\cdot 2}\iff 6k\equiv 2\iff k\equiv 2\,\pmod{\! 5}$
$x\equiv 7\,\pmod{\! 11}\iff 15m+8\equiv 7\,\pmod{\! 11}$
$\iff 15m\equiv 4m\equiv -1\equiv -12\stackrel{:4}\iff m\equiv -3\equiv 8\,\pmod{\! 11}$
I.e. $m=11a+8$, then $x=15(11a+8)+8=165a+128$.
In the same way we could've used Extended Euclidean or Gauss' algorithm to solve $4m\equiv -1\,\pmod{\! 11}$, like above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ From the post Evaluating limit (iterated sine function) and some discussions inside, one can collect the
following three limits
\begin{eqnarray*}
\lim_{x\rightarrow 0 }\frac{\sin _{n}x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{x-\sin _{n}x}{x^{3}} &=&\frac{n}{6} \\
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}}{x^{5}} &=&%
\frac{n^{2}}{24}-\frac{n}{30}.
\end{eqnarray*}
where $\sin _{n}x=\sin (\sin \cdots (\sin x)),\ n$ times composition.
So the next question would be, what is the following limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{%
n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}
\end{equation*}
and what are those corresponding limits after order 7 $?$
| Continuing in the same spirit as in my answer to the first question, we can probly admit that the limit could be expressed as a cubic polynomial in $n$. Going on with the expansions (tiring !!), I arrived to $${\sin _{n}(x)-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}=-\Big(\frac{5 n^3}{432}-\frac{n^2}{45}+\frac{41 n}{3780}\Big)x^7+\cdots$$ which gives for the fourth limit $$L_4=-\frac{5 n^3}{432}+\frac{n^2}{45}-\frac{41 n}{3780}$$ Continuing with the next limit (very tiring !!) $$L_5=\frac{35 n^4}{10368}-\frac{71 n^3}{6480}+\frac{67 n^2}{5670}-\frac{4 n}{945}$$
At this point, I give up (hoping that I did not make any mistake with all these developments).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A limit problem: $\lim\limits_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$ I need help in solving the limit below:
$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n} }{1+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n} }$$
What I've done is to simplify the upper part to: $$\frac{2^{n+1}-1}{2^n}$$
Any hints or solutions will be greatly appreciated.
| Hint: Simplify the denominator to
$$\frac{\frac{1}{3^{n+1}}-1}{\frac{1}{3}-1}=\frac{3^{n+1}-1}{2\cdot 3^n}$$
| {
"language": "en",
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"answer_count": 5,
"answer_id": 2
} |
What am I doing wrong? - Change of basis matrix Problem: Let $\alpha$ be the standard basis of $\mathbb{R}^3$ and let $\beta = \left\{(1,0,0), (1,1,0), (1,1,1)\right\}$ be another basis. Consider the linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by \begin{align*} T(x,y,z) = (x+2y+z, -y, x+4z). \end{align*}
Find the change of basis matrix from $\alpha$ to $\beta$ and the change of basis matrix from $\beta$ to $\alpha$.
Attempt at solution: For the change of basis matrix from $\alpha$ to $\beta$ I did: $T(1,0,0) = (1,0,1), T(1,1,0) = (3,-1,1), T(1,1,1) = (4,-1,5)$. Then I wrote this in the columns of a matrix: \begin{align*} P = \begin{pmatrix} 1 & 3 & 4 \\ 0 & -1 & -1 \\ 1 & 1 & 5 \end{pmatrix}. \end{align*} So this would be the change of basis matrix from $\alpha$ to $\beta$.
Now, I wanted to check if it was correct by verifying a theorem I recall: "Let $P$ be the change of basis matrix from $\alpha$ to $\beta$. Then for each vector $v$ we have \begin{align*} &(i) P[v]_{\beta} = [v]_{\alpha} \\ & (ii) P^{-1} [v]_{\alpha} = [v]_{\beta},\end{align*} where $[v]_{\beta}$ represents the coordinates (written as a column vector) of an arbitrary vector with respect to the basis $\beta$.
