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Symmetry of function defined by integral Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as
$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
One can use, for example, the Residue Theorem to show that
$$ f(\alpha, \beta) = \frac{\pi \sin{\left (\pi \alpha \beta\right )}}{ \sin{\left (\pi \alpha\right )} \, \sin{\left (\pi \beta\right )}} $$
Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$. My question is, can one see this symmetry directly from the integral expression?
| Very interesting question! But, alas, not an answer. Only few representations for the integral obtained. One of them evaluated to the form claimed in the question.
First, transform the integral into a form, symmetric under $\alpha \mapsto -\alpha$:
$$
\int_0^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x = \int_0^1 \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x + \int_1^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x
$$
Make a change of variables $x \to x^{-1}$ in the last integral to obtain:
$$
f(\alpha,\beta) = \int_0^1 \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1}
$$
Now, making a change of variables $x = \exp(-t)$ we have:
$$
f(\alpha,\beta) = \int_0^\infty \frac{\cosh(\alpha t)}{\cosh(t) + \cos(\beta \pi)} \mathrm{d} t \tag{2}
$$
Using
$$
\int_0^\infty \exp\left(-u \left( \cosh t + \cos \pi \beta \right) \right) \mathrm{d}u = \frac{1}{\cosh(t) + \cos(\beta \pi)}
$$
and the integral representation of the modified Bessel function of the second kind:
$$
\int_0^\infty \cosh(\alpha t) \exp\left( - u \cosh t \right) \mathrm{d}t = K_\alpha(u)
$$
we arrive at a compact representation:
$$
f(\alpha,\beta) = \int_0^\infty K_\alpha(u) \mathrm{e}^{-u \cos\left(\pi \beta\right)} \mathrm{d} u \tag{3}
$$
expanding the exponential into series and using $\int_0^\infty u^n K_\alpha(u) \mathrm{d} u = 2^{n-1} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right)$ we get:
$$
f(\alpha,\beta) = \sum_{n=0}^\infty \frac{2^{n-1}}{n!} \left(-\cos \pi \beta\right)^{n} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right) \tag{4}
$$
summing over even and over odd integers:
$$
f(\alpha, \beta) = \frac{\pi}{2} \frac{ \cos\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \cos \left( \frac{\pi \alpha}{2} \right)} - \frac{\pi}{2} \frac{ \sin\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \sin \left( \frac{\pi \alpha}{2} \right)} = \pi \frac{\sin \left( \alpha \left( \frac{\pi}{2} - \arcsin \cos(\pi \beta) \right) \right)}{ | \sin \pi \beta | \sin(\pi \alpha)}
$$
Now $\frac{\pi}{2} - \arcsin \cos(\pi \beta) = \arccos \cos(\pi \beta) = \pi | \beta |$ for $-1<\beta<1$. Thus, restoring parity, we recover the OP's expression:
$$
f(\alpha, \beta) = \pi \frac{ \sin(\pi \alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} = \frac{\operatorname{sinc}(\pi \alpha \beta)}{\operatorname{sinc}(\pi \alpha) \operatorname{sinc}(\pi \beta)} \tag{5}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the volume obtained by rotating the following around the x axis? The original problem is:
"Find the volume of the solid obtained by rotating about the x axis the region enclosed by the curves $y = \frac{9}{x^2 + 9},y=0,x=0,\,$and $x = 3$"
I set up the following integral $$81\pi\int_0^3\frac{1}{(x^2 + 9)^2}dx$$ using the cylinder method (I believe it's called like that) and when I calculated it using a computer I obtained the correct answer but I have been having difficulties in solving it manually. I tried the shell method as well and I didn't see it any easier to solve but I may be wrong of course.
| We want to find
$$\int \frac{dx}{(x^2+9)^2}.$$
Let $x=3u$. Apart from a constant factor, we end up with
$$\int\frac{du}{(u^2+1)^2}.$$
Let $u=\tan t$ (we could have gone more directly, by letting $x=3\tan t$). Then $du=\sec^2 t\,dt$, and we end up with
$$\int \frac{\sec^2 t}{\sec^4 t}\,dt.$$
This is the familiar integral
$$\int \cos^2 t\,dt.$$
A common approach to this is to use the fact that $\cos 2t=2\cos^2 t-1$.
e can save time if we make the substitutions on the definite integral, that is, substitute for the endpoints.
Another way: Consider $\displaystyle\int \frac{dx}{x^2+9}$. Attack this by integration by parts, letting $du=dx$ and $v=\frac{1}{9+x^2}$. Then we can take $u=x$. Also, $dv=\frac{-2x\,dx}{(x^2+9)^2}$. Thus
$$\int \frac{dx}{x^2+9}=\frac{x}{x^2+9}+\int\frac{2x^2\,dx}{(x^2+9)^2}.$$
But $2x^2=2x^2+18-18$. Thus
$$\int\frac{dx}{x^2+9}=\frac{x}{x^2+9}+2\int\frac{dx}{x^2+9}-18\int\frac{dx}{(9+x^2)^2}.$$
We will be finished if we can find $\displaystyle\int\frac{dx}{x^2+9}$. The substitution $x=3w$ turns this into a familiar integral.
Remark: In the second approach, we obtained a reduction formula. A very similar trick expresses $\displaystyle\int \frac{dx}{(x^2+9)^{n+1}}$ in terms of
$\displaystyle\int \frac{dx}{(x^2+9)^{n}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of the series $\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $
Study the convergence of the next series: $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $$
My solution: since $$\frac{(2n)!!}{(2n+2)!!} \leq \frac{(2n)!!}{(2n+1)!!}$$
forall $n \in \mathbb{N}$ and since $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+2)!!} = \sum_{n=1}^\infty \frac{1}{2n+2} = \infty$$
then the first series diverges. Is there anything wrong? Thanks in advance.
| Yes. What you have done is correct. You can attempt to quantify the divergence a bit better. Note that $(2n)!! = 2^n n!$ and $(2n+1)!! = \dfrac{(2n+2)!}{2^{n+1} (n+1)!}$. Hence,
$$\dfrac{(2n)!!}{(2n+1)!!} = 2^{2n+1} \dfrac{(n+1)! n!}{(2n+2)!} = \dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)}$$
We have that $\dbinom{2n+2}{n+1} \sim \dfrac{4^{n+1}}{ \sqrt{\pi(n+1)}}$. Hence,
$$\dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)} \sim \dfrac{2^{2n+1} \sqrt{\pi}}{4^{n+1} \sqrt{n+1}} = \dfrac{\sqrt{\pi}}2 \cdot \dfrac1{\sqrt{n+1}}$$
Hence, $$\sum_{n=1}^{\infty}\dfrac{(2n)!!}{(2n+1)!!} \,\,\,\,\,\,\,\,\text{diverges as} \,\,\,\,\,\, \dfrac{\sqrt{\pi}}2 \cdot \sum_{n=1}^{\infty}\dfrac1{\sqrt{n+1}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| This is a standard / common problem.
Hint:
$$\prod_{i=1}^{98} \frac {i}{i+1} = \frac {1}{99}$$
$$\sqrt{\frac {1}{99}} \times \frac {99}{100} = \frac {\sqrt{99}}{100} < \frac {1}{10}$$
Hint elaborated slightly:
Let $A = \prod_{i=1}^{49} \frac {2i-1}{2i}$ and $B= \prod_{i=1}^{49} \frac {2i}{2i+1}$. It is clear that $A<B$.
The question wants us to show that $A \times \frac {99}{100} < \frac {1}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/274753",
"timestamp": "2023-03-29T00:00:00",
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$ A = x + y + z$, number of solutions in $Z$ if $x, y, z$ are bounded in intervals For the equation $x + y = A$, it's easy, when you notice that when iterating over all possible $x$, the number of solutions for $y$ is $0$ at the beginning, then increases by $1$, then stays constant, then decreases by $1$, and at the end $0$. This can be calculated in $O(1)$.
$l_1 < a < u_1$, $l_2 < b < u_2$, $l_3 < c < u_3$.
I'm looking for a general answer for all $u_i$, $l_i$, $A$ ($i = 1, 2, 3$).
I assume this problem is just as easy, but it's hard to find the formula, since i have to think in $3$ dimensions instead of $2$.
example: $2 < x < 5$, $1 < y < 5$, $3 < z < 7$, $A = 11$, the answer is 5.
| The number of non-negative integer solutions to $x+y+z=11$ where $2<x<5, 1<y<5, 3<z<7$ is equal to the number of integer solutions to $a+b+c=2$ where $0\leq a\leq 1$, $0\leq b\leq 2$, $0\leq c\leq 2$ by setting $a=x-3$, $b=y-2$, $c=z-4$.
In this case it is easy to count the number of solutions: $\binom{2+3-1}{3-1}-1=\binom{4}{2}-1=5$, since there is only one 'bad' option: $a=2,b=0,c=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Optimization of $2x+3y+z$ under the constraint $x^2+ y^2+ z^2= 1$
Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the
maximum and minimum values of $2x + 3y + z$ .
How can I able to solve the problem? Thanks for your help.
| Using the Cauchy-Schwarz inequality, $2x+3y+z\leq \sqrt{2^2+3^2+1^2}\sqrt{x^2+y^2+z^2}=\sqrt{14}$ which is the maximum value.Since $x$,$y$ and $z$ are reals, the minimum value occurs when $x$,$y$ and $z$ are negative which can be used to infer that the minimum value is $-\sqrt{14}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| @RonGordon last sum becomes:
\begin{align}
&\sum_{k = 1}^{\infty}\left\{
1 + \left(k + 1\right)^{2}\log\left(1 - {1 \over \left[k + 1\right]^2}\right) \right\}=
\sum_{k = 1}^{\infty}\left\{
1 - \left(k + 1\right)^{2}
\int_{0}^{1}{{\rm d}x \over \left(k + 1\right)^{2} - x}\right\}
\\[3mm]&=
-\left[\int_{0}^{1}\sum_{k = 1}^{\infty}{1 \over \left(k + 1\right)^{2} - x}\right]
\,x\,{\rm d}x
=
-\left[\int_{0}^{1}\sum_{k = 0}^{\infty}
{1 \over \left(k + 2 + x^{1/2}\right)\left(k + 2 - x^{1/2}\right)}\right]
\,x\,{\rm d}x
\\[3mm]&=
\int_{0}^{1}{
\Psi\left(2 - x^{1/2}\right) - \Psi\left(2 + x^{1/2}\right) \over 2x^{1/2}}\,x
\,{\rm d}x
=
\underbrace{\int_{0}^{1}
\color{#c00000}{\left\lbrack\Psi\left(2 - x\right) - \Psi\left(2 + x\right)\right\rbrack}\,x^{2}\,{\rm d}x}
_{\displaystyle{\log\left(\pi\right) - {3 \over 2}}}
\end{align}
since
\begin{align}
&\color{#c00000}{\Psi\left(2 - x\right) - \Psi\left(2 + x\right)}
=\left[\Psi\left(1 - x\right) + {1 \over 1 - x}\right]
-
\left[\Psi\left(x\right) + {1 \over 1 + x} + {1 \over x}\right]
\\[3mm]&=\left[\Psi\left(1 - x\right) - \Psi\left(x\right) - {1 \over x}\right]
+{2x \over 1 - x^{2}}
=\color{#c00000}{\left[\pi\cot\left(\pi x\right) - {1 \over x}\right]
+ {2x \over 1 - x^{2}}}
\end{align}
Sorry. It was too long for a comment !!!
| {
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"timestamp": "2023-03-29T00:00:00",
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation}
\int \frac{1}{ax + b} dx = \frac{1}{a} \ln{|ax + b|} + C
\end{equation}
the following solution is obtained
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2}\ln{|-2x|} + C
\end{equation}
Why are these solutions different? Which is correct?
The second solution can be simplified
\begin{equation}
-\frac{1}{2}\ln{|-2x|} + C = -\frac{1}{2}\ln{|-2|} -\frac{1}{2}\ln{|x|} + C= -\ln{\frac{1}{\sqrt{2}}} - \frac{1}{2}\ln{|x|} + C
\end{equation}
but they still differ.
| They are both correct!
$$
-\frac{1}{2}\ln|-2x| + C = -\frac{1}{2}(\ln 2 + \ln |x|) + C = -\frac{1}{2}\ln |x| + (C - \frac{1}{2} \ln 2)
$$
The constant of integration is what "differs" here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| I want to give a complete answer.
Assume by the sake of contradiction that $$\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}<1.\tag{1}$$
By Cauchy Schwarz it follows from $(1)$ that $$\frac{3}{abc}>\frac{3}{abc}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\geq \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2.\tag{2}$$
From $(2)$ we recover two facts
*
*By $AM-GM$ mean $$\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2\geq 3\left(\frac{1}{a^2bc}+\frac{1}{ab^2c}+\frac{1}{abc^2}\right),$$ therefore, combined with $(2)$, we obtain on one hand $$1>\frac{1}{a}+\frac{1}{b}+\frac{1}{c},\tag{3}$$ which implies that $\min\{a,b,c\}>1$ (remember that they are all positive).
*On the other hand we can also derive from $(2)$ and $(3)$ the following: $$\frac{1}{\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}>\frac{a+b+c}{3}>1,\tag{4}$$
which in the end leads to an absurd with respect to the hypothesis of the problem.
Then $(1)$ must be false and the problem is solved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_0^\infty \frac{\sqrt{x}}{x^{2}+1}\, \mbox{d} x$ I've been trying to integrate the following
$$\int_{0}^{\infty} \frac{\sqrt{x}}{x^{2}+1} \mbox{d} x$$
on half an annulus in the upper half plane. I keep getting $\frac{\pi}{\sqrt{2}}\ i$, which doesn't give with the numerical approximations I get using WolframAlpha.
How should I attack this problem?
| One way to attack this particular problem is to make the substitution $x=t^2$, as in Ahlfors, Complex Analysis, Third Edition, p. 159; the integral becomes
$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} $$
We may now use a semicircular contour $C$ in the half plane $\Re{z} \ge 0$ rather than the annulus as we have disposed of the branch point at the origin. We write
$$ \oint_C dz \: \frac{z^2}{z^4+1} = i 2 \pi \left ( \mathrm{Res}_{z=e^{i \frac{\pi}{4}}} \frac{z^2}{z^4+1} + \mathrm{Res}_{z=e^{i \frac{3 \pi}{4}}} \frac{z^2}{z^4+1} \right ) $$
$$\begin{align} \mathrm{Res}_{z=e^{i \frac{\pi}{4}}} \frac{z^2}{z^4+1} &= \frac{e^{i \frac{2 \pi}{4}}}{\left (e^{i \frac{\pi}{4}}-e^{i \frac{3 \pi}{4}}\right )\left (e^{i \frac{\pi}{4}}-e^{i \frac{-3 \pi}{4}}\right )\left (e^{i \frac{\pi}{4}}-e^{i \frac{-\pi}{4}}\right )} \\ &= \frac{i}{i (1-i)(2) \left (2 i \sin{\frac{\pi}{4}} \right )} \\ &= -i \frac{1+i}{4 \sqrt{2}} \end{align}$$
Similarly,
$$ \mathrm{Res}_{z=e^{i \frac{3 \pi}{4}}} = -i \frac{1-i}{4 \sqrt{2}} $$
Therefore,
$$ \oint_C dz \: \frac{z^2}{z^4+1} = i 2 \pi (-i) \frac{1}{2 \sqrt{2}} = \frac{\pi}{\sqrt{2}} $$
Now, about the contour $C$ which has a large radius of, say, $R$:
$$ \oint_C dz \: \frac{z^2}{z^4+1} = \int_{C_R} dz \: \frac{z^2}{z^4+1} + \int_{-R}^R dt \: \frac{t^2}{t^4+1} $$
In the limit as $R \rightarrow \infty$,
$$\int_{C_R} dz \: \frac{z^2}{z^4+1} \sim \frac{\pi}{R} $$
and therefore vanishes. We may then conclude that
$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} = \int_0^{\infty} \frac{\sqrt{x}}{x^2+1} = \frac{\pi}{\sqrt{2}} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating integrals such as $\int \frac{1+\cos^{2}x}{1+\cos2x}$ We started integrals not too long ago, I understand it for the most part but I always have a problem figuring out how to solve ones involving trig identities. Like this:
$$\int \frac{1+\cos^{2}x}{1+\cos2x}$$
Indefinite integral of $$\frac{ 1 + \cos^2(x)}{ 1 + \cos(2x) }.$$
I tried changing the denominator to $2\cos^2(x)$ but I still can't make a u substitution.
| Write $$2\cos^2x = 1+ \cos2x$$
$$\cos^2x= \frac{1+ \cos2x}{2}$$
$$=∫1+\frac{1+ \cos2x}{2} * \frac{1}{1+\cos2x} $$
$$=\int\frac{1}{1+\cos2x}+\int\frac{1}{2}$$
Use $$u=2x$$
$$=\frac{1}{2}[\int\frac{1}{1+\cos(u)}]+\frac{1}{2}x$$
$$=\frac{1}{2}[\int\frac{1-cos(u)}{1-\cos^2u}]$$
$$=\frac{1}{2}[\int\frac{1-cos(u)}{\sin^2u}]$$
$$=\frac{1}{2}[\int\frac{1 }{\sin^2u}du -\int\frac{\cos u}{\sin^2u}du]$$
Use $$v=\sin u$$
$$=\frac{1}{2}[-\cot u + \frac{1}{\sin u}]$$
$$=\frac{1}{2}[-\cot 2x + \frac{1}{\sin 2x}]$$
$$=\frac{1}{2}\tan x$$
$$=\frac{1}{2}\tan x+\frac{1}{2}x$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
| I couldn't follow what you were doing, but I do see that you have the right answer to first order. I can, however, point to a simpler way based on Laplace's Method.
First of all, write the integral as an exponential whose maximum is at $x=0$:
$$I_n = \int_0^{\pi/4} dx \: e^{n \log{\tan{\left ( \frac{\pi}{4} - x \right )}}} $$
As $n \rightarrow \infty$, the integral is dominated by its contributions from where the integrand is a maximum, or at $x=0$. We may then approximate $\log{\tan{\left ( \frac{\pi}{4} - x \right )}}$ by its lower-order terms in its Taylor expansion about $x=0$.
To first order, note that $\log{\tan{\left ( \frac{\pi}{4} - x \right )}} \sim -2 x$ as $x \rightarrow \infty$. We may then write
$$I_n \sim \int_0^{\pi/4} dx \: e^{-2 n x} \: \: (n \rightarrow \infty)$$
We may also may as well send that upper integration limit to $\infty$, as, again, the contributions to the integral away from $x=0$ are exponentially small. The leading order term in the expansion is then
$$I_n \sim \frac{1}{2 n} \: \: (n \rightarrow \infty)$$
You may also derive higher order terms from the higher-order terms in the Taylor expansion. The integrals that result will come from Taylor expanding the exponential piece containing these higher-order terms, which will result in Gamma-function-like integrals which will bring in decreasing powers of $n$ in the asymptotic expansion.