So I picked an arbitry vector $v = (10,8,3) = 2(1,0,0) + 5(1,1,0) + 3(1,1,1)$. Hence its coordinates with respect to $\beta$ are \begin{align*} [v]_{\beta} = \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix}. \end{align*} So I did \begin{align*} \begin{pmatrix} 1 & 3 & 4 \\ 0 & -1 & -1 \\ 1 & 1 & 5 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 29 \\ -8 \\ 10 \end{pmatrix}. \end{align*} But $(v) = (10,8,3) \neq 29(1,0,0) -8 (0,1,0) + 10 (0,0,1)$ ?? So now I'm confused and I don't know what I did wrong. Does the theorem hold only under special conditions, or is my change of basis matrix wrong?
| I'm not sure where exactly $T$ plays a role in this problem. To find $P$, write the vectors in $\alpha$ in terms of $\beta$
$$
\alpha_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}_\beta,
\alpha_2 = \begin{pmatrix}-1\\1\\0\end{pmatrix}_\beta,
\alpha_3 = \begin{pmatrix}0\\-1\\1\end{pmatrix}_\beta
$$
and so
$$
P = \begin{pmatrix}1&-1&0\\0&1&-1\\0&0&1\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.
| Instead of using power series, which feels like a way of hiding the use of derivatives and, ultimately, L'Hôpital, we can use a trigonometric identity and take limits.
Since
$$
\frac{4\sin(x)-2\sin(2x)}{2\sin(2x)-\sin(4x)}
=\frac1{2(1+\cos(x))\cos(x)}\tag{1}
$$
we have
$$
\lim_{x\to0}\frac{2^{k+2}\sin\left(\frac{x}{2^{k+2}}\right)-2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)}{2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)-2^k\sin\left(\frac{x}{2^k}\right)}
=\frac14\tag{2}
$$
Taking the product of $(2)$ from $k=0$ to $k=n-1$, we get
$$
\lim_{x\to0}\frac{2^{n+1}\sin\left(\frac{x}{2^{n+1}}\right)-2^n\sin\left(\frac{x}{2^n}\right)}{2\sin\left(\frac x2\right)-\sin\left(x\right)}
=\frac1{4^n}\tag{3}
$$
Summing $(3)$ from $n=0$ to $n=\infty$, we get
$$
\lim_{x\to0}\frac{x-\sin(x)}{2\sin\left(\frac x2\right)-\sin\left(x\right)}=\frac43\tag{4}
$$
Applying $(1)$ gives
$$
\begin{align}
\lim_{x\to0}\frac{2\sin\left(\frac x2\right)-\sin\left(x\right)}{2\sin(x)-\sin(2x)}
&=\lim_{x\to0}\frac12\frac{4\sin\left(\frac x2\right)-2\sin\left(x\right)}{2\sin(x)-\sin(2x)}\\
&=\frac18\tag{5}
\end{align}
$$
Multiplying $(4)$ and $(5)$, and taking the reciprocal, yields
$$
\begin{align}
\lim_{x\to0}\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}=6\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
$
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
| Case $x>3$ :
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \geq 2x(x-3) \Rightarrow x^2-x-12 \geq 2x^2-6x \\ \Rightarrow x^2-5x+12 \leq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So we reject this case. } $
Case $x<3$ :
*
*Subcase $0<x<3$ :
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \leq 2x(x-3) \Rightarrow x^2-x-12 \leq 2x^2-6x \\ \Rightarrow x^2-5x+12 \geq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So } \checkmark $
*Subcase $x<0$ :
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2+x-12 \leq 2x(x-3) \Rightarrow x^2+x-12 \leq 2x^2-6x \\ \Rightarrow x^2-7x+12 \geq 0 $$ $\Delta=49-48=1>0 \\ x_{1,2}=\frac{7 \pm 1}{2}= 3 \text{ or } 4 $
For $x \in (-\infty, 3] \cup [4, +\infty)$ we have that $x^2-7x+12 \geq 0$ but since we have that $x<0$ we conclude that in the case $x<0$ $x^2-7x+12 \geq 0$.