EDIT
I will do out the next order. To second order:
$$\log{\tan{\left ( \frac{\pi}{4} - x \right )}} =-2 x -\frac{4}{3} x^3 +O(x^4) $$
The corrected expansion is then
$$I_n \sim \int_0^{\infty} dx \: e^{-2 n x} \left ( 1 - \frac{4}{3} n x^3 \right ) \: \: (n \rightarrow \infty)$$
Doing this next integral as well, we find that
$$I_n \sim \frac{1}{2 n} - \frac{1}{2 n^3} \: \: (n \rightarrow \infty)$$
in agreement with your result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
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Floor function inequality I am racking my brain trying to get this problem solved and I can't seem to break it...
Let $m, n$ be positive integers, with $m > 1$. Prove
$$\left\lfloor\frac{n}m\right\rfloor+\left\lfloor\frac{n+1}m\right\rfloor\le\left\lfloor\frac{2n}m\right\rfloor$$
I started trying to use the inequalities that say
$$\lfloor x\rfloor + \lfloor y\rfloor + \lfloor x+y\rfloor \le \lfloor 2x\rfloor + \lfloor 2y\rfloor\;.$$
From there I rewrote both sides with the notion that for a real number $y$,
$y = \lfloor y\rfloor + \{y\}$, where $\{y\}$ is the fractional part, but I keep going in circles and can not get to the desired result...
| Let $a={\sf floor}(\frac{n}{m})$ and $b={\sf floor}(\frac{n+1}{m})$.
We have $a \leq \frac{n}{m}$ and $b \leq \frac{n+1}{m}$, so $a+b \leq \frac{2n+1}{m}$. If this were an equality, it would imply that both $a =\frac{n}{m}$ and $b = \frac{n+1}{m}$, so that $m$ would divide both $n$ and $n+1$, and hence divide $n+1-n=1$ also, which is absurd.
So we have the strict inequality $a+b \lt \frac{2n+1}{m}$, yielding $m(a+b) \lt 2n+1$. Since $m,a+b$ and $2n+1$ are all integers, we deduce $m(a+b) \leq 2n$. So $a+b$ is an integer lower than $\frac{2n}{m}$ ; it is therefore smaller than the floor of $\frac{2n}{m}$ as wished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
General expression for exponentiating a matrix? $A= \begin{pmatrix}
\cos(x) & \sin(x) \\
-\sin(x) & \cos(x) \\
\end{pmatrix}$
I found that
$A^2 = \begin{pmatrix}
\cos^2(x)-\sin^2(x) & 2\cos(x)\sin(x) \\
-2\cos(x)\sin(x) & \cos^2(x)-\sin^2(x) \\
\end{pmatrix}$
$A^2 = \begin{pmatrix}
\cos(2x) & \sin(2x) \\
-\sin(2x) & \cos(2x) \\
\end{pmatrix}$
and
$A^3 = \begin{pmatrix}
\cos^2(x)[\cos^2(x)-3\sin^2(x)] & \sin(x)[3\cos^2(x)\sin^2(x)] \\
-\sin(x)[3\cos^2(x)+\sin^2(x)] & \cos(x)[-3\sin^2(x)+\cos^2(x)] \\
\end{pmatrix}$
$A^3 = \begin{pmatrix}
\cos(2x)cos(x) - \sin(x)\sin(2x) & \sin(x)\cos(2x) + \cos(x)\sin(2x) \\
-\cos(x)sin(2x)-\sin(x)\cos(2x) & \cos(x) + \cos(2x) -\sin(x)\sin(2x) \\
\end{pmatrix}$
but I am unable to find a general solution. I probably need to perform some algebraic manipulations or whatnot.
| Hint: this is rotation by the angle x
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with simple integral I'm trying to solve this simple integral:
$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$
Here's what I have done so far:
*
*$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$
*$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)^2}t dt$
*$\displaystyle \int (t^2 - 1)^2 dt = \frac15 t^5 - \frac23 t^3 + t + C$
*$\displaystyle \frac15 t^5 - \frac23 t^3 + t + C = \frac15 \sqrt{(x + 1)^5} - \frac23 \sqrt{(x + 1)^3} + \sqrt{x + 1} + C$
WolframAlpha tells me steps 1 and 3 are right so the mistake must be somewhere in steps 2 and 4, but I really can't see it.
| There is a (slightly) more obvious way of solving it: rewrite the numerator as $x^2+1-1$ and then the whole integral as a sum of two integrals:
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} + \int \frac{dx}{\sqrt{x+1}}
$$
The second integral is easy, the first one is
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} =\int \frac{(x+1)(x-1)dx}{\sqrt{x+1}}=\int x \sqrt{x+1}dx-\int \sqrt{x+1}dx
$$
The second integral is also easy, and the integrand in the first one should be rewritten and $(x+1-1)\sqrt{x+1}$ and the rest is easy. This is a bit too straightforward, I admit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| For the first one, induction readily gives you the answer.
For $k=1$, we have $2^3 \vert 3^{2^1} - 1$.Assume that it is true for some $k=m$ i.e. $2^{m+2} \vert 3^{2^m} - 1$ i.e. $3^{2^m}-1 = M \cdot 2^{m+2}$We have that $$3^{2^{m+1}}-1 = \left(3^{2^m}+1\right)\left(3^{2^m}-1\right) =\left(3^{2^m}+1\right) \cdot M \cdot 2^{m+2}$$But we also have that$\left(3^{2^m}+1\right) $ to be even and hence $\left(3^{2^m}+1\right) = 2 N$.Hence, we get that$$3^{2^{m+1}}-1 = 2^{m+3}MN$$i.e. $2^{m+3} \vert 3^{2^{m+1}}-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$ $a, b, c > 0$ such that $abc=1$, prove inequality (Lagrang not allowed here because I am still in grade $9^{th}$):
$$P=\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$$
I have tried to use AM-GM, that I have:
$(a+b^2+c^2)(a+1+1)\ge (a+b+c)^2$, so $$P \le \frac{a(a+2)+b(b+2)+c(c+2)}{(a+b+c)^2}$$ but if $$\frac{a(a+2)+b(b+2)+c(c+2)}{(a+b+c)^2} \le 1 \Rightarrow ab + bc + ca \ge a+b+c \ (\text{not true})$$
P/S: Can someone tell me what good inequalities books that I should buy ?
| Let $a = \frac{x^2}{yz}$ and so forth. Expanding (*):
$$(x^4 yz + y^6 + z^6)(y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - x^4yz (y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - y^4xz (x^4yz + y^6 + z^6)(z^4 xy + x^6 + y^6) - z^4xy(x^4yz + y^6 + z^6)(y^4 xz + x^6 + z^6) $$
at quickmath.com gives (**)
$$x^{12}y^6 + x^{12}z^6 + y^{12}x^6 + y^{12}z^6 + z^{12}x^6 + z^{12}y^6 - x^2y^5z^{11} - x^2y^{11}z^5 - y^2x^5z^{11} - y^2x^{11}z^5 - z^2x^5y^{11} - z^2x^{11}y^5$$
The given inequality is equivalent to $(*) \ge 0$, same as $(**) \ge 0$, which follows from Muirhead inequality and $[12,6,0] \succ [11,5,2]$.
An aesthetically more appealing solution is note that
$$\frac{a}{a+b^2+c^2} \leq \frac{a^r}{a^r+b^r+c^r}$$
for $r = 4/3$. To show this, note that after clearing denominator we need to show
$$\begin{eqnarray} a(b^r+c^r) \leq a^r(b^2+c^2)\\
\Leftrightarrow (bc)^{r-1}(b^r+c^r) = a^{1-r}(b^r+c^r) \leq b^2+c^2 \end{eqnarray}$$
Now $(bc)^{1/3}(b^{4/3} + c^{4/3}) \leq b^2 + c^2$ follows from AM-GM and power mean inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Least Natural number $n$ which has $18$ divisors The Least Natural number $n$ which has $18$ divisors
My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$
Now How can I solve it
Thanks
| We know the number of divisors of $\prod_{1\le r\le m}p_i^{m_i}$ is $\prod_{1\le r\le m}(m_i+1)$ where $p_i$s are distinct primes, and power $m$s positive integers
As $18=2\cdot3^2$, $m$ can not be $\ge4$
If $m=1$, $m_1+1=18\implies m_1=17\implies n_\text{min}=2^{17}$
If $m=2$, $(m_1+1)(m_2+1)=18=6\cdot3=2\cdot9$
So, either $m_1=2$, $m_2=5\implies n_\text{min}=2^5 3^2=32\cdot9=288<2^{17}$
or $m_1=1$,$m_2=8 \implies n_\text{min}=2^8 3=768$
If $m=3$, $m_1,m_2.m_3$ can only be $1,2,2$ so in this case, $n_\text{min}=2^2 3^2 5=180<288$
So, the minimum natural number with $18$ divisors is $180.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the identity $1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$ Here is the problem:
$$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
Can you help me prove that this is an trigonometric identity?
| $$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
$$\cos{2\alpha}=2cos^{2}{\alpha}-1$$
$$\cos{2\cdot \left(45^{\circ}-\frac{x}{2}\right)}=\cos{\left(90^{\circ}-x\right)}=\cos{90^{\circ}\cos{x}}+\sin{90^{\circ}\sin{x}}=0+1\cdot \sin{x}=\sin{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Change of Variable (Double Integral) I have been trying to integrate the two following integrands; $$\int \int_{D}(x^{2}+y^{2})dxdy$$ where $D=\{{x^{2}+xy+y^{2}\leq 1}\}$ and $$\int \int_{D}\sqrt{x^{2}+y^{2}}dxdy$$ where $D=\{{x^{2}+y^{2}\leq x}\}$. Now, I have been tempted to use change of variables (polar coordinates) in both cases. For example, in the first case, I completed the square so: $$D=\{{(x+1/2y)^{2}+3/4y^{2}\leq 1}\}$$ I then set $u=r\cos{t}-(1/2)r\sin{t}$ and $v=r\sin{t}$, performed partial differentiation, formed the jacobin matrix, and calculated the determinant. I then set the new boundaries $0\leq r\leq$1 and $0\leq t \leq 2\pi$ before calculating the whole expression. Now, the answer I obtain is $(53\pi)/96$, which is blatantly wrong as the textbook gives $(4\pi)/(3\sqrt{3})$. Since my approach to the second problem is similar, I fear, that the answer I obtain there is wrong as well. I would be exceedingly grateful if you could help me.
| First, do coordinates transform
$$
x = \frac u{\sqrt 3} - v; \qquad y = \frac u{\sqrt 3} + v
$$
It will reduce your domain $D$ to the inner region of the circle:
$$
x^2+xy+y^2= \frac {u^2}3 - 2\frac{uv}{\sqrt 3} + v^2 + \frac {u^2}3 - v^2 + \frac{u^2}3 + 2 \frac{uv}{\sqrt 3} + v^2 = u^2 + v^2 \leq 1
$$
Now change the system to "polar":
$$
u = r \cos \phi; \qquad v = r \sin \phi
$$
so $D = \left \{ \left ( r, \phi\right )\ \left |\ 0 \le r \le 1, 0 \le \phi \le 2\pi \right .\right \}$
As for the integrand
$$
x^2+y^2 = \frac {u^2}3 - 2\frac{uv}{\sqrt 3} + v^2 + \frac{u^2}3 + 2 \frac{uv}{\sqrt 3} + v^2 = \frac {2u^2}3 + 2v^2
$$
Also $dx\ dy = J_1\ du\ dv$, where $J$ is a Jacobian of first $(x,y) \rightarrow (u, v)$ transformation. On the other hand $du\ dv = J_2\ dr\ d\phi$, where $J_2$ is a Jacobian of second, $(u, v) \rightarrow (r, \phi)$, transformation, and we know that $J_2 = r$. Let's find $J_1$:
$$
J_1 = \text{det} \left [
\begin{array}{cc}
\frac 1{\sqrt 3} & -1 \\ \frac 1{\sqrt 3} & 1
\end{array}
\right ] = \frac 2{\sqrt 3}
$$
so $dx\ dy = \frac {2r}{\sqrt 3} dr\ d\phi$.
And finally
$$
I_1 = \frac 2{\sqrt 3}\int_0^{2\pi} \int_0^1 \left ( \frac {2r^2 \cos^2 \phi}3 + 2 r^2 \sin^2 \phi\right) r dr d\phi = \frac 4{3\sqrt 3}\int_0^{2\pi} \left ( \left . \frac {r^4}4 \right |_0^1\right )\left( cos^2 \phi + 3 \sin^2 \phi \right)d\phi = \frac 1{3 \sqrt 3} \int_0^{2\pi} \left ( 2 - \cos 2\phi\right ) = \frac {4\pi}{3\sqrt 3}
$$
Same can be done for 2 part of the problem.
| {
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Assume $n$ is even. Prove that $323$ divides $20^n+16^n-3^n-1$. I'm unclear what is the best method to teach this with minimum math experience.
| Since $n$ is even:
$20^n + 16^n - 3^n - 1 \equiv 1^n + (-3)^n - 3^n - 1 = 1 + 3^n - 3^n - 1 = 0 \bmod 19$
Also:
$20^n + 16^n - 3^n - 1 \equiv 3^n + (-1)^n - 3^n - 1 = 3^n + 1 - 3^n - 1 = 0 \bmod 17$
So the quantity is divisible by $17\times 19 = 323$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| Let $a=\cos\alpha,b=\cos\beta$
So, the the given expression $$=\sqrt{(1-ab)^2-(1-a^2)(1-b^2)}=\sqrt{1-2ab+a^2b^2-(1-a^2-b^2+a^2b^2)}=\sqrt{a^2+b^2-2ab}=|a-b|=|\cos\alpha-\cos\beta|$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to measure the streakedness of numerical data? Would anyone know how to use C/C++ to calculate the streakedness of data? The definition of streakedness is how many deviations away from the mean(i.e running average a numerical data streak. Thank you for your help.
[EDIT] From our company's chief software architect, here is the requirement for a statistical measure. Could someone please define a statistical formula based onour architect's definition of data streakedness? -- February 19th 2013 8:45 AM
Equal numbers are a streak. $1,2,3,3,3,4,5$ has a streak of $7$.
Case A: $1,2,3,4,5,6,7,8,9,10,11,12,13$ has a longest streak of $13$.
Case B: $1,2,3,4,5,6,7,3,8,9,10,11,12$ has a longest streak of $7$, a second smaller streak of $6$.
Case C: $1,2,3,4,5,6,7,1,2,3,4,5,6$ has a longest streak of $7$, and a second smaller streak of $6$.
Case D: $1,2,3,4,5,6,7,1,2,3,1,2,1$ has a longest streak of $7$, a second smaller streak of $3$, and a third smallest streak of $2$
Case E: $1,2,3,4,5,6,7,6,5,4,1,2,3$ has a longest streak of $7$, and a second smaller streak of $3$.
Case F: $1,2,3,4,5,6,7,6,5,4,3,2,1$ has a longest streak of $7$, and no smaller streaks.
The cases A – F are ordered in ‘most sorted to least sorted’, but all have the same length longest streak. Using the averages of streak length is not appropriate: | $\text{Average} = 13/1 = 13$
B: $\text{Average} = (7+6)/2 = 6.5$
C: $\text{Average} = (7+6)/2 = 6.5$
D: $\text{Average} = (7+3+2)/3 = 4$
E: $\text{Average} = (7+3)/2 = 5$
F: $\text{Average} = 7/1 = 7$
Factoring in non-streaks (counting them as 1’s): | {
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A question on summation notation and Raabe's test So I have a sum
$$\sum_{n=1}^\infty\frac{2\cdot4\cdot...\cdot2n}{5\cdot7\cdot...\cdot(2n+3)}$$
which I thought meant $$\sum_{n=1}^\infty\frac{2n}{2n+3}$$ which is trivially convergent by the limit test, and by Raabe's Test (which i'm currently practicing). But Taylor and Mann's book says this series is convergent. So I'm assuming These two series are not equivalent. If that is the case, how do I go about testing for convergence using Raabe's Test?
| The series
$$
\sum_{n=1}^\infty\frac{2n}{2n+3}\tag{1}
$$
diverges because the terms don't tend to $0$.
The ratio of the terms of the series
$$
\sum_{n=1}^\infty\frac{2\cdot4\cdot6\cdots2n}{5\cdot7\cdot9\cdots(2n+3)}\tag{2}
$$
is
$$
\frac{2n}{2n+3}\tag{3}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}\frac{2n}{2n+3}
&=\lim_{n\to\infty}\frac{2}{2+3/n}\\
&=1\tag{4}
\end{align}
$$
and
$$
\begin{align}
\lim_{n\to\infty}n\left(\frac{2n}{2n+3}-1\right)
&=\lim_{n\to\infty}\frac{-3n}{2n+3}\\
&=\lim_{n\to\infty}\frac{-3}{2+3/n}\\
&=-\frac32\\
&\lt-1\tag{5}
\end{align}
$$
$(4)$ and $(5)$, by Raabe's Test, indicate convergence.
As a check, we can also use the AM-GM Inequality to show
$$
\sqrt{(2n+2)(2n+4)}\le2n+3\tag{6}
$$
Therefore
$$
\begin{align}
\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{5}\cdot\color{#00A000}{7}\cdot\color{#0000FF}{9}\cdots\color{#C08000}{(2n+3)}}
&\le\sqrt{\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{4}\cdot\color{#00A000}{6}\cdot\color{#0000FF}{8}\cdots\color{#C08000}{(2n+2)}}}\sqrt{\frac{2\cdot4\cdot6\cdots2n}{\color{#C00000}{6}\cdot\color{#00A000}{8}\cdot\color{#0000FF}{10}\cdots\color{#C08000}{(2n+4)}}}\\
&=\sqrt{\frac{2}{2n+2}}\sqrt{\frac{2\cdot4}{(2n+2)(2n+4)}}\\
&\le\frac{\sqrt2}{(n+1)^{3/2}}\tag{7}
\end{align}
$$
then apply the $p$-test and comparison test to show that $(2)$ converges.
| {
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Definite integration evaluation of $\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$ OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\frac{2I}{\pi}=\int_{0}^{\pi}\frac{1}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What should I do ahead to get the value I need? Any tips? (Thanks in advance)
| I prefer to the following method:
\begin{align*}
I
:= \int_{0}^{\pi} \frac{x}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \, dx
&= \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x )^2} \, \sec^2 x \, dx.
\end{align*}
Now we make the substitution $b \tan x \mapsto a \tan x$. Then
\begin{align*}
I
&= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} \frac{b^2 + a^2\tan^2 x}{(1 + \tan^2 x )^2} \, \sec^2 x \, dx \\
&= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} ( b^2 \cos^2 x + a^2\sin^2 x ) \, dx \\
&= \frac{\pi}{(ab)^3} \cdot \frac{\pi}{4} \left( a^2 + b^2 \right)
= \frac{\pi^2(a^2 + b^2)}{4(ab)^3}.
\end{align*}
| {
"language": "en",
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What are conditions on $x$ that make roots of $y^2 + 2xy - 2 = 0$ greater than 1 in absolute value? Original question: How do we solve $\sqrt{x^2+2}<x-1$? If $x>1$, the solution is $x<-0.5$ which does not make sense. What if $x<1$?
Update: Thank you for all answers. Actually, the original question was: "What are conditions on $x$ that make the roots of the following equation in absolute value strictly greater than one?"
$$y^2 + 2xy - 2 = 0$$
The discriminant is $D = 4x^2 +8$ and so solutions are $y_1 = -x-\sqrt{x^2+2}$ and $y_2 = -x+\sqrt{x^2+2}$. That's how I arrived to the original inequality. One condition would be $-0.5<x<0.5$. However, I discarded cases when there are complex roots.
| If you're only interested in real solutions then firstly you must have $x^2 + 2 > 0$, so $x^2 > -2$ but $x^2 > 0$ for all $x \in \mathbb{R}$ so you must have $x>0$.
Assuming $x\geq1$ and squaring both sides gives $x^2 +2 < (x-1)^2$ expanding gives $x^2 + 2 < x^2 - 2x + 1$ which can be written as $1 < -2x$ and clearly there are no solutions to this since $x\geq1$. So if a solution $x$ exists, we must have $0<x<1$ but no such solution can exist because when $0<x<1$, $\sqrt{x^2+2} > 0$ yet the other side of the inequality, $x-1<0$. So no such solution in $\mathbb{R}$ exists.
| {
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"source": "stackexchange",
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What is the remainder when $x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$ is divided by: $x^2-8x+12$. What is the remainder when: $$x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$$ is divided by: $$x^2-8x+12$$
$x^2-8x+12$ $\leftrightarrow (x-2)(x-6)$
This gives me the polynomial: $(x-2)(x-6)k(x) + r(x)$, where $r(x)$ is the remainder.
The expression gives me the remainder when $x=2$ or $x = 6$
By replacing $x$ with $2$ into the original polynomial you get: $$x^{100} - 4x^{100}+3x^{100}-3x^{10}+12x^{10}-9x^{10}+3x^{2}-29x + 41 = 3x^2-29x + 41$$
By plugging in $2$ I get: $12-58 + 41 = -5$.
Hence $r(x)$ should be: $-5$.
This is apparently the wrong answer. What did I do wrong?
Thank you kindly for your help!
| It is worth noting that
$$
\begin{multline}
x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41\\
=x^{98}(x^2-8x+12)-3x^8(x^2-8x+12)+3(x^2-8x+12)-5x+5
\end{multline}
$$
so that the remainder is $-5x+5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$ T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix}
$$
Then the standard matrix for T is...
I'm not exactly sure how to approach this problem.
Could anyone explain how to solve this problem?
| You can put into a matrix given vectors and their images. If you then do elementary row operations, this property is not changed. (After each step you have in each row a vector and its image. This is because of linearity.)
If you manage to obtain the identity matrix on the left, then you know the images of the vectors from the standard basis, which is sufficient to obtain the matrix of your linear transformation. (Since you're using column vectors, the result is the transpose of the matrix on the right. It is worth mentioning that some authors prefer row vectors - in such case the matrix wouldn't be transposed. See also the Wikipedia article on Row and column vectors.)
$
\left(\begin{array}{ccc|ccc}
-2 & 3 & -4 & 5 & 3 & 14\\
3 & -2 & 3 & -4 & 6 &-14\\
-4 & -5 & 5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 1 & -1 & 1 & 9 & 0\\
3 & -2 & 3 & -4 & 6 &-14\\
-4 & -5 & 5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 1 & -1 & 1 & 9 & 0\\
0 & -5 & 6 & -7 &-21 &-14\\
-4 & -5 & 5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 1 & -1 & 1 & 9 & 0\\
0 & -5 & 6 & -7 &-21 &-14\\
0 & -1 & 1 & -2 & -4 &-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 5 &-2\\
0 & -5 & 6 & -7 &-21 &-14\\
0 & -1 & 1 & -2 & -4 &-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 5 &-2\\
0 & 1 & -1 & 2 & 4 & 2\\
0 & -5 & 6 & -7 &-21 &-14
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 5 &-2\\
0 & 1 & -1 & 2 & 4 & 2\\
0 & 0 & 1 & 3 & -1 &-4
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 5 &-2\\
0 & 1 & 0 & 5 & 3 &-2\\
0 & 0 & 1 & 3 & -1 &-4
\end{array}\right)
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Help with particular solution to solving $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$. I asked a question about how to solve: $$z^4−2z^3+9z^2−14z+14 = 0$$
When all you know is that there is a root with the real part of 1. I was given great answers and you can find the question here.
(If you feel that this question is redundant or that I should rather update my old question, please feel free to delete this one).
After working some more with this question I think I have found out that the particular solution that I should have been using is the following:
I know that there must be a root: $1 + yi$ and its complex conjugate $1 -yi$.
$z-(1+yi)$ and $z + (1-yi)$ must therefore both be factors to the equation above.
There should therefore be a factor: $(z-1-yi)(z - 1+yi)=z^2-2z+(1+y^2)$.
If I divide the polynomial:
$$(z^4 - 2z^3 + 9z^2 - 14z + 14)$$
With: $$(z^2-2z+(1+y^2))$$ I get the quotient:
$$z^2 + (8-y^2)$$
And the remainder: $$2z(8-y^2)+6-7y^2+y^4 \longleftrightarrow 2z(8-y^2)+6+(-7+y^2)y^2$$
My strategy: when: $2z(8-y^2)+6+(-7+y^2)y^2 = 0$. I will have the possible values of $y$ and the rest of the roots can be found when the quotient $z^2 + (8-y^2) = 0$.
First of all, is this a viable strategy for solving this problem?
Secondly how do I procede and solve for when $2z(8-y^2)+6+(-7+y^2)y^2 = 0$ ?
Thank you!
| Right as far as it goes. Just that you made a mistake: The remainder is $(2 - 2 y^2) z + (y^4 - 7 y^2 + 6)$.
Now, for the proposed factor really to be a factor, this remainder must be the zero polynomial, that is:
$$
\begin{align*}
2 - 2 y^2 &= 0 \\
y^4 - 7 y^2 + 6 &= 0
\end{align*}
$$
From the first one you get $y = 1$, and that checks with the second one, so your roots are $1 \pm i$, and the factor is $z^2 - 2 z + 2$. The other one is $z^2 + 7$, with respective roots $\pm i \sqrt{7}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim\limits_{n\rightarrow \infty}\int_1^3\frac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and evaluate it. I am trying to show that $\lim_{n\rightarrow \infty}\int_1^3\dfrac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and what its value is. I know that to do this I must show that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$ and that each $\dfrac{nx^{99}+5}{x^3+nx^{66}}$ is integrable on $[1,3]$ and the rest will follow. I am having a difficult time showing that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$. So far my proof for uniform convergence is as follows.
Let $\epsilon > 0$, choose $N\in \mathbb{N}$ such that $\left|\frac{4}{N+1}\right|<\epsilon$. Then $n\geq N$ implies
\begin{align*}
\left|\frac{nx^{99}+5}{x^3+nx^{66}}-x^{33}\right|&=\left|\frac{nx^{99}+5-x^{33}(x^3+nx^{66})}{x^3+nx^{66}}\right|\\
&=\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\\
&\leq\left|\frac{4}{n+1}\right|\\
&\leq\left|\frac{4}{N+1}\right|<\epsilon
\end{align*}
and so we have uniform convergence on $[1,3]$.
| The integrand simplifies to $\dfrac{nx^{96}}{1+nx^{63}}+\dfrac{5}{x^3+nx^{66}}$.
The second term is less than $\dfrac{5}{n}$ on our interval, so it is harmless.
For the first term, divide. We get $x^{33}-\dfrac{x^{33}}{1+nx^{63}}$. The function $\dfrac{x^{33}}{1+nx^{63}}$ is less than $\dfrac{1}{n}$ on our interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help understanding the solution to #83 Section 9.1 Calculus 9e, by Larson.
Find $a_n$ of $1, \; \frac{-1}{1\times 3}, \; \frac{1}{1\times3\times5}, \frac{-1}{1\times 3\times 5 \times 7}$
I solved it as $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$
The solution guide says: $\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n\right)!} $
I am unable to figure out how $2^n$ gets into the numerator.
if you multiply $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$ by $\dfrac{2n}{2n}$ results in $\dfrac{\left(-1\right)^{n-1}2n}{\left(2n\right)!}$
Please, help me understand how $2n$ goes to $2^n n!$
Have learned from the comments that the denomator is $\left(2n-1\right)!!$ Sorry, still not sure how to solve this problem. This is my first time learning calculus and haven't seen any double factorials. Hate this about this Larson Calculus book, great problems, but they are like Sherlock Novels, always some missing knowledge not showen that is needed to solve the later problems.
| It is $(2n-1)!!$, not $(2n-1)!$. Use that $(2n)!!=2^n n!$
If you don't know: $(2n-1)!!=(2n-1)(2n-3)\cdot\ldots \cdot3 \cdot 1$ and $(2n)!!=(2n)(2n-2)\cdot \ldots 4 \cdot 2$
Hint: multiple and divide by $(2n)!!$, note that $(2n-1)!! \cdot (2n)!!= (2n)!$
Full solution:
$\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!!}=\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!!} \cdot \dfrac{(2n)!!}{(2n)!!} = \dfrac{\left(-1\right)^{n-1}(2n)!!}{\left(2n-1\right)!!(2n)!!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n-1\right)!!(2n)!!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{(2n)!}$
Proof that $(2n)!!=2^n n!$:
\begin{align*}(2n)!!&=(2n)(2n-2)\cdot \ldots \cdot 4 \cdot 2\\ &= (2n)(2(n-1))\cdot \ldots \cdot (2\cdot 2) \cdot (2\cdot 1)\\
&=2^n (n(n-1)\cdot \ldots \cdot 2 \cdot 1) = 2^n n!\end{align*}
Proof that $(2n-1)!! \cdot (2n)!!=(2n)!$:
$\underbrace{(2n-1)(2n-3)\cdot\ldots \cdot3 \cdot 1}_{(2n-1)!!}\cdot \underbrace{ (2n)(2n-2)\cdot \ldots 4 \cdot 2}_{(2n)!!}= (2n)(2n-1)\cdot \ldots \cdot 4 \cdot 3 \cdot 2 \cdot 1=(2n)!$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.
My Try:: Let $x,y,z$ be the roots of the quadratic equation
$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$
Let $xyz = p, $ Then let $f(t) = t^3-3t^2-9t-p.$
Now we will check the extermes value of $f(t)$
$f^{'}(t) = 3t^2-6t-9 = 3(t^2-2t-3)$
$f^{''}(t) = 3(2t-3)$
Now for Max. or Min. , $f^{'}(t)=0$ or $t=-1\;,t=3$
So $f^{''}(-1) = -15=-$(ve)
so Local Max. at $t=-1$
and $f^{''}(3) = 9 = +$(ve)
So Local Min. at $t=3$
Now My question is How can I check Range of $z$ using Graph of cubic equatio using Derivatives
Thanks
| Equating the values of $x$
$$3-y-z=-\frac{9+yz}{y+z}\implies y^2+y(z-3)+z^2-3z-9=0$$
As $y$ is real, the discriminant $(z-3)^2-4\cdot1\cdot(z^2-3z-9)$ must be $\ge0$
$$\implies z^2-2z-15\le0\implies (z-5)(z+3)\le0$$
As $(x-a)(x-b)\le 0$ for $a\le b\implies a\le x\le b$
So,$-3\le z\le 5$
Observe that the given conditions are symmetric wrt $x,y,z$
So, $x,y,z$ will have the same range.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get the sum of the values in a $N \times N$ table? How to get the sum of the values in a $N \times N$ table (without adding repeating products such as $6 \times 7$ and $7 \times 6$ twice and without counting perfect squares)?
Figured out that
$1 \cdot 0+2 \cdot 1+3 \cdot (1+2)+4 \cdot (1+2+3)+\dots+n\cdot (1+2+3+\dots+(n-1))=1\dbinom{1}{2}+2\dbinom{2}{2}+3\dbinom{3}{2}+4\dbinom{4}{2}+\dots+n\dbinom{n}{2}$
At this point I'm completely stuck. What do I do to get an exact number e. g. for $n=50$, $n=100$ etc.?
| If I understand correctly, you want
$$\begin{align*}
\sum_{i=1}^{N-1}\sum_{k=i+1}^Nik&=\sum_{k=2}^N\sum_{i=1}^{k-1}ik\\
&=\sum_{k=2}^Nk\sum_{i=1}^{k-1}i\\
&=\sum_{k=2}^Nk\left(\frac{k(k-1)}2\right)\\
&=\frac12\sum_{k=2}^N\left(k^3-k^2\right)\\
&=\frac12\sum_{k=2}^Nk^3-\frac12\sum_{k=2}^Nk^2\\
&=\frac12\left(\frac14N^2(N+1)^2-1\right)-\frac12\left(\frac16N(N+1)(2N+1)-1\right)\\
&=\frac18N^2(N+1)^2-\frac1{12}N(N+1)(2N+1)\\
&=\frac{N(N+1)}{24}\Big(3N(N+1)-2(2N+1)\Big)\\
&=\frac1{24}N(N-1)(N+1)(3N+2)\;,
\end{align*}$$
if I made no careless algebraic errors. This is the sum of all products of unordered pairs of integers from the set $\{1,\dots,N\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.
Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.
Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.
In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.
Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.
Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.
Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.
We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
| Here is a more geometric flavored proof:
$x^2+y^2-3=0 \iff x^2+y^2 = \sqrt{3}^2$ is a circle with radius $\sqrt{3}$ centered at the origin. Think of the points along the circle as polar coordinates $(r, \theta)$, i.e.
\begin{equation}(\sqrt{3}, \theta) \text{ where } 0 \leq \theta \leq 2\pi \end{equation}
The formulas for converting a polar coordinate into a cartesian coordinate is just right triangle trigonometry:
\begin{align*}
x &= r\cos\theta \\
y &= r\sin\theta
\end{align*}
Since $r=\sqrt{3}$, we have that
\begin{align*}
x &= \sqrt{3}\cos\theta \\
y &= \sqrt{3}\sin\theta
\end{align*}
Then you can say that $\sqrt{3}$ multiplied by any number between $-1$ and $1$ is irrational.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$
Then I am really stuck. Could someone help me?
| $$I=\int_0^{\pi} \frac{-x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(x-\pi)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)$$
$$\Rightarrow I=\frac{\pi}{2}\int_0^{\pi}\frac{-\sin x}{1+\cos^2 x}\,dx$$
Let $t=\cos x:$
$$I=\frac{\pi}{2}\int_{-1}^{1}-\frac{1}{1+t^2}\,dt=-\frac{\pi^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
$\mathcal{D}$-classes Let $$\alpha = \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right) \in \mathcal{T}_3\text{.}$$
(a) Show that the $\mathcal{D}$-class of $\alpha$ contains all those elements of $\mathcal{T}_3$ which have the same rank (cardinality of their image) as $\alpha$.
(b) Show that the $\mathcal{D}$-class of $\alpha$ contains 3 $\mathcal{R}$-classes and 3 $\mathcal{L}$-classes and that it has 2 element $\mathcal{H}$-classes.
How to do (b)?
From (a) we have that $D_{\alpha} = \{ \beta \in \mathcal{T}_3 : | \text{Im}( \beta )| = 2 \}$.
Explicitly then:
$$D_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&1&2
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&2&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&3&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&1&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&1
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&2&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&3&2
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&3
\end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&3
\end{array}\right) \right\}$$
| I've had to change accounts as I can't access my email account so I can't reply to any posts because I'm not using the account I originally asked the question from!
So:
$\beta \in R_{\alpha} \iff \text{Ker}(\beta)$ has classes $\{1\},\{2,3\}$ and $\beta \in L_{\alpha} \iff \text{Im}(\beta) = \{1,3\}$
$$R_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&2&2
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&1&1
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&1&1
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
2&3&3
\end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
3&2&2
\end{array}\right) \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Fourier series of function $f(x)=0$ if $-\pi$$f(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\
\sin(x) & \text{if }0<x<\pi.
\end{cases}$$
My attempt:
I went the route of expanding this function with a complex Fourier series.
$$f(x) = \sum_{n=-\infty}^{+\infty} C_{n}e^{inx}$$
$$C_{n} = \frac {1}{2\pi} \int_{0}^{\pi} \frac {e^{ix}-e^{-ix}}{2i} e^{-inx} \,\mathrm dx = \frac {1}{\pi}\left(\frac {1}{1-n^2}\right)$$
because only even $n$ terms survive, odd $n$ are 0
$$
C_0 = \frac {1}{2\pi} \int_{0}^{\pi} \sin(x)\, \mathrm dx = \frac {1}{\pi}
$$
so
$$
f(x) = \frac{1}{\pi} + \frac {1}{\pi} \left(\frac {e^{i2x}}{1-2^2} + \frac {e^{i4x}}{1-4^2}+\frac {e^{i6x}}{1-6^2}+\cdots\right) + \frac {1}{\pi} \left(\frac {e^{-i2x}}{1-2^2} + \frac {e^{-i4x}}{1-4^2}+\frac {e^{-i6x}}{1-6^2}+\cdots\right)
$$
In sine and cosine terms,
$$
f(x) = \frac{1}{\pi} + \frac {2}{\pi} \left(\frac {\cos(2x)}{1-2^2} + \frac {\cos(4x)}{1-4^2}+\frac {\cos(6x)}{1-6^2}+\cdots\right)
$$
But the answer in my book is given as
$$
f(x) = \frac{1}{\pi} + \frac{1}{2} \sin(x)+ \frac {2}{\pi} \left(\frac {\cos(2x)}{2^2-1} + \frac {\cos(4x)}{4^2-1}+\frac {\cos(6x)}{6^2-1}+\dotsb\right)$$
I don't understand how there is a sine term and the denominator of the cosines has $-1$.
| Since the function $\phi(t)$ is defined on $-L=-\pi<t<\pi=L$ , the Fourier series can be expressed as shown below :
The numerical tests of the formula are well consistent with a good accuracy.