In conclusion, it stands that $\frac{x^2-|x|-12}{x-3} \geq 2x$ for $x<3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2}
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$?
| Set $x=1/h$ to get $$\lim_{h\to0^+}\dfrac{\sqrt{1+h}(h-\sqrt{2+3h})}{7h^2-6h+4}=\dfrac{\sqrt{1+0}(0-\sqrt{2+3\cdot0})}{7\cdot0^2-6\cdot0+4}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why am I getting two different answers for this diff. equation? Completing the square vs quad form. With this differential equation, after seperating and integrating, and using the initial condition to solve for C, and then substituting that value of C into the general solution, I must solve for Y to get the particular solution. Why am I getting two different solutions, where in method 1 I complete the square, and method 2, use the quadratic formula?
The ODE is:
$\frac{dy}{dx} = \frac{2x}{1+2y}$
The initial condition is:
$y(2) = 0$
Separating and integrating, I get the general soln.:
$y + y^2 = x^2 + c$
Plugging in the initial condition:
$0 + 0^2 = (2)^2 + c$
$c = -4$
Now, plugging in (-4) into the general solution:
$y + y^2 = x^2 - 4$
Now, method 1, completing the square:
$y^2 + y = x^2 - 4$
$(y + \frac{1}{2})^2 = x^2 - 4 + \frac{1}{4}$
$y = -\frac{1}{2} ^+_- (4x^2 - 15)^{\frac{1}{2}}$
Now, method 2, quad. formula:
$y = \frac{-1 ^+_- (1^2 - 4(1)(-x^2+4))^{\frac{1}{2}}}{2}$
$y = -\frac{1}{2} ^+_- \frac{1}{2}(4x^2-15)^{\frac{1}{2}}$
So, what did I do wrong with completing the square?
The quad. formula is the same answer from the back of the
book. There is a 1/2 in front of the square root, whereas with
completing the square, the 1/2 is not there. What did I do wrong?
| $$\left( y+ \frac{1}{2}\right)^2= x^2-\frac{15}{4} \Rightarrow y+ \frac{1}{2}= \pm \sqrt{\left( x^2-\frac{15}{4}\right) }\Rightarrow y=-\frac{1}{2} \pm \sqrt{\left( x^2-\frac{15}{4}\right)} \\=-\frac{1}{2} \pm \sqrt{ \frac{1}{4}\left( 4x^2-15\right)}= -\frac{1}{2} \pm \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$$
EDIT: But since only $y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$ satisfies the initial condition $y(2)=0$, we reject $y(x)= -\frac{1}{2} - \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$ and so our only solution is:
$$y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative of the matrix exponential with respect to its matrix argument I was trying to find the Frechet derivative of $f = \exp(X)$, where $X \in \mathbb{R}^{n\times n}$ is positive definite. I thought it ought to be $\exp(X)$.
I see results where the derivative is with respect to a scalar argument, but this question has not been asked before.
I tried to see if I could find $Df_X$ starting with
$Df_X[H] = \exp(X+H) - \exp(X)$. If I can show that the right hand side evaluates to $I + XH + X^2H/2 + \ldots = \exp(X)H$, I am done.
After I use the series definition, however, I am lost because I see no reason to assume that $A$ and $H$ commute.
Please help.
EDIT
Following the suggestion in the comment, I try to compute the Gateaux derivative as $\exp(X + tH)$ by writing down the first few terms.
$\exp(X+tH) = I + (X + tH) + (X^2 + tXH + tHX + t^2H^2)/2 + \cdots$
$\dfrac{d}{dt}\exp(X+tH)\Big|_{t=0} = H + (XH+HX)/2 + \cdots$
And now am stuck again. It seems the expression on the right cannot be rearranged to give what I want.