On the figure below, small values of $m$ are taken in order to make clear the deviations in case of series limited to not enough terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
$T(n) = 2T(n/2) +\lg(n!) $ asymptotic bounds $T(n) = 2T(n/2) + \lg(n!)$
What are the upper and lower bounds of this equation?
| $$\log(n!) \sim n\log(n)$$
Hence, we have
$$T(n) \sim 2T(n/2) + n \log(n)$$
Setting $n=2^k$, and calling $T(2^k) = g(k)$, we get that
\begin{align}
g(k) & \sim 2g(k-1) + k2^k \log(2) \sim k 2^k \log(2) + 2((k-1)2^{k-1} \log(2) + 2g(k-2))\\
& \sim k 2^k \log(2) + (k-1)2^k \log(2) + 4 g(k-2)\\
& \sim k 2^k \log(2) + (k-1)2^k \log(2) + 4 ((k-2)2^{k-2} \log(2) + 2g(k-3))\\
& \sim k 2^k \log(2) + (k-1)2^k \log(2) + (k-2)2^{k} \log(2) + 8g(k-3)\\
& \sim k 2^k \log(2) + (k-1)2^k \log(2) + (k-2)2^{k} \log(2) + \cdots + 2^k \log(2) + 2^kg(0)\\
& \sim k(k+1) 2^{k-1} \log(2) + 2^k g(0)\\
& \sim (k+1) 2^{k-1} \log(2^k)
\end{align}
Hence,
$$T(n) \sim \dfrac{n\log^2(n)}{2 \log 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
On prime numbers let $q$ be a prime
let $p = 2^q -1 $
is p must be prime always for any prime q ?
is this is true always ?
or it is false for some prime q ?
if it is false , give an example to show that there is a prime q such that
$2^q -1$ is not a prime
thanx
| Here's some example computations to suggest a way you might try to prove that $q$ must be prime:
$$x^2 - 1 = (x - 1)(x + 1)$$
So:
$$x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$$
And so:
$$2^6 - 1 = 63 = (2^3 - 1)(2^3 + 1) = 7 \cdot 9$$
Or:
$$x^3 - 1 = (x - 1)(x^2 + x + 1)$$
So:
$$x^{15} - 1 = (x^5)^3 - 1 = (x^5 - 1)(x^{10} + x^5 + 1)$$
And so:
$$2^{15} - 1 = 32767 = (2^5 - 1)(x^{10} + 2^5 + 1) = 31 \cdot 1057$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove $\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e$ I need to prove that $$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e.$$ I tried to activate the identity of $\sin^2x+\cos^2x=1$ but I'm still stuck with $$\left(\cfrac{1}{\sin\frac1x-\cos\frac1x}\right)^x.$$
Can I get a hint?
| My attempted solution without using asymptoptic arguments:
$$
\sin{\frac{1}{x} } = \frac{1}{x} - \frac{1}{3!x^3}+\frac{1}{5!x^5}-\ldots \\
\cos{\frac{1}{x} } = 1 -\frac{1}{2!x^2} + \frac{1}{4!x^4} -\ldots\\
e^{x^{-1}}- \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = 2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right) \Longrightarrow \\
e^{x^{-1}}+ \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = -2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
\left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x = \\ \lim_{x\to\infty}\left( e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\right)^x = \\ \lim_{x\to\infty} \left(e^{x^{-1}} \right)^x\left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \\ e\lim_{x\to\infty} \left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \ldots = e
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
$(x-a)(x-b)(x-c)(x-d)=ex$ We can verify that $x=125,162,343$ are the roots of equation $(x-105)(x-210)(x-315)=2584x$.
My question is,Could you find five positive integers $a,b,c,d,e$, which $(x-a)(x-b)(x-c)(x-d)=ex$ has four positive integer roots?
Thanks in advance.
| Thank you for your attention, some one has offered me some examples:
$(x-5)(x-6)(x-16)(x-18)=84x,x=4,9,12,20;$
$(x-3)(x-4)(x-16)(x-22)=280x,x=2,8,11,24;$
$(x-3)(x-6)(x-32)(x-44)=2520x,x=2,11,24,48.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
How to calculate volume of $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ area by tripple integral? I am calculating volume of body that is defined by $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ to do this I have two possible ways:
*
*$$\int\limits_0^2\int\limits_0^{2-x}\int\limits_0^{2-x-y}1\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}dx-\int\limits_1^2\int\limits_1^{2-x}\int\limits_1^{2-x-y}1\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x=\frac{2}{3}$$ where the first integral is volume without $z=1$ and second integral is the part over $z=1$.
*I expect that $z = 0, x+y+z=2, x = 0, y = 0$ is half of cube $2\times2\times2$. Volume of that cube is $8$ so volume of half is $4$. The part over $z=1$ is half of cube $1\times1\times1$. Volume of this cube is 1 and volume of half is $\frac{1}{2}$. So volume I trying to find is $4-\frac{1}{2}=\frac{7}{2}$
At least one of (1), (2) must by wrong. What is right volume of area defined by $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ ?
| If you look at a cross-section of the volume in a plane $z=\text{constant}$, then the region is a right triangle of area
$$\frac{1}{2} (2-z)^2$$
The volume is then
$$\int_0^1 dz \: \frac{1}{2} (2-z)^2 = \left [ \frac{1}{6} (2-z)^3\right ]_1^0 = \frac{7}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$ when $z_k$ solves $z^3+8=0$ The problem in the book is
Compute $\int_0^\infty \frac{dx}{x^3+8}$.
I set up the keyhole contour, apply the residue theorem, and go through the tedious algebra. I get stuck in doing so, but looking at the solution, I see the following remark:
Note then that $\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$.
Here, $z_k$ represents one of the poles of the function $\frac{\log z}{z^3+8}$. What I want to know is how the author makes this claim.
I am trying to compute the residues using the simple formula
$$\textrm{Res}(f(z); z_k) = \frac{A(z_k)}{B'(z_k)}$$
when $z_k$ is a simple pole. In this case, letting $B(z) = z^3+8$, I have the following:
$$z_k = -2, 1\pm\sqrt{3}i.$$
This yields the following residues
$$\begin{align*}
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=-2\right) &= \frac{1}{12}\left(\log 2 + i\pi\right), \\
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=1+\sqrt{3}i\right) &= \frac{2}{12(-1+\sqrt{3}i)}\left(\log 2 + \frac{i\pi}{3}\right), \\
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=1-\sqrt{3}i\right) &= \frac{2}{12(-1-\sqrt{3}i)}\left(\log 2 - \frac{i\pi}{3}\right).
\end{align*}
$$
However, when I sum the results, I get a non-zero real and imaginary part. What am I doing wrong?
| OK, I am convinced that your mistake lies in calling one of the poles $z=e^{-i \pi/3}$. For your keyhole contour, it should be $z=e^{i 5 \pi/3}$.
To see this, I am just going to write down the equation you get after applying the contour and the residues:
$$-\int_0^{\infty} \frac{dx}{x^3+8} = \frac{\log{2}+i \pi/3}{12 e^{i 2 \pi/3}} + \frac{\log{2}+i \pi}{12} + \frac{\log{2}+i 5\pi/3}{12 e^{i 4 \pi/3}}$$
You can show that the $\log{2}$'s cancel. The rest of it is
$$\frac{i \pi}{12} \frac{1}{3} [ e^{-i 2 \pi/3} + 3 + 5 e^{i 2 \pi/3}] = \frac{i \pi}{12} \frac{1}{3} (-3+3+i 2 \sqrt{3}) = -\frac{\pi}{6 \sqrt{3}}$$
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?
| $$
\underbrace{\vphantom{\frac11}+1}_{\text{length }1}
\underbrace{-\frac12-\frac13}_{\text{length }2}
\underbrace{+\frac14+\frac15+\frac16}_{\text{length }3}
\underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4}
\underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots
$$
The absolute values of the terms of the same-sign block of length $n$ are from
$$
\dfrac1{n(n-1)/2+1}\quad\text{to}\quad\dfrac1{n(n+1)/2}
$$
and the sum of the block must satisfy
$$
\frac2{n+1}=\frac{n}{n(n+1)/2}\le(-1)^{n-1}\text{sum}\le\frac{n}{n(n-1)/2+1}\lt\frac2{n-1}
$$
which tends to $0$.
The absolute value of the sum of the same-sign block of length $n$ and the same-sign block of length $n+1$ is at most
$$
\frac2{n-1}-\frac2{n+2}=\frac6{(n+2)(n-1)}
$$
Thus, the sum of pairs of blocks converge absolutely and the blocks converge to $0$. Thus, the full series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 2
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| It's fairly simple. Look at it this way:
$$
\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \frac{\sum_{k=1}^i 1}{2^i} = \sum_{i=1}^\infty \sum_{k=1}^i \frac{1}{2^i}
$$
From here, we just change the order of addition. Rather than adding along $k$, and then $i$, we add along $j=i-k$, and then along $k$. This turns our double sum into
$$
\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{k=1}^\infty \sum_{j=0}^\infty \frac{1}{2^{j+k}} = \sum_{k=1}^\infty \frac{1}{2^k} \sum_{j=0}^\infty \frac{1}{2^j} = 1\cdot 2 = 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 9
} |
4-th derivative of $(1+x+x^2) / (1-x+x^2) $ using Taylor polynomial for $1/(1-x)$ Using $n$-th Taylor polynomial for $f_1(x)=\frac{1}{1-x}$ with center in $0$, find $4$-th derivative of $f_2(x)=\frac{1+x+x^2}{1-x+x^2}$ in the point $0$ without calculating it's $1$,$2$ or $3$ derivative.
I'm looking for hints, it's a homework. I've tried using the Taylor expansion with Lagrange rest, but it won't work because of restriction of calculating $1-3$ derivatives. Also I don't see connection between $f_1$ and $f_2$
Thanks in advance for help!
| $$f_2(x)=\frac{1+x+x^2}{1-x+x^2}=\frac{(1+x+x^2)(1+x)}{(1-x+x^2)(1+x)}=\\
=\frac{(1+x+x^2)(1+x)}{1+x^3}=
=\frac{1+2x+2x^2+x^3}{1+x^3}=\\
=1+2\frac{1+x}{1+x^3}=1+2 \frac{1}{1+x^3}+2x\frac{1}{1+x^3}.$$
Added:
From Taylor's polynomial for
$$\frac{1}{1-x}=\sum\limits_{k=1}^{n}{\;x^k}+o(x^n)$$
we have
$$\frac{1}{1+x^3}=\sum\limits_{k=1}^{n}{(-1)^k (x^3)^k}+o((x^3)^n)=\sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+o(x^{3n}),$$
so
$$f_2(x)=1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+2x \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}=\\
=1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}} + 2\sum\limits_{k=1}^{n}{(-1)^k x^{3k+1}}+o(x^{3n+1}).$$
In this expansion coefficient $a_4$ at $x^4$ equals $-2.$ On the other hand, $a_4=\frac{f_2^{(4)}(0)}{4!},$ therefore,
$$f_2^{(4)}(0)=(-2)\cdot 4!=-48.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$
| Hint:
$$\tan(2x) = \tan(x) \quad \Rightarrow \\
\frac{\sin(2x)}{\cos(2x)} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\sin(x)\cos(x)}{2\cos^2(x) -1} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\cos^2(x)}{2\cos^2(x) - 1} = 1 \quad \text{ or }\quad \sin(x) = 0.
$$
It looks like you got the "other solution $\sin(x) = 0$. Of course you can't have $0 = -1$. All that means is just that $2\cos^2(x)$ is never equal to $2\cos^2(x) = 1$. So that specific equation does not have any solution. But again, there is the possibility that $\sin(x)= 0$ (which is equivalent to $\tan(x) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
| Why not just do it this way:
For $x + 4 > 0$ the following is legal:
$$
\begin{align*}
\frac{4}{x + 4} &\le 2 \\
\frac{x + 4}{4} &\ge \frac{1}{2} \\
x + 4 &\ge 2 \\
x &\ge -2
\end{align*}
$$
If $x + 4 < 0$, i.e. $x < -4$, the inequality is satisfied.
The solution is $(-\infty, -4) \cup [-2, \infty)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^9-x^5+x-2=0$ , how to prove $\sqrt[11]{3}If $x^9-x^5+x-2=0$ is known,
how to prove $\sqrt[11]{3}<x<\sqrt[10]{3}$ ，$\sqrt[7]{2}<x<\sqrt[6]{2}$。
| Hint: $x^9-x^5+x=2 \implies \dfrac{((x^4)^3+1)}{x^4+1}=\dfrac{2}{x}$
Spoiler:
Wolframalpha gives only real solution to this, i.e $x \approx 1.112$, which is in between $3^{\frac{1}{11}}$ and $3^{\frac{1}{10}}$ OR(as Thomas Andrews says) $2^{\frac{1}{6}}$ and $3^{\frac{1}{7}}$ .
| {
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Prove: $\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ How can I show:
$$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
I've tried multiplying with its "conjugate" but that doesn't seem to help that much.
Thanks!
| $\displaystyle(x+1)^{1/3}-x^{1/3}=\frac{1}{3c^{2/3}},c\in(x,x+1)$(By mean value theorem)
$\displaystyle \frac{1}{3(x+1)^{2/3}}\le\frac{1}{3c^{2/3}}\le\frac{1}{3x^{2/3}},\forall c\in(x,x+1)$
So we have ,
$\displaystyle\frac{1}{3(x+1)^{2/3}}\le(x+1)^{1/3}-x^{1/3}\le \frac{1}{3x^{2/3}}$
$\Rightarrow \displaystyle\frac{x^{2/3}}{3(x+1)^{2/3}}\le x^{2/3}((x+1)^{1/3}-x^{1/3})\le \frac{1}{3}$
$\Rightarrow \displaystyle\frac{1}{3(1+\frac{1}{x})^{2/3}}\le x^{2/3}((x+1)^{1/3}-x^{1/3})\le \frac{1}{3}$
Taking limit as $x\to \infty$ we have,
$\Rightarrow \displaystyle\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}\le \lim_{x\to \infty}x^{2/3}((x+1)^{1/3}-x^{1/3})\le \lim_{x\to \infty}\frac{1}{3}$
Using squeeze theorem, $\Rightarrow \lim_{x\to \infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ (As $\lim_{x\to \infty}\frac{1}{3(1+\frac{1}{x})^{2/3}}=1/3=\lim_{x\to \infty}\frac{1}{3}$ )
| {
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$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
| We know $$f(x-1)=a(x-1)^2+b(x-1)+c\quad\text{and}\quad f(x-2)=a(x-2)^2+b(x-2)+c.$$ Expand these terms, add them, and combine like terms via powers of $x$. Now you can get three equations in three variables by equating the coefficients of the left with the right since you know $$f(x-1)+f(x-2)=5x^2-2x+9.$$
| {
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Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$
I have also tried to use the squeeze theorem with two other equations and obtained different values:
$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$
| $$ \displaystyle \lim_{(x,y)\to(0,0)} \frac{(x^2+y^2)}{(x^2-y^2)}+ix $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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2 test for convergence problems: $\sum^{\infty}_{k=1}5^k/(3^k+4^k)$and $\sum^{\infty}_{n=1}\tan\left(1/n\right)$ For the first problem: $$\sum^{\infty}_{k=1}\frac{5^k}{3^k+4^k}$$
I tried to take the ratio test but was having trouble simplifying $$\frac{3^k + 4^k}{3^{k+1} + 4^{k+1}}$$
For the second problem: $$\sum^{\infty}_{n=1}\tan\left(\frac{1}{n}\right)$$
I don't know how to approach it.
Thanks for the help as always
| For the first problem
$$\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}$$
$$=\frac{5^{k+1}(3^k+4^k)}{5^k(3^{k+1}+4^{k+1})}$$
$$=5\cdot\frac{\left(\frac34\right)^k+1}{3\cdot\left(\frac34\right)^k+4}$$
So, using Ratio Test: $$\lim_{k\to\infty}\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}=\frac54$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the limiting distribution of a $3 \times 3$ Markov chain This is a question from a book.
Find $\lim_{n\rightarrow \theta}P^n$ where $$P=\begin{pmatrix}0 & 1 & 0\\
\frac{1}{6} & \frac{1}{2} & \frac{1}{3}\\
0 & \frac{2}{3} & \frac{1}{3}
\end{pmatrix}$$
I assumed that there was a typo and in fact what is being asked was $\lim_{n\rightarrow \infty}P^n$. I then proceeded to solve
$$\begin{pmatrix}\pi_{0} & \pi_{1} & \pi_{2}\end{pmatrix}\begin{pmatrix}0 & 1 & 0\\
\frac{1}{6} & \frac{1}{2} & \frac{1}{3}\\
0 & \frac{2}{3} & \frac{1}{3}
\end{pmatrix}=\begin{pmatrix}\pi_{0} & \pi_{1} & \pi_{2}\end{pmatrix}
$$
which converts to the system of equations
$$\begin{align} 0\pi_{0} + \frac{1}{6}\pi_{1} + 0\pi_{2} &= \pi_{0} \\
1\pi_{0} + \frac{1}{2}\pi_{1} + \frac{2}{3}\pi_{2} &= \pi_{1} \\
0\pi_{0} + \frac{2}{3}\pi_{1} + \frac{1}{3}\pi_{2} &= \pi_{2} \\
\pi_0 + \pi_1 + \pi_2 &= 1 \end{align}$$
which doesn't seem to be consistent, since ignoring the final line gives $\pi_0=\pi_1=\pi_2=0$ when I row-reduce. However, the textbook does give a solution of $\pi^T=(0.4, 0.45, 0.15)$. What am I doing wrong?
| There are two issues:
*
*Your third equation has a wrong coefficient. It should be $0\pi_{0}+\color{red}{\frac{1}{3}}\pi_{1}+\frac{1}{3}\pi_{2}=\pi_{2}$.
*The textbook answer doesn't seem to be correct. The solution should be $(\pi_0,\,\pi_1,\,\pi_2)=(0.1,\, 0.6,\, 0.3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $\sum_{n=1}^{\infty}\frac{1}{n k^n}$ How to evaluate this series for $k > 1$?
$$\sum_{n=1}^{\infty}\frac{1}{n k^n}$$
For $k = 2$, I tried to evaluate $\displaystyle \sum_{n = 0}^\infty \int_{1}^{2} x^{-(n+1)}dx = \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx$ $\displaystyle = \int_{1}^{2}\frac{1}{x-1}dx - \int_{1}^{2}\frac 1 x dx = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n2^n}\right)$
But both $\displaystyle \int_{1}^{2}\frac{1}{x-1}dx$ and $\displaystyle \sum_{n=1}^\infty\frac1n$ diverges. The answer is $\ln 2$, are these divergent terms equal?
| $$\frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{nk^{n-1}} =\frac{1}{k} \sum_{n=1}^{\infty} \int^1_0 (x/k)^{n-1} dx=\frac{1}{k} \int^1_0 \sum_{n=1}^{\infty} (x/k)^{n-1} dx = \int^1_0 \frac{1}{k-x} dx= \log \left( \frac{k}{k-1} \right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Circles of radius $2$ passing through origin with centers on $x=1$
There are two circles of radius $2$ that have centers on the line $x=1$
and pass through the origin. Find their equations.
Please explain to me what the problem is really saying.
| A circle with centre $(a,b)$ and radius $r>0$ has equation $(x-a)^2+(y-b)^2=r^2$. Since these circles have centre on the line $x=1$ and radius $2$, they have equation $(x-1)^2+(y-b)^2=4$. Since the circles pass through the point $(0,0)$, we have $(0-1)^2+(0-b)^2=4$, so $b^2+1=4$. Solving for $b$ gives $b=\pm\sqrt 3$, so the circles are given by $(x-1)^2+(y-\sqrt 3)^2=4$ and $(x-1)^2+(y+\sqrt 3)^2=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| $k(k+1) = \frac{1}{3}((k+1)^3-k^3-1)$. So all the $k$'s cancel, except the first and last. We get:
$\sum_1^n k(k+1) = \frac{1}{3}\sum_1^n ((k+1)^3-k^3-1) = \frac{1}{3}((n+1)^3-n-1) = \frac{1}{3}n(n+1)(n+2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| Starter: Since $a^3-b^3=(a-b)(a^2+ab+b^2)$, and by Fermat'little theorem, we conclude that, under our condition, $a\equiv b\pmod 3$.
Suppose the contrary: $a, b$ are relatively prime to $3$. So $a=3k\pm1, b=3l\pm1$. Then $a^2=9k^2\pm6k+1$,
$b^2=9l^2\pm6l+1$,
$ab=9kl\pm3(k+l)+1$.
Summing together, we find: $a^2+ab+b^2=9(k^2+l^2+kl\pm k\pm l)+3$, which is not divisible by $9$, a contradiction. Therefore $a$ and $b$ are both divisible by $3$.
If any mistakes are presented, tell me. Thanks in advance.
| {
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Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
$c$.
I have no idea how to solve these types of problems. Can anybody help me please?
| Because the digits sum to a multiple of $3$, a block of $3$ identical digits is divisible by $3$.
$3$ of those blocks of $3$ digits, having been divided by $3$, is also divisible by $3$.
$3$ of those blocks of $3^2$ digits, having been divided by $3^2$, is also divisible by $3$.
$3$ of those blocks of $3^3$ digits, having been divided by $3^3$, is also divisible by $3$.
and so on. For example,
$\begin{align}
&111/3\\
=&037
\end{align}$
$\begin{align}
&111\ 111\ 111/3^2\\
=&037\ 037\ 037/3\\
=&012\ 345\ 679
\end{align}$
$\begin{align}
&111111111\ 111111111\ 111111111/3^3\\
=&037037037\ 037037037\ 037037037/3^2\\
=&012345679\ 012345679\ 012345679/3\\
=&004115226\ 337448559\ 670781893
\end{align}$
| {
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Identity involving partial sums of Fourier series Suppose $f$ is a continuous periodic function and $S_Nf(x) = \sum^N_{n=−N} \hat f(n) e^{inx}$, where $$\hat f(n)= \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx} dx.$$
How can I show that $$\sum_{j=0}^{N-1}S_jf(x)= \int_{-\pi}^{\pi} \frac{\sin^2(\frac{1}{2}Ny)}{\sin^2(\frac{1}{2}y)}f(x-y)dy?$$
Supposedly it can be done using routine trigonometric manipulation, but I don't see it right away. Thank you in advance.
| This relies on switching the order of summation and integration. For one particular value of $S_k$:
$$\begin{align}S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \sum_{n=-k}^k e^{i k (x-x')} \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1)(x-x')} - e^{-i k (x-x')}}{e^{i (x-x')} -1}\\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}}{\sin{\left[\frac{1}{2}(x-x')\right]}} \end{align}$$
Now we want to evaluate a sum over $k$ of $S_k$:
$$\begin{align}\sum_{k=0}^{N-1} S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{1}{\sin{\left[\frac{1}{2}(x-x')\right]}}\sum_{k=0}^{N-1}\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}\end{align}$$
Now
$$\begin{align}\sum_{k=0}^{N-1}\sin{\left[\left(k+\frac{1}{2}\right)(x-x')\right]}&= \Im{\left[\sum_{k=0}^{N-1}e^{i\left[\left(k+\frac{1}{2}\right)(x-x')\right]}\right]} \\ &= \Im{\left[e^{i\left[\frac{1}{2}(x-x')\right]} \sum_{k=0}^{N-1}e^{i\left[k(x-x')\right]}\right]}\\ &=\Im{\left[e^{i\left[\frac{1}{2}(x-x')\right]}\frac{e^{i N (x-x')}-1}{e^{i(x-x')}-1}\right]}\\ &= \Im{\left[e^{i N (x-x')/2} \frac{\sin{[N (x-x')/2]}}{\sin{[(x-x')/2}]} \right]} \\ &= \frac{\sin^2{[N (x-x')/2]}}{\sin{[(x-x')/2}]}\end{align}$$
Therefore
$$\sum_{k=0}^{N-1} S_k = \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{\sin^2{[N (x-x')/2]}}{\sin^2{[(x-x')/2}]}$$
The stated result follows, save for the factor of $2 \pi$.
| {
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Simple Modulo Question $6x = 9 \pmod{11}$ I am trying to solve $6x = 9\pmod{11}$
The solution suggests notice that $2*6 = 12 = 1 \pmod {11}$
$$6x = 9 \pmod {11}\\12x = 18 \pmod {11}\\x = 7 \pmod{11}$$
I don't get the final step from $12x = 18\pmod{11}$ to the solution.
| The fact that $2\cdot 6 \equiv 12 \equiv 1 \pmod {11}$ tells you that $2$ is the multiplicative inverse of $6 \pmod{11}$, i.e., $\;2\cdot 6 = 12 \equiv 1 \pmod{11}$
$$2\cdot 6 x\equiv 2\cdot 9 \pmod{11} \implies 1\cdot x \equiv 18 \pmod{11}$$ $$\iff \underbrace{x\equiv 18 - 11 \pmod{11} \iff x\equiv 7 \pmod{11}}_{\large *}$$
$(*)$ In short, we see that $18 \equiv 7 \pmod{11},\quad \text{since} 11 \mid (18 - 7) \;\text{i.e.} \;\; 11\mid 11$
Further notes:
It's also true that while $x = 7$ is a solution, $7$ here represents the residue class modulo $11$, which we can denote by $[7]$. $$x = [7] = \{11n + 7\mid n \in \mathbb Z\}$$
There are 11 residue classes modulo $11$:
$\{[0], [1], [2], \cdots , [10]\}$.
For $[k]$ modulo $11$, $0 \leq k \leq 10$, the elements of the class include all integers satisfying: $[k] = \{11n + k\mid n \in \mathbb Z\}$.
| {
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Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| You can decompose your integrand as follows:
$$ \frac{ x^2 + 2x + 5 - x - 1 - 1}{x^2 + 2x + 5} = 1 - \frac{x + 1}{x^2 + 2x + 5} - \frac{1}{(x+1)^2 + 4}$$
You can integrate the first term directly, the second term after the substitution $u = x^2 + 2x + 5$, and the third term by recalling that $(\arctan{x})' = 1/(x^2 + 1)$, and then using another substitution to make the expression look like the derivative of $\arctan$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Basic calculation with root roots and power Sorry for the boring question but I just need someone to remind me the way to calculate this:
$\displaystyle \left(\frac{a}{2}\right)^2 + x^2 = a^2$ (i used carrot sign cause i dont know how to do factorial in my mac keyboard)
The answer is $\displaystyle \frac{\sqrt {3}}{2}a$ but i forgot how to calculate this type of exercise, can someone be kind enough and help?
tnx
| $$\left(\frac a2\right)^2=\frac{a^2}{2^2}=\frac{a^2}4.$$ Thus, using the difference of squares identity $$y^2-z^2=(y-z)(y+z),$$ the following are equivalent: $$\left(\frac a2\right)^2+x^2=a^2\\x^2+\frac{a^2}4-a^2=0\\x^2-\frac{3a^2}4=0\\x^2-\left(\frac{a\sqrt{3}}2\right)^2=0\\\left(x-\frac{a\sqrt{3}}2\right)\left(x+\frac{a\sqrt{3}}2\right)=0.$$ Now what can you conclude from that? (Incidentally, the answer you mention is incorrect.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Riemann-Stieltjes integration with floor function Evaluate $$\int_{\frac{2}{3}}^8 f(x)d\alpha(x)$$ where $\alpha$ is continuous and $f$ is the floor function, that is $f(x)$ is the greatest integer less than or equal $x$.
| $$I = \int_{2/3}^8 \lfloor x\rfloor d \alpha(x) = \int_{2/3}^1 \lfloor x\rfloor d \alpha(x) + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x) = 0 + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x)$$
Now $\lfloor x \rfloor = k$ for $x \in [k,k+1)$. Hence, we get that
\begin{align}
I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\
& = \sum_{k=2}^7 \alpha(k) (-k + k-1) - \alpha(1) + 7 \alpha(8)\\
& = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k)
\end{align}
EDIT
To make the last step clear, let us explicitly write it out and see.
\begin{align}
I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\
& = 1 \cdot(\alpha(2) - \alpha(1)) + 2 \cdot(\alpha(3) - \alpha(2)) + 3 \cdot(\alpha(3) - \alpha(2)) + 4 \cdot(\alpha(4) - \alpha(3))\\
& + 5 \cdot(\alpha(5) - \alpha(4)) + 6 \cdot(\alpha(6) - \alpha(5)) + 7 \cdot(\alpha(8) - \alpha(7))\\
& = -\alpha(1) + (1-2) \cdot \alpha(2) + (2-3) \cdot \alpha(3) + (3-4) \cdot \alpha(4) + (4-5) \cdot \alpha(5) + (5-6) \cdot \alpha(6)\\
& + (6-7) \cdot \alpha(7) + 8 \cdot \alpha(8)\\
& = -\alpha(1) - \alpha(2) - \alpha(3) - \alpha(4) - \alpha(5) - \alpha(6) - \alpha(7) + 7 \alpha(8)\\
& = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
| First,
$$
a^2+3b^2\geq 0
$$
Now,
$$
a^2+2ab+b^2+a^2-2ab+b^2\geq a^2-b^2
$$
Thus,
$$
(a+b)^2+(a-b)^2\geq (a+b)(a-b)
$$
Let $a+b=x, a-b=y$, i.e., $a=\frac{x+y}{2}, b=\frac{x-y}{2}$ which is one-to-one correspondence between $(x,y)$ and $(a,b)$, then,
$$
x^2+y^2\geq xy
$$
Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 25,
"answer_id": 11
} |
How can I determine the Jordan Form of a matrix? How can I go about proving that the characteristic polynomial, minimal polynomial, and the dim(eigenspace) is enough to determine the Jordan Form of a matrix for n<7?
| Let's give it a try:
$$A=\begin{pmatrix}\color{red}0&\color{red}1&\color{red}0&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}1&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}0&0&0&0&0\\
0&0&0&\color{red}0&\color{red}1&0&0\\
0&0&0&\color{red}0&\color{red}0&0&0\\
0&0&0&0&0&1&0\\
0&0&0&0&0&0&1\end{pmatrix}\;,\;\;\;
B=\begin{pmatrix}\color{blue}0&\color{blue}1&\color{blue}0&\color{blue}0&0&0&0&\\
\color{blue}0&\color{blue}0&\color{blue}1&\color{blue}0&0&0&0\\
\color{blue}0&\color{blue}0&\color{blue}0&\color{blue}1&0&0&0\\
\color{blue}0&\color{blue}0&\color{blue}0&\color{blue}0&0&0&0\\
0&0&0&0&\color{blue}0&0&0\\
0&0&0&0&0&1&0\\
0&0&0&0&0&0&1\end{pmatrix}$$
Clearly $\,A\nsim B\,$ since one has Jordan zero blocks of size $\,3,2\,$ resp. (in red) and the other one two Jordan zero blocks of size $\,4,1\,$ resp. (in blue) .
Both matrices have $\,x^5(x-1)^2\,$ as characteristic polynomial, both have $\,x^4(x-1)\,$ as minimal polynomial and both have zero eigenspace of dimension $\,5\,$ (check this!)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\circ}}<2$,then $a=1$
then $$\dfrac{b^2}{16}(1-\cos{(2c)})+\dfrac{b}{4}\sin{c}=1-\sin{50^{0}}$$
so $b=4$
then we have $\sin{c}-\cos{(2c)}=-\sin{50^{0}}$
then my question: How can prove this $c$ must equality 10?
Thank you everyone: yesterday,when I go to bed, I have consider this:let $f(c)=\sin{c}-\cos{(2c)}$.then we have $f(10)=\sin{10}-\cos{20}=\cos{80}-\cos{20}=2\sin{\dfrac{80-20}{2}}\sin{\dfrac{80+20}{2}}=\sin{50}$, by other hand, we have $f'(c)=\cos{c}+2\sin{2c}>0,0<c<\dfrac{\pi}{2}$,so if we $f(c)=f(10)$,we must $c=10$
| Well the most forward way I think of is using the fact that $cos(2c)=1-2sin^2(c)$. This gives you some equation like:
$2X^2+X-(1+sin(50°))=0$ where $X=sin(c)$
Once you have $sin(c)$ you can get $c$ quite easily given your condition over $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
An equation to solve in natural numbers Some clues needed for the equation
$$(3+\sqrt{2})^x-(5-3\sqrt{2})^y=6+9\sqrt{2}, \quad x,y \in \mathbb{N}$$
| Since you want to solve in natural numbers, notice that $0 < 5 - 3\sqrt{2} < 1$. This implies that $ 0 < ( 5 - 3 \sqrt{2} ) ^ y < 1$, and hence
$$6 + 9 \sqrt{2} < (3 + \sqrt{2} )^x < 7 + 9 \sqrt{2} $$
This forces $x = 2$, from which $y = 1$ (from Brian).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?
I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. Are there some good shortcuts?
This is not homework; I'm preparing for an examination.
| $x^3=\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)^3$
$=\left(\sqrt[3]{3}\right)^3+\left(\frac{1}{\sqrt[3]{3}}\right)^3+3\cdot\sqrt[3]{3}\cdot\frac{1}{\sqrt[3]{3}}\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)$
$=3+\frac13+3\cdot\sqrt[3]{3}\cdot \frac{1}{\sqrt[3]{3}}\cdot x$ as $\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}=x$
$=\frac{10}3+3x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$ Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$
I thought about maybe breaking the polynomial in two different fractions in order to make the sum more manageable and reduce it to something similar to $\lim_{n\to\infty}(1+{1\over1!}+{1\over2!}+...+{1\over n!})$, but didn't manage
| Express $$n^4+2n^3-3n^2-8n-3=(n+2)(n+1)n(n-1)+B(n+2)(n+1)n+C(n+2)(n+1)+D(n+2)+E--->(1)$$
So that $$T_n=\frac{n^4+2n^3-3n^2-8n-3}{(n+2)!}=\frac1{(n-2)!}+\frac B{(n-1)!}+\frac C{(n)!}+\frac D{(n+1)!}+\frac E{(n+2)!}$$
Putting $n=-2$ in $(1), E=2^4+2\cdot2^3-3\cdot2^2-8\cdot2-3=1$
Similarly, putting $n=-1,0,1$ we can find $D=0,C=-2,B=0$ .
$$\implies T_n=\frac{n^4+2n^3-3n^2-8n-3}{(n+2)!}=\frac1{(n-2)!}-\frac 2{n!} +\frac 1{(n+2)!}$$
Putting $n=2, T_2=\frac1{0!}-\frac 2{2!} +\frac 1{4!}$
Putting $n=3, T_3=\frac1{1!}-\frac 2{3!} +\frac 1{5!}$
Putting $n=4, T_4=\frac1{2!}-\frac 2{4!} +\frac 1{6!}$
$$\cdots$$
So, the sum will be $$\sum_{0\le r<\infty}\frac1{r!}-2\sum_{2\le s<\infty}\frac1{s!}+\sum_{4\le t<\infty}\frac1{t!}$$
$=\sum_{0\le r<\infty}\frac1{r!}-2\left(\sum_{0\le s<\infty}\frac1{s!}-\frac1{0!}-\frac1{1!}\right)+\sum_{0\le t<\infty}\frac1{t!}-\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\right)$
$$=e-2e+e-\{-2\left(\frac1{0!}+\frac1{1!}\right)+\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}\right)\}=-\frac34$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex analysis graphing an ellipse. Give a geometric argument that $|z-4i|+|z+4i|=10$.
My attempt: $|z-4i|$ represents the distance from an arbitrary point to the coordinate $(0,4i)$ while $|z+4i|$ is the distance from an arbitrary point to the coordinate $(0,-4i)$.
Then stuck; I don't know how this relates to an ellipse...Thanks for the help.
| An ellipse is a set of points such that the sum of the distances from each point in the set to a pair of fixed points is a constant.
Here is a derivation of the equation of an ellipse from this definition:
Imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable. First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic. Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$. We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that a critical point is a minimum I am solving a problem, which is
"Find the point of the paraboloid $P:=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2=z\}$ which is the nearest to the point $(1,1,\frac12)$."
I have already determined (using the Lagrange multipliers method) that $p=(2^{-\frac23},2^{-\frac23},2^{-\frac13})$ is a critical point of $f(x,y,z)=(x-1)^2+(y-1)^2+(z-\frac12)^2$ (which is the square of the distance function) with the restraint $g(x,y,z)=0$, where $g(x,y,z)=x^2+y^2-z$.
I need some help in order to prove that $p$ is the minimum of $f|_P$.
If possible, give me only a hint, as it is homework and I only want some idea to finish the problem by myself.
EDIT: As requested, I'm showing my calculations so far.
A critical point of $f|_P$ is, by the Lagrange multipliers method, one which satisfies $g(x,y,z)=0$ (that is, $x^2+y^2=z$) and $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ for some $\lambda \in \mathbb{R}$.
We have $\nabla f(x,y,z)=(2x-2,2y-2,2z-1)$ and $\nabla g(x,y,z)= (2x,2y,-1)$, which yields the system
$$(2x-2=\lambda 2x)\wedge(2y-2=\lambda 2y)\wedge(2z-1=-\lambda),$$
which is equivalent to
$$(x=\tfrac{1}{1-\lambda})\wedge (y=\tfrac{1}{1-\lambda}) \wedge(z=\tfrac{1-\lambda}{2}).$$
So we get $x=y$ and $2x^2=z$ (because $x^2+y^2=z$), which gives $\frac{2}{(1-\lambda)^2}=\frac{1-\lambda}{2}$, that is, $\lambda = 1-2^\frac23$.
This gives us $x=y=2^{-\frac23}$ and $z=2^{-\frac13}$.
| Prof. Cook is using the multivariate index $D$ applied to the distance-squared function for points on the paraboloid measured from $(1,1,\frac{1}{2})$. It's not the easiest thing to read, but it does establish that the point you found is at the minimal distance from the point external to the paraboloid.