I think it is the derivative of the trace of the exponential, not the exponential itself that yields $\exp(X)$
| Let $f : \mathbb R^{n \times n} \to \mathbb R^{n \times n}$ be defined by $f (X) = \exp(X)$. Hence,
$\begin{array}{rl} f (X + h V) &= \exp(X + h V)\\ &= I_n + (X + h V) + \frac{1}{2!} (X + h V)^2 + \frac{1}{3!} (X + h V)^3 + \cdots\\ &= I_n + X + h V + \frac{1}{2!} X^2 + \frac{h}{2!} (X V + V X) + \frac{h^2}{2!} V^2 + \frac{1}{3!} X^3 + \\ &\,\,\,\,\,+ \frac{h}{3!} (X^2 V + X V X + V X^2) + \frac{h^2}{3!} (X V^2 + V X V + V^2 X) + \frac{h^3}{3!} V^3 + \cdots\\ &= f (X) + h \left( V + \frac{1}{2!} (X V + V X) + \frac{1}{3!} (X^2 V + X V X + V X^2) + \cdots\right) + \cdots\end{array}$
Thus, the directional derivative of $f$ in the direction of $V$ at $X$ is given by
$$\begin{array}{rl} D_V f (X) &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( f (X + h V) - f (X) \right)\\ &= V + \frac{1}{2!} (X V + V X) + \frac{1}{3!} (X^2 V + X V X + V X^2) + \cdots\end{array}$$
We then write
$$D_V f (X) = M_0 + \frac{1}{2!} M_1 + \frac{1}{3!} M_2 + \frac{1}{4!} M_3 + \cdots$$
where
$$\begin{array}{rl} M_0 &= V\\ M_1 &= X V + V X =: \{X,V\}\\ M_2 &= X^2 V + X V X + V X^2\\ &\vdots\\ M_k &= \displaystyle\sum_{i=0}^k X^{k-i} V X^i\end{array}$$
| {
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"url": "https://math.stackexchange.com/questions/1361636",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Solve equation: $25x+9\sqrt{9x^2-4}=\frac{2}{x}+\frac{18x}{x^2+1}$ Solve equation: $25x+9\sqrt{9x^2-4}=\dfrac{2}{x}+\dfrac{18x}{x^2+1}$
I used wolframalpha.com and got the only solution $x=-\dfrac{1}{\sqrt2}$
And this is my try:
Domain: $|x|\ge\dfrac{2}{3}$
If $x\ge\dfrac{2}{3}$, we have:
$\dfrac{2}{x}+\dfrac{18x}{x^2+1}\le3+\dfrac{18x}{2x}=12$ and $25x+9\sqrt{9x^2-4}\ge25x\ge\dfrac{50}{3}$ (no solution)
If $x\le\dfrac{-2}{3}$,... (I have no idea in this case)
| rewrite your equation in the form
$$9\sqrt{9x^2-4}=\left(\frac{2}{x}+\frac{18x}{x^2+1}-25x\right)^2$$
this can be written in the form
$$4(-1+2x^2)(1+78x^2+117x^4+13x^6)=0$$
solving this we get $$x=\pm\frac{1}{\sqrt{2}}$$
only the number with $$x=-\frac{1}{\sqrt{2}}$$ is the searched solution.
| {
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"url": "https://math.stackexchange.com/questions/1364057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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This is equation is giving me issues $x^2 - 6x + 15 = 0 $ I was given this equation $x^2 - 6x + 15 = 0 $
I tried to look for numbers whose sum is big and product of ac and i could not find any. I tried using the quadratic formula
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
and this is what i got
$x = \frac{6\pm\sqrt{-24}}{2}$
I just dont know what to do from here: any help would be appreacited.
| So you need to find the roots as follow:
Note that:
$$ i^2 = -1$$
$$ x = \frac{6\pm \sqrt{-24}}{2}=\frac{6\pm i\sqrt{24}}{2} $$
$$ x= \frac{6\pm 2i\sqrt{6}}{2} = 3\pm i\sqrt{6}$$
Further explanation ($i^2 = -1$):
let $a = \sqrt{-24}$
$a^2 = -24= 24 \times -1= 24i^2$
$a =\pm \sqrt{24} \times \sqrt{i^2}= \pm i\sqrt{24} = \pm i \times \sqrt{4} \times \sqrt{6} = \pm2i\sqrt{6}$
| {
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"url": "https://math.stackexchange.com/questions/1364683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof question: Prove that 2^(odd integer) + 5^(odd integer) + 2 is a multiple of 3, and 4^(any integer) + 1 can be expressed as 5n, 5n + 1 or 5n + 2. I had been working on the claim in the above question for sometime now. Statistically speaking, it works. For example:
$$\left(2^1 \right)+\left(5^1\right)+2=9,\\
\left(2^3\right)+\left(5^1\right)+2=15,\\ \ldots$$
It works with all the integers that I tested with. Can we give a simple proof for this one?