Alternatively, you could put your results for $x$ and $y$ ($x = y = \frac{1}{2z}$), into the distance squared formula to reduce it to a function of $z$ alone:
$$s^2 = (\frac{1}{2z} - 1)^2 + (\frac{1}{2z} - 1)^2 + (z - \frac{1}{2})^2 = \frac{1}{2z^2} - \frac{2}{z} + 2 + z^2 - z + \frac{1}{4}.$$
The second derivative is $2 - \frac{4}{z^3} + \frac{3}{z^4}$, which is positive for $z = 2^{-1/3}$, so $s^2$ is minimized for that value of $z$ (and thus for the point you found). Since the distance $s$ is non-negative, finding the minimum for $s^2$ suffices.
EDIT: D'oh -- it would have been even easier to use $y=x$ and $z=2x^2$ to write
$$s^2 = (x - 1)^2 + (x - 1)^2 + (2x^2 - \frac{1}{2})^2 = 4x^4 - 4x + \frac{9}{4} \Rightarrow \frac{d^2}{dx^2}(s^2) = 48x^2 , $$which is plainly positive for $x = 2^{-2/3}$; it is also easier to see that $\frac{d}{dx}(s^2) = 16x^3 - 4 = 0$ there. (I'd been working with $z$ as the principal variable the first time through...)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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To find the logarithm of $1728$ to the base $2 \sqrt{3}$
Find the logarithm of: $1728$ to base $2\sqrt{3}$.
Let, $\log_{2\sqrt{3}} 1728 = y$, then
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
&= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should I do next to find the logarithm of $1728$?
| Or: $1728 = 12^{3}$ and $12 = (2\sqrt{3})^{2},$ so $1728 = (2\sqrt{3})^{6}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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series involving $\log \left(\tanh\frac{\pi k}{2} \right)$ I found an interesting series
$$\sum_{k=1}^\infty \log \left(\tanh \frac{\pi k}{2} \right)=\log(\vartheta_4(e^{-\pi}))=\log \left(\frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{4}}\Gamma \left( \frac{3}{4}\right)} \right)$$
*
*Does anybody know how to approach this series using Jacobi Theta Function?
*Also, can any one suggest any good papers/books on Jacobi theta functions and Jacobi Elliptic functions?
Thank you very much!
| It may be of interest to note that the sum
$$ g(x) = \sum_{k=1}^\infty \log\tanh (kx)$$
is harmonic and may be evaluated using Mellin transforms, yielding an asymptotic expansion about zero.
The Mellin transform $f^*(s)$ of the base function
$$ f(x) = \log\tanh x$$
may be computed as follows
\begin{align}
f^*(s) & = \mathfrak{M}(f(x); s) =
\int_0^\infty \log \frac{e^x-e^{-x}}{e^x+e^{-x}} x^{s-1} \, dx
= \int_0^\infty \log \left(1 - 2\frac{e^{-x}}{e^x+e^{-x}}\right) x^{s-1} dx \\[6pt]
& = \int_0^\infty \log \left(1 - 2\frac{e^{-2x}}{1+e^{-2x}}\right) x^{s-1} dx
= - \int_0^\infty \sum_{q\ge 1} \frac{1}{q} 2^q e^{-2qx}\left(\frac{1}{1+e^{-2x}}\right)^q x^{s-1} dx \\[6pt]
& = -\sum_{q\ge 1} \frac{1}{q} 2^q \int_0^\infty e^{-2qx}
\sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} e^{-2mx} x^{s-1} dx \\[6pt]
& = -\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m}
\int_0^\infty e^{-2qx} e^{-2mx} x^{s-1} dx \\[6pt]
& =-\Gamma(s)\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m}
\frac{1}{(2m+2q)^s} \\[6pt]
& = -\frac{\Gamma(s)}{2^s} \sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \frac{1}{(m+q)^s}.
\end{align}
To complete this calculation, ask about the coefficient of
$$\frac{1}{n^s} = \frac{1}{(m+q)^s}.$$
It is given by
$$\sum_{m=0}^{n-1} \frac{1}{n-m} 2^{n-m} (-1)^m \binom{n-1}{m} =
- \frac{1}{n} (-1)^n +
\frac{1}{n} \sum_{m=0}^n 2^{n-m} (-1)^m \binom{n}{m} =
\frac{1}{n} \left(1-(-1)^n\right).$$
It follows that the double sum is
$$\sum_{n\ge 1} \frac{1-(-1)^n}{n^{s+1}}=
2 \sum_{m\ge 0} \frac{1}{(2m+1)^{s+1}} =
2 \zeta(s+1) \left(1 - \frac{1}{2^{s+1}}\right) =
\zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$
This gives the following for $f^*(s):$
$$ f^*(s) = -\frac{\Gamma(s)}{2^s} \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$
Now introduce $g(x)$, the harmonic sum we are trying to calculate, so that
$$ g(x) = \sum_{k=1}^\infty f(kx).$$
The Mellin transform of $g(x)$ is then given by
$$ g^*(s) = \mathfrak{M}(g(x); s) =
-\frac{\Gamma(s)}{2^s} \zeta(s) \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$
The zeros of the two zeta function terms cancel the poles of the gamma function, so that inverting $g^*(s)$ we only have two terms that contribute, namely
$$\operatorname{Res}(g^*(s) x^{-s}; s=1) = -1/8\,{\frac {{\pi }^{2}}{x}}$$ and
$$\operatorname{Res}(g^*(s) x^{-s}; s=0) =
1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right).$$
This yields that in a neighborhood of zero
$$ g(x) \sim 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right) -1/8\,{\frac {{\pi }^{2}}{x}}$$
Setting $x=\frac{\pi}{2}$, we obtain that
$$g\left(\frac{\pi}{2}\right) \sim \log 2 - \frac{\pi}{4},$$
which produces only three good digits. On the other hand, for e.g. $x=\frac{1}{4}$, we get
$$g\left(\frac{1}{4}\right) \sim 3/2\,\log \left( 2 \right) +1/2\,\log \left( \pi \right) -1/2\,{\pi }^{2} \sim -3.322716487,$$
which has nine good digits.
For e.g. $x=\frac{\pi}{16}$, we get
$$g\left(\frac{\pi}{16}\right) \sim 5/2\,\log \left( 2 \right) -2\,\pi
\sim -4.5503173557797232034$$
which has 20 good digits.
It seems quite intriguing to ask whether this expansion can also be derived directly from properties of the Jacobi theta function without using Mellin transforms.
| {
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"answer_id": 0
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Value of summation $ij$ What is the value of $\sum \limits_{1 \le i < j \le 10} ij$?
| $$\sum_{1\le r\le n}r\left(\sum_{r+1\le s\le n}s\right)$$
$$=\sum_{1\le r\le n}r\{\sum_{1\le s\le n}s-\sum_{1\le s\le r}s\}$$
$$=\sum_{1\le r\le n}r\{\frac{n(n+1)}2-\frac{r(r+1)}2\}$$
$$=\frac{n(n+1)}2\sum_{1\le r\le n}r- \frac12\{\sum_{1\le r\le n}r^3+\sum_{1\le r\le n}r^2\}$$
Now, you know
$\sum_{1\le r\le n}r=\frac{n(n+1)}2,$
$\sum_{1\le r\le n}r^2=\frac{n(n+1)(2n+1)}6,$
$\sum_{1\le r\le n}r^3=\left(\frac{n(n+1)}2\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/363276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$.
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.
What I did so far is :
plot the y var from $x^2-y^2=1 \implies y=\sqrt{1-x^2}$
Create a distance equation : $d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$ d = \sqrt{x^2+(1-y)^2}$
$ d = \sqrt{2-2\sqrt{1-x^2}}$
$\displaystyle \frac d{dx} (d) =\frac x{\sqrt{1-x^2}\sqrt{2-2\sqrt{1-x^2}}}$
I need to find the max or min?
Any suggestions? thanks!
| Your approach is fine. It is a little easier to minimize $d^2$ than $d$ as that gets rid of one square root. You lost a sign when you went from $d=\sqrt{x^2+(1-y)^2}$ to $d=\sqrt{2-2\sqrt{1-x^2}}$ as the $y^2$ term is positive. Plotting the curve shows that the distance to either branch from $(0,1)$ is the same, so we can use the positive square root $x=\sqrt{1+y^2}$. We have $$d^2=x^2+(1-y)^2=2-2y+2y^2\\\frac {d(d^2)}{dy}=-2+4y$$ which we set to zero and find $y=\frac 12, x=\sqrt{\frac 54}$. The squared distance is then $d^2=\frac 54+\frac 14=\frac 32$ and the linear distance is $d=\sqrt \frac 32$. This must be a minimum, as there are points on the hyperbola very far away.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Evaluate $ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $ using substitution $t=\frac{\pi}{2}-x$ This problem is given on a sample test for my calculus two class. $$ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $$ I can find the value of this integral using other substitutions which lead to partial fractions but the prof added a hint to use the substitution $$t=\frac{\pi}{2}-x $$ so I've been trying to figure out how to do it his way but am pretty lost. Any ideas?
| There is an easy way to solve. Let
$$ A=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}dx, B=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx. $$
Then
$$ A+B=\frac{\pi}{2} $$
and
$$ A-B=\int_0^{\frac{\pi}{2}}\frac{\sin x-\cos x}{\sin x+\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{-d(\sin x+\cos x)}{\sin x+\cos x}=-\ln(\sin x+\cos x)\big|_0^{\frac{\pi}{2}}=0. $$
From this, we have
$$ A=B=\frac{\pi}{4}. $$
Generally, I can use this method to compute (for $a,b>0$)
$$ A=\int_0^{\frac{\pi}{2}}\frac{\sin x}{a\sin x+b\cos x}dx, B=\int_0^{\frac{\pi}{2}}\frac{\cos x}{a\sin x+b\cos x}dx. $$
It is easy to see
$$ aA+bB=\frac{\pi}{2} $$
and
\begin{eqnarray*}
aB-bA&=&\int_0^{\frac{\pi}{2}}\frac{-b\sin x+a\cos x}{a\sin x+b\cos x}dx\\
&=&\int_0^{\frac{\pi}{2}}\frac{d(a\sin x+b\cos x)}{a\sin x+a\cos x}\\
&=&\ln(a\sin x+b\cos x)\big|_0^{\frac{\pi}{2}}\\
&=&\ln a-\ln b.
\end{eqnarray*}
From this, we have
$$ A=\frac{a\pi-2b\ln\frac{a}{b}}{2(a^2+b^2)},B=\frac{b\pi+2a\ln\frac{a}{b}}{2(a^2+b^2)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$ I know how to expand the binomial expansion but I have no idea how to do this second part.
Hence Express $$(3 - \sqrt 2)^4$$ in the form of $$P+Q\sqrt 2$$
Where P and Q are integers, and then how to state the values of P and Q.
| Here is the binomial formula with two complex numbers $a$ and $b$ :
$$(a+b)^n = \sum_{k=0}^n {n \choose k}\ a^{n-k} b^k$$
For $n=4$ you get :
$$\begin{array}{rcl} (a+b)^4 & = & \sum_{k=0}^4 {4 \choose k}\ a^{4-k} b^k \\ & =& {4 \choose 0}\ a^4b^0+{4 \choose 1}\ a^3 b^1+{4\choose 2}\ a^2b^2+{4 \choose 3}\ a^1b^3+{4\choose 4}\ a^0 b^4 \\
&=& a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{array}$$
Now choose $a=3$ and $b=-\sqrt 2$.
You have $a^2=9$; $a^3=27$; and $a^4=81$.
You have $b^2=\left(-\sqrt 2\right)^2=2$; $b^3=b^2b=-2\sqrt 2$ and $b^4=\left(b^2\right)^2=2^2=4$.
Using the above formula, you get :
$$\begin{array}{rcl}\left(3-\sqrt 2\right)^2 & =& 81+4\times 27\times\left(-\sqrt 2\right)+6\times 9\times 2+4\times 3\times \left(-2\sqrt 2\right)+4 \\
& = & 81-108\sqrt 2+108-24\sqrt 2+4 \\
& = & 193-132\sqrt 2\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Series of Functions - Pointwise and Uniform Convergence. I was hoping for some help for the following questions.
Prove that the series $\sum_{n=1}^\infty x^n(1-x)$ converges pointwise but not uniformly on $[0,1]$.
Prove that the series $\sum_{n=1}^\infty (-1)^nx^n(1-x)$ converges uniformly on $[0,1]$.
| $f(x)=\sum_{n=1}^\infty x^n(1-x)=x-x^2+x^2-x^3+x^3-x^4+$..
Note that $f(x)=x $ except $f(1)=0$ and $|S_N-x|=x^{N+1}$ which has supremum $1$ on $[0,1)$
for all $N$.So we don't have uniform convergence .[Pointwise convergence follows from
$x^{N+1} \to 0$ as $ N\to \infty$]
Or you may use the fact that uniform convergence of cont. functions is cont.
$g(x)=\sum_{n=1}^\infty (-1)^nx^n(1-x)$
$=-x+x^2+x^2-x^3+x^4-x^3+x^4-x^5..$
$= -x+2x^2( 1-x+x^2-x^3+...) =-x+2x^2\frac1 {1+x} = \frac {x^2-x}{1+x} $
We again calculate $|S_{2N}-g(x)|=[-x+2x^2-2x^3+...+2x^{2N}-x^{2N+1} ] - \frac {x^2-x}{1+x} = $
$=\frac {x^{2N+1}-x^{2N+2}}{1+x} < min (\frac {1-x}{1+x} , x^{2N+1})$ and supremum of this function approaches $0$ as $N \to \infty . $
[In view of $\lim_{x\to 1} \frac {1-x}{1+x}=0 $ and $x^k $ is uniformly convergent to $0$ on $[0,a] $ for $a<1 $]
Similarly , $|S_{2N+1}-g(x)|$ approaches $0$ as $N \to \infty$.
So we have uniform convergence to $g(x)$ on $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Deriving Closed Form for a Recursion via Generating Functions Consider (1) $a_{n+2} = 2a_{n+1} - a_n + 4n3^n$ with $a_0 = a_1 = 1$.
Using generating functions and setting $A(x) = \sum a_nx^n$ we obtain
$$\begin{align*}&\quad\sum a_{n+2}x^{n+2} = \sum2a_{n+1}x^{n+2} - \sum a_nx^{n+2} + \sum 4n3^nx^{n+2}\\ &\implies [A(x) - a_0 - a_1x] = 2x[A(x)-a_0] - x^2A(x) + \sum_n 4n3^nx^{n+2}\end{align*}$$
Is this correct so far? Is there always a best way to go about rearranging the obtained generating function, or does it vary from problem to problem? Further, is it simpler to use this method here or to instead obtain a particular solution through undetermined coefficients? Any help is much appreciated.
Is it possible to decompose $4n3^nx^{n+2}$ into $x^2$$\sum 4n \times 1/(1-3x)$ and does this help?
| Use Wilf's techniques (see "generatingfunctionology"). Your recurrence is:
$$
a_{n + 2} = 2 a{n + 1} - a_n + 4 n 3^n \qquad a_0 = a_1 = 1
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognizing the resulting sums gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 2 \frac{A(z) - a_0}{z} - A(z) + 4 z \frac{d}{d z} \frac{1}{1 - 3 z}
$$
This gives:
$$
A(z) = \frac{1 - 7 z + 15 z^2 + 3 z^3}{1 - 8 z + 22 z^2 - 24 z^3 + 9 z^4}
= \frac{1}{(1 - 3 z)^2} - 4 \cdot \frac{1}{1 - 3 z}
+ \frac{1}{1 - z} + 3 \cdot \frac{1}{(1 - z)^2}
$$
As we have:
$$
(1 - u)^{-m} = \sum_{n \ge 0} \binom{-m}{n} (-u)^n
= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} u^n
$$
the expression for $A(z)$ gives directly:
$$
a_n = \binom{n + 1}{1} 3^n - 4 \cdot 3^n + 1 + 3 \binom{n + 1}{1}
= (n - 3) \cdot 3^n + 3 n + 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
$\sum_{m=1}^\infty \frac {2^{\widehat m}+2^{-\widehat m}}{2^{m}} =3$ For any positive integer $n$ , let $\widehat n$ denote the integer nearest to $\sqrt n$. Then how to prove that $$\sum_{m=1}^\infty \frac {2^{\widehat m}+2^{-\widehat m}}{2^{m}} =3$$?
| Let $f(m)=m-\widehat m, g(m)=m+\widehat m$, then
\begin{equation}
\sum_{m=1}^{\infty}{\frac{2^{\widehat m}+2^{-\widehat m}}{2^{m}}}=\sum_{m=1}^{\infty}{\left(\left(\frac{1}{2}\right)^{m-\widehat m}+\left(\frac{1}{2}\right)^{m+\widehat m}\right)}=\sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^{f(m)}}+\sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^{g(m)}}
\end{equation}
Note that
\begin{equation}
\widehat{m+1}-\widehat{m}=
\begin{cases}
1 & \text{if} \, m=k^2+k, \, k \in \mathbb{Z}^+\\
0 & \text{otherwise}
\end{cases}
\end{equation}
Thus
\begin{equation}
f(m+1)-f(m)=1-(\widehat{m+1}-\widehat{m})=
\begin{cases}
0 & \text{if} \, m=k^2+k, \, k \in \mathbb{Z}^+\\
1 & \text{otherwise}
\end{cases}
\end{equation}
\begin{equation}
g(m+1)-g(m)=1+(\widehat{m+1}-\widehat{m})=
\begin{cases}
2 & \text{if} \, m=k^2+k, \, k \in \mathbb{Z}^+\\
1 & \text{otherwise}
\end{cases}
\end{equation}
Also $f(1)=0, g(1)=2$. Therefore the sequence $f(m)$ has $f(k^2+k)=k^2, k \geq 1$ occurring twice, and all other non-negative integers occurring once, while the sequence $g(m)$ omits $g(k^2+k)+1=(k+1)^2, k \geq 1$, and contains all other positive integers $ \geq 2$ exactly once.
Thus
\begin{align}
& \sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^{f(m)}}+\sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^{g(m)}} \\
& =\left(\sum_{m=0}^{\infty}{\left(\frac{1}{2}\right)^m}+\sum_{m=k^2, k \geq 1}{\left(\frac{1}{2}\right)^m}\right)+\left(\sum_{m=2}^{\infty}{\left(\frac{1}{2}\right)^m}-\sum_{m=k^2, k \geq 2}{\left(\frac{1}{2}\right)^m}\right) \\
& =\left(\frac{1}{2}\right)^0+2\sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^m} \\
&=3
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/372178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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What's the value of $ y^{(n)}$when $ y=\frac{x^n}{(x+1)^2(x+2)^2}$ What's the value of $\displaystyle y^{(n)}$when $\displaystyle y=\frac{x^n}{(x+1)^2(x+2)^2}$?
My Try:Let $\displaystyle y_n=\frac{x^n}{(x+1)^2(x+2)^2}$,so $\displaystyle y_n=xy_{n-1}$.According to Leibniz's formula,$$y_n^{(n)}=ny_{n-1}^{(n-1)}+xy_{n-1}^{(n)}$$.But I don't konw how to achieve it.
| You can prove inductively that $$y_n = p_n(x) + (-1)^n\left(\frac{-(n+2)}{x+1}+\frac{1}{(x+1)^2} + \frac{2^{n+1}-n2^{n-1}}{x+2} + \frac{2^n}{(x+2)^2}\right)$$
Where $p_n$ is a polynomial of degree less than $n$.
That would let you compute $y_n^{(n)}$. It doesn't appear to give a pretty result, however.
I suspect Andre is right, and the real problem was to compute $y^{(n)}$ at $x=0$.
I arrived at this by noting that if $y_n = p_n(x)+\frac{a_n}{x+1}+\frac{b_n}{(x+1)^2}+\frac{c_n}{x+2}+\frac{d_n}{(x+2)^2}$ we can use the fact that $y_{n+1}=xy_n$ to get a recursive definition of $(a_{n+1},b_{n+1},c_{n+1},d_{n+1})$ in terms of $(a_n,b_n,c_n,d_n)$:$$a_{n+1}=b_n-a_n, b_{n+1}=-b_n, c_{n+1}=d_n-2c_n, d_{n+1}=-2d_n$$ Then I solved the recursion. It is nice that $a_i, b_i$ are never affected by values $c_i,d_i$, so the problem splits to solving two $2$-dimensional linear recurrences, and the matrices are fairly easy:
$$\begin{pmatrix}a_{n}\\b_n\end{pmatrix} = \begin{pmatrix}-1&1\\0&-1\end{pmatrix}^n\begin{pmatrix}a_{0}\\b_0\end{pmatrix}$$
and:
$$\begin{pmatrix}c_{n}\\d_n\end{pmatrix} = \begin{pmatrix}-2&1\\0&-2\end{pmatrix}^n\begin{pmatrix}c_{0}\\d_0\end{pmatrix}$$
Then use that $\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}^n=\begin{pmatrix}1&n\alpha\\0&1\end{pmatrix}$. Finally, find $a_0=-2,b_0=1,c_0=2,d_0=1$ to get the above expression.
I'm wondering if there is a fun way to do this using $$h(x,z)=\sum_{n=0}^\infty y_nz^n = \frac{1}{(1-xz)(1+x)^2(2+x)^2}$$ and computing the partial fractions for this relative to $x$, yielding functions:
$$h(x,z)=\frac{a(z)}{1+x} + \frac{b(z)}{(1+x)^2} + \frac{c(z)}{x+2}+\frac{d(z)}{(x+2)^2} + \frac{f(z)}{1-xz}$$
Where $a(z),b(z),c(z),d(z), \text{ and } f(z)$ are rational functions of $z$. Knowing the power series for these functions then would give us the power series the $n$th partial derivative of $h$ relative to $x$.
It turns out there is a simple relationship between these functions and the sequences $a_i,b_i,c_i,d_i$, namely, $a(z)=\sum_{i} a_iz^i$, and the same for $b,c,d.$
For example, Wolfram Alpha gives me $a(z)=\frac{-z-2}{(1+z)^2}$, which is $\sum_{n=0}^\infty (-1)^{n+1}(n+2)z^n$, and $a_n=(-1)^{n+1}(n+2)$ was just what I got above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
| Here's a way of finding, at the very least, a large class of rational solutions. It seems plausible to me that these are all the rational solutions, but I don't actually have a proof yet...
Say we want to solve $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$ for some fixed $n$. The left-hand side looks an awful lot like the root of a depressed cubic (as it would be given by Cardano's formula). So let's try to build some specific depressed cubic having $n$ as a root, where the cubic formula realizes $n$ as $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$.
The depressed cubics having $n$ as a root all take the following form:
$$(x-n)(x^2+nx+b) = x^3 + (b-n^2)x-nb$$
where $b$ is arbitrary. If we want to apply the cubic formula to such a polynomial and come up with the root $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$, we must have:
\begin{eqnarray}
p&=&\frac{nb}{2}\\
3q^2&=& \frac{(nb)^2}{4}+\frac{(-n^2+b)^3}{27}\\
&=&\frac{b^3}{27}+\frac{5b^2n^2}{36}+\frac{bn^4}{9}-\frac{n^6}{27}\\
&=&\frac{1}{108}(4b-n^2)(b+2n^2)^2
\end{eqnarray}
(where I cheated and used Wolfram Alpha to do the last factorization :)).
So the $p$ that arises here will be rational iff $b$ is; the $q$ that arises will be rational iff $4b-n^2$ is a perfect rational square (since $3 * 108=324$ is a perfect square). That is, we can choose rational $n$ and $m$ and set $m^2=4b-n^2$, and then we will be able to find rational $p,q$ via the above formulae, where
$(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is a root of the cubic
$$
(x-n)\left(x^2+nx+\frac{m^2+n^2}{4}\right)=(x-n)\left(\left(x+\frac{n}{2}\right)^2+\left(\frac{m}{2}\right)^2\right) \, .
$$
The quadratic factor of this cubic manifestly does not have real roots unless $m=0$; since $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is real, it must therefore be equal to $n$ whenever $m \neq 0$.
To summarize, we have found a two-parameter family of rational solutions to the general equation $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$. One of those parameters is $n$ itself; if the other is $m$, we can substitute $b=\frac{m^2+n^2}{4}$ into the above relations to get
\begin{eqnarray}
p&=&n\left(\frac{m^2+n^2}{8}\right)\\
q&=&m\left(\frac{m^2+9n^2}{72}\right) \, .
\end{eqnarray}
To make sure I didn't make any algebra errors, I randomly picked $n=5$, $m=27$ to try out. These give $(p,q)=\left(\frac{1885}{4},\frac{1431}{4}\right)$, and indeed Wolfram Alpha confirms that
$$
\left(\frac{1885}{4}+\frac{1431}{4} \sqrt{3}\right)^{1/3}+\left(\frac{1885}{4}-\frac{1431}{4} \sqrt{3}\right)^{1/3}=5 \, .
$$
| {
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"url": "https://math.stackexchange.com/questions/374619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
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How do you determine the local extrema points for $y=\sqrt{3}\cos(3x)+\sin(3x)$ $$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the derivative for when it's value is $0$ and got $x=\dfrac{\pi}{2}$ . I then used this critical point and subdivided the interval. I found that there were 3 points of local extrema after doing all my work, local maximum $x=0, \dfrac{2\pi}{3}$ and local minimum $x=\dfrac{\pi}{2}$. However according to the online homework, there were 4 different points of extrema. Local maximum $x=\dfrac{\pi}{18},\dfrac{2\pi}{3}$ and local minimum $x=0,\dfrac{7\pi}{18}$. I am really confused as to how there are 4 points of local extrema, did I leave out an answer somewhere? I am also confused as to how they got $x=\dfrac{\pi}{18},\dfrac{7\pi}{18}$ as points of local extrema. Could someone explain this to me?
| $ y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3}) $
$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2} $, another 2 are the end points ie:$x=0$ or $x=\dfrac{3 \pi}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Analytic geometry straight line problem Prove that two straight lines represented by the equation $x^3+y^3+bx^2y+cxy^2=0$ will be at right angles if $b+c=-2$
I didn't know that even straight lines like planes can be represented by a combined equation. can someone please explain how this happens and how to find the individual lines so that the angle between them may be determined.
Thanks
| Suppose you have two lines through the origin, one of slope $\alpha$ and the other of slope $\beta$:
\begin{align*}
l_1:\quad y &= \alpha x & \alpha x - y &= 0 \\
l_2:\quad y &= \beta x & \beta x - y &= 0
\end{align*}
Then you can multiply (powers of) these euqations to obtain a combined equation, which will be zero if and only if one of its factor polynomials is zero:
\begin{align*}
l_1^2 \cdot l_2&: &
\alpha^{2} \beta x^{3}
- (\alpha^{2} + 2 \, \alpha \beta) x^{2} y
+ (\beta + 2 \, \alpha) x y^{2}
- y^{3}
&= 0\\
l_1 \cdot l_2^2&: &
\alpha \beta^{2} x^{3}
- (\beta^{2} + 2 \, \alpha \beta) x^{2} y
+ (\alpha + 2 \, \beta) x y^{2}
- y^{3}
&= 0
\end{align*}
But these equations don't have the required form. It is enough to concentrate on the first equation, since $l_1$ and $l_2$ are structurally equivalent, so you might as well swap them. To turn $l_1^2\cdot l_2$ into the required form, you should multiply it with $-1$ in order to get $+y^3$ instead of $-y^3$. Then you get a monomial of $+x^3$ exactly if $\alpha^2\beta=-1$. So we now want
\begin{align*}
\alpha^2\beta &= -1 &
\\
\alpha^2 + 2\alpha\beta &= b &
\\
\beta + 2\alpha &= -c
\\
b + c &= -2
\end{align*}
Typing this in e.g. Wolfram Alpha you will get one real and two complex solutions:
\begin{align*}
b &= -1 & b &= -1-2i & b &= -1+2i \\
c &= -1 & c &= -1+2i & c &= -1-2i \\
\alpha &= 1 & \alpha &= -i & \alpha &= i \\
\beta &= -1 & \beta &= 1 & \beta &= 1 \\
\alpha\beta &= -1 & \alpha\beta &= -i & \alpha\beta &= i
\end{align*}
The real solution satisifes your requirement: the slopes multiply to $-1$, so the lines are orthogonal. Moreover, there is only a single pair of lines with that property. The complex solutions don't satisfy your requirement, but you'll probably want to exclude these from your consideration in any case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3}\frac{5}{6}\frac{9}{10} \ldots..\frac{(n-1)(n+2)}{n(n+1)}$. now not sure what to do next ?Any idea?
| $$
\begin{align}
\prod_{j=2}^n\left(1-\frac2{j(j+1)}\right)
&=\prod_{j=2}^n\left(\frac{(j-1)(j+2)}{j(j+1)}\right)\\
&=\color{#00A000}{\prod_{j=2}^n\frac{j-1}{j}}
\color{#C00000}{\prod_{j=2}^n\frac{j+2}{j+1}}\\
&=\color{#00A000}{\frac1n}\color{#C00000}{\frac{n+2}{3}}\\
&=\frac13\frac{n+2}{n}
\end{align}
$$
Taking it to the limit, we get
$$
\prod_{j=2}^\infty\left(1-\frac2{j(j+1)}\right)=\frac13
$$
The product of the squares is the square of the product (to get the answer to the modified question).
| {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
| The Newton-Raphson method for $x=\left(\frac{2}{5}\right)^{2/5}$ is $x_{n+1}=\frac45x_n+\frac{4}{125x_n^4}$. Take $x_0=1$ and assume that each $x_n\geq\left(\frac{2}{5}\right)^{2/5}$. Compute the iterations up to $x_5=0.693145$, using a sufficient level of precision, around $6$ decimals. Finally, compute the upper bound of $e^{x_5}$ with the first $8$ terms of its Taylor series plus another $8$th term.
$$1.999997=\sum_{n=0}^8\frac{(x_5)^n}{n!}+\frac{(x_5)^8}{8!}\geq e^{x_5}\geq e^{\left[\left(\frac25\right)^{2/5}\right]}$$
Therefore, $\ln(2)\geq\left(\frac25\right)^{2/5}$. The exercise is a question of how few arithmetic operations you need to reach the inequality. This method has a total of $36$ multiplications, $16$ divisions and $14$ additions *, which is feasible to do within an hour or so, despite being rather tedious.
* Each Newton-Raphson iteration takes $5$ multiplications, $2$ divisions and $1$ addition. For $n>0$, each term in the series takes at most $2$ multiplications, assuming the previous power of $x_5$ and factorial were stored. Each term also requires $1$ division and $1$ addition. The $n=0$ and repeated $n=8$ terms of the series are single additions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "95",
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"answer_id": 7
} |
What did I do wrong in u-substitution The problem: find the indefinite integral of $x(x-1)^2$
I used u-substitution, $u = x-1, x = u+1, du = dx$.
which gave me $(u+1)u^2$. I distributed and got $u^3 + u^2$, and took the integral to get $[(u^4)/4] + [(u^3)/3]$ replacing $u$ gave me an answer of $[((x-1)^4)/4] + [((x-1)^3)/3] + C$.
The answer sheet solved by distributing before integrating rather than u substitution and got $(x^4)/4 - (2x^3)/3 + (x^2)/2 + C$.
I graphed both integrals thinking they would be equivalent $\pm C$, but they don't appear to be, did I do something wrong?
| You did nothing "wrong" with your u-substitution. Your evaluation of the indefinite integral is correct. To see this,
Suggestion:
Expand the binomials in your answer: Expand $(x-1)^4$ and $(x - 1)^3$ in the numerators, respectively, simplify, and take into account the constant value (absorbed by the constant of integration)...The answers will match, up to the constant of integration.
$$\dfrac{(x-1)^4}{4} + \dfrac{(x-1)^3}{3} + C \quad = \quad \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + \left(\dfrac{1}{12} + C\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the following: $$\sum_{k=1}^{\infty }\frac{1}{(2k-1)^{2}}=\frac{\pi ^{2}}{8}$$
I don't really know how to prove this, will assuming that $$cos(x)=\sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k}}{(2k)!}$$ help?
| If you must use $\cos (x)$, you can focus on zeros of it. $(...,-\frac{(2n-1)\pi}{2},....,-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},...,\frac{(2n-1)\pi}{2},..)$ where $n$ is a positive integer
Hint:
$$\cos x = \cdots\left(1+\frac{x}{\frac{3\pi}{2}}\right)\left(1+\frac{x}{\frac{\pi}{2}}\right)\left(1-\frac{x}{\frac{\pi}{2}}\right)\left(1-\frac{x}{\frac{3\pi}{2}}\right)\cdots =\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\cdots$$
Take derivatives of both side and divide to $\cos (x)$ .
Please let me know if you cannot go forward.
EDIT:
$$\frac{\cos' x}{\cos x} =\frac{-\sin x}{\cos x} =\cfrac{\left(-\frac{2x}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots+\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(-\frac{2x}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots}{\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots}$$
$$\frac{\cos' x}{\cos x} =\frac{-\sin x}{\cos x} =-\frac{8x}{\pi^2}[\cfrac{\left(\frac{1}{1}\right)}{\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)}+\cfrac{\left(\frac{1}{3^2}\right)}{\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)}+....]$$
Is it clear what I mean now? I believe you can handle after that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| You did everything thus far correctly, I just pick up with where you left off in the second line:
$$\begin{align}(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)
& = \dfrac{1 - 2 \cos A + \cos^2 A + \sin^2 A}{\sin A(1 - \cos A)} \\ \\
& = {1 \color{blue}{\bf + \cos^2 A} -2\cos A + 1 \color{blue}{\bf - \cos^2A} \over \sin A(1-\cos A)} \\ \\
& = \dfrac{2 - 2\cos A}{\sin A(1 - \cos A)}\\ \\
& = \dfrac{2\color{red}{\bf (1-\cos A)}}{\sin A\color{red}{\bf (1 - \cos A)}}\\ \\
& = \frac{2}{\sin A} \\ \\
& = 2 \csc A
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to derive $\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$ Consider the following formula to calculate a trigonometric function:
$$\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$$ $[x]$ denotes the integer part of $x$. The formula is valid for $n=0,2,4,6,...$
I'm curious how this formula is derived? Is there any analytical method or is it a pure guessing?
| First, observe that
$$\cos\frac{(n+6)\pi}{3}=\cos\left(\frac{n\pi}{3}+2\pi\right)=\cos\frac{n\pi}{3}$$
and
$$(-1)^{\lfloor\frac{(n+6)+1}{3}\rfloor}=(-1)^{\lfloor\frac{n+1}{3}\rfloor+2}=(-1)^{\lfloor\frac{n+1}{3}\rfloor}$$
Hence we only need to consider $n=0$, $n=2$, and $n=4$. These cases are trivial to check to be true; I will do it for $n=2$:
$$\cos\frac{2\pi}{3}=-\frac{1}{2}=\frac{1+3(-1)^{\lfloor\frac{2+1}{3}\rfloor}}{4}$$
Edit:
I misunderstood the question. To derive this formula, we just have to find an expression $E(n)$ such that $E(0)=1$, $E(2)=E(4)=-\frac{1}{2}$, and $E(n+6)=E(n)$. The key here is $(-1)^{\lfloor\frac{n+1}{3}\rfloor}$, which is $6$-periodic and alternates between $1$, $-1$, and $-1$. Going further, the formula
$$\frac{1+3(-1)^{\lfloor\frac{2n+1}{3}\rfloor}}{4}+((-1)^{\lfloor\frac{3n+1}{3}\rfloor}-1)\frac{(-1)^{\lfloor\frac{2n+2}{3}\rfloor}}{2}$$
is valid for $n=0,1,2,3,\dots$
| {
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Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$
It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
| HINT:
To find $\sum_{1\le r\le n}r^m$ we can utilize the identity
$$(r+1)^{m+1}-r^{m+1}$$
$$=\sum_{1\le t\le m+1}\binom {m+1}t r^{m+1-t}=(m+1)r^m+\sum_{2\le t\le m+1}\binom {m+1}t r^{m+1-t}$$ and need to know $\sum_{1\le r\le n}r^s$ for $0\le s\le m-1$
For $m=2,$ $$(r+1)^3-r^3=3r^2+\sum_{2\le t\le 3}\binom 3t r^{3-t}=3r^2+3r+1$$
Put $r=1,2,3,\cdots,n-1,n$ and add to get $$(n+1)^3-1^3=3\sum_{1\le r\le n}r^2+3\sum_{1\le r\le n}r+\sum_{1\le r\le n}1$$
Now, we know $\sum_{1\le r\le n}r=\frac{n(n+1)}2$ and $\sum_{1\le r\le n}1=n$
| {
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"timestamp": "2023-03-29T00:00:00",
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On the Hurwitz Zeta Function In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like
$\zeta(2k+1,\frac{1}{4})=C(k)\zeta(2k+1)$. With $C(k)$ some elementary function of $k$.