The second part is like this:
$$\left(4^1\right)+1=5,$$
$$\left(4^2\right)+1=17 = 15+2,$$
$$\left(4^3\right)+1=65,$$
$$\left(4^4\right)+1=257 = 255+2,\\ \ldots$$
So, the numbers are all expressed as $5n$, $5n+1$ or $5n+2$. Is there a generalised proof for this observation?
Thanks.
| Anyone who has studied the $3x + 1$ problem with any degree of detail is aware of this property of numbers of the form $2^n$: either $2^n - 1$ is a multiple of $3$, or $2^n + 1$ is.
Obviously, $2^n$ itself can't be a multiple of $3$ (as long as $n$ is an integer). If $2^n = 3m - 1$, then $2^{n + 1} = (3 - 1)(3m - 1) = 9m - 3 - 3m + 1 = 3k + 1$, where $k = 2m - 1$.
But if $2^n = 3m + 1$, then $2^{n + 1} = (3 - 1)(3m + 1) = 9m + 3 - 3m - 1 = 3k - 1$, where $k = 2m + 1$.
So the powers of $2$ alternate this relationship with multiples of $3$, and since $1$ is odd in $2^1 = 2 = 3 - 1$ and $2$ is even in $2^2 = 4 = 3 + 1$, it follows that all odd-indexed powers of $2$ are one less than a multiple of $3$ and all even-indexed powers of $2$ are one more than a multiple of $3$. In the notation of congruences, we have $2^n \equiv 2 \pmod 3$ if $n$ is odd and $2^n \equiv 1 \pmod 3$ if $n$ is even.
Since $5 \equiv 2 \pmod 3$, we can pretty much repeat the same steps to show that the powers of $5$ also alternate this relationship to $3$ with the same correspondence between odd and even $n$. So if $m$ and $n$ are both odd integers, then $2^m + 5^n + 2 \equiv 2 + 2 + 2 = 6 \equiv 0 \pmod 3$.
The case of $4^n + 1$ is very similar and you should be able to figure out along similar lines.
| {
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"timestamp": "2023-03-29T00:00:00",
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a log inequality Can anyone offer some guidance on proving the following inequality? Define $\Lambda_1(a)=-a\log a$ and $\Lambda_2(a,b)=-(a+b)\log(a+b)$. Then if $a$, $b$, $c$, and $d$ are non-negative numbers summing to one, the following holds:
\begin{align}
\Lambda_2(a,b)+\Lambda_2(b,c)+\Lambda_2(c,d)+\Lambda_2(d,a)\geq \Lambda_1(a)+\Lambda_1(b)+\Lambda_1(c)+\Lambda_1(d).
\end{align}
I've tested a bunch of cases in Mathematica, so I'm pretty certain it's true. The concavity of $\Lambda_1$ gives an upper bound on the left-hand-side, so that doesn't seem to be the right approach. It's also straightforward to show that equality holds if $a=d$ and $b=c$, but I don't think that has much to do with the general case. I assume this would follow quickly from the right log inequality, so even just the name of such an inequality would be helpful.
| By weighted AM/GM,
\begin{align*}
&\left(\frac{(a+b)(d+a)}{a}\right)^a
\left(\frac{(b+c)(a+b)}{b}\right)^b
\left(\frac{(c+d)(b+c)}{c}\right)^c
\left(\frac{(d+a)(c+d)}{d}\right)^d \\
&\le
a\cdot\frac{(a+b)(d+a)}{a}
+b\cdot\frac{(b+c)(a+b)}{b}
+c\cdot\frac{(c+d)(b+c)}{c}
+d\cdot\frac{(d+a)(c+d)}{d}
\\
&=
(a+b)(d+a)
+(b+c)(a+b)
+(c+d)(b+c)
+(d+a)(c+d)
\\
&=
\big((a+b)+(c+d)\big)
\big((d+a)+(b+c)\big)
\\
&= 1
\end{align*}
Rearranging,
$$ \frac1{a^a b^b c^c d^d} \le \frac1{(a+b)^{a+b} (b+c)^{b+c} (c+d)^{c+d} (d+a)^{d+a}} $$
Taking logs yields the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\
2\theta = 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\
\theta = k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
| HINT
If $z = a+bi$ note that
$$
z^3 = a^3 - 3ab^2 + 3a^2bi - b^3i = (a^3 - 3ab^2) + (3a^2b-b^3)i
$$
so
$$
0 = \bar{z} + z^3 = (a^3 - 3ab^2+a) + (3a^2b-b^3-b)i,
$$
which implies you have 2 equations in 2 unknowns.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root. Let $W(x)$ be a polynomial of degree> 2 having at least three different real roots. How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root?
Whether the assumption of the theorem can be weakened?
| By assumption (a,b,c being different reals)
$$W(x)=(x-a)(x-b)(x-c)Q(x)$$
then
$$W'(x)=((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b))Q(x)+(x-a)(x-b)(x-c)Q'(x) \\
=(3x^2-2(a+b+c)x+ab+bc+ac) Q(x)+(x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc) Q'(x)$$
and
$$W''(x)=(2(x-c)+2(x-a)+2(x-b))Q(x)+(x-a)(x-b)(x-c)Q''(x)+2((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)) Q'(x) \\
=(6x-2(a+b+c)) Q(x) + (x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc) Q'' + 2 (3x^2-2(a+b+c)x+ab+bc+ac) Q'(x)$$
Since we do not know Q any root of $P(x)$ must be a root of the "prefactor" polynomials
$$ 3x^2-2(a+b+c)x+ab+bc+ac) = 0 \\
x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc = 0\\
6x-2(a+b+c) = 0 $$
And here I ran out of time and motivation.. Also no guarantee that any of that is correct, but should suffice
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int \frac{\arctan x}{x^4} dx$? I have written the integral as $\int x^{-4} \arctan x dx$. Then, by applying by parts, I got $-3\dfrac{\arctan x}{x^3} + 3\int \dfrac{1}{x^3(1 + x^2)} dx$. Now, how can I solve the later integral? Is there any other trick to do this?
|
$$\int\frac{\arctan x}{x^4}dx$$
By parts:
$$=-\frac{\arctan x}{3x^3}+\frac{1}{3}\int\frac{1}{x^3(x^2+1)}dx$$
$u=x^2,\;du=2xdx$
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\frac{1}{u^2(u+1)}du$$
Partial fractions
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\bigg(\frac{1}{u^2}+\frac{1}{u+1}-\frac{1}{u}\bigg)du$$
$$=-\frac{1}{6x^3}(x^3\ln(x^2)-x^3\ln(x^2+1)+x+2\arctan (x))+C$$
Which is equivalent for restricted $x$ values to:
$$\boxed{\color{red}{-\frac{1}{6x^3}(2x^3\ln(x)-x^3\ln(x^2+1)+x+2\arctan(x))+C)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding Linear Combination of Polynomials I am stuck on a question involving finding the greatest common divisor of polynomials and then solving to find the linear combination of them yielding the greatest common divisor. My work thus far is as follows:
Define: $f(x) = x^3-3x+3$, $g(x) = x^2 - 4$.
Execute the Euclidean Algorithm for Polynomials as follows:
$x^3-3x+3 = x(x^2 -4)+(x+3)$
$x^2-4 = (x-3)(x+3) +5$
$x+3 = \frac x5(5) +3$
$5 = 1(3) +2$
$3 = 1(2) +1$
$2 = 1(2) +0$
So we know that $\gcd(f(x),g(x)) = 1$. Now, my objective is to find polynomials $s(x), t(x)$ such that $f(x)s(x) + g(x)t(x) = 1$.