After some basic calculations I found the following
$\zeta(2k+1,\frac{1}{4})=2^{2k+1}(2^{2k+1}-1)\zeta(2k+1)-\zeta(2k+1,\frac{3}{4})$
but I don't know what to do about $\zeta(2k+1,\frac{3}{4})$. Trying the same method on this function, I stumble upon $\zeta(2k+1,\frac{1}{8})$. And I'm going in a loop getting nothing like my prof. said. Also I've looked through books and articles on the Hurwitz function and found nothing.
| \begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{4}) \\
\zeta (2n+1,\frac{3}{4})%
\end{array}%
\right\} =2^{2n}(2^{2n+1}-1){\zeta }(2n+1) \\
\pm \frac{1}{2\pi }\left( 2n+2+4^{2n+2}\right) {\zeta }(2n+2)-2\sum
\limits_{l=0}^{n-1}4^{2n-2l}{{{\zeta }(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(1)}
\end{equation}
\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{3}) \\
\zeta (2n+1,\frac{2}{3})%
\end{array}%
\right\} =\frac{{3^{2n+1}-1}}{2}{\zeta (2n+1)} \\
\pm \frac{\sqrt{3}}{2\pi }\left( \left( 2n+2+3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}3^{2n-2l}{{{\zeta }(2n-2l)\zeta }}%
(2l+2)\right)%
\end{array}
\tag*{(2)}
\end{equation}
\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{6}) \\
\zeta (2n+1,\frac{5}{6})%
\end{array}%
\right\} =\frac{{6^{2n+1}-{3^{2n+1}}-{{2^{2n+1}}+1}}}{2}{\zeta (2n+1)} \\
\pm \frac{1}{2\sqrt{3}\pi }\left( 6^{2n+2}-3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}\left( 6^{2n-2l}-3^{2n-2l}\right) {{{\zeta }%
(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(3)}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
| I tried to generalize, but equation $(2)$ shows why $x=\frac12$ is needed to make a simple equation.
Note that
$$
\begin{align}
\sum_{k=0}^{n+1}x^k\binom{n+k+1}{k}
&=\sum_{k=0}^{n+1}x^k\left[\binom{n+k}{k}+\binom{n+k}{k-1}\right]\\
&=\sum_{k=0}^nx^k\binom{n+k}{k}+x^{n+1}\binom{2n+1}{n+1}\\
&+\sum_{k=0}^{n+1}x^{k+1}\binom{n+k+1}{k}-x^{n+2}\binom{2n+2}{n+1}\tag{1}
\end{align}
$$
Moving terms around and multiplying by $(1-x)^n$, we have
$$
\begin{align}
\hspace{-1cm}(1-x)^{n+1}\sum_{k=0}^{n+1}x^k\binom{n+k+1}{k}
&=(1-x)^n\sum_{k=0}^nx^k\binom{n+k}{k}\\
&+(1-x)^n\left[x^{n+1}\binom{2n+1}{n}-x^{n+2}\binom{2n+2}{n+1}\right]\\
&=(1-x)^n\sum_{k=0}^nx^k\binom{n+k}{k}\\
&+(1-x)^n\left[\frac12x^{n+1}-x^{n+2}\right]\binom{2n+2}{n+1}\tag{2}
\end{align}
$$
since $\binom{2n+1}{n+1}=\binom{2n+1}{n}$ and $\binom{2n+1}{n+1}+\binom{2n+1}{n}=\binom{2n+2}{n+1}$.
Setting $x=\frac12$ in $(2)$ yields
$$
\begin{align}
2^{-n-1}\sum_{k=0}^{n+1}2^{-k}\binom{n+k+1}{k}
&=2^{-n}\sum_{k=0}^n2^{-k}\binom{n+k}{k}\\
&\vdots\\
&=2^{-0}\sum_{k=0}^02^{-k}\binom{0+k}{k}\\[9pt]
&=1\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=0}^n2^{-k}\binom{n+k}{k}=2^n\tag{4}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving recurrence equation using generating functions $$a_{0} = 0, a_{1} = 1, a_{2} = 2, a_{3} = 6$$
$$a_{n} = a_{n+3} - a_{n+2}$$
$\sum = a_{n}x^{n}$
$A(x) = \frac{A(X) - x - 2x^{2}}{x^{3}} - \frac{A(x) - x}{x^{2}}$
$A(x) = \frac{x^{3} + x - 1}{-x^{2} - x}$
Is this the correct solution? The answer is different as far as I know, why is that so? Is it needed to replace the recurrence equation with minuses after n's? (well, all examples I've seen are with minuses there) And if so, how can I do this?
| If we make a blanket assumption that $a_n=0$ for all $n<0$, we can write the recurrence as
$$a_n=a_{n-1}+a_{n-3}+[n=1]+[n=2]+4[n=3]\;,\tag{1}$$
where the last three terms are Iverson brackets, and it will be correct for all $n\ge 0$.
If you’ve not seen this idea before, try evaluating $(1)$ for $n=0,1,2,3$ on the assumption mentioned at the beginning: $$\begin{align*}&a_0=a_{-1}+a_{-3}+0+0+4\cdot0=0\\&a_1=a_0+a_{-2}+1+0+4\cdot0=1\\&a_2=a_1+a_{-1}+0+1+4\cdot0=2\,\text{ and}\\&a_3=a_2+a_0+0+0+4\cdot1=6\;.\end{align*}$$
Multiplying through by $x^n$ and summing over $n\ge 0$, we get the generating function:
$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_{n\ge 0}\Big(a_{n-1}+a_{n-3}+[n=1]+[n=2]+4[n=3]\Big)x^n\\
&=\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}a_{n-3}x^n+x+x^2+4x^3\\
&=xA(x)+x^3A(x)+x+x^2+4x^3\;,
\end{align*}$$
so
$$A(x)=\frac{x+x^2+4x^3}{1-x-x^3}\;.$$
I find this method the easiest way to account for the initial values correctly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving ODE using frobenius method. 3 coefficients I'm trying to learn frobenius method by solving some problems (ODEs).
For example:
$$xy''+(2x+1)y'+(x+1)y=0$$
Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$. Then, I took derivatives and put into the equation:
$$\sum\limits_{n=0}^\infty a_n(n+r)^2x^{n+r-1}+2\sum\limits_{n=0}^\infty a_n(n+r+1)x^{n+r}+\sum\limits_{n=0}^\infty a_nx^{n+r+1}=0$$
After I shifted to make their orders same:
$$\sum\limits_{k=-2}^\infty a_{k+2}(k+r+2)^2x^{k+r+1}+2\sum\limits_{k=-1}^\infty a_{k+1}(k+r+2)x^{k+r+1}+\sum\limits_{k=0}^\infty a_kx^{k+r+1}$$
And if I leave first $k=-2, k=-1$ parts, I can find relationship among 3 coefficient:
$$\sum\limits_{k=0}^\infty [x^{k+r+1}(a_{k+2}(k+r+2)^2+a_{k+1}(k+r+2)+a_k)]=0$$
Now, here I could find relationship with $a_{k+2},a_{k+1},a_k$. But, in this method we should find proportionality between 2 coefficients, not 3. For example, this: Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$
Can you, please, suggest a solution?
| You state:
After I shifted to make their orders same:
$$
\sum_{k = -2}^\infty a_{k + 2}(k + r + 2)^2 x^{k + r + 1} + 2 \sum_{k = -1}^\infty a_{k + 1}(k + r + 2)x^{k + r + 1}+\sum_{k = 0}^\infty a_k x^{k + r + 1}
$$
What you do is the following. Given that
$$
\sum_{k = 0}^\infty a_k (k + r)^2 x^{k + r - 1} + \sum_{k = 0}^\infty a_k(2 k + 2 r + 1)x^{k + r} + \sum_{k = 0}^\infty a_k x^{k + r + 1}
$$
(note that your second sum was incorrectly calculated)
you need to separate the necessary terms of the sums in order to group the powers of $x$ correctly, i.e:
\begin{multline}
\sum_{n = 0}^\infty a_n (n + r)^2 x^{n + r - 1} + 2 \sum_{n = 0}^\infty a_n (n + r + 1) x^{n + r} + \sum_{n = 0}^\infty a_n x^{n + r + 1} = \\
a_0 r^2 x^{r-1} + a_1 (r + 1)^2 x^r + \sum_{n=2}^\infty a_n (n + r)^2 x^{n + r - 1} + (2 r + 1) a_0 x^r + \\ \sum\limits_{n = 1}^\infty a_n (2 n + 2 r+ 1)x^{n + r} +\sum_{n = 0}^\infty a_n x^{n + r + 1} = 0
\end{multline}
Regrouping orders, you have
\begin{multline}
a_0 r^2 x^{r-1} + [a_1 (r + 1) + a_0 (2 r + 1)] x^r + \\ \sum_{k = 0}^\infty \left\{ a_{k + 2}(k + r + 2)^2 + 2 a_{k + 1} (k + r + 2) + a_k \right\} x^{k + r + 1} = 0
\end{multline}
Each power of $x$ needs to vanish, hence $r^2 = 0$. This is the indicial polynomial (details here). This means that $r = 0$ and
\begin{align}
a_1 + a_0 &= 0 \\
a_{k + 2}(k + 2)^2 + 2 a_{k + 1} (2 k + 3) + a_k &= 0
\end{align}
which closes the recurrence relation. The first tree terms are
\begin{align}
a_1 &= -a_0\\
a_2 &= \frac{1}{2!}a_0\\
a_3 &= -\frac{1}{3!}a_0
\end{align}
and it's clear that a relationship is forming. By induction, the whole solution can be computed.
Note that, assuming that $y$ is somehow well behaved, for $x \sim 0$,
$$
x y'' + (2x + 1) y' + (x + 1) y = 0 \quad \sim \quad y' + y = 0.
$$
Proposing the anzats $y(x) = e^{-x} z(x)$ and substituting in the original ode,
$$
x y'' + (2x + 1) y' + (x + 1) y = e^{-x}\left(x z'' + z'\right) = 0,
$$
and it's easily verified that $z = c_1 \log x + c_2$. Hence
$$
y(x) = e^{-x}\left(c_1 \log x + c_2\right)
$$
Cool trick ha?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr
& = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr
& = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr
& = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$
The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
| Why don't you put $u = 2x + 1$ so that $du = 2 \,dx$? Then we'd have $$\int \log (2x + 1) \, dx = \frac {1} {2} \int \log u \, du = \frac {1} {2} (u \log u - u) = \frac {1} {2} (2x + 1) \log (2x + 1) - x - \frac {1} {2} + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Polynomials - The sum of two roots If the sum of two roots of
$$x^4 + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation
| Let those two roots be $a,-a$ and the other two roots are $b,c$
$$\text{Now, }(x-a)(x+a)(x-b)(x-c)=(x^2-a^2)\{x^2-(b+c)x+bc\}$$
$$=x^4-x^3(b+c)+(bc-a^2)x^2+a^2(b+c)x-a^2bc$$
Comparing the coefficients of $x^3$ of this with that of $x^4 + 2x^3 - 8x^2 - 18x - 9 = 0,$
$b+c=-2$
Comparing the coefficients of $x,a^2(b+c)=-18\implies 2a^2=18\implies a=\pm3$
Comparing the constants, $a^2bc=9\implies bc=1$
So, $b,c$ are the roots of the equation $t^2-(-2)t+(1)=0\implies t=-1,-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How prove this $\sum_{n=1}^{\infty}\frac{\zeta_{2}}{n^4}=\zeta^2(3)-\frac{1}{3}\zeta(6)$ show that
$$\sum_{n=1}^{\infty}\dfrac{\zeta_{2}}{n^4}=\zeta^2(3)-\dfrac{1}{3}\zeta(6)$$
where
$$\zeta_{m}=\sum_{k=1}^{n}\dfrac{1}{k^m},\zeta(m)=\sum_{k=1}^{\infty}\dfrac{1}{k^m}$$
is true？
because This result is my frend tell me.
This problem have someone research it?Thank you
my some idea:
$$\zeta^3(3)=\left(\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^3}\right)^2=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\dfrac{1}{(k+1)^3(n-k+1)^3}$$
and use
$$\dfrac{1}{(k+1)(n-k+1)}=\dfrac{1}{n+2}\left(\dfrac{1}{k+1}+\dfrac{1}{n-k+1}\right)$$
and $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
and
$$\sum_{n=1}^{\infty}\dfrac{H_{n}}{(n+1)^5}=\dfrac{1}{2}\left(5\zeta(6)-2\zeta(2)\zeta(4)-\zeta^2(3)\right)$$
But is very ugly, someone have other nice methods? Thank you .
| Consider $$f(z) = \frac{\Big(\psi_{1}(-z) \Big)^{2}}{z^{\color{red}{3}}}$$
(where $\psi_{1}(z)$ is the trigamma function) and integrate around a circle centered at the origin of complex plane of radius $N + \frac{1}{2}$ where $N$ is a positive integer.
Then $$\lim_{N \to \infty } \int_{|z| = N+\frac{1}{2}} f(z) \ dz = 0 = 2 \pi i \left(\text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res}[f(z),n] \right) .$$
For $ n \ge 1 $,
$$ \begin{align} f(z) &= \frac{1}{z^{3}} \Bigg[\frac{1}{(z-n)^{2}}+ \Big(H_{n}^{(2)} + \zeta(2) \Big) - 2 \Big(H_{n}^{(3)}-\zeta(3)\Big) (z-n) + \mathcal{O}\Big((z-n)^{2}\Big) \Bigg]^{2} \\ &= \frac{1}{z^{3}} \Bigg(\frac{1}{(z-n)^{4}} + \frac{2 H_{n}^{(2)}+2 \zeta(2)}{(z-n)^{2}} - \frac{4H_{n}^{(3)} -4 \zeta(3)}{z-n} + \mathcal{O}(1) \Bigg) \end{align}$$
where I'm using the notation $$H_{n}^{(m)} = \sum_{k=1}^{n} \frac{1}{k^{m}} .$$
Therefore,
$$ \begin{align} \text{Res} [f(z),n] &= \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{1}{(z-n)^{4}},n \Bigg] + \text{Res} \Bigg[ \frac{1}{z^{3}} \frac{2 H_{n}^{(2)} + 2 \zeta(2)}{(z-n)^{2}},n\Bigg] + \text{Res} \Bigg[ \frac{4 \zeta(3) - 4 H_{n}^{(3)}}{z-n},n\Bigg] \\ &= -\frac{10}{n^{6}} - \frac{6 H_{n}^{(2)}}{n^{4}} - \frac{6 \zeta(2)}{n^{4}} + \frac{4 \zeta(3)}{n^{3}}- \frac{4 H_{n}^{(3)}}{n^{3}} . \end{align}$$
And at the origin,
$$ \begin{align} f(z) &= \frac{1}{z^{3}} \Big( \frac{1}{z^{2}} + \zeta(2) + 2 \zeta(3) z + 3 \zeta(4) z^{2} + 4 \zeta(5) z^{3} + 5 \zeta(6)z^{4} + \mathcal{O}(z^{5}) \Big)^{2} \\ &= \frac{1}{n^{7}} + \frac{2 \zeta(3)}{z^{5}} + \frac{2 \zeta(3)}{z^{4}} + \frac{6 \zeta(4) + \zeta^{2}(z)}{z^{3}} + \frac{9 \zeta(5) + 4 \zeta(2) \zeta(3)}{z^{2}} \\ &+ \frac{10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3)}{z} + \mathcal{O}(1) . \end{align}$$
So
$$ \text{Res}[f(z),0] = 10 \zeta(6) + 6 \zeta(2) \zeta(4)+ 4 \zeta^{2}(3) .$$
Summing up all the residues we have
$$ -10 \zeta(6) - 6 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} - 6 \zeta(2) \zeta(4) + 4 \zeta^{2}(3) - 4 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}} + 10 \zeta(6) + 6 \zeta(2) \zeta(4) + 4 \zeta^{2}(3) =0$$
$$ \implies 6 \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} = 8 \zeta^{2}(3) - 4 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{3}} .$$
In general, $$\sum_{n=1}^{\infty} \frac{H_{n}^{(m)}}{n^{m}} = \frac{\zeta^{2}(m) + \zeta(2m)}{2} .$$
Therefore,
$$ \begin{align} \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}} &= \frac{1}{6} \Bigg( 8 \zeta^{2}(3)- 4 \Big( \frac{\zeta^{2}(3)+\zeta(6)}{2} \Big) \Bigg) \\ &= \zeta^{2}(3) - \frac{\zeta(6)}{3} . \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Compute limit of the sequence $x_n$ given by $x_{n+2}=-\frac{1}{2}(x_{n+1}-x_n^2)^2+x_n^4$ Let $(x_n)$ be a real sequence such that $x_0=a\in\mathbb{R},x_1=b\in\mathbb{R},x_{n+2}=-\dfrac{1}{2}\left(x_{n+1}-x_{n}^2\right)^2+x_{n}^4\;\forall n\in\mathbb{N} $ and $|x_n|\leq \dfrac{3}{4},\forall n\in\mathbb{N}$. The sequence $x_n$ is convergence or not and compute $\lim x_n$ (if exists)?
I think we could use $\inf$ and $\sup$. But I am not sure
| As Greg Martin stated in comment, the limit satisfies the equation $\displaystyle L=-\frac{1}{2}(L-L^2)^2+L^4$, not hard to find(by rearranging and factoring the equation) the solutions of the equation are $0,1,-1,-2$.
Because $\displaystyle |x_n|\leq \frac{3}{4}$ the sequence cannot converge to $1, -1, -2$. The sequence can only converge to $0$. It remains to prove the sequence indeed converges to 0, which can be done by estimating an descending upper bound of $|x_n|$
To do the estimation, we start by proving a lemma, for real numbers $\alpha$ and $\beta$ we have $|\frac{1}{2}\alpha^2+\alpha\beta-\frac{1}{2}\beta^2|\leq \frac{\sqrt2}{2}(\alpha^2+\beta^2)$.
To prove it, let $\gamma^2=\alpha^2+\beta^2$, we can parametrize $\alpha$ and $\beta$ to get $\alpha=\gamma \cos(\theta),\beta=\gamma \sin(\theta)$, which means $$\frac{1}{2}\alpha^2+\alpha\beta-\frac{1}{2}\beta^2=\frac{\gamma^2}{2}(\cos(2\theta)+sin(2\theta))=\frac{\sqrt2\gamma^2}{2}(sin(2\theta+\pi/4))$$ and therefore the inequality.
Back to the question, if $|x_n|\leq c$ and $|x_{n+1}|\leq c$, $\frac{3}{4}\geq c\geq 0$, by the lemma(set $\alpha=x_n^2$,$\beta=x_{n+1}$) $|x_{n+2}|=|-\frac{1}{2}(x_{n+1}-x_n^2)^2+x_n^4|=|\frac{1}{2}x_n^4+x_n^2x_{n+1}-\frac{1}{2}x_{n+1}^2|\leq|\frac{\sqrt2}{2}(x_n^4+x_{n+1}^2)|\leq\frac{\sqrt2}{2}(c^4+c^2)=\frac{\sqrt2}{2}(c^3+c)c\leq \frac{\sqrt2}{2}((\frac{3}{4})^3+(\frac{3}{4}))c<0.83c$
Therefore we get $|x_1|\leq\frac{3}{4},|x_2|\leq\frac{3}{4},|x_3|\leq(0.83)\frac{3}{4},|x_4|\leq(0.83)\frac{3}{4},|x_5|\leq(0.83)^2\frac{3}{4}......|x_{2n}|\leq(0.83)^{n-1}\frac{3}{4}$, where $(0.83)^n \rightarrow 0$ as $n \rightarrow \infty$, by comparison we conclude the original sequence converges to 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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