I have tried using back substitution, a method involving matrix multiplication of coefficients, and a method using continuous fractions. None of these have worked- all returning incorrect answers, two of which were the same. Does anyone have any suggestions as to where I am going wrong, or a simpler way to finish this computation?
Note: it is not specified in my book which field these polynomials are defined over. I am assuming the reals.
| The first two lines are fine. Then you are finished, though you may want to write $x+3=\frac{x+3}{5}\cdot 5+0$.
Work backwards from the second line. We get $5=x^2-4 -(x-3)(x+3)$. But from the first line we have $x+3=x^3-3x+3-x(x^2-4)$. Substituting we get
$$5=x^2-4-(x-3)(x^3-3x+3-x(x^2-4)).$$
This simplifies to
$$5=(3-x)(x^3-3x+3)+(x^2-3x+1)(x^2-4).$$
If you want a $1$ on the left-hand side, multiply through by $\frac{1}{5}$. If you are working in a field that does not contain the rationals (but is not of characteristic $5$), instead of multiplying by $\frac{1}{5}$, multiply by the inverse of $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $p_{ij}^{(n)}$ for the transition matrix
Let
$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}$$
find $p_{11}^{(n)},p_{12}^{(n)},p_{13}^{(n)}$
Since the characteristic function is $$-\mu^3+\frac{4}{3}\mu^2-\frac{1}{3}\mu=0$$
the eigenvalues are $$\mu_0=1,\mu_1=\frac{1}{3},\mu_2=0$$
Here the problem starts, from what I understood from the book, we need to found three constants such that $$p_{ij}^{(n)}=\mu_0^nA+\mu_1^n+\mu_2^nC=A+\left(\frac{1}{3}\right)^nB$$
But how can I find the values of A and B?
| Why not just computing $P^2$, then interpolating? We have:
$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad P^2=\begin{bmatrix}\frac{1}{3}&\frac{2}{9}&\frac{4}{9}\\\frac{1}{3}&\frac{4}{9}&\frac{2}{9}\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}$$
so $p_{11}^{(1)}=A+\frac{B}{3}=\frac{1}{3}$ and $p_{11}^{(2)}=A+\frac{B}{9}=\frac{1}{3}$ give $A=\frac{1}{3},B=0$ and $p_{11}^{(n)}=\frac{1}{3}$.
Similarly,
$$ p_{12}^{(n)}=\frac{1}{3}-\frac{1}{3^n}, \qquad p_{13}^{(n)}=\frac{1}{3}+\frac{1}{3^n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| Consider $$(1+x+x^2+x^3+\cdots)(1+x+^2+x^3+\cdots)$$ Coefficient of $x^k$ in this expansion is just $k+1.$ Because $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Therefore $$(1+x+x^2+x^3+\cdots)^2=1+2x+3x^2+4x^3+\cdots.$$
Now evaluate (you can use the same procedure) $$(1-2x+x^2)(1+2x+3x^2+4x^3+\cdots)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Prove that $\alpha + \beta=\frac {\pi}{2}$ It is given that-
(1) $0<\alpha,\beta<90$.
(2) $\sin^2\alpha+\sin^ 2\beta=\sin(\alpha+\beta).$
Prove that $\alpha + \beta=\frac {\pi}{2}$
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\sin^2A+\sin^2B=1-(\cos^2A-\sin^2B)=1-\cos(A+B)\cos(A-B)$$
$$\implies\sin(A+B)+\cos(A+B)\cos(A-B)=1$$
Using Weierstrass substitution for $\sin(A+B),\cos(A+B)$ and setting $\tan\dfrac{A+B}2=t,$
$$[1+\cos(A-B)]t^2-2t+1-\cos(A-B)=0$$
Using Weierstrass substitution for $\cos(A-B),$
$$t=1,\tan^2\dfrac{A-B}2$$
If $\tan\dfrac{A+B}2=1,\dfrac{A+B}2=n\pi+\dfrac\pi2\iff A+B=?$ where $n$
But I'm not sure about $\tan\dfrac{A+B}2=\tan^2\dfrac{A-B}2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